Reviewer copy generated from the built lesson pages and the test JSON · 61 lessons · 3265 steps · 761 figures · 989 lesson MCQ · 153 numeric · 168 lesson FRQ · 27 worked examples · 5 practice sets · 5 tests · 140 test/practice MCQ · 20 test/practice FRQ · 139 videos

APBIO-U05-L01 Twenty-three pairs

Topic 5.1 · Meiosis · 60 steps

Left: a photograph of a real human karyotype from a male, 46 stained, banded chromosomes laid out in pairs from the longest to the shortest, the last pair, X and Y, unequal in length. Right: 46 chromosomes drawn as rods sorted by size into 23 pairs in three rows, the longest pair at the top left; in every pair one rod is lighter and one darker; in the last pair, at the bottom right, the lighter rod is much shorter than the darker one
Left: a photograph of a real human karyotype from a male, 46 stained, banded chromosomes laid out in pairs from the longest to the shortest, the last pair, X and Y, unequal in length. Right: 46 chromosomes drawn as rods sorted by size into 23 pairs in three rows, the longest pair at the top left; in every pair one rod is lighter and one darker; in the last pair, at the bottom right, the lighter rod is much shorter than the darker one

Photo: National Human Genome Research Institute, Wikimedia Commons, public domain (resized).

Here are the 46 chromosomes from one cell of a child, sorted by size. They come in twos: two of every length, 23 pairs.

This karyotype is from a male. The last pair, X and Y, differ in length; every other pair matches.

What are the two chromosomes of a pair, and do they carry the same genes?

Unit 5 · Heredity

1One from each parent

2

Video: Watch: One from each parent

The 46 chromosomes sorting themselves into 23 pairs; one pair pulled out and its two members traced back, one to the mother and one to the father.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01a.mp4

3

What are the two chromosomes of a pair to each other? They are the same length, and they carry the same genes in the same order.

4

One of the two came from the mother and one from the father. Every one of the 23 pairs is read the same way: two chromosomes of one length make one pair.

5

Forty-six is too many to follow one at a time, so we draw a model cell with only four chromosomes: two long and two short.

Four rod-shaped chromosomes standing in a row, sorted by size: two long ones on the left, one lighter and one darker, and two short ones on the right, one lighter and one darker; each has a small centromere dot
Four rod-shaped chromosomes standing in a row, sorted by size: two long ones on the left, one lighter and one darker, and two short ones on the right, one lighter and one darker; each has a small centromere dot
6

Sort the four by size and they fall into two matching pairs: the two long ones match each other, and the two short ones match each other. The child’s 46 sort the same way, into 23 matching pairs.

7
Check q1

Every chromosome carries genes along its length.

What does one gene carry?

  1. A. ✓ The instructions for making one protein
  2. B. The instructions for making one whole chromosome
    A chromosome carries hundreds of genes.
    Each gene carries the instructions for one protein.

Why: A gene is one stretch of DNA.
That stretch carries the instructions for making one protein.

8

The two chromosomes of a matching pair carry the same genes in the same order: gene 1, gene 2 and gene 3 sit at the same positions on both.

The two long chromosomes side by side, each carrying three gene positions marked 1, 2 and 3 at the same heights on both
The two long chromosomes side by side, each carrying three gene positions marked 1, 2 and 3 at the same heights on both
9

One chromosome of each pair came from the child’s mother and the other from the child’s father. In every drawing here, the mother’s chromosome is the darker one and the father’s is the lighter one.

The two long chromosomes side by side, the darker one labelled from the mother and the lighter one labelled from the father
The two long chromosomes side by side, the darker one labelled from the mother and the lighter one labelled from the father
10

When two chromosomes match like this, the same length and the same genes in the same order, one from each parent, we call them a , because homologous means matching in structure.

11

Each chromosome of a homologous pair is a homolog of the other: the long darker chromosome is the homolog of the long lighter one.

12

What you are expected to know Pick out the two members of a homologous pair on a drawing of a cell’s chromosomes sorted by size: the same length, the same genes in the same order, one from each parent.

13
Check q2

Here are the four chromosomes of one model cell, shuffled and numbered 1 to 4.

Four rod-shaped chromosomes in a row, numbered 1 to 4, each with a centromere dot; chromosomes 1 and 4 are darker, chromosomes 2 and 3 lighter
Four rod-shaped chromosomes in a row, numbered 1 to 4, each with a centromere dot; chromosomes 1 and 4 are darker, chromosomes 2 and 3 lighter

Which chromosome is the homolog of chromosome 1?

  1. A. Chromosome 2
    Chromosome 2 is short and chromosome 1 is long.
    Chromosomes of different lengths carry different genes.
  2. B. ✓ Chromosome 3
  3. C. Chromosome 4
    Chromosome 4 is short and chromosome 1 is long.
    Chromosomes of different lengths carry different genes.

Why: The homolog of a chromosome is the same length and carries the same genes in the same order.
Chromosome 3 is the only other long chromosome.
So chromosome 3 is the homolog of chromosome 1.

14
Check q3

A plant’s body cells each hold six chromosomes. Here are the six from one cell, sorted by size.

Six rod-shaped chromosomes standing in a row, sorted by size from the longest on the left to the shortest on the right, each with a centromere dot; shades alternate darker and lighter
Six rod-shaped chromosomes standing in a row, sorted by size from the longest on the left to the shortest on the right, each with a centromere dot; shades alternate darker and lighter

How many homologous pairs do the six chromosomes form?

  1. A. 2
    The six chromosomes come in three lengths, two of each.
    Two chromosomes of the same length are one homologous pair.
  2. B. ✓ 3
  3. C. 6
    Six chromosomes is six chromosomes, not six pairs.
    A homologous pair is two chromosomes of the same length.

Why: Chromosomes of the same length carry the same genes in the same order, one from each parent.
The six come in three lengths, two of each length.
Each length is one homologous pair, so the cell holds three homologous pairs.

15Quick quiz: homologous pair mixed practice

16
Check q4

A body cell’s chromosomes are sorted by size.

Which of the following is a homologous pair?

  1. A. ✓ Two chromosomes of the same length, with the same genes in the same order
  2. B. Two identical sister chromatids joined at one centromere
    Two sister chromatids joined at one centromere are one copied chromosome, not two chromosomes.
  3. C. Two chromosomes of different lengths, one from each parent
    Chromosomes of different lengths carry different genes, so they are not a pair.

Why: A homologous pair is two separate chromosomes of the same length.
The two carry the same genes in the same order, one from each parent.

17
Check q5

Here is a drawing from one cell of a mouse.

Two rods side by side, one darker and one lighter, each with a centromere dot
Two rods side by side, one darker and one lighter, each with a centromere dot

Is this a homologous pair?

  1. A. ✓ Yes
  2. B. No
    The two rods are the same length, and each has its own centromere: two separate chromosomes carrying the same genes, one from each parent.

Why: The two rods are the same length, so they carry the same genes in the same order.
One is darker, from the mother, and one is lighter, from the father.
Each rod has its own centromere, so they are two separate chromosomes.
So they are a homologous pair.

18
Check q6

Here is a drawing from one cell of a pea plant.

Two rods side by side, one darker and one lighter, each with a centromere dot
Two rods side by side, one darker and one lighter, each with a centromere dot

Is this a homologous pair?

  1. A. Yes
    One rod is long and the other is short.
    Chromosomes of different lengths carry different genes.
  2. B. ✓ No

Why: One rod is long and the other is short.
Chromosomes of different lengths carry different genes.
The two members of a homologous pair carry the same genes in the same order, so they are the same length.
So these two are not a homologous pair.

19
Check q7

Here is a drawing from one cell of a fruit fly.

Two rods side by side, one darker and one lighter, each with a centromere dot
Two rods side by side, one darker and one lighter, each with a centromere dot

Is this a homologous pair?

  1. A. Yes
    The lighter rod is clearly shorter than the darker rod.
    Two chromosomes of different lengths carry different genes.
  2. B. ✓ No

Why: The lighter rod is clearly shorter than the darker rod.
Chromosomes of different lengths carry different genes.
The two members of a homologous pair are the same length.
So these two are not a homologous pair.

20
Check q8

Here is a drawing from one cell of a dog.

Two rods side by side, one darker and one lighter, each with a centromere dot
Two rods side by side, one darker and one lighter, each with a centromere dot

Is this a homologous pair?

  1. A. ✓ Yes
  2. B. No
    A short pair is still a pair.
    The two rods are the same length, and each has its own centromere.

Why: The two rods are the same length, so they carry the same genes in the same order.
One is darker, from the mother, and one is lighter, from the father.
Each rod has its own centromere, so they are two separate chromosomes.
So they are a homologous pair.

21
Check q9

Here is a drawing from one cell of a frog.

One rod on its own, with a centromere dot
One rod on its own, with a centromere dot

Is this a homologous pair?

  1. A. Yes
    The drawing shows one rod with one centromere: one chromosome.
    A homologous pair is two chromosomes.
  2. B. ✓ No

Why: The drawing shows one rod with one centromere, so it shows one chromosome.
A homologous pair is two chromosomes of the same length.
So one chromosome on its own is not a homologous pair.

22
Practice writing an answer

A mouse’s body cells each hold 40 chromosomes. Sorted by size, the 40 chromosomes show 20 different lengths, with two chromosomes of each length.

(a) Explain how the sorted lengths show that the mouse’s 40 chromosomes form 20 homologous pairs. (1 pt)

Model answer The two members of a homologous pair carry the same genes in the same order, so they are the same length.
The 40 chromosomes sort into 20 lengths with two chromosomes of each length.
So each length is one homologous pair, and 20 lengths are 20 homologous pairs.
Rubric
  • Award 1 point for: the two chromosomes of a homologous pair are the same length (because they carry the same genes in the same order), so two chromosomes of each of the 20 lengths make 20 homologous pairs.

23A pair is not a copied chromosome

24

Video: Watch: A pair is not a copied chromosome

Two rods side by side against one X: the centromeres counted, one X collapsing to one chromosome, two X’s of the same length standing as a pair.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01b.mp4

25
Check q10

This drawing shows one X shape. The X is two sister chromatids joined where the arms cross. To count chromosomes, count centromeres.

An X shape with a centromere dot where its arms cross
An X shape with a centromere dot where its arms cross

How many chromosomes does the drawing show?

  1. A. ✓ One
  2. B. Two
    The X has one centromere, and one centromere is one chromosome.

Why: The X is two identical sister chromatids joined at one centromere, so it is one chromosome.

26

So a copied chromosome, two sister chromatids joined at one centromere, is one chromosome, not a pair. A homologous pair is two separate chromosomes with two centromeres.

Left: a homologous pair, two separate rods of the same length each with its own centromere dot. Right: one copied chromosome, an X of two sister chromatids joined at one centromere
Left: a homologous pair, two separate rods of the same length each with its own centromere dot. Right: one copied chromosome, an X of two sister chromatids joined at one centromere
27

For example, here are two rods of the same length, each with its own centromere. This is a homologous pair, because it has two centromeres and its two chromosomes are the same length.

Two rods of the same length side by side, one lighter and one darker, each with its own centromere dot
Two rods of the same length side by side, one lighter and one darker, each with its own centromere dot
28

But here is one X. This is not a homologous pair, because it has one centromere: it is one copied chromosome.

One X-shaped chromosome with a single centromere dot at the crossing
One X-shaped chromosome with a single centromere dot at the crossing
29

And here are two X’s of the same length. This is a homologous pair, because it has two centromeres and its two chromosomes are the same length.

Two X-shaped chromosomes of the same length side by side, one lighter and one darker, each with its own centromere dot
Two X-shaped chromosomes of the same length side by side, one lighter and one darker, each with its own centromere dot
30

But here are a long X and a short X. This is not a homologous pair, because its two chromosomes are different lengths, so they carry different genes.

A long X-shaped chromosome and a short X-shaped chromosome side by side, one lighter and one darker, each with its own centromere dot
A long X-shaped chromosome and a short X-shaped chromosome side by side, one lighter and one darker, each with its own centromere dot
31

To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

32

Two centromeres are two chromosomes. Those two chromosomes are a homologous pair only if they are the same length.

33

Here is a table of the four drawings: what you see, how many centromeres, how many chromosomes, and whether it is a homologous pair.

A table of the four drawings: what you see, how many centromeres, how many chromosomes, and whether it is a homologous pair
A table of the four drawings: what you see, how many centromeres, how many chromosomes, and whether it is a homologous pair
34

What you are expected to know Tell a homologous pair, two chromosomes with two centromeres, from one copied chromosome, two sister chromatids with one centromere, on a drawing.

35Quick quiz: pair or copied chromosome? mixed practice

36
Check q11

Here is a drawing from one cell of a grasshopper.

An X shape with a centromere dot where its arms cross
An X shape with a centromere dot where its arms cross

Which of the following does the drawing show?

  1. A. A homologous pair
    A homologous pair has two centromeres.
    This X has one centromere.
  2. B. ✓ One copied chromosome
  3. C. Neither
    One centromere is one chromosome, so the X is one copied chromosome.

Why: The X has one centromere.
One centromere is one chromosome, however many arms it has.
So the X is one copied chromosome: two sister chromatids joined together.

37
Check q12

Here is a drawing from one cell of a pea plant.

Two X shapes side by side, one darker and one lighter, each with a centromere dot where its arms cross
Two X shapes side by side, one darker and one lighter, each with a centromere dot where its arms cross

Which of the following does the drawing show?

  1. A. ✓ A homologous pair
  2. B. One copied chromosome
    Each X has its own centromere, so the drawing shows two chromosomes, not one.
  3. C. Neither
    The two X’s have two centromeres and are the same length, which is a homologous pair.

Why: Each X has its own centromere, so the drawing shows two chromosomes.
Each X is one copied chromosome: two sister chromatids, one chromosome.
The two chromosomes are the same length, one from each parent.
So they are a homologous pair.

38
Check q13

Here is a drawing from one cell of a cat.

Two rods side by side, one darker and one lighter, each with a centromere dot
Two rods side by side, one darker and one lighter, each with a centromere dot

Which of the following does the drawing show?

  1. A. ✓ A homologous pair
  2. B. One copied chromosome
    Each rod has its own centromere, so the drawing shows two chromosomes, not one.
  3. C. Neither
    The two rods have two centromeres and are the same length, which is a homologous pair.

Why: Each rod has its own centromere, so the drawing shows two chromosomes.
The two rods are the same length, one from each parent.
So they are a homologous pair.

39
Check q14

Here is a drawing from one cell of a horse.

Two rods side by side, one darker and one lighter, each with a centromere dot
Two rods side by side, one darker and one lighter, each with a centromere dot

Which of the following does the drawing show?

  1. A. A homologous pair
    The two rods are different lengths, so they carry different genes.
  2. B. One copied chromosome
    The drawing has two centromeres, so it shows two chromosomes, not one copied chromosome.
  3. C. ✓ Neither

Why: Each rod has its own centromere, so the drawing shows two chromosomes.
One rod is short and one is long, so the two carry different genes.
So the two are neither a homologous pair nor one copied chromosome.

40
Check q15

Here is a drawing from one cell of a salmon.

Two X shapes side by side, one lighter and one darker, each with a centromere dot where its arms cross
Two X shapes side by side, one lighter and one darker, each with a centromere dot where its arms cross

Which of the following does the drawing show?

  1. A. A homologous pair
    The two X’s are different lengths, so they carry different genes.
  2. B. One copied chromosome
    The drawing has two centromeres, so it shows two chromosomes, not one copied chromosome.
  3. C. ✓ Neither

Why: Each X has its own centromere, so the drawing shows two chromosomes.
One X is long and one is short, so the two carry different genes.
So the two are neither a homologous pair nor one copied chromosome.

41
Check q16

A student looks at the drawing below and says: “This X is a homologous pair, and its two arms are the two homologs.”

An X shape with a centromere dot where its arms cross
An X shape with a centromere dot where its arms cross

Is the student correct?

  1. A. ✓ No: the X is one copied chromosome, and its two arms are sister chromatids
  2. B. Yes: the two arms of the X are the two homologs of a pair
    The X has one centromere, so it is one chromosome.
    Its two arms are two identical copies of that chromosome.

Why: The X has one centromere.
One centromere is one chromosome.
So the X is one copied chromosome, and its two arms are its two sister chromatids.
A homologous pair is two chromosomes with two centromeres.

42Same genes, different versions

43

Video: Watch: Same genes, different versions

One gene position marked on both homologs of a pea plant’s pair; the two boxes reading purple and white; the next position reading round on both.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01c.mp4

44

Take a pea plant’s homologous pair instead of the child’s, because its flower-color gene is easy to follow. Both homologs carry the flower-color gene, at the same position on each.

A pea plant's homologous pair, two rods side by side, with a ring around the same position on each rod, labelled the flower-color gene
A pea plant's homologous pair, two rods side by side, with a ring around the same position on each rod, labelled the flower-color gene
45

The two copies of the gene need not say the same thing. Here the homolog from the mother carries a version that gives purple flowers, and the homolog from the father carries a version that gives white flowers.

The same pair with a box at the flower-color position on each rod: the darker homolog from the mother reads purple, the lighter homolog from the father reads white
The same pair with a box at the flower-color position on each rod: the darker homolog from the mother reads purple, the lighter homolog from the father reads white
46

When a gene comes in more than one version, we call each version an , from a Greek word meaning one another, because the alleles are alternatives to one another at the same position.

47
Check q17

Two alleles of one gene are two slightly different stretches of DNA.

Where does a DNA molecule carry its information?

  1. A. In the number of its strands
    Every DNA molecule has two strands, whatever it codes for.
  2. B. ✓ In the order of its bases

Why: The information in DNA is in the order of its bases.

48

Two alleles of one gene differ somewhere in that order of bases, so a cell reading them can make two slightly different versions of the protein.

49

At the next position along, the seed-shape gene, both homologs of this plant carry the same allele. Same allele or different alleles, the gene at that position is the same gene on both homologs.

The same pair with two positions boxed on each rod: at the flower-color position the boxes read purple and white; at the seed-shape position below it both boxes read round
The same pair with two positions boxed on each rod: at the flower-color position the boxes read purple and white; at the seed-shape position below it both boxes read round
50

Two different genes, at two different positions, are never two alleles. Alleles are versions of one gene, at one position.

51

What you are expected to know Say, for a marked position on a drawn homologous pair, whether the two homologs carry the same allele or two different alleles of that gene.

52
Check q18

In mice, one gene sets coat color and another gene, further along the same chromosome, sets tail length. One mouse’s homologous pair is drawn below with the version each homolog carries at the two positions.

A mouse's homologous pair, two rods side by side, with two boxed positions on each: at the coat-color position the darker rod reads black and the lighter rod reads brown; at the tail-length position both rods read long
A mouse's homologous pair, two rods side by side, with two boxed positions on each: at the coat-color position the darker rod reads black and the lighter rod reads brown; at the tail-length position both rods read long

Which two things are alleles of one gene?

  1. A. The coat-color gene and its brown version
    The coat-color gene is the gene, and brown is one version of it.
    Alleles are two or more versions of one gene.
  2. B. The two homologs of the pair
    The two homologs are whole chromosomes, and each carries many genes.
    An allele is one version of one gene.
  3. C. The black coat version and the long tail version
    Black coat is a version of the coat-color gene; long tail is a version of the tail-length gene.
    Two different genes, so not alleles of one gene.
  4. D. ✓ The black version and the brown version

Why: Alleles are versions of one gene at one position.
At the coat-color position this pair carries two different versions, black and brown.
So the black version and the brown version are alleles of the coat-color gene.

53
Check q19

The mouse’s homologous pair drawn below carries, on each homolog, a version of the tail-length gene at the tail-length position.

A mouse's homologous pair, two rods side by side, with two boxed positions on each: at the coat-color position the darker rod reads black and the lighter rod reads brown; at the tail-length position both rods read long
A mouse's homologous pair, two rods side by side, with two boxed positions on each: at the coat-color position the darker rod reads black and the lighter rod reads brown; at the tail-length position both rods read long

Which of the following does the pair carry at the tail-length position?

  1. A. Two different alleles
    Both boxes at the tail-length position read long.
    Two different alleles would read two different words, as black and brown do at the coat-color position.
  2. B. ✓ The same allele

Why: Both boxes at the tail-length position read long.
So both homologs carry the same version of the tail-length gene: the same allele.

54

Back to the child’s 46 chromosomes sorted by size: two of every length, 23 pairs.

55

The two chromosomes of each pair are a homologous pair: the same length, the same genes in the same order, one from the mother and one from the father.

56

At any one gene position, the two homologs may carry the same allele or two different alleles.

57Quick quiz: allele mixed practice

58
Check q20

A pea plant’s flower-color gene comes in a purple version and a white version.

Which of the following is an allele?

  1. A. One chromosome of a homologous pair
    A chromosome carries many genes.
    An allele is one version of one of those genes.
  2. B. Two different genes at two different positions
    Two different genes are two genes.
    Alleles are versions of one gene at one position.
  3. C. ✓ One version of one gene

Why: A gene can come in more than one version.
Each version of that one gene is an allele.

59
Practice writing an answer

In pea plants, the seed-color gene comes in two versions. One version gives yellow seeds and the other gives green seeds. The two versions sit at the same position on the two homologs of a pair.

(a) Explain why the yellow version and the green version are alleles of one gene. (1 pt)

Model answer The yellow version and the green version sit at the same position on the two homologs.
The yellow version and the green version are both versions of one gene, the seed-color gene.
A version of one gene is an allele.
So the yellow version and the green version are two alleles of the seed-color gene.
Rubric
  • Award 1 point for: both versions are versions of the same gene at the same position (the seed-color gene), and a version of one gene is an allele.

Glossary

homologous pair
Two chromosomes of the same length that carry the same genes in the same order, one inherited from the mother and one from the father. Each chromosome is a homolog of the other. A human body cell’s 46 chromosomes form 23 homologous pairs.
allele
One version of a gene. The two homologs of a pair carry the same gene at the same position, and may carry the same allele or two different alleles of it.

APBIO-U05-L01B Two sets or one

Topic 5.1 · Meiosis · 35 steps

Three simple figures: a tall mother on the left, a tall father on the right and a shorter child between them; above each the count 46 chromosomes in every cell; beneath, the sum 46 plus 46 with a question mark
Three simple figures: a tall mother on the left, a tall father on the right and a shorter child between them; above each the count 46 chromosomes in every cell; beneath, the sum 46 plus 46 with a question mark

Suppose you sort the 23 chromosomes a sperm cell carries by size. You get one of every length and no pairs at all.

Sort the 46 chromosomes of a cheek cell the same way and you get two of every length. The sperm holds one of every pair; the cheek cell holds two of every pair. How many complete sets of chromosomes does each cell hold?

Unit 5 · Heredity

1One of every pair: a set

2

Video: Watch: One of every pair: a set

The child’s 46 sorting into the mother’s 23 and the father’s 23, one of every pair in each; then any 23 picked at random, and a pair’s genes going missing.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Ba.mp4

3

How many sets of chromosomes does a cell carry? One chromosome of every pair, every gene once, is a complete set.

4

A body cell carries two complete sets, one from each parent. A sperm or an egg carries one complete set.

5

Sort a cell’s chromosomes by size. Two of every length is two sets, and one of every length is one set.

6
Check q1

A human body cell’s 46 chromosomes sort by size into 23 pairs.

Which of the following describes the two chromosomes of one pair?

  1. A. ✓ Two homologs, one from each parent
  2. B. Two sister chromatids of one copied chromosome
    Two sister chromatids are joined at one centromere and are one chromosome.
    The two chromosomes of a pair have two centromeres.

Why: The two chromosomes of a pair are the same length and carry the same genes in the same order.
One came from the mother and one from the father.
So the two are homologs.

7

Count the child’s 46 chromosomes as pairs: 23 homologous pairs, and in every pair one homolog came from the mother and one from the father.

The model cell's four chromosomes arranged as two pairs, a long pair and a short pair; the darker member of each pair is the mother's and the lighter member of each is the father's
The model cell's four chromosomes arranged as two pairs, a long pair and a short pair; the darker member of each pair is the mother's and the lighter member of each is the father's
8

So the 23 chromosomes from the mother are one of every pair, every gene once. When a group of chromosomes holds one of every pair, we call it a complete , and the child’s 46 are two sets.

9

Any 23 of the 46 is not a set. Two from one pair and none from another leaves the missing pair’s genes out.

Left: a long chromosome and a short one, labelled a complete set, every gene once. Right: two long chromosomes and no short one, labelled not a set, the short chromosome's genes missing
Left: a long chromosome and a short one, labelled a complete set, every gene once. Right: two long chromosomes and no short one, labelled not a set, the short chromosome's genes missing
10

What you are expected to know Say whether a group of chromosomes is a complete chromosome set: one chromosome of every pair, every gene once.

11
Check q2

A plant’s body cells each hold four chromosomes, two long and two short. Drawings 1 to 3 below each show chromosomes taken from one of the plant’s cells.

Three boxed drawings numbered 1 to 3. Drawing 1: two rods, one darker and one lighter. Drawing 2: two rods, one lighter and one darker. Drawing 3: three rods, two darker and one lighter
Three boxed drawings numbered 1 to 3. Drawing 1: two rods, one darker and one lighter. Drawing 2: two rods, one lighter and one darker. Drawing 3: three rods, two darker and one lighter

Which drawing shows one complete chromosome set?

  1. A. Drawing 1
    Drawing 1 shows two long chromosomes and no short one.
    The short chromosome’s genes are missing, so the two are not a complete set.
  2. B. ✓ Drawing 2
  3. C. Drawing 3
    Drawing 3 shows one long chromosome and two short ones.
    A complete set is one of every pair, and a second short chromosome is one too many.

Why: A complete chromosome set is one chromosome of every pair, every gene once.
The plant has a long pair and a short pair.
Drawing 2 shows one long chromosome and one short one: one of each pair.
So drawing 2 shows one complete set.

12
Check q3

A horse’s body cells each hold 64 chromosomes, 32 homologous pairs. A student says: “Take any 32 of the 64 and you have one complete chromosome set.”

Is the student correct?

  1. A. ✓ No: any 32 might hold both chromosomes of one pair and neither chromosome of another
  2. B. Yes: 32 chromosomes is half of the horse cell’s 64, so any 32 of them make one complete set
    Half the count is not the test.
    A complete set is one chromosome of every one of the 32 pairs.

Why: A complete chromosome set is one chromosome of every pair, every gene once.
Any 32 of the 64 might hold both chromosomes of one pair and neither chromosome of another.
Then that other pair’s genes are missing.
So any 32 is not a complete set.

13Quick quiz: chromosome set mixed practice

14
Check q4

A body cell’s chromosomes are sorted by size into homologous pairs.

Which of the following is one complete chromosome set?

  1. A. Half of the cell’s chromosomes, whichever half
    Half the chromosomes might hold two of one pair and none of another.
    A set is one of every pair.
  2. B. ✓ One chromosome of every homologous pair
  3. C. Both chromosomes of one homologous pair
    Both chromosomes of one pair carry the same genes twice.
    A set holds every gene once, one chromosome of every pair.

Why: A complete chromosome set is one chromosome of every homologous pair.
One of every pair carries every gene once.

15
Practice writing an answer

A plant’s body cells each hold 12 chromosomes, in six pairs. One cell from the plant holds six chromosomes: two of one length, two of another and two of a third. The cell does not hold a complete chromosome set.

(a) Explain how the chromosome lengths show that this six-chromosome cell lacks a complete set. (1 pt)

Model answer A complete chromosome set is one chromosome of every pair, every gene once.
The plant’s six pairs are six lengths, and this cell holds two chromosomes of three lengths and none of the other three.
So three pairs are missing, and the cell does not hold a complete set, however the count comes out.
Rubric
  • Award 1 point for: a complete set holds one chromosome of each pair (one of each of the six lengths); this cell holds two of three lengths and none of the other three, so it is not one complete set.

16Diploid or haploid

17

Video: Watch: Diploid or haploid

The model cell with two sets, 2n = 4, beside a cell with one set, n = 2; then the sperm’s 23 and the cheek cell’s 46 labelled n and 2n.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Bb.mp4

18

When a cell holds two complete sets, so that every chromosome has a homolog, we call it , because di- means two, and we write it 2n.

19

The letter n stands for the number of chromosomes in one set. For a human, n=23 and 2n=46; for the model cell, n=2 and 2n=4.

20

This n counts the chromosomes in one set. It is a different n from the sample size n in statistics, which counts readings.

21

When a cell holds one complete set, one of every chromosome and no homologs, we call it , because haplo- means single, and we write it n. The exam sometimes writes the same thing as 1n.

Left: a cell holding four chromosomes, a long pair and a short pair, labelled diploid, two complete sets. Right: a smaller cell holding one long and one short chromosome, labelled haploid, one complete set
Left: a cell holding four chromosomes, a long pair and a short pair, labelled diploid, two complete sets. Right: a smaller cell holding one long and one short chromosome, labelled haploid, one complete set
22

What you are expected to know Say whether a cell is diploid or haploid from a drawing or a stated count.

23

Back to the sperm cell’s 23 chromosomes sorted by size, one of every length, and the cheek cell’s 46, two of every length. The sperm cell holds one of every pair: one complete set, so it is haploid, n=23.

24

The cheek cell holds two of every pair: two complete sets, so it is diploid, 2n=46.

25Quick quiz: diploid or haploid mixed practice

26
Check q5

A cell’s chromosomes are counted and sorted by size.

Which of the following describes a diploid cell?

  1. A. A cell holding as many chromosomes as a body cell, whichever chromosomes they are
    A count alone is not the test: those chromosomes might hold two of one pair and none of another.
    A diploid cell holds two complete sets, two of every pair.
  2. B. A cell holding one complete chromosome set
    A cell holding one complete set is haploid.
  3. C. ✓ A cell holding two complete chromosome sets

Why: A diploid cell holds two complete chromosome sets, one from each parent.
So every chromosome in a diploid cell has a homolog.

27
Check q6

A cell’s chromosomes are counted and sorted by size.

Which of the following describes a haploid cell?

  1. A. ✓ A cell holding one complete chromosome set
  2. B. A cell holding two complete chromosome sets, one from each parent
    A cell holding two complete sets is diploid.
  3. C. A cell holding half of a chromosome set
    Half of a set leaves genes missing.
    A haploid cell holds one whole set, one chromosome of every pair.

Why: A haploid cell holds one complete chromosome set: one of every chromosome and no homologs.

28
Check q7

Here is a cell from a plant whose body cells hold four chromosomes, two long and two short.

A cell holding two rods side by side, one darker and one lighter
A cell holding two rods side by side, one darker and one lighter

Which of the following describes the cell?

  1. A. Diploid
    A diploid cell holds two complete sets: two long chromosomes and two short ones.
    This cell has no short chromosome.
  2. B. Haploid
    A haploid cell holds one of every pair: one long and one short.
    This cell has two long and no short.
  3. C. ✓ Not a complete set

Why: The cell holds two long chromosomes and no short one.
A complete set is one chromosome of every pair.
This cell has two of the long pair and none of the short pair, so the short chromosome’s genes are missing.
So the cell does not hold a complete set.

29
Check q8

Here is a cell from a plant whose body cells hold four chromosomes, two long and two short.

A small cell holding two rods side by side, one darker and one lighter
A small cell holding two rods side by side, one darker and one lighter

Which of the following describes the cell?

  1. A. Diploid
    A diploid cell holds two of every chromosome, so every chromosome has a homolog.
    Here the long chromosome has no partner and neither does the short one.
  2. B. ✓ Haploid
  3. C. Not a complete set
    The cell holds one long and one short chromosome, which is one of every pair, so every gene is present once.
    That is a complete set.

Why: The cell holds one long chromosome and one short one: one of every pair, and no homologs.
One of every pair is one complete set.
A cell with one complete set is haploid.

30
Check q9

Here is a cell from a plant whose body cells hold four chromosomes, two long and two short.

A cell holding four rods in a row, two darker and two lighter
A cell holding four rods in a row, two darker and two lighter

Which of the following describes the cell?

  1. A. ✓ Diploid
  2. B. Haploid
    A haploid cell holds one of every pair.
    This cell holds two long and two short, so it has two of every pair.
  3. C. Not a complete set
    The cell has both a long chromosome and a short one, so no gene is missing; it holds every pair twice over.

Why: The cell holds two long chromosomes and two short ones: two of every pair, so every chromosome has a homolog.
Two complete sets make a cell diploid.

31
Check q10

A gorilla’s body cells each hold 48 chromosomes. A cell in a gorilla’s testis holds 24 chromosomes, one of each length.

Which of the following describes the cell?

  1. A. Diploid
    Two complete sets of a gorilla’s chromosomes are 48, two of every length.
    This cell has 24, one of every length, which is one set.
  2. B. ✓ Haploid
  3. C. Not a complete set
    24 chromosomes of 24 different lengths means no length is missing and none appears twice.
    That is one chromosome of every pair, which is exactly a complete set.

Why: A gorilla’s 48 chromosomes are 24 homologous pairs, two of every length.
Twenty-four chromosomes of 24 different lengths is one of every pair.
One of every pair is one complete set, so the cell is haploid.

32
Check q11

A rabbit’s body cells each hold 44 chromosomes, two of each length.

Which of the following describes one of the rabbit’s body cells?

  1. A. ✓ Diploid
  2. B. Haploid
    A haploid cell holds one of each length, 22 for a rabbit.
    This cell holds two of each length, 44.
  3. C. Not a complete set
    Two of each length means every pair is present, twice over, so no gene is missing.

Why: Two of each length means every chromosome has a homolog.
A rabbit’s 44 are 22 pairs, two complete sets.
A cell holding two complete sets is diploid.

33
Check q12

A dog’s body cells each hold 78 chromosomes. A cell in a dog’s ovary holds 39 chromosomes; sorted by size, they show 39 different lengths, one chromosome of each length.

Which of the following describes the cell?

  1. A. Diploid
    Two complete sets of a dog’s chromosomes are 78, two of every pair.
    Thirty-nine chromosomes, one of every pair, is one set.
  2. B. ✓ Haploid
  3. C. Not a complete set
    No length appears twice and none is missing, so the cell holds one of every pair: a complete set.

Why: A dog’s 78 chromosomes are 39 homologous pairs, two of every length.
Thirty-nine chromosomes of 39 different lengths is one of every pair.
One of every pair is one complete set, so the cell is haploid.

34
Practice writing an answer

A cat’s body cells each hold 38 chromosomes: 19 different lengths, two chromosomes of each length. A cell in a cat’s ovary holds 19 chromosomes, one chromosome of each length.

(a) Explain how the chromosome lengths show that the ovary cell is haploid. (1 pt)

Model answer A complete chromosome set is one chromosome of every pair, one of each length.
The ovary cell holds 19 chromosomes of 19 different lengths, so it holds one of every pair.
One of every pair is one complete set.
A cell holding one complete set is haploid.
Rubric
  • Award 1 point for: one chromosome of each of the 19 lengths is one of every pair, which is one complete set, and a cell with one complete set is haploid.

Glossary

chromosome set
One chromosome of every homologous pair: every gene once. A human set is 23 chromosomes; the model cell’s set is one long and one short.
diploid (2n)
Holding two complete chromosome sets, one from each parent, so every chromosome has a homolog. A human body cell is diploid, 2n = 46.
haploid (n)
Holding one complete chromosome set, one of every chromosome and no homologs. A human sperm or egg is haploid, n = 23; the exam also writes 1n.

APBIO-U05-L01C Forty-six, not ninety-two

Topic 5.1 · Meiosis · 51 steps

Left: a photograph of a family, a baby lying between its mother and its father, all three looking at each other. Right: three boxes reading 46, 46 and 92 with a question mark, labelled the mother's cells, the father's cells and the child's cells, with a plus sign and an equals sign between them; beneath, the line the child's cells hold 46
Left: a photograph of a family, a baby lying between its mother and its father, all three looking at each other. Right: three boxes reading 46, 46 and 92 with a question mark, labelled the mother's cells, the father's cells and the child's cells, with a plus sign and an equals sign between them; beneath, the line the child's cells hold 46

Photo: sheldonl, Pixabay via Wikimedia Commons, CC0 (resized).

Here is a family: a child between two parents. Every cell of each of them holds 46 chromosomes.

The child grew from one cell, made from one cell of the mother and one cell of the father. Add them: 92. But the child’s cells hold 46. Why not 92?

Unit 5 · Heredity

1The sperm and the egg

2

Video: Watch: The sperm and the egg

The child traced back to one cell made from two; a sperm and an egg drawn side by side, each with one of every pair, 23 in a human.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Ca.mp4

3

Why does a child not end up with twice the parents’ chromosomes? The child began as one cell made from two: a sperm from the father and an egg from the mother.

4

The sperm and the egg each carry one complete set, 23 chromosomes. The two fuse into one cell that carries two sets, 46.

5

If the sperm and the egg carried two sets each, the count would double every generation. So a division that halves the count must come before the two cells fuse.

6
Check q1

A human sperm cell holds 23 chromosomes, one of every pair.

Which of the following describes the sperm cell?

  1. A. Diploid
    A diploid cell holds two complete sets, two of every pair: 46 in a human.
  2. B. ✓ Haploid

Why: One of every pair is one complete chromosome set.
A cell holding one complete set is haploid.

7

So the sperm and the egg are each haploid: one of every pair, 23 chromosomes, one complete set.

A small sperm cell with a tail, holding one long and one short chromosome, and a larger egg cell holding one long and one short chromosome; count labels read 2 for each in the model and 23 for each in a human
A small sperm cell with a tail, holding one long and one short chromosome, and a larger egg cell holding one long and one short chromosome; count labels read 2 for each in the model and 23 for each in a human
8

When a haploid cell joins with another to start a new organism, we call it a , from a Greek word for a marriage partner. The sperm and the egg are the two gametes.

9

What you are expected to know Name the two gametes, the sperm and the egg, and say that each is haploid.

10
Check q2

Cells are taken from a frog and from a tadpole.

Which of the following cells is a gamete?

  1. A. A frog’s skin cell
    A skin cell is a body cell with two complete sets.
    It does not join with another cell to start a new frog.
  2. B. ✓ A frog’s egg
  3. C. A tadpole’s muscle cell
    A muscle cell is a body cell with two complete sets.
    It does not join with another cell to start a new frog.

Why: A gamete is a haploid cell that joins with another to start a new organism.
A frog’s egg holds one complete set, and a sperm joins it to start a new frog.
So the egg is a gamete.

11Quick quiz: gamete mixed practice

12
Check q3

A new animal starts from two cells joining.

Which of the following is a gamete?

  1. A. ✓ A haploid cell that joins with another to start a new organism
  2. B. The diploid cell made when two gametes fuse into one
    The cell made when two gametes fuse is the start of the new organism, not one of the two cells that joined.
  3. C. Any body cell that divides by mitosis to make two
    Body cells divide by mitosis and hold two complete sets; a gamete holds one.

Why: A gamete is a haploid cell, one complete set.
It joins with another gamete to start a new organism.
The sperm and the egg are the two gametes.

13
Practice writing an answer

A sea urchin releases sperm into the sea, and a second sea urchin releases eggs. Each sperm and each egg holds one complete set of the sea urchin’s chromosomes. A sperm joins an egg, and a new sea urchin develops from the cell they make.

(a) Explain why the sea urchin’s sperm and eggs are gametes. (1 pt)

Model answer A gamete is a haploid cell that joins with another cell to start a new organism.
Each sperm and each egg holds one complete set, so each is haploid.
The sperm joins the egg, and a new sea urchin develops from the cell they make.
So the sperm and the egg are gametes.
Rubric
  • Award 1 point for: each is haploid (one complete set) and joins with the other to start a new organism, which is what a gamete is.

14Fertilization and the zygote

15

Video: Watch: Fertilization and the zygote

The sperm and the egg fusing into one cell with 46; that cell dividing by mitosis again and again into a body.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Cb.mp4

16

The sperm and the egg fuse into a single cell. When two gametes fuse, we call it .

A small sperm cell holding one long and one short chromosome, a larger egg cell holding one long and one short, an arrow, and a zygote holding all four; count labels read 2, 2 and 4 for the model and 23, 23 and 46 for a human
A small sperm cell holding one long and one short chromosome, a larger egg cell holding one long and one short, an arrow, and a zygote holding all four; count labels read 2, 2 and 4 for the model and 23, 23 and 46 for a human
17

When fertilization has made one cell from the two gametes, we call that cell a , from a Greek word meaning yoked together: two joined as one.

18

The zygote holds both sets, the sperm’s and the egg’s, so it is diploid: 46 chromosomes in a human, 4 in the model cell.

19
Check q4

A mouse’s sperm holds 20 chromosomes and a mouse’s egg holds 20.

How many chromosomes does the zygote hold?

  1. A. 20
    The zygote holds both gametes’ sets: the sperm’s 20 and the egg’s 20.
  2. B. ✓ 40
  3. C. 80
    Fertilization puts the two gametes’ sets together once, two sets of 20.
    Nothing is copied at fertilization.

Why: The sperm brings one set of 20 chromosomes.
The egg brings one set of 20 chromosomes.
Fertilization puts both sets in one cell.
So the zygote holds 40 chromosomes, two sets of 20.

20
Check q5

A zygote with 46 chromosomes divides by mitosis.

How many chromosomes does each of the two new cells hold?

  1. A. ✓ 46
  2. B. 23
    Before mitosis the cell copies every chromosome once.
    Mitosis then sends one copy of each to each new cell, so the count does not halve.

Why: Mitosis gives two cells with the same chromosomes as the parent cell, so each new cell holds 46.

21

The zygote divides by mitosis again and again, so every body cell of the child carries the zygote’s 46.

22

When a new organism starts from the fusion of two gametes, one from each parent, we call it , because two parents each contribute one set.

23

What you are expected to know Name fertilization and the zygote in a described case of sexual reproduction, and say that the zygote is diploid.

24
Check q6

A frog’s body cells each hold 26 chromosomes. In a pond, frog sperm reach frog eggs, and tadpoles later hatch.

Which cell is the zygote?

  1. A. The egg, before the sperm reaches it
    The egg before fertilization is a gamete: haploid, with 13 chromosomes.
    Fertilization has not happened yet, so there is no zygote yet.
  2. B. The sperm, once it has entered the egg
    The sperm is a gamete, one of the two cells that fuse.
    The zygote is the one cell the sperm and the egg make together.
  3. C. Each cell of the tadpole’s body
    The tadpole’s body cells grew from the zygote by mitosis.
    They are the zygote’s descendants, not the zygote.
  4. D. ✓ The single cell made when the sperm and the egg fuse

Why: Fertilization is the fusion of the two gametes.
The single cell it makes is called the zygote.
Here that cell is diploid, with 26 chromosomes, and every body cell of the tadpole grows from it by mitosis.

25Quick quiz: fertilization, zygote and sexual reproduction mixed practice

26
Check q7

A sperm and an egg are in the same place.

Which of the following events is fertilization?

  1. A. ✓ The two gametes fusing into one cell
  2. B. The zygote dividing by mitosis
    The zygote dividing by mitosis is how the body grows, after fertilization has made the zygote.
  3. C. A body cell dividing into two
    A body cell dividing by mitosis makes two body cells; no gametes fuse.

Why: Fertilization is the fusion of two gametes, a sperm and an egg, into a single cell.

27
Check q8

A sperm has fused with an egg.

Which of the following is the zygote?

  1. A. The haploid cell that joined with the egg at fertilization
    The haploid cell that joined with the egg is the sperm, a gamete.
  2. B. ✓ The single cell made when the sperm and the egg fused
  3. C. Any body cell of the new organism, made by mitosis
    The body cells grew from the zygote by mitosis; they are its descendants.

Why: Fertilization makes one cell from the two gametes.
That single cell, holding both sets, is the zygote.

28
Check q9

A new organism is starting.

Which of the following describes sexual reproduction?

  1. A. A new organism growing from one cell of one parent by mitosis
    One parent’s cell dividing by mitosis gives a copy of that parent; no gametes fuse.
  2. B. A new organism growing all of its body cells by mitosis after it starts
    Every organism grows its body cells by mitosis, after it has started.
    Sexual reproduction is how the organism starts.
  3. C. ✓ A new organism starting from two gametes fusing, one from each parent

Why: In sexual reproduction two gametes fuse, one from each parent.
Each parent contributes one chromosome set to the new organism.

29
Practice writing an answer

In a pond, a frog’s sperm reaches a frog’s egg and the two fuse. The single cell they make divides again and again, and a tadpole grows from it. Every body cell of the tadpole holds 26 chromosomes.

(a) Identify the two gametes and the zygote in this case. (1 pt)

Model answer The two gametes are the frog’s sperm and the frog’s egg.
The zygote is the single cell made when the sperm and the egg fuse.
Rubric
  • Award 1 point for: gametes = the sperm and the egg; zygote = the single cell made by their fusion.

(b) Explain why every body cell of the tadpole holds two complete chromosome sets. (1 pt)

Model answer The sperm brings one complete set and the egg brings one complete set.
Fertilization puts the two sets into the zygote, so the zygote is diploid.
The zygote divides by mitosis, and mitosis gives each new cell the same chromosomes as the parent cell.
So every body cell of the tadpole holds two complete sets.
Rubric
  • Award 1 point for: the zygote receives one set from each gamete (diploid), and mitosis passes the same chromosomes to every body cell.

30Why gametes must be haploid

31

Video: Watch: Why gametes must be haploid

The count doubling every generation, 92, 184, 368, if gametes carried 46; then a sperm and an egg each bringing one set and 23 with 23 settling at 46.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Cc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L01Cc.mp4

32

Suppose gametes carried two sets, 46 each, like a body cell. The zygote would hold 92, the next generation’s zygotes 184, and the one after 368: the count would double every generation.

Four boxes in a row joined by arrows, reading 46 for the parents' cells, then 92 for their child, then 184 for the grandchild, then 368; a caption says the count would double every generation
Four boxes in a row joined by arrows, reading 46 for the parents' cells, then 92 for their child, then 184 for the grandchild, then 368; a caption says the count would double every generation
33

Because gametes are haploid, the two single sets they bring make one double set: 23 from the sperm with 23 from the egg, and the zygote has 46, every generation.

A loop: a body cell with 46, an arrow labelled halving to a gamete with 23, an arrow labelled fertilization joining it with a second gamete with 23 to a zygote with 46; a caption says 46 every generation
A loop: a body cell with 46, an arrow labelled halving to a gamete with 23, an arrow labelled fertilization joining it with a second gamete with 23 to a zygote with 46; a caption says 46 every generation
34

So a division that halves the count, from two sets to one, must come before fertilization.

35

Fertilization then puts two single sets back together. The halving and the fertilization cancel each other, so the count stays at 46.

36

What you are expected to know Predict what would happen to the chromosome count over generations if gametes carried two sets, and explain why a halving division must come before fertilization.

37
Check q10

Imagine human sperm and eggs each carried 46 chromosomes, like a body cell, and a sperm and an egg still fused at fertilization.

How many chromosomes would the zygote hold?

  1. A. 23
    Nothing in this case halves the count; each gamete brings 46.
  2. B. 46
    Fertilization adds the two gametes’ chromosomes together, 46 from each.
  3. C. ✓ 92

Why: The sperm would bring 46 chromosomes and the egg would bring 46.
Fertilization puts both into one cell.
So the zygote would hold 92 chromosomes, twice 46.

38
Practice writing an answer

A dog’s sperm carries 39 chromosomes and a dog’s egg carries 39. The puppy that grows from their zygote has 78 chromosomes in every body cell, the same count as each parent.

(a) Explain how these counts show that a division halving the chromosome count happens before fertilization. (1 pt)

Model answer Each parent’s body cells hold 78 chromosomes, two complete sets.
The sperm and the egg each carry 39, one complete set, so the count was halved before the gametes were finished.
Fertilization puts the two single sets together, 39 from each gamete, making 78.
So the halving before fertilization and the joining at fertilization cancel, and the puppy has 78, not 156.
Rubric
  • Award 1 point for: body cells hold 78 but each gamete carries 39, so the count was halved before fertilization; fertilization then restores 78 (39 + 39), so the count stays at 78.
39
Check q11

A student says: “Gametes could carry 46 chromosomes each. The zygote would hold 92 and simply throw the extra 46 away.”

Is the student correct?

  1. A. Yes: a zygote holding 92 chromosomes would keep 46 of them and throw the other 46 away
    The zygote throws nothing away: it keeps both sets it received.
    Mitosis then gives every body cell the same chromosomes.
  2. B. ✓ No: the zygote keeps every chromosome it receives and passes them on by mitosis

Why: The zygote keeps both sets it receives at fertilization.
Mitosis then gives every body cell the same chromosomes as the zygote.
So a zygote holding 92 would give body cells holding 92, and the count would double every generation.

40

Back to the family: a child between two parents, 46 chromosomes in every cell of each of them. The child began as one cell made from the mother’s egg and the father’s sperm.

41

Each gamete brought 23 chromosomes, one complete set. Fertilization put the two single sets together, and every body cell of the child grew from that zygote by mitosis, so the child has 46, not 92.

42Mixed practice mixed practice

43
Check q12

A human egg holds 23 chromosomes, one of every pair.

Which of the following describes the egg?

  1. A. ✓ A haploid gamete
  2. B. A diploid gamete
    One of every pair is one complete set, so the egg is haploid, not diploid.
  3. C. A haploid zygote
    The egg is one of the two cells that fuse; the zygote is the cell their fusion makes.

Why: The egg holds one complete set, so it is haploid.
The egg joins with a sperm to start a new organism, so it is a gamete.

44
Check q13

A grasshopper’s body cells each hold 24 chromosomes.

What are the chromosome counts of a grasshopper sperm, a grasshopper body cell and a grasshopper zygote, in that order?

  1. A. 24, 24 and 48
    A sperm is haploid, one set of 12, and a zygote holds two sets, 24, not four.
  2. B. ✓ 12, 24 and 24
  3. C. 12, 12 and 24
    A body cell is diploid, 24, like every cell that grew from the zygote by mitosis.
  4. D. 24, 12 and 24
    The sperm, a gamete, is the haploid cell with 12, and body cells are diploid with 24.

Why: A gamete carries one set, 12.
A body cell carries two sets, 24.
The zygote is made from two gametes, so it carries two sets, 24.
Every body cell grows from the zygote.

45
Check q14

In a tide pool, sea urchins release sperm and eggs into the water.

Which of the following events is fertilization?

  1. A. An egg being released into the water
    An egg in the water is still a gamete on its own; nothing has fused.
  2. B. The zygote dividing by mitosis
    The zygote divides by mitosis after fertilization has made it.
  3. C. ✓ A sperm fusing with an egg

Why: Fertilization is the fusion of two gametes.
A sperm fusing with an egg is that fusion.

46
Check q15

A salmon’s zygote holds 58 chromosomes and divides by mitosis many times as the fish grows.

How many chromosomes does each body cell of the salmon hold?

  1. A. 29
    Mitosis never halves the count; each new cell gets one copy of every chromosome.
  2. B. ✓ 58
  3. C. 116
    Before mitosis the cell copies every chromosome once.
    Mitosis then sends one copy of each to each new cell, so the count does not double.

Why: Mitosis gives each new cell the same chromosomes as the parent cell.
The zygote holds 58, so every body cell holds 58.

47
Check q16

A fish’s body cells each hold 50 chromosomes, and so do the body cells of its parents and of its offspring.

Which statement explains why the count stays at 50 from one generation to the next?

  1. A. ✓ Halving before fertilization cancels the doubling at fertilization
  2. B. Body cells discard one set soon after fertilization
    No set is discarded; the zygote keeps both sets it received and every body cell inherits them by mitosis.
  3. C. Cells keep the count at 50 by copying only 25 chromosomes
    Copying, in S phase, makes sister chromatids of every chromosome and never changes the count.
  4. D. Fertilization halves the count that two diploid gametes bring
    Fertilization adds the two gametes’ sets together; it never halves anything.

Why: Gametes carry one set, 25 chromosomes, because a halving division comes before fertilization.
Fertilization then joins the two single sets into a double set of 50.
So the halving and the doubling cancel each other, and the count stays at 50.

48
Check q17

Imagine a species whose gametes were made by mitosis, with no halving of the chromosome count, and whose eggs and sperm still fused at fertilization.

What would happen to the chromosome count of the body cells over three generations?

  1. A. The count would double once and then hold steady
    The doubling would happen again at every fertilization, not once; each generation’s gametes would carry the doubled count into the next zygote.
  2. B. The count would halve every generation
    Nothing in this species halves the count; mitosis keeps it and fertilization adds two gametes’ sets together.
  3. C. ✓ The count would double every generation
  4. D. The count would stay the same
    Fertilization never halves the count; it joins two sets into one cell.

Why: Fertilization adds the two gametes’ chromosome sets together.
With no halving division before it, every generation’s zygote would hold twice its parents’ count.
So the count would double every generation.

49
Check q18 numeric entry

A donkey’s body cells each hold 62 chromosomes, two complete sets. A donkey’s sperm holds one complete set.

Calculate the number of chromosomes in a donkey’s sperm.

Answer: 31  (tolerance ±0)

Working
Write down the values in the question:
2n=62
Write down the equation:
n=2n2
Substitute the values into the equation:
n=2n2
n=622=31
50
Practice writing an answer

A rabbit’s body cells each hold 44 chromosomes, and so do the body cells of its parents and of its young.

(a) Explain how this steady count of 44 demonstrates that a halving division comes before fertilization. (1 pt)

Model answer If gametes carried 44 chromosomes each, the zygote would hold 88 and the count would double every generation.
The count stays at 44, so each gamete must carry 22, one complete set.
The halving from 44 to 22 happens before fertilization, and fertilization then joins 22 from each gamete to make 44.
So the halving and the joining cancel each other, and every generation holds 44.
Rubric
  • Award 1 point for: a steady count means each gamete carries half (22), so the count is halved before fertilization and fertilization restores 44 (22 + 22); without the halving the count would double every generation.

Glossary

gamete
A haploid cell made to join with another and start a new organism: the sperm and the egg.
fertilization
The fusion of two gametes, a sperm and an egg, into a single cell.
zygote
The single diploid cell made by fertilization, from which every body cell of the new organism grows by mitosis.
sexual reproduction
Starting a new organism from the fusion of two gametes, one from each parent, so that the offspring carries one chromosome set from each.

APBIO-U05-L02 Which division halves the count?

Topic 5.1 · Meiosis · 65 steps

An oval cell holding four X-shaped chromosomes: a long dark X, a long light X, a short dark X and a short light X, each two chromatids joined at a white centromere dot
An oval cell holding four X-shaped chromosomes: a long dark X, a long light X, a short dark X and a short light X, each two chromatids joined at a white centromere dot

Here is the model cell after it has copied its DNA: four X-shaped chromosomes, two long and two short, each made of two sister chromatids.

It is about to divide twice, and each of the four cells it makes will hold two chromosomes. At which of the two divisions does the count halve?

Unit 5 · Heredity

1Copied once, divided twice

2

Video: Watch: Copied once, divided twice

The copied model cell: four X’s, eight chromatids. One copying, then two divisions in a row, meiosis I and meiosis II, with no copying between them; one diploid cell becomes four haploid cells.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L02a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L02a.mp4

3

How does one cell with four chromosomes become four cells with two? The cell copies its DNA once and then divides twice.

4

The first division pulls apart the two members of each homologous pair. The second division pulls apart the two sister chromatids of each chromosome.

5

The chromosome count halves at the first division. The first division is the one that sends the two members of each pair to different cells.

6

Once you know which partners part in which division, you can name the division in any drawing of a dividing cell.

7
Check q1

Before a cell divides, it copies its DNA once, in S phase.

What does that copying turn each chromosome into?

  1. A. ✓ Two identical sister chromatids joined at one centromere
  2. B. Two separate chromosomes, each with its own centromere
    Copying does not part the two copies.
    The two copies stay joined at one centromere until a division parts them.

Why: Copying turns every chromosome into two identical sister chromatids joined at one centromere.

8

The four X’s in the picture are the model cell’s four chromosomes after the copying. Each X is two sister chromatids joined at one centromere: four chromosomes, eight chromatids.

An oval cell holding four X-shaped chromosomes, two long and two short, one dark and one light of each length; a caption reads four chromosomes, eight chromatids
An oval cell holding four X-shaped chromosomes, two long and two short, one dark and one light of each length; a caption reads four chromosomes, eight chromatids
9

In every drawing in this unit, the chromosome from the mother is drawn dark and the chromosome from the father is drawn light.

10

In mitosis, one copying is followed by one division. The two cells it makes each hold the same four chromosomes as the parent cell.

11

Now consider the cells that make gametes, in the ovaries and the testes. They copy their DNA once and then divide twice in a row, making four cells.

Left: a cell with an arrow to two cells, labelled mitosis, one copying then one division. Right: a cell with an arrow to two cells and then arrows to four cells, labelled meiosis, one copying then two divisions
Left: a cell with an arrow to two cells, labelled mitosis, one copying then one division. Right: a cell with an arrow to two cells and then arrows to four cells, labelled meiosis, one copying then two divisions
12

When a diploid cell copies its DNA once and then divides twice into four haploid cells, the two divisions together are called , from a Greek word meaning to make smaller, because the chromosome count comes out halved.

13

The first of the two divisions is called and the second is called meiosis II. The cell does not copy its DNA between them.

14

Meiosis happens only in the cells that make gametes. Every other division in the body is mitosis.

15

What you are expected to know Describe meiosis as one DNA copying followed by two divisions, meiosis I and then meiosis II, that turn one diploid cell into four haploid cells.

16Quick quiz: meiosis, meiosis I, meiosis II mixed practice

17
Check q2

A cell goes through meiosis.

What is meiosis?

  1. A. One DNA copying, then one division
    One DNA copying and one division is mitosis, which makes two cells with the parent cell’s count.
  2. B. ✓ One DNA copying, then two divisions in a row
  3. C. Two DNA copyings, then two divisions in a row
    Meiosis copies the DNA once, before meiosis I, and never between the two divisions.

Why: Meiosis is one DNA copying followed by two divisions in a row, meiosis I and then meiosis II.

18
Check q3

A cell goes through meiosis.

What is meiosis I?

  1. A. ✓ The first division
  2. B. The second division
    The second division of meiosis is meiosis II.
  3. C. The DNA copying before the divisions
    The DNA copying happens in S phase, before meiosis I begins.

Why: Meiosis I is the first of the two divisions of meiosis.

19
Check q4

A cell has finished meiosis I and is about to begin meiosis II.

How many times does the cell copy its DNA between meiosis I and meiosis II?

  1. A. ✓ Never
  2. B. Once
    The one DNA copying of meiosis happens in S phase, before meiosis I.
    Nothing is copied between the two divisions.
  3. C. Twice
    Meiosis has one DNA copying in all, in S phase before meiosis I.

Why: Meiosis copies the DNA once, in S phase, before meiosis I.
Meiosis II follows meiosis I with no copying between them.
So the cell copies its DNA zero times between the two divisions.

20
Practice writing an answer

A cell goes through meiosis.

(a) State what meiosis is. (1 pt)

Model answer Meiosis is one DNA copying followed by two divisions in a row, which turns one diploid cell into four haploid cells.
Rubric
  • Award 1 point for: one DNA copying followed by two divisions in a row (one diploid cell to four haploid cells).

(b) State what meiosis I and meiosis II are. (1 pt)

Model answer Meiosis I is the first of the two divisions of meiosis and meiosis II is the second, with no DNA copying between them.
Rubric
  • Award 1 point for: meiosis I is the first division and meiosis II the second (no DNA copying between them).
21
Check q5

Suppose a cell in the anther of a lily, the part of the flower that makes pollen, goes through meiosis.

How many cells does the one cell make?

  1. A. Two
    One division makes two cells.
    Meiosis is two divisions in a row, so each of the two cells divides again.
  2. B. ✓ Four

Why: Meiosis is two divisions in a row.
The first division makes two cells.
Each of the two divides again.
So the one cell makes four cells.

22
Check q6

Suppose a diploid cell in the anther of a lily, the part of the flower that makes pollen, goes through meiosis and makes four cells.

Which of the following describes the four cells?

  1. A. ✓ Haploid
  2. B. Diploid
    The parent cell was diploid, with two chromosome sets.
    Meiosis turns one diploid cell into four haploid cells, with one set each.

Why: The lily cell began diploid.
Meiosis turns one diploid cell into four haploid cells.
So each of the four cells is haploid.

23
Practice writing an answer

A cell in the anther of a lily, the part of the flower that makes pollen, goes through meiosis and makes four cells.

(a) Explain how the four cells show that meiosis is two divisions in a row. (1 pt)

Model answer One division splits one cell into two cells.
Two cells become four cells only if each of them divides again.
So four cells from one cell means the cell divided twice.
Rubric
  • Award 1 point for: one division makes two cells, so four cells from one cell needs a second division.

24Which partners part

25

Video: Watch: Which partners part

The copied model cell in meiosis I: whole X’s move apart, the dark member of each pair to one pole and the light member to the other. One of the two new cells in meiosis II: each X splits at its centromere and the two chromatids move apart as V’s.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L02b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L02b.mp4

26
Check q7

Two chromosomes of the same length lie side by side in a cell, one from the mother and one from the father.

What are the two chromosomes called?

  1. A. Sister chromatids
    Sister chromatids are the two identical copies of one chromosome, joined at one centromere.
    Two chromosomes of the same length from the two parents are a homologous pair.
  2. B. ✓ A homologous pair

Why: One chromosome from each parent, the same length and the same genes in the same order, is a homologous pair.

27

Suppose the copied model cell begins meiosis I. Meiosis I pulls apart the two members of each homologous pair.

The model cell with a pole at each end: the long dark X and the short light X are moving toward the left pole, the long light X and the short dark X toward the right pole; every chromosome is still a whole X
The model cell with a pole at each end: the long dark X and the short light X are moving toward the left pole, the long light X and the short dark X toward the right pole; every chromosome is still a whole X
28

The dark long chromosome moves to one pole, and the light long chromosome moves to the other pole. The two short chromosomes part the same way.

29

Each chromosome moving in meiosis I stays a whole X. Its two sister chromatids are still joined at the centromere.

30

Now consider one of the two cells meiosis I made, as it begins meiosis II. Meiosis II pulls apart the two sister chromatids of each chromosome, exactly as mitosis does.

One of the two cells from meiosis I, smaller, with a pole at each end: the two chromatids of the long chromosome move apart as V shapes, one to each pole, and so do the two chromatids of the short chromosome
One of the two cells from meiosis I, smaller, with a pole at each end: the two chromatids of the long chromosome move apart as V shapes, one to each pole, and so do the two chromatids of the short chromosome
31

Each new cell receives one chromatid of every chromosome. The chromatids move as V’s, with the centromere leading.

32

Meiosis I parts the two members of each homologous pair. Meiosis II parts the two sister chromatids of each chromosome.

33

Mitosis splits every chromosome at its centromere. Meiosis I splits none of them, and that is the difference between meiosis I and mitosis.

34

Here is a table of the two divisions and the partners each one parts.

A table with two columns, meiosis I and meiosis II, and three rows: what parts (homologous pairs; sister chromatids), what moves to each pole (whole X's; single V's), whether the centromere splits (no; yes)
A table with two columns, meiosis I and meiosis II, and three rows: what parts (homologous pairs; sister chromatids), what moves to each pole (whole X's; single V's), whether the centromere splits (no; yes)
35

What you are expected to know Identify which partners part in each division: the two members of a homologous pair in meiosis I, the two sister chromatids of a chromosome in meiosis II.

36
Check q8

Two cells from a fruit fly are drawn below, numbered 1 and 2, each frozen while its chromosomes move toward the poles. Only some of the chromosomes are drawn.

Two boxed drawings numbered 1 and 2, each a cell with a pole at each end and four chromosome pieces moving, two toward each pole, each piece with a fiber to the pole on its side; in each drawing two pieces are darker and two lighter
Two boxed drawings numbered 1 and 2, each a cell with a pole at each end and four chromosome pieces moving, two toward each pole, each piece with a fiber to the pole on its side; in each drawing two pieces are darker and two lighter

Which drawing shows meiosis I?

  1. A. ✓ Drawing 1
  2. B. Drawing 2
    In drawing 2 single V-shaped chromatids move apart, one chromatid of each chromosome to each pole.
    Sister chromatids parting is meiosis II.

Why: In drawing 1 whole X-shaped chromosomes move apart, each still two sister chromatids.
Whole chromosomes moving apart means the two members of each homologous pair are parting.
That is meiosis I.

37Quick quiz: meiosis I or meiosis II? mixed practice

38
Check q9

In a cell from a hamster, the two sister chromatids of every chromosome move to opposite poles.

Which division is this?

  1. A. Meiosis I
    Meiosis I never splits a chromosome at its centromere; whole chromosomes move.
    The division that parts sister chromatids is meiosis II.
  2. B. ✓ Meiosis II

Why: Sister chromatids are the two copies joined at one centromere.
Meiosis II is the division that parts sister chromatids, as mitosis does.
So this is meiosis II.

39
Check q10

In a cell from a hamster, the two members of every homologous pair move to opposite poles.

Which division is this?

  1. A. ✓ Meiosis I
  2. B. Meiosis II
    Meiosis II parts the two sister chromatids of each chromosome, not the two members of a homologous pair.
    The division that parts homologous pairs is meiosis I.

Why: Meiosis I is the division that parts the two members of each homologous pair.
So this is meiosis I.

40
Check q11

In a cell from a moth, whole X-shaped chromosomes, each still two sister chromatids, move toward the poles.

Which division is this?

  1. A. ✓ Meiosis I
  2. B. Meiosis II
    At meiosis II each centromere splits and single chromatids move, as V shapes.
    Whole X’s moving is meiosis I.

Why: A whole X moving means nothing split at the centromere.
Meiosis I moves whole chromosomes, one member of each homologous pair to each pole.
So this is meiosis I.

41
Check q12

In a cell from a grasshopper, each new cell receives one chromatid of every chromosome.

Which division is this?

  1. A. Meiosis I
    At meiosis I each new cell receives one whole chromosome of every pair, still two chromatids.
    One chromatid of every chromosome is the outcome of meiosis II.
  2. B. ✓ Meiosis II

Why: One chromatid of every chromosome reaches each new cell only when the sister chromatids part.
Sister chromatids part at meiosis II.
So this is meiosis II.

42
Check q13

Here is one dividing cell from a fruit fly, drawn with only some of its chromosomes.

A cell with a pole at each end and four chromosome pieces moving, two toward each pole, each with a fiber to the pole on its side; two pieces are darker and two lighter
A cell with a pole at each end and four chromosome pieces moving, two toward each pole, each with a fiber to the pole on its side; two pieces are darker and two lighter

Which division is this?

  1. A. Meiosis I
    In meiosis I whole X’s move, each still two sister chromatids.
    Here single V-shaped chromatids move, so the centromeres have split: meiosis II.
  2. B. ✓ Meiosis II

Why: Single V-shaped chromatids are moving toward the poles.
Single chromatids move only after the two sister chromatids part.
Sister chromatids part at meiosis II.
So this is meiosis II.

43
Check q14

Here is one dividing cell from a beetle, drawn with only some of its chromosomes.

A cell with a pole at each end and four chromosome pieces moving, two toward each pole, each with a fiber to the pole on its side; two pieces are darker and two lighter
A cell with a pole at each end and four chromosome pieces moving, two toward each pole, each with a fiber to the pole on its side; two pieces are darker and two lighter

Which division is this?

  1. A. ✓ Meiosis I
  2. B. Meiosis II
    In meiosis II single V-shaped chromatids move.
    Here whole X’s move, each still two sister chromatids: meiosis I.

Why: Whole X-shaped chromosomes are moving toward the poles.
Whole chromosomes move apart when the two members of each homologous pair part.
Homologous pairs part at meiosis I.
So this is meiosis I.

44
Check q15

Two cells from a fruit fly are drawn below, numbered 1 and 2, each frozen while its chromosomes move toward the poles. Only some of the chromosomes are drawn. One of the two cells is in meiosis I.

Two boxed drawings numbered 1 and 2, each a cell with a pole at each end and four chromosome pieces moving, two toward each pole, each piece with a fiber to the pole on its side; in each drawing two pieces are darker and two lighter
Two boxed drawings numbered 1 and 2, each a cell with a pole at each end and four chromosome pieces moving, two toward each pole, each piece with a fiber to the pole on its side; in each drawing two pieces are darker and two lighter

What in that cell’s drawing shows meiosis I rather than meiosis II?

  1. A. The chromosomes are moving toward the two poles
    Chromosomes move toward the poles in both divisions.
    What tells them apart is whether whole X’s or single chromatids move.
  2. B. Two chromosomes are going to each of the two poles
    Two chromosomes go to each pole in drawing 2 as well, once the sister chromatids of two chromosomes have parted.
    A count per pole cannot tell the two divisions apart.
  3. C. Single chromatids are moving apart
    Single chromatids moving apart is meiosis II, which is drawing 2.
    In drawing 1 every moving chromosome is still a whole X.
  4. D. ✓ Whole chromosomes of two chromatids are moving apart

Why: In meiosis I the two members of each homologous pair part as whole chromosomes, each still two sister chromatids.
Drawing 1 shows whole X’s moving, so it is meiosis I.
Drawing 2, with single chromatids moving, is meiosis II.

45Where the count halves

46

Video: Watch: Where the count halves

The model cell dividing twice with the count written under each cell as it changes: four centromeres, then two in each cell after meiosis I, then still two in each cell after meiosis II. Count centromeres; the count halves once.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L02c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L02c.mp4

47
Check q16

This drawing shows one X shape. The X is two sister chromatids joined where the arms cross. To count chromosomes, count centromeres.

An X shape with a centromere dot where its arms cross
An X shape with a centromere dot where its arms cross

How many chromosomes does the drawing show?

  1. A. ✓ One
  2. B. Two
    The two sister chromatids share one centromere.
    To count chromosomes, count centromeres: one centromere is one chromosome.

Why: To count chromosomes, count centromeres.
The two sister chromatids are joined at one centromere.
So the drawing shows one chromosome.

48

To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

49

Before meiosis I the model cell holds four centromeres. So it holds four chromosomes.

50

After meiosis I each cell holds one member of every homologous pair: one long chromosome and one short chromosome. Each cell holds two centromeres, so it holds two chromosomes.

51

The count halved, from four to two, for one reason. The two members of each homologous pair went to different cells.

52

Now consider meiosis II. Each new cell receives one chromatid of each of the two chromosomes, so it holds two centromeres and two chromosomes.

Three stages in a row: a cell with four X's and the count 4, an arrow labelled meiosis I to two cells each with one long and one short X and the count 2, an arrow labelled meiosis II to four small cells each with one long and one short rod and the count 2
Three stages in a row: a cell with four X's and the count 4, an arrow labelled meiosis I to two cells each with one long and one short X and the count 2, an arrow labelled meiosis II to four small cells each with one long and one short rod and the count 2
53

The count did not change at meiosis II. Each chromatid was already counted inside its chromosome, as one of two sister chromatids sharing one centromere.

54

The count halves once, at meiosis I. In a human the counts are 46 chromosomes, then 23, then 23.

55

Suppose the count halved again at meiosis II. Each human gamete would then hold 11.5 chromosomes, and a cell cannot hold half a chromosome.

56

What you are expected to know Explain why the chromosome count halves at meiosis I and stays the same at meiosis II.

57
Check q17 numeric entry

A plant’s body cells each hold 12 chromosomes. One of its cells goes through meiosis. To count chromosomes, count centromeres.

Calculate the number of chromosomes in each cell after meiosis I.

Answer: 6  (tolerance ±0)

Working
Write down the values in the question:
chromosomes before meiosis = 12
Write down the equation:
chromosomes after meiosis I=chromosomes before meiosis2
Substitute the values into the equation:
chromosomes after meiosis I=122=6
58
Check q18 numeric entry

A plant’s body cells each hold 12 chromosomes. One of its cells has been through meiosis I and now finishes meiosis II.

Calculate the number of chromosomes in each of the four cells after meiosis II.

Answer: 6  (tolerance ±0)

Working
Write down the values in the question:
chromosomes after meiosis I = 6
Write down the equation:
chromosomes after meiosis II=chromosomes after meiosis I
Substitute the values into the equation:
chromosomes after meiosis II=6
59
Practice writing an answer

A plant’s body cells each hold 12 chromosomes. One of its cells goes through meiosis. After meiosis I each cell holds 6 chromosomes, and after meiosis II each cell still holds 6 chromosomes.

(a) Explain why the count falls from 12 to 6 at meiosis I. (1 pt)

Model answer Meiosis I pulls apart the two members of each homologous pair.
One member of every pair goes to each new cell.
So each new cell holds half the centromeres.
To count chromosomes, count centromeres.
So each new cell holds half the chromosomes: 6.
Rubric
  • Award 1 point for: one member of each homologous pair goes to each new cell, so each new cell holds half the centromeres and half the chromosomes.

(b) Explain why the count stays at 6 through meiosis II. (1 pt)

Model answer Meiosis II pulls apart the two sister chromatids of each chromosome.
Each new cell receives one chromatid of every chromosome.
The two sister chromatids shared one centromere.
So each chromatid was already counted inside its chromosome.
So each new cell still holds 6 chromosomes.
Rubric
  • Award 1 point for: sister chromatids part, and each chromatid was already counted as part of its chromosome (one centromere each), so the count stays the same.
60
Check q19

A cat’s body cells each hold 38 chromosomes. A student says: “The chromosome count halves at both divisions of meiosis, so each of the cat’s gametes holds 9.5 chromosomes.”

Is the student correct?

  1. A. Yes: each gamete holds 9.5 chromosomes
    A cell cannot hold half a chromosome.
    The count halves only at meiosis I, when the two members of each homologous pair part.
  2. B. ✓ No: the count halves once, so each gamete holds 19 chromosomes

Why: The count halves at meiosis I, when the two members of each homologous pair go to different cells.
Meiosis II parts sister chromatids, and each chromatid was already counted inside its chromosome.
So the count halves once: each gamete holds 19 chromosomes.

61
Check q20 numeric entry

A horse’s body cells each hold 64 chromosomes. One of its cells goes through meiosis.

Calculate the number of chromosomes in each of the four cells after meiosis II.

Answer: 32  (tolerance ±0)

Working
Write down the values in the question:
chromosomes before meiosis = 64
Write down the equation (the count halves once, at meiosis I, and stays the same at meiosis II):
chromosomes after meiosis II=chromosomes before meiosis2
Substitute the values into the equation:
chromosomes after meiosis II=642=32
62

Back to the model cell: four X-shaped chromosomes, two long and two short, each two sister chromatids, about to divide twice.

An oval cell holding four X-shaped chromosomes, two long and two short, one dark and one light of each length; a caption reads four chromosomes, eight chromatids
An oval cell holding four X-shaped chromosomes, two long and two short, one dark and one light of each length; a caption reads four chromosomes, eight chromatids
63

At meiosis I the two members of each homologous pair went to different cells: four chromosomes became two.

64

At meiosis II the two sister chromatids of each chromosome parted, and each cell kept one of everything it had: still two chromosomes.

Glossary

meiosis
One DNA copying followed by two divisions in a row, meiosis I and meiosis II, turning one diploid cell into four haploid cells; the division that makes gametes.
meiosis I and meiosis II
The two divisions of meiosis. Meiosis I parts the two members of each homologous pair and halves the chromosome count; meiosis II parts the two sister chromatids of each chromosome and leaves the count unchanged. No DNA copying happens between them.

APBIO-U05-L03 The pairs find each other

Topic 5.1 · Meiosis · 71 steps

An oval cell in which a long dark X and a long light X drift toward each other from the left half, and a short dark X and a short light X drift toward each other from the right half; small arrows show the motion
An oval cell in which a long dark X and a long light X drift toward each other from the left half, and a short dark X and a short light X drift toward each other from the right half; small arrows show the motion

Inside the model cell, the two long chromosomes are drifting toward each other, and so are the two short ones.

In mitosis the chromosomes never did this. What happens when the two long chromosomes reach each other, and how will the pairs line up when the spindle pulls the chromosomes to the middle of the cell?

Unit 5 · Heredity

1Prophase I: the pairs come together

2

Video: Watch: Prophase I, the pairs come together

The model cell entering meiosis I: chromosomes condensing, as in mitosis. Then the dark long X and the light long X slide together along their whole length, and the two short X’s do the same. While the pairs lie together, the spindle grows and the envelope breaks down.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L03a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L03a.mp4

3

What is different about the start of meiosis I? The chromosomes condense, exactly as in mitosis.

4

Then the two homologs of every homologous pair come together and lie side by side along their whole length.

5

Where an arm of one homolog crosses an arm of the other, the two homologs are held together at the crossing point.

6

While the homologs lie paired, the spindle grows and the nuclear envelope breaks down, as in mitosis.

7

The spindle then moves each homologous pair to the equator as a pair. At the equator the pairs stand in a double row.

8

The double row of pairs is how you tell meiosis I from mitosis at a glance.

9
Check q1

A body cell begins mitosis and enters prophase.

Which of the following happens in prophase?

  1. A. The two sister chromatids of each chromosome separate and move apart
    Sister chromatids separate in anaphase.
    In prophase the chromosomes condense into rods and the nuclear envelope breaks down.
  2. B. ✓ The chromosomes condense into rods and the nuclear envelope breaks down
  3. C. The cell copies its DNA, so each chromosome becomes two chromatids
    The cell copies its DNA in S phase, before mitosis begins.
    In prophase the chromosomes condense into rods and the nuclear envelope breaks down.

Why: Prophase is the first stage of mitosis.
The copied chromosomes condense into rods, the spindle begins to grow between the poles, and the nuclear envelope breaks down.

10

The first stage of meiosis I begins as prophase of mitosis does: the chromosomes condense inside the nuclear envelope.

The model cell with a pole at each end and a short spindle fiber beginning to grow from each pole; a dashed nuclear envelope is still whole, and inside it four condensed X-shaped chromosomes lie apart, two long and two short, one dark and one light of each length
The model cell with a pole at each end and a short spindle fiber beginning to grow from each pole; a dashed nuclear envelope is still whole, and inside it four condensed X-shaped chromosomes lie apart, two long and two short, one dark and one light of each length
11

Then comes something mitosis never does. The two homologs of each homologous pair come together and lie side by side along their whole length.

The model cell with the long dark X and the long light X lying together so that their arms cross, and the short dark X and the short light X lying together the same way
The model cell with the long dark X and the long light X lying together so that their arms cross, and the short dark X and the short light X lying together the same way
12

In the model cell the dark long X and the light long X lie together, and the dark short X and the light short X lie together.

13

When the two homologs of a homologous pair lie together along their whole length like this, the pairing is called , from a Greek word meaning a joining together.

14

This stage is the prophase of the first division, meiosis I. So it is called .

15

Homologs pair only in prophase I. Mitosis has no synapsis, and neither does meiosis II.

16

What you are expected to know Describe synapsis: in prophase I the two homologs of each homologous pair come together and lie side by side along their whole length, which mitosis never does.

17
Check q2

Suppose a cell in a rat’s testis begins prophase I.

What do the two homologs of each homologous pair do?

  1. A. The two homologs move to opposite poles of the cell
    In prophase I nothing moves to the poles yet.
    The two homologs of each pair come together and lie side by side along their whole length.
  2. B. The two homologs stay apart from each other, as in mitosis
    Mitosis never pairs the homologs.
    In prophase I the two homologs of each pair come together and lie side by side along their whole length.
  3. C. ✓ The two homologs lie side by side along their whole length

Why: In prophase I each homologous pair comes together.
The two homologs lie side by side along their whole length: synapsis.

18
Check q3 numeric entry

A mosquito’s body cells each hold 6 chromosomes. One of these cells is in prophase I, and every chromosome has come together with its one partner of the same length.

Calculate the number of homologous pairs in the cell.

Answer: 3  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 6
chromosomes in each homologous pair = 2
Write down the equation:
homologous pairs=chromosomeschromosomes in each pair
Substitute the values into the equation:
homologous pairs=62=3

19Quick quiz: synapsis mixed practice

20
Check q4

A cell is in prophase I.

What is synapsis?

  1. A. ✓ The two homologs of a homologous pair lying side by side along their whole length
  2. B. The two sister chromatids of a chromosome parting at the centromere
    Sister chromatids part in mitosis and in meiosis II, never in synapsis.
    Synapsis is the two homologs of a pair lying side by side along their length.
  3. C. The chromosomes standing in a single row along the equator
    A single row along the equator is metaphase of mitosis.
    Synapsis is the two homologs of a pair lying side by side along their length.

Why: Synapsis is the pairing of the two homologs of a homologous pair, side by side along their whole length, in prophase I.

21
Check q5

A cell is dividing.

In which of the following does synapsis happen?

  1. A. Mitosis
    Mitosis never pairs the homologs.
    Synapsis happens in prophase I, the first stage of meiosis I.
  2. B. ✓ Meiosis I
  3. C. Meiosis II
    In meiosis II the chromosomes lie single, and nothing pairs.
    Synapsis happens in prophase I, the first stage of meiosis I.

Why: Homologs pair only in prophase I.
Prophase I is the first stage of meiosis I.

22
Practice writing an answer

A cell is in prophase I.

(a) State what synapsis is. (1 pt)

Model answer Synapsis is the two homologs of a homologous pair coming together and lying side by side along their whole length.
Rubric
  • Award 1 point for: the two homologs of a homologous pair lie side by side along their whole length (in prophase I).

23Held where the arms cross

24

Video: Watch: Held where the arms cross

A close view of one paired homologous pair: an arm of the dark homolog crosses an arm of the light homolog and the two are held there. Then the whole model cell at the end of prophase I: both pairs paired and held at their chiasmata, the spindle reaching in, the envelope gone.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L03b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L03b.mp4

25

Look closely at the long homologous pair lying together.

A close view of the paired long chromosomes: a dark X and a light X lying together with their arms crossing; a dashed ring marks one point where an upper arm of the dark X crosses an upper arm of the light X
A close view of the paired long chromosomes: a dark X and a light X lying together with their arms crossing; a dashed ring marks one point where an upper arm of the dark X crosses an upper arm of the light X
26

At one or more points, an arm of one homolog crosses an arm of the other homolog. At each crossing point the two homologs are held to each other.

27

When two paired homologs are held together at a crossing point like this, the point is called a , after the Greek letter chi, written as an X, because the crossing looks like an X.

28

Two or more crossing points are called chiasmata.

29

By the end of prophase I the nuclear envelope is gone and the spindle reaches in from both poles. Both pairs of the model cell lie paired, each held at its chiasmata.

The model cell at the end of prophase I: both pairs lie together, each with a dashed ring at a crossing point, spindle fibers reach in from both poles and there is no nuclear envelope
The model cell at the end of prophase I: both pairs lie together, each with a dashed ring at a crossing point, spindle fibers reach in from both poles and there is no nuclear envelope
30

What you are expected to know Identify a chiasma: the crossing point at which the two paired homologs of a homologous pair are held to each other in prophase I.

31Quick quiz: chiasma mixed practice

32
Check q6

A cell is in prophase I.

What is a chiasma?

  1. A. Where two sister chromatids are joined to each other
    Two sister chromatids are joined to each other at the centromere.
    A chiasma is a crossing point between the two homologs of a pair.
  2. B. The plane halfway between the two poles of the cell
    The plane halfway between the poles is the equator.
    A chiasma is a crossing point between the two homologs of a pair.
  3. C. ✓ Where two paired homologs cross and are held to each other

Why: A chiasma is a crossing point at which two paired homologs are held to each other, where an arm of one crosses an arm of the other.

33
Check q7

A cell is in prophase I.

Which two things does a chiasma hold together?

  1. A. ✓ The two homologs of a homologous pair
  2. B. The two sister chromatids of one chromosome
    The two sister chromatids of one chromosome are held together at the centromere.
    A chiasma holds the two homologs of a pair together.
  3. C. A chromosome and a spindle fiber
    A spindle fiber attaches to a chromosome at its centromere.
    A chiasma holds the two homologs of a pair together.

Why: A chiasma is where an arm of one homolog crosses an arm of the other homolog.
The two homologs are held to each other there.

34
Practice writing an answer

A cell is in prophase I.

(a) State what a chiasma is. (1 pt)

Model answer A chiasma is a crossing point at which the two paired homologs of a homologous pair are held to each other.
Rubric
  • Award 1 point for: a crossing point at which the two paired homologs are held to each other (where an arm of one crosses an arm of the other).

35Quick quiz: prophase I or prophase of mitosis? mixed practice

36
Check q8

Here is one dividing cell from a grasshopper, drawn with only some of its chromosomes.

A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three dark and three light, two at the upper left, two at the center and two at the upper right, a small dashed ring over each place; no nuclear envelope
A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three dark and three light, two at the upper left, two at the center and two at the upper right, a small dashed ring over each place; no nuclear envelope

Which stage is the cell in?

  1. A. Prophase of mitosis
    Mitosis never pairs the homologs.
    Here each chromosome lies beside its partner of the same length, so the cell is in prophase I.
  2. B. ✓ Prophase I

Why: The chromosomes lie in groups of two, a dark X and a light X of the same length side by side.
Two chromosomes of the same length, one from each parent, are a homologous pair.
Homologs lie side by side only in prophase I.

37
Check q9

Here is one dividing cell from a mouse, drawn with only some of its chromosomes.

A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three dark and three light, at six different places across the cell, some high and some low; no nuclear envelope
A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three dark and three light, at six different places across the cell, some high and some low; no nuclear envelope

Which stage is the cell in?

  1. A. ✓ Prophase of mitosis
  2. B. Prophase I
    In prophase I each chromosome lies beside its homolog.
    Here every chromosome lies apart from the others, so the cell is in prophase of mitosis.

Why: Every chromosome lies on its own, apart from every other.
Homologs lie side by side in prophase I.
Chromosomes lying apart, with no pairing, is prophase of mitosis.

38
Check q10

Here is one dividing cell from a fruit fly, drawn with only some of its chromosomes.

A cell with a pole at each end and one spindle fiber reaching in from each; four X-shaped chromosomes of two lengths, two dark and two light, two at the upper left and two at the lower right, a small dashed ring over each place; no nuclear envelope
A cell with a pole at each end and one spindle fiber reaching in from each; four X-shaped chromosomes of two lengths, two dark and two light, two at the upper left and two at the lower right, a small dashed ring over each place; no nuclear envelope

Which stage is the cell in?

  1. A. Prophase of mitosis
    Mitosis never pairs the homologs.
    Here each chromosome lies beside its partner of the same length, held where their arms cross, so the cell is in prophase I.
  2. B. ✓ Prophase I

Why: The chromosomes lie in groups of two, a dark X and a light X of the same length side by side.
Each group is held where two arms cross: a chiasma.
Paired homologs held at chiasmata is prophase I.

39
Check q11

Here is one dividing cell from an onion, drawn with only some of its chromosomes.

A cell with a pole at each end and one spindle fiber reaching in from each; four X-shaped chromosomes of two lengths, two dark and two light, at four different places across the cell, alternately high and low; no nuclear envelope
A cell with a pole at each end and one spindle fiber reaching in from each; four X-shaped chromosomes of two lengths, two dark and two light, at four different places across the cell, alternately high and low; no nuclear envelope

Which stage is the cell in?

  1. A. ✓ Prophase of mitosis
  2. B. Prophase I
    In prophase I each chromosome lies beside its homolog.
    Here every chromosome lies apart from the others, so the cell is in prophase of mitosis.

Why: Every chromosome lies on its own, apart from every other.
Homologs lie side by side in prophase I.
Chromosomes lying apart, with no pairing, is prophase of mitosis.

40
Check q12

Here is one dividing cell from a lily, drawn with only some of its chromosomes.

A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three dark and three light, two at the upper left, two low in the middle and two at the upper right, a small dashed ring over each place; no nuclear envelope
A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three dark and three light, two at the upper left, two low in the middle and two at the upper right, a small dashed ring over each place; no nuclear envelope

Which stage is the cell in?

  1. A. Prophase of mitosis
    Mitosis never pairs the homologs.
    Here each chromosome lies beside its partner of the same length, held where their arms cross, so the cell is in prophase I.
  2. B. ✓ Prophase I

Why: The chromosomes lie in groups of two, a dark X and a light X of the same length side by side.
Each group is held where two arms cross: a chiasma.
Paired homologs held at chiasmata is prophase I.

41Metaphase I: pairs at the equator

42

Video: Watch: Metaphase I, pairs at the equator

Fibers from the left pole attach to the dark homolog of each pair and fibers from the right pole to the light homolog; the spindle moves each pair to the equator as one unit; the double row of meiosis I beside the single row of mitosis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L03c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L03c.mp4

43
Check q13

A body cell is in metaphase of mitosis.

Where do its chromosomes stand?

  1. A. ✓ In a single row along the equator
  2. B. Scattered through the cell
    Scattered chromosomes are prophase.
    In metaphase the spindle has pulled every chromosome into a single row along the equator.
  3. C. At the two poles of the cell
    Chromosomes reach the poles after anaphase.
    In metaphase every chromosome stands in a single row along the equator.

Why: In metaphase, fibers from both poles have attached to every chromosome and pulled the chromosomes into a single row along the equator, the plane halfway between the poles.

44

In metaphase of mitosis the chromosomes stand in a single row along the equator.

45

In meiosis I the spindle moves each homologous pair as a pair.

The model cell with both pairs still short of the vertical middle line, the long pair to its left and the short pair to its right: in the long pair a fiber from the left pole is attached to the centromere of the dark homolog and a fiber from the right pole to the centromere of the light homolog; in the short pair the fiber from the left pole is attached to the light homolog and the fiber from the right pole to the dark homolog
The model cell with both pairs still short of the vertical middle line, the long pair to its left and the short pair to its right: in the long pair a fiber from the left pole is attached to the centromere of the dark homolog and a fiber from the right pole to the centromere of the light homolog; in the short pair the fiber from the left pole is attached to the light homolog and the fiber from the right pole to the dark homolog
46

Fibers from one pole attach to one homolog’s centromere, and fibers from the other pole attach to the other homolog’s centromere. So the two homologs of each pair face opposite poles.

47

The pairs come to rest at the equator side by side. In the model cell, the long pair and the short pair sit across the middle as a double row, two chromosomes wide.

The model cell at metaphase I: the long pair and the short pair sit across the middle of the cell, each pair two chromosomes wide, one homolog facing each pole
The model cell at metaphase I: the long pair and the short pair sit across the middle of the cell, each pair two chromosomes wide, one homolog facing each pole
48

This stage is the metaphase of the first division, meiosis I. So it is called .

49

The exam’s word for the equator is the metaphase plate. Both names mean the plane halfway between the poles.

50

A double row of homologous pairs is metaphase I. A single row of chromosomes, one wide, is metaphase of mitosis.

Left: the model cell at metaphase I, two pairs across the middle, two chromosomes wide. Right: the same cell at metaphase of mitosis, four single X's in one column across the middle
Left: the model cell at metaphase I, two pairs across the middle, two chromosomes wide. Right: the same cell at metaphase of mitosis, four single X's in one column across the middle
51

A drawing with four single X’s in one row is metaphase of mitosis, not metaphase I.

52

Here is a table of the two kinds of metaphase and what stands at the equator in each.

A table with two columns, metaphase I and metaphase of mitosis, and three rows: what stands at the equator (homologous pairs; single chromosomes), how many chromosomes wide the row is (two; one), the row (a double row; a single row)
A table with two columns, metaphase I and metaphase of mitosis, and three rows: what stands at the equator (homologous pairs; single chromosomes), how many chromosomes wide the row is (two; one), the row (a double row; a single row)
53

What you are expected to know Identify metaphase I from a drawing: homologous pairs, not single chromosomes, lined up at the equator in a double row, the two homologs of each pair facing opposite poles.

54
Check q14

Two cells from an insect whose body cells hold six chromosomes are drawn below, numbered 1 and 2. In both, the chromosomes sit across the middle of the cell with fibers attached.

Two boxed drawings numbered 1 and 2, each a cell with a pole at each end and six X-shaped chromosomes at the middle. Drawing 1: the six X's stand one above another at the middle line, alternately a little left and a little right of it, three dark and three light, with fibers reaching them from both poles. Drawing 2: the six X's stand at the middle line in two columns, three a little left of it and three a little right, one dark and one light at each height, with fibers reaching them from both poles
Two boxed drawings numbered 1 and 2, each a cell with a pole at each end and six X-shaped chromosomes at the middle. Drawing 1: the six X's stand one above another at the middle line, alternately a little left and a little right of it, three dark and three light, with fibers reaching them from both poles. Drawing 2: the six X's stand at the middle line in two columns, three a little left of it and three a little right, one dark and one light at each height, with fibers reaching them from both poles

Which drawing shows metaphase I?

  1. A. Drawing 1
    Every chromosome sits on the equator in metaphase of mitosis too.
    Drawing 1 shows six single chromosomes in one row, which is mitosis.
  2. B. ✓ Drawing 2
  3. C. Both
    Fibers reach every chromosome in both kinds of metaphase.
    What names metaphase I is pairs at the equator, which only drawing 2 shows.

Why: Metaphase I is homologous pairs, not single chromosomes, lined up at the equator, the two homologs of each pair facing opposite poles.
Drawing 2 shows three pairs side by side.
Drawing 1 is a single row, metaphase of mitosis.

55
Practice writing an answer

Two cells from an insect whose body cells hold six chromosomes are drawn below, numbered 1 and 2. In both, the chromosomes sit across the middle of the cell with fibers attached.

Two boxed drawings numbered 1 and 2, each a cell with a pole at each end and six X-shaped chromosomes at the middle. Drawing 1: the six X's stand one above another at the middle line, alternately a little left and a little right of it, three dark and three light, with fibers reaching them from both poles. Drawing 2: the six X's stand at the middle line in two columns, three a little left of it and three a little right, one dark and one light at each height, with fibers reaching them from both poles
Two boxed drawings numbered 1 and 2, each a cell with a pole at each end and six X-shaped chromosomes at the middle. Drawing 1: the six X's stand one above another at the middle line, alternately a little left and a little right of it, three dark and three light, with fibers reaching them from both poles. Drawing 2: the six X's stand at the middle line in two columns, three a little left of it and three a little right, one dark and one light at each height, with fibers reaching them from both poles

(a) Explain why the row in drawing 2 is two chromosomes wide and the row in drawing 1 is one chromosome wide. (2 pt)

Model answer In drawing 2 the two homologs of each pair came together in prophase I.
Fibers from opposite poles attached to the two homologs.
So the spindle moved each pair to the equator as a pair.
The two homologs stand side by side: two chromosomes wide.
Drawing 1 is mitosis, which never pairs the homologs.
Fibers from both poles attached to each chromosome on its own.
So each chromosome stands alone: one chromosome wide.
Rubric
  • Award 1 point for: in drawing 2 the homologs paired in prophase I and the spindle moved each pair to the equator as a pair, so the two homologs of each pair stand side by side (meiosis I).
  • Award 1 point for: in drawing 1 the homologs never paired (mitosis), so fibers from both poles attached to each chromosome on its own and each stands alone in the row.
56

Back to the model cell as meiosis I began: the two long chromosomes were drifting toward each other, and so were the two short chromosomes.

An oval cell in which a long dark X and a long light X drift toward each other in the left half, and a short dark X and a short light X drift toward each other in the right half; small arrows show the motion
An oval cell in which a long dark X and a long light X drift toward each other in the left half, and a short dark X and a short light X drift toward each other in the right half; small arrows show the motion
57

The two long chromosomes came together and lay side by side along their whole length, held where their arms crossed.

58

The spindle then moved each homologous pair to the equator as a pair.

59

So the drawing of metaphase I shows two pairs across the middle, not four single chromosomes.

The model cell at metaphase I: the long pair and the short pair sit across the middle of the cell, each pair two chromosomes wide, one homolog facing each pole
The model cell at metaphase I: the long pair and the short pair sit across the middle of the cell, each pair two chromosomes wide, one homolog facing each pole

60Quick quiz: metaphase I mixed practice

61
Check q15

A cell is in meiosis I.

What is metaphase I?

  1. A. Single chromosomes standing in a single row, each chromosome on its own
    A single row of single chromosomes is metaphase of mitosis.
    Metaphase I is homologous pairs in a double row at the equator.
  2. B. ✓ Homologous pairs standing in a double row, the two homologs facing opposite poles
  3. C. The two homologs of each pair lying side by side along their whole length, in no row
    Homologs lying side by side in no row is synapsis, in prophase I.
    Metaphase I is homologous pairs in a double row at the equator.

Why: In meiosis I the spindle moves each homologous pair to the equator as a pair.
So the pairs stand at the equator in a double row, the two homologs of each pair facing opposite poles: metaphase I.

62
Check q16

In a cell from a dog, homologous pairs stand at the equator in a row two chromosomes wide, the two homologs of each pair facing opposite poles.

Which stage is the cell in?

  1. A. Prophase I
    In prophase I the pairs lie where they formed, with no row at the equator.
    Pairs in a double row at the equator is metaphase I.
  2. B. Metaphase of mitosis
    In metaphase of mitosis single chromosomes stand in a row one wide.
    Pairs in a double row at the equator is metaphase I.
  3. C. ✓ Metaphase I

Why: Homologous pairs at the equator in a double row, the two homologs of each pair facing opposite poles, is metaphase I.

63
Check q17

In a cell from a horse, single chromosomes stand in one row along the equator, one chromosome wide.

Is the cell in metaphase I?

  1. A. Yes
    In metaphase I homologous pairs stand at the equator, two chromosomes wide.
    A single row, one wide, is metaphase of mitosis.
  2. B. ✓ No

Why: Metaphase I is a double row of homologous pairs.
A single row of single chromosomes, one wide, is metaphase of mitosis.
So the cell is in metaphase of mitosis.

64
Practice writing an answer

A cell is in metaphase I.

(a) State what metaphase I is. (1 pt)

Model answer Metaphase I is the stage of meiosis I in which the homologous pairs stand at the equator in a double row, the two homologs of each pair facing opposite poles.
Rubric
  • Award 1 point for: homologous pairs, not single chromosomes, lined up at the equator in a double row, the two homologs of each pair facing opposite poles.

65Mixed practice mixed practice

66
Check q18

In a cell from a dog, every chromosome lies pressed against one other chromosome of the same length, the two touching from end to end. No row has formed.

Which stage is the cell in?

  1. A. ✓ Prophase I
  2. B. Metaphase I
    At metaphase I the pairs stand in a row at the equator.
    Homologs lying side by side, with no row, is prophase I.
  3. C. Prophase of mitosis
    Mitosis never pairs the homologs.
    Homologs lying side by side along their length is synapsis, in prophase I.

Why: Synapsis, the two homologs of each pair lying side by side along their length, happens only in prophase I.

67
Check q19

In a cell in prophase I, an arm of one homolog crosses an arm of the other homolog, and the two homologs are held to each other there.

What is the crossing point called?

  1. A. A centromere
    The centromere joins the two sister chromatids of one chromosome.
    The crossing point between two paired homologs is a chiasma.
  2. B. ✓ A chiasma
  3. C. A pole
    A pole is one of the two ends of the cell.
    The crossing point between two paired homologs is a chiasma.

Why: A crossing point at which two paired homologs are held to each other is a chiasma.

68
Check q20

A cell is in metaphase I.

Which way do the two homologs of each homologous pair face?

  1. A. ✓ Opposite poles
  2. B. The same pole
    Fibers from one pole attach to one homolog and fibers from the other pole attach to the other homolog.
    So the two homologs face opposite poles.

Why: Fibers from one pole attach to one homolog’s centromere and fibers from the other pole to the other homolog’s.
So the two homologs of each pair face opposite poles.

69
Check q21

Here is one dividing cell from a moth, drawn with only some of its chromosomes.

A cell with a pole at each end; six X-shaped chromosomes of three lengths stand one above another at the vertical middle line, alternately a little left and a little right of it, three dark and three light, each with a fiber from each pole
A cell with a pole at each end; six X-shaped chromosomes of three lengths stand one above another at the vertical middle line, alternately a little left and a little right of it, three dark and three light, each with a fiber from each pole

Which stage is the cell in?

  1. A. Prophase I
    In prophase I the chromosomes lie paired, away from any row.
    Here they stand in one row at the equator.
  2. B. Metaphase I
    Metaphase I is a double row of homologous pairs, two chromosomes wide.
    Here the row is one chromosome wide: metaphase of mitosis.
  3. C. ✓ Metaphase of mitosis

Why: The chromosomes stand in one row along the equator, one chromosome wide.
Each chromosome stands alone, with no homolog beside it.
A single row of single chromosomes is metaphase of mitosis.

70
Practice writing an answer

A cell from a grasshopper is drawn below, with only some of its chromosomes. Its chromosomes are condensed and there is no nuclear envelope. The cell is in prophase I.

A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three dark and three light, two at the upper left, two at the center and two at the upper right, a small dashed ring over each place; no nuclear envelope
A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three dark and three light, two at the upper left, two at the center and two at the upper right, a small dashed ring over each place; no nuclear envelope

(a) Explain how the drawing shows that the cell is in meiosis I rather than mitosis. (1 pt)

Model answer The chromosomes lie in three groups of two.
In each group a dark X and a light X of the same length lie side by side.
Two chromosomes of the same length, one from each parent, are a homologous pair.
Homologs lie side by side, in synapsis, only in prophase I of meiosis I.
Mitosis never pairs the homologs.
So the cell is in meiosis I.
Rubric
  • Award 1 point for: the homologs lie paired side by side along their length (synapsis), which happens only in prophase I of meiosis I; mitosis never pairs the homologs.

Glossary

synapsis
The pairing of the two homologs of a homologous pair side by side along their whole length, in prophase I.
prophase I
The first stage of meiosis I. The chromosomes condense, as in mitosis. Then the two homologs of each homologous pair come together and lie side by side along their whole length (synapsis). While they lie paired, the spindle grows and the nuclear envelope breaks down.
chiasma (plural chiasmata)
A crossing point at which two paired homologs are seen held to each other in prophase I, where an arm of one crosses an arm of the other.
metaphase I
The stage of meiosis I in which homologous pairs, not single chromosomes, stand at the equator in a double row, the two homologs of each pair facing opposite poles.

APBIO-U05-L04 Pulled apart, still doubled

Topic 5.1 · Meiosis · 50 steps

An oval cell with a pole at each end; at the vertical middle line two homologous pairs sit side by side, a long pair above a short pair, one homolog of each pair facing each pole: the darker long homolog faces the left pole and the darker short homolog faces the right pole; a spindle fiber runs from each pole to each centromere
An oval cell with a pole at each end; at the vertical middle line two homologous pairs sit side by side, a long pair above a short pair, one homolog of each pair facing each pole: the darker long homolog faces the left pole and the darker short homolog faces the right pole; a spindle fiber runs from each pole to each centromere

Here is the model cell at metaphase I: its two pairs sit at the equator, and spindle fibers from the two poles are attached to every pair, one homolog to each pole.

When the fibers pull, what will move: whole X-shaped chromosomes, or their halves?

Unit 5 · Heredity

1Anaphase I: whole chromosomes move

2

Video: Watch: Pulled apart, still doubled

Anaphase I as motion: the fibers shorten and one whole X of each pair moves to each pole, every X still two chromatids joined at its centromere.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L04a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L04a.mp4

3

What does the pull of meiosis I move? Whole X-shaped chromosomes move.

4

The spindle pulls the two members of each homologous pair to opposite poles. So each pole receives one chromosome of every pair, half of the chromosomes.

5

Nothing breaks at the centromere. So every chromosome arriving at a pole is still two sister chromatids.

6

Then the cell divides into two. Each new cell has one long and one short chromosome, each still doubled.

7

That is why a second division has something left to part.

8
Check q1

In anaphase of mitosis, the spindle pulls partners apart to opposite poles.

Which partners part in anaphase of mitosis?

  1. A. ✓ The two sister chromatids of each chromosome
  2. B. The two homologs of each pair
    Mitosis never pairs the homologs.
    So it has no pairs to part.

Why: In anaphase of mitosis the connection at each centromere breaks.
Then the spindle pulls the two sister chromatids to opposite poles.

9

Now the fibers of the model cell shorten. Nothing breaks at the centromere.

10

Fibers from the left pole are attached to one homolog of each pair. Fibers from the right pole are attached to the other homolog.

11

So the shortening fibers pull whole chromosomes. One homolog of each pair moves toward each pole.

The model cell with the two pairs just pulled apart: the darker long X and the lighter short X have moved a little toward the left pole, the lighter long X and the darker short X a little toward the right, each still a whole X with a fiber to its centromere
The model cell with the two pairs just pulled apart: the darker long X and the lighter short X have moved a little toward the left pole, the lighter long X and the darker short X a little toward the right, each still a whole X with a fiber to its centromere
12

In the model cell, one long X and one short X reach each pole. Every chromosome arriving at a pole is still an X: both sister chromatids joined at its centromere.

The model cell at anaphase I: the darker long X and the lighter short X near the left pole, the lighter long X and the darker short X near the right pole, every chromosome still two chromatids joined at its centromere
The model cell at anaphase I: the darker long X and the lighter short X near the left pole, the lighter long X and the darker short X near the right pole, every chromosome still two chromatids joined at its centromere
13

To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

14

The model cell had 4 chromosomes. Each pole now has 2: the chromosome number has halved.

15

Because this is the anaphase of the first division, we call it .

16

At anaphase I the spindle pulls the two members of each homologous pair to opposite poles, so each cell receives one chromosome of every pair and the chromosome number halves.

17

At anaphase I nothing splits at the centromere. Whole copied chromosomes move.

18

What you are expected to know Describe anaphase I: the spindle pulls the two members of each homologous pair to opposite poles, and every chromosome that arrives at a pole is still two sister chromatids joined at its centromere.

19
Check q2

A cell from a lily’s anther, the part of the flower that makes pollen, has reached anaphase I.

Which of the following moves toward each pole?

  1. A. Single chromatids, the two sister chromatids of each chromosome parted
    At anaphase I nothing splits at the centromere.
    The two sister chromatids stay joined.
    So whole chromosomes move.
  2. B. Whole homologous pairs, both members of each pair together
    The two members of a pair face opposite poles at metaphase I.
    The spindle pulls them apart, one member to each pole.
  3. C. ✓ Whole chromosomes, each still two joined sister chromatids

Why: Fibers from one pole are attached to one homolog of each pair, and fibers from the other pole to the other homolog.
Nothing breaks at the centromere.
So the fibers pull whole chromosomes, one member of each pair to each pole.

20
Practice writing an answer

A cell from an onion flower bud is drawn below, frozen part way through a division. Only some of its chromosomes are drawn. The drawing records anaphase I.

A cell with a pole at each end; four X-shaped chromosomes, two on the left half and two on the right half, each with a fiber running to the pole on its side; on each side one X is long and one is short; the long X is lighter on the left and darker on the right, the short X darker on the left and lighter on the right
A cell with a pole at each end; four X-shaped chromosomes, two on the left half and two on the right half, each with a fiber running to the pole on its side; on each side one X is long and one is short; the long X is lighter on the left and darker on the right, the short X darker on the left and lighter on the right

(a) Explain how the drawing shows anaphase I rather than anaphase of mitosis. (1 pt)

Model answer Every chromosome moving toward a pole is still a whole X: two sister chromatids joined at one centromere.
In anaphase of mitosis the connection at each centromere has broken.
So single chromatids move.
Here the sister chromatids are still joined.
So the two members of each homologous pair are parting: anaphase I.
Rubric
  • Award 1 point for: the moving chromosomes are still whole (two sister chromatids joined at the centromere), so homologs are parting; in anaphase of mitosis the sister chromatids have separated and single chromatids move.
21
Check q3

A student watches a cell at anaphase I and says: “Each centromere has split, so the sister chromatids are moving to opposite poles.”

Is the student correct?

  1. A. Yes, each centromere has split
    At anaphase I nothing splits at the centromere.
    Every chromosome reaching a pole is still two sister chromatids joined at one centromere.
  2. B. ✓ No, each centromere is still intact

Why: At anaphase I the spindle pulls the two members of each homologous pair apart.
Nothing splits at the centromere.
So whole chromosomes move, each still two sister chromatids.

22Quick quiz: anaphase I or anaphase of mitosis? mixed practice

23
Check q4

A cell from a grasshopper’s testis is drawn below. Only some of its chromosomes are drawn.

A cell with a pole at each end; six X shapes in two columns of three, one column left of the middle and one right of the middle, a white dot at the crossing of each X and a line from the pole on its side to that dot; in each column one X is long, one middle-length and one short
A cell with a pole at each end; six X shapes in two columns of three, one column left of the middle and one right of the middle, a white dot at the crossing of each X and a line from the pole on its side to that dot; in each column one X is long, one middle-length and one short

Which stage is the cell in?

  1. A. ✓ Anaphase I
  2. B. Anaphase of mitosis
    Every moving chromosome here is still a whole X, two sister chromatids joined at one centromere.
    Whole chromosomes moving apart is anaphase I.

Why: Each moving chromosome is still an X: two sister chromatids joined at one centromere.
Nothing has split at the centromere.
Whole chromosomes moving apart, one member of each pair each way, is anaphase I.

24
Check q5

A cell from a mouse is drawn below. Only some of its chromosomes are drawn.

A cell with a pole at each end; eight V shapes in two columns of four, the left column with its points facing the left pole and the right column with its points facing the right pole, a white dot at each point and a line from the pole on its side to that dot; in each column two V's are long and two short
A cell with a pole at each end; eight V shapes in two columns of four, the left column with its points facing the left pole and the right column with its points facing the right pole, a white dot at each point and a line from the pole on its side to that dot; in each column two V's are long and two short

Which stage is the cell in?

  1. A. Anaphase I
    Each moving shape here is a single chromatid, a V with its centromere leading.
    Single chromatids moving apart is anaphase of mitosis.
  2. B. ✓ Anaphase of mitosis

Why: Each moving shape is a single chromatid: one arm each side of a leading centromere.
The connection at each centromere has broken.
Single chromatids moving apart is anaphase of mitosis.

25
Check q6

A cell from a beetle is drawn below. Only one pair of its chromosomes is drawn.

A cell with a pole at each end; four V shapes of the same length in two columns of two, the left column with its points facing the left pole and the right column with its points facing the right pole, a white dot at each point and a line from the pole on its side to that dot; in each column one V is darker and one lighter
A cell with a pole at each end; four V shapes of the same length in two columns of two, the left column with its points facing the left pole and the right column with its points facing the right pole, a white dot at each point and a line from the pole on its side to that dot; in each column one V is darker and one lighter

Which stage is the cell in?

  1. A. Anaphase I
    Each moving shape here is a single chromatid, a V with its centromere leading.
    Single chromatids moving apart is anaphase of mitosis.
  2. B. ✓ Anaphase of mitosis

Why: Each moving shape is a single chromatid: one arm each side of a leading centromere.
The connection at each centromere has broken.
Single chromatids moving apart is anaphase of mitosis.

26
Check q7

A cell from a fruit fly’s ovary is drawn below. Only one pair of its chromosomes is drawn.

A cell with a pole at each end; two X shapes of the same length, one left of the middle and one right of the middle, a white dot at the crossing of each and a line from the pole on its side to that dot; the left X is lighter and the right X darker
A cell with a pole at each end; two X shapes of the same length, one left of the middle and one right of the middle, a white dot at the crossing of each and a line from the pole on its side to that dot; the left X is lighter and the right X darker

Which stage is the cell in?

  1. A. ✓ Anaphase I
  2. B. Anaphase of mitosis
    Both moving chromosomes here are still whole X’s, two sister chromatids joined at one centromere.
    Whole chromosomes moving apart is anaphase I.

Why: Each moving chromosome is still an X: two sister chromatids joined at one centromere.
The two members of the pair are moving to opposite poles.
Whole chromosomes moving apart is anaphase I.

27
Check q8

A cell from a rye plant’s anther, the part of the flower that makes pollen, is drawn below. Only some of its chromosomes are drawn.

A cell with a pole at each end; four X shapes in two columns of two, one column left of the middle and one right of the middle, a white dot at the crossing of each X and a line from the pole on its side to that dot; in each column one X is long and one short; on the left the long X is darker and the short X lighter, on the right the long X is lighter and the short X darker
A cell with a pole at each end; four X shapes in two columns of two, one column left of the middle and one right of the middle, a white dot at the crossing of each X and a line from the pole on its side to that dot; in each column one X is long and one short; on the left the long X is darker and the short X lighter, on the right the long X is lighter and the short X darker

Which stage is the cell in?

  1. A. ✓ Anaphase I
  2. B. Anaphase of mitosis
    Every moving chromosome here is still a whole X, two sister chromatids joined at one centromere.
    Whole chromosomes moving apart is anaphase I.

Why: Each moving chromosome is still an X: two sister chromatids joined at one centromere.
One long and one short chromosome move toward each pole.
Whole chromosomes moving apart is anaphase I.

28
Check q9

A cell is in meiosis I.

What is anaphase I?

  1. A. The stage in which the spindle pulls the two sister chromatids of each chromosome to opposite poles
    At anaphase I nothing splits at the centromere.
    The sister chromatids part later, at meiosis II.
  2. B. ✓ The stage in which the spindle pulls the two members of each homologous pair to opposite poles
  3. C. The stage in which the homologous pairs line up across the equator
    The homologous pairs line up across the equator at metaphase I.
    Anaphase I is the pull that follows.

Why: At anaphase I the spindle pulls the two members of each homologous pair to opposite poles.
Every chromosome that arrives is still two sister chromatids joined at its centromere.

29
Practice writing an answer

A cell is in meiosis I.

(a) State what anaphase I is. (1 pt)

Model answer Anaphase I is the stage of meiosis I in which the spindle pulls the two members of each homologous pair to opposite poles, each chromosome still two sister chromatids joined at its centromere.
Rubric
  • Award 1 point for: the stage of meiosis I in which the two members of each homologous pair are pulled to opposite poles (each chromosome still two sister chromatids).

30Telophase I and cytokinesis: two cells, still doubled

31

Video: Watch: Two cells, still doubled

The spindle gone, an envelope closing around each set, the furrow pinching the cell in two; the two cells shown with their doubled chromosomes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L04b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L04b.mp4

32
Check q10

In mitosis, the separated chromatids have arrived at the two poles.

Which of the following happens next?

  1. A. The DNA of every chromosome is copied again
    The DNA was copied in S phase, before mitosis began.
    After anaphase, a nuclear envelope forms around each set.
  2. B. The sister chromatids of every chromosome separate
    The sister chromatids separated at the start of anaphase.
    After they arrive at the poles, a nuclear envelope forms around each set.
  3. C. ✓ A nuclear envelope forms around each set of chromosomes

Why: After anaphase the chromosomes have arrived at the poles.
The spindle breaks down, and a nuclear envelope forms around each set: telophase.
Then cytokinesis divides the cytoplasm.

33

After anaphase of mitosis, three things happen in order:
1 the spindle breaks down
2 a nuclear envelope forms around each set of chromosomes
3 a cleavage furrow (animal cell) or a cell plate (plant cell) divides the cytoplasm

34

Exactly the same three things happen now, after anaphase I.

The model cell pinching in at the middle: at each side a dashed envelope forms around one long X and one short X; the spindle is gone
The model cell pinching in at the middle: at each side a dashed envelope forms around one long X and one short X; the spindle is gone
35

Each of the two cells has one chromosome of every pair: one long and one short.

Two separate cells side by side, each holding one long X and one short X inside a dashed envelope; in the left cell the long X is darker and the short X lighter, in the right cell the long X is lighter and the short X darker
Two separate cells side by side, each holding one long X and one short X inside a dashed envelope; in the left cell the long X is darker and the short X lighter, in the right cell the long X is lighter and the short X darker
36

Each of those chromosomes is still two sister chromatids joined at one centromere, an X.

37

One of every chromosome is one complete set. So the two cells are already haploid, even though every chromosome in them is still doubled.

38

Cytokinesis happens here, at the end of meiosis I. It happens again at the end of meiosis II: two cells now, four later.

39

Because this is the telophase of the first division, we call it .

40

What you are expected to know Describe the end of meiosis I: the spindle breaks down, a nuclear envelope forms around each set, the cell divides, and each of the two cells has one still-doubled chromosome of every pair.

41
Check q11 numeric entry

A plant’s body cells each have 6 chromosomes. One of its cells has just finished meiosis I and cytokinesis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each of the two cells.

Answer: 3  (tolerance ±0)

Working
Write down the values in the question:
chromosomes before meiosis I = 6
Write down the equation:
chromosomes after meiosis I=chromosomes before meiosis I2
Substitute the values into the equation:
chromosomes after meiosis I=62=3
42
Check q12

A plant’s cell has just finished meiosis I and cytokinesis. Look at one chromosome in one of the two new cells.

How many chromatids does that chromosome have?

  1. A. One
    Nothing splits at the centromere in meiosis I.
    Every chromosome in the two cells is still an X: two sister chromatids joined at one centromere.
  2. B. ✓ Two

Why: Nothing splits at the centromere in meiosis I.
So every chromosome in the two cells is still two sister chromatids joined at one centromere.
The chromatids part only at meiosis II.

43
Check q13

A cell from a fruit fly, whose body cells have 8 chromosomes, has just finished meiosis I and cytokinesis. Each of the two new cells has 4 chromosomes, and each of those chromosomes is still two sister chromatids.

Is each new cell haploid or diploid?

  1. A. ✓ Haploid
  2. B. Diploid
    A diploid cell has two of every chromosome, a homolog for each.
    Each new cell has one of every pair.
    So it is haploid.

Why: To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.
Each new cell has one chromosome of every pair, 4 of the 8.
One of every chromosome is one complete set.
So each new cell is haploid.

44
Practice writing an answer

Suppose a cell from a rye plant, whose body cells have 14 chromosomes, has just finished meiosis I and cytokinesis. Each of the two new cells has 7 chromosomes, and every one of those chromosomes is still two sister chromatids. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

(a) Explain how these two cells show that the chromosome number halves at meiosis I even though no chromosome has split at its centromere. (2 pt)

Model answer At anaphase I the spindle pulled the two members of each homologous pair to opposite poles.
So each new cell received one chromosome of every pair: 7 of the 14.
Nothing split at the centromere.
So each of the 7 chromosomes is still two sister chromatids joined at one centromere.
Each X counts as one chromosome.
So the number halved because the pairs parted, not because any chromosome split.
Rubric
  • Award 1 point for: each new cell received one member of every homologous pair (7 of the 14), because the homologs parted at anaphase I.
  • Award 1 point for: an X of two sister chromatids counts as one chromosome, so the chromosomes are doubled and the number has still halved.
45

Back to the model cell at metaphase I, its two pairs at the equator, each homolog attached to a fiber from one pole and its partner to a fiber from the other. When the fibers pulled, whole X-shaped chromosomes moved: one member of each pair to each pole.

46

Nothing split at the centromere. Each new cell has one long and one short chromosome, and each is still doubled.

47Quick quiz: telophase I mixed practice

48
Check q14

A cell is at the end of meiosis I.

Which of the following happens at telophase I?

  1. A. The homologous pairs come together along their length
    The homologous pairs come together in prophase I.
    At telophase I the two sets have already parted and an envelope forms around each.
  2. B. ✓ A nuclear envelope forms around each set of chromosomes
  3. C. The homologous pairs line up across the equator
    The homologous pairs line up across the equator at metaphase I.
    At telophase I the two sets sit at the poles and an envelope forms around each.

Why: At telophase I the spindle breaks down.
A nuclear envelope forms around each set of chromosomes.
Then cytokinesis divides the cell in two.

49
Practice writing an answer

A cell has reached the end of meiosis I.

(a) State what happens at telophase I. (1 pt)

Model answer At telophase I the spindle breaks down.
A nuclear envelope forms around each set of chromosomes.
Then cytokinesis divides the cell in two.
Each of the two cells has one chromosome of every pair, each still two sister chromatids.
Rubric
  • Award 1 point for: a nuclear envelope forms around each set of chromosomes at the two poles and the cell divides in two. Accept, but do not require: the spindle breaks down; each of the two cells holds one still-doubled chromosome of every pair.

Glossary

anaphase I
The stage of meiosis I in which the spindle pulls the two members of each homologous pair to opposite poles. Nothing splits at the centromere, so every chromosome arriving at a pole is still two sister chromatids.
telophase I
The end of meiosis I: the spindle breaks down, a nuclear envelope forms around each set of chromosomes and the cell divides. Each of the two cells holds one still-doubled chromosome of every pair.

APBIO-U05-L04B Which stage of meiosis I?

Topic 5.1 · Meiosis · 27 steps

Four small drawings of the model cell in a row, numbered 1 to 4: in drawing 1 whole X's move toward the poles, two each way; in drawing 2 two pairs sit side by side across the middle; in drawing 3 two cells are forming, each with two X's; in drawing 4 the homologs lie paired with arms crossing
Four small drawings of the model cell in a row, numbered 1 to 4: in drawing 1 whole X's move toward the poles, two each way; in drawing 2 two pairs sit side by side across the middle; in drawing 3 two cells are forming, each with two X's; in drawing 4 the homologs lie paired with arms crossing

Here are four drawings of the model cell, each frozen at one moment of meiosis I, in shuffled order.

Which stage is each one, and what in the drawing tells you?

Unit 5 · Heredity

1Name the stage from what the cell shows

2

Video: Watch: Which stage of meiosis I?

The four shuffled frames sorted into order as the viewer watches, with the tell-tale feature of each circled: paired homologs, a double row, whole X’s moving, two cells forming.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L04Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L04Ba.mp4

3

How do you name the stage of a cell in meiosis I? You look at what the chromosomes are doing, never at their shape alone.

4

Every chromosome is an X in all four stages. So the X alone names nothing.

5

The four stages of meiosis I happen in this order: prophase I, metaphase I, anaphase I and telophase I. Each stage has one feature that names it.

Four boxed drawings of the model cell in order, labelled prophase I, metaphase I, anaphase I and telophase I: paired homologs with arms crossing; two pairs across the middle; whole X's moving to the poles; two forming cells each with two X's
Four boxed drawings of the model cell in order, labelled prophase I, metaphase I, anaphase I and telophase I: paired homologs with arms crossing; two pairs across the middle; whole X's moving to the poles; two forming cells each with two X's
6

In drawing 1, whole X-shaped chromosomes are moving toward the poles, one long and one short each way. Whole chromosomes moving name it: anaphase I.

The model cell with whole X's moving toward the poles: a long and a short darker X toward the left pole, a long and a short lighter X toward the right
The model cell with whole X's moving toward the poles: a long and a short darker X toward the left pole, a long and a short lighter X toward the right
7

In drawing 2, two pairs sit side by side across the middle of the cell, with fibers from both poles. Pairs at the equator name it: metaphase I.

The model cell with two pairs sitting side by side across the middle, fibers from both poles
The model cell with two pairs sitting side by side across the middle, fibers from both poles
8

In drawing 3, two groups of X’s sit at the poles, an envelope forming around each, and a furrow pinches between them. Two forming cells name it: telophase I.

The model cell pinched in the middle, a dashed envelope forming around one long and one short X at each side
The model cell pinched in the middle, a dashed envelope forming around one long and one short X at each side
9

In drawing 4, the two homologs of each pair lie together, an arm of one crossing an arm of the other, with no envelope. Paired homologs name it: prophase I.

The model cell with the two long X's lying together and the two short X's lying together, arms crossing, spindle reaching in, no envelope
The model cell with the two long X's lying together and the two short X's lying together, arms crossing, spindle reaching in, no envelope
10

Drawing 4 and drawing 2 both show the homologs of each pair side by side.

11

In drawing 2 the pairs stand at the equator with fibers from both poles. That row names metaphase I.

12

Drawing 1 and drawing 3 both show two groups of X’s.

13

In drawing 1 the groups are still moving, with fibers attached. In drawing 3 an envelope is forming around each group, and a furrow pinches between them.

14

Here is a table of the four stages of meiosis I, what the chromosomes are doing in each, and the one feature that names it.

A table of the four stages of meiosis I, what the chromosomes are doing in each, and the one feature that names it
15

What you are expected to know Name prophase I, metaphase I, anaphase I or telophase I from a drawn or a described cell, by what its chromosomes are doing.

16
Check q1

In a cell from a mouse’s ovary, the homologous pairs sit side by side across the middle of the cell, one homolog of each pair facing each pole, with a fiber from each pole to each pair.

Which stage of meiosis I is the cell in?

  1. A. Prophase I
    In prophase I the paired homologs lie scattered through the cell with no row.
    Here every pair stands across the middle with fibers from both poles.
  2. B. ✓ Metaphase I
  3. C. Anaphase I
    In anaphase I the two members of each pair have parted and are moving toward the poles.
    Here every pair still sits together at the middle.
  4. D. Telophase I
    In telophase I the chromosomes sit at the two ends and the cell is pinching in two.
    Here the cell is whole and the pairs sit at its middle.

Why: The pairs stand side by side across the middle of the cell, one homolog of each pair facing each pole.
A double row of pairs at the equator names metaphase I.

17
Check q2

In a cell from a grasshopper’s testis, the two homologs of each pair lie together along their length, an arm of one crossing an arm of the other, and there is no nuclear envelope.

Which stage of meiosis I is the cell in?

  1. A. ✓ Prophase I
  2. B. Metaphase I
    At metaphase I the pairs stand in a row across the middle of the cell.
    Here the pairs lie together but in no row.
  3. C. Anaphase I
    In anaphase I the two members of each pair have parted and whole X’s move toward the poles.
    Here each pair still lies together.
  4. D. Telophase I
    In telophase I two cells are forming with the chromosomes gathered at the poles.
    Here there is one cell with paired chromosomes inside it.

Why: The two homologs of each pair lie together along their length, held where an arm of one crosses an arm of the other.
Paired homologs name prophase I.

18

Back to the four drawings of the model cell, each frozen at one moment of meiosis I and shuffled: here they are again, numbered 1 to 4. Each one is placed by what its chromosomes are doing.

Four boxed drawings of the model cell numbered 1 to 4: whole X's moving to the poles; two pairs across the middle; two forming cells each with two X's; the homologs lying paired
Four boxed drawings of the model cell numbered 1 to 4: whole X's moving to the poles; two pairs across the middle; two forming cells each with two X's; the homologs lying paired
19

Whole X’s moving name drawing 1 anaphase I. Pairs at the equator name drawing 2 metaphase I.

20

Two forming cells name drawing 3 telophase I. Paired homologs name drawing 4 prophase I.

21Quick quiz: which stage of meiosis I? mixed practice

22
Check q3

A cell from a plant whose body cells hold six chromosomes is drawn below.

A cell narrowed at its middle into two lobes; in each lobe a dashed oval encloses three X-shaped chromosomes of three different lengths standing side by side, the longest nearest the end of the cell; in the left lobe the long and the short X are darker and the middle one lighter, in the right lobe the long and the short are lighter and the middle one darker
A cell narrowed at its middle into two lobes; in each lobe a dashed oval encloses three X-shaped chromosomes of three different lengths standing side by side, the longest nearest the end of the cell; in the left lobe the long and the short X are darker and the middle one lighter, in the right lobe the long and the short are lighter and the middle one darker

Which stage of meiosis I is the cell in?

  1. A. Prophase I
    In prophase I the homologs lie paired inside one undivided cell.
    Here the cell is pinching in two and the chromosomes are gathered at the poles.
  2. B. Metaphase I
    At metaphase I the pairs sit across the middle of one cell.
    Here the chromosomes are at the two ends and a furrow divides the cell.
  3. C. Anaphase I
    In anaphase I the chromosomes are still moving through one cell with fibers attached.
    Here the spindle is gone, envelopes are forming and the cell is pinching in two.
  4. D. ✓ Telophase I

Why: Two groups of X’s sit at the poles, an envelope forming around each, and a furrow pinches between them.
Two cells forming, each chromosome still doubled, name telophase I.

23
Check q4

A cell from a plant whose body cells hold six chromosomes is drawn below.

A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three darker and three lighter, two at the upper left, two low in the middle and two at the upper right, a small dashed ring over each place; no envelope
A cell with a pole at each end and one spindle fiber reaching in from each; six X-shaped chromosomes of three lengths, three darker and three lighter, two at the upper left, two low in the middle and two at the upper right, a small dashed ring over each place; no envelope

Which stage of meiosis I is the cell in?

  1. A. ✓ Prophase I
  2. B. Metaphase I
    At metaphase I the pairs sit in a row across the middle of the cell.
    Here they lie scattered, still paired, with the spindle only reaching in.
  3. C. Anaphase I
    In anaphase I the homologs have parted and whole X’s move toward the poles.
    Here each pair still lies together.
  4. D. Telophase I
    Telophase I shows two forming cells with the chromosomes gathered at the poles.
    Here there is one cell with paired chromosomes scattered inside it.

Why: The homologs lie paired along their length, arms crossing, with the spindle reaching in and no envelope.
Paired homologs name prophase I.

24
Check q5

A cell from a plant whose body cells hold six chromosomes is drawn below.

A cell with a pole at each end; six X-shaped chromosomes, three on the left half and three on the right half, one long, one middle-length and one short on each side, each with a fiber running to the pole on its side; on the left the long and the short X are darker and the middle one lighter, on the right the long and the short are lighter and the middle one darker
A cell with a pole at each end; six X-shaped chromosomes, three on the left half and three on the right half, one long, one middle-length and one short on each side, each with a fiber running to the pole on its side; on the left the long and the short X are darker and the middle one lighter, on the right the long and the short are lighter and the middle one darker

Which stage of meiosis I is the cell in?

  1. A. Prophase I
    In prophase I the homologs lie paired.
    Here they have parted and are moving toward the poles.
  2. B. Metaphase I
    At metaphase I the pairs sit still at the equator.
    Here the chromosomes have left the middle and are moving toward the poles.
  3. C. ✓ Anaphase I
  4. D. Telophase I
    In telophase I the chromosomes have arrived at the poles, envelopes are forming and the cell is pinching in two.
    Here the chromosomes are still moving with fibers attached.

Why: Whole X-shaped chromosomes are moving toward the poles, one of each pair each way.
Whole chromosomes moving name anaphase I.

25
Check q6

A cell from a plant whose body cells hold six chromosomes is drawn below.

A cell with a pole at each end; six X-shaped chromosomes of three lengths stand at the vertical middle line in two columns, three a little left of it and three a little right, one darker and one lighter at each height, one above another; a fiber reaches from the left pole to each X on the left and from the right pole to each X on the right
A cell with a pole at each end; six X-shaped chromosomes of three lengths stand at the vertical middle line in two columns, three a little left of it and three a little right, one darker and one lighter at each height, one above another; a fiber reaches from the left pole to each X on the left and from the right pole to each X on the right

Which stage of meiosis I is the cell in?

  1. A. Prophase I
    In prophase I the paired homologs lie scattered through the cell.
    Here every pair sits at the middle line with fibers attached from both poles.
  2. B. ✓ Metaphase I
  3. C. Anaphase I
    In anaphase I the homologs have parted and are moving toward the poles.
    Here every pair still sits together at the middle of the cell.
  4. D. Telophase I
    In telophase I the chromosomes have reached the poles and the cell is pinching in two.
    Here the cell is whole and the pairs sit at its middle.

Why: Three pairs sit across the middle of the cell, side by side, each pair two chromosomes wide.
A fiber runs from one pole to one homolog and from the other pole to the other.
Pairs at the equator name metaphase I.

26
Check q7

A cell from a mouse is drawn below. Only some of the chromosomes are drawn.

A cell narrowed at its middle into two lobes; in each lobe a dashed oval encloses one long X and one short X; in the left lobe the long X is darker and the short X lighter, in the right lobe the long X is lighter and the short X darker
A cell narrowed at its middle into two lobes; in each lobe a dashed oval encloses one long X and one short X; in the left lobe the long X is darker and the short X lighter, in the right lobe the long X is lighter and the short X darker

Which stage of meiosis I is the cell in?

  1. A. Prophase I
    In prophase I the homologs lie paired inside one undivided cell.
    Here the cell is pinching in two and an envelope is forming around each group.
  2. B. Metaphase I
    At metaphase I the pairs sit across the middle of one cell.
    Here the chromosomes sit at the two ends and a furrow divides the cell.
  3. C. Anaphase I
    In anaphase I the chromosomes are still moving through one cell with fibers attached.
    Here the spindle is gone, envelopes are forming and the cell is pinching in two.
  4. D. ✓ Telophase I

Why: Two groups of X’s sit at the poles, an envelope is forming around each group, and a furrow pinches between them.
Two cells forming, each chromosome still an X, name telophase I.

APBIO-U05-L05 The second division

Topic 5.1 · Meiosis · 49 steps

An oval cell holding one long X-shaped chromosome above one short X-shaped chromosome, each two arms crossing at a small dot; a dashed nuclear envelope surrounds them
An oval cell holding one long X-shaped chromosome above one short X-shaped chromosome, each two arms crossing at a small dot; a dashed nuclear envelope surrounds them

Here is one of the two cells from meiosis I. It holds one long chromosome and one short one. Each chromosome is still two sister chromatids joined at its centromere.

The cell is going to divide again. It has not copied its DNA. What will it do to the two X-shaped chromosomes, and what comes out?

Unit 5 · Heredity

1The mitosis stages, repeated by a haploid cell

2

Video: Watch: The second division

The mitosis stage sequence replayed on the haploid cell: a new spindle, one row, the two chromatids of each X parting as V's, then two cells of single rods.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L05a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L05a.mp4

3

What does the second division do? Meiosis II is the four stages of mitosis, repeated by a haploid cell.

4

A spindle forms. The chromosomes line up in one row at the equator, each on its own.

5

Then the spindle pulls the two sister chromatids of every chromosome apart, to opposite poles. The cell divides.

6

The DNA was not copied between the two divisions. So meiosis ends with four haploid cells whose chromosomes are single chromatids.

7
Check q1

A skin cell is dividing by mitosis. It has reached anaphase, and the connection at each centromere has broken.

What does the spindle pull to opposite poles?

  1. A. The two members of each homologous pair
    Mitosis never pairs homologous chromosomes.
    At anaphase of mitosis the connection at each centromere has broken, so the spindle pulls single chromatids.
  2. B. Whole chromosomes, each still two sister chromatids
    The connection at each centromere has already broken.
    So the spindle pulls single chromatids, one of each chromosome, to each pole.
  3. C. ✓ The two sister chromatids of each chromosome

Why: In mitosis the connection at each centromere breaks at anaphase.
So the two sister chromatids of each chromosome separate.
The spindle then pulls them to opposite poles.

8

The second division repeats those four stages of mitosis. Because they are the stages of the second division, they are called .

9

First, a new spindle forms in each of the two cells from meiosis I. Each chromosome attaches to fibers from both poles at its centromere.

A small haploid cell with a pole at each end and a new spindle reaching in; one long X and one short X lie apart from each other, and a fiber from each pole reaches toward the centromere of each X
A small haploid cell with a pole at each end and a new spindle reaching in; one long X and one short X lie apart from each other, and a fiber from each pole reaches toward the centromere of each X
10

A haploid cell has no homologous pairs. So each chromosome stays on its own; nothing pairs up.

11

Next, the fibers pull the chromosomes into one row at the equator, as in metaphase of mitosis.

The small haploid cell at metaphase II: the long X above the short X in one column across the middle, a fiber from each pole to each centromere
The small haploid cell at metaphase II: the long X above the short X in one column across the middle, a fiber from each pole to each centromere
12

Metaphase I had a double row of homologous pairs. Metaphase II has one row of chromosomes on their own.

13

At anaphase II the connection at each centromere breaks. The spindle pulls the two sister chromatids of each chromosome to opposite poles.

The small haploid cell at anaphase II: the two chromatids of the long chromosome move apart as V shapes toward the two poles, and so do the two chromatids of the short chromosome
The small haploid cell at anaphase II: the two chromatids of the long chromosome move apart as V shapes toward the two poles, and so do the two chromatids of the short chromosome
14

So each cell receives one chromatid of every chromosome. The chromosome number stays the same.

15

To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

16

Each separated chromatid has its own centromere. So each cell still holds as many chromosomes as before.

17

Last, a nuclear envelope re-forms around each group of chromosomes. The chromosomes uncoil.

The small cell pinching in two, a dashed envelope forming at each side around one long rod and one short rod
The small cell pinching in two, a dashed envelope forming at each side around one long rod and one short rod
18

Each cell then divides in two. From the two cells of meiosis I come four cells.

19

Only the dividing cell differs from mitosis: it is haploid, with no homologous pairs.

20

What you are expected to know Describe meiosis II as the four stages of mitosis, repeated by each haploid cell from meiosis I, with the two sister chromatids of every chromosome pulled apart at anaphase II.

21
Check q2

One of the two cells from a grasshopper’s meiosis I has reached anaphase II.

What arrives at each pole?

  1. A. One whole chromosome of every homologous pair, still two chromatids
    Homologous pairs parted at anaphase I.
    A cell in meiosis II has no pairs; at anaphase II the two sister chromatids of each chromosome part.
  2. B. ✓ One chromatid of every chromosome
  3. C. Both chromatids, still joined at their centromere
    At anaphase II the connection at each centromere breaks.
    The two sister chromatids of each chromosome go to opposite poles, so nothing arrives still joined.

Why: At anaphase II the connection at each centromere breaks.
The spindle pulls the two sister chromatids of each chromosome to opposite poles.
So each pole receives one chromatid of every chromosome.

22
Check q3

A student says: “Before meiosis II begins, the cell copies its DNA again.”

Is the student correct?

  1. A. ✓ No — the cell does not copy its DNA between meiosis I and meiosis II
  2. B. Yes — a cell always copies its DNA before it divides
    Every chromosome entering meiosis II is already two sister chromatids, from the copying in S phase.
    No second copying happens.

Why: The cell copied its DNA once, in S phase before meiosis I.
Each chromosome left meiosis I still two sister chromatids.
So the cell enters meiosis II with those chromosomes and copies nothing.

23
Check q4

An insect’s body cells hold six chromosomes. One of the two cells its meiosis I produced is drawn below, part way through meiosis II.

A small oval cell with a dot at each end; three X shapes of three different lengths stand one above another down the middle of the cell; lines reach from the dot at each end of the cell to the crossing point of every X
A small oval cell with a dot at each end; three X shapes of three different lengths stand one above another down the middle of the cell; lines reach from the dot at each end of the cell to the crossing point of every X

Which stage of meiosis II is the cell in?

  1. A. Prophase II
    In prophase II the chromosomes lie scattered, only attaching to a forming spindle.
    Here they stand in one row across the middle, with fibers from both poles.
  2. B. ✓ Metaphase II
  3. C. Anaphase II
    At anaphase II the sister chromatids have parted and single chromatids are moving toward the poles.
    Here every chromosome is still an X at the equator.
  4. D. Telophase II
    At telophase II the chromosomes have reached the poles and envelopes are forming.
    Here the chromosomes sit in a row at the middle.

Why: Three X-shaped chromosomes stand in one row at the equator, one deep, with fibers from both poles.
One row of chromosomes on their own, before their chromatids part, is metaphase II.

24Quick quiz: prophase II, metaphase II, anaphase II, telophase II mixed practice

25
Check q5

Here is a cell from a lizard’s ovary, part way through meiosis II. Only some of its chromosomes are drawn.

A small cell pinched in at its middle; on each side a dashed oval holds three straight rods of three different lengths, each rod with a small dot part way along it
A small cell pinched in at its middle; on each side a dashed oval holds three straight rods of three different lengths, each rod with a small dot part way along it

Which stage of meiosis II is the cell in?

  1. A. Prophase II
    In prophase II the chromosomes are still X shapes attaching to a new spindle.
    Here every chromosome is a single rod inside a forming envelope.
  2. B. Metaphase II
    In metaphase II the X-shaped chromosomes stand in one row at the equator.
    Here the chromosomes sit at the two ends, in forming envelopes.
  3. C. Anaphase II
    In anaphase II the chromatids are still moving toward the poles, as V shapes.
    Here they have arrived, and an envelope is forming around each group.
  4. D. ✓ Telophase II

Why: The cell is pinching in two.
A nuclear envelope is forming around each group, and every chromosome is a single rod.
Envelopes re-forming around single chromatids is telophase II.

26
Check q6

Here is a cell from a moth’s testis, part way through meiosis II. Only some of its chromosomes are drawn.

A small oval cell with a dot at each end; two V shapes point toward the left end and two V shapes point toward the right end, one long and one short on each side, with a line from the nearer end dot to the point of each V
A small oval cell with a dot at each end; two V shapes point toward the left end and two V shapes point toward the right end, one long and one short on each side, with a line from the nearer end dot to the point of each V

Which stage of meiosis II is the cell in?

  1. A. Prophase II
    In prophase II each chromosome is still a whole X.
    Here the chromatids have parted and are moving as V shapes.
  2. B. Metaphase II
    In metaphase II the whole X shapes stand in one row at the equator.
    Here the chromatids have parted and left the equator.
  3. C. ✓ Anaphase II
  4. D. Telophase II
    In telophase II the chromatids have reached the poles and envelopes are forming.
    Here the V shapes are still moving, with fibers pulling them.

Why: The chromatids have parted and move toward the two poles as V shapes, the centromere leading.
Sister chromatids moving apart is anaphase II.

27
Check q7

Here is a cell from a newt’s ovary, part way through meiosis II. Only some of its chromosomes are drawn.

A small oval cell with a dot at each end; three X shapes of three different lengths lie scattered at different heights and positions inside the cell; two short lines reach in from the end dots; no oval is drawn around the X shapes
A small oval cell with a dot at each end; three X shapes of three different lengths lie scattered at different heights and positions inside the cell; two short lines reach in from the end dots; no oval is drawn around the X shapes

Which stage of meiosis II is the cell in?

  1. A. ✓ Prophase II
  2. B. Metaphase II
    In metaphase II the X shapes stand in one row at the equator.
    Here they lie scattered, with the spindle only starting to reach them.
  3. C. Anaphase II
    In anaphase II the chromatids have parted into V shapes.
    Here every chromosome is still a whole X.
  4. D. Telophase II
    In telophase II the chromosomes sit at the poles in forming envelopes.
    Here they lie scattered, with no envelope.

Why: Whole X-shaped chromosomes lie scattered, and a new spindle is reaching in from the poles.
No chromosome has a partner, so this is a haploid cell.
A new spindle reaching scattered chromosomes is prophase II.

28
Check q8

Here is a cell from a grass plant’s anther, part way through meiosis II. Only some of its chromosomes are drawn.

A small oval cell with a dot at each end; four X shapes of four different lengths stand one above another in a straight column down the middle of the cell; lines reach from the dot at each end to the crossing point of every X
A small oval cell with a dot at each end; four X shapes of four different lengths stand one above another in a straight column down the middle of the cell; lines reach from the dot at each end to the crossing point of every X

Which stage of meiosis II is the cell in?

  1. A. Prophase II
    In prophase II the chromosomes lie scattered.
    Here they stand in one straight row with fibers from both poles.
  2. B. ✓ Metaphase II
  3. C. Anaphase II
    In anaphase II the chromatids have parted and move as V shapes.
    Here every chromosome is still a whole X standing at the equator.
  4. D. Telophase II
    In telophase II the chromosomes sit at the poles in forming envelopes.
    Here they stand in a row at the middle.

Why: Four X-shaped chromosomes stand in one row at the equator, one deep, with a fiber from each pole to every centromere.
One row of whole chromosomes is metaphase II.

29
Check q9

Here is a cell from a frog’s testis, part way through meiosis II. Only some of its chromosomes are drawn.

A small oval cell with a dot at each end; three V shapes of three lengths point toward the left end and three point toward the right end, with a line from the nearer end dot to the point of each V
A small oval cell with a dot at each end; three V shapes of three lengths point toward the left end and three point toward the right end, with a line from the nearer end dot to the point of each V

Which stage of meiosis II is the cell in?

  1. A. Prophase II
    In prophase II each chromosome is a whole X.
    Here the chromatids have parted into V shapes moving toward the poles.
  2. B. Metaphase II
    In metaphase II the X shapes stand in one row at the equator.
    Here the chromatids have parted and are moving away from it.
  3. C. ✓ Anaphase II
  4. D. Telophase II
    In telophase II an envelope forms around each group at the poles.
    Here the V shapes are still on their way, with fibers pulling them.

Why: Three V shapes move toward each pole, the centromere leading, with fibers still attached.
Sister chromatids moving apart is anaphase II.

30
Practice writing an answer

A cell from a pea plant’s anther, the part of the flower that makes pollen, has finished meiosis I. Each of the two cells it made holds 7 chromosomes, every chromosome still two sister chromatids. No DNA is copied before meiosis II.

(a) Explain how meiosis II turns each of these two cells into two cells that each hold 7 chromosomes, every chromosome a single chromatid. (2 pt)

Model answer At metaphase II the 7 chromosomes stand in one row at the equator, each on its own.
At anaphase II the connection at each centromere breaks.
The spindle pulls the two sister chromatids of each chromosome to opposite poles.
So each pole receives one chromatid of every chromosome.
A nuclear envelope forms around each group, and the cell divides in two.
Each separated chromatid has its own centromere.
So each new cell holds 7 chromosomes, each a single chromatid.
Rubric
  • Award 1 point for: at anaphase II the spindle pulls the two sister chromatids of each chromosome to opposite poles, so each pole receives one chromatid of every chromosome (the chromosomes having stood in one row, each on its own, at metaphase II).
  • Award 1 point for: a nuclear envelope forms around each group and the cell divides; each separated chromatid has its own centromere, so each new cell holds 7 chromosomes, every one a single chromatid.

31Four haploid cells

32

Video: Watch: Four haploid cells

The four finished cells side by side: one long and one short single chromatid in each, one chromosome of every pair, and in an animal the gametes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L05b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L05b.mp4

33

So meiosis ends with four cells. Each cell holds one chromosome of every homologous pair.

Four small round cells in a row, each holding one long rod and one short rod with a small dot on each; a label over the first two cells reads from one of the two cells of meiosis I, and a label over the last two reads from the other cell
Four small round cells in a row, each holding one long rod and one short rod with a small dot on each; a label over the first two cells reads from one of the two cells of meiosis I, and a label over the last two reads from the other cell
34

Each of those chromosomes is now a single chromatid. In the model cell, that is one long rod and one short rod.

35

Each of the four cells is haploid: one complete set, one chromosome of every pair. Here is the count for a human cell.

36
Worked example

A human body cell holds 46 chromosomes. How many chromosomes does each of the four cells hold when meiosis ends?

Write down the values in the question:
chromosomes in the body cell = 46
Write down the equation:
pairs=chromosomes in the body cell2
chromosomes in each cell after meiosis=pairs
Substitute the values into the equation:
pairs=462=23
chromosomes in each cell after meiosis=23
37

Every one of those 23 chromosomes is a single chromatid. So each of the four cells holds 23 chromatids as well.

38

In an animal the four cells are, or become, the gametes: sperm in a testis, eggs in an ovary. The four cells are also called daughter cells.

39

The four cells are not all identical to one another. None is identical to the cell that began meiosis.

40

Each cell holds one member of each homologous pair. The two members of a pair may carry different alleles.

41

Plants and fungi add one step: their four haploid cells first grow into other cells, and only those later cells make the gametes.

42

What you are expected to know State the outcome of meiosis: four haploid cells whose chromosomes are single chromatids, which in an animal become the gametes.

43
Check q10

A cell in a frog’s ovary completes meiosis.

How many cells does meiosis make from the one cell?

  1. A. Two
    Meiosis I makes two cells.
    Then each of the two divides again in meiosis II, so four cells come out.
  2. B. ✓ Four
  3. C. Eight
    Eight cells would need a third division.
    Meiosis has two divisions: one cell becomes two, then the two become four.

Why: Meiosis I divides the one cell into two.
Meiosis II divides each of the two into two more.
So meiosis makes four cells.

44
Check q11

Here is one chromosome from one of the four cells a newt’s meiosis made.

What is the chromosome?

  1. A. ✓ A single chromatid
  2. B. Two sister chromatids joined at one centromere
    Anaphase II parted the two sister chromatids of every chromosome.
    So a chromosome in a finished cell is a single chromatid.

Why: At anaphase II the spindle pulled the two sister chromatids of every chromosome to opposite poles.
So every chromosome in the four finished cells is a single chromatid.

45
Check q12

A fruit fly’s body cells hold eight chromosomes. One of its cells completes meiosis.

What does each of the four cells hold?

  1. A. ✓ Four single-chromatid chromosomes, one of every pair
  2. B. Four chromosomes, each still two sister chromatids
    Anaphase II parted the sister chromatids.
    So every chromosome in the finished cells is a single chromatid.
  3. C. Two single-chromatid chromosomes, half of a set
    The count halves once, at meiosis I, from eight to four.
    Meiosis II parts chromatids and leaves four chromosomes in each cell.
  4. D. Eight single-chromatid chromosomes, two of every kind
    Meiosis I sent one member of every pair to each cell.
    So each of the four holds four chromosomes, not eight.

Why: Meiosis I gives each cell one of every pair: four chromosomes from eight.
Meiosis II parts the sister chromatids.
So each of the four cells ends with four single-chromatid chromosomes, a complete haploid set.

46
Check q13 numeric entry

A sheep’s body cells hold 54 chromosomes. One cell completes meiosis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each of the four cells.

Answer: 27  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the body cell = 54
Write down the equation:
pairs=chromosomes in the body cell2
chromosomes in each cell after meiosis=pairs
Substitute the values into the equation:
pairs=542=27
chromosomes in each cell after meiosis=27
47

Back to the one cell from meiosis I: one long chromosome and one short, each still two sister chromatids joined at its centromere, and no copying to come.

48

Its spindle pulled the two sister chromatids of each chromosome apart, and it divided in two. The other cell from meiosis I did the same, so four haploid cells came out, each holding one long and one short single chromatid.

Glossary

prophase II, metaphase II, anaphase II, telophase II
The four stages of meiosis II: the same sequence as mitosis, repeated by each haploid cell from meiosis I. At anaphase II the two sister chromatids of each chromosome go to opposite poles, so the chromosome number stays the same.

APBIO-U05-L05B Count the DNA, not the chromosomes

Topic 5.1 · Meiosis · 51 steps

A cell on the left labelled 6 pg; then an arrow to one larger cell with a blank label, an arrow to two cells with blank labels, and an arrow to four small cells with blank labels; captions read before copying, after S phase, after meiosis I, after meiosis II
A cell on the left labelled 6 pg; then an arrow to one larger cell with a blank label, an arrow to two cells with blank labels, and an arrow to four small cells with blank labels; captions read before copying, after S phase, after meiosis I, after meiosis II

Here is one cell, and the amount of DNA it holds: 6 picograms (pg), measured before the cell copies its DNA.

Beside it stand the cells it will become: one cell after the copying, two after meiosis I, four after meiosis II. Each cell has an empty label. How much DNA does each of those cells hold?

Unit 5 · Heredity

1Doubled once, halved twice

2

Video: Watch: Doubled once, halved twice

The DNA per cell as a bar that doubles once at S phase and then halves twice, at meiosis I and at meiosis II, with no copying between the two divisions; beside it the chromosome count, halving only once.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L05Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L05Ba.mp4

3

Why does the DNA per cell halve twice, while the chromosome number halves only once? The answer is in when the DNA is copied, and in how a chromosome is counted.

4

The cell copies its DNA once, in S phase before meiosis I. Between meiosis I and meiosis II the cell copies nothing.

5

So the DNA per cell doubles once and is halved twice. The 6 pg cell becomes 12 pg, then 6 pg in each cell after meiosis I, then 3 pg in each cell after meiosis II.

6

A chromosome is counted by its centromere. So the chromosome number halves only when the homologous pairs part, at meiosis I.

7
Check q1

A liver cell holds 4 pg of DNA before S phase. In S phase it copies its DNA.

How much DNA does the liver cell hold after S phase?

  1. A. 2 pg
    Copying adds DNA; nothing in S phase removes any.
  2. B. 4 pg
    Copying makes a second copy of every chromosome, so the amount cannot stay the same.
  3. C. ✓ 8 pg

Why: Copying turns every chromosome into two sister chromatids.
So the DNA in the cell doubles, from 4 pg to 8 pg.

8

Copying doubles the DNA in any cell.

Two bars: DNA per cell before copying, 6 pg, and after S phase, 12 pg, twice as tall
Two bars: DNA per cell before copying, 6 pg, and after S phase, 12 pg, twice as tall
9

Meiosis I sends one chromosome of every pair, still two sister chromatids, to each of the two cells. So each of the two cells holds half of the DNA the copied cell held.

10

Between meiosis I and meiosis II the cell does not copy its DNA. The two cells enter meiosis II with the DNA they left meiosis I with.

11

Meiosis II parts the sister chromatids. So each of the four cells holds half of what its parent cell held.

Four bars: 6 pg before copying, 12 pg after S phase, 6 pg per cell after meiosis I, 3 pg per cell after meiosis II
Four bars: 6 pg before copying, 12 pg after S phase, 6 pg per cell after meiosis I, 3 pg per cell after meiosis II
12

Each of the four cells holds 25% of the copied amount, and half of the amount before copying. The DNA was doubled once and halved twice.

13

Chromosome number and DNA amount are two different counts.

14

To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

15

After meiosis I each cell holds half the chromosomes and half the DNA. Yet each of its chromosomes is still two sister chromatids.

Left: a cell after meiosis I with a dashed nuclear envelope holding one long X and one short X side by side, labelled two chromosomes, each still two chromatids. Right: a smaller cell after meiosis II with a dashed nuclear envelope holding one long rod and one short rod, labelled two chromosomes, each a single chromatid
Left: a cell after meiosis I with a dashed nuclear envelope holding one long X and one short X side by side, labelled two chromosomes, each still two chromatids. Right: a smaller cell after meiosis II with a dashed nuclear envelope holding one long rod and one short rod, labelled two chromosomes, each a single chromatid
16

After meiosis II each cell holds the same number of chromosomes as after meiosis I, and half the DNA. Now each chromosome is a single chromatid.

17

Here is a table comparing the model cell after meiosis I with the model cell after meiosis II: chromosomes per cell, chromatids per chromosome, and DNA per cell.

A table comparing the model cell after meiosis I with the model cell after meiosis II: chromosomes per cell 2 and 2; chromatids per chromosome 2 and 1; DNA per cell as a percentage of the copied cell's 50% and 25%
A table comparing the model cell after meiosis I with the model cell after meiosis II: chromosomes per cell 2 and 2; chromatids per chromosome 2 and 1; DNA per cell as a percentage of the copied cell's 50% and 25%
18

What you are expected to know Explain why the DNA per cell halves twice while the chromosome number halves once: the DNA is copied once, never between the two divisions, and a chromosome is counted by its centromere whether it is one chromatid or two.

19
Check q2

A cell in a newt’s testis has finished meiosis I and is about to begin meiosis II.

Does the cell copy its DNA before meiosis II begins?

  1. A. Yes
    The DNA is copied once, in S phase before meiosis I.
    The cell enters meiosis II with the chromosomes it left meiosis I with, each still two sister chromatids.
  2. B. ✓ No

Why: The DNA is copied once, in S phase before meiosis I.
Between meiosis I and meiosis II nothing is copied.
So the cell begins meiosis II with the DNA it left meiosis I with.

20
Practice writing an answer

A cell in a lizard’s ovary holds 14 pg of DNA before it copies its DNA. After meiosis I, each of the two cells holds 14 pg. After meiosis II, each of the four cells holds 7 pg.

(a) Explain how these measurements show that no DNA was copied between meiosis I and meiosis II. (1 pt)

Model answer Copying doubles the DNA in a cell.
If the cell had copied its DNA again before meiosis II, each cell would have entered meiosis II with 28 pg and left it with 14 pg.
Each cell left meiosis II with 7 pg, half of what it left meiosis I with.
So no DNA was copied between meiosis I and meiosis II.
Working
Write down the values in the question:
DNA before copying = 14 pg
DNA in each cell after meiosis I = 14 pg
DNA in each cell after meiosis II = 7 pg
Write down the equation:
after S phase=2×before copying
after meiosis I=after S phase2
if copied again: start of meiosis II=2×after meiosis I
after meiosis II=start of meiosis II2
Substitute the values into the equation:
after S phase=2×14pg=28pg
after meiosis I=28pg2=14pg
if copied again: start of meiosis II=2×14pg=28pg
if copied again: after meiosis II=28pg2=14pg
measured after meiosis II=7pg
Rubric
  • Award 1 point for: a second copying would have left 14 pg in each cell after meiosis II (28 pg at the start of meiosis II, halved); the measured 7 pg is half of the 14 pg after meiosis I, so no second copying happened.
21
Check q3

A student says: “After meiosis I each cell has half the chromosomes and half the DNA, so each chromosome must now be a single chromatid.”

Is the student correct?

  1. A. Yes — half the DNA means half of every chromosome
    Meiosis I parts homologous pairs, not sister chromatids.
    Each cell gets one whole doubled chromosome of every pair, so it holds half the chromosomes and half the DNA.
  2. B. ✓ No — every chromosome after meiosis I is still two sister chromatids

Why: At anaphase I the spindle pulls the two members of each homologous pair apart.
Each cell receives one whole chromosome of every pair, still two sister chromatids.
Half the chromosomes carry half the DNA.
So the DNA halved while every chromosome stayed doubled.

22
Check q4

A cell in a moth’s ovary has finished meiosis I. Its chromosome count has halved, yet every chromosome is still two sister chromatids.

Why has the count halved?

  1. A. ✓ Each cell received one whole chromosome of every pair, one centromere each
  2. B. Meiosis I parted the sister chromatids, and each chromatid now counts as a chromosome
    Sister chromatids part at anaphase II, not at anaphase I.
    At anaphase I whole chromosomes move, each still two sister chromatids.
  3. C. Meiosis I removed half of the DNA from every chromosome in the cell
    No DNA is removed from a chromosome.
    The count halved because each cell received one whole chromosome of every pair.

Why: At anaphase I the spindle pulls the two members of each homologous pair to opposite poles.
Each cell receives one whole chromosome of every pair, still two sister chromatids.
Count centromeres: one of every pair is half the count.

23Calculate the DNA in each cell

24

Video: Watch: Calculate the DNA in each cell

The calculation for the 6 pg cell written out step by step, the four bars filling in, then the same four amounts written in C: 2C, 4C, 2C, 1C.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L05Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L05Bb.mp4

25

Take the cell that holds 6 pg of DNA before it copies its DNA. Here is the calculation for the three kinds of cell it becomes.

26
Worked example

A cell holds 6 pg of DNA before it copies its DNA. How much DNA does each cell hold after S phase, after meiosis I and after meiosis II?

Write down the values in the question:
DNA before copying = 6 pg
Write down the equation:
after S phase=2×before copying
after meiosis I=after S phase2
after meiosis II=after meiosis I2
Substitute the values into the equation:
after S phase=2×6pg=12pg
after meiosis I=12pg2=6pg
after meiosis II=6pg2=3pg
27

Exam questions often measure DNA in sets rather than in picograms. Biologists take the amount of DNA in one haploid set, every chromosome a single chromatid, as the unit.

28

That unit is called : 1C is the DNA in one haploid set of single-chromatid chromosomes. A gamete holds exactly one such set, so a gamete holds 1C.

29
Worked example

A body cell holds two haploid sets of chromosomes, 2C of DNA, before it copies its DNA. How much DNA, in C, does each cell hold after S phase, after meiosis I and after meiosis II?

Write down the values in the question:
DNA before copying = 2C (two haploid sets)
Write down the equation:
after S phase=2×before copying
after meiosis I=after S phase2
after meiosis II=after meiosis I2
Substitute the values into the equation:
after S phase=2×2C=4C
after meiosis I=4C2=2C
after meiosis II=2C2=1C
30

Here are the same four amounts as bars, in C.

Four bars labelled in C: 2C before copying, 4C after S phase, 2C per cell after meiosis I, 1C per cell after meiosis II
Four bars labelled in C: 2C before copying, 4C after S phase, 2C per cell after meiosis I, 1C per cell after meiosis II
31

What you are expected to know Calculate the DNA per cell after S phase, after meiosis I and after meiosis II, in picograms or in C, from the amount before copying.

32
Check q5 numeric entry

A cell holds 8 pg of DNA before it copies its DNA, and then goes through meiosis.

Calculate the DNA in each cell after meiosis II.

Part 1. Calculate the DNA in the cell after S phase.

Answer: 16 pg  (tolerance ±0.01)

Working
Double the amount before copying:
after S phase=2×8pg=16pg

Part 2. Calculate the DNA in each cell after meiosis I.

Answer: 8 pg  (tolerance ±0.01)

Working
Halve the amount after S phase:
after meiosis I=16pg2=8pg

Answer: 4 pg  (tolerance ±0.01)

Working
Write down the values in the question:
DNA before copying = 8 pg
Write down the equation:
after S phase=2×before copying
after meiosis I=after S phase2
after meiosis II=after meiosis I2
Substitute the values into the equation:
after S phase=2×8pg=16pg
after meiosis I=16pg2=8pg
after meiosis II=8pg2=4pg
33
Check q6 numeric entry

A cell that has already copied its DNA holds 18 pg. It goes through meiosis.

Calculate the DNA in each cell after meiosis II.

Answer: 4.5 pg  (tolerance ±0.01)

Working
Write down the values in the question:
DNA after S phase = 18 pg
Write down the equation:
after meiosis I=after S phase2
after meiosis II=after meiosis I2
Substitute the values into the equation:
after meiosis I=18pg2=9pg
after meiosis II=9pg2=4.5pg
34
Check q7 numeric entry

A cell in a salmon’s ovary holds 4C of DNA after copying it.

Calculate the DNA in each cell after meiosis I, in C.

Answer: 2 C  (tolerance ±0.01)

Working
Write down the values in the question:
DNA after S phase = 4C
Write down the equation:
after meiosis I=after S phase2
Substitute the values into the equation:
after meiosis I=4C2=2C
35

Back to the one cell with 6 pg of DNA before it copies its DNA, and the cells it becomes: one after S phase, two after meiosis I, four after meiosis II, each with an empty label.

A cell on the left labelled 6 pg before copying; an arrow to one larger cell labelled 12 pg after S phase; an arrow to two cells labelled 6 pg each after meiosis I; an arrow to four small cells labelled 3 pg each after meiosis II
A cell on the left labelled 6 pg before copying; an arrow to one larger cell labelled 12 pg after S phase; an arrow to two cells labelled 6 pg each after meiosis I; an arrow to four small cells labelled 3 pg each after meiosis II
36

The labels now read 12 pg after S phase, 6 pg in each cell after meiosis I and 3 pg in each cell after meiosis II.

37

The DNA was doubled once and halved twice, and never copied between the two divisions. In C, the same four labels read 2C, 4C, 2C and 1C.

38Quick quiz: C, the DNA in one haploid set mixed practice

39
Check q8

A biologist writes the DNA in a cell as 2C.

What is 1C?

  1. A. ✓ The DNA in one haploid set of single-chromatid chromosomes
  2. B. The DNA in one body cell before it copies its DNA
    A body cell before copying holds two haploid sets.
    So a body cell before copying holds 2C.
  3. C. The DNA in one chromosome of a haploid set
    A haploid set is many chromosomes, one of every pair.
    1C is the DNA in the whole set, not in one chromosome.

Why: 1C is the amount of DNA in one haploid set of chromosomes, every chromosome a single chromatid.

40
Practice writing an answer

DNA amounts in cells are often written in C.

(a) State what 1C is. (1 pt)

Model answer 1C is the amount of DNA in one haploid set of chromosomes, every chromosome a single chromatid.
Rubric
  • Award 1 point for: the DNA in one haploid set (one of every chromosome), each chromosome a single chromatid.
41
Check q9 numeric entry

A cell in a toad’s ovary holds 2C of DNA after meiosis I.

Calculate the DNA in each cell after meiosis II.

Answer: 1 C  (tolerance ±0.01)

Working
Write down the values in the question:
DNA after meiosis I = 2C
Halve the amount after meiosis I:
after meiosis II=2C2=1C
42
Check q10 numeric entry

A body cell of a newt holds 2C of DNA before it copies its DNA.

Calculate the DNA in the cell after S phase.

Answer: 4 C  (tolerance ±0.01)

Working
Write down the values in the question:
DNA before copying = 2C
Double the amount before copying:
after S phase=2×2C=4C
43
Check q11 numeric entry

A cell in a trout’s testis has just finished meiosis I and holds 2C of DNA. It enters meiosis II.

Calculate the DNA in the cell at the start of meiosis II.

Answer: 2 C  (tolerance ±0.01)

Working
Write down the values in the question:
DNA after meiosis I = 2C
No DNA is copied between meiosis I and meiosis II:
start of meiosis II=after meiosis I=2C
44
Check q12 numeric entry

A cell in a moth’s ovary holds 4C of DNA after S phase.

Calculate the DNA in each cell after meiosis I.

Answer: 2 C  (tolerance ±0.01)

Working
Write down the values in the question:
DNA after S phase = 4C
Halve the amount after S phase:
after meiosis I=4C2=2C
45
Check q13 numeric entry

A sperm of a bull holds 1C of DNA.

Calculate the DNA in one of the bull’s body cells before it copies its DNA.

Answer: 2 C  (tolerance ±0.01)

Working
Write down the values in the question:
DNA in a sperm (after meiosis II) = 1C
A body cell holds two haploid sets:
before copying=2×1C=2C

46Mixed practice mixed practice

47
Check q14 numeric entry

A cell in a frog’s testis holds 10 pg of DNA before it copies its DNA.

Calculate the DNA in each cell after meiosis I.

Answer: 10 pg  (tolerance ±0.01)

Working
Write down the values in the question:
DNA before copying = 10 pg
Write down the equation:
after S phase=2×before copying
after meiosis I=after S phase2
Substitute the values into the equation:
after S phase=2×10pg=20pg
after meiosis I=20pg2=10pg
48
Check q15

A student claims that a cell copies its DNA again between meiosis I and meiosis II.

Which of the following observations shows that the claim is wrong?

  1. A. Each cell after meiosis I holds half the DNA of the copied cell
    The halving at meiosis I happens whether or not the cell copies its DNA afterward.
    So it cannot show what happened in the gap between the divisions.
  2. B. ✓ Each cell after meiosis II holds half the DNA of a cell after meiosis I
  3. C. The chromosome count is the same before and after meiosis II
    The chromosome count depends on centromeres, not on the amount of DNA.
    It would stay the same through meiosis II even after a second copying.
  4. D. Every chromosome is two sister chromatids at the moment meiosis I ends
    Doubled chromosomes at the end of meiosis I are what the first copying left.
    They do not show whether a second copying followed.

Why: If the cell copied its DNA between the divisions, each cell would enter meiosis II with twice the DNA it left meiosis I with.
Then the four cells would each hold as much as the two.
Measured DNA halves at meiosis II, so there was no second copying.

49
Check q16 numeric entry

A gamete of a fish holds 3 pg of DNA.

Calculate the DNA in one of the fish’s body cells before it copies its DNA.

Answer: 6 pg  (tolerance ±0.01)

Working
Write down the values in the question:
DNA in a gamete (after meiosis II) = 3 pg
Write down the equation:
before copying=2×after meiosis II
Substitute the values into the equation:
before copying=2×3pg=6pg
50
Check q17

Two cells from one plant are measured. Cell 1 holds 2C of DNA and 12 chromosomes, each still two sister chromatids. Cell 2 holds 2C of DNA and 24 chromosomes, each a single chromatid.

Which statement describes the two cells?

  1. A. ✓ Cell 1 has just finished meiosis I; cell 2 is a body cell before copying
  2. B. Cell 1 is a body cell before copying; cell 2 has just finished meiosis I
    A body cell before copying holds single-chromatid chromosomes, and cell 1’s are doubled.
    A cell after meiosis I holds doubled chromosomes, and cell 2’s are single.
  3. C. Both cells have just finished meiosis II
    A cell after meiosis II holds 1C, not 2C.
    Its chromosomes are single chromatids, which does not fit cell 1.
  4. D. Both cells have just finished meiosis I
    A cell that has just finished meiosis I holds chromosomes that are still two sister chromatids.
    Cell 2’s are single chromatids.

Why: After meiosis I a cell holds 2C on half the chromosome count, each chromosome two chromatids.
A body cell before copying holds 2C on the full count, each chromosome one chromatid.
So the chromatid count places each cell: cell 1 after meiosis I, cell 2 before copying.

Glossary

C (DNA amount)
The amount of DNA in one haploid set of chromosomes, every chromosome a single chromatid, used as a unit: a body cell before copying holds 2C, a copied cell 4C, a cell after meiosis I 2C, and a gamete 1C.

APBIO-U05-L06 Meiosis I or meiosis II?

Topic 5.1 · Meiosis · 37 steps

Eight tiny drawings of a cell in two rows of four, numbered 1 to 8 in shuffled order. Three drawings show one large oval cell, three show one small oval cell, and two show two cells side by side; in some the chromosomes are X shapes, in some V shapes and in some rods, standing at the middle, moving toward the two ends, or lying in two groups
Eight tiny drawings of a cell in two rows of four, numbered 1 to 8 in shuffled order. Three drawings show one large oval cell, three show one small oval cell, and two show two cells side by side; in some the chromosomes are X shapes, in some V shapes and in some rods, standing at the middle, moving toward the two ends, or lying in two groups

Here are eight drawings of the model cell, shuffled: one frozen moment from every stage of meiosis I and meiosis II.

Which division is each drawing from, and which stage?

Unit 5 · Heredity

1Meiosis I or meiosis II?

2

Video: Watch: Meiosis I or meiosis II?

The eight frames sorted into two rows as the viewer watches, with the tell-tale feature circled on each: homologous pairs or single chromosomes at the equator, whole chromosomes or single chromatids moving.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L06a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L06a.mp4

3
Check q1

Meiosis has two divisions, meiosis I and meiosis II.

Which partners does meiosis I pull apart?

  1. A. ✓ The two members of each homologous pair
  2. B. The two sister chromatids of each chromosome
    Meiosis II pulls apart the two sister chromatids of each chromosome.
    Meiosis I pulls apart the two members of each homologous pair.

Why: Meiosis I pulls apart the two members of each homologous pair.
Meiosis II pulls apart the two sister chromatids of each chromosome.

4

Here is a drawing of the eight stages of meiosis in the model cell, in order. The four stages of meiosis I are across the top, and the four stages of meiosis II are across the bottom.

Eight boxed drawings in two rows, labelled: prophase I, metaphase I, anaphase I, telophase I across the top; prophase II, metaphase II, anaphase II, telophase II across the bottom
Eight boxed drawings in two rows, labelled: prophase I, metaphase I, anaphase I, telophase I across the top; prophase II, metaphase II, anaphase II, telophase II across the bottom
5

Both rows use the same four names: prophase, metaphase, anaphase and telophase. So a stage name alone does not tell you which division a drawing shows.

6

Two questions about the chromosomes place the division:
1 At the equator, do the chromosomes stand in homologous pairs, or singly?
2 Moving to the poles, are they whole X-shaped chromosomes, or single chromatids?

7

For example, here the chromosomes stand at the equator in homologous pairs. This cell is in meiosis I, because homologous pairs stand at the equator only in meiosis I.

The large model cell with a pole at each end; four X-shaped chromosomes stand at the vertical middle line as two pairs, two X’s of the same length side by side in each pair, one pair above the other, with a fiber from the left pole to the left X and from the right pole to the right X of each pair; caption: homologous pairs at the equator: meiosis I
The large model cell with a pole at each end; four X-shaped chromosomes stand at the vertical middle line as two pairs, two X’s of the same length side by side in each pair, one pair above the other, with a fiber from the left pole to the left X and from the right pole to the right X of each pair; caption: homologous pairs at the equator: meiosis I
8

But here the chromosomes stand at the equator singly, in one row. This cell is in meiosis II, because single chromosomes stand at the equator only in meiosis II.

A small cell with two single X-shaped chromosomes, one long and one short, in one row across the middle with fibers from both poles; caption: single chromosomes at the equator: meiosis II
A small cell with two single X-shaped chromosomes, one long and one short, in one row across the middle with fibers from both poles; caption: single chromosomes at the equator: meiosis II
9

Here whole X-shaped chromosomes are moving to the poles. This cell is in meiosis I, because whole chromosomes move to the poles only in meiosis I.

The large model cell with whole X-shaped chromosomes moving toward the two poles, two each way, each with a fiber to its pole; caption: whole chromosomes moving to the poles: meiosis I
The large model cell with whole X-shaped chromosomes moving toward the two poles, two each way, each with a fiber to its pole; caption: whole chromosomes moving to the poles: meiosis I
10

But here single chromatids are moving to the poles. This cell is in meiosis II, because single chromatids move to the poles only in meiosis II.

A small cell with V-shaped single chromatids moving toward the two poles, two each way, each with a fiber to its pole; caption: single chromatids moving to the poles: meiosis II
A small cell with V-shaped single chromatids moving toward the two poles, two each way, each with a fiber to its pole; caption: single chromatids moving to the poles: meiosis II
11

The X shape alone never places a drawing. Every chromosome is an X from prophase I until the sister chromatids part at anaphase II.

12

Here is a table of the eight stages, comparing what you see at each stage of meiosis I with what you see at the same-named stage of meiosis II.

A table with three columns, stage, meiosis I and meiosis II, and four rows: prophase, homologous pairs form against single X-shaped chromosomes attach to a new spindle; metaphase, pairs at the equator, two wide against single X-shaped chromosomes in one row; anaphase, whole X-shaped chromosomes move to the poles against single chromatids move to the poles; telophase, two cells form, X-shaped chromosomes inside against four cells form, rods inside
A table with three columns, stage, meiosis I and meiosis II, and four rows: prophase, homologous pairs form against single X-shaped chromosomes attach to a new spindle; metaphase, pairs at the equator, two wide against single X-shaped chromosomes in one row; anaphase, whole X-shaped chromosomes move to the poles against single chromatids move to the poles; telophase, two cells form, X-shaped chromosomes inside against four cells form, rods inside
13

What you are expected to know Identify from a drawing whether a cell is in meiosis I or in meiosis II.

14

What you are expected to know State the feature that places it: homologous pairs or single chromosomes at the equator, whole chromosomes or single chromatids moving to the poles.

15Quick quiz: meiosis I or meiosis II? mixed practice

16
Check q2

A fruit fly’s body cells hold eight chromosomes. A cell from its testis is drawn below.

A small cell with a pole at each end; eight V-shaped pieces, four on the left half pointing left and four on the right half pointing right, each with a fiber to the pole on its side; on each side the four pieces are of four different lengths
A small cell with a pole at each end; eight V-shaped pieces, four on the left half pointing left and four on the right half pointing right, each with a fiber to the pole on its side; on each side the four pieces are of four different lengths

Which division is the cell in?

  1. A. Meiosis I
    In meiosis I whole X-shaped chromosomes move to the poles.
    Here the moving pieces are V-shaped single chromatids.
  2. B. ✓ Meiosis II

Why: V-shaped single chromatids are moving to the poles.
Sister chromatids part only in meiosis II.
So the cell is in meiosis II.

17
Check q3

A plant’s body cells hold six chromosomes. A cell from one of its flowers, dividing by meiosis, is drawn below.

A cell with a pole at each end; six X-shaped chromosomes, three on the left half and three on the right half, one long, one middle-length and one short on each side, each with a fiber running to the pole on its side
A cell with a pole at each end; six X-shaped chromosomes, three on the left half and three on the right half, one long, one middle-length and one short on each side, each with a fiber running to the pole on its side

Which division is the cell in?

  1. A. ✓ Meiosis I
  2. B. Meiosis II
    In meiosis II single chromatids move to the poles.
    Here every moving chromosome is still a whole X.

Why: Whole X-shaped chromosomes are moving to the poles, each still two sister chromatids.
Whole chromosomes move to the poles only in meiosis I, when the two chromosomes of each homologous pair part.
So the cell is in meiosis I.

18
Check q4

A plant’s body cells hold six chromosomes. A cell from one of its flowers, dividing by meiosis, is drawn below.

A cell with a pole at each end; six X-shaped chromosomes stand at the vertical middle line, three just left of the line and three just right of it, at three heights; a fiber joins the left pole to each X on the left and the right pole to each X on the right
A cell with a pole at each end; six X-shaped chromosomes stand at the vertical middle line, three just left of the line and three just right of it, at three heights; a fiber joins the left pole to each X on the left and the right pole to each X on the right

Which division is the cell in?

  1. A. ✓ Meiosis I
  2. B. Meiosis II
    In meiosis II single chromosomes stand at the equator in one row.
    Here the chromosomes stand in pairs, two X’s of the same length side by side.

Why: The chromosomes stand at the equator in homologous pairs, two X’s of the same length side by side.
Homologous pairs stand at the equator only in meiosis I.
So the cell is in meiosis I.

19
Check q5

A fruit fly’s body cells hold eight chromosomes. A cell from one of its testes is drawn below.

A small cell with a pole at each end; four X-shaped chromosomes of four different lengths stand one above another down the vertical middle line, each with a fiber from each pole
A small cell with a pole at each end; four X-shaped chromosomes of four different lengths stand one above another down the vertical middle line, each with a fiber from each pole

Which division is the cell in?

  1. A. Meiosis I
    In meiosis I the chromosomes stand at the equator in homologous pairs, two wide.
    Here four single chromosomes stand in one row.
  2. B. ✓ Meiosis II

Why: Four single chromosomes stand in one row at the equator, with no pairs.
Single chromosomes stand at the equator only in meiosis II.
So the cell is in meiosis II.

20
Check q6

A plant’s body cells hold six chromosomes. A dividing cell from one of its flowers, in meiosis, is drawn below.

A cell narrowed at its middle into two lobes; in each lobe a dashed oval encloses three X-shaped chromosomes of three different lengths standing side by side
A cell narrowed at its middle into two lobes; in each lobe a dashed oval encloses three X-shaped chromosomes of three different lengths standing side by side

Which division is the cell in?

  1. A. ✓ Meiosis I
  2. B. Meiosis II
    At the end of meiosis II every chromosome in the forming cells is a single rod.
    Here every chromosome is still an X.

Why: Two cells are forming.
Every chromosome in them is still an X of two sister chromatids.
Sister chromatids part only in meiosis II, so here they are still joined.
So the cell is in meiosis I.

21
Check q7

A fruit fly’s body cells hold eight chromosomes. A dividing cell from its testis is drawn below.

A small cell narrowed at its middle into two lobes; in each lobe a dashed oval encloses four straight rods of four different lengths standing side by side
A small cell narrowed at its middle into two lobes; in each lobe a dashed oval encloses four straight rods of four different lengths standing side by side

Which division is the cell in?

  1. A. Meiosis I
    At the end of meiosis I every chromosome is still an X of two sister chromatids.
    Here every chromosome is a single rod.
  2. B. ✓ Meiosis II

Why: Two cells are forming.
Every chromosome in them is a single rod: one chromatid.
Sister chromatids part only in meiosis II.
So the cell is in meiosis II.

22Name the stage as well

23

Video: Watch: Name the stage as well

One drawing at a time: the division is placed first, then what the chromosomes are doing names the stage, and the two labels are written under the drawing.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L06b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L06b.mp4

24

Once you have placed the division, what the chromosomes are doing names the stage, just as it does in mitosis.

25

1 attaching to a forming spindle: prophase
2 standing at the equator: metaphase
3 moving to the poles: anaphase
4 gathered at the poles while the cell divides: telophase

26

So every drawing gets two labels. The stage comes from what the chromosomes are doing, and the division comes from whether they are homologous pairs or single chromosomes, whole chromosomes or single chromatids.

27

For example, here single X-shaped chromosomes stand in one row at the equator of a small cell. Single chromosomes at the equator is meiosis II, and standing at the equator is metaphase, so the drawing is metaphase II.

A small cell with two single X-shaped chromosomes, one long and one short, in one row across the middle with fibers from both poles; caption: single chromosomes at the equator: meiosis II
A small cell with two single X-shaped chromosomes, one long and one short, in one row across the middle with fibers from both poles; caption: single chromosomes at the equator: meiosis II
28

What you are expected to know Name both the stage and the division of a drawn cell in meiosis, for example metaphase II or anaphase I.

29
Check q8

A plant’s body cells hold six chromosomes. A cell from one of its flowers, dividing by meiosis, is drawn below.

A small cell with a pole at each end; three X-shaped chromosomes of three different lengths stand one above another down the middle of the cell, a long X at the top, a middle-length X in the middle and a short X at the bottom; fibers run from the poles to the chromosomes
A small cell with a pole at each end; three X-shaped chromosomes of three different lengths stand one above another down the middle of the cell, a long X at the top, a middle-length X in the middle and a short X at the bottom; fibers run from the poles to the chromosomes

Which stage is the cell in?

  1. A. Metaphase I
    Metaphase I is a double row of homologous pairs at the equator.
    Here three single chromosomes stand in one row, half the plant’s six.
  2. B. ✓ Metaphase II
  3. C. Prophase II
    In prophase II the chromosomes are still scattered, attaching to a forming spindle.
    Here the chromosomes stand in a row at the equator with fibers from both poles.
  4. D. Anaphase II
    At anaphase II single chromatids are moving toward the poles.
    Here every chromosome is still a whole X standing at the equator.

Why: Three single X-shaped chromosomes stand in one row at the equator, half the plant’s six.
Single chromosomes standing at the equator is metaphase II.

30
Check q9

A plant’s body cells hold six chromosomes. Cells from one of its flowers are drawn below at the moment one stage of meiosis has just ended.

Four small circles in a row; inside each, a dashed circle encloses three straight rods of three different lengths standing side by side
Four small circles in a row; inside each, a dashed circle encloses three straight rods of three different lengths standing side by side

Which stage has just finished?

  1. A. Telophase I and cytokinesis
    Telophase I ends with two cells whose chromosomes are still X’s.
    Here four cells have formed, and each chromosome is a single rod.
  2. B. Anaphase II
    At anaphase II the chromatids are still moving inside undivided cells.
    Here the cells have divided, and a nuclear envelope has formed in each.
  3. C. Anaphase I
    Anaphase I moves whole X-shaped chromosomes apart inside one cell.
    Only two cells follow anaphase I; here four have formed.
  4. D. ✓ Telophase II and cytokinesis

Why: Four cells have formed.
Each cell holds three single-chromatid chromosomes inside a nuclear envelope.
Four finished cells with single chromatids is the end of meiosis: telophase II and cytokinesis have just finished.

31
Check q10

A plant’s body cells hold six chromosomes. A cell from one of its flowers, dividing by meiosis, is drawn below.

A small cell with a pole at each end; six V-shaped pieces, three on the left half pointing left and three on the right half pointing right, each with a fiber to the pole on its side; on each side one piece is long, one middle-length and one short
A small cell with a pole at each end; six V-shaped pieces, three on the left half pointing left and three on the right half pointing right, each with a fiber to the pole on its side; on each side one piece is long, one middle-length and one short

Which stage is the cell in?

  1. A. Anaphase I
    Anaphase I moves whole X-shaped chromosomes.
    Here the moving pieces are V-shaped single chromatids.
  2. B. Metaphase II
    At metaphase II the chromosomes stand still in a row at the equator.
    Here the chromosomes are moving toward the poles.
  3. C. Telophase I
    At telophase I the cell is dividing and its chromosomes are still X’s.
    Here single chromatids are moving in one undivided cell.
  4. D. ✓ Anaphase II

Why: V-shaped single chromatids are moving toward the poles.
Sister chromatids part only in anaphase II.
So the cell is in anaphase II.

32
Check q11

A plant’s body cells hold six chromosomes. A cell from one of its flowers, dividing by meiosis, is drawn below.

A small cell with a pole at each end and one fiber reaching in from each; three X-shaped chromosomes of three different lengths lie apart from one another, a long X at the upper left, a middle-length X low in the middle and a short X at the upper right
A small cell with a pole at each end and one fiber reaching in from each; three X-shaped chromosomes of three different lengths lie apart from one another, a long X at the upper left, a middle-length X low in the middle and a short X at the upper right

Which stage is the cell in?

  1. A. Prophase I
    In prophase I the homologous pairs lie together in a cell holding all six chromosomes.
    Here three single chromosomes lie apart in a small cell.
  2. B. Metaphase I
    Metaphase I is a double row of homologous pairs at the equator.
    Here the chromosomes are scattered and single.
  3. C. ✓ Prophase II
  4. D. Metaphase II
    At metaphase II the chromosomes stand in one row at the equator.
    Here the chromosomes are scattered, attaching to the forming spindle.

Why: Three single X-shaped chromosomes, half the plant’s six, lie apart and are attaching to a new spindle.
Single chromosomes attaching to a new spindle is prophase II.

33

Here again are the eight shuffled drawings of the model cell, one frozen moment from every stage of meiosis I and meiosis II, numbered as before.

Eight boxed drawings of the model cell in two rows of four, numbered 1 to 8 in the shuffled order of the opening page: 1 single chromosomes in one row in a small cell; 2 whole X's moving apart in a large cell; 3 two small cells with rods; 4 paired homologs in a large cell; 5 V-shaped chromatids moving apart in a small cell; 6 pairs across the middle of a large cell; 7 two single X's attaching to a spindle in a small cell; 8 two forming cells each with two X's
Eight boxed drawings of the model cell in two rows of four, numbered 1 to 8 in the shuffled order of the opening page: 1 single chromosomes in one row in a small cell; 2 whole X's moving apart in a large cell; 3 two small cells with rods; 4 paired homologs in a large cell; 5 V-shaped chromatids moving apart in a small cell; 6 pairs across the middle of a large cell; 7 two single X's attaching to a spindle in a small cell; 8 two forming cells each with two X's
34

Drawing 6 has homologous pairs at the equator, so it is meiosis I. Standing at the equator is metaphase, so drawing 6 is metaphase I.

35

Drawing 5 has single chromatids moving to the poles, so it is meiosis II. Moving to the poles is anaphase, so drawing 5 is anaphase II.

36

You place every one of the eight drawings the same way: homologous pairs or single chromosomes at the equator, whole chromosomes or single chromatids moving to the poles, and then what the chromosomes are doing names the stage.

APBIO-U05-L06B Count the chromosomes and chromatids

Topic 5.1 · Meiosis · 33 steps

A cell holding eight X-shaped chromosomes of four lengths, one darker and one lighter of each length, scattered; captions read 2n = 24, after S phase, and eight of the 24 chromosomes are drawn; beside it a table with rows after S phase, after meiosis I and after meiosis II and columns chromosomes per cell and chromatids per cell, all six cells blank
A cell holding eight X-shaped chromosomes of four lengths, one darker and one lighter of each length, scattered; captions read 2n = 24, after S phase, and eight of the 24 chromosomes are drawn; beside it a table with rows after S phase, after meiosis I and after meiosis II and columns chromosomes per cell and chromatids per cell, all six cells blank

Here is a cell from a plant whose body cells hold 24 chromosomes, drawn after S phase with only eight of its chromosomes shown, and beside it a table: chromosomes and chromatids per cell after S phase, after meiosis I and after meiosis II. The six cells of the table are empty.

What numbers go in them?

Unit 5 · Heredity

1Count centromeres

2

Video: Watch: Count the chromosomes and chromatids

The counting table filling in for the model cell and then for 2n = 24, one cell at a time, with the centromeres counted on the drawing as each number is written.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L06Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L06Ba.mp4

3

How many chromosomes and how many chromatids does a cell hold after S phase, after meiosis I and after meiosis II? The two counts follow two different rules.

4

To count chromosomes, count centromeres. To count chromatids, count two for every chromosome until meiosis II parts the sister chromatids.

5

So copying changes only the chromatid count. The chromosome count halves once, at meiosis I.

6
Check q1

A chromosome has been copied. It is now two sister chromatids joined at one centromere.

How many chromosomes is it?

  1. A. ✓ One
  2. B. Two
    The copied chromosome has one centromere.
    One centromere is one chromosome.

Why: Count centromeres.
The copied chromosome has one centromere, so it is one chromosome, made of two chromatids.

7

To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Left: one rod-shaped chromosome with one centromere dot, labelled one centromere: one chromosome, one chromatid. Right: one X-shaped chromosome with one centromere dot, labelled one centromere: one chromosome, two sister chromatids
Left: one rod-shaped chromosome with one centromere dot, labelled one centromere: one chromosome, one chromatid. Right: one X-shaped chromosome with one centromere dot, labelled one centromere: one chromosome, two sister chromatids
8

So copying doubles the chromatids, not the chromosomes.

9

After S phase a cell still holds 2n chromosomes. Each chromosome is two sister chromatids, so the cell holds twice as many chromatids as chromosomes.

The copied model cell with four X's; two count boxes read 4 chromosomes and 8 chromatids; caption: after S phase, 2n chromosomes and twice as many chromatids
The copied model cell with four X's; two count boxes read 4 chromosomes and 8 chromatids; caption: after S phase, 2n chromosomes and twice as many chromatids
10

After meiosis I each cell holds n chromosomes, one from every homologous pair. Each chromosome is still two sister chromatids, so each cell holds 2×n chromatids.

11

After meiosis II each cell holds n chromosomes, and each chromosome is now a single chromatid. So each cell holds n chromatids.

12

Here is a table of the model cell’s counts: chromosomes and chromatids per cell after S phase, after meiosis I and after meiosis II.

A table for the model cell: after S phase 4 chromosomes and 8 chromatids; after meiosis I 2 and 4 per cell; after meiosis II 2 and 2 per cell
A table for the model cell: after S phase 4 chromosomes and 8 chromatids; after meiosis I 2 and 4 per cell; after meiosis II 2 and 2 per cell
13

The copying shows only in the chromatid column. The halving shows once, in the chromosome column, at meiosis I.

14
Worked example

A plant’s body cells hold 24 chromosomes. How many chromosomes and how many chromatids does each cell hold after S phase, after meiosis I and after meiosis II?

Write down the values in the question:
2n=24
n=12
Write down the equation:
chromosomes after S phase=2n
chromatids after S phase=2×2n
chromosomes after meiosis I=n
chromatids after meiosis I=2×n
chromosomes after meiosis II=n
chromatids after meiosis II=n
Substitute the values into the equation:
chromosomes after S phase=24
chromatids after S phase=2×24=48
chromosomes after meiosis I=12
chromatids after meiosis I=2×12=24
chromosomes after meiosis II=12
chromatids after meiosis II=12
15

What you are expected to know Calculate the number of chromosomes and the number of chromatids per cell after S phase, after meiosis I and after meiosis II, for a stated 2n.

16
Check q2 numeric entry

A dog’s body cells hold 78 chromosomes. One cell copies its DNA and then divides by meiosis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids in each cell after meiosis II.

Part 1. Calculate the number of chromatids in the cell after S phase.

Answer: 156  (tolerance ±0)

Working
Double the chromosome count, because every chromosome is now two chromatids:
chromatids after S phase=2×78=156

Part 2. Calculate the number of chromosomes in each cell after meiosis I.

Answer: 39  (tolerance ±0)

Working
Halve the count, because each cell receives one chromosome of every homologous pair:
chromosomes after meiosis I=782=39

Part 3. Calculate the number of chromatids in each cell after meiosis I.

Answer: 78  (tolerance ±0)

Working
Double the chromosome count, because each chromosome is still two chromatids:
chromatids after meiosis I=2×39=78

Answer: 39  (tolerance ±0)

Working
Write down the values in the question:
2n=78
n=39
Write down the equation:
chromosomes after meiosis II=n
chromatids after meiosis II=n
Substitute the values into the equation:
chromosomes after meiosis II=39
chromatids after meiosis II=39
17
Check q3 numeric entry

A fruit fly’s body cells hold 8 chromosomes. One cell copies its DNA and then divides by meiosis I.

Calculate the number of chromatids in each cell after meiosis I.

Answer: 8  (tolerance ±0)

Working
Write down the values in the question:
2n=8
n=4
Write down the equation:
chromosomes after meiosis I=n
chromatids after meiosis I=2×n
Substitute the values into the equation:
chromatids after meiosis I=2×4=8
18
Check q4 numeric entry

A human sperm holds 23 chromosomes.

Calculate the number of chromatids in a cell in a human testis after S phase, before meiosis I begins.

Answer: 92  (tolerance ±0)

Working
Write down the values in the question:
n=23
2n=46
Write down the equation:
chromatids after S phase=2×2n
Substitute the values into the equation:
chromatids after S phase=2×46=92
19

Back to the plant whose body cells hold 24 chromosomes, drawn after S phase, with the empty table beside it. Here is that table filled in.

The opening table filled in for the plant with 24 chromosomes: after S phase 24 chromosomes and 48 chromatids; after meiosis I 12 and 24 per cell; after meiosis II 12 and 12 per cell
The opening table filled in for the plant with 24 chromosomes: after S phase 24 chromosomes and 48 chromatids; after meiosis I 12 and 24 per cell; after meiosis II 12 and 12 per cell
20

The chromosome column comes from counting centromeres. The copying shows only in the chromatid column, and the halving shows once, at meiosis I.

21Quick quiz: counts mixed practice

22
Check q5 numeric entry

An onion’s body cells hold 16 chromosomes. One cell copies its DNA and then divides by meiosis.

Calculate the number of chromatids in the cell after S phase.

Answer: 32  (tolerance ±0)

Working
Write down the values in the question:
2n=16
Write down the equation:
chromatids after S phase=2×2n
Substitute the values into the equation:
chromatids after S phase=2×16=32
23
Check q6 numeric entry

An onion’s body cells hold 16 chromosomes. One cell copies its DNA and then divides by meiosis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each cell after meiosis I.

Answer: 8  (tolerance ±0)

Working
Write down the values in the question:
2n=16
Write down the equation:
chromosomes after meiosis I=n=2n2
Substitute the values into the equation:
chromosomes after meiosis I=162=8
24
Check q7 numeric entry

An onion’s body cells hold 16 chromosomes. One cell copies its DNA and then divides by meiosis.

Calculate the number of chromatids in each cell after meiosis I.

Answer: 16  (tolerance ±0)

Working
Write down the values in the question:
2n=16
n=8
Write down the equation:
chromatids after meiosis I=2×n
Substitute the values into the equation:
chromatids after meiosis I=2×8=16
25
Check q8 numeric entry

An onion’s body cells hold 16 chromosomes. One cell copies its DNA and then divides by meiosis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each cell after meiosis II.

Answer: 8  (tolerance ±0)

Working
Write down the values in the question:
2n=16
Write down the equation:
chromosomes after meiosis II=n=2n2
Substitute the values into the equation:
chromosomes after meiosis II=162=8
26
Check q9 numeric entry

An onion’s body cells hold 16 chromosomes. One cell copies its DNA and then divides by meiosis.

Calculate the number of chromatids in each cell after meiosis II.

Answer: 8  (tolerance ±0)

Working
Write down the values in the question:
2n=16
n=8
Write down the equation:
chromatids after meiosis II=n
Substitute the values into the equation:
chromatids after meiosis II=8

27Mixed practice mixed practice

28
Check q10 numeric entry

A pea plant’s body cells hold 14 chromosomes. One cell copies its DNA and then divides by meiosis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each cell after meiosis II.

Answer: 7  (tolerance ±0)

Working
Write down the values in the question:
2n=14
Write down the equation:
chromosomes after meiosis II=n=2n2
Substitute the values into the equation:
chromosomes after meiosis II=142=7
29
Check q11 numeric entry

An elephant’s sperm holds 28 chromosomes. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids in one of the elephant’s body cells after S phase.

Answer: 112  (tolerance ±0)

Working
Write down the values in the question:
n=28
2n=2×28=56
Write down the equation:
chromatids after S phase=2×2n
Substitute the values into the equation:
chromatids after S phase=2×56=112
30
Check q12

A horse’s body cells hold 64 chromosomes. One of its cells holds 32 chromosomes and 64 chromatids. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Which stage has this cell just finished?

  1. A. S phase
    After S phase the cell still holds all 64 chromosomes, as 128 chromatids.
  2. B. ✓ Meiosis I
  3. C. Meiosis II
    After meiosis II each chromosome is a single chromatid, so the chromatid count equals the chromosome count: 32 and 32.
  4. D. Mitosis
    Mitosis gives cells with the parent’s 64 chromosomes, each a single chromatid.

Why: The cell holds 32 chromosomes, half of 64, so the homologous pairs have parted.
Each chromosome is still two chromatids, 64 in all, so the sister chromatids have not parted.
Pairs parted and sister chromatids still joined is the end of meiosis I.

31
Check q13

A sunflower’s body cells hold 34 chromosomes. One of its cells holds 17 chromosomes and 17 chromatids. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Which stage has this cell just finished?

  1. A. S phase
    After S phase the cell still holds all 34 chromosomes, as 68 chromatids.
  2. B. Mitosis
    Mitosis gives cells with the parent’s 34 chromosomes, each a single chromatid.
  3. C. Meiosis I
    After meiosis I each chromosome is still two chromatids, so 17 chromosomes would be 34 chromatids.
  4. D. ✓ Meiosis II

Why: The cell holds 17 chromosomes, half of 34, so the homologous pairs have parted.
Each chromosome is one chromatid, 17 in all, so the sister chromatids have parted too.
Pairs parted and sister chromatids parted is the end of meiosis II.

32
Check q14

A student counts the chromosomes in a cell that has just finished meiosis I and gets 20. The student says the cell must have held 40 chromatids after S phase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Which of the following did the cell hold after S phase?

  1. A. 40 chromosomes and 20 chromatids
    Copying never leaves a chromosome with half a chromatid.
    After S phase each of the 40 chromosomes is two chromatids.
  2. B. 40 chromosomes and 40 chromatids
    After S phase every chromosome is two chromatids, so 40 chromosomes carry 80 chromatids.
  3. C. ✓ 40 chromosomes and 80 chromatids
  4. D. 80 chromosomes and 80 chromatids
    Copying doubles the chromatids, not the chromosomes.
    20 chromosomes after meiosis I means 40 before it, and copying left those 40 as 40.

Why: Twenty chromosomes after meiosis I means the cell held 40 chromosomes before copying.
Copying leaves the chromosome count at 40.
After S phase each of those 40 chromosomes was two chromatids.
So the cell held 40 chromosomes and 80 chromatids.

APBIO-U05-L07 Two divisions, side by side

Topic 5.1 · Meiosis · 43 steps

Two oval cells side by side, each with a pole at each end and spindle fibers. In the left cell four X-shaped chromosomes stand in one column across the middle. In the right cell the four X's stand as two pairs across the middle, each pair two chromosomes wide
Two oval cells side by side, each with a pole at each end and spindle fibers. In the left cell four X-shaped chromosomes stand in one column across the middle. In the right cell the four X's stand as two pairs across the middle, each pair two chromosomes wide

Here are two cells from the same body, drawn side by side. On the left, a skin cell is dividing to close a cut. On the right, a cell in an ovary is dividing to make eggs. Both cells have spindles, and both started with copied chromosomes.

What do the two divisions share, and where do they part ways?

Unit 5 · Heredity

1What the two divisions share

2

Video: Watch: What the two divisions share

The skin cell and the ovary cell dividing in parallel: one copying in S phase, one spindle each, the same four stage names appearing under both.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L07a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L07a.mp4

3

How do mitosis and meiosis compare? Mitosis and meiosis both copy the DNA once, both build a spindle, and both use the same four stage names.

4

Mitosis is one division. Mitosis makes two cells with the parent cell’s chromosome number, each identical to the parent cell.

5

Meiosis is two divisions. Meiosis makes four cells with half the parent cell’s chromosome number, none identical to the parent cell.

6
Check q1

Both a skin cell and a cell in an ovary spend time in S phase before they divide.

What happens in S phase?

  1. A. ✓ The cell copies every one of its DNA molecules
  2. B. The cell pulls its sister chromatids apart
    Sister chromatids part later, at anaphase.
    In S phase the cell copies its DNA.
  3. C. The cell splits into two cells
    The cell splits at the end of division, after the stages.
    In S phase the cell copies its DNA.

Why: S phase is the copying stage of interphase.
The cell copies every DNA molecule, so each chromosome becomes two sister chromatids.

7

Both cells copied their DNA once, in S phase, before dividing.

8

So every chromosome in both cells is two sister chromatids joined at one centromere.

9

Both cells build a spindle from two poles. In both cells the spindle moves the chromosomes.

Two cells at metaphase side by side, both with a pole at each end, spindle fibers and X-shaped chromosomes; the left is a single column of four X's, labelled mitosis, metaphase; the right is two pairs, labelled meiosis, metaphase I
Two cells at metaphase side by side, both with a pole at each end, spindle fibers and X-shaped chromosomes; the left is a single column of four X's, labelled mitosis, metaphase; the right is two pairs, labelled meiosis, metaphase I
10

Both divisions have stages named prophase, metaphase, anaphase and telophase. Both divisions end with cytokinesis.

11

In the cell dividing by meiosis, each stage name carries a I or a II, because meiosis has two divisions.

12

So a drawing of a spindle pulling chromosomes could be either division. To place it, look at what stands at the equator: homologous pairs or single chromosomes.

13

What you are expected to know Identify the features mitosis and meiosis share: one DNA copying in S phase, a spindle that moves the chromosomes, and the four stage names.

14
Check q2

A cell from a lily is dividing. Its chromosomes are X-shaped, and fibers from two poles are attached to them, so it could be in either mitosis or meiosis.

Which of the following is true of the cell, whichever division it is in?

  1. A. The cell will produce four cells by the end
    Only meiosis ends with four cells.
    A cell in mitosis ends with two.
  2. B. ✓ The cell copied its DNA once before meiosis or mitosis began
  3. C. The cell will produce two cells identical to itself
    Two cells identical to the parent is the outcome of mitosis alone.
    A cell in meiosis makes four haploid cells, none identical to it.
  4. D. The cell’s chromosome number will stay the same
    The chromosome number stays the same only in mitosis.
    Meiosis halves it.

Why: Whichever division the cell is in, it copied its DNA once in S phase first.
That copying is why every chromosome is an X.
Four cells and a halved count belong to meiosis alone; two identical cells and an unchanged count belong to mitosis alone.

15Quick quiz: mitosis or meiosis? mixed practice

16
Check q3

A plant’s body cells hold six chromosomes. One of its cells is drawn below.

A cell with a pole at each end; six X-shaped chromosomes stand at the vertical middle line, three just left of the line and three just right of it, at three heights; a fiber joins the left pole to each X on the left and the right pole to each X on the right
A cell with a pole at each end; six X-shaped chromosomes stand at the vertical middle line, three just left of the line and three just right of it, at three heights; a fiber joins the left pole to each X on the left and the right pole to each X on the right

Which division is the cell in?

  1. A. Mitosis
    In mitosis the chromosomes stand at the equator singly, in one row.
    Here the chromosomes stand in pairs, two X’s of the same length side by side.
  2. B. ✓ Meiosis

Why: The chromosomes stand at the equator in homologous pairs.
Homologous pairs form only in meiosis.
So the cell is in meiosis.

17
Check q4

A dog’s body cells hold 78 chromosomes. One of its cells divides and makes two cells with 78 chromosomes each.

Which division was it?

  1. A. ✓ Mitosis
  2. B. Meiosis
    Meiosis makes four cells, each with half the parent’s chromosomes: 39.
    Two cells with the parent’s 78 is mitosis.

Why: The division made two cells.
Each new cell has 78 chromosomes, the same as the parent.
Two cells with the parent’s count is mitosis.

18
Check q5

A plant’s body cells hold six chromosomes. One of its cells is drawn below.

A cell with a pole at each end; twelve V-shaped pieces, six on the left half pointing left and six on the right half pointing right, each with a fiber to the pole on its side; on each side the six pieces come in three lengths, two of each length
A cell with a pole at each end; twelve V-shaped pieces, six on the left half pointing left and six on the right half pointing right, each with a fiber to the pole on its side; on each side the six pieces come in three lengths, two of each length

Which division is the cell in?

  1. A. ✓ Mitosis
  2. B. Meiosis
    In anaphase II of meiosis the cell is haploid: in this plant, three chromatids move to each pole.
    Here six chromatids move to each pole, the plant’s full six.

Why: Single chromatids are moving to the poles, six each way.
Six is the plant’s full count, so this cell is not haploid.
Sister chromatids parting in a cell with the full count is mitosis.

19
Check q6

A dog’s body cells hold 78 chromosomes. One of its cells divides and makes four cells with 39 chromosomes each.

Which division was it?

  1. A. Mitosis
    Mitosis makes two cells, each with the parent’s 78 chromosomes.
    Four cells with half the count is meiosis.
  2. B. ✓ Meiosis

Why: The division made four cells.
Each new cell has 39 chromosomes, half the parent’s 78.
Four cells with half the count is meiosis.

20
Check q7

A plant’s body cells hold six chromosomes. One of its cells is drawn below.

A cell with a pole at each end and one fiber reaching in from each; six X-shaped chromosomes lie in the cell, two at the upper left, two low in the middle and two at the upper right
A cell with a pole at each end and one fiber reaching in from each; six X-shaped chromosomes lie in the cell, two at the upper left, two low in the middle and two at the upper right

Which division is the cell in?

  1. A. Mitosis
    In mitosis the chromosomes never pair up.
    Here each X lies against a partner of the same length: three pairs.
  2. B. ✓ Meiosis

Why: Each X-shaped chromosome lies against another X of the same length.
Homologous pairs form only in meiosis, in prophase I.
So the cell is in meiosis.

21Where they part ways

22

Video: Watch: Where they part ways

The two cells dividing in parallel while a two-column table builds one row at a time beneath them; the row that differs lights up as the ovary cell's homologous pairs part first.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L07b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L07b.mp4

23

Here is a table comparing mitosis and meiosis, one row per difference. It fills in one row at a time, starting with the number of divisions.

A table comparing mitosis and meiosis, with columns mitosis and meiosis and 1 of six rows filled: divisions, one against two: meiosis I and meiosis II
A table comparing mitosis and meiosis, with columns mitosis and meiosis and 1 of six rows filled: divisions, one against two: meiosis I and meiosis II
24

Mitosis is one division. Meiosis is two divisions, meiosis I and then meiosis II.

25

Mitosis produces two cells. Meiosis produces four cells.

A table comparing mitosis and meiosis, with columns mitosis and meiosis and 2 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four
A table comparing mitosis and meiosis, with columns mitosis and meiosis and 2 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four
26

The skin cell’s two cells each hold 46, the same as the skin cell, so they are diploid. The ovary cell’s four cells each hold 23, half the ovary cell’s 46, so they are haploid.

A table comparing mitosis and meiosis, with columns mitosis and meiosis and 3 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four; chromosome number, 2n, the same as the parent against n, half the parent's
A table comparing mitosis and meiosis, with columns mitosis and meiosis and 3 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four; chromosome number, 2n, the same as the parent against n, half the parent's
27

Are the new cells identical to the parent? In mitosis, yes.

28

Every chromosome was copied once in S phase into two identical sister chromatids, and anaphase sent one chromatid of every chromosome to each pole.

29

So each daughter cell receives one complete copy of the genome. The skin cell’s two cells carry the same chromosomes as the skin cell, all 46.

30

In meiosis, no. Each of the four cells holds one chromosome of each homologous pair.

31

The two chromosomes of a homologous pair may carry different alleles. So the four cells differ from one another, and none matches the ovary cell.

A table comparing mitosis and meiosis, with columns mitosis and meiosis and 4 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four; chromosome number, 2n, the same as the parent against n, half the parent's; identical to the parent?, yes against no
A table comparing mitosis and meiosis, with columns mitosis and meiosis and 4 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four; chromosome number, 2n, the same as the parent against n, half the parent's; identical to the parent?, yes against no
32

In mitosis, homologous pairs never form. In meiosis, homologous pairs form in prophase I.

33

Because the pairs formed, meiosis I can part the two chromosomes of each homologous pair instead of the sister chromatids.

A table comparing mitosis and meiosis, with columns mitosis and meiosis and 5 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four; chromosome number, 2n, the same as the parent against n, half the parent's; identical to the parent?, yes against no; do homologous pairs form?, never against in prophase I
A table comparing mitosis and meiosis, with columns mitosis and meiosis and 5 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four; chromosome number, 2n, the same as the parent against n, half the parent's; identical to the parent?, yes against no; do homologous pairs form?, never against in prophase I
34

Mitosis grows and repairs the body: one cell becomes two alike. Meiosis makes gametes.

A table comparing mitosis and meiosis, with columns mitosis and meiosis and 6 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four; chromosome number, 2n, the same as the parent against n, half the parent's; identical to the parent?, yes against no; do homologous pairs form?, never against in prophase I; what it is for, growth and repair against making gametes
A table comparing mitosis and meiosis, with columns mitosis and meiosis and 6 of six rows filled: divisions, one against two: meiosis I and meiosis II; cells produced, two against four; chromosome number, 2n, the same as the parent against n, half the parent's; identical to the parent?, yes against no; do homologous pairs form?, never against in prophase I; what it is for, growth and repair against making gametes
35

The same organism does both. The skin cell and the ovary cell come from one body.

36

Whether a cell divides by mitosis or by meiosis depends on which cell it is, not on which organism it comes from.

37

What you are expected to know Compare mitosis and meiosis row by row: divisions, cells produced, chromosome number, identical to the parent or not, whether homologous pairs form, and what each is for.

38
Check q8

A cell in a mouse’s skin divides and makes two cells with 40 chromosomes each, the same as the skin cell. A cell in the mouse’s testis divides and makes four cells with 20 chromosomes each.

Which of the following happened in the testis cell’s division alone?

  1. A. The chromosomes condensed into X shapes before they moved
    Chromosomes condense into X shapes in mitosis as well; the skin cell’s did too.
  2. B. The spindle pulled sister chromatids apart
    The spindle pulls sister chromatids apart in mitosis as well.
    Meiosis does it in its second division.
  3. C. ✓ Homologous pairs lay together before lining up at the equator
  4. D. The cytoplasm divided by cytokinesis at the end
    Both divisions end with cytokinesis; meiosis has it twice.

Why: Four cells with half the count is meiosis.
Only meiosis forms homologous pairs, in prophase I, before they line up.
Condensing, parting sister chromatids and cytokinesis happen in both divisions.

39
Practice writing an answer

Two cells from one human body, whose body cells hold 46 chromosomes, are drawn below at metaphase, labelled cell 1 and cell 2. Cell 1 came from the skin and cell 2 from an ovary. Only a few of the chromosomes are drawn, as X shapes.

Two cells drawn side by side, labelled cell 1 and cell 2, both from one human body. Each cell has a pole at each end and spindle fibers, and four X-shaped chromosomes stand across its middle
Two cells drawn side by side, labelled cell 1 and cell 2, both from one human body. Each cell has a pole at each end and spindle fibers, and four X-shaped chromosomes stand across its middle

(a) Describe one feature of each drawing that shows which division the cell is in. (1 pt)

Model answer In cell 1 the chromosomes stand in a single row at the equator, each on its own, not paired.
Cell 1 holds a dark X and a light X of each length: both chromosomes of each pair, unpaired.
Only mitosis lines up both chromosomes of every pair without pairing them, so cell 1 is in mitosis.
In cell 2 the chromosomes stand in homologous pairs at the equator, two wide.
Pairs at the equator is the mark of meiosis I.
Rubric
  • Award 1 point for: cell 1, the chromosomes stand in a single row at the equator, each on its own and not paired (mitosis); AND cell 2, the chromosomes stand in homologous pairs at the equator, two wide (meiosis I).
  • Also accept for cell 1: a dark X and a light X of each length are both present but not paired, so the cell holds both chromosomes of each pair and is not in meiosis II.
  • Do not award: the skin or ovary source alone; the task asks for a feature of the drawing.

Slip Reading the X shapes as the sign of one division. Both cells have copied chromosomes; what differs is pairs against single chromosomes at the equator.

(b) Describe one way the cells that cell 2 finally produces differ from cell 2 itself. (1 pt)

Model answer Each cell that cell 2 produces is haploid: it holds one chromosome of each homologous pair, where cell 2 holds two of each pair.
The two chromosomes of a homologous pair may carry different alleles.
So none of the cells carries the same alleles as cell 2.
Rubric
  • Award 1 point for: the cells are haploid, one of each homologous pair, while cell 2 is diploid, two of each pair; or: they do not carry the same alleles as cell 2 because each holds only one member of each pair; or: each of their chromosomes is a single chromatid, where every chromosome of cell 2 is two sister chromatids.

Slip Saying the cells are identical copies of cell 2. Each cell holds one member of each pair, and the members of a pair may differ.

(c) Predict the number of cells each division finally produces and the chromosome number of each. (1 pt)

Model answer Cell 1 produces two cells with 46 chromosomes each.
Cell 2 produces four cells with 23 chromosomes each.
Rubric
  • Award 1 point for: two cells of 46 from cell 1 and four cells of 23 from cell 2.

Slip Halving the count for both, or giving cell 2 two cells. Meiosis halves the count and divides twice.

(d) Explain why the cells that cell 1 produces are genetically identical to cell 1. (1 pt)

Model answer Every chromosome of cell 1 was copied once in S phase into two identical sister chromatids.
At anaphase the spindle sends one chromatid of every chromosome to each pole.
So each daughter cell receives one complete copy of the genome, the same 46 chromosomes as cell 1.
Rubric
  • Award 1 point for: the sister chromatids are identical copies, and anaphase sends one chromatid of every chromosome to each pole, so each cell gets one complete copy of the genome.

Slip Saying the cells are identical because mitosis is for growth. The purpose does not explain it; the identical copies and their separation do.

40

Back to the skin cell and the ovary cell from one body: the skin cell dividing to close a cut, the ovary cell dividing to make eggs.

41

The skin cell divides once, by mitosis, and makes two cells, each with 46, identical to itself. The ovary cell divides twice, by meiosis, and makes four cells, each with 23, none identical to it or to one another.

42

The copying, the spindle and the stage names were the same in both. What differed is that the ovary cell’s homologous pairs found each other and parted first.

APBIO-U05-P51 Practice questions: Topic 5.1

Topic 5.1 · Meiosis · 10 MCQ · 2 FRQ · for APBIO-U05-T51

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation. The first free-response question walks you through one cell’s meiosis one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. In every drawing, a chromosome from the mother is dark and one from the father is light.

Video: Watch first: Meiosis, summed up

One copying and two divisions; the pairs part in meiosis I and the count halves; the copies part in meiosis II; four haploid cells; the eight stages by what the cell shows; mitosis beside meiosis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T51-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T51-summary.mp4

Q1 P51-q01

An animal’s body cells hold six chromosomes. The six from one cell are drawn below and numbered 1 to 6; the mother’s chromosomes are dark and the father’s are light.

Six chromosomes from one of the animal’s cells, numbered 1 to 6.
Six chromosomes from one of the animal’s cells, numbered 1 to 6.

Which two chromosomes are a homologous pair?

  1. A. Chromosomes 4 and 5
    4 and 5 are different lengths, so they carry different genes.
  2. B. Chromosomes 1 and 2
    1 and 2 are different lengths.
  3. C. ✓ Chromosomes 2 and 5
  4. D. Chromosomes 1 and 4
    1 and 4 are the longest and the shortest, different genes.

Why: A homologous pair is two chromosomes of the same length carrying the same genes, one from each parent: 2 and 5, the two medium chromosomes, one dark and one light.

Q2 P51-q02

A red deer’s body cells each hold 68 chromosomes. A student says that any 34 of the 68 make one complete chromosome set.

Which of the following statements about the student’s claim is correct?

  1. A. The student is right: a set is any half of the chromosomes
    Any 34 might hold both members of one pair and neither member of another.
    Then that other pair’s genes are missing, so the 34 are not a set.
  2. B. The student is wrong: a set is all 68 chromosomes
    All 68 chromosomes are both members of every pair: two complete sets.
    One set is one chromosome of every pair, 34 chosen that way.
  3. C. The student is right: any 34 chromosomes carry every gene once
    Only one chromosome of every pair carries every gene once.
    Any 34 might hold both members of one pair and neither member of another.
  4. D. ✓ The student is wrong: a set is one chromosome of every pair, not any 34

Why: A complete chromosome set is one chromosome of every pair, every gene once.
Any 34 of the 68 might hold both members of one pair and neither member of another.
Then that pair’s genes are missing.
So the student is wrong: a set is 34 chosen one from every pair.

Q3 P51-q03

In a salamander, a sperm and an egg fuse.

Which term names the fusion of the two gametes?

  1. A. Meiosis
    Meiosis is the division that made the gametes, before they fuse.
  2. B. Mitosis
    Mitosis is the division that later builds the body from the zygote.
  3. C. ✓ Fertilization
  4. D. Cytokinesis
    Cytokinesis is the splitting of one cell into two at the end of a division; fertilization joins two cells into one.

Why: Fertilization is the fusion of two gametes, a sperm and an egg, into a single cell, the zygote.

Q4 P51-q04

A snail’s body cells hold 34 chromosomes. Its gametes hold 17 and its zygote holds 34, so its offspring’s body cells hold 34 as well.

Which statement explains why the count stays at 34 from one generation to the next?

  1. A. ✓ A halving division before fertilization cancels the doubling at fertilization
  2. B. Fertilization keeps only one gamete’s set and discards the other
    Fertilization fuses two gametes, so the zygote holds both gametes’ sets together and discards nothing.
    The gametes hold 17 each, so the zygote holds 34.
  3. C. Fertilization halves the two full sets the gametes bring to one set
    Fertilization adds the two gametes’ sets together, so the count goes up at fertilization, never down; the gametes already hold 17 because meiosis halved the count first.
  4. D. The zygote’s first mitosis halves the count back to 34
    Before each mitosis the cell copies every chromosome once.
    Mitosis then keeps the count the same, so every body cell holds 34.
    The halving happened earlier, in meiosis.

Why: Meiosis halves the count before fertilization, so each gamete carries one set, 17.
Fertilization joins two single sets into a double set, 34.
So the halving and the doubling cancel, and the count stays at 34.

Q5 P51-q05

One homologous pair from a locust cell in prophase I is drawn below. Two points on the drawing are marked J and K.

One homologous pair from a locust cell in prophase I, with points J and K marked.
One homologous pair from a locust cell in prophase I, with points J and K marked.

Which of the following names the points J and K?

  1. A. J a centromere, K a centromere
    K is where the two sister chromatids of one homolog are joined: a centromere.
    J is where an arm of one homolog crosses an arm of the other: a chiasma.
  2. B. J a centromere, K a chiasma
    J is the crossing point between the two homologs, a chiasma.
    K joins the two sister chromatids of one homolog, so K is a centromere.
  3. C. ✓ J a chiasma, K a centromere
  4. D. J a chiasma, K a chiasma
    K joins the two sister chromatids of one homolog: a centromere.
    A chiasma is a crossing point between the two homologs, and only J is one.

Why: In prophase I the two homologs of a pair lie side by side.
An arm of one crosses an arm of the other, and the two are held there: that crossing point, J, is a chiasma.
K sits where the two sister chromatids of one homolog are joined: a centromere.

Q6 P51-q06

A moss’s cells that go through meiosis start with 40 chromosomes. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

What is the chromosome count per cell after meiosis I?

  1. A. 40
    Meiosis I parts the homologous pairs, so the count halves.
  2. B. ✓ 20
  3. C. 10
    10 would be two halvings; the count halves once, at meiosis I.
  4. D. 80
    Copying doubles the chromatids, not the chromosomes, and meiosis I halves the count.

Why: Each cell after meiosis I receives one member of every pair: half of 40, 20.

Q7 P51-q07

An insect’s body cells hold eight chromosomes. Two of its cells are drawn below, numbered 1 and 2, each with its chromosomes across the middle of the cell. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Two of the insect’s cells, numbered 1 and 2, with chromosomes across the middle.
Two of the insect’s cells, numbered 1 and 2, with chromosomes across the middle.

Which drawing shows metaphase II?

  1. A. Drawing 1 only
    In drawing 1 the chromosomes stand in pairs, two wide, and all eight are present.
    Pairs at the equator is metaphase I.
  2. B. Both drawings
    Drawing 1 has pairs at the equator and the full eight; drawing 2 has single chromosomes in one row and only four, so only drawing 2 is metaphase II.
  3. C. Neither drawing
    Drawing 2 shows four single chromosomes in one row at the equator, half the insect’s eight.
    A single row of half the count is metaphase II.
  4. D. ✓ Drawing 2 only

Why: Metaphase II is single chromosomes, not pairs, in one row at the equator of a haploid cell: drawing 2, with four singles in a row, half of eight.
Drawing 1, with four pairs two wide, is metaphase I.

Q8 P51-q08

One of the two cells from a mouse’s meiosis I has begun to divide again.

Which statement describes this division?

  1. A. ✓ The same four stages as mitosis, in a haploid cell
  2. B. A new pairing of homologs followed by their parting
    A haploid cell has no homologs to pair; meiosis II parts sister chromatids.
  3. C. One more DNA copying followed by one division
    No DNA copying happens between the two divisions.
  4. D. A division with no spindle, because the cell is haploid
    Meiosis II builds a spindle and uses it to part the sister chromatids.

Why: In meiosis II, each haploid cell from meiosis I passes through prophase, metaphase, anaphase and telophase; only the cell differs from mitosis.

Q9 P51-q09

A cell in a turtle’s skin and a cell in its testis both divide.

What does each division produce?

  1. A. ✓ Skin cell: two identical diploid cells; testis cell: four haploid cells
  2. B. Skin cell: four haploid cells; testis cell: two diploid cells
    The skin cell divides by mitosis, which gives two diploid cells, and the testis cell divides by meiosis, which gives four haploid cells.
  3. C. Skin cell: two haploid cells; testis cell: four diploid cells
    Mitosis keeps the diploid count and meiosis halves it.
  4. D. Both: two identical diploid cells
    The testis cell makes gametes by meiosis, four haploid cells.

Why: Mitosis, in the skin, gives two cells identical to the parent, diploid; meiosis, in the testis, gives four haploid cells, none identical.

Q10 P51-q10

A female cod releases her eggs into the sea, and a male cod releases his sperm over them. Each egg that a sperm enters grows into a young cod.

Which of the following are the gametes?

  1. A. The female’s and the male’s body cells
    Body cells hold two complete sets and divide by mitosis to build the body.
    A gamete is a haploid cell that joins with another to start a new organism.
  2. B. ✓ The eggs and the sperm
  3. C. The eggs that sperm have entered
    An egg that a sperm has entered is the zygote: the single cell the two gametes fused into.
  4. D. The cells of the young cod
    The young cod’s cells grew from the zygote by mitosis.
    Each of those cells holds two complete sets.

Why: A gamete is a haploid cell that joins with another to start a new organism.
Each egg and each sperm holds one complete set.
A sperm joins an egg, and a new cod starts from the joined cell.
So the eggs and the sperm are the gametes.

FRQ 1 P51-frq1 · Conceptual Analysis scaffolded

A radish’s body cells each hold 18 chromosomes. One cell in its anther, the part of the flower that makes pollen, copies its DNA and goes through meiosis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

(a) Calculate the number of homologous pairs in one of the radish’s body cells. (1 pt)

Frame The cell holds 18 chromosomes, and a homologous pair is … chromosomes, so it has … pairs.

Hint How many chromosomes make up one homologous pair?

Answer: 9  (tolerance ±0)

Model answer The cell holds 18 chromosomes, and a homologous pair is 2 chromosomes, so it has 9 pairs.
Working
Write down the values in the question:
tex:2n = 18
Write down the equation:
tex:\text{pairs} = n = \frac{2n}{2}
Substitute the values into the equation:
tex:\text{pairs} = \frac{18}{2} = 9
Rubric
  • Award 1 point for: 9 pairs.

(b) Calculate the number of chromatids in the cell after it copies its DNA. (1 pt)

Frame After the copying, each chromosome is … chromatids, so the cell holds … chromatids.

Hint What does copying do to each chromosome, and how many chromosomes are there?

Answer: 36  (tolerance ±0)

Model answer After the copying, each chromosome is 2 chromatids, so the cell holds 36 chromatids.
Working
Write down the values in the question:
tex:2n = 18
Write down the equation:
tex:\text{chromatids after copying} = 2 \times 2n
Substitute the values into the equation:
tex:\text{chromatids} = 2 \times 18 = 36
Rubric
  • Award 1 point for: 36 chromatids.

(c) Calculate the number of chromosomes in each cell after meiosis I. (1 pt)

Frame Each cell holds … chromosomes.

Hint Which partners part in meiosis I?

Answer: 9  (tolerance ±0)

Model answer Each cell holds 9 chromosomes.
Working
Write down the values in the question:
tex:2n = 18
Write down the equation:
tex:\text{chromosomes after meiosis I} = n = \frac{2n}{2}
Substitute the values into the equation:
tex:\text{chromosomes after meiosis I} = \frac{18}{2} = 9
Rubric
  • Award 1 point for: 9 chromosomes.

(d) Explain why each cell after meiosis I holds twice as many chromatids as chromosomes. (1 pt)

Frame Each chromosome is …, so each cell has two chromatids for every …

Hint Count the centromeres of one chromosome after meiosis I.

Model answer Each chromosome is still two sister chromatids joined at one centromere.
Chromosomes are counted by centromere.
So each cell has two chromatids for every chromosome: 18 chromatids and 9 chromosomes.
Rubric
  • Award 1 point for: the reason — each chromosome is two sister chromatids joined at one centromere, and chromosomes are counted by centromere.

Slip Saying meiosis I split the centromeres. Nothing splits at the centromere in meiosis I.

(e) Calculate the number of chromosomes in each cell after meiosis II. (1 pt)

Frame Meiosis II parts the …, so each cell holds … chromosomes, each a single chromatid.

Hint Which partners part in meiosis II, and were they already counted inside their chromosome?

Answer: 9  (tolerance ±0)

Model answer Meiosis II parts the sister chromatids, so each cell still holds 9 chromosomes, each a single chromatid.
Working
Write down the values in the question:
tex:n = 9
Write down the equation:
tex:\text{chromosomes after meiosis II} = n
Substitute the values into the equation:
tex:\text{chromosomes after meiosis II} = 9
Rubric
  • Award 1 point for: 9 chromosomes.

(f) Explain your answers to (c) and (e). (1 pt)

Frame At meiosis I each cell receives …, so the count …; at meiosis II the two … of one chromosome part, and each was already counted as …, so the count …

Hint Count centromeres: what does each cell receive in each division?

Model answer At meiosis I each cell receives one member of every homologous pair, so the count halves from 18 chromosomes to 9.
At meiosis II the two sister chromatids of each chromosome part, and each chromatid was already counted inside its chromosome, so each cell keeps 9.
Rubric
  • Award 1 point for: BOTH mechanisms — meiosis I gives each cell one member of every pair, so the count halves; AND meiosis II parts sister chromatids, each already counted as one chromosome, so the count stays.
  • Do not award: one mechanism alone.

Slip Saying the count halves twice. A second halving would give three and a half chromosomes.

FRQ 2 P51-frq2 · Conceptual Analysis

Two cells from an insect, whose body cells hold six chromosomes, are drawn below, numbered 1 and 2. Both are in meiosis, frozen while their chromosomes move toward the poles. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Two of the insect’s cells, numbered 1 and 2, frozen while their chromosomes move toward the poles.
Two of the insect’s cells, numbered 1 and 2, frozen while their chromosomes move toward the poles.

(a) Identify the stage each cell is in. (1 pt)

Model answer Cell 1 is in anaphase II; cell 2 is in anaphase I.
Rubric
  • Award 1 point for: cell 1 anaphase II and cell 2 anaphase I.
  • Do not award: anaphase of mitosis for cell 1; three chromatids reach each pole, half the insect’s six, so the cell is haploid and in meiosis II.

Slip Swapping the two. Whole X’s moving is meiosis I; single chromatids moving is meiosis II.

(b) Describe what the spindle pulls apart in each cell. (1 pt)

Model answer In cell 2 the spindle pulls the two members of each homologous pair to opposite poles, each still two sister chromatids.
In cell 1 the spindle pulls the two sister chromatids of each chromosome to opposite poles.
Rubric
  • Award 1 point for: homologs in cell 2 (anaphase I), sister chromatids in cell 1 (anaphase II).

Slip Saying sister chromatids part in both. At anaphase I nothing splits at the centromere.

(c) Predict the chromosome count in each of the two cells that cell 2 produces. (1 pt)

Answer: 3  (tolerance ±0)

Model answer 3 chromosomes in each, one of every pair, each still two chromatids.
Working
Write down the values in the question:
tex:2n = 6
Write down the equation:
tex:\text{chromosomes after meiosis I} = n = \frac{2n}{2}
Substitute the values into the equation:
tex:\text{chromosomes after meiosis I} = \frac{6}{2} = 3
Rubric
  • Award 1 point for: 3.

(d) Evaluate the claim that the four cells at the end of this insect’s meiosis are genetically identical to the cell that began it. (1 pt)

Model answer The claim is not supported.
Each of the four cells holds one member of each homologous pair, not both.
The two members of a pair may carry different alleles.
So each cell carries only one allele of each pair, while the starting cell carried both.
Therefore the four cells differ genetically from the cell that began the meiosis.
Rubric
  • Award 1 point for: the judgement (the claim is not supported) AND the ground (each cell receives one member of each pair, and the members of a pair may carry different alleles).
  • Accept as the ground, any one of: each end cell holds one member of each pair, so half the chromosomes; the members of a pair may carry different alleles; crossing over gave new allele combinations.

Slip Saying the claim is wrong with no ground. Evaluate needs the judgement and the reason the biology gives for it.

APBIO-U05-T51 End-of-topic test: Meiosis

Topic 5.1 · Meiosis · 19 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation, then open the scoring guide and mark your own work against it. In every drawing, a chromosome from the mother is dark and one from the father is light.

Q1 T51-q01

A midge’s body cells hold six chromosomes. The six from one cell are drawn below and numbered 1 to 6. In every drawing here the mother’s chromosomes are dark and the father’s are light.

Six chromosomes from one midge cell, numbered 1 to 6.
Six chromosomes from one midge cell, numbered 1 to 6.

Which two chromosomes are a homologous pair?

  1. A. Chromosomes 1 and 2
    Chromosomes 1 and 2 are different lengths, so they carry different genes; the two members of a pair are the same length.
  2. B. ✓ Chromosomes 1 and 4
  3. C. Chromosomes 1 and 6
    Chromosomes 1 and 6 are different lengths and so carry different genes.
  4. D. Chromosomes 3 and 6
    Chromosomes 3 and 6 are different lengths and so carry different genes; sharing a parent does not make two chromosomes a pair.

Why: A homologous pair is two chromosomes of the same length, carrying the same genes in the same order, one from each parent: 1 and 4 are the two long chromosomes, one dark and one light.

Q2 T51-q02

In guinea pigs, a gene sets fur color. One guinea pig’s homologous pair is drawn below: at the fur-color position, the homolog from the mother carries the black version of the gene and the homolog from the father carries the cream version.

One guinea pig’s homologous pair with the fur-color position boxed: black on the maternal homolog, cream on the paternal.
One guinea pig’s homologous pair with the fur-color position boxed: black on the maternal homolog, cream on the paternal.

What are the black version and the cream version?

  1. A. Two different genes, one for black fur and one for cream fur
    Both versions sit at the same position and set the same feature, fur color, so they are versions of one gene, not two genes.
  2. B. Two sister chromatids of one chromosome
    Sister chromatids are identical copies made in S phase; these two versions sit on two different homologs and differ.
  3. C. ✓ Two alleles of the fur-color gene
  4. D. Two complete chromosome sets
    A chromosome set is one of every pair, many chromosomes; these are two versions of a single gene.

Why: Alleles are versions of one gene at one position on the two homologs; black and cream are two alleles of the fur-color gene.

Q3 T51-q03

A chimpanzee’s body cells each hold 48 chromosomes.

Which of the following cells is haploid?

  1. A. ✓ A cell with 24 chromosomes, one of every pair
  2. B. A cell holding both members of 12 pairs and neither member of the other 12
    Both members of 12 pairs and neither member of the other 12 is 24 chromosomes but not one of every pair.
    A haploid cell is one complete set.
  3. C. A cell with 48 chromosomes, two of every length
    Two of every length is both members of every pair.
    So this cell holds two complete sets, and it is diploid, a body cell.
  4. D. A cell with 96 chromosomes, four of every length
    No chimpanzee cell holds 96 chromosomes.
    A body cell after S phase holds 96 chromatids, but each chromosome is still one chromosome, so the cell still holds 48.

Why: A chimpanzee’s 48 chromosomes are 24 homologous pairs.
A haploid cell holds one complete set: one member of every pair, 24 chromosomes, every gene once.
Twenty-four chromosomes that are both members of 12 pairs are not a set, and 48 of two of every length are two sets.

Q4 T51-q04

An axolotl’s body cells hold 28 chromosomes. An egg and a sperm fuse.

Which statement about the zygote is correct?

  1. A. It holds 14 chromosomes, one set from each parent
    Each gamete brings one set of 14. Fertilization adds the two sets together.
    So the zygote holds 28, not 14.
  2. B. It holds 56 chromosomes, both parents’ full sets
    A gamete is haploid, one set of 14, not a full body-cell count of 28.
    Two sets of 14 make 28.
  3. C. It is the egg before the sperm enters it
    The egg before the sperm enters it is a gamete, haploid.
    The zygote is the single cell that the fusion makes.
  4. D. ✓ It holds 28 chromosomes and divides by mitosis to build the body

Why: The egg and the sperm each carry one set of 14 chromosomes.
Fertilization fuses them into one cell, the zygote, which holds both sets, 28.
The zygote then divides by mitosis, so every body cell holds 28.

Q5 T51-q05

An iguana’s body cells hold 36 chromosomes and its gametes hold 18.

Why must the chromosome count be halved before fertilization?

  1. A. Because a gamete is too small to hold a full set of chromosomes
    A gamete’s size has nothing to do with the count.
    The count is halved so that fertilization can add two half sets back to one full count.
  2. B. Because mitosis after fertilization would halve the count back down anyway
    Before each mitosis the cell copies every chromosome once.
    Mitosis then keeps the count the same in every cell it makes.
    Nothing after fertilization halves the count.
  3. C. Because the zygote keeps only one of the two parents’ sets
    The zygote keeps both sets, one from each gamete.
    That is why each gamete must bring half.
  4. D. ✓ Because fertilization adds the two sets together, so two full sets would give 72

Why: Fertilization fuses two gametes, so the zygote holds both gametes’ sets added together.
Two gametes of 18 make a zygote of 36.
If each gamete carried a full 36, the zygote would hold 72, and the count would double every generation.
So a halving division must come before fertilization.

Q6 T51-q06

One of the two cells from a badger’s meiosis I begins meiosis II. A student says that the cell’s homologous pairs come together again before meiosis II can part them.

Which of the following statements about the student’s claim is correct?

  1. A. The student is right: the pairs come together before each of the two divisions
    Meiosis I sent the two members of every pair to different cells.
    So a cell in meiosis II holds one member of each pair, and no pair to bring together.
  2. B. ✓ The student is wrong: the cell is haploid, so it holds no homologous pairs to bring together
  3. C. The student is right: meiosis II is a repeat of meiosis I
    Meiosis I parts the homologous pairs.
    A cell in meiosis II is haploid, with no pairs, so meiosis II repeats the stages of mitosis, not meiosis I.
  4. D. The student is wrong: the cell still holds every pair, and the pairs do not need to come together again
    Meiosis I sent one member of every pair to each cell.
    So the cell in meiosis II holds one member of each pair, not the pair.

Why: Meiosis I sends the two members of every pair to different cells.
So each cell in meiosis II holds one member of each pair: it is haploid.
A haploid cell has no pairs to bring together.
So the student is wrong: meiosis II repeats mitosis’s stages on a haploid cell.

Q7 T51-q07

A cell from a shrew is drawn below, frozen part way through a division.

A shrew cell, frozen part way through a division.
A shrew cell, frozen part way through a division.

Which stage is the cell in?

  1. A. ✓ Anaphase I
  2. B. Anaphase II
    At anaphase II the moving pieces are single chromatids, V-shaped.
    Every moving piece here is a whole X of two chromatids.
  3. C. Anaphase of mitosis
    At anaphase of mitosis the centromeres have split and single chromatids move.
    Here every moving piece is a whole X, so no centromere has split.
  4. D. Metaphase I
    At metaphase I the chromosomes stand still at the equator, in pairs.
    Here the chromosomes have left the equator and move toward the poles.

Why: Whole X-shaped chromosomes, each still two sister chromatids, are moving toward the two poles.
Whole chromosomes part only at anaphase I, when the two members of each homologous pair separate.
At anaphase II and at anaphase of mitosis the moving pieces are single chromatids.

Q8 T51-q08

A plant’s body cells each hold 20 chromosomes. One cell goes through meiosis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

What is the chromosome count per cell after meiosis I, and after meiosis II?

  1. A. 20 after meiosis I, and 10 after meiosis II
    Meiosis I parts the homologous pairs, so the count halves there, from 20 chromosomes to 10.
  2. B. 10 after meiosis I, and 5 after meiosis II
    Meiosis II parts sister chromatids, and each chromatid was already counted inside its chromosome, so the count stays at 10.
  3. C. ✓ 10 after meiosis I, and 10 after meiosis II
  4. D. 20 after meiosis I, and 20 after meiosis II
    Meiosis I sends one member of every pair to each cell, halving the count from 20 chromosomes to 10.

Why: The count halves once, at meiosis I, when each cell receives one member of every pair: 20 to 10.
Meiosis II parts sister chromatids, so each cell keeps 10.

Q9 T51-q09

A cell from a species of sedge whose body cells hold six chromosomes is drawn below. Its chromosomes are condensed, and there is no nuclear envelope.

A sedge cell, part way through a division.
A sedge cell, part way through a division.

Which feature of the drawing belongs to prophase I alone, so that it rules out prophase of mitosis?

  1. A. The chromosomes have condensed
    The chromosomes condense in prophase of mitosis as well.
    So condensed chromosomes are true of both prophases, and they name neither.
  2. B. The nuclear envelope has broken down
    The nuclear envelope breaks down in prophase of mitosis as well.
    So a missing envelope is true of both prophases, and it names neither.
  3. C. Spindle fibers reach in from the two poles
    The spindle forms and its fibers reach in during prophase of mitosis as well.
    So fibers reaching in are true of both prophases, and they name neither.
  4. D. ✓ The homologs lie paired along their length

Why: Condensed chromosomes, a broken-down envelope and a forming spindle happen in both prophases.
Homologs pairing along their length, synapsis, happens only in prophase I. So the paired homologs are the one feature that shows prophase I.

Q10 T51-q10

A mite’s body cells hold four chromosomes. Two of its cells are drawn below, numbered 1 and 2, each frozen part way through a division. All the chromosomes are drawn. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Two mite cells, numbered 1 and 2, each frozen part way through a division.
Two mite cells, numbered 1 and 2, each frozen part way through a division.

Which drawing shows anaphase I?

  1. A. Drawing 1 only
    In drawing 1 single chromatids move, so every centromere has split: not anaphase I.
    In drawing 2 whole X’s move: anaphase I.
  2. B. ✓ Drawing 2 only
  3. C. Both drawings
    In drawing 1 single chromatids move, so every centromere has split.
    At anaphase I no centromere splits: whole X’s move, as in drawing 2 only.
  4. D. Neither drawing
    In drawing 2 every moving piece is a whole X of two chromatids, so no centromere has split.
    Whole chromosomes part only at anaphase I.

Why: In drawing 1 single chromatids move, four toward each pole.
Four is the mite’s full count, so the cell is diploid: anaphase of mitosis.
In drawing 2 whole X’s of two chromatids move, two toward each pole.
Whole chromosomes part only at anaphase I, so drawing 2 alone shows it.

Q11 T51-q11

In a cell from a leek flower at anaphase I, the spindle fibers shorten and chromosomes move toward the two poles.

Which partners are being pulled apart?

  1. A. ✓ The two members of each homologous pair
  2. B. The two sister chromatids of each chromosome
    Sister chromatids part in meiosis II, and in mitosis; at anaphase I nothing splits at the centromere, and what separates is the two members of each homologous pair.
  3. C. The two members of each pair and the two chromatids of each member, all at once
    The two divisions do different jobs: meiosis I parts the members of each pair, and only in meiosis II do the sister chromatids part.
  4. D. Nothing within a pair: each whole pair is pulled to one pole or the other
    The two members of a pair face opposite poles at metaphase I, and at anaphase I the spindle pulls them apart, one to each pole.

Why: At anaphase I the spindle pulls the two members of each homologous pair to opposite poles; nothing splits at the centromere, so the sister chromatids of each chromosome stay joined until meiosis II.

Q12 T51-q12

A cell from a bean plant whose body cells hold 22 chromosomes has just finished meiosis I and cytokinesis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

How many chromatids does each of the two cells hold?

  1. A. 11
    11 is the number of chromosomes in each cell.
    Nothing split at the centromere in meiosis I, so each of those 11 is still two sister chromatids.
  2. B. ✓ 22
  3. C. 44
    44 is the chromatid count of the parent cell after S phase.
    Meiosis I sent one member of every pair to each cell, so each cell holds half of them.
  4. D. 88
    88 would need a second copying after meiosis I.
    No copying happens between the two divisions.

Why: After meiosis I each cell holds one member of every pair, 11 chromosomes.
Nothing split at the centromere in meiosis I, so each of those chromosomes is still two sister chromatids.
So each cell holds twice 11 chromatids; the working below gives 22.

Q13 T51-q13

A cell from an animal whose body cells hold six chromosomes is drawn below, part way through meiosis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

An animal cell, part way through a division.
An animal cell, part way through a division.

Which stage is the cell in?

  1. A. Metaphase I
    Metaphase I is a double row of homologous pairs in a diploid cell of six; here three single chromosomes stand in one row, half the animal’s six.
  2. B. Prophase II
    In prophase II the chromosomes are still scattered and attaching to a forming spindle; here they stand in a row at the equator with fibers from both poles.
  3. C. ✓ Metaphase II
  4. D. Anaphase II
    At anaphase II single chromatids are moving toward the poles; here every chromosome is still a whole X standing at the equator.

Why: Three single X-shaped chromosomes in one row at the equator of a haploid cell, half the animal’s six, is metaphase II.

Q14 T51-q19

A cell from a cicada is drawn below, part way through a division.

A cicada cell, part way through a division.
A cicada cell, part way through a division.

Which stage is the cell in?

  1. A. Anaphase I
    At anaphase I the chromosomes are still moving toward the poles and the cell has not begun to pinch; here a furrow crosses the cell and an envelope is re-forming.
  2. B. ✓ Telophase I
  3. C. Telophase II
    At telophase II every chromosome is a single chromatid, a rod; here every chromosome is still an X of two chromatids, so only the first division has happened.
  4. D. Telophase of mitosis
    At telophase of mitosis the sister chromatids have parted and single chromatids sit in each forming cell; here the chromosomes are still X-shaped.

Why: A furrow across the cell, an envelope re-forming on each side and every chromosome still an X of two chromatids: two cells are forming after the first division, telophase I.

Q15 T51-q14

A cell in a quail’s ovary holds 8 picograms (pg) of DNA before it copies its DNA. It then goes through meiosis.

How much DNA does each cell hold after meiosis I?

  1. A. 16 pg
    16 pg is the amount after the copying, before either division; meiosis I then halves it.
  2. B. ✓ 8 pg
  3. C. 4 pg
    4 pg would be the amount after both divisions; after meiosis I only one halving has happened.
  4. D. 2 pg
    2 pg would need three halvings; the DNA is copied once to 16 pg and halved twice, to 8 pg and then 4 pg.

Why: The copying doubles the DNA, so the cell holds twice the 8 pg.
Meiosis I sends one member of every pair to each cell, so each cell holds half of that.
The doubling and the halving cancel; the working below gives 8 pg.

Q16 T51-q15

Three drawings of chromosomes from a hedgehog’s cells are numbered 1 to 3 below. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Three drawings from a hedgehog’s cells, numbered 1 to 3.
Three drawings from a hedgehog’s cells, numbered 1 to 3.

Which of the drawings show a homologous pair?

  1. A. Drawing 1 only
    Drawing 1 has one centromere, so it is one chromosome: two sister chromatids, not a pair.
    Drawing 2’s two rods have two centromeres and the same length: a pair.
  2. B. Drawing 2 only
    Drawing 3’s two X’s have two centromeres and the same length, so they are a pair.
    Copying a chromosome does not stop it being one member of a pair.
  3. C. ✓ Drawings 2 and 3
  4. D. Drawings 1, 2 and 3
    Drawing 1 has one centromere, so it is one copied chromosome, however many arms it has.
    A pair is two chromosomes, so a pair has two centromeres.

Why: A homologous pair is two chromosomes of the same length, one from each parent.
Two chromosomes means two centromeres.
Drawings 2 and 3 each show two centromeres on chromosomes of the same length.
Drawing 1 shows one centromere: one chromosome of two sister chromatids.

Q17 T51-q16

Each of the four cells at the end of meiosis in a tulip holds 12 single-chromatid chromosomes. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

How many chromosomes does a tulip body cell hold, and how many homologous pairs?

  1. A. 12 chromosomes, 6 pairs
    12 is the count of one of the four cells, which holds one member of every pair.
    A body cell holds both members of every pair, twice as many.
  2. B. 48 chromosomes, 24 pairs
    48 doubles the count twice.
    The four cells hold one of every pair, 12, so a body cell holds two of every pair, 24.
  3. C. ✓ 24 chromosomes, 12 pairs
  4. D. 12 chromosomes, 12 pairs
    12 is one cell's count, and a pair is two chromosomes, so 12 chromosomes make 6 pairs; a body cell holds both members of every pair: 24.

Why: Each of the four cells is haploid: it holds one member of every homologous pair.
So the 12 chromosomes are 12 pairs’ single members.
A body cell holds both members of every pair: 12 pairs, 24 chromosomes.

Q18 T51-q17

A fish’s body cells each hold 16 chromosomes. One cell copies its DNA. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

How many chromatids does the cell hold after it copies its DNA?

  1. A. ✓ 32
  2. B. 16
    16 is the number of chromosomes; after the copying every chromosome is two chromatids.
  3. C. 8
    8 would be a haploid set of chromosomes, which copying does not make; copying doubles the chromatids.
  4. D. 64
    64 would need every chromosome to become four chromatids; one copying makes two.

Why: The cell copies every chromosome once.
So each of the 16 chromosomes becomes two sister chromatids.
The cell holds twice 16 chromatids; the working below gives 32.

Q19 T51-q18

A cell in a wheat plant’s root divides by mitosis, and a cell in its anther, the part of the flower that makes pollen, divides by meiosis.

Which feature do the two divisions share?

  1. A. Homologous pairs come together before the chromosomes part
    Homologous pairs come together only in meiosis, in prophase I.
    Mitosis never pairs the homologs.
  2. B. Four cells are produced
    Mitosis produces two cells.
    Only meiosis, with its two divisions in a row, produces four.
  3. C. The chromosome number halves
    The chromosome number halves only in meiosis, at meiosis I.
    Mitosis keeps the parent cell’s count.
  4. D. ✓ The DNA is copied once before the division begins

Why: Both divisions begin after one DNA copying in S phase, so every chromosome enters either division as two sister chromatids.
Pairing of homologs, four cells and a halved count belong to meiosis alone.

FRQ 1 T51-frq1 · Scientific Investigation

Two cells from one tomato plant, whose body cells hold 24 chromosomes, are drawn below at metaphase, labeled cell 1 and cell 2. Both cells hold all 24 chromosomes, but only a few of them are drawn, as X shapes. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Cell 1, left, and cell 2, right, each at metaphase.
Cell 1, left, and cell 2, right, each at metaphase.

(a) Determine whether each cell is in mitosis, meiosis I or meiosis II, using the drawings and the chromosome count. (2 pt)

Model answer Cell 1’s chromosomes stand in a single row at the equator, one deep.
A single row is what mitosis and meiosis II both show.
Cell 1 holds all 24 chromosomes, and a cell in meiosis II holds only 12, so cell 1 is in mitosis.
Cell 2’s chromosomes stand in homologous pairs at the equator, two wide.
Pairs at the equator is what only meiosis I shows, so cell 2 is in meiosis I.
Rubric
  • Award 1 point for: cell 1 in mitosis, with the ground: a single row at the equator AND the full count of 24 chromosomes (a cell in meiosis II holds 12).
  • Award 1 point for: cell 2 in meiosis I, with the ground: homologous pairs at the equator, two wide.
  • Do not award: the number of chromosomes drawn as the feature for either cell; only a few are drawn.
  • Do not award: a single row alone for cell 1; metaphase II also shows a single row.

Slip Naming mitosis for cell 1 from the single row alone. Metaphase II also shows a single row; the full count of 24, where a cell in meiosis II holds 12, is what settles mitosis.

(b) Calculate the number of chromatids in cell 1 at the stage drawn. (1 pt)

Answer: 48  (tolerance ±0)

Model answer 48 chromatids.
Working
Write down the values in the question:
tex:2n = 24
Write down the equation:
tex:\text{chromatids at metaphase} = 2 \times 2n
Substitute the values into the equation:
tex:\text{chromatids} = 2 \times 24 = 48
Rubric
  • Award 1 point for: 48 chromatids.

(c) Predict, for each cell, how many cells there will be, and the chromosome number of each, when the cell has finished every division it will go through. (1 pt)

Model answer Cell 1 produces two cells with 24 chromosomes each.
Cell 2 produces four cells with 12 chromosomes each.
Rubric
  • Award 1 point for: cell 1 two cells of 24 chromosomes AND cell 2 four cells of 12 chromosomes, the counts when all division is over.
  • Do not award: two cells of 12 for cell 2; that is the end of meiosis I, and meiosis II still follows.

Slip Stopping cell 2 at the end of meiosis I (two cells of 12), or halving cell 1’s count. Meiosis divides twice; mitosis keeps the count.

(d) Explain, for each cell, whether the daughter cells match it genetically, and why. (2 pt)

Model answer Every chromosome of cell 1 was copied in S phase into two identical sister chromatids.
At anaphase the spindle sends one chromatid of every chromosome to each pole.
So each daughter cell receives the same 24 chromosomes as cell 1: the daughter cells match it.
Each of cell 2’s four cells holds one member of each pair.
The two members of a pair may carry different alleles.
So cell 2’s cells do not match cell 2.
Rubric
  • Award 1 point for: cell 1’s daughter cells match it, because the sister chromatids are identical copies and each daughter cell receives one of every chromosome.
  • Award 1 point for: cell 2’s cells do not match it, because each receives only one member of each pair (it is haploid), and the two members may carry different alleles. Accept the haploid ground on its own.

Slip Saying only that cell 2’s cells have fewer chromosomes, without saying what is missing. What is missing is the other member of each pair, and the two members may carry different alleles.

FRQ 2 T51-frq2 · Conceptual Analysis

The DNA in one cell of a plant, whose body cells hold 12 chromosomes, was measured before the cell copied its DNA, and again in one cell at the end of meiosis. The table below gives the two measurements in picograms (pg).

DNA per cell, in picograms, before copying and after meiosis II.
DNA per cell, in picograms, before copying and after meiosis II.

(a) Calculate the DNA per cell after the cell copies its DNA. (1 pt)

Answer: 8 pg  (tolerance ±0)

Model answer 8 pg.
Working
Write down the values in the question:
tex:\text{DNA before copying} = 4\,\text{pg}
Write down the equation:
tex:\text{DNA after copying} = 2 \times \text{DNA before copying}
Substitute the values into the equation:
tex:\text{DNA after copying} = 2 \times 4\,\text{pg} = 8\,\text{pg}
Rubric
  • Award 1 point for: 8 pg.

(b) Explain why the chromosome count stays at 12 after the cell copies its DNA. (1 pt)

Model answer The copying makes two identical sister chromatids of every chromosome.
The two sister chromatids stay joined at one centromere.
Chromosomes are counted by centromere.
So the cell still holds 12 chromosomes, each now two chromatids.
Rubric
  • Award 1 point for: the reason — each chromosome is now two sister chromatids joined at one centromere, and chromosomes are counted by centromere, so the count stays 12.

Slip Saying the cell has 24 chromosomes after the copying. Copying doubles the chromatids, not the centromeres.

(c) Calculate the number of chromatids in one cell after meiosis I. (1 pt)

Answer: 12  (tolerance ±0)

Model answer 12 chromatids: 6 chromosomes, each still two chromatids.
Working
Write down the values in the question:
tex:2n = 12
tex:n = 6
Write down the equation:
tex:\text{chromatids after meiosis I} = 2 \times n
Substitute the values into the equation:
tex:\text{chromatids after meiosis I} = 2 \times 6 = 12
Rubric
  • Award 1 point for: 12 chromatids WITH the working: 6 chromosomes after meiosis I, each of two chromatids, 6 × 2 = 12.
  • Do not award: 12 with no working.

(d) Evaluate the claim that the plant cell copied its DNA again between meiosis I and meiosis II, using the measured 2 pg. (1 pt)

Model answer After meiosis I each cell holds 4 pg, half of the 8 pg.
If the DNA had been copied again, each cell would have gone from 4 pg to 8 pg.
Meiosis II would then have halved that 8 pg, leaving 4 pg in each of the four cells.
The measured value is 2 pg, half of 4 pg.
So the claim is not supported: no copying happened between the two divisions.
Rubric
  • Award 1 point for: the judgement (the claim is not supported) AND the ground (a second copying would leave 4 pg per cell after meiosis II; the measured 2 pg is half of that).

Slip Predicting 8 pg for a second copying. A second copying would double 4 pg to 8 pg, but meiosis II then halves it again to 4 pg.

APBIO-U05-L08 Shuffled pairs

Topic 5.2 · Meiosis and Genetic Diversity · 78 steps

Left: a photograph of two young brothers sitting side by side, both smiling. Right: an oval cell at metaphase I with two homologous pairs across the middle, the dark member of each pair facing the left pole and the light member facing the right pole
Left: a photograph of two young brothers sitting side by side, both smiling. Right: an oval cell at metaphase I with two homologous pairs across the middle, the dark member of each pair facing the left pole and the light member facing the right pole

Photo: Gary Ward, U.S. Navy, via Wikimedia Commons, public domain (cropped and resized).

Here are two brothers with the same parents. Each got 23 chromosomes from their mother and 23 from their father, and they are not alike. Beside them is the model cell at metaphase I: in each pair, the mother’s chromosome is dark and the father’s is light.

Both brothers got one set from each of the same two people. Where does the difference between them come from?

Unit 5 · Heredity

1A gamete's set is a mixture

2

Video: Watch: A gamete's set is a mixture

The model cell’s two pairs, the maternal chromosome of each drawn dark and the paternal light; one gamete forming with the maternal long chromosome and the paternal short chromosome, a mixture.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08a.mp4

3

Why are two children of the same parents not alike? Every gamete carries one chromosome of each homologous pair.

4

Which member it carries, the mother’s or the father’s, is settled pair by pair.

5

At metaphase I each homologous pair lines up facing either way on its own, regardless of the other pairs.

6

With two pairs a parent makes four kinds of gamete. With 23 pairs the kinds number in the millions.

7

Each brother grew from a different pair of gametes. So each brother carries a different mixture of the same two parents’ chromosomes.

8
Check q1

Meiosis ends with four haploid cells.

How many chromosomes of every homologous pair does each of the four cells hold?

  1. A. None
    Each of the four cells holds one complete set: one chromosome of every pair.
  2. B. ✓ One
  3. C. Two
    Two chromosomes of a pair is a diploid cell’s count.
    Meiosis I sends the two members of each pair to different cells.

Why: Each of the four cells is haploid: one complete set.
One complete set is one chromosome of every homologous pair.

9

The chromosome of a pair that came from the mother is called the .

10

The chromosome of a pair that came from the father is called the paternal chromosome.

11

In every drawing in this unit, the maternal chromosome is drawn dark and the paternal chromosome is drawn light.

12

Meiosis I sends one member of every homologous pair to each cell. So each gamete receives one chromosome of every pair: the maternal one or the paternal one.

13

Which member a gamete gets is settled pair by pair. A gamete from the model cell might carry the maternal long chromosome and the paternal short chromosome.

Two finished cells side by side, each a circle with a dashed inner ring: the left holds a long dark rod and a short light rod; the right holds a long light rod and a short dark rod
Two finished cells side by side, each a circle with a dashed inner ring: the left holds a long dark rod and a short light rod; the right holds a long light rod and a short dark rod
14

The maternal chromosomes do not all travel together into one gamete. The paternal chromosomes do not all travel together into another gamete.

15

So a gamete’s set is usually a mixture of maternal and paternal chromosomes.

16

What you are expected to know State what a gamete’s set is: one chromosome of every homologous pair, the maternal or the paternal one, settled pair by pair, so usually a mixture.

17
Check q2

Three finished cells are drawn below, numbered 1, 2 and 3. The model cell that made them has one long homologous pair and one short homologous pair.

Three circles numbered 1, 2 and 3, each with a dashed inner ring. Circle 1 holds two long rods, one dark and one light. Circle 2 holds one long dark rod and one short light rod. Circle 3 holds four rods: a long dark, a long light, a short dark and a short light
Three circles numbered 1, 2 and 3, each with a dashed inner ring. Circle 1 holds two long rods, one dark and one light. Circle 2 holds one long dark rod and one short light rod. Circle 3 holds four rods: a long dark, a long light, a short dark and a short light

Which of the drawn cells could be a gamete made by the model cell?

  1. A. Cell 1
    Cell 1 holds both members of the long pair and no short chromosome.
    A gamete receives one member of every pair.
  2. B. ✓ Cell 2
  3. C. Cell 3
    Cell 3 holds all four chromosomes, both members of both pairs.
    A gamete is haploid: one member of each pair, two chromosomes.

Why: A gamete receives one chromosome of every homologous pair.
Cell 2 holds one long chromosome and one short chromosome: the maternal long and the paternal short.
So cell 2 could be a gamete made by the model cell.

18
Check q3

A student looks at gametes from a plant with three homologous pairs and says: “Every gamete carries only maternal chromosomes or only paternal chromosomes.”

Is the student correct?

  1. A. ✓ No: a gamete usually carries some maternal and some paternal chromosomes
  2. B. Yes: a gamete carries all maternal or all paternal chromosomes
    Which member of each pair a gamete receives is settled pair by pair.
    So a gamete usually carries some maternal and some paternal chromosomes.

Why: Meiosis I sends one member of every pair to each cell.
Which member goes to which cell is settled pair by pair.
So a gamete usually carries a mixture of maternal and paternal chromosomes.

19
Practice writing an answer

A plant’s body cells hold six chromosomes: a long pair, a medium pair and a short pair. One of its gametes carries the maternal long chromosome, the paternal medium chromosome and the maternal short chromosome.

(a) Explain how this gamete’s set shows that which member of a pair a gamete receives is settled pair by pair. (1 pt)

Frame The set shows this because …

Model answer The set shows this because the gamete holds one chromosome of every pair.
The long pair sent its maternal member to this gamete.
The medium pair sent its paternal member to the same gamete.
So the long pair’s choice did not fix the medium pair’s choice.
Therefore which member a gamete receives is settled pair by pair.
Rubric
  • Award 1 point for: one pair sent its maternal member and another pair sent its paternal member to the same gamete, so one pair’s choice does not fix another pair’s (settled pair by pair).

20Quick quiz: maternal chromosome, paternal chromosome mixed practice

21
Check q4

A homologous pair is one chromosome from each parent.

Which chromosome of the pair is the maternal chromosome?

  1. A. ✓ The member that came from the mother
  2. B. The member that came from the father
    The chromosome that came from the father is the paternal chromosome.

Why: Maternal means from the mother.
The maternal chromosome of a pair is the member that came from the mother.

22
Check q5

A homologous pair is one chromosome from each parent.

Which chromosome of the pair is the paternal chromosome?

  1. A. The member that came from the mother
    The chromosome that came from the mother is the maternal chromosome.
  2. B. ✓ The member that came from the father

Why: Paternal means from the father.
The paternal chromosome of a pair is the member that came from the father.

23
Practice writing an answer

A homologous pair is one chromosome from each parent.

(a) State what the maternal chromosome and the paternal chromosome of a homologous pair are. (1 pt)

Model answer The maternal chromosome is the member of the pair that came from the mother, and the paternal chromosome is the member that came from the father.
Rubric
  • Award 1 point for: the maternal chromosome came from the mother and the paternal chromosome came from the father.
24
Check q6

In the drawing below, dark marks the chromosome from the mother and light marks the chromosome from the father. An arrow marks one chromosome of a homologous pair.

Two rods of the same length side by side, the left one dark and the right one light, each with a white centromere dot; an arrow from above points at the left rod
Two rods of the same length side by side, the left one dark and the right one light, each with a white centromere dot; an arrow from above points at the left rod

Which is the arrowed chromosome?

  1. A. ✓ The maternal chromosome
  2. B. The paternal chromosome
    The arrowed chromosome is the dark one.
    In the drawing the dark chromosome is the maternal chromosome.

Why: The arrowed chromosome is dark.
In the drawing the dark chromosome is the one from the mother.
So the arrowed chromosome is the maternal chromosome.

25
Check q7

In the drawing below, dark marks the chromosome from the mother and light marks the chromosome from the father. An arrow marks one chromosome of a homologous pair.

Two rods of the same length side by side, the left one dark and the right one light, each with a white centromere dot; an arrow from above points at the right rod
Two rods of the same length side by side, the left one dark and the right one light, each with a white centromere dot; an arrow from above points at the right rod

Which is the arrowed chromosome?

  1. A. The maternal chromosome
    The arrowed chromosome is the light one.
    In the drawing the light chromosome is the paternal chromosome.
  2. B. ✓ The paternal chromosome

Why: The arrowed chromosome is light.
In the drawing the light chromosome is the one from the father.
So the arrowed chromosome is the paternal chromosome.

26
Check q8

In the drawing below, dark marks the chromosome from the mother and light marks the chromosome from the father. An arrow marks one chromosome of a gamete.

A circle with a dashed inner ring holding a long dark rod and a short light rod; an arrow from above points at the short rod
A circle with a dashed inner ring holding a long dark rod and a short light rod; an arrow from above points at the short rod

Which is the arrowed chromosome?

  1. A. The maternal chromosome of the short pair
    The arrowed chromosome is the light one.
    In the drawing the light chromosome is the paternal chromosome.
  2. B. ✓ The paternal chromosome of the short pair

Why: The arrowed chromosome is the short light rod.
In the drawing the light chromosome is the one from the father.
So the gamete carries the paternal chromosome of the short pair.

27
Check q9

In the drawing below, dark marks the chromosome from the mother and light marks the chromosome from the father. An arrow marks one chromosome of a gamete.

A circle with a dashed inner ring holding a long light rod and a short dark rod; an arrow from above points at the short rod
A circle with a dashed inner ring holding a long light rod and a short dark rod; an arrow from above points at the short rod

Which is the arrowed chromosome?

  1. A. ✓ The maternal chromosome of the short pair
  2. B. The paternal chromosome of the short pair
    The arrowed chromosome is the dark one.
    In the drawing the dark chromosome is the maternal chromosome.

Why: The arrowed chromosome is the short dark rod.
In the drawing the dark chromosome is the one from the mother.
So the gamete carries the maternal chromosome of the short pair.

28
Check q10

In the drawing below, dark marks the chromosome from the mother and light marks the chromosome from the father. An arrow marks one chromosome of a short homologous pair.

Two short rods of the same length side by side, the left one light and the right one dark, each with a white centromere dot; an arrow from above points at the right rod
Two short rods of the same length side by side, the left one light and the right one dark, each with a white centromere dot; an arrow from above points at the right rod

Which is the arrowed chromosome?

  1. A. ✓ The maternal chromosome
  2. B. The paternal chromosome
    The arrowed chromosome is the dark one.
    In the drawing the dark chromosome is the maternal chromosome, whichever side it stands on.

Why: The arrowed chromosome is dark.
In the drawing the dark chromosome is the one from the mother.
So the arrowed chromosome is the maternal chromosome.

29Each pair lines up on its own

30

Video: Watch: Each pair lines up on its own

The two pairs at the equator of the model cell; the short pair turning round while the long pair stays; the two orientations side by side, each as likely as the other.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08b.mp4

31
Check q11

A cell with two homologous pairs is at metaphase I.

How do the two members of each homologous pair sit at the equator?

  1. A. One behind the other, both facing the same pole
    Fibers from one pole attach to one member’s centromere.
    Fibers from the other pole attach to the other’s.
    So the two members face opposite poles.
  2. B. ✓ Side by side, facing opposite poles

Why: At metaphase I the two members of each homologous pair sit side by side at the equator and face opposite poles.

32

Suppose the model cell is at metaphase I. The long pair sits at the equator with its maternal chromosome facing the left pole and its paternal chromosome facing the right pole.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X faces the left pole and the light X the right; in the short pair the dark X faces the left pole and the light X the right; labels name a pole, a centromere and a spindle fiber
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X faces the left pole and the light X the right; in the short pair the dark X faces the left pole and the light X the right; labels name a pole, a centromere and a spindle fiber
33

Which way round the short pair sits is not fixed. Its maternal chromosome may face the left pole, or it may face the right pole.

34

Here are both drawn side by side: the short pair facing one way, and facing the other way. In the left cell the maternal short chromosome faces the left pole.

Two cells at metaphase I side by side. In the left cell the dark member of both pairs faces the left pole. In the right cell the long pair is the same but the short pair is turned round, its light member facing the left pole and its dark member the right
Two cells at metaphase I side by side. In the left cell the dark member of both pairs faces the left pole. In the right cell the long pair is the same but the short pair is turned round, its light member facing the left pole and its dark member the right
35

In the right cell the maternal short chromosome faces the right pole. In a parent’s many dividing cells, about half have the short pair facing each way.

36

Each homologous pair lines up at metaphase I facing either way regardless of the other pairs, so which member of each pair a gamete receives is decided pair by pair.

37

When each homologous pair faces either way on its own like this, it is called : each pair’s orientation is independent of the others. The exam calls the same thing the random assortment of chromosomes.

38

Which way one pair faces has no effect on the next pair. With the long pair facing its maternal chromosome left, the short pair is as likely to face either way.

39

Anaphase I then pulls each chromosome to the pole it faces. So the way the pairs faced decides which chromosomes each new cell receives.

40

What you are expected to know Explain independent orientation: each homologous pair faces either way at metaphase I on its own, so which member of each pair a gamete receives is decided pair by pair.

41
Check q12

Suppose a cell with two homologous pairs is at metaphase I. The long pair faces its maternal chromosome toward the left pole.

Which way can the short pair face its maternal chromosome?

  1. A. Toward the left pole only
    Which way the long pair faces has no effect on the short pair.
    The short pair faces either way on its own.
  2. B. Toward the right pole only
    Nothing turns the short pair the opposite way to the long pair.
    The short pair faces either way on its own.
  3. C. ✓ Toward either pole

Why: Each homologous pair lines up at metaphase I on its own.
The long pair’s orientation does not set the short pair’s orientation.
So the short pair can face its maternal chromosome toward either pole.

42
Practice writing an answer

In a cricket cell at metaphase I, the long pair faces its maternal chromosome toward the left pole.

(a) Explain why the short pair is as likely to face its maternal chromosome toward either pole. (1 pt)

Frame The short pair is as likely to face either way because …

Model answer The short pair is as likely to face either way because each homologous pair lines up at metaphase I on its own.
Nothing links the short pair to the long pair.
So the long pair’s orientation does not set the short pair’s orientation.
Therefore the short pair faces its maternal chromosome left in about half of cells and right in the other half.
Rubric
  • Award 1 point for: each pair orients independently of the others at metaphase I, so the long pair’s orientation does not fix the short pair’s.
43
Check q13

In a grasshopper cell at metaphase I, the long pair faces its maternal chromosome toward the left pole. A student says: “So the short pair must face its maternal chromosome toward the left pole too.”

Is the student correct?

  1. A. ✓ No: the short pair is as likely to face either pole
  2. B. Yes: the maternal chromosomes all face the same pole
    Each pair lines up facing either way regardless of the other pairs.
    So the short pair is as likely to face either pole.

Why: Which way one pair faces has no effect on the next pair.
So the long pair facing left does not turn the short pair left.
The short pair is as likely to face either pole.

44Quick quiz: independent orientation mixed practice

45
Check q14

A cell with several homologous pairs is at metaphase I.

What is independent orientation?

  1. A. The two members of a homologous pair face opposite poles
    The two members of a pair facing opposite poles is true of every pair at metaphase I.
    Independent orientation is about how one pair faces relative to the other pairs.
  2. B. ✓ Each homologous pair faces either way regardless of the other pairs
  3. C. Each chromosome lines up on its own at metaphase II
    At metaphase II the cell has no homologous pairs.
    Independent orientation is about homologous pairs at metaphase I.

Why: Independent orientation is each homologous pair facing either way at metaphase I regardless of the other pairs.

46
Check q15

The exam has its own name for independent orientation.

Which of the following is the exam’s name for it?

  1. A. ✓ The random assortment of chromosomes
  2. B. The pairing of homologs in synapsis
    Synapsis is the two homologs of a pair lying together along their whole length in prophase I, before metaphase I.
  3. C. The splitting of the cell in cytokinesis
    Cytokinesis is the cell splitting in two after a division.

Why: The exam calls independent orientation the random assortment of chromosomes.

47
Practice writing an answer

A cell with several homologous pairs is at metaphase I.

(a) State what independent orientation is. (1 pt)

Model answer Independent orientation is each homologous pair lining up at metaphase I facing either way regardless of the other pairs, so which member of each pair a gamete receives is decided pair by pair.
Rubric
  • Award 1 point for: each homologous pair faces either way at metaphase I regardless of (independently of) the other pairs.
48
Check q16

A cell with three homologous pairs is at metaphase I. The long pair faces its maternal chromosome toward the left pole.

Does the long pair’s orientation fix which way the medium pair faces?

  1. A. Yes
    Each pair faces either way regardless of the other pairs.
    The long pair’s orientation has no effect on the medium pair.
  2. B. ✓ No

Why: Each homologous pair lines up on its own.
So the long pair’s orientation does not fix the medium pair’s orientation.

49
Check q17

A cell with three homologous pairs is at metaphase I. The medium pair faces its paternal chromosome toward the left pole.

Does the medium pair’s orientation fix which way the short pair faces?

  1. A. Yes
    Which way one pair faces has no effect on the next pair.
    So the short pair is as likely to face either way.
  2. B. ✓ No

Why: Each homologous pair lines up on its own.
The medium pair’s orientation has no effect on the short pair.
So the short pair is as likely to face its paternal chromosome toward either pole.

50
Check q18

A gamete from a cell with three homologous pairs carries one chromosome of each pair.

Did each pair decide on its own which member it sent to this gamete?

  1. A. ✓ Yes
  2. B. No
    Each pair faced either way at metaphase I on its own.
    So which member each pair sent to the gamete was decided pair by pair.

Why: Each homologous pair faced either way at metaphase I regardless of the other pairs.
So which member of each pair the gamete received was decided pair by pair.

51Four kinds of gamete from two pairs

52

Video: Watch: Four kinds of gamete from two pairs

Each orientation of the model cell pulled apart at anaphase I; the four kinds of gamete appearing below, two from each orientation.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08c.mp4

53

Here are the two cells again, the short pair facing one way and then the other. Take the left cell first: both maternal chromosomes face the left pole.

Two cells at metaphase I side by side. In the left cell the dark member of both pairs faces the left pole. In the right cell the long pair is the same but the short pair is turned round, its light member facing the left pole and its dark member the right
Two cells at metaphase I side by side. In the left cell the dark member of both pairs faces the left pole. In the right cell the long pair is the same but the short pair is turned round, its light member facing the left pole and its dark member the right
54

In the left cell, the left pole receives the maternal long chromosome and the maternal short chromosome. The right pole receives the paternal long chromosome and the paternal short chromosome.

55

In the right cell, the left pole receives the maternal long chromosome and the paternal short chromosome. The right pole receives the paternal long chromosome and the maternal short chromosome.

56

Each orientation gives two kinds of gamete. So two pairs give four kinds of gamete.

Four finished cells in a row, each a circle with a dashed inner ring holding one long and one short rod: dark long with dark short; light long with light short; dark long with light short; light long with dark short
Four finished cells in a row, each a circle with a dashed inner ring holding one long and one short rod: dark long with dark short; light long with light short; dark long with light short; light long with dark short
57

Here are the four kinds as a list:

  1. maternal long with maternal short
  2. paternal long with paternal short
  3. maternal long with paternal short
  4. paternal long with maternal short

58

One cell, once its pairs have lined up, gives only two of the four kinds of gamete. The other two kinds come from cells whose short pair faced the other way.

59

In a parent’s many dividing cells, about half have the short pair facing each way. So a parent makes all four kinds of gamete, in about equal numbers.

60

What you are expected to know List the four kinds of gamete two homologous pairs can give: maternal with maternal, paternal with paternal, and the two mixed sets.

61
Check q19

The cell drawn below is at metaphase I with one long pair and one short pair. The maternal chromosome of each pair is dark and the paternal chromosome is light. Meiosis then finishes normally.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the dark X is on the left and the light X on the right
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the dark X is on the left and the light X on the right

Can this cell make a gamete carrying the maternal long chromosome and the maternal short chromosome?

  1. A. ✓ Yes
  2. B. No
    In this cell the maternal long and the maternal short chromosome both face the left pole.
    So the left pole receives both, and a gamete carries both.

Why: Each pole receives the chromosome of each pair facing it.
Here the maternal long and the maternal short chromosome both face the left pole.
So a gamete from this cell can carry the maternal long with the maternal short.

62
Check q20

The cell drawn below is at metaphase I with one long pair and one short pair. The maternal chromosome of each pair is dark and the paternal chromosome is light. Meiosis then finishes normally.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the dark X is on the left and the light X on the right
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the dark X is on the left and the light X on the right

Can this cell make a gamete carrying the maternal long chromosome and the paternal short chromosome?

  1. A. Yes
    In this cell the maternal short chromosome faces the same pole as the maternal long chromosome.
    So the paternal short chromosome goes to the other pole, with the paternal long.
  2. B. ✓ No

Why: Each pole receives the chromosome of each pair facing it.
Here the maternal long and the maternal short chromosome both face the left pole.
So no gamete from this cell carries the maternal long with the paternal short.

63
Check q21

The cell drawn below is at metaphase I with one long pair and one short pair. The maternal chromosome of each pair is dark and the paternal chromosome is light. Meiosis then finishes normally.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the light X is on the left and the dark X on the right
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the light X is on the left and the dark X on the right

Can this cell make a gamete carrying the maternal long chromosome and the paternal short chromosome?

  1. A. ✓ Yes
  2. B. No
    In this cell the maternal long chromosome faces the left pole and so does the paternal short chromosome.
    So the left pole receives both, and a gamete carries both.

Why: Each pole receives the chromosome of each pair facing it.
Here the maternal long and the paternal short chromosome both face the left pole.
So a gamete from this cell can carry the maternal long with the paternal short.

64
Check q22

The cell drawn below is at metaphase I with one long pair and one short pair. The maternal chromosome of each pair is dark and the paternal chromosome is light. Meiosis then finishes normally.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the light X is on the left and the dark X on the right
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the light X is on the left and the dark X on the right

Can this cell make a gamete carrying the maternal long chromosome and the maternal short chromosome?

  1. A. Yes
    In this cell the maternal short chromosome faces the right pole, away from the maternal long chromosome.
    So the two maternal chromosomes go to different cells.
  2. B. ✓ No

Why: Each pole receives the chromosome of each pair facing it.
Here the maternal long chromosome faces the left pole and the maternal short chromosome faces the right pole.
So no gamete from this cell carries both maternal chromosomes.

65
Check q23

The cell drawn below is at metaphase I with one long pair and one short pair. The maternal chromosome of each pair is dark and the paternal chromosome is light. Meiosis then finishes normally.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the light X is on the left and the dark X on the right
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the light X is on the left and the dark X on the right

Can this cell make a gamete carrying the maternal long chromosome and the paternal long chromosome?

  1. A. Yes
    The two members of a homologous pair face opposite poles, so they go to different cells.
    No gamete carries both long chromosomes.
  2. B. ✓ No

Why: The maternal long and the paternal long chromosome are the two members of one homologous pair.
The two members of a pair face opposite poles and go to different cells.
So no gamete carries both.

66
Check q24

A cell with one long pair and one short pair is drawn below at metaphase I; in each pair the maternal chromosome is dark and the paternal chromosome is light. Meiosis then finishes normally.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the dark X is on the left and the light X on the right
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the dark X is on the left and the light X on the right

Which of the following two kinds of gamete does this one cell produce?

  1. A. ✓ Maternal long with maternal short, and paternal long with paternal short
  2. B. Maternal long with paternal long, and maternal short with paternal short
    The two members of a pair go to opposite poles.
    So no gamete carries both longs or both shorts.
  3. C. Maternal long with paternal short, and paternal long with maternal short
    In this cell the dark, maternal chromosome of the short pair faces the same pole as the dark, maternal long chromosome.
  4. D. All four kinds: every combination of one long and one short
    One cell, once its pairs have lined up, gives only two kinds.
    The other two kinds come from cells whose short pair faced the other way.

Why: Each pole receives the chromosome of each pair facing it.
Here the left pole gets both maternal chromosomes.
The right pole gets both paternal chromosomes.
So this cell’s four gametes are of those two kinds.

67

Back to the two brothers with the same parents, and the model cell at metaphase I with the mother’s chromosome of each pair dark and the father’s light.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X faces the left pole and the light X the right; in the short pair the dark X faces the left pole and the light X the right; labels name a pole, a centromere and a spindle fiber
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X faces the left pole and the light X the right; in the short pair the dark X faces the left pole and the light X the right; labels name a pole, a centromere and a spindle fiber
68

In each parent, every pair faced one way or the other on its own. With two pairs that is four kinds of gamete.

69

Each brother grew from a different pair of gametes. So each brother carries a different mixture of the same two parents’ chromosomes.

70Mixed practice mixed practice

71
Check q25

A cell with one long pair and one short pair is drawn below at metaphase I; in each pair the maternal chromosome is dark and the paternal chromosome is light. Meiosis then finishes normally.

An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the light X is on the left and the dark X on the right
An oval cell with a pole at each end and two homologous pairs across the middle in a double row: in the long pair the dark X is on the left and the light X on the right; in the short pair the light X is on the left and the dark X on the right

Which of the following two kinds of gamete does this one cell produce?

  1. A. Maternal long with maternal short, and paternal long with paternal short
    In this cell the short pair faces the other way: its light, paternal chromosome faces the same pole as the dark, maternal long chromosome.
  2. B. ✓ Maternal long with paternal short, and paternal long with maternal short
  3. C. Maternal long with paternal long, and maternal short with paternal short
    The two members of a pair go to opposite poles.
    So no gamete carries both longs or both shorts.
  4. D. All four kinds: every combination of one long and one short
    One cell, once its pairs have lined up, gives only two kinds of gamete.
    The other two kinds come from cells whose short pair faced the other way.

Why: Each pole receives the chromosome of each pair facing it.
Here the left pole gets the maternal long and the paternal short.
The right pole gets the paternal long and the maternal short.
So this cell’s four gametes are of those two kinds.

72
Check q26

A plant’s body cells hold six chromosomes: a long pair, a medium pair and a short pair, one member of each pair from each parent.

Which of the following sets could one of the plant’s gametes carry?

  1. A. The maternal long and the paternal long only
    The two long chromosomes are the two members of one pair, so a gamete gets only one of them.
    A gamete also needs one medium and one short chromosome.
  2. B. Both members of the long pair and the maternal short
    A gamete receives one member of every pair, never both.
    A gamete also needs one member of the medium pair.
  3. C. ✓ The paternal long, the maternal medium and the paternal short
  4. D. All six chromosomes, three maternal and three paternal
    Six chromosomes is the count in a body cell.
    A gamete is haploid: one of each pair, three chromosomes.

Why: A gamete carries one chromosome of every pair.
Which member it carries, maternal or paternal, is settled pair by pair.
So one of each of the three pairs, mixed, is a possible set.

73
Check q27

A cell with three homologous pairs is at metaphase I. The long pair faces its maternal chromosome toward the left pole, and the medium pair faces its maternal chromosome toward the right pole.

Which way does the short pair face its maternal chromosome?

  1. A. Toward the left pole
    The long pair’s orientation has no effect on the short pair.
    Each pair faces either way on its own.
  2. B. Toward the right pole
    The medium pair’s orientation has no effect on the short pair.
    Each pair faces either way on its own.
  3. C. ✓ Toward either pole

Why: Each homologous pair lines up at metaphase I regardless of the other pairs.
Neither the long pair nor the medium pair sets the short pair’s orientation.
So the short pair faces either pole, as likely one as the other.

74
Check q28

A gamete from a plant with three homologous pairs carries the paternal long, the paternal medium and the paternal short chromosome.

Is this set of chromosomes possible?

  1. A. ✓ Yes
  2. B. No
    A gamete receives one member of every pair, and each pair sends its member on its own.
    All three pairs may send their paternal member to the same gamete.

Why: Which member each pair sends to a gamete is settled pair by pair.
So all three pairs may send their paternal member to the same gamete.
The all-paternal set is one of the possible sets.

75
Check q29

A parent’s cells have two homologous pairs. Each pair faces either way at metaphase I on its own.

How many kinds of gamete can the parent make in all?

  1. A. Two
    Two kinds come from one orientation of the short pair.
    The other orientation gives two more kinds.
  2. B. ✓ Four
  3. C. Eight
    Each of the two pairs faces either way: two choices twice.
    That gives four kinds of gamete, not eight.

Why: Two pairs, each facing either way, give four kinds of gamete: maternal with maternal, paternal with paternal, and the two mixed sets.

76
Check q30

A student says: “At metaphase I all the maternal chromosomes face the same pole, so a gamete gets all of them together.”

Is the student correct?

  1. A. ✓ No: each homologous pair faces either way on its own
  2. B. Yes: the maternal chromosomes face one pole together
    Each homologous pair faces either way at metaphase I regardless of the other pairs.
    So the maternal chromosomes do not all face the same pole.

Why: Each homologous pair lines up at metaphase I on its own.
So some pairs face their maternal chromosome toward one pole and some toward the other.
A gamete usually receives a mixture of maternal and paternal chromosomes.

77
Practice writing an answer

Two brothers have the same parents. Each brother got 23 chromosomes from their mother and 23 from their father, yet the two brothers are not alike.

(a) Explain how independent orientation makes the two brothers’ sets of chromosomes differ. (2 pt)

Frame The brothers’ sets differ because …

Model answer The brothers’ sets differ because each brother grew from one gamete of the mother and one gamete of the father.
In each parent, every homologous pair faced either way at metaphase I on its own.
So each gamete carried a different mixture of that parent’s chromosomes.
The two gametes that made one brother differed from the two that made the other.
So the brothers carry different mixtures of the same two parents’ chromosomes.
Rubric
  • Award 1 point for: in each parent, each homologous pair faced either way at metaphase I independently of the other pairs, so different gametes carry different mixtures of that parent’s chromosomes.
  • Award 1 point for: each brother grew from a different pair of gametes, so the two brothers received different mixtures of the same two parents’ chromosomes.

Glossary

maternal chromosome and paternal chromosome
In a homologous pair, the maternal chromosome came from the mother and the paternal chromosome came from the father. A gamete receives one of the two, settled pair by pair. In this unit's drawings the maternal chromosome is dark and the paternal chromosome is light.
independent orientation (random assortment of chromosomes)
Each homologous pair lines up at metaphase I facing either way regardless of the other pairs, so which member of each pair a gamete receives is decided pair by pair. The exam calls it the random assortment of chromosomes.

APBIO-U05-L08B How many kinds of gamete?

Topic 5.2 · Meiosis and Genetic Diversity · 45 steps

Four gametes in a row, each a circle holding one long and one short rod, in the four combinations of dark and light: dark long with dark short, light long with light short, dark long with light short, light long with dark short; to the right the words 23 pairs and a question mark
Four gametes in a row, each a circle holding one long and one short rod, in the four combinations of dark and light: dark long with dark short, light long with light short, dark long with light short, light long with dark short; to the right the words 23 pairs and a question mark

Here are the four kinds of gamete the model cell can make from its two pairs, each pair facing either way on its own at metaphase I.

A human cell has 23 pairs. How many kinds of gamete can one person make this way?

Unit 5 · Heredity

1Every pair doubles the count

2

Video: Watch: Every pair doubles the count

The doubling tree growing one pair at a time, 2, 4, 8, 16, with the count written beside each level, and the leap to 23 pairs.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08Ba.mp4

3

How many kinds of gamete can one parent make by independent orientation alone? Every pair doubles the count.

4

One pair gives 2 kinds, two pairs give 4 kinds, three pairs give 8 kinds.

5

So the number of kinds is 2 multiplied by itself once for every pair.

6

A person has 23 pairs. So one person makes millions of kinds of gamete.

7

Other events in meiosis add more kinds. So this count is the count from independent orientation alone.

8

Reading the rule backwards, a species whose gametes come in 16 kinds by independent orientation alone has 4 pairs.

9
Check q1

A parent’s cells have two homologous pairs, a long pair and a short pair. Each pair faces either way at metaphase I on its own.

How many kinds of gamete can the parent make in all?

  1. A. Eight
    Each of the two pairs faces either way: two choices, twice.
    That gives four kinds, not eight.
  2. B. ✓ Four
  3. C. Three
    The two mixed sets are two different kinds: maternal long with paternal short, and paternal long with maternal short.
    With the two unmixed sets, that is four kinds.

Why: Two pairs, each facing either way, give four kinds of gamete: maternal with maternal, paternal with paternal, and the two mixed sets.

10

One pair facing either way gives two kinds of gamete: one carrying the maternal chromosome, one carrying the paternal chromosome.

Two finished cells side by side, each a circle with a dashed inner ring: the left holds one long dark rod, the right one long light rod
Two finished cells side by side, each a circle with a dashed inner ring: the left holds one long dark rod, the right one long light rod
11

Add a second pair. Each of the two kinds of gamete can go with either member of the second pair.

12

So the kinds double, from two to four.

13

Every pair you add faces either way, whichever way the other pairs face. So every pair you add doubles the count again.

A branching tree read left to right: one circle splits into two for the first pair, each of those into two for the second pair making four, each of those into two for the third pair making eight, and each of those into two for the fourth pair making sixteen; the captions under the five columns read start 1, one pair 2, two pairs 4, three pairs 8, four pairs 16
A branching tree read left to right: one circle splits into two for the first pair, each of those into two for the second pair making four, each of those into two for the third pair making eight, and each of those into two for the fourth pair making sixteen; the captions under the five columns read start 1, one pair 2, two pairs 4, three pairs 8, four pairs 16
14

Three pairs give eight kinds of gamete, and four pairs give sixteen kinds of gamete.

15

So for n pairs the number of kinds is 2 multiplied by itself n times, written 2n.

16

The kinds multiply across pairs. They never add: three pairs give 8 kinds, not 6.

17
Worked example

A human cell has 23 pairs of chromosomes. How many kinds of gamete can one person make by independent orientation alone?

Write down the values in the question:
n=23
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=223=8,388,608
That is about 8.4 million kinds of gamete from one person.
18

Exam questions that ask for 2n mean this count, from how the pairs face at metaphase I.

19

Other events in meiosis add still more kinds. So 2n is the smallest the number can be.

20

What you are expected to know Calculate the number of kinds of gamete independent orientation gives for a stated number of pairs, 2n.

21
Check q2 numeric entry

A fruit fly’s body cells hold 8 chromosomes, in 4 pairs.

Calculate the number of kinds of gamete a fruit fly can make by independent orientation alone.

Answer: 16  (tolerance ±0)

Working
Write down the values in the question:
n=4
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=24=16
22
Check q3 numeric entry

A mosquito’s body cells hold 6 chromosomes, in 3 pairs.

Calculate the number of kinds of gamete a mosquito can make by independent orientation alone.

Answer: 8  (tolerance ±0)

Working
Write down the values in the question:
n=3
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=23=8

23Quick quiz: how many kinds of gamete? mixed practice

24
Check q4 numeric entry

An animal has 1 pair of chromosomes.

Calculate the number of kinds of gamete it can make by independent orientation alone.

Answer: 2  (tolerance ±0)

Working
Write down the values in the question:
n=1
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=21=2
25
Check q5 numeric entry

An insect has 2 pairs of chromosomes.

Calculate the number of kinds of gamete it can make by independent orientation alone.

Answer: 4  (tolerance ±0)

Working
Write down the values in the question:
n=2
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=22=4
26
Check q6 numeric entry

An insect has 6 pairs of chromosomes.

Calculate the number of kinds of gamete it can make by independent orientation alone.

Answer: 64  (tolerance ±0)

Working
Write down the values in the question:
n=6
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=26=64
27
Check q7 numeric entry

A pea plant’s body cells hold 14 chromosomes.

Calculate the number of kinds of gamete a pea plant can make by independent orientation alone.

Answer: 128  (tolerance ±0)

Working
Write down the values in the question:
2n=14
n=7
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=27=128
28
Check q8 numeric entry

An animal’s body cells hold 18 chromosomes.

Calculate the number of kinds of gamete the animal can make by independent orientation alone.

Answer: 512  (tolerance ±0)

Working
Write down the values in the question:
2n=18
n=9
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=29=512

29From the kinds back to the pairs

30

Video: Watch: From the kinds back to the pairs

The doubling tree read from right to left: 16 kinds of gamete traced back through 8, 4 and 2 to the one cell, one doubling undone per pair, so 16 kinds means 4 pairs.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L08Bb.mp4

31

Now consider the rule read backwards. Suppose a plant makes 16 kinds of gamete by independent orientation alone.

32

Here is the doubling tree again. Every pair doubled the count once: 1, then 2, then 4, then 8, then 16.

A branching tree read left to right: one circle splits into two for the first pair, each of those into two for the second pair making four, each of those into two for the third pair making eight, and each of those into two for the fourth pair making sixteen; the captions under the five columns read start 1, one pair 2, two pairs 4, three pairs 8, four pairs 16
A branching tree read left to right: one circle splits into two for the first pair, each of those into two for the second pair making four, each of those into two for the third pair making eight, and each of those into two for the fourth pair making sixteen; the captions under the five columns read start 1, one pair 2, two pairs 4, three pairs 8, four pairs 16
33

Four doublings reach 16. So the plant has 4 pairs of chromosomes.

34

To find the number of pairs, count how many times 2 is multiplied by itself to reach the number of kinds. That count is n, the number of pairs.

35
Worked example

A plant makes 16 kinds of gamete by independent orientation alone. How many pairs of chromosomes does the plant have?

Write down the values in the question:
kinds of gamete=16
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
2n=16=2×2×2×2=24
n=4 pairs
36

What you are expected to know Calculate the number of homologous pairs from the number of gamete kinds independent orientation gives, by counting how many times 2 is multiplied by itself to reach that number.

37
Check q9 numeric entry

An animal can make 256 kinds of gamete by independent orientation alone.

Calculate the number of pairs of chromosomes the animal has.

Answer: 8  (tolerance ±0)

Working
Write down the values in the question:
kinds of gamete=256
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
2n=256=28
n=8
38
Check q10 numeric entry

A plant can make 64 kinds of gamete by independent orientation alone.

Calculate the number of pairs of chromosomes the plant has.

Answer: 6  (tolerance ±0)

Working
Write down the values in the question:
kinds of gamete=64
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
2n=64=26
n=6
39

Back to the four kinds of gamete the model cell makes from its two pairs, each pair facing either way on its own at metaphase I.

Four finished cells in a row, each a circle with a dashed inner ring holding one long and one short rod: dark long with dark short; light long with light short; dark long with light short; light long with dark short
Four finished cells in a row, each a circle with a dashed inner ring holding one long and one short rod: dark long with dark short; light long with light short; dark long with light short; light long with dark short
40

Two pairs give 22=4 kinds. One person’s 23 pairs, each facing either way on its own, give 223 kinds of gamete, about 8.4 million.

41Mixed practice mixed practice

42
Practice writing an answer

An insect has 3 pairs of chromosomes, so it makes 8 kinds of gamete by independent orientation alone.

(a) Explain why each extra pair doubles the number of kinds of gamete. (1 pt)

Frame Each extra pair doubles the number of kinds because …

Model answer Each extra pair doubles the number of kinds because every pair faces either way at metaphase I on its own.
Take any one arrangement of the pairs already counted.
The new pair can face one way or the other way with that arrangement.
So every kind of gamete already counted becomes two kinds.
Therefore the number of kinds doubles with each pair added: two for one pair, four for two pairs, eight for three pairs.
Rubric
  • Award 1 point for: each pair faces either way independently of the others, so every existing kind of gamete becomes two kinds and the count doubles.
43
Check q11 numeric entry

A plant can make 32 kinds of gamete by independent orientation alone.

Calculate the number of pairs of chromosomes the plant has.

Answer: 5  (tolerance ±0)

Working
Write down the values in the question:
kinds of gamete=32
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
2n=32=2×2×2×2×2=25
n=5
44
Check q12 numeric entry

An animal’s body cells hold 20 chromosomes.

Calculate the number of kinds of gamete the animal can make by independent orientation alone.

Answer: 1024  (tolerance ±0)

Working
Write down the values in the question:
2n=20
n=10
Write down the equation:
kinds of gamete=2n
Substitute the values into the equation:
kinds of gamete=210=1,024

APBIO-U05-L09 Swapped segments

Topic 5.2 · Meiosis and Genetic Diversity · 59 steps

Left: a dark X and a light X lying paired with their upper inner arms crossing inside a small dashed ring, labelled before. Right: the same two X's drawn apart, the upper inner tip of the dark X now light and the upper inner tip of the light X now dark, labelled after
Left: a dark X and a light X lying paired with their upper inner arms crossing inside a small dashed ring, labelled before. Right: the same two X's drawn apart, the upper inner tip of the dark X now light and the upper inner tip of the light X now dark, labelled after

Back in prophase I, here is the long pair of the model cell lying paired, the maternal chromosome dark and the paternal chromosome light, held at a crossing point. Beside it is the same pair after the two homologs have parted: one dark chromatid now ends in a light tip, and one light chromatid ends in a dark tip.

What happened at the crossing point?

Unit 5 · Heredity

1One chromatid from each homolog

2

Video: Watch: One chromatid from each homolog

A close view of the paired long pair held at a chiasma. The four chromatids are picked out one at a time: the two sister chromatids of the dark homolog, then one dark chromatid beside one light chromatid. A swap between the two sisters is shown changing nothing.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L09a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L09a.mp4

3

What happens at a chiasma? Each homolog of the paired pair is two sister chromatids.

4

At the crossing point, one chromatid of the maternal homolog and one chromatid of the paternal homolog break at the same place and exchange the pieces.

5

Sister chromatids are identical copies. A swap between two sister chromatids would change nothing.

6

A swap between a maternal chromatid and a paternal chromatid moves a piece of the mother’s chromosome onto the father’s. It moves a piece of the father’s chromosome onto the mother’s.

7

That exchange is where the dark chromatid with the light tip came from.

8
Check q1

A chromosome was copied before the cell began to divide.

What are its two sister chromatids?

  1. A. ✓ Two identical copies of one chromosome, joined at the centromere
  2. B. The two homologs of a homologous pair, one from each parent
    The two homologs of a pair came from two parents and may carry different alleles.
    Sister chromatids are the two identical copies of one chromosome, joined at the centromere.
  3. C. Two chromosomes of different lengths that carry different genes
    Chromosomes of different lengths carry different genes and are not copies.
    Sister chromatids are the two identical copies of one chromosome, joined at the centromere.

Why: Before division the cell copied every chromosome.
The two copies stay joined at the centromere.
The two joined copies are called sister chromatids, and they carry the same alleles at every position.

9

Look closely at the long homologous pair lying paired in prophase I, held at a chiasma.

A close view of the paired long chromosomes: a dark X and a light X lying together with their arms crossing; a dashed ring marks the point where an upper arm of the dark X crosses an upper arm of the light X
A close view of the paired long chromosomes: a dark X and a light X lying together with their arms crossing; a dashed ring marks the point where an upper arm of the dark X crosses an upper arm of the light X
10

Each homolog was copied before meiosis began. So each homolog is two sister chromatids joined at one centromere.

11

The paired pair therefore holds four chromatids: two on the maternal homolog and two on the paternal homolog.

The same paired pair with its four chromatids numbered at their outer tips: 1 at the upper-left tip and 2 at the lower-left tip of the dark X, 3 at the lower-right tip and 4 at the upper-right tip of the light X; the upper inner arms cross inside the dashed ring
The same paired pair with its four chromatids numbered at their outer tips: 1 at the upper-left tip and 2 at the lower-left tip of the dark X, 3 at the lower-right tip and 4 at the upper-right tip of the light X; the upper inner arms cross inside the dashed ring
12

The four chromatids are numbered 1 to 4 in the drawing.

13

Chromatids 1 and 2 are the maternal homolog’s two chromatids. Chromatids 3 and 4 are the paternal homolog’s two chromatids.

14

Chromatids 1 and 2 are the two copies of the maternal chromosome. They are sister chromatids.

15

Chromatids 2 and 3 are one chromatid of the maternal homolog and one chromatid of the paternal homolog. They are not copies of each other.

16

Two chromatids that come one from each homolog of a homologous pair are called , because they are not the two sister chromatids of one chromosome.

17

Any chromatid of the maternal homolog and any chromatid of the paternal homolog are non-sister chromatids: 1 and 3, 1 and 4, 2 and 3, and 2 and 4.

18

Sister chromatids carry the same alleles at every position. So a piece swapped between two sister chromatids is identical to the piece it replaces.

19

A swap between two sister chromatids would change nothing.

Left: one dark X with its two chromatids numbered 1 at the upper-left tip and 2 at the lower-left tip, labelled sister chromatids, identical copies. Right: a dark X and a light X side by side and apart, the dark X numbered 1 at its upper-left tip and the light X numbered 3 at its lower-right tip, labelled non-sister chromatids, one from each homolog
Left: one dark X with its two chromatids numbered 1 at the upper-left tip and 2 at the lower-left tip, labelled sister chromatids, identical copies. Right: a dark X and a light X side by side and apart, the dark X numbered 1 at its upper-left tip and the light X numbered 3 at its lower-right tip, labelled non-sister chromatids, one from each homolog
20

Two non-sister chromatids are one from the mother’s chromosome and one from the father’s chromosome. At a position, the two may carry different alleles.

21

So a swap between two non-sister chromatids can change which alleles each chromatid carries.

22

What you are expected to know Identify non-sister chromatids on a paired homologous pair: one chromatid of the maternal homolog and one chromatid of the paternal homolog.

23
Check q2

Suppose two chromatids of a paired homologous pair swapped matching pieces.

Which swap could change the alleles a chromatid carries?

  1. A. A swap between either kind of chromatid
    Sister chromatids carry the same alleles, so a swap between them changes nothing.
    Only a swap between non-sister chromatids can change what a chromatid carries.
  2. B. A swap between two sister chromatids only
    Sister chromatids carry the same alleles at every position.
    A piece swapped between them is identical to the piece it replaces.
  3. C. ✓ A swap between two non-sister chromatids only

Why: Sister chromatids are identical copies, so a swap between them changes nothing.
Non-sister chromatids come from two parents and may carry different alleles.
So only a swap between non-sister chromatids can change what a chromatid carries.

24Quick quiz: non-sister chromatids mixed practice

25
Check q3

A homologous pair lies paired in prophase I.

What are non-sister chromatids?

  1. A. The two chromatids of one copied chromosome
    The two chromatids of one copied chromosome are sister chromatids.
    Non-sister chromatids are one from each homolog of the pair.
  2. B. ✓ Two chromatids, one from each homolog of the pair
  3. C. Two chromatids from chromosomes of different lengths
    Chromosomes of different lengths are not a homologous pair.
    Non-sister chromatids are one from each homolog of one pair.

Why: Each homolog of the pair is two sister chromatids.
One chromatid of the maternal homolog and one chromatid of the paternal homolog are non-sister chromatids.

26
Practice writing an answer

A homologous pair lies paired in prophase I. Each homolog is two sister chromatids.

(a) State what non-sister chromatids are. (1 pt)

Model answer Non-sister chromatids are two chromatids that come one from each homolog of a homologous pair.
Rubric
  • Award 1 point for: two chromatids, one from each homolog of a homologous pair (not the two copies of one chromosome).

(b) Explain why a swap of matching pieces between two sister chromatids would change nothing. (1 pt)

Model answer Sister chromatids are the two copies made when the chromosome was copied.
So they carry the same alleles at every position.
Therefore the piece one gives is identical to the piece it receives, and each chromatid still reads exactly as before.
Rubric
  • Award 1 point for: sister chromatids are identical copies, so exchanged pieces carry the same alleles and neither chromatid changes.
27
Check q4

A homologous pair is drawn below in prophase I, its four chromatids numbered 1 to 4.

Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4
Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4

Which of the following describes chromatids 1 and 2?

  1. A. ✓ Sister chromatids
  2. B. Non-sister chromatids
    Chromatids 1 and 2 are the two chromatids of one homolog, the dark X.
    The two chromatids of one homolog are sister chromatids.

Why: Chromatids 1 and 2 are the two chromatids of the dark homolog.
The two chromatids of one copied chromosome are sister chromatids.

28
Check q5

A homologous pair is drawn below in prophase I, its four chromatids numbered 1 to 4.

Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4
Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4

Which of the following describes chromatids 2 and 3?

  1. A. Sister chromatids
    Chromatid 2 belongs to the dark X and chromatid 3 to the light X.
    One chromatid from each homolog makes them non-sister chromatids.
  2. B. ✓ Non-sister chromatids

Why: Chromatid 2 is a chromatid of the dark homolog, and chromatid 3 is a chromatid of the light homolog.
One chromatid from each homolog: non-sister chromatids.

29
Check q6

A homologous pair is drawn below in prophase I, its four chromatids numbered 1 to 4.

Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4
Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4

Which of the following describes chromatids 1 and 4?

  1. A. Sister chromatids
    Chromatid 1 belongs to the dark X and chromatid 4 to the light X.
    One chromatid from each homolog makes them non-sister chromatids.
  2. B. ✓ Non-sister chromatids

Why: Chromatid 1 is a chromatid of the dark homolog, and chromatid 4 is a chromatid of the light homolog.
One chromatid from each homolog: non-sister chromatids.

30
Check q7

A homologous pair is drawn below in prophase I, its four chromatids numbered 1 to 4.

Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4
Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4

Which of the following describes chromatids 3 and 4?

  1. A. ✓ Sister chromatids
  2. B. Non-sister chromatids
    Chromatids 3 and 4 are the two chromatids of one homolog, the light X.
    The two chromatids of one homolog are sister chromatids.

Why: Chromatids 3 and 4 are the two chromatids of the light homolog.
The two chromatids of one copied chromosome are sister chromatids.

31
Check q8

A homologous pair is drawn below in prophase I, its four chromatids numbered 1 to 4.

Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4
Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4

Which of the following describes chromatids 2 and 4?

  1. A. Sister chromatids
    Chromatid 2 belongs to the dark X and chromatid 4 to the light X.
    One chromatid from each homolog makes them non-sister chromatids.
  2. B. ✓ Non-sister chromatids

Why: Chromatid 2 is a chromatid of the dark homolog, and chromatid 4 is a chromatid of the light homolog.
One chromatid from each homolog: non-sister chromatids.

32Broken at the same point and exchanged

33

Video: Watch: Broken at the same point and exchanged

One crossover happening at the chiasma of the long pair: a dark chromatid and a light chromatid break at the same point, and each piece joins onto the other chromatid. Then the homologs part, and the dark chromatid is seen ending in a light tip and the light chromatid in a dark tip.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L09b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L09b.mp4

34

Now look at the chiasma of the long pair. Two non-sister chromatids cross there: one dark and one light.

Left: the paired long pair, a dark X and a light X with their upper inner arms crossing inside a dashed ring. Right: the same two X's drawn apart; the upper inner tip of the dark X is now light, and the upper inner tip of the light X is now dark
Left: the paired long pair, a dark X and a light X with their upper inner arms crossing inside a dashed ring. Right: the same two X's drawn apart; the upper inner tip of the dark X is now light, and the upper inner tip of the light X is now dark
35

At the crossing point the two non-sister chromatids break at the same place.

36

Each broken piece then joins onto the other chromatid, in the place of the piece that left.

37

So the dark chromatid now ends in a piece of the light chromatid, and the light chromatid ends in a piece of the dark chromatid.

38

When two non-sister chromatids break at the same point and exchange the pieces, the exchange is called . The name says what the two chromatids do: they cross over each other at the chiasma.

39

The exam writes it as crossing over (recombination).

40

Crossing over happens in prophase I, while the two homologs lie paired.

41

The piece that moved carries the alleles at its positions.

42

So each of the two chromatids now carries a combination of alleles that neither homolog had. The combination is new; the alleles are not.

43

Crossing over moves alleles that already exist. It makes no new allele.

44

A new allele comes only from a change in the DNA itself, a mutation.

45

What you are expected to know Describe crossing over: in prophase I, two non-sister chromatids break at the same point at a chiasma and exchange the pieces, so each carries a new combination of the same alleles.

46
Check q9

Suppose the two homologs of a pair have parted after prophase I. One dark chromatid, from the maternal homolog, now ends in a light tip.

What is the light tip?

  1. A. A piece of its sister chromatid, joined on at the centromere
    Sister chromatids are identical, so a piece from the sister would read exactly as the piece it replaced.
    The tip came from a non-sister chromatid of the paternal homolog.
  2. B. A new piece of DNA the cell made in prophase I
    The cell copies its DNA in S phase, before meiosis, and makes no new piece in prophase I.
    The tip is a piece of a paternal chromatid.
  3. C. ✓ A piece of a paternal chromatid, joined on at the chiasma

Why: At the chiasma a maternal chromatid and a paternal chromatid broke at the same point.
Each piece joined onto the other chromatid.
So the maternal chromatid now ends in a piece of the paternal chromatid: the light tip.

47
Practice writing an answer

The two homologs of a pair have parted after prophase I. One dark chromatid, from the maternal homolog, now ends in a light tip.

(a) Explain how the dark chromatid came to end in a light tip. (2 pt)

Model answer In prophase I the two homologs lay paired, held at a chiasma.
At the chiasma a chromatid of the maternal homolog and a chromatid of the paternal homolog broke at the same point.
The two chromatids exchanged the broken pieces.
So the dark chromatid received the light piece and now ends in a light tip.
Rubric
  • Award 1 point for: two non-sister chromatids, one maternal and one paternal, broke at the same point at the chiasma in prophase I.
  • Award 1 point for: the two chromatids exchanged the pieces, so the maternal chromatid now carries the paternal piece.
48
Check q10

A student says: “Crossing over makes new alleles.”

Is the student correct?

  1. A. Yes
    Crossing over moves alleles that already exist onto another chromatid.
    It changes no DNA sequence, so it makes no new allele.
  2. B. ✓ No

Why: Crossing over moves alleles that already exist into new combinations.
Only a change in the DNA itself, a mutation, makes a new allele.

49
Check q11

A homologous pair is drawn below in prophase I, its four chromatids numbered 1 to 4. The chiasma drawn joins chromatids 2 and 3.

Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4
Two X shapes, one dark and one light, lying together with their upper inner arms crossing inside a dashed ring; the four outer tips carry the numbers 1 to 4

Which other two chromatids could also exchange pieces by crossing over?

  1. A. Chromatids 1 and 2
    Chromatids 1 and 2 are the two sister chromatids of the dark homolog.
    They are identical, so a swap between them would change nothing.
  2. B. ✓ Chromatids 1 and 4
  3. C. Chromatids 3 and 4
    Chromatids 3 and 4 are the two sister chromatids of the light homolog.
    They are identical, so a swap between them would change nothing.

Why: Crossing over is between non-sister chromatids, one from each homolog.
Chromatid 1 belongs to the dark homolog and chromatid 4 to the light homolog, so they could exchange pieces.
Chromatids 1 and 2 are sisters, and so are 3 and 4.

50

Back to the long pair of the model cell in prophase I: the maternal chromosome dark, the paternal chromosome light, lying paired and held at a crossing point.

Left: the paired long pair, a dark X and a light X with their upper inner arms crossing inside a dashed ring. Right: the same two X's drawn apart; the upper inner tip of the dark X is now light, and the upper inner tip of the light X is now dark
Left: the paired long pair, a dark X and a light X with their upper inner arms crossing inside a dashed ring. Right: the same two X's drawn apart; the upper inner tip of the dark X is now light, and the upper inner tip of the light X is now dark
51

At the crossing point a dark chromatid and a light chromatid broke at the same place.

52

Each piece joined onto the other chromatid.

53

So after the homologs parted, one dark chromatid ended in a light tip, and one light chromatid ended in a dark tip. Each of those two chromatids carries a new combination of the same alleles.

54Quick quiz: crossing over (recombination) mixed practice

55
Check q12

A cell is in prophase I.

What is crossing over?

  1. A. ✓ Two non-sister chromatids breaking at the same point and exchanging the pieces
  2. B. Two sister chromatids parting from each other at the centromere
    Sister chromatids part in anaphase II, and nothing is exchanged.
    Crossing over is two non-sister chromatids breaking at the same point and exchanging the pieces.
  3. C. Two homologs coming together and lying side by side along their length
    Two homologs lying side by side along their length is synapsis.
    Crossing over is two non-sister chromatids breaking at the same point and exchanging the pieces.

Why: At a chiasma in prophase I, two non-sister chromatids break at the same point.
They exchange the pieces.
That exchange is crossing over.

56
Check q13

A cell is in meiosis.

In which stage does crossing over happen?

  1. A. ✓ Prophase I
  2. B. Metaphase I
    At metaphase I the pairs stand at the equator, and the exchange has already happened.
    Crossing over happens in prophase I, while the homologs lie paired.
  3. C. Anaphase II
    In anaphase II the sister chromatids part, and nothing is exchanged.
    Crossing over happens in prophase I, while the homologs lie paired.

Why: The homologs lie paired along their whole length only in prophase I.
The exchange happens then, at a chiasma.
So crossing over happens in prophase I.

57
Check q14

After meiosis in a yeast cell, one of the four cells carries an allele whose DNA sequence differs by one base from the allele on either homolog of the parent cell.

Which of the following events made the new allele?

  1. A. Crossing over between non-sister chromatids
    Crossing over moves alleles that already exist into new combinations.
    It never changes an allele’s sequence.
  2. B. Independent orientation of the pair
    Independent orientation decides which whole chromosome a gamete receives.
    It changes no DNA.
  3. C. ✓ A mutation in the DNA of one chromatid

Why: Crossing over and independent orientation shuffle alleles that already exist.
A sequence found on neither homolog is a new allele.
Only a mutation makes one.

58
Practice writing an answer

A cell is in prophase I.

(a) State what crossing over is. (1 pt)

Model answer Crossing over is two non-sister chromatids, one from each homolog, breaking at the same point in prophase I and exchanging the pieces.
Rubric
  • Award 1 point for: two non-sister chromatids (one from each homolog) break at the same point (at a chiasma, in prophase I) and exchange the pieces.

Glossary

non-sister chromatids
Two chromatids, one from each homolog of a homologous pair. Crossing over happens between them, never between the two sister chromatids of one chromosome.
crossing over (recombination)
In prophase I, while the homologs are paired, a chromatid of one homolog and a chromatid of the other break at the same point and exchange the pieces, so those chromatids carry combinations of alleles that neither homolog had.

APBIO-U05-L09B New combinations of old alleles

Topic 5.2 · Meiosis and Genetic Diversity · 57 steps

A dark X and a light X drawn apart. On the dark X both upper arms carry a box reading A and, nearer the tip, a box reading B. On the light X both upper arms carry a and, nearer the tip, b
A dark X and a light X drawn apart. On the dark X both upper arms carry a box reading A and, nearer the tip, a box reading B. On the light X both upper arms carry a and, nearer the tip, b

Now consider the long pair of the model cell with letters on it. The maternal homolog carries allele A at one position and allele B further along the same arm, nearer the tip; the paternal homolog carries a and b at the same two positions. Before crossing over, both maternal chromatids read AB and both paternal chromatids read ab. One crossover falls between the two positions.

What do the four chromatids read afterwards, and has any allele changed?

Unit 5 · Heredity

1Four chromatids after one crossover

2

Video: Watch: Four chromatids after one crossover

The lettered long pair: A and B on both chromatids of the dark homolog, a and b on both chromatids of the light homolog. One crossover between the two positions; the two pieces carrying B and b change places. The four chromatids are read out, AB, Ab, aB and ab, and the letters are checked one by one: none has changed.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L09Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L09Ba.mp4

3

What does crossing over change, and what does it leave alone?

4

After one crossover between two positions, two chromatids still read AB and ab, the combinations the parent’s two chromosomes had.

5

The other two chromatids read Ab and aB, combinations neither chromosome had.

6

Not one allele has changed: A is still A, and b is still b. Only the combinations are new.

7

So a gamete can now carry A with b, although no chromosome in the parent did. That is how crossing over adds variety to the gametes.

8
Check q1

At one gene position, the maternal homolog of a pair carries A and the paternal homolog carries a.

What are A and a?

  1. A. ✓ Two alleles of one gene
  2. B. Two different genes
    Two different genes sit at two different positions.
    A and a sit at one position, so they are two alleles of one gene.
  3. C. Two sister chromatids
    Sister chromatids are the two copies of one chromosome.
    A and a are two versions of the gene at one position: two alleles.

Why: A and a sit at the same position on the two homologs.
Versions of one gene at one position are alleles of that gene.

9
Check q2

A cell is in prophase I, its homologs paired.

Which two chromatids exchange pieces in crossing over?

  1. A. The two sister chromatids of one homolog
    Sister chromatids are identical copies, so a swap between them would change nothing.
    Crossing over is between non-sister chromatids.
  2. B. Two chromatids from two different homologous pairs
    Crossing over happens at a chiasma, which holds the two homologs of one pair.
    Crossing over is between non-sister chromatids of one pair.
  3. C. ✓ Two non-sister chromatids, one from each homolog

Why: At a chiasma a chromatid of one homolog and a chromatid of the other homolog break at the same point and exchange the pieces.
Those two are non-sister chromatids.

10

Here is the long pair of the model cell in prophase I, drawn apart so that the letters can be read. Two gene positions on one arm carry letters.

The long pair drawn apart: a dark X whose two upper arms each carry a box reading A and, nearer the tip, a box reading B; a light X whose two upper arms each carry a and, nearer the tip, b
The long pair drawn apart: a dark X whose two upper arms each carry a box reading A and, nearer the tip, a box reading B; a light X whose two upper arms each carry a and, nearer the tip, b
11

At the first position the maternal homolog carries allele A, and the paternal homolog carries allele a.

12

At the second position, nearer the tip, the maternal homolog carries allele B, and the paternal homolog carries allele b.

13

Each homolog is two sister chromatids, identical copies of one chromosome.

14

So both maternal chromatids read AB, and both paternal chromatids read ab.

15

Now suppose one crossover happens between the two positions.

The same pair after one crossover between the two positions: the right upper arm of the dark X now ends in a light piece carrying b, and the left upper arm of the light X ends in a dark piece carrying B; the other two chromatids are unchanged
The same pair after one crossover between the two positions: the right upper arm of the dark X now ends in a light piece carrying b, and the left upper arm of the light X ends in a dark piece carrying B; the other two chromatids are unchanged
16

One maternal chromatid and one paternal chromatid break between the positions and exchange their tip pieces, the pieces beyond the break.

17

The tip piece of the maternal chromatid carries B, and the tip piece of the paternal chromatid carries b. So the exchange carries B and b across.

18

The maternal chromatid that took part now reads Ab. The paternal chromatid that took part now reads aB.

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The other two chromatids took no part in the crossover. One still reads AB, and the other still reads ab.

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So the four chromatids read AB, Ab, aB and ab.

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No allele has changed: A is still A, and b is still b.

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Only the combinations are new: Ab and aB are combinations neither homolog had.

23

The pieces that move carry only the alleles beyond the break. A crossover in a different place moves the alleles beyond that place.

24

After meiosis, each of the four gametes receives one of these four chromatids as its chromosome. So a gamete can now carry A with b, although no chromosome in the parent did.

Four gametes in a row, each holding one chromosome drawn as a rod with two allele boxes: AB, Ab, aB and ab; the Ab rod is dark above its break and light below it, the aB rod light above and dark below
Four gametes in a row, each holding one chromosome drawn as a rod with two allele boxes: AB, Ab, aB and ab; the Ab rod is dark above its break and light below it, the aB rod light above and dark below
25

What you are expected to know Predict the four chromatids after one crossover between two marked positions: two read as the homologs did, two read new combinations of the same alleles, and no allele has changed.

26
Check q3

Suppose that in a rat, the maternal homolog of a pair carries D at one position and E at the next position along the same arm, D nearer the centromere, and the paternal homolog carries d and e. One crossover happens between the two positions, and one maternal chromatid takes part.

What does that maternal chromatid read afterward?

  1. A. D with E
    D with E is what the maternal chromatid read before the crossover.
    The crossover replaced its piece carrying E with the paternal piece carrying e.
  2. B. ✓ D with e
  3. C. d with e
    d with e is what a paternal chromatid reads.
    The maternal chromatid kept D and received e: D with e.

Why: The break fell between the two positions.
The maternal chromatid kept its piece carrying D.
It received the paternal piece carrying e in place of its own piece carrying E.
So it reads D with e.

27
Check q4 numeric entry

Suppose that in a horse, the maternal homolog of a pair carries S and T at two positions on one arm, and the paternal homolog carries s and t. One crossover happens between the two positions.

How many of the four chromatids now read a combination that neither homolog had?

Answer: 2  (tolerance ±0)

Working
Write down the values in the question:
chromatids on the pair = 4
crossovers = 1
chromatids that take part in one crossover = 2, one from each homolog
Write down the equation:
chromatids with a new combination=chromatids that took part in the crossover
Substitute the values into the equation:
chromatids with a new combination=2
28
Practice writing an answer

In a rat, the maternal homolog of a pair carries D at one position and E at the next position along the same arm, and the paternal homolog carries d and e. One crossover happens between the two positions. Afterward, one chromatid reads D with e.

(a) Explain how the chromatid reading D with e came to carry those two alleles. (2 pt)

Model answer Before the crossover the maternal chromatid read D with E.
In prophase I it broke between the two positions, at the same point as a paternal chromatid.
It kept its own piece carrying D.
It received the paternal piece carrying e in place of its piece carrying E.
So it now reads D with e.
Rubric
  • Award 1 point for: a maternal chromatid and a paternal chromatid broke at the same point between the two positions and exchanged the pieces beyond the break.
  • Award 1 point for: the maternal chromatid kept D and received the paternal piece carrying e, so it reads D with e.

(b) Explain why the chromatid carries no new allele. (1 pt)

Model answer D was on the maternal homolog before the crossover, and e was on the paternal homolog.
The crossover moved the piece carrying e onto the maternal chromatid.
It changed no DNA sequence.
So D and e are alleles the parent already had, in a combination neither homolog had.
Rubric
  • Award 1 point for: both alleles were already on the parent’s homologs; the crossover moved a piece and changed no DNA sequence, so the combination is new but the alleles are not.
29
Check q5

Suppose that in a dog, the maternal homolog of a pair carries G at one position and H at the next position along the same arm, and the paternal homolog carries g and h. One crossover happens beyond both positions, nearer the tip than H and h.

Which combinations do the four chromatids read afterward?

  1. A. ✓ GH and gh only
  2. B. Gh and gH only
    Gh and gH would need the break to fall between the two positions.
    A break beyond both positions moves a piece carrying neither letter.
  3. C. GH, Gh, gH and gh
    Two chromatids took part, and the pieces they exchanged carried no letter.
    Every chromatid still reads GH or gh.

Why: The pieces that move carry only the alleles beyond the break.
The break fell beyond both positions, so the pieces carried neither letter.
Every chromatid still reads GH or gh.

30Old combination or new

31

Video: Watch: Old combination or new

The four chromatids AB, Ab, aB and ab lined up as four gametes’ chromosomes. AB and ab are marked as the combinations the parent’s homologs carried; Ab and aB as the combinations neither carried. Then a fresh pair with other letters, and one gamete classed at a time against the parent’s two combinations.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L09Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L09Bb.mp4

32

Look again at the four chromatids after the crossover: AB, Ab, aB and ab. After meiosis, each is the chromosome of one gamete.

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AB and ab are the two combinations the parent’s homologs carried before the crossover.

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When a chromatid keeps the combination its parent homolog carried, the combination is called the . AB and ab are parental.

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Ab and aB are combinations neither of the parent’s homologs carried.

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When a chromatid carries a combination of alleles that neither of the parent’s homologs had, the chromatid is called , because its alleles have been re-combined. Ab and aB are recombinant.

Four gametes in a row, each holding one chromosome drawn as a rod with two allele boxes: AB, Ab, aB and ab; the two recombinant rods change shade below their break; AB and ab are labelled parental, Ab and aB recombinant
Four gametes in a row, each holding one chromosome drawn as a rod with two allele boxes: AB, Ab, aB and ab; the two recombinant rods change shade below their break; AB and ab are labelled parental, Ab and aB recombinant
37

A gamete that receives a recombinant chromatid is a recombinant gamete: it carries A with b, or a with B.

38

To class a chromatid or a gamete, compare its combination with the two combinations the parent’s homologs carried.

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A combination the same as one of them is parental. A combination the same as neither is recombinant.

40

Without the parent’s two combinations, a chromatid cannot be classed.

41

What you are expected to know Classify a chromatid or a gamete as parental or recombinant by comparing its combination with the two combinations the parent’s homologs carried.

42
Check q6

Suppose that in a mouse, the maternal homolog of a pair carries M and N at two positions on one arm, and the paternal homolog carries m and n. One crossover happens between the two positions.

Which of the following chromatids is recombinant afterward?

  1. A. A chromatid reading M with N
    M with N is the combination the maternal homolog carried, so a chromatid that keeps it is parental.
  2. B. ✓ A chromatid reading M with n
  3. C. A chromatid reading m with n
    m with n is the combination the paternal homolog carried, so a chromatid that keeps it is parental.

Why: The parent’s homologs carried MN and mn.
M with n is a combination neither carried, so the chromatid is recombinant.

43

Back to the long pair with its letters, before the crossover: both maternal chromatids read AB, and both paternal chromatids read ab.

The long pair drawn apart: a dark X whose two upper arms each carry a box reading A and, nearer the tip, a box reading B; a light X whose two upper arms each carry a and, nearer the tip, b
The long pair drawn apart: a dark X whose two upper arms each carry a box reading A and, nearer the tip, a box reading B; a light X whose two upper arms each carry a and, nearer the tip, b
44

One crossover between the two positions moved the piece carrying B onto a paternal chromatid and the piece carrying b onto a maternal chromatid.

45

So the four chromatids read AB, Ab, aB and ab: two parental and two recombinant.

The same pair after one crossover between the two positions: the right upper arm of the dark X now ends in a light piece carrying b, and the left upper arm of the light X ends in a dark piece carrying B; the other two chromatids are unchanged
The same pair after one crossover between the two positions: the right upper arm of the dark X now ends in a light piece carrying b, and the left upper arm of the light X ends in a dark piece carrying B; the other two chromatids are unchanged
46

No allele changed. A is still A, and b is still b; only the combinations are new.

47Quick quiz: parental combination, recombinant mixed practice

48
Check q7

A chromatid is described as carrying the parental combination.

What is a parental combination?

  1. A. A combination of alleles that neither of the parent’s homologs carried
    A combination neither homolog carried is recombinant.
    The parental combination is the one a parent homolog carried.
  2. B. The combination of alleles the parent’s two homologs carry together
    The two homologs each carry their own combination.
    The parental combination is the one a single parent homolog carried.
  3. C. ✓ The combination of alleles one of the parent’s homologs carried

Why: Before the crossover each parent homolog carried one combination of alleles.
A chromatid that keeps that combination has the parental combination.

49
Check q8

A gamete’s chromosome is recombinant.

What does recombinant mean?

  1. A. ✓ The chromosome carries a combination of alleles that neither of the parent’s homologs had
  2. B. The chromosome carries an allele that neither of the parent’s homologs had
    An allele neither homolog had would be a new allele, made by mutation.
    Recombinant means a new combination of the parent’s own alleles.
  3. C. The chromosome carries the same combination as one of the parent’s homologs
    The combination one homolog had is the parental combination.
    Recombinant means a combination neither homolog had.

Why: Crossing over moved a piece from one non-sister chromatid to the other.
The chromatid now carries alleles from both homologs, a combination neither homolog had: recombinant.

50
Check q9

A parent’s homologs carried AB and ab. A student looks at a recombinant chromatid reading Ab and says: “This chromatid carries a new allele.”

Is the student correct?

  1. A. Yes
    A and b were both on the parent’s homologs before the crossover.
    The chromatid carries a new combination, not a new allele.
  2. B. ✓ No

Why: A was on the maternal homolog, and b was on the paternal homolog.
The crossover moved the piece carrying b onto the maternal chromatid.
Ab is a new combination of alleles the parent already had.

51
Practice writing an answer

A parent’s homologs carried AB and ab. After one crossover between the two positions, the four chromatids read AB, Ab, aB and ab.

(a) State what a parental combination is. (1 pt)

Model answer A parental combination is the combination of alleles one of the parent’s homologs carried: here AB or ab.
Rubric
  • Award 1 point for: the combination of alleles a parent homolog carried (AB or ab here).

(b) State what a recombinant chromatid is. (1 pt)

Model answer A recombinant chromatid is a chromatid that carries a combination of alleles neither of the parent’s homologs had: here Ab or aB.
Rubric
  • Award 1 point for: a chromatid carrying a combination of alleles neither parent homolog had (Ab or aB here).
52
Check q10

A rabbit’s homologous pair is drawn below: the maternal homolog carries K and L, the paternal homolog k and l. After meiosis, one gamete carries K with l.

A homologous pair drawn apart: a dark X whose two upper arms each carry a box reading K and, nearer the tip, a box reading L; a light X whose two upper arms each carry k and, nearer the tip, l
A homologous pair drawn apart: a dark X whose two upper arms each carry a box reading K and, nearer the tip, a box reading L; a light X whose two upper arms each carry k and, nearer the tip, l

Is the gamete’s chromosome parental or recombinant?

  1. A. Parental
    Neither homolog carried K with l: one carried KL and the other kl.
  2. B. ✓ Recombinant

Why: The parent’s two homologs carried KL and kl.
K with l is a combination neither had, so the chromosome is recombinant.

53
Check q11

A rabbit’s homologous pair is drawn below: the maternal homolog carries K and L, the paternal homolog k and l. After meiosis, one gamete carries k with l.

A homologous pair drawn apart: a dark X whose two upper arms each carry a box reading K and, nearer the tip, a box reading L; a light X whose two upper arms each carry k and, nearer the tip, l
A homologous pair drawn apart: a dark X whose two upper arms each carry a box reading K and, nearer the tip, a box reading L; a light X whose two upper arms each carry k and, nearer the tip, l

Is the gamete’s chromosome parental or recombinant?

  1. A. ✓ Parental
  2. B. Recombinant
    The paternal homolog carried exactly k with l.

Why: The paternal homolog carried k with l.
The gamete keeps that combination, so the chromosome is parental.

54
Check q12

A hamster’s gamete carries M with n. Nobody has recorded what the parent’s two homologs carried.

Is the gamete’s chromosome parental or recombinant?

  1. A. Parental
    Parental means the combination a parent homolog carried.
    Nothing here says what the parent’s homologs carried.
  2. B. Recombinant
    A parent homolog could itself have carried M with n.
  3. C. ✓ Cannot tell

Why: Parental and recombinant are judged against the combinations the parent’s two homologs carried.
Those are not given, so the chromosome cannot be classed.

55
Check q13

In a goat, the maternal homolog of a pair carries R and V, and the paternal homolog carries r and v. After meiosis, a gamete carries R with V.

Is the gamete’s chromosome parental or recombinant?

  1. A. ✓ Parental
  2. B. Recombinant
    The maternal homolog carried exactly R with V.

Why: The maternal homolog carried R with V.
The gamete keeps that combination, so the chromosome is parental.

56
Check q14

In a beetle, the maternal homolog of a pair carries W and z, and the paternal homolog carries w and Z. After meiosis, a gamete carries W with Z.

Is the gamete’s chromosome parental or recombinant?

  1. A. Parental
    The parent’s homologs carried W with z and w with Z.
    Neither carried W with Z.
  2. B. ✓ Recombinant

Why: The parent’s two homologs carried Wz and wZ.
W with Z is a combination neither had, so the chromosome is recombinant.

Glossary

parental combination
The combination of alleles a parent homolog carried, kept by the chromatids that did not take part in a crossover: AB and ab when the homologs carried AB and ab.
recombinant
A chromatid, or a gamete, that carries a combination of alleles neither parent homolog had: Ab and aB when the homologs carried AB and ab.

APBIO-U05-L10 Which egg fuses with which sperm

Topic 5.2 · Meiosis and Genetic Diversity · 62 steps

Left: a photograph looking down into a steel hatchery bucket holding thousands of orange salmon eggs, with the male's white sperm fluid poured over them. Right: a bucket drawn in outline holding scattered large circles, the eggs, and small circles with short tails, the sperm; a caption reads eggs from one female, sperm from one male
Left: a photograph looking down into a steel hatchery bucket holding thousands of orange salmon eggs, with the male's white sperm fluid poured over them. Right: a bucket drawn in outline holding scattered large circles, the eggs, and small circles with short tails, the sperm; a caption reads eggs from one female, sperm from one male

Photo: Cheri Anderson, U.S. Fish and Wildlife Service, Wikimedia Commons, public domain (resized).

Here is a bucket at a salmon hatchery: eggs from one female and sperm from one male, mixed together. Every egg carries a different combination of her chromosomes, and every sperm a different combination of his.

Now any egg can fuse with any sperm. Where does the variety among the young salmon come from?

Unit 5 · Heredity

1Any egg with any sperm

2

Video: Watch: Any egg with any sperm

The kinds of egg and the kinds of sperm as two columns, and a line drawn between every egg and every sperm.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10a.mp4

3

Where does the variety among one pair’s young come from? Two shuffles happened while meiosis made the gametes: crossing over and independent orientation.

4

A third shuffle happens in the bucket, when the gametes fuse.

5
Check q1

A sperm fuses with an egg.

Which of the following is the name for this event?

  1. A. ✓ Fertilization
  2. B. Meiosis
    Meiosis is the division that makes the sperm and the egg.
    Meiosis does not join them.
  3. C. Mitosis
    Mitosis divides one cell into two copies.
    Mitosis joins nothing.

Why: Two gametes fusing into one cell is called fertilization.

6

Meiosis makes the gametes first. Fertilization happens afterward: two finished gametes fuse into one cell, the zygote.

7

In the bucket, which egg fuses with which sperm is not fixed. Any egg can fuse with any sperm.

Four large circles in a column on the left, the kinds of egg, and four small circles with short tails in a column on the right, the kinds of sperm; a line joins every egg to every sperm, sixteen lines in all; the caption reads any egg can fuse with any sperm
Four large circles in a column on the left, the kinds of egg, and four small circles with short tails in a column on the right, the kinds of sperm; a line joins every egg to every sperm, sixteen lines in all; the caption reads any egg can fuse with any sperm
8

When any of one parent’s gametes can fuse with any of the other parent’s, we call it , because which egg fuses with which sperm is left to chance.

9

Random fertilization changes nothing inside the egg or the sperm. It only decides which egg and which sperm are joined.

10

So the zygote’s combination of chromosomes is new: one of her many kinds of egg joined with one of his many kinds of sperm.

11

What you are expected to know Describe random fertilization: any egg can fuse with any sperm, in a separate event after meiosis has finished.

12
Check q2

A sea urchin reproduces sexually.

When does random fertilization happen?

  1. A. Before meiosis begins
    Before meiosis there are no gametes to fuse.
  2. B. Between meiosis I and meiosis II
    Between the two divisions the cells are still inside the parent.
    They are not yet gametes.
  3. C. ✓ After meiosis has finished

Why: Fertilization is not a stage of meiosis.
Meiosis makes the gametes first.
Random fertilization is which finished egg happens to fuse with which finished sperm afterward.

13
Check q3

A female frog sheds her eggs into a pond, and a male frog sheds his sperm over them. A student says: “Each egg lets in only the sperm whose set of chromosomes matches its own, so the egg picks its sperm.”

Is the student correct?

  1. A. Yes: an egg fuses only with a sperm that carries a matching set of chromosomes
    A sperm does not read the egg’s chromosomes.
    Which sperm fuses with an egg is left to chance.
  2. B. ✓ No: any of the sperm can fuse with any of the eggs

Why: Any of his sperm can fuse with any of her eggs.
Which egg fuses with which sperm is left to chance.
That is random fertilization.

14Quick quiz: random fertilization mixed practice

15
Check q4

Three events add variety to the young of two parents.

Which of the following is random fertilization?

  1. A. ✓ Any egg can fuse with any sperm
  2. B. Each homologous pair faces either way at metaphase I
    Each homologous pair facing either way at metaphase I is independent orientation.
  3. C. Two non-sister chromatids exchange pieces
    Two non-sister chromatids exchanging pieces is crossing over.

Why: Random fertilization is any egg fusing with any sperm.
Which egg fuses with which sperm is left to chance.

16
Check q5

An egg and a sperm fuse into a zygote.

Does random fertilization change the chromosomes inside the egg or the sperm?

  1. A. Yes
    Fertilization joins the egg’s set to the sperm’s set.
    Neither set is changed.
  2. B. ✓ No

Why: Random fertilization only decides which egg and which sperm are joined.
The chromosomes inside each gamete stay as meiosis left them.

17
Practice writing an answer

A female trout sheds 2,000 eggs into a tank, and a hatchery worker pours a male trout’s sperm over them. Every egg carries a different combination of her chromosomes, and every sperm a different combination of his.

(a) Explain why each zygote in the tank carries a combination of chromosomes that no single gamete carried. (1 pt)

Model answer Each egg carries one of the female’s many kinds of set.
Each sperm carries one of the male’s many kinds of set.
Any egg can fuse with any sperm.
So each zygote joins one of her kinds of egg with one of his kinds of sperm, a pairing that no single gamete carried.
Rubric
  • Award 1 point for: any egg can fuse with any sperm (random fertilization), so the zygote joins one of her kinds of egg with one of his kinds of sperm.

Slip Saying that fertilization changes the chromosomes inside the gametes. Fertilization joins two sets; it changes neither.

18Her kinds times his kinds

19

Video: Watch: Her kinds times his kinds

The grid of egg kinds against sperm kinds filling cell by cell, then the working for the model cell and for a human.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10b.mp4

20
Check q6

A parent’s cells have 3 pairs of chromosomes.

Which of the following is the number of kinds of gamete independent orientation alone can give this parent?

  1. A. 6
    Each homologous pair doubles the count.
    The counts multiply across pairs; they never add.
  2. B. ✓ 8
  3. C. 9
    Each homologous pair doubles the count.
    Three pairs do not multiply three by three.

Why: One pair gives 2 kinds.
Each added pair doubles the count.
kinds of gamete=23=8

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How many kinds of zygote can two parents make? Count her kinds of egg, count his kinds of sperm, and multiply.

22

Take the model cell, with two pairs of chromosomes. Independent orientation gives the female four kinds of egg and the male four kinds of sperm.

23

Here is a grid of the four kinds of egg against the four kinds of sperm. Each empty box stands for one egg joined with one sperm: one kind of zygote.

A four-by-four grid. Along the top edge the four kinds of sperm, each drawn as a long rod and a short rod, dark or light; down the left edge the four kinds of egg, drawn the same way; the sixteen empty boxes are the sixteen kinds of zygote, one egg joined with one sperm
A four-by-four grid. Along the top edge the four kinds of sperm, each drawn as a long rod and a short rod, dark or light; down the left edge the four kinds of egg, drawn the same way; the sixteen empty boxes are the sixteen kinds of zygote, one egg joined with one sperm
24
Worked example

Two parents each have 2 pairs of chromosomes. How many kinds of zygote can they make, counting independent orientation and random fertilization only?

Write down the values in the question:
n=2
Write down the equation:
kinds of gamete per parent=2n
kinds of zygote=kinds of egg×kinds of sperm
Substitute the values into the equation:
kinds of gamete per parent=22=4
kinds of zygote=4×4=16
25

The kinds multiply. Every kind of egg can fuse with every kind of sperm, so the kinds of zygote are the kinds of egg multiplied by the kinds of sperm.

26

Now consider a human, with 23 pairs of chromosomes. Independent orientation alone gives each parent about 8.4 million kinds of gamete.

27
Worked example

Two people each have 23 pairs of chromosomes. How many kinds of zygote can they make, counting independent orientation and random fertilization only?

Write down the values in the question:
n=23
Write down the equation:
kinds of gamete per parent=2n
kinds of zygote=kinds of egg×kinds of sperm
Substitute the values into the equation:
kinds of gamete per parent=223≈8,400,000
kinds of zygote≈8,400,000×8,400,000≈7.0×1013
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So two people can make more than 70 trillion kinds of zygote by independent orientation and random fertilization alone. Crossing over adds still more kinds on top.

29

What you are expected to know Calculate the number of kinds of zygote two parents can make: the kinds of egg multiplied by the kinds of sperm.

30
Check q7 numeric entry

An insect has 7 pairs of chromosomes.

Calculate the number of kinds of zygote two of these insects can make, counting independent orientation and random fertilization only.

Part 1. How many kinds of gamete can one of these insects make by independent orientation alone?

Answer: 128  (tolerance ±0)

Working
Substitute n = 7 into the equation:
kinds of gamete per parent=27=128

Answer: 16384  (tolerance ±0)

Working
Write down the values in the question:
n=7
Write down the equation:
kinds of gamete per parent=2n
kinds of zygote=kinds of egg×kinds of sperm
Substitute the values into the equation:
kinds of gamete per parent=27=128
kinds of zygote=128×128=16,384
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Check q8 numeric entry

An animal has 9 pairs of chromosomes.

Calculate the number of kinds of zygote two of these animals can make, counting independent orientation and random fertilization only.

Answer: 262144  (tolerance ±0)

Working
Write down the values in the question:
n=9
Write down the equation:
kinds of gamete per parent=2n
kinds of zygote=kinds of egg×kinds of sperm
Substitute the values into the equation:
kinds of gamete per parent=29=512
kinds of zygote=512×512=262,144

32Three shuffles

33

Video: Watch: Three shuffles

The three shuffles as three labelled panels that stay on screen, then the table comparing them row by row.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10c.mp4

34
Check q9

At a salmon hatchery, the eggs from one female were collected in a bucket. The eggs differed from one another before any sperm reached them.

Which two events made the eggs differ?

  1. A. ✓ Independent orientation at metaphase I and crossing over in prophase I
  2. B. Fertilization and the first mitosis of the zygote
    Fertilization and mitosis happen after the egg is finished.
    They cannot make eggs differ from one another.

Why: Each homologous pair faced either way at metaphase I: independent orientation.
Non-sister chromatids exchanged pieces in prophase I: crossing over.

35

Random fertilization is the third shuffle. Here are the three shuffles, one drawing each: crossing over, independent orientation and random fertilization.

Three boxed panels. Left: one chromatid drawn as a rod, dark in its upper part and light in its lower part, labelled crossing over. Middle: a gamete holding a dark long chromosome and a light short one, labelled independent orientation. Right: an egg and a sperm fusing, labelled random fertilization
Three boxed panels. Left: one chromatid drawn as a rod, dark in its upper part and light in its lower part, labelled crossing over. Middle: a gamete holding a dark long chromosome and a light short one, labelled independent orientation. Right: an egg and a sperm fusing, labelled random fertilization
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Here is a table comparing crossing over, independent orientation and random fertilization: when each happens, what each shuffles, and one example of each.

ShuffleWhen it happensWhat it shufflesOne example
Crossing overprophase I, while the homologs are pairedpieces of chromatids: which alleles sit together along one chromosomethe mother’s allele beside the father’s on one chromatid
Independent orientationmetaphase I, as each homologous pair lines upwhole chromosomes: which member of each pair a gamete getsthe maternal long chromosome with the paternal short one in one gamete
Random fertilizationafter meiosis, when two gametes fusewhole sets: which egg is joined with which spermthis egg fused with this sperm

the three shuffles of sexual reproduction

A table with four columns, shuffle, when it happens, what it shuffles and one example, and three rows: crossing over, independent orientation, random fertilization
37

The variety of alleles and allele combinations among the individuals of a population is called ; the exam also calls it genetic diversity.

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Sexual reproduction makes genetic variation at every generation, through the three shuffles: crossing over, independent orientation and random fertilization.

39

Each shuffle is an event with a name: which non-sister chromatids exchanged pieces, which way each homologous pair faced, which egg fused with which sperm.

40

What you are expected to know Classify a described source of variation as crossing over, independent orientation or random fertilization.

41

Back to the bucket at the salmon hatchery: eggs from one female and sperm from one male, mixed together.

Four large circles in a column on the left, the kinds of egg, and four small circles with short tails in a column on the right, the kinds of sperm; a line joins every egg to every sperm, sixteen lines in all; the caption reads any egg can fuse with any sperm
Four large circles in a column on the left, the kinds of egg, and four small circles with short tails in a column on the right, the kinds of sperm; a line joins every egg to every sperm, sixteen lines in all; the caption reads any egg can fuse with any sperm
42

Crossing over and independent orientation had already made her eggs differ from one another. The same two shuffles had made his sperm differ from one another.

43

In the bucket, any egg can fuse with any sperm. So the kinds of young salmon are her kinds of egg multiplied by his kinds of sperm.

44

That is where the variety among the young salmon comes from.

45Quick quiz: genetic variation (genetic diversity) mixed practice

46
Check q10

A zebrafish egg carries the maternal copy of chromosome 3 and the paternal copy of chromosome 7.

Which shuffle put that combination together?

  1. A. Crossing over
    Crossing over changes which alleles sit together along one chromosome.
    Here two whole chromosomes of different pairs have been combined.
  2. B. ✓ Independent orientation
  3. C. Random fertilization
    The combination is inside one egg, before any sperm has fused with it.

Why: Which member of each pair a gamete receives is decided pair by pair at metaphase I.
The maternal copy of one pair with the paternal copy of another is independent orientation.

47
Check q11

One chromosome in a mouse sperm carries the mother’s allele of a gene and the father’s allele of the next gene along it.

Which shuffle made that chromosome?

  1. A. ✓ Crossing over
  2. B. Independent orientation
    Independent orientation moves whole chromosomes.
    It cannot put two parents’ alleles on one chromosome.
  3. C. Random fertilization
    The chromatid is inside a sperm that has not yet fused with an egg.

Why: Only crossing over puts pieces of both homologs on one chromatid.
Two non-sister chromatids broke at the same point and exchanged pieces.
So one chromatid, now this sperm’s chromosome, carries the mother’s allele at one position and the father’s at the next.

48
Check q12

A female frog sheds thousands of eggs. One particular sperm reaches one particular egg and fuses with it.

Which shuffle is this?

  1. A. Crossing over
    Crossing over happens in prophase I, long before any sperm arrives.
  2. B. Independent orientation
    Independent orientation happens at metaphase I, when the gametes are made.
  3. C. ✓ Random fertilization

Why: Which egg fuses with which sperm is left to chance.
That is random fertilization.

49
Check q13

In a barley gamete, the tip of the maternal copy of one chromosome is a piece of the paternal copy.

Which shuffle made that chromosome?

  1. A. ✓ Crossing over
  2. B. Independent orientation
    Independent orientation never puts a piece of one homolog onto the other.
  3. C. Random fertilization
    The chromosome is inside a gamete that has not yet fused with another.

Why: Two non-sister chromatids broke at the same point and exchanged pieces.
So a piece of the paternal copy now sits on the maternal copy.
That is crossing over.

50
Check q14

In a cricket cell at metaphase I, the long pair faces its maternal member toward the left pole and the short pair faces its maternal member toward the right pole.

Which shuffle is at work in this cell?

  1. A. Crossing over
    No pieces are being exchanged here.
    Whole pairs are facing the poles.
  2. B. ✓ Independent orientation
  3. C. Random fertilization
    This cell is still in meiosis I; no gametes exist yet.

Why: Each homologous pair is facing either way on its own at metaphase I.
That is independent orientation.

51
Check q15

A hen’s eggs come in many kinds. One of those eggs fused with a sperm, and a chick grew from the zygote.

Which shuffle decided which kind of egg the chick grew from?

  1. A. Crossing over
    Crossing over made the eggs differ from one another in prophase I.
    It did not decide which egg fused with the sperm.
  2. B. Independent orientation
    Independent orientation gave the eggs their different sets at metaphase I.
    It did not decide which egg fused with the sperm.
  3. C. ✓ Random fertilization

Why: Which egg fuses with which sperm is left to chance.
That is random fertilization.

52
Check q16

Biologists study the individuals of a population of wild mice.

Which of the following is the population’s genetic variation?

  1. A. The number of kinds of gamete one mouse can make
    The number of kinds of gamete is one parent’s count.
    Genetic variation is a property of the whole population.
  2. B. The exchange of pieces between two non-sister chromatids
    The exchange of pieces between non-sister chromatids is crossing over, one of the shuffles that makes genetic variation.
  3. C. ✓ The variety of alleles and allele combinations among the mice

Why: Genetic variation is the variety of alleles and allele combinations among the individuals of a population.

53
Practice writing an answer

A population of wild strawberry plants reproduces sexually. Its plants show many combinations of flower color, leaf shape and fruit size.

(a) Explain how the many combinations of flower color, leaf shape and fruit size in this population demonstrate genetic variation. (1 pt)

Model answer Each plant’s flower color, leaf shape and fruit size come from the alleles it carries.
The plants differ in these traits, so they carry different alleles and different combinations of alleles.
That variety of alleles and allele combinations among the plants is genetic variation.
Rubric
  • Award 1 point for: the differing traits show that the plants carry different alleles and allele combinations, which is genetic variation.

Slip Saying only that the plants look different. Genetic variation is the variety of alleles and allele combinations, not the look of the plants.

54Mixed practice mixed practice

55
Check q17

A female sea star sheds eggs into the sea, and a male sea star sheds sperm nearby. One egg carries the maternal copy of the longest chromosome.

Which sperm can fuse with this egg?

  1. A. ✓ Any sperm at all
  2. B. Only a sperm carrying the paternal copy, so that the zygote gets one of each
    A sperm does not read the egg’s chromosomes.
    The zygote gets one copy from the egg and one from the sperm, whichever copies they carry.
  3. C. Only a sperm carrying the maternal copy, so that the two copies match
    A sperm does not read the egg’s chromosomes.
    Any sperm can fuse with the egg, whichever copy it carries.

Why: Any egg can fuse with any sperm.
Which egg fuses with which sperm is left to chance.
That is random fertilization.

56
Check q18

In an animal, independent orientation gives each parent 64 kinds of gamete.

What does random fertilization add to the variety of the young?

  1. A. Nothing: the young come in 64 kinds, just like the gametes
    A zygote joins two gametes.
    So the kinds of zygote are the kinds of egg multiplied by the kinds of sperm, far more than 64.
  2. B. Twice the variety: the young come in 128 kinds
    The kinds of egg are multiplied by the kinds of sperm, not doubled.
  3. C. ✓ Many times the variety: the young come in 4,096 kinds
  4. D. New alleles that no gamete carried
    Fertilization joins the alleles the gametes already carry.
    It makes no new allele.

Why: Any of the 64 kinds of egg can fuse with any of the 64 kinds of sperm.
kinds of zygote=64×64=4,096

57
Check q19

A horse’s egg cell carries 32 chromosomes, one from each of the mother’s 32 homologous pairs: some are maternal copies and some are paternal copies.

Which shuffle decided which copy of each pair the egg carries?

  1. A. Crossing over
    Crossing over exchanges pieces between non-sister chromatids.
    It does not decide which whole chromosome a gamete gets.
  2. B. ✓ Independent orientation
  3. C. Random fertilization
    The egg has not yet fused with a sperm.

Why: Each homologous pair faced either way at metaphase I, on its own.
So which member of each pair the egg received was decided pair by pair.
That is independent orientation.

58
Check q20

A cell in a plant’s anther divides by meiosis, and the gametes it makes later take part in fertilization.

Which of the following happens after meiosis has finished?

  1. A. Crossing over
    Crossing over happens in prophase I, inside meiosis.
  2. B. Independent orientation
    Independent orientation happens at metaphase I, inside meiosis.
  3. C. ✓ Random fertilization

Why: Meiosis makes the gametes first.
Only then can any egg fuse with any sperm.
So random fertilization happens after meiosis has finished.

59
Check q21

Two puppies from the same litter have the same mother and father.

Which statement explains why the two puppies carry different combinations of alleles?

  1. A. ✓ The two puppies grew from two different eggs, each fusing with a different sperm
  2. B. Meiosis made every egg and every sperm different, so which egg fused with which sperm made no difference
    Meiosis did make every egg and every sperm different.
    But a puppy is one egg joined with one sperm, so which egg fused with which sperm decides its mix.
  3. C. The mother’s eggs are alike, so the differences came from the father’s sperm
    Independent orientation and crossing over happen in the mother’s meiosis as well as the father’s.
    So her eggs differ from one another as much as his sperm do.
  4. D. Independent orientation in a puppy’s body cells after birth gave it a different mix of the parents’ chromosomes
    Independent orientation happens at metaphase I, when the gametes are made.
    Before each mitosis a body cell copies every chromosome once.
    Mitosis then passes the zygote’s chromosomes on unchanged.

Why: Independent orientation and crossing over made every egg and every sperm a different mix.
Random fertilization joined a different egg with a different sperm for each puppy.

60
Check q22 numeric entry

A plant has 5 pairs of chromosomes.

Calculate the number of kinds of zygote two of these plants can make, counting independent orientation and random fertilization only.

Answer: 1024  (tolerance ±0)

Working
Write down the values in the question:
n=5
Write down the equation:
kinds of gamete per parent=2n
kinds of zygote=kinds of egg×kinds of sperm
Substitute the values into the equation:
kinds of gamete per parent=25=32
kinds of zygote=32×32=1,024
61
Practice writing an answer

At a trout hatchery, a worker mixes eggs from one female with sperm from one male in a tank. The young trout that hatch differ from one another. Trout have many pairs of chromosomes; count only three of one female's pairs, so that her gametes come in 8 kinds by independent orientation alone.

(a) Describe how independent orientation gives this female eggs of more than one kind. (1 pt)

Model answer Each homologous pair lined up at metaphase I facing either way regardless of the other pairs.
So which member of each pair an egg received was decided pair by pair.
Different eggs received different mixes of her maternal and paternal chromosomes.
Rubric
  • Award 1 point for: each pair facing either way on its own at metaphase I (independent orientation), so eggs differ in which member of each pair they carry.

Slip Describing crossing over instead. Crossing over changes which alleles sit together along one chromosome; independent orientation decides which whole member of each pair the egg gets.

(b) Describe how crossing over gives a chromatid a new combination of alleles. (1 pt)

Model answer In prophase I, while the homologs are paired, a chromatid of one homolog and a chromatid of the other break at the same point.
The two non-sister chromatids exchange the pieces.
So those chromatids carry combinations of alleles that neither homolog had.
Rubric
  • Award 1 point for: non-sister chromatids breaking at the same point and exchanging pieces in prophase I, so a chromatid carries alleles from both homologs.

Slip Saying sister chromatids swap. Sister chromatids are identical copies; a swap between them changes nothing.

(c) Calculate the number of kinds of zygote this female and a male whose gametes also come in 8 kinds can make, counting independent orientation and random fertilization only. (1 pt)

Answer: 64  (tolerance ±0)

Model answer 64 kinds of zygote.
Working
Write down the values in the question:
kinds of egg = 8
kinds of sperm = 8
Write down the equation:
kinds of zygote=kinds of egg×kinds of sperm
Substitute the values into the equation:
kinds of zygote=8×8=64
Rubric
  • Award 1 point for: 64.

(d) Explain why mixing the eggs and sperm in one tank adds variation beyond what meiosis made. (1 pt)

Model answer Meiosis made every egg and every sperm a different combination.
In the tank, any egg can fuse with any sperm.
So each zygote is a new combination: one of her kinds of egg with one of his kinds of sperm.
The kinds of zygote are the kinds of egg multiplied by the kinds of sperm.
Rubric
  • Award 1 point for: random fertilization, any egg with any sperm, so the combinations multiply beyond the gametes’ own variety.

Slip Saying fertilization makes the gametes vary. The gametes varied already; fertilization multiplies the combinations by joining them at random.

Glossary

random fertilization
Any of one parent’s gametes can fuse with any of the other’s, so the zygote’s combination of chromosomes is new; the kinds of zygote are the kinds of egg multiplied by the kinds of sperm.
genetic variation (genetic diversity)
The variety of alleles and allele combinations among the individuals of a population. In sexual reproduction it comes from crossing over, independent orientation and random fertilization.

APBIO-U05-L10B A question you could test

Topic 5.2 · Meiosis and Genetic Diversity · 38 steps

Two cards side by side. The left card is labelled observation and reads: cells of strain 1 show more crossovers than cells of strain 2. The right card is labelled a question you could test and is blank with a question mark
Two cards side by side. The left card is labelled observation and reads: cells of strain 1 show more crossovers than cells of strain 2. The right card is labelled a question you could test and is blank with a question mark

Here is an observation from a corn field, written on a card: cells of one strain of corn show more crossovers in prophase I than cells of another strain.

What question could you test about it, and what makes a question testable?

Unit 5 · Heredity

1Name what you would change and what you would measure

2

Video: Watch: Name what you would change and what you would measure

The observation card turning into a question card as its two parts, what is changed and what is measured, are written in; then five questions judged one after another.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10Ba.mp4

3

What makes a question one an experiment can answer? An experiment changes one thing and measures another, with everything else kept the same.

4
Check q1

An experiment changes one variable and measures another, with everything else kept the same.

What is the variable the experimenter changes called?

  1. A. ✓ The independent variable
  2. B. The dependent variable
    The dependent variable is the one measured.
    Its value depends on the change.

Why: The experimenter changes the independent variable and measures the dependent variable.

5
Check q2

An experiment changes one factor and measures one result.

What does the null hypothesis state?

  1. A. That the tested factor changes the measured result
    A prediction that the factor changes the result is the ordinary hypothesis.
    The null hypothesis says the factor makes no difference.
  2. B. That the measured result was recorded without error
    The null hypothesis is a statement about the tested factor, not about how carefully the result was recorded.
  3. C. ✓ That the tested factor makes no difference to the measured result

Why: The null hypothesis is the statement that the tested factor makes no difference to the measured result.

6

So a testable question names both: what you would change and what you would measure. A question that names neither, or only one, cannot be set up as an experiment.

7

For example, “Why does crossing over happen?” is not testable. It names nothing to change and nothing to measure.

8

But “Does the number of DNA breaks in a strain change the mean number of crossovers per cell?” is testable. It names what would be changed, the number of DNA breaks, and what would be measured, the crossovers per cell.

A question card reading: does the number of DNA breaks in a strain change the mean number of crossovers per cell? Beneath it two boxes: what is changed, the number of DNA breaks; what is measured, crossovers per cell
A question card reading: does the number of DNA breaks in a strain change the mean number of crossovers per cell? Beneath it two boxes: what is changed, the number of DNA breaks; what is measured, crossovers per cell
9

But “What is the mean number of crossovers per cell in this strain?” is not testable. It names what would be measured, the crossovers per cell, but nothing to change.

10

And “Does raising the temperature change the mean number of crossovers per cell?” is testable. It names what would be changed, the temperature, and what would be measured, the crossovers per cell.

11

But “Is crossing over important?” is not testable. It names nothing to change and nothing to measure.

12

Here is a table comparing these questions, from “Why does crossing over happen?” to “Is crossing over important?”: what each would change, what each would measure, and whether an experiment could answer it.

QuestionWhat would be changedWhat would be measuredTestable?
Why does crossing over happen?——No
Does the number of DNA breaks in a strain change the mean number of crossovers per cell?the number of DNA breakscrossovers per cellYes
What is the mean number of crossovers per cell in this strain?—crossovers per cellNo
Does raising the temperature change the mean number of crossovers per cell?the temperaturecrossovers per cellYes
Is crossing over important?——No

a question is testable when both middle columns are filled

A table with four columns, the question, what would be changed, what would be measured and testable, and one row per question judged above, with a dash where a question names nothing
13

A testable question has a null hypothesis too. For the corn question it is that the number of DNA breaks makes no difference to the crossovers per cell.

14

What you are expected to know Judge whether a question is testable: it names what would be changed and what would be measured.

15Quick quiz: testable or not? mixed practice

16
Check q3

“Does the amount of light a pea plant grows in change the mean number of crossovers per cell?”

Is this a question an experiment could answer, by changing one thing and measuring another?

  1. A. ✓ Yes
  2. B. No
    The question names what is changed, the amount of light, and what is measured, the mean number of crossovers per cell.
    So it is testable.

Why: The question names what is changed, the amount of light, and what is measured, crossovers per cell.
A question that names both is testable.

17
Check q4

“Why do some chromosomes have more chiasmata than others?”

Is this a question an experiment could answer, by changing one thing and measuring another?

  1. A. Yes
    “Why” names nothing to change and nothing to measure.
    So no experiment can answer the question as written.
  2. B. ✓ No

Why: The question names nothing to change and nothing to measure.
So it is not testable as written.

18
Check q5

“Does the amount of water a wheat plant gets change the mean number of chiasmata per cell?”

Is this a question an experiment could answer, by changing one thing and measuring another?

  1. A. ✓ Yes
  2. B. No
    The question names what is changed, the amount of water, and what is measured, the mean number of chiasmata per cell.
    So it is testable.

Why: The question names what is changed, the amount of water, and what is measured, chiasmata per cell.
A question that names both is testable.

19
Check q6

“Does the dose of X-rays a female fruit fly receives change the fraction of her eggs that carry an extra chromosome?”

Is this a question an experiment could answer, by changing one thing and measuring another?

  1. A. ✓ Yes
  2. B. No
    The question names what is changed, the dose of X-rays, and what is measured, the fraction of eggs with an extra chromosome.
    So it is testable.

Why: The question names what is changed, the dose of X-rays, and what is measured, the fraction of eggs with an extra chromosome.
So it is testable.

20
Check q7

“Why do some plants make more pollen than others?”

Is this a question an experiment could answer, by changing one thing and measuring another?

  1. A. Yes
    “Why” names nothing to change and nothing to measure.
    So no experiment can answer the question as written.
  2. B. ✓ No

Why: The question names nothing to change and nothing to measure.
So it is not testable as written.

21
Check q8

“What fraction of a locust’s sperm carry an extra chromosome?”

Is this a question an experiment could answer, by changing one thing and measuring another?

  1. A. Yes
    The fraction can be counted, but the question changes nothing.
    It is one measurement to make, not an experiment.
  2. B. ✓ No

Why: The question names what to measure but nothing to change.
It asks for one measurement, so it is not testable as written.

22
Check q9

“Does the dose of a drug change the number of crossovers per cell in yeast?”

Which of the following is the variable that is changed?

  1. A. The number of crossovers per cell
    The number of crossovers per cell is the result the experimenter counts.
  2. B. The yeast species
    The yeast species stays the same throughout.
  3. C. ✓ The dose of the drug

Why: The experimenter sets the dose of the drug and then counts crossovers.
So the dose is the variable that is changed.

23
Check q10

“Does the amount of nitrogen in the soil change the number of chiasmata per cell in barley?”

Which of the following is the variable that is measured?

  1. A. The barley variety
    The barley variety stays the same throughout.
  2. B. ✓ The number of chiasmata per cell
  3. C. The amount of nitrogen in the soil
    The experimenter sets the amount of nitrogen.

Why: The experimenter sets the amount of nitrogen and then counts chiasmata.
So the number of chiasmata per cell is the variable that is measured.

24Write the question

25

Video: Watch: Write the question

A second observation card, then the three steps written in one at a time: the change, the measurement, the question.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L10Bb.mp4

26

How do you write a testable question from an observation? Now consider a second observation: a protein sits at a spot on a chromosome where crossovers are common.

27

To write the question, follow three steps:

  1. 1. Name what you would change: here, whether the protein is present or removed.
  2. 2. Name what you would measure: here, the number of crossovers at that spot.
  3. 3. Write “Does [what you would change] change [what you would measure]?”

28

So the question reads: “Does removing the protein change the number of crossovers at that spot?”

A question card reading: does removing the protein change the number of crossovers at that spot? Beneath it two boxes: what is changed, the protein present or removed; what is measured, crossovers at the spot
A question card reading: does removing the protein change the number of crossovers at that spot? Beneath it two boxes: what is changed, the protein present or removed; what is measured, crossovers at the spot
29

The question names both things, so an experiment can answer it: remove the protein from some cells, leave it in others, and count the crossovers at the spot in each group.

30

What you are expected to know Write a testable question from an observation about meiosis, naming what you would change and what you would measure.

31
Check q11

A student notices that pollen from plants grown in a hot greenhouse contains more recombinant gametes than pollen from plants grown outdoors.

Which of the following questions about this observation is testable as written?

  1. A. Why does heat matter so much to meiosis in plants grown in a greenhouse?
    The question names nothing to change and nothing to measure.
    “Why” alone cannot be set up as an experiment.
  2. B. What fraction of the gametes in a plant’s pollen are recombinant?
    The question names something to measure but nothing to change.
    It asks for one measurement, with no two conditions to compare.
  3. C. Is a hot greenhouse better for the plants than growing outdoors?
    “Better for the plants” names nothing an experiment can measure.
  4. D. ✓ Does a higher growing temperature raise the fraction of recombinant gametes?

Why: A testable question names what would be changed and what would be measured.
This one changes the growing temperature and measures the fraction of recombinant gametes.
The other three name one of the two or neither.

32
Check q12

A gardener notices that in a species of lily, cells in older plants show more failures of a pair to separate at anaphase I than cells in young plants.

Which of the following questions turns this observation into one an experiment could answer?

  1. A. Why do older lilies make more mistakes in meiosis than young ones?
    “Why” names nothing to change and nothing to measure.
  2. B. In what fraction of a lily’s cells does a pair fail to separate at anaphase I?
    The question names what to measure but nothing to change.
    Old and young plants are not compared.
  3. C. ✓ Does plant age change the fraction of cells in which a pair fails to separate?
  4. D. Is meiosis a more complicated process in old plants than in young?
    “More complicated” cannot be measured, so the question names nothing to measure.

Why: Plant age is what would be changed.
The fraction of anaphase I cells in which a pair fails to separate is what would be measured.
A question that names both can be tested.
The other three name one of the two or neither.

33
Practice writing an answer

A grower notices that pollen from apple trees growing in shade contains fewer recombinant gametes than pollen from apple trees growing in full sun.

(a) Identify what you would change to test this observation. (1 pt)

Model answer I would change the amount of sunlight the apple trees grow in: shade or full sun.
Rubric
  • Award 1 point for: the amount of sunlight (shade or full sun) as the thing changed.

Slip Naming the recombinant gametes as the thing changed. The grower sets the light; the gametes change as a result.

(b) Identify what you would measure. (1 pt)

Model answer I would measure the fraction of recombinant gametes in each tree’s pollen.
Rubric
  • Award 1 point for: the fraction (or number) of recombinant gametes in the pollen as the thing measured.

Slip Naming the sunlight as the thing measured. The sunlight is set by the grower; the gametes are counted.

(c) Write a testable question about this observation. (1 pt)

Model answer Does the amount of sunlight an apple tree grows in change the fraction of recombinant gametes in its pollen?
Rubric
  • Award 1 point for: a question that names the amount of sunlight as what would be changed and the fraction of recombinant gametes as what would be measured.

Slip Writing “Why do shaded trees make fewer recombinant gametes?” A why-question names nothing to change and nothing to measure.

34
Practice writing an answer

A biologist notices that rye plants grown at 30 °C show more chiasmata per cell than rye plants grown at 20 °C.

(a) Write a testable question about this observation. (1 pt)

Model answer Does growing rye plants at a higher temperature change the number of chiasmata per cell?
Rubric
  • Award 1 point for: a question that names the growing temperature as what would be changed and the number of chiasmata per cell as what would be measured.

Slip Writing “What is the number of chiasmata per cell at 30 °C?” A single measurement names nothing to change.

35

Back to the card from the corn field: cells of one strain of corn show more crossovers in prophase I than cells of another strain.

A question card reading: does the number of DNA breaks in a strain change the mean number of crossovers per cell? Beneath it two boxes: what is changed, the number of DNA breaks; what is measured, crossovers per cell
A question card reading: does the number of DNA breaks in a strain change the mean number of crossovers per cell? Beneath it two boxes: what is changed, the number of DNA breaks; what is measured, crossovers per cell
36

The observation became a question you could test the moment it named two things.

37

It named what to change, the number of DNA breaks. It named what to measure, the crossovers per cell.

APBIO-U05-L11 When a pair fails to separate

Topic 5.2 · Meiosis and Genetic Diversity · 70 steps

An oval cell with a pole at each end: both long X-shaped chromosomes, one dark and one light, are moving toward the left pole, each with a fiber from that pole, while the dark short X moves left and the light short X moves right
An oval cell with a pole at each end: both long X-shaped chromosomes, one dark and one light, are moving toward the left pole, each with a fiber from that pole, while the dark short X moves left and the light short X moves right

Here is the model cell in anaphase I, but this time both long chromosomes are moving to the same pole. The short pair is parting normally, and nothing else has gone wrong.

What will the four cells at the end of meiosis hold?

Unit 5 · Heredity

1When a pair fails to separate

2

Video: Watch: When a pair fails to separate

The long pair failing to part in the model cell at anaphase I, both long chromosomes drawn to one pole; then the two cells, one with a chromosome too many and one with one too few.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11a.mp4

3

What happens when one pair fails to part?

4

In meiosis I, both members of a homologous pair can go to one pole together. In meiosis II, both sister chromatids of one chromosome can go to one pole together.

5

A failure in meiosis I sends both chromosomes of the pair into one cell and none into the other.

6

Meiosis II then passes that fault into all four cells: two cells with one chromosome too many, two with one too few.

7

A failure in meiosis II touches only one of the two cells. So two of the four cells are normal, one is over and one is under.

8
Check q1

A cell is at anaphase I, and the spindle is pulling partners apart.

Which partners does the spindle pull apart at anaphase I?

  1. A. ✓ The two homologs of each pair
  2. B. The two sister chromatids of each chromosome
    Sister chromatids part at anaphase II.
    Anaphase I parts the homologs.

Why: At anaphase I the spindle pulls the two members of each homologous pair to opposite poles.

9

Here the long pair has not parted: both long chromosomes are moving to the left pole, each pulled by a fiber from that pole.

The model cell at anaphase I: both long X-shaped chromosomes, one dark and one light, move toward the left pole, each on a fiber from that pole; the dark short X moves left and the light short X moves right, each on a fiber from its pole
The model cell at anaphase I: both long X-shaped chromosomes, one dark and one light, move toward the left pole, each on a fiber from that pole; the dark short X moves left and the light short X moves right, each on a fiber from its pole
10

The short pair parts normally, one short chromosome to each pole. Nothing else has gone wrong.

11

When a homologous pair, or in meiosis II a pair of sister chromatids, fails to separate and both go to the same pole, we call it . Disjunction is the parting; non- says it did not happen.

12
Check q2

In mitosis, the M checkpoint holds the cell at metaphase.

What does the M checkpoint check before it lets anaphase begin?

  1. A. That the DNA of every chromosome has been copied
    The DNA is copied in S phase, before mitosis begins.
    The M checkpoint checks the attachment of every chromosome to the spindle.
  2. B. That the cell has grown to twice its size
    Growth is checked before S phase, at the G1 checkpoint.
    The M checkpoint checks the attachment of every chromosome to the spindle.
  3. C. ✓ That every chromosome is attached to fibers from both poles

Why: The M checkpoint checks that every chromosome is attached to spindle fibers from both poles.
Only then does anaphase begin.

13

In meiosis I the M checkpoint holds the cell until every homologous pair is attached, one member to each pole. When that hold fails, the unparted pair goes through.

14

The cell that receives both members of the pair has one chromosome too many. The cell at the other pole has one chromosome too few.

Two cells after a failed meiosis I: the left cell holds both long X's, one dark and one light, and the dark short X, count 3; the right cell holds the light short X only, count 1
Two cells after a failed meiosis I: the left cell holds both long X's, one dark and one light, and the dark short X, count 3; the right cell holds the light short X only, count 1
15

A haploid cell holds exactly one of every chromosome. So neither cell is haploid any more.

16

Here one pair fails while every other pair separates normally. Two pairs can fail in one meiosis, but that is rare.

17

What you are expected to know Describe nondisjunction: a homologous pair, or a pair of sister chromatids, fails to separate, so one cell gains a chromosome and another loses one.

18
Check q3

A cell is at anaphase I. Both members of one homologous pair move to the left pole, and every other pair parts normally.

Which of the following does the cell at the left pole receive?

  1. A. One chromosome of every pair, the failed pair included
    Both members of the failed pair went to the left pole.
    So the left cell holds that pair’s two chromosomes as well as one of every other pair.
  2. B. ✓ Both chromosomes of the failed pair and one of every other pair
  3. C. Two chromosomes of every pair, the failed pair included
    Only one pair failed.
    Every other pair parted, one chromosome to each pole.

Why: Every normal pair sent one chromosome to the left pole.
The failed pair sent both.
So the left cell receives both chromosomes of the failed pair and one of every other pair.

19
Practice writing an answer

A cell is at anaphase I. Both members of one homologous pair move to the left pole, and every other pair parts normally. Meiosis I then finishes and the cell divides in two.

(a) Describe how the chromosome count of each of the two cells compares with the normal haploid count, and explain why. (1 pt)

Model answer Both cells hold the wrong number of chromosomes because a haploid cell holds exactly one chromosome of every pair.
Both members of the failed pair went to the left pole.
So the left cell holds two chromosomes of that pair: one too many.
The right cell received no chromosome of that pair: one too few.
So each cell holds a number that is one away from a full set.
Rubric
  • Award 1 point for: the left cell holds both chromosomes of the failed pair (one too many) and the right cell holds none of it (one too few), so neither holds exactly one chromosome of every pair.
20
Check q4

A student says: “When both members of one pair go to the same pole in meiosis I, every other pair in that cell does the same.”

Is the student correct?

  1. A. Yes
    Nondisjunction is one pair’s failure.
    The other pairs separate normally, one member to each pole.
  2. B. ✓ No

Why: Nondisjunction is one pair’s failure to separate.
The rest of the chromosomes separate normally.

21Quick quiz: nondisjunction mixed practice

22
Check q5

A cell is dividing by meiosis.

What is nondisjunction?

  1. A. Two non-sister chromatids exchanging pieces at a chiasma in prophase I
    Two non-sister chromatids exchanging pieces is crossing over.
    Nondisjunction is a pair that fails to separate.
  2. B. A homologous pair facing either pole at metaphase I, on its own
    A pair facing either pole on its own is independent orientation.
    Nondisjunction is a pair that fails to separate.
  3. C. ✓ A pair of homologs, or of sister chromatids, failing to separate, both going to one pole

Why: Nondisjunction is a homologous pair, or a pair of sister chromatids, failing to separate.
Both go to the same pole, so one cell gains a chromosome and another loses one.

23
Practice writing an answer

A cell is dividing by meiosis.

(a) State what nondisjunction is. (1 pt)

Model answer Nondisjunction is the failure of a homologous pair, in meiosis I, or of a pair of sister chromatids, in meiosis II, to separate, so that both go to the same pole.
Rubric
  • Award 1 point for: a homologous pair (meiosis I) or a pair of sister chromatids (meiosis II) fails to separate and both go to the same pole. Accept with or without: one cell gains a chromosome and another loses one.
24
Check q6

A cell from a lily’s anther, the part of the flower that makes pollen, is drawn below at anaphase I. Only some of its chromosomes are drawn.

A cell with a pole at each end; four X shapes: one long X left of the middle and one long X right of the middle, each with a line from the pole on its side to its center dot; two short X's both on the left half, one low near the outline and one nearer the middle, each with a line from the left pole to its center dot
A cell with a pole at each end; four X shapes: one long X left of the middle and one long X right of the middle, each with a line from the pole on its side to its center dot; two short X's both on the left half, one low near the outline and one nearer the middle, each with a line from the left pole to its center dot

Has nondisjunction happened in this cell?

  1. A. ✓ Yes
  2. B. No
    Both members of the short pair are moving to the left pole.
    A pair whose two members go to one pole has failed to separate: nondisjunction.

Why: The long pair has parted, one X to each pole.
Both short X’s are moving to the left pole, each on a fiber from that pole.
A pair whose two members go to one pole has failed to separate: nondisjunction.

25
Check q7

A cell from a grasshopper’s testis is drawn below at anaphase I. Only some of its chromosomes are drawn.

A cell with a pole at each end; four X shapes in two columns of two, one column left of the middle and one right of the middle, a white dot at the crossing of each X and a line from the pole on its side to that dot; in each column one X is long and one short
A cell with a pole at each end; four X shapes in two columns of two, one column left of the middle and one right of the middle, a white dot at the crossing of each X and a line from the pole on its side to that dot; in each column one X is long and one short

Has nondisjunction happened in this cell?

  1. A. Yes
    Each pair has parted, one member to each pole.
    Nondisjunction is a pair whose two members go to one pole, and no pair here does.
  2. B. ✓ No

Why: The two long X’s are moving to opposite poles, and so are the two short X’s.
Every pair has parted, one member to each pole.
So no nondisjunction has happened.

26
Check q8

A cell from a mouse is drawn below, part way through meiosis II. Only some of its chromosomes are drawn.

A small cell with a pole at each end; one long X shape left of the middle with a line from the left pole to its center dot; two short V shapes lower down, one with its point toward the left pole and one with its point toward the right pole, each with a line from the pole on its side to its point
A small cell with a pole at each end; one long X shape left of the middle with a line from the left pole to its center dot; two short V shapes lower down, one with its point toward the left pole and one with its point toward the right pole, each with a line from the pole on its side to its point

Has nondisjunction happened in this cell?

  1. A. ✓ Yes
  2. B. No
    One chromosome is moving whole to the left pole, both its chromatids still joined.
    Two sister chromatids that go to one pole together have failed to separate: nondisjunction.

Why: The short chromosome’s two chromatids have parted and move as V’s to opposite poles.
The long chromosome moves whole to the left pole, both its chromatids still joined.
Two sister chromatids going to one pole together is nondisjunction.

27
Check q9

A cell from a rye plant’s anther, the part of the flower that makes pollen, is drawn below at anaphase I. Only some of its chromosomes are drawn.

A cell with a pole at each end; six X shapes: a long X and a short X on the left half and a long X and a short X on the right half, each with a line from the pole on its side to its center dot; two middle-length X's both on the left half, one below the middle line and one above it, each with a line from the left pole to its center dot
A cell with a pole at each end; six X shapes: a long X and a short X on the left half and a long X and a short X on the right half, each with a line from the pole on its side to its center dot; two middle-length X's both on the left half, one below the middle line and one above it, each with a line from the left pole to its center dot

Has nondisjunction happened in this cell?

  1. A. ✓ Yes
  2. B. No
    Both members of the middle-length pair are moving to the left pole.
    A pair whose two members go to one pole has failed to separate: nondisjunction.

Why: The long pair and the short pair have each parted, one X to each pole.
Both middle-length X’s are moving to the left pole, each on a fiber from that pole.
A pair whose two members go to one pole has failed to separate: nondisjunction.

28
Check q10

A cell from a newt is drawn below, part way through meiosis II. Only some of its chromosomes are drawn.

A small cell with a pole at each end; four V shapes in two columns of two, the left column with its points toward the left pole and the right column with its points toward the right pole, a white dot at each point and a line from the pole on its side to that dot; in each column one V is long and one short
A small cell with a pole at each end; four V shapes in two columns of two, the left column with its points toward the left pole and the right column with its points toward the right pole, a white dot at each point and a line from the pole on its side to that dot; in each column one V is long and one short

Has nondisjunction happened in this cell?

  1. A. Yes
    The two chromatids of each chromosome are moving to opposite poles as V’s.
    Nondisjunction is two sister chromatids going to one pole together, and none here do.
  2. B. ✓ No

Why: Each chromosome’s two sister chromatids have parted and move as V’s to opposite poles.
Every pair of sister chromatids has separated.
So no nondisjunction has happened.

29The four cells after a failure in meiosis I

30

Video: Watch: Two over, two under

The two cells of the failed meiosis I each dividing again, the four finished cells drawn with their counts appearing under them: 3, 3, 1, 1; then the human 24, 24, 22, 22.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11b.mp4

31
Check q11

One of the two cells from meiosis I begins meiosis II.

Which partners does the spindle pull apart at anaphase II?

  1. A. The two members of each homologous pair
    The homologous pairs parted at anaphase I.
    A cell in meiosis II has no pairs left to part.
  2. B. ✓ The two sister chromatids of each chromosome

Why: At anaphase II the connection at each centromere breaks.
The spindle pulls the two sister chromatids of each chromosome to opposite poles.

32

Follow the failure in meiosis I. The left cell received both long chromosomes and one short: 3 chromosomes, one too many.

33

The right cell received one short chromosome and no long: 1 chromosome, one too few.

34

Meiosis II then parts the sister chromatids normally in both cells. The left cell gives two cells of 3, and the right cell gives two cells of 1.

Four finished cells in a row with counts 3, 3, 1, 1: the first two each hold two long rods, one dark and one light, and one dark short rod; the last two each hold one light short rod
Four finished cells in a row with counts 3, 3, 1, 1: the first two each hold two long rods, one dark and one light, and one dark short rod; the last two each hold one light short rod
35

All four cells are abnormal: n+1, n+1, n−1, n−1.

36

In a human, n=23. A failure in meiosis I gives four gametes of 24, 24, 22 and 22 chromosomes.

37

A failure never gives one gamete with an extra chromosome on its own. Where one cell gained a chromosome, the cell at the other pole lost it.

38

What you are expected to know Predict the chromosome counts of the four cells when one homologous pair fails to separate in meiosis I.

39
Check q12

A plant’s body cells hold 14 chromosomes. In one cell, one homologous pair undergoes nondisjunction in meiosis I, and meiosis II is normal.

How many of the four cells hold the normal count of 7?

  1. A. ✓ None
  2. B. Two
    A failure in meiosis I sends both members of the pair into one cell and none into the other.
    Meiosis II then passes that fault to all four cells.
  3. C. Four
    The pair failed, so one cell of meiosis I is over and the other is under.
    Meiosis II passes those counts to all four cells.

Why: Both members of the pair went into one of the two cells of meiosis I, and none into the other.
Each of those two cells passes its count to two cells in meiosis II.
So all four cells are over or under, and none is normal.

40
Check q13 numeric entry

An insect’s body cells hold 18 chromosomes. In one cell, one homologous pair undergoes nondisjunction in meiosis I; in meiosis II the sister chromatids part normally.

Calculate the chromosome count of the cell that received both members of the pair after meiosis I.

Part 1. What is the normal haploid count, n, for this insect?

Answer: 9  (tolerance ±0)

Working
Halve the body cell’s count:
2n=18
n=9

Part 2. How many chromosomes does the cell at the other pole hold?

Answer: 8  (tolerance ±0)

Working
Take the missing member from the haploid count:
n−1=9−1=8

Answer: 10  (tolerance ±0)

Working
Write down the values in the question:
2n=18
n=9
the failure is in meiosis I
Write down the equation:
cell that received both members=n+1
cell at the other pole=n−1
Substitute the values into the equation:
n+1=9+1=10
n−1=9−1=8
the four gametes: 10, 10, 8, 8
41
Check q14

An animal’s body cells hold 20 chromosomes. In one cell, one homologous pair undergoes nondisjunction in meiosis I, and meiosis II is normal.

Which of the following are the chromosome counts of the four cells produced?

  1. A. 10, 10, 10 and 10
    One pair failed, so at least one cell is over and one is under.
  2. B. 11, 9, 10 and 10
    A failure in meiosis I reaches all four cells.
    Two normal cells would need a normal meiosis I.
  3. C. ✓ 11, 11, 9 and 9
  4. D. 11, 11, 11 and 9
    The cell that gained a chromosome has a partner cell that lost it.
    Two 11s need two 9s.

Why: The haploid count is 10.
The cell that received both members of the pair holds 11, and the cell at the other pole holds 9.
Meiosis II passes each count to two cells.
So the four cells hold 11, 11, 9 and 9; the working is below.

42The four cells after a failure in meiosis II

43

Video: Watch: One over, one under, two normal

A normal meiosis I, then one of the two cells failing at anaphase II, the whole X moving to one pole; the four finished cells with their counts appearing: 3, 1, 2, 2; then the human 24, 22, 23, 23.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11c.mp4

44

Now consider a failure in meiosis II instead. Meiosis I is normal, so each of the two cells holds one long and one short chromosome.

45

In one of the two cells, the two sister chromatids of the long chromosome fail to part at anaphase II. The whole X moves to one pole.

Two small cells side by side at anaphase II. In the left cell the long chromosome is still a whole X moving toward the left pole on one fiber from that pole, while the short chromosome's two chromatids part as V shapes toward the two poles. In the right cell the two chromatids of each chromosome part as V shapes toward the two poles
Two small cells side by side at anaphase II. In the left cell the long chromosome is still a whole X moving toward the left pole on one fiber from that pole, while the short chromosome's two chromatids part as V shapes toward the two poles. In the right cell the two chromatids of each chromosome part as V shapes toward the two poles
46

That cell gives one cell with both long chromatids and a short: 3 chromosomes, one too many. It gives one cell with a short chromosome only: 1 chromosome, one too few.

47

The other cell divides normally into two cells of 2.

Four finished cells in a row with counts 3, 1, 2, 2: the first holds two dark long rods and one light short rod, the second one light short rod, the third and fourth one light long rod and one dark short rod each
Four finished cells in a row with counts 3, 1, 2, 2: the first holds two dark long rods and one light short rod, the second one light short rod, the third and fourth one light long rod and one dark short rod each
48

So the four cells hold n+1, n−1, n, n: two abnormal, two normal.

49

Here is a table comparing the two failures in a human, where n=23: the division that failed, and the four gametes’ counts.

A table: a failure in meiosis I gives four human gametes of 24, 24, 22 and 22 chromosomes; a failure in meiosis II gives 24, 22, 23 and 23
A table: a failure in meiosis I gives four human gametes of 24, 24, 22 and 22 chromosomes; a failure in meiosis II gives 24, 22, 23 and 23
50

What you are expected to know Predict the chromosome counts of the four cells when one pair of sister chromatids fails to separate in meiosis II.

51
Check q15

An animal’s body cells hold 16 chromosomes. In one cell, meiosis I is normal. Then, in one of the two cells, the sister chromatids of one chromosome undergo nondisjunction in meiosis II.

How many of the four cells hold the normal count of 8?

  1. A. None
    The other cell of meiosis I divides normally.
    It gives two cells with the normal count.
  2. B. ✓ Two
  3. C. Four
    One cell failed in meiosis II.
    It gives one cell over and one cell under.

Why: Meiosis I was normal, so both of its cells held the normal count of 8.
One of them failed in meiosis II and gave one cell of 9 and one of 7.
The other divided normally and gave two cells of 8.

52
Check q16

A plant’s body cells hold 12 chromosomes. In one cell, meiosis I is normal. Then, in one of the two cells, the sister chromatids of one chromosome undergo nondisjunction in meiosis II.

Which of the following are the chromosome counts of the four cells produced?

  1. A. 6, 6, 6 and 6
    One chromosome’s chromatids failed to part, so one cell is over and one is under.
  2. B. ✓ 7, 5, 6 and 6
  3. C. 7, 6, 6 and 6
    The cell that gained a chromosome is matched by the cell at the other pole, which lost it.
    So a 7 is always matched by a 5.
  4. D. 7, 7, 5 and 5
    A failure in meiosis I would leave all four cells abnormal.
    Here meiosis I was normal, so the other cell gives two normal cells.

Why: The haploid count is 6.
The cell whose chromatids failed to part gives one cell of 7 and one of 5.
The other cell from meiosis I gives two normal cells of 6.
So the four cells hold 7, 5, 6 and 6; the working is below.

53

Back to the model cell at anaphase I: both long chromosomes moving to the left pole, each on a fiber from that pole, while the short pair parts normally.

The model cell at anaphase I: both long X-shaped chromosomes, one dark and one light, move toward the left pole, each on a fiber from that pole; the dark short X moves left and the light short X moves right, each on a fiber from its pole
The model cell at anaphase I: both long X-shaped chromosomes, one dark and one light, move toward the left pole, each on a fiber from that pole; the dark short X moves left and the light short X moves right, each on a fiber from its pole
54

That failure gave two cells with 3 chromosomes and two cells with 1: two over, two under.

55

Had the failure come in meiosis II instead, in one of the two cells, one cell would be over, one under, and two normal.

56Quick quiz: which cells are abnormal? mixed practice

57
Check q17

A plant’s body cells hold 4 chromosomes. The four cells from one meiosis are drawn below, with each cell’s chromosome count written under it.

Four cells in a row with the counts 3, 3, 1 and 1 written under them; the first two hold three rods each, the last two one rod each
Four cells in a row with the counts 3, 3, 1 and 1 written under them; the first two hold three rods each, the last two one rod each

Which of the four cells are abnormal?

  1. A. None of the four cells
    A normal cell from this meiosis holds 2 chromosomes.
    No cell here holds 2.
  2. B. The two cells with 3
    The two cells with 1 chromosome are one under each.
    A normal cell holds 2, so they are abnormal too.
  3. C. ✓ All four of the cells

Why: A body cell holds 4 chromosomes, so a normal cell from meiosis holds 2.
The two cells with 3 are one over each, and the two cells with 1 are one under each.
So all four cells are abnormal.

58
Check q18

A plant’s body cells hold 6 chromosomes. The four cells from one meiosis are drawn below, with each cell’s chromosome count written under it.

Four cells in a row with the counts 4, 4, 2 and 2 written under them; the first two hold four rods each, the last two hold two rods each
Four cells in a row with the counts 4, 4, 2 and 2 written under them; the first two hold four rods each, the last two hold two rods each

Which of the four cells are abnormal?

  1. A. None of the four cells
    A normal cell from this meiosis holds 3 chromosomes.
    No cell here holds 3.
  2. B. The two cells with 4
    The two cells with 2 chromosomes are one under each.
    A normal cell holds 3, so they are abnormal too.
  3. C. ✓ All four of the cells

Why: A body cell holds 6 chromosomes, so a normal cell from meiosis holds 3.
The two cells with 4 are one over each, and the two cells with 2 are one under each.
So all four cells are abnormal.

59
Check q19

An insect’s body cells hold 6 chromosomes. The four cells from one meiosis are drawn below, with each cell’s chromosome count written under it.

Four cells in a row with the counts 3, 3, 3 and 3 written under them; each holds three rods of three different lengths
Four cells in a row with the counts 3, 3, 3 and 3 written under them; each holds three rods of three different lengths

Which of the four cells are abnormal?

  1. A. ✓ None of the four cells
  2. B. Two of the four cells
    A normal cell from this meiosis holds 3 chromosomes.
    Every cell here holds 3.
  3. C. All four of the cells
    A normal cell from this meiosis holds 3 chromosomes.
    Every cell here holds 3.

Why: A body cell holds 6 chromosomes, so a normal cell from meiosis holds 3.
Every one of the four cells holds 3.
So no cell is abnormal.

60
Check q20

An animal’s body cells hold 4 chromosomes. The four cells from one meiosis are drawn below, with each cell’s chromosome count written under it.

Four cells in a row with the counts 2, 2, 3 and 1 written under them; the first two hold two rods each, the third three rods, the fourth one rod
Four cells in a row with the counts 2, 2, 3 and 1 written under them; the first two hold two rods each, the third three rods, the fourth one rod

Which of the four cells are abnormal?

  1. A. None of the four cells
    A normal cell from this meiosis holds 2 chromosomes.
    The cell with 3 and the cell with 1 do not.
  2. B. ✓ The cells with 3 and with 1
  3. C. All four of the cells
    A normal cell from this meiosis holds 2 chromosomes.
    Two of the four cells hold exactly 2.

Why: A body cell holds 4 chromosomes, so a normal cell from meiosis holds 2.
The cell with 3 is one over and the cell with 1 is one under.
The two cells with 2 are normal.

61
Check q21

An animal’s body cells hold 6 chromosomes. The four cells from one meiosis are drawn below, with each cell’s chromosome count written under it.

Four cells in a row with the counts 3, 4, 3 and 2 written under them; the first and third hold three rods each, the second four rods, the fourth two rods
Four cells in a row with the counts 3, 4, 3 and 2 written under them; the first and third hold three rods each, the second four rods, the fourth two rods

Which of the four cells are abnormal?

  1. A. None of the four cells
    A normal cell from this meiosis holds 3 chromosomes.
    The cell with 4 and the cell with 2 do not.
  2. B. ✓ The cells with 4 and with 2
  3. C. All four of the cells
    A normal cell from this meiosis holds 3 chromosomes.
    Two of the four cells hold exactly 3.

Why: A body cell holds 6 chromosomes, so a normal cell from meiosis holds 3.
The cell with 4 is one over and the cell with 2 is one under.
The two cells with 3 are normal.

62Mixed practice mixed practice

63
Check q22

A plant cell is drawn below, part way through meiosis. Only some of its chromosomes are drawn.

A small cell with a pole at each end; one long X shape right of the middle with a line from the right pole to its center dot; two short V shapes lower down, one with its point toward the left pole and one with its point toward the right pole, each with a line from the pole on its side to its point
A small cell with a pole at each end; one long X shape right of the middle with a line from the right pole to its center dot; two short V shapes lower down, one with its point toward the left pole and one with its point toward the right pole, each with a line from the pole on its side to its point

Which of the following has happened in this cell?

  1. A. Crossing over
    Crossing over happens in prophase I, while the homologs are paired.
    Here a whole chromosome, both its chromatids, went to one pole instead of splitting.
  2. B. Independent orientation
    Independent orientation is how a homologous pair sits at metaphase I.
    This cell has no pairs, and a whole chromosome went one way instead of splitting.
  3. C. Nondisjunction in meiosis I
    Chromatids parting as V’s toward the poles is anaphase II.
    The homologs parted earlier, in meiosis I.
  4. D. ✓ Nondisjunction in meiosis II

Why: The short chromosome’s two chromatids have parted and move as V’s to opposite poles: anaphase II.
The long chromosome moved whole toward one pole, so both its chromatids went one way.
That is nondisjunction in meiosis II.

64
Check q23

In one cell, both members of one homologous pair go to the left pole at anaphase I, and every other pair parts normally. Meiosis I finishes and the cell divides in two.

Which of the two cells is haploid?

  1. A. The cell at the left pole only
    The left cell holds both chromosomes of the failed pair: one too many.
    A haploid cell holds exactly one of every chromosome.
  2. B. The cell at the right pole only
    The right cell holds no chromosome of the failed pair: one too few.
    A haploid cell holds exactly one of every chromosome.
  3. C. ✓ Neither cell

Why: A haploid cell holds exactly one of every chromosome.
The left cell holds two of the failed pair, and the right cell holds none of it.
So neither cell is haploid.

65
Check q24

An insect’s body cells hold 6 chromosomes. In one cell, meiosis I is normal; then, in one of the two cells, the sister chromatids of one chromosome undergo nondisjunction in meiosis II.

Which of the following are the chromosome counts of the four cells produced?

  1. A. 3, 3, 3 and 3
    One cell did fail, so its two cells are one over and one under.
  2. B. 4, 3, 3 and 3
    The cell that gained a chromosome is matched by the cell at the other pole, which lost it.
    So a 4 is always matched by a 2.
  3. C. 4, 4, 2 and 2
    Here meiosis I was normal, so the other cell’s two cells are normal.
  4. D. ✓ 4, 2, 3 and 3

Why: The haploid count for this insect is 3.
The cell whose chromatids failed to part gives one cell of 4 and one of 2.
The other cell from meiosis I gives two normal cells of 3.
So the four cells hold 4, 2, 3 and 3; the working is below.

66
Check q25

A student says: “When the two sister chromatids of one chromosome undergo nondisjunction in meiosis II, all four cells end up abnormal.”

Is the student correct?

  1. A. ✓ No
  2. B. Yes
    A failure in meiosis II happens in one of the two cells.
    The other cell divides normally and gives two normal cells.

Why: Meiosis I was normal, so both of its cells held the normal count.
Only one of them fails in meiosis II, giving one cell over and one under.
The other gives two normal cells.

67
Check q26

One of the four cells from a meiosis in a plant holds one chromosome too many.

Which of the following must be true of the other three cells?

  1. A. ✓ At least one of them holds one chromosome too few
  2. B. All three hold the normal count
    Where one cell gained a chromosome, the cell at the other pole lost it.
    So at least one other cell is one under.
  3. C. Two of them hold one chromosome too many
    A cell over is always matched by a cell under.
    Two cells over come from a failure in meiosis I, and then two cells are under too.

Why: In a failure to separate, both partners go to one pole.
The cell at that pole gains one chromosome, and the cell at the other pole loses it.
So a cell with one too many always has a partner cell with one too few.

68
Check q27

A cell is in meiosis II, and nondisjunction happens at anaphase II.

Which partners went to the same pole?

  1. A. The two homologs of one pair
    The homologs of every pair parted in meiosis I.
    A cell in meiosis II has no pairs, so its only partners are sister chromatids.
  2. B. ✓ The two sister chromatids of one chromosome

Why: In meiosis II each chromosome is two sister chromatids joined at one centromere.
At anaphase II those two sister chromatids are the partners that part.
So a nondisjunction at anaphase II sends two sister chromatids to the same pole.

69
Practice writing an answer

In the plant cell drawn below, one chromosome moved whole to one pole while the other chromosome’s two chromatids parted.

A small cell with a pole at each end; one long X shape right of the middle with a line from the right pole to its center dot; two short V shapes lower down, one with its point toward the left pole and one with its point toward the right pole, each with a line from the pole on its side to its point
A small cell with a pole at each end; one long X shape right of the middle with a line from the right pole to its center dot; two short V shapes lower down, one with its point toward the left pole and one with its point toward the right pole, each with a line from the pole on its side to its point

(a) Explain why the two partners that failed to part in this cell are sister chromatids. (1 pt)

Frame The partners that failed to part are sister chromatids because …

Model answer The partners that failed to part are sister chromatids because this cell is in meiosis II.
In meiosis I the two homologs of every pair went to opposite poles, so this cell holds no pairs.
Each chromosome in this cell is still two sister chromatids joined at one centromere.
At anaphase II those two sister chromatids are the partners that part.
So the whole chromosome that moved to one pole is two sister chromatids that failed to part.
Rubric
  • Award 1 point for: the cell is in meiosis II, so the homologs have already parted and the only partners left to separate are the two sister chromatids of one chromosome.

Glossary

nondisjunction
The failure of a homologous pair (in meiosis I) or of a pair of sister chromatids (in meiosis II) to separate, so both go to the same pole: one cell gains a chromosome and another loses one, and the gametes are no longer haploid.

APBIO-U05-L11B Which division failed?

Topic 5.2 · Meiosis and Genetic Diversity · 28 steps

Four circles in a row, each a cell, with the counts 24, 22, 23 and 23 written beneath them
Four circles in a row, each a cell, with the counts 24, 22, 23 and 23 written beneath them

Here are the four cells one meiosis produced in a human, with their chromosome counts written under them: 24, 22, 23 and 23. Somewhere, one pair failed to separate.

Which division did it fail in, and how do the counts tell you?

Unit 5 · Heredity

1Count the normal cells

2

Video: Watch: Which division failed?

The 24/22/23/23 and 24/24/22/22 patterns side by side, the normal cells marked as they are counted, and each pattern traced back to the division that made it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11Ba.mp4

3
Check q1

In one cell of an animal, one homologous pair fails to separate in meiosis I. Meiosis II is normal.

How many of the four cells hold the normal chromosome count?

  1. A. ✓ None
  2. B. Two
    A failure in meiosis I sends both members of the pair into one cell and none into the other.
    Meiosis II then passes that fault to all four cells.
  3. C. Four
    The pair failed, so at least one cell is over and one is under.
    Here the fault reaches all four cells.

Why: Both members of the pair go into one of the two cells of meiosis I, and none into the other.
Each of those two cells passes its count to two cells in meiosis II.
So all four cells are over or under, and none is normal.

4

How do you read a set of four cells backwards to the division that failed? Count the cells with the normal number.

5

Here is a table comparing the two failures: the division that failed, the four cells’ counts, and how many of the cells are normal.

A table: a failure in meiosis I gives n plus 1, n plus 1, n minus 1, n minus 1, no cell normal; a failure in meiosis II gives n plus 1, n minus 1, n, n, two cells normal
A table: a failure in meiosis I gives n plus 1, n plus 1, n minus 1, n minus 1, no cell normal; a failure in meiosis II gives n plus 1, n minus 1, n, n, two cells normal
6

If none of the four cells is normal, the pair failed in meiosis I. A failure in meiosis I reaches every cell.

7

If two of the four cells are normal, the pair failed in meiosis II. A failure in meiosis II touches only one of the two cells.

8

Now consider 24, 22, 23 and 23 in a human, where the normal count is 23. Two cells are normal.

9

So meiosis I parted every pair correctly. One of the two cells then failed in meiosis II, giving the 24 and the 22.

Four cells with counts 24, 22, 23 and 23; the two cells of 23 are marked normal, the 24 one over and the 22 one under; a caption reads two normal cells: meiosis II
Four cells with counts 24, 22, 23 and 23; the two cells of 23 are marked normal, the 24 one over and the 22 one under; a caption reads two normal cells: meiosis II
10

Now consider 24, 24, 22 and 22 in a human. No cell is normal.

11

So both members of one pair went to one cell in meiosis I. Meiosis II then passed that error on to two cells each way.

Four cells with counts 24, 24, 22 and 22, the first two marked one over and the last two one under; a caption reads no normal cell: meiosis I
Four cells with counts 24, 24, 22 and 22, the first two marked one over and the last two one under; a caption reads no normal cell: meiosis I
12

Whichever division failed, the counts balance: for every chromosome one cell gained, another cell lost one.

13

So a set with a cell over and no cell under cannot come from one failure.

14

What you are expected to know Identify, from the four cells’ chromosome counts, whether the pair failed in meiosis I or in meiosis II.

15
Check q2

A horse’s body cells hold 64 chromosomes. The four cells from one meiosis are drawn below with their counts.

Four cells with counts 33, 33, 31 and 31 written beneath them, and no other marks
Four cells with counts 33, 33, 31 and 31 written beneath them, and no other marks

Which division did the failure happen in?

  1. A. ✓ Meiosis I
  2. B. Meiosis II
    A failure in meiosis II leaves two cells with the normal count.
    Here no cell has the normal count of 32: two are over and two are under.

Why: The normal count is 32. No cell has it: 33, 33, 31 and 31 are all abnormal.
So both cells of meiosis I were already wrong, and the pair failed in meiosis I.

16
Practice writing an answer

A horse’s body cells hold 64 chromosomes. The four cells from one meiosis hold 33, 33, 31 and 31.

(a) Identify the division in which the failure happened, and justify the answer using the four counts. (1 pt)

Model answer The counts show that the failure was in meiosis I because the normal count is 32, and no cell has it.
A failure in meiosis II leaves two cells with the normal count, because the other cell of meiosis I divides normally.
Here every cell is over or under.
So both cells of meiosis I were already wrong.
Therefore the pair failed in meiosis I.
Rubric
  • Award 1 point for: no cell has the normal count of 32, so both products of the first division were already wrong; a failure in meiosis II leaves two normal cells.
17
Check q3

An animal’s body cells hold 20 chromosomes. A student reports that the four cells from one meiosis, with one failure to separate, hold 11, 10, 10 and 10.

Which of the following is wrong with the report?

  1. A. The four counts should add up to 20, the number in one body cell
    Four cells of 10 hold 40 chromosomes between them, twice the body cell’s 20, whether or not a failure happened.
  2. B. A failure changes a count by two, so the over cell should hold 12
    A failure moves one chromosome to the wrong pole.
    So the over cell holds one extra, 11, and the under cell one fewer, 9.
  3. C. Two cells with the normal count of 10 can never follow a failure to separate
    A failure in meiosis II leaves two normal cells.
    What a failure never leaves is a cell over with no cell under.
  4. D. ✓ One cell is over with no cell under, yet what one cell gains another loses

Why: Where one cell gained a chromosome, the cell at the other pole lost it.
So an 11 must be matched by a 9.
The set 11, 10, 10, 10 has a cell over and no cell under, so it cannot come from one failure.

18

Back to the four human cells from one meiosis: 24, 22, 23 and 23 chromosomes. Two cells hold the normal 23, so meiosis I parted every pair correctly.

Four cells with counts 24, 22, 23 and 23; the two cells of 23 are marked normal, the 24 one over and the 22 one under; a caption reads two normal cells: meiosis II
Four cells with counts 24, 22, 23 and 23; the two cells of 23 are marked normal, the 24 one over and the 22 one under; a caption reads two normal cells: meiosis II
19

One of the two cells then failed in meiosis II. That cell gave the 24 and the 22: one over, one under.

20Quick quiz: which division failed? mixed practice

21
Check q4

A plant’s body cells hold 16 chromosomes. The four cells from one meiosis are drawn below with their counts.

Four cells with counts 9, 7, 8 and 8 written beneath them, and no other marks
Four cells with counts 9, 7, 8 and 8 written beneath them, and no other marks

Which division did the failure happen in?

  1. A. Meiosis I
    A failure in meiosis I leaves no cell with the normal count.
    Here two cells have the normal count of 8.
  2. B. ✓ Meiosis II
  3. C. Neither: the counts cannot happen
    The counts balance: the cell that gained a chromosome, 9, is matched by the cell that lost one, 7.

Why: The normal count is 8. Two cells have it.
So meiosis I parted every pair, and one of the two cells then failed in meiosis II, giving the 9 and the 7.

22
Check q5

An animal’s body cells hold 22 chromosomes. The four cells from one meiosis are drawn below with their counts.

Four cells with counts 12, 11, 11 and 11 written beneath them, and no other marks
Four cells with counts 12, 11, 11 and 11 written beneath them, and no other marks

Which division did the failure happen in?

  1. A. Meiosis I
    A failure in meiosis I leaves no cell with the normal count of 11.
    Here three cells have it.
  2. B. Meiosis II
    A failure in meiosis II leaves one cell over and one cell under.
    Here one cell is over and no cell is under.
  3. C. ✓ Neither: the counts cannot happen

Why: One cell is over, 12, and no cell is under.
But a chromosome one cell gains is lost by the cell at the other pole.
So 12, 11, 11, 11 cannot come from one failure.

23
Check q6

An animal’s body cells hold 26 chromosomes. The four cells from one meiosis are drawn below with their counts.

Four cells with counts 14, 14, 12 and 12 written beneath them, and no other marks
Four cells with counts 14, 14, 12 and 12 written beneath them, and no other marks

Which division did the failure happen in?

  1. A. ✓ Meiosis I
  2. B. Meiosis II
    A failure in meiosis II leaves two cells with the normal count.
    Here no cell has the normal count of 13.
  3. C. Neither: the counts cannot happen
    The counts balance: the two cells that gained a chromosome, 14, are matched by the two that lost one, 12.

Why: The normal count is 13. No cell has it: two are over and two are under.
So the pair failed in meiosis I.

24
Check q7

A plant’s body cells hold 10 chromosomes. The four cells from one meiosis are drawn below with their counts.

Four cells with counts 6, 4, 5 and 5 written beneath them, and no other marks
Four cells with counts 6, 4, 5 and 5 written beneath them, and no other marks

Which division did the failure happen in?

  1. A. Meiosis I
    A failure in meiosis I leaves no cell with the normal count.
    Here two cells have the normal count of 5.
  2. B. ✓ Meiosis II
  3. C. Neither: the counts cannot happen
    The counts balance: the cell that gained a chromosome, 6, is matched by the cell that lost one, 4.

Why: The normal count is 5. Two cells have it.
So meiosis I parted every pair, and one of the two cells then failed in meiosis II, giving the 6 and the 4.

25
Check q8

An animal’s body cells hold 28 chromosomes. The four cells from one meiosis are drawn below with their counts.

Four cells with counts 15, 13, 14 and 14 written beneath them, and no other marks
Four cells with counts 15, 13, 14 and 14 written beneath them, and no other marks

Which division did the failure happen in?

  1. A. Meiosis I
    A failure in meiosis I leaves no cell with the normal count.
    Here two cells have the normal count of 14.
  2. B. ✓ Meiosis II
  3. C. Neither: the counts cannot happen
    The counts balance: the cell that gained a chromosome, 15, is matched by the cell that lost one, 13.

Why: The normal count is 14. Two cells have it.
So meiosis I parted every pair, and one of the two cells then failed in meiosis II, giving the 15 and the 13.

26Mixed practice mixed practice

27
Practice writing an answer

A human body cell holds 46 chromosomes. The four cells one meiosis produced are drawn below with their chromosome counts.

Four cells with counts 24, 22, 23 and 23 written beneath them, and no other marks
Four cells with counts 24, 22, 23 and 23 written beneath them, and no other marks

(a) Determine the division in which a pair failed to separate, using the counts. (1 pt)

Model answer The failure was in meiosis II.
Two of the four cells have the normal count of 23.
So meiosis I parted every pair correctly, and both cells it made were normal.
Only one of those two cells then failed, in meiosis II, giving the 24 and the 22.
Rubric
  • Award 1 point for: the decision (meiosis II) AND the ground (two cells are normal, 23, which a failure in meiosis I could not leave).

Slip Choosing meiosis I because there is a 24 and a 22. Both patterns have a cell over and a cell under; the two normal cells are what place the failure.

(b) Describe what happened at anaphase in the division that failed. (1 pt)

Model answer At anaphase II, in one of the two cells, the two sister chromatids of one chromosome did not separate.
Both went to the same pole.
So one cell received an extra chromosome and the other received none of that chromosome.
Rubric
  • Award 1 point for: at anaphase II the two sister chromatids of one chromosome went to the same pole (nondisjunction).

Slip Saying two homologs failed to part. Homologs part in meiosis I; in meiosis II the partners are sister chromatids.

(c) Predict the four counts for a human cell in which both members of one homologous pair go to the same pole in meiosis I, and meiosis II is normal. (1 pt)

Model answer 24, 24, 22 and 22.
Both members of one pair go to one cell in meiosis I.
So one cell has 24 chromosomes and the other has 22.
Meiosis II then parts sister chromatids in each cell, so it gives two cells of 24 and two cells of 22.
Rubric
  • Award 1 point for: 24, 24, 22, 22.

Slip Giving 24, 22, 23, 23. That is the meiosis II pattern; a meiosis I failure leaves no normal cell.

(d) Explain why the set 24, 23, 23 and 23 is impossible after one failure to separate. (1 pt)

Model answer In a failure to separate, both partners go to one pole.
So the cell at that pole gains one chromosome, and the cell at the other pole loses that chromosome.
Therefore a 24 must be matched by a 22 from the same division.
The set 24, 23, 23, 23 has a cell over and no cell under, so one failure cannot make it.
Rubric
  • Award 1 point for: the counts must balance, a cell over matched by a cell under, and 24, 23, 23, 23 has no cell under.

Slip Saying the total must be 92 and this set is 93. The total is a consequence of the balance; the reason is that the chromosome gained by one cell was lost by another.

APBIO-U05-L11C Three copies of one chromosome

Topic 5.2 · Meiosis and Genetic Diversity · 61 steps

A large circle labelled egg, 24 chromosomes; a small circle with a tail labelled sperm, 23 chromosomes; an arrow to a larger circle labelled zygote with a question mark
A large circle labelled egg, 24 chromosomes; a small circle with a tail labelled sperm, 23 chromosomes; an arrow to a larger circle labelled zygote with a question mark

Here is an egg with 24 chromosomes, one chromosome too many, about to be fertilized by a sperm with the normal 23.

What will the zygote hold, and what happens to it?

Unit 5 · Heredity

1The zygote’s chromosome count

2

Video: Watch: Three copies of one chromosome

The 24-chromosome egg and the normal sperm fusing into a zygote of 47, the three copies of one chromosome drawn side by side; then the 22-chromosome egg giving 45.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11Ca.mp4

3
Check q1

A human sperm with 23 chromosomes fertilizes an egg with 23 chromosomes.

How many chromosomes does the zygote hold?

  1. A. 23
    Fertilization joins the two sets.
    The zygote holds the sperm’s 23 and the egg’s 23.
  2. B. ✓ 46
  3. C. 92
    Each gamete carries one set of 23, not two.
    So the zygote holds 46.

Why: Fertilization joins the sperm’s set to the egg’s set.
The zygote holds the sperm’s 23 and the egg’s 23: 46 chromosomes.

4

What becomes of a gamete with the wrong chromosome count? Fertilization joins the two gametes’ sets, whatever each set holds.

5

Here the egg holds 24 chromosomes: two copies of one chromosome and one copy of every other. The sperm holds the normal 23.

6

The zygote holds the egg’s 24 and the sperm’s 23: 47 chromosomes.

An egg holding two short rods and one long rod, count 24; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding three short rods and two long rods, count 47
An egg holding two short rods and one long rod, count 24; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding three short rods and two long rods, count 47
7

So the zygote holds two copies of every chromosome and three copies of one.

8

Now consider an egg with 22 chromosomes: no copy of one chromosome. A normal sperm brings one copy of every chromosome.

9

The zygote holds the egg’s 22 and the sperm’s 23: 45 chromosomes. It holds two copies of every chromosome and one copy of one.

An egg holding no short rod and one long rod, count 22; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding one short rod and two long rods, count 45
An egg holding no short rod and one long rod, count 22; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding one short rod and two long rods, count 45
10

A gamete with one chromosome too many makes a zygote with one chromosome too many. A gamete with one chromosome too few makes a zygote with one chromosome too few.

11

What you are expected to know Calculate the zygote’s chromosome count when a gamete with one chromosome too many, or one too few, fuses with a normal gamete.

12
Check q2

A human sperm with 22 chromosomes fertilizes an egg with the normal 23.

Which of the following does the zygote hold?

  1. A. 44 chromosomes
    The egg is normal, so only the sperm is short.
    The zygote is short by exactly one chromosome.
  2. B. ✓ 45 chromosomes
  3. C. 46 chromosomes
    Fertilization joins the two sets as they are.
    Nothing makes up for the sperm’s missing chromosome.
  4. D. 47 chromosomes
    A gamete with one chromosome too few brings the zygote down by one, not up.

Why: The zygote holds the sperm’s 22 and the egg’s 23.
That is 45 chromosomes in all.
Every chromosome is present twice except one, which is present once.

13
Check q3 numeric entry

A dog’s body cells hold 78 chromosomes. An egg with 40 chromosomes is fertilized by a sperm with the normal 39.

Calculate the number of chromosomes in the zygote.

Answer: 79  (tolerance ±0)

Working
Write down the values in the question:
egg = 40
sperm = 39
Write down the equation:
zygote=egg+sperm
Substitute the values into the equation:
zygote=40+39=79
14
Check q4 numeric entry

A cat’s body cells hold 38 chromosomes. A sperm with 18 chromosomes fertilizes an egg with the normal 19.

Calculate the number of chromosomes in the zygote.

Answer: 37  (tolerance ±0)

Working
Write down the values in the question:
egg = 19
sperm = 18
Write down the equation:
zygote=egg+sperm
Substitute the values into the equation:
zygote=19+18=37
15
Check q5 numeric entry

A gamete carrying two copies of one chromosome fuses with a normal gamete.

How many copies of that chromosome does the zygote have?

Answer: 3  (tolerance ±0)

Working
Write down the values in the question:
copies in the abnormal gamete = 2
copies in the normal gamete = 1
Write down the equation:
copies in the zygote=copies from one gamete+copies from the other
Substitute the values into the equation:
copies in the zygote=2+1=3
16
Check q6

A pig’s body cells hold 38 chromosomes. A zygote holds 37.

Which of the following gametes made this zygote, with a normal gamete?

  1. A. ✓ A gamete with one chromosome too few
  2. B. A gamete with one chromosome too many
    A gamete with one chromosome too many would make a zygote of 39.
    This zygote has 37, one too few.

Why: The zygote is one short of the body cell’s 38.
The normal gamete brought its full 19.
So the other gamete brought 18, one chromosome too few.

17Three copies, or one

18

Video: Watch: Naming the zygote’s count

The zygote of 47 beside the zygote of 45, the odd chromosome ringed in each; the names trisomy and monosomy written under them; trisomy 21 named.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11Cb.mp4

19

Here is the zygote of 47 again: two copies of every chromosome and three copies of one.

An egg holding two short rods and one long rod, count 24; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding three short rods and two long rods, count 47
An egg holding two short rods and one long rod, count 24; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding three short rods and two long rods, count 47
20

When a zygote carries three copies of one chromosome, 2n+1, we call it a , because tri- means three and -somy comes from an old word for a chromosome, a body in the cell.

21

Here is the zygote of 45 again: two copies of every chromosome and one copy of one.

An egg holding no short rod and one long rod, count 22; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding one short rod and two long rods, count 45
An egg holding no short rod and one long rod, count 22; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding one short rod and two long rods, count 45
22

When a zygote carries one copy of one chromosome, 2n−1, we call it a , because mono- means one.

23

Most zygotes with an extra or a missing chromosome do not develop.

24

The ones seen at birth are the exceptions. Trisomy 21, three copies of chromosome 21, is the most common trisomy seen at birth.

25

Every body cell of the child grows from the zygote by mitosis. So every body cell of a child with trisomy 21 carries three copies of chromosome 21.

26

What you are expected to know Name a zygote as a trisomy or a monosomy from its chromosome count and the body cell’s count.

27
Check q7

An animal’s body cells hold 50 chromosomes. A zygote holds 49 chromosomes.

Which of the following is the zygote?

  1. A. A trisomy
    A trisomy holds one chromosome more than a body cell: three copies of one chromosome.
    This zygote holds one chromosome fewer.
  2. B. ✓ A monosomy
  3. C. A normal zygote
    A normal zygote holds the body cell’s count, 50.
    This zygote holds one chromosome fewer.

Why: A body cell holds 50 chromosomes and this zygote holds 49: one fewer.
One fewer means one copy of one chromosome.
One copy of one chromosome is a monosomy.

28
Check q8

A plant’s body cells hold 24 chromosomes. A zygote holds 25 chromosomes.

Which of the following is the zygote?

  1. A. ✓ A trisomy
  2. B. A monosomy
    A monosomy holds one chromosome fewer than a body cell: one copy of one chromosome.
    This zygote holds one chromosome more.
  3. C. A normal zygote
    A normal zygote holds the body cell’s count, 24.
    This zygote holds one chromosome more.

Why: A body cell holds 24 chromosomes and this zygote holds 25: one more.
One more means three copies of one chromosome.
Three copies of one chromosome is a trisomy.

29
Check q9

A fish’s body cells hold 48 chromosomes. A zygote holds 48 chromosomes.

Which of the following is the zygote?

  1. A. A trisomy
    A trisomy holds one chromosome more than a body cell.
    This zygote holds the body cell’s count.
  2. B. A monosomy
    A monosomy holds one chromosome fewer than a body cell.
    This zygote holds the body cell’s count.
  3. C. ✓ A normal zygote

Why: A body cell holds 48 chromosomes and this zygote holds 48.
Two copies of every chromosome is the normal count.
So this zygote is normal.

30
Check q10

A human’s body cells hold 46 chromosomes. A zygote holds 45 chromosomes.

Which of the following is the zygote?

  1. A. A trisomy
    A trisomy holds one chromosome more than a body cell.
    This zygote holds one chromosome fewer.
  2. B. ✓ A monosomy
  3. C. A normal zygote
    A normal zygote holds the body cell’s count, 46.
    This zygote holds one chromosome fewer.

Why: A body cell holds 46 chromosomes and this zygote holds 45: one fewer.
One fewer means one copy of one chromosome.
One copy of one chromosome is a monosomy.

31
Check q11

An insect’s body cells hold 12 chromosomes. A zygote holds 13 chromosomes.

Which of the following is the zygote?

  1. A. ✓ A trisomy
  2. B. A monosomy
    A monosomy holds one chromosome fewer than a body cell.
    This zygote holds one chromosome more.
  3. C. A normal zygote
    A normal zygote holds the body cell’s count, 12.
    This zygote holds one chromosome more.

Why: A body cell holds 12 chromosomes and this zygote holds 13: one more.
One more means three copies of one chromosome.
Three copies of one chromosome is a trisomy.

32Quick quiz: trisomy, monosomy mixed practice

33
Check q12

A zygote has just formed.

What is a trisomy?

  1. A. A zygote missing one copy of a chromosome
    A zygote missing one copy of a chromosome is a monosomy.
    A trisomy has three copies of one chromosome.
  2. B. ✓ A zygote with three copies of one chromosome
  3. C. A zygote with three copies of every chromosome
    A trisomy has three copies of one chromosome only.
    Every other chromosome is present twice.

Why: A trisomy is a zygote with three copies of one chromosome.
A gamete carrying two copies of that chromosome fused with a normal gamete carrying one.

34
Practice writing an answer

A zygote has just formed.

(a) State what a monosomy is. (1 pt)

Model answer A monosomy is a zygote with one copy of one chromosome, made when a gamete missing that chromosome fuses with a normal gamete.
Rubric
  • Award 1 point for: a zygote with one copy of one chromosome. Accept with or without: made when a gamete missing that chromosome fuses with a normal gamete.
35
Check q13

A dog’s body cells hold 78 chromosomes. A zygote holds 79 chromosomes, one more than a body cell.

Which of the following is the zygote?

  1. A. ✓ A trisomy
  2. B. A monosomy
    A monosomy holds one chromosome fewer than a body cell.
    This zygote holds one more, so it has three copies of one chromosome.

Why: One chromosome more than the body cell’s count means three copies of one chromosome.
Three copies of one chromosome is a trisomy.

36
Check q14

A cat’s body cells hold 38 chromosomes. A zygote holds 37 chromosomes, one fewer than a body cell.

Which of the following is the zygote?

  1. A. A trisomy
    A trisomy holds one chromosome more than a body cell.
    This zygote holds one fewer, so it has one copy of one chromosome.
  2. B. ✓ A monosomy

Why: One chromosome fewer than the body cell’s count means one copy of one chromosome.
One copy of one chromosome is a monosomy.

37Why the extra chromosome matters

38

Video: Watch: Three copies of every gene

Two copies of a chromosome beside three copies, each copy making its product; the product box growing with the copy count.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11Cc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L11Cc.mp4

39
Check q15

A cell is using one of its genes to make that gene’s protein.

What is this called?

  1. A. ✓ Gene expression
  2. B. A mutation
    A mutation is a change in the base sequence of a gene’s DNA.
    Making a protein from a gene changes no sequence.
  3. C. Fertilization
    Fertilization is two gametes fusing into one cell.
    Making a protein from a gene is gene expression.

Why: A cell is expressing a gene when it uses that gene’s instructions to make the protein.
Making the protein from its gene is gene expression.

40

Why does one extra chromosome change development so much that most such zygotes do not develop?

41

Now consider a zygote with trisomy 21. Every gene on chromosome 21 is present three times in each of its cells, instead of two.

42

A cell makes a gene’s product from each copy of the gene it holds. So a cell with three copies of a gene makes more of that gene’s product than a cell with two copies.

Left: two copies of chromosome 21 drawn as two rods under the label chromosome 21, with an arrow to a box labelled product, the usual amount. Right: three copies drawn as three rods under the same label, with an arrow to a taller box labelled product, more
Left: two copies of chromosome 21 drawn as two rods under the label chromosome 21, with an arrow to a box labelled product, the usual amount. Right: three copies drawn as three rods under the same label, with an arrow to a taller box labelled product, more
43

So the cells of this zygote make more of each of those genes’ products.

44

The three copies of chromosome 21 are ordinary, undamaged copies. The extra amount of each product, not damage to any gene, is what changes development.

45

What you are expected to know Explain why a zygote with an extra chromosome develops differently: every gene on that chromosome is present three times, so its cells make more of each of those genes’ products.

46
Check q16

A zygote has trisomy 21: three copies of chromosome 21.

How many copies of each gene on chromosome 21 does each of its cells hold?

  1. A. One
    Each copy of chromosome 21 carries a copy of every gene on it.
    Three copies of the chromosome mean three copies of each of its genes.
  2. B. Two
    Two copies of each gene is the normal count, from two copies of the chromosome.
    This zygote holds three copies of chromosome 21.
  3. C. ✓ Three

Why: Each copy of chromosome 21 carries a copy of every gene on it.
This zygote holds three copies of chromosome 21.
So each cell holds three copies of each gene on it.

47
Practice writing an answer

A child has trisomy 21: three copies of chromosome 21 in every body cell.

(a) Explain how the extra copy of chromosome 21 changes what the child’s cells make. (1 pt)

Frame The extra copy of chromosome 21 changes what the cells make because …

Model answer The extra copy of chromosome 21 changes what the cells make because every gene on chromosome 21 is present three times instead of two.
A cell makes a gene’s product from each copy of the gene.
So three copies of a gene make more of that gene’s product than two copies.
Therefore the child’s cells make more of each of those genes’ products.
Rubric
  • Award 1 point for: every gene on chromosome 21 is present three times instead of two, so the cells make more of each of those genes’ products.
48
Check q17

A student says: “The extra chromosome 21 must be a damaged copy, and the damage is what changes development.”

Is the student correct?

  1. A. Yes, the extra copy is damaged
    All three copies of chromosome 21 are ordinary, undamaged copies.
    What changed is the number of copies, so the cells make more of each gene’s product.
  2. B. ✓ No, the extra copy is an ordinary chromosome 21

Why: The extra chromosome 21 is an ordinary, undamaged copy.
Every gene on it is present three times instead of two.
So the cells make more of each of those genes’ products, and that amount changes development.

49
Check q18

A zygote has trisomy 18: three copies of chromosome 18. Compare one of its cells with a cell holding two copies of chromosome 18.

How much of each chromosome 18 gene’s product does the trisomy 18 cell make?

  1. A. Less
    A cell makes a gene’s product from each copy of the gene.
    Three copies make more than two, not less.
  2. B. The same amount
    A cell makes a gene’s product from each copy of the gene.
    Three copies make more than two.
  3. C. ✓ More

Why: Every gene on chromosome 18 is present three times in the trisomy 18 cell, instead of two.
A cell makes a gene’s product from each copy of the gene.
So the trisomy 18 cell makes more of each of those genes’ products.

50

Back to the egg with 24 chromosomes, one chromosome too many, and the sperm with the normal 23. Fertilization joined the two sets, so the zygote holds 47 chromosomes: three copies of one chromosome, a trisomy.

An egg holding two short rods and one long rod, count 24; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding three short rods and two long rods, count 47
An egg holding two short rods and one long rod, count 24; a plus sign; a sperm holding one short rod and one long rod, count 23; an arrow to a zygote holding three short rods and two long rods, count 47
51

Every body cell that grows from that zygote by mitosis carries the extra copy. So every gene on that chromosome is present three times.

52

Each of those cells makes more of each of those genes’ products.

53Mixed practice mixed practice

54
Check q19

A fish’s body cells hold 48 chromosomes. An egg with 25 chromosomes is fertilized by a sperm with the normal 24.

How many chromosomes does the zygote hold?

  1. A. 47
    A gamete with one chromosome too many brings the zygote up by one, not down.
  2. B. 48
    Fertilization joins the two sets as they are.
    The egg’s extra chromosome is not lost.
  3. C. ✓ 49

Why: The zygote holds the egg’s 25 and the sperm’s 24.
That is 49 chromosomes: one more than a body cell’s 48.

55
Check q20

A child has trisomy 21: three copies of chromosome 21 in every body cell.

Which of the following gametes could have produced this zygote?

  1. A. ✓ A gamete with 24 chromosomes, carrying two copies of chromosome 21
  2. B. A gamete with 22 chromosomes, missing chromosome 21
    A gamete missing chromosome 21 would give a zygote with one copy of it.
    The child has three.
  3. C. A sperm carrying all 46 of the father’s chromosomes
    A 46-chromosome sperm would give a zygote with 69 chromosomes: three copies of every chromosome, not of one.
  4. D. A gamete with 23 chromosomes, one of them a chromosome 21 carrying a new allele
    A new allele changes a sequence, not the number of copies.
    A 23-chromosome gamete brings one copy of chromosome 21, so the zygote has two.

Why: Three copies of one chromosome come from a gamete that carried two copies of it.
That is a 24-chromosome gamete, made by a nondisjunction.
It joined a normal gamete with one copy.

56
Check q21

A mouse’s body cells hold 40 chromosomes. A zygote holds 39 chromosomes.

Which of the following is the zygote?

  1. A. A trisomy
    A trisomy holds one chromosome more than a body cell.
    This zygote holds one fewer.
  2. B. ✓ A monosomy

Why: A body cell holds 40 chromosomes and this zygote holds 39: one fewer.
One fewer means one copy of one chromosome.
One copy of one chromosome is a monosomy.

57
Check q22

A zygote with an extra copy of chromosome 21 develops differently from one with two copies.

Which of the following explains why the extra chromosome changes development?

  1. A. The extra chromosome 21 switches off the genes on the other two copies
    All three copies are ordinary working copies.
    Three copies of every gene on chromosome 21 make more of each product than two, and that amount is what changes development.
  2. B. The child received one more chromosome from one parent than from the other
    The gamete with two copies did come from one parent.
    But that says where the extra chromosome came from, not why it changes development.
  3. C. The extra copy carries alleles neither parent had, so the child shows new traits
    The alleles on the extra copy are ordinary alleles the parent carried.
    What changed is the number of copies.
  4. D. ✓ Every gene on chromosome 21 is present three times, so more product is made

Why: An extra whole chromosome means every gene on it is present three times.
Three copies make more of a gene’s product than two.
So the amounts in the cells change.
The chromosome itself is undamaged.

58
Check q23

A human zygote holds 47 chromosomes. It divides by mitosis again and again to build the body.

How many chromosomes does each body cell hold?

  1. A. 23
    Before each mitosis the cell copies every chromosome once.
    Mitosis then gives each new cell the full count.
    A body cell holds what the zygote held.
  2. B. 46
    Mitosis gives each new cell exactly what the zygote held.
    Nothing removes the extra chromosome.
  3. C. ✓ 47

Why: Every body cell grows from the zygote by mitosis.
Mitosis gives each new cell the same chromosomes the parent cell held.
So every body cell holds the zygote’s 47.

59
Check q24

In a human, a normal gamete holds 23 chromosomes.

Which of the following pairs of gametes makes a monosomy?

  1. A. ✓ An egg with 22 chromosomes and a sperm with 23
  2. B. An egg with 23 chromosomes and a sperm with 23
    Two normal gametes make a normal zygote of 46: two copies of every chromosome.
  3. C. An egg with 24 chromosomes and a sperm with 23
    An egg with 24 chromosomes carries two copies of one chromosome.
    With the sperm’s one copy, the zygote has three copies: a trisomy.

Why: An egg with 22 chromosomes carries no copy of one chromosome.
The sperm brings one copy of every chromosome.
So the zygote holds one copy of that chromosome: a monosomy.

60
Practice writing an answer

A human egg with 22 chromosomes is fertilized by a sperm with the normal 23. The zygote holds 45 chromosomes.

(a) Explain why this zygote has only one copy of one of its chromosomes. (1 pt)

Frame The zygote has only one copy of one chromosome because …

Model answer The zygote has only one copy of one chromosome because the egg was missing that chromosome.
An egg with 22 chromosomes carries no copy of one chromosome.
The sperm is normal, so it carries one copy of every chromosome.
Fertilization joins the two sets as they are.
So the zygote gets the sperm’s one copy of that chromosome and no copy from the egg: one copy, a monosomy.
Rubric
  • Award 1 point for: the egg carried no copy of that chromosome and the sperm carried one, and fertilization adds the two sets, so the zygote has one copy (a monosomy).

Glossary

trisomy
A zygote, or the person who grows from it, with three copies of one chromosome, 2n + 1, made when a gamete carrying two copies of that chromosome fuses with a normal gamete. Trisomy 21 is three copies of chromosome 21.
monosomy
A zygote with one copy of one chromosome, 2n − 1, made when a gamete missing that chromosome fuses with a normal gamete.

APBIO-U05-P52 Practice questions: Topic 5.2

Topic 5.2 · Meiosis and Genetic Diversity · 10 MCQ · 2 FRQ · for APBIO-U05-T52

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation. The first free-response question walks you through one nondisjunction one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it.

Video: Watch first: Meiosis and genetic diversity, summed up

Pairs that line up on their own and the two to the power of the pairs; non-sister chromatids trading pieces; any egg with any sperm; when a pair fails to separate, and how the counts say which division failed.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T52-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T52-summary.mp4

Q1 P52-q01

A gamete from a dragonfly carries one chromosome of every pair.

Which statement describes where the chromosomes in that gamete came from?

  1. A. All from the dragonfly’s mother, or all from its father
    Each pair orients on its own, so a gamete is not dealt one parent's whole set; an all-maternal gamete is one rare outcome among many, not the rule.
  2. B. ✓ Some from the mother and some from the father, settled pair by pair
  3. C. One of every pair from the mother and one of every pair from the father
    A gamete holds one member of each pair, not both; two of every pair is the diploid set.
  4. D. Whichever parent contributed the larger set
    Both parents contributed one set of the same size; a gamete takes one member of each pair from either.

Why: Each gamete receives one chromosome of every pair, the maternal one or the paternal one, settled pair by pair, so its set is usually a mixture.

Q2 P52-q02

An animal’s body cells hold 10 pairs of chromosomes.

How many kinds of gamete can one of these animals make by independent orientation alone?

  1. A. 20
    20 adds two for each pair.
    The kinds double with each pair, so ten pairs give 2¹⁰.
  2. B. ✓ 1,024
  3. C. 20,480
    20,480 multiplies 1,024 by the 20 chromosomes.
    The count of kinds is 2ⁿ for n pairs and nothing more.
  4. D. 1,048,576
    1,048,576 is 2²⁰, doubling once for each chromosome.
    Each pair, not each chromosome, doubles the count, and 20 chromosomes make 10 pairs.

Why: Each homologous pair faces either way at metaphase I on its own.
So each pair doubles the number of kinds of gamete.
Ten pairs double the count ten times, which the working below gives as 1,024.

Q3 P52-q03

In prophase I in a bee, two chromatids break at the same point and exchange pieces.

Which two chromatids are they?

  1. A. Two sister chromatids of the maternal homolog
    Sister chromatids are identical copies; a swap between them changes nothing.
  2. B. One chromatid of each of two different chromosome pairs
    Crossing over happens between the two paired homologs of one pair, not between different pairs.
  3. C. ✓ One chromatid of the maternal homolog and one of the paternal homolog
  4. D. Two sister chromatids of the paternal homolog
    The paternal homolog’s two chromatids are sisters, identical copies.

Why: Crossing over is between non-sister chromatids: one from each of the two paired homologs.

Q4 P52-q04

In a squirrel, the maternal homolog of a pair carries R at one position and T at the next position along the same arm; the paternal homolog carries r and t. After one crossover between the two positions, the four chromatids read R with T, R with t, r with T and r with t.

Which two are the parental combinations?

  1. A. ✓ R with T and r with t
  2. B. R with t and r with T
    R with t and r with T are the combinations neither homolog carried: recombinant.
  3. C. R with T and R with t
    R with t is a combination neither homolog carried.
    The parental combinations are what each homolog carried before the crossover.
  4. D. r with T and r with t
    r with T is a combination neither homolog carried, so it is recombinant.
    Only r with t of this pair is parental.

Why: The parental combinations are the two the homologs carried before the crossover: R with T and r with t.
R with t and r with T are recombinant.

Q5 P52-q05

In a species of animal, each parent makes 16 kinds of gamete by independent orientation.

How many kinds of zygote can two of these animals make, counting independent orientation and random fertilization only?

  1. A. 16
    16 is one parent’s gamete kinds; the zygote joins one of hers with one of his.
  2. B. 32
    32 adds the two parents’ kinds; the kinds multiply.
  3. C. ✓ 256
  4. D. 65,536
    65,536 is 256 multiplied by 256, as if each parent made 256 kinds.

Why: Any of her 16 kinds of egg can fuse with any of his 16 kinds of sperm.
So the kinds of zygote are the kinds of egg times the kinds of sperm.
The working below gives 256.

Q6 P52-q06

One chromatid in a guinea pig’s egg carries the mother’s allele of one gene and the father’s allele of the next gene along the same chromosome.

In which stage of meiosis did that chromatid form?

  1. A. Metaphase I
    At metaphase I the pairs line up and face the poles.
    Lining up moves whole chromosomes and changes nothing along a chromatid.
  2. B. Anaphase II
    Anaphase II parts the two sister chromatids of each chromosome and changes nothing along either.
  3. C. At fertilization
    Fertilization joins a finished egg with a finished sperm.
    The chromatid is inside the egg before any sperm reaches it.
  4. D. ✓ Prophase I

Why: A chromatid with alleles from both parents was made by crossing over.
Crossing over happens in prophase I, when the homologs lie paired and two non-sister chromatids break at the same point and exchange pieces.
So the chromatid formed in prophase I.

Q7 P52-q07

An animal’s body cells hold 30 chromosomes. In one cell, one homologous pair undergoes nondisjunction in meiosis I; meiosis II is normal.

What are the chromosome counts of the four cells produced?

  1. A. 16, 14, 15 and 15
    One over, one under and two normal is the meiosis II pattern; a failure in meiosis I leaves all four abnormal.
  2. B. ✓ 16, 16, 14 and 14
  3. C. 15, 15, 15 and 15
    One pair did fail, so two cells are over and two under.
  4. D. 30, 30, 15 and 15
    No gamete keeps the full 30; meiosis I halves the count and the failure changes it by one either way.

Why: The haploid count for this animal is 15.
In meiosis I both members of one pair go to one cell, so that cell holds 16 and the other 14.
Meiosis II copies each count into two gametes.
So the four cells hold 16, 16, 14 and 14.

Q8 P52-q08

An insect’s body cells hold 14 chromosomes. The four cells from one meiosis hold 8, 6, 7 and 7.

Which division did the failure happen in?

  1. A. Meiosis I
    A failure in meiosis I leaves no cell with the normal count.
    Here two cells hold the normal 7, so meiosis I parted every pair correctly.
  2. B. Either division
    A failure in meiosis I leaves all four cells abnormal; a failure in meiosis II leaves two normal cells, and two cells here hold the normal 7.
  3. C. ✓ Meiosis II
  4. D. Neither: the counts cannot happen
    The counts can happen.
    They add up to 28, the total the four cells must hold, and the cell over is matched by the cell under.

Why: Two normal cells mean meiosis I parted every pair; one of the two cells then failed in meiosis II, giving the 8 and the 6.

Q9 P52-q09

A sheep’s body cells hold 54 chromosomes. A sheep egg with 28 chromosomes is fertilized by a normal sperm with 27.

What does the zygote hold?

  1. A. 54 chromosomes, because the sperm corrects the egg
    Fertilization joins the two sets as they are; nothing removes the egg’s extra chromosome.
  2. B. 56 chromosomes: four copies of one of them
    The egg brings one extra copy, not two, so the zygote has three copies of that chromosome, not four.
  3. C. 53 chromosomes: one copy of one of them, a monosomy
    An egg with one chromosome too many raises the count, to 55, not lowers it.
  4. D. ✓ 55 chromosomes: three copies of one of them, a trisomy

Why: The zygote holds the egg’s 28 and the sperm’s 27: 55 in all, two copies of every chromosome and three copies of one, a trisomy.

Q10 P52-q10

One parent of a species of animal makes 2,048 kinds of gamete by independent orientation alone.

How many pairs of chromosomes does this animal have?

  1. A. ✓ 11
  2. B. 22
    22 would be the number of chromosomes in a body cell, two for each pair.
    The count of doublings that reach 2,048 is the number of pairs.
  3. C. 1,024
    1,024 is 2,048 halved once.
    The number of pairs is how many times 2 is multiplied by itself to reach 2,048, not half of 2,048.
  4. D. 4,096
    4,096 doubles 2,048 once more, as if a twelfth pair were added.
    The question asks how many doublings reach 2,048.

Why: Every pair doubles the number of kinds of gamete.
So the number of pairs is how many times 2 is multiplied by itself to reach 2,048.
The working below gives 11.

FRQ 1 P52-frq1 · Conceptual Analysis scaffolded

A plant’s body cells each hold 8 chromosomes. One cell in its anther, the part of the flower that makes pollen, goes through meiosis.

(a) Calculate the number of chromosomes in each of the four cells when meiosis is normal. (1 pt)

Frame Each of the four cells holds … chromosomes, one of every pair.

Hint How many pairs does a cell with 8 chromosomes have, and how many of each pair does a gamete get?

Answer: 4  (tolerance ±0)

Model answer Each of the four cells holds 4 chromosomes, one of every pair.
Working
Write down the values in the question:
tex:2n = 8
Write down the equation:
tex:\text{chromosomes per gamete} = n = \frac{2n}{2}
Substitute the values into the equation:
tex:\text{chromosomes per gamete} = \frac{8}{2} = 4
Rubric
  • Award 1 point for: 4.

(b) Predict the four counts if one homologous pair undergoes nondisjunction in meiosis I and meiosis II is normal. (1 pt)

Frame The cell that receives both members holds …, the other holds …, and meiosis II copies each into two cells: …, …, …, ….

Hint In meiosis I both members of one pair go to one cell; what does meiosis II then do to each of the two cells?

Model answer The cell that receives both members holds 5, the other holds 3, and meiosis II copies each into two cells: 5, 5, 3, 3.
Rubric
  • Award 1 point for: 5, 5, 3, 3.

Slip Writing 5, 3, 4, 4. That is the pattern of a failure in meiosis II.

(c) Predict the four counts if instead meiosis I is normal and, in one of the two cells, the sister chromatids of one chromosome undergo nondisjunction in meiosis II. (1 pt)

Frame The cell that failed gives … and …; the other cell gives … and ….

Hint Only one of the two cells from meiosis I fails; what does the other cell give?

Model answer The cell that failed gives 5 and 3; the other cell gives 4 and 4, the normal count.
Rubric
  • Award 1 point for: 5, 3, 4, 4.

Slip Writing 5, 5, 3, 3. Only one cell failed, so two of the four are normal.

(d) A set of four cells from a second plant, whose body cells hold 34 chromosomes, holds 18, 18, 16 and 16. Determine the division in which the failure happened, using the counts. (1 pt)

Frame The failure was in meiosis …, because … of the four cells hold the normal count of …

Hint Compare this set with the two patterns you predicted in (b) and (c).

Model answer The normal count for this plant is 17.
None of the four cells holds 17, so all four are abnormal.
Only a failure in meiosis I makes both cells it produces abnormal.
So the failure was in meiosis I.
Meiosis II then copied each count into two cells: 18, 18, 16 and 16.
Rubric
  • Award 1 point for: the decision (meiosis I) AND the ground it rests on (no cell has the normal count of 17, so all four cells are abnormal).

Slip Choosing meiosis II because a cell is over and a cell is under. Both patterns have that; the cells with the normal count place the failure.

(e) Return to the first plant, with 8 chromosomes. Predict the chromosome number of the zygote made when one of the gametes with the extra chromosome from part (b) fuses with a normal gamete, and identify it as a trisomy or a monosomy. (1 pt)

Frame The zygote holds … from the abnormal gamete and … from the normal one: … chromosomes, a ….

Hint Add the abnormal gamete’s count to a normal gamete’s count, then ask how many copies of the extra chromosome the zygote holds.

Model answer The zygote holds 5 from the abnormal gamete and 4 from the normal one: 9 chromosomes, a trisomy, with three copies of one chromosome.
Rubric
  • Award 1 point for: 9, a trisomy.

Slip Calling it a monosomy. An extra chromosome in a gamete gives three copies in the zygote.

FRQ 2 P52-frq2 · Scientific Investigation

Observation: in a species of sunflower, cells from plants grown in salty soil show fewer chiasmata in prophase I than cells from plants grown in ordinary soil.

(a) Construct a testable question about this observation that names what would be changed and what would be measured. (1 pt)

Model answer Does the amount of salt in the soil change the mean number of chiasmata per cell in prophase I?
The amount of salt in the soil is changed and the number of chiasmata per cell is measured.
Rubric
  • Award 1 point for: a question naming the amount of salt in the soil as what is changed and the number of chiasmata per cell as what is measured.

Slip Asking why salty soil matters. A “why” question names nothing to change or measure.

(b) Identify the independent variable and the dependent variable in your question. (1 pt)

Model answer The independent variable is the amount of salt in the soil; the dependent variable is the mean number of chiasmata per cell.
Rubric
  • Award 1 point for: independent variable the amount of salt in the soil; dependent variable chiasmata per cell.

Slip Swapping the two. What the experimenter changes is independent; what is measured is dependent.

(c) State the null hypothesis for the experiment. (1 pt)

Model answer The amount of salt in the soil makes no difference to the mean number of chiasmata per cell; any difference between the groups is due to chance.
Rubric
  • Award 1 point for: no difference in chiasmata per cell between the salt treatments (any difference due to chance).

Slip Stating that salty soil lowers the number of chiasmata. That is the alternative hypothesis, the prediction, not the null.

(d) Predict what fewer chiasmata per cell would mean for the fraction of recombinant gametes the salty-soil plants make, and explain why. (1 pt)

Model answer Fewer chiasmata mean fewer crossovers, so fewer chromatids carry a combination of alleles neither homolog had, and a smaller fraction of the gametes are recombinant.
Rubric
  • Award 1 point for: a smaller fraction of recombinant gametes, because each chiasma is where non-sister chromatids exchanged pieces.

Slip Saying the alleles themselves change. Crossing over recombines existing alleles; the number of alleles is unchanged.

APBIO-U05-T52 End-of-topic test: Meiosis and Genetic Diversity

Topic 5.2 · Meiosis and Genetic Diversity · 18 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation, then open the scoring guide and mark your own work against it. In every drawing, a chromosome from the mother is dark and one from the father is light.

Q1 T52-q01

A plant’s body cells hold six chromosomes: a long pair, a medium pair and a short pair. A gamete from this plant carries the paternal long, the maternal medium and the maternal short chromosome.

Which statement about this gamete is correct?

  1. A. It is impossible: a gamete carries all of one parent’s chromosomes
    Each pair faces either way at metaphase I on its own, so nothing forces all of one parent’s chromosomes into one gamete.
  2. B. ✓ It is a possible gamete: one member of each pair, each from either parent
  3. C. It is a possible gamete: crossing over swapped the paternal long chromosome into a maternal set
    Crossing over swaps pieces between two chromatids and moves no whole chromosome.
    At metaphase I the long pair faced its paternal member toward this gamete’s pole.
  4. D. It is impossible: a gamete carries the maternal member of every pair
    Each pair faces either way on its own, so a gamete can carry the paternal member of one pair and the maternal member of another.

Why: A gamete carries one chromosome of every pair.
Which member it carries, maternal or paternal, is settled pair by pair at metaphase I. So one paternal and two maternal chromosomes, one of each pair, is a possible gamete.

Q2 T52-q02

A cell from a midge with a long pair, a medium pair and a short pair is drawn below at metaphase I. In the drawing the mother’s chromosomes are dark and the father’s are light. Meiosis then finishes normally.

A midge cell at metaphase I: three pairs across the middle, labelled long, medium and short at the left; each pair is a dark X and a light X side by side, with a fiber from each pole to the member on its side.
A midge cell at metaphase I: three pairs across the middle, labelled long, medium and short at the left; each pair is a dark X and a light X side by side, with a fiber from each pole to the member on its side.

Which set of chromosomes does the cell that forms at the left pole receive?

  1. A. All three maternal
    The medium pair faces the other way from the long pair: its light, paternal member faces the left pole, so the left cell receives the paternal medium.
  2. B. Maternal long, maternal medium, paternal short
    In the drawing the medium pair’s light member faces the left pole and the short pair’s dark member faces the left pole, the other way round from this option.
  3. C. ✓ Maternal long, paternal medium, maternal short
  4. D. Paternal long, maternal medium, paternal short
    This is the set the right pole receives.
    Each pole receives the member of each pair that faces it, and the members facing the left pole are dark, light, dark.

Why: Each pole receives the member of each pair that faces it.
Facing the left pole: the dark long, the light medium and the dark short.
Dark is maternal and light is paternal.
So the left cell receives the maternal long, the paternal medium and the maternal short.

Q3 T52-q03

An animal’s body cells hold 5 pairs of chromosomes.

How many kinds of gamete can one of these animals make by independent orientation alone?

  1. A. 10
    10 is two added for each pair; the kinds multiply across pairs, doubling with each.
  2. B. 16
    16 is the count for four pairs; the fifth pair doubles it again.
  3. C. 25
    25 is five multiplied by five; each pair offers two choices, not five.
  4. D. ✓ 32

Why: Each homologous pair faces either way at metaphase I on its own.
So each pair doubles the number of kinds of gamete.
Five pairs double the count five times; the working below gives 32.

Q4 T52-q04

The drawing below shows a homologous pair from a salamander in prophase I, its two homologs set apart so that the four chromatids can be numbered 1 to 4. An arm of chromatid 2 crosses an arm of chromatid 3 at a chiasma.

A homologous pair from a salamander in prophase I, the two homologs drawn apart: a dark X with chromatids 1 and 2 and a light X with chromatids 3 and 4, each numbered at its upper tip; one arm of chromatid 2 crosses one arm of chromatid 3.
A homologous pair from a salamander in prophase I, the two homologs drawn apart: a dark X with chromatids 1 and 2 and a light X with chromatids 3 and 4, each numbered at its upper tip; one arm of chromatid 2 crosses one arm of chromatid 3.

Which of the following pairs of chromatids could also exchange pieces by crossing over?

  1. A. Chromatids 1 and 2
    Chromatids 1 and 2 are the two sister chromatids of the dark homolog, identical copies.
    A swap between them would change nothing.
  2. B. ✓ Chromatids 1 and 4
  3. C. Chromatids 3 and 4
    Chromatids 3 and 4 are the two sister chromatids of the light homolog, identical copies.
    A swap between them would change nothing.
  4. D. Any two of the four
    Chromatids 1 and 2 are sisters, and so are 3 and 4.
    Crossing over is between non-sister chromatids only, one from each homolog.

Why: Each homolog of the pair is two sister chromatids.
Crossing over is between non-sister chromatids: one chromatid of the dark homolog and one of the light homolog.
Chromatid 1 belongs to the dark homolog and chromatid 4 to the light homolog, so 1 and 4 could also exchange pieces.

Q5 T52-q05

A cell from a snail goes through meiosis. One of the four cells it produces carries an allele whose DNA sequence differs by one base from both of the parent cell’s alleles. A student says crossing over made the new allele.

Which statement about the student’s claim is correct?

  1. A. ✓ The student is wrong: crossing over moves alleles the parent already had into new combinations
  2. B. The student is right: crossing over rewrites the DNA sequence at the point where the two chromatids break
    Crossing over rewrites no DNA sequence: two non-sister chromatids break at the same point and exchange pieces, so every allele keeps its own sequence and moves to a new partner.
  3. C. The student is wrong: independent orientation, not crossing over, made the new allele at metaphase I
    Independent orientation changes no DNA sequence either.
    It decides which whole chromosome of each pair a gamete receives.
    The alleles on that chromosome are the parent’s own.
  4. D. The student is right, because a new allele can only arise in prophase I, while the homologs are paired
    A new allele can arise whenever a cell copies its DNA, not only in prophase I; prophase I is when crossing over happens, and crossing over changes no sequence.

Why: Crossing over moves alleles the parent already had into new combinations.
It changes no allele’s sequence.
A sequence found on neither of the parent’s homologs is a new allele.
Only a mutation, a change in the DNA itself, makes a new allele.
So the student is wrong.

Q6 T52-q06

In a gerbil, the maternal homolog of a pair carries E at one position and G at the next position along the same arm; the paternal homolog carries e and g. One crossover happens between the two positions. The four chromatids afterward are drawn below.

The gerbil’s pair after one crossover, the two homologs drawn apart: the four chromatids read E with G, E with g, e with g and e with G.
The gerbil’s pair after one crossover, the two homologs drawn apart: the four chromatids read E with G, E with g, e with g and e with G.

Which chromatid is recombinant?

  1. A. A chromatid reading E with G
    E with G is the combination the maternal homolog carried, so a chromatid that keeps it is parental.
  2. B. A chromatid reading e with g
    e with g is the combination the paternal homolog carried, so a chromatid that keeps it is parental.
  3. C. ✓ A chromatid reading E with g
  4. D. All four chromatids
    One crossover involves one chromatid from each homolog.
    The other two chromatids keep their parental combinations.

Why: The parent’s homologs carried E with G and e with g.
A recombinant chromatid carries a combination neither homolog had.
E with g is such a combination, and so is e with G.
E with G and e with g are the parental combinations.

Q7 T52-q07

In a species of animal, each parent makes 256 kinds of gamete by independent orientation.

How many kinds of zygote can two of these animals make, counting independent orientation and random fertilization only?

  1. A. 256
    256 is the number of kinds of one parent’s gametes.
    A zygote joins one of her kinds of egg with one of his kinds of sperm, so the kinds multiply.
  2. B. 512
    512 adds the two parents’ kinds.
    Any egg can fuse with any sperm, so the kinds multiply.
  3. C. ✓ 65,536
  4. D. 131,072
    131,072 is 65,536 doubled, as if random fertilization added one more pair.
    It multiplies the kinds of egg by the kinds of sperm.

Why: Any of her 256 kinds of egg can fuse with any of his 256 kinds of sperm.
So the kinds of zygote are the kinds of egg times the kinds of sperm.
The working below gives 65,536.

Q8 T52-q08

A female herring sheds thousands of eggs into the sea and a male herring sheds sperm over them; one sperm enters each egg.

Which statement describes random fertilization in this spawning?

  1. A. Each egg received its chromosomes by chance inside the female, before it was shed
    Which chromosomes an egg received was settled inside the female, during her meiosis.
    Random fertilization happens afterward, in the sea, when one sperm enters one egg.
  2. B. The sperm and the egg exchange pieces of their chromosomes at the moment they fuse
    Crossing over is an exchange between non-sister chromatids in prophase I, inside one parent; fertilization fuses two finished gametes and exchanges nothing.
  3. C. The sperm enters the egg before the egg’s meiosis has begun, and starts that meiosis
    The egg’s meiosis began inside the female long before the egg was shed, so the sperm’s entry does not start it.
  4. D. ✓ Which sperm enters each egg is chance, so each zygote pairs one kind of egg with one kind of sperm

Why: Meiosis in each parent made many kinds of egg and many kinds of sperm.
In the sea, which sperm reaches which egg is chance.
So each zygote pairs one of her kinds of egg with one of his kinds of sperm, and that chance pairing is random fertilization.

Q9 T52-q09

A perch’s gametes and zygotes carry new combinations of its parents’ chromosomes and alleles.

Which of the following was produced by independent orientation alone?

  1. A. ✓ An egg carrying the maternal copy of chromosome 2 and the paternal copy of chromosome 5
  2. B. A chromatid carrying the mother’s allele of one gene and the father’s allele of the next gene along the same chromosome
    Two alleles from different parents on one chromatid come only from non-sister chromatids that swapped pieces in prophase I; independent orientation moves whole chromosomes and changes nothing along a chromatid.
  3. C. A zygote made of one of the mother’s kinds of egg and one of the father’s kinds of sperm
    That zygote was made by random fertilization.
    It joins one finished egg with one finished sperm, after meiosis.
    Independent orientation acts inside one parent’s meiosis, at metaphase I.
  4. D. A sperm carrying an allele with a base that neither parent carried
    That allele was made by mutation.
    A base neither parent carried is a change in the DNA itself.
    Independent orientation reshuffles whole chromosomes the parents already had.

Why: Independent orientation decides, pair by pair at metaphase I, which member of each pair a gamete receives.
So the maternal copy of one pair with the paternal copy of another, in one egg, is its work alone.

Q10 T52-q10

A pair of zebra finches raise four clutches over two years. No two of their chicks carry the same combination of the parents’ alleles.

Which statement explains why the chicks differ?

  1. A. The parents’ alleles changed by mutation between one clutch and the next
    Mutation is rare and makes a new allele; the chicks differ in which of the parents’ existing alleles they carry, settled when each egg fused with its sperm.
  2. B. The chicks of one clutch grew from one egg and one sperm, and crossing over in their own cells made them differ
    Each chick grew from its own egg and its own sperm; crossing over happens in meiosis, when gametes are made, and a chick’s body cells divide by mitosis, not meiosis.
  3. C. The female’s eggs were all alike, so the chicks differ only in which sperm reached each egg
    Independent orientation and crossing over happen in the female’s meiosis as well as the male’s, so her eggs differ from one another as much as his sperm do.
  4. D. ✓ Each chick grew from a different egg and a different sperm, each carrying its own mix of alleles

Why: Independent orientation and crossing over made every egg and every sperm a different mix of the parents’ alleles.
Random fertilization then joined a different egg with a different sperm for each chick.
So no two chicks carry the same combination.

Q11 T52-q11

A student notices that voles fed a low-protein diet show fewer chiasmata per cell in prophase I than voles fed an ordinary diet.

Which of the following questions about this observation is testable as written?

  1. A. Why does protein matter so much to meiosis in voles?
    A “why” question names nothing to change and nothing to measure.
  2. B. Is a low-protein diet bad for the voles?
    “bad for the voles” names nothing that could be measured.
  3. C. What is the purpose of chiasmata in a vole’s meiosis in prophase I?
    “purpose” is not something an experiment changes or measures.
  4. D. ✓ Does a low-protein diet lower the number of chiasmata per cell?

Why: A testable question names what would be changed, the diet, and what would be measured, the number of chiasmata per cell.
The other three questions name neither.

Q12 T52-q12

A plant’s body cells hold 42 chromosomes. In one cell going through meiosis, the two cells that meiosis I produces hold 22 and 20 chromosomes.

Which of the following explains these counts?

  1. A. ✓ Nondisjunction in meiosis I
  2. B. Nondisjunction in meiosis II
    A failure in meiosis II leaves the two cells from meiosis I at the normal 21 each.
    Here they hold 22 and 20, so the failure came earlier.
  3. C. Crossing over in prophase I
    Crossing over exchanges pieces between non-sister chromatids in prophase I.
    It moves no whole chromosome, so it changes no count.
  4. D. Independent orientation at metaphase I
    Independent orientation is which way each pair faces at metaphase I.
    Whichever way a pair faces, each cell still receives one member of it, so both cells hold 21.

Why: Meiosis I should send one member of every pair to each cell, 21 and 21.
One cell holds 22 and the other 20, so both members of one pair went to one cell.
A pair that fails to separate is nondisjunction, and here it happened in meiosis I.

Q13 T52-q13

An insect’s body cells hold 24 chromosomes. In one cell, drawn below at anaphase I, both members of one homologous pair move to the same pole; meiosis II is then normal. Only two of the cell’s twelve pairs are drawn.

An insect cell at anaphase I, two of its twelve pairs drawn: both members of the long pair move toward the left pole; the short pair parts normally.
An insect cell at anaphase I, two of its twelve pairs drawn: both members of the long pair move toward the left pole; the short pair parts normally.

What are the chromosome counts of the four cells produced?

  1. A. 13, 11, 12 and 12
    In meiosis I both members of the failing pair go to one cell, so both cells it makes are abnormal: 13 and 11.
    Meiosis II keeps each cell’s count.
  2. B. ✓ 13, 13, 11 and 11
  3. C. 13, 12, 12 and 12
    A chromosome gained by one cell is lost by the cell at the other pole; there is always a cell under as well as over.
  4. D. 24, 24, 12 and 12
    No gamete keeps the full 24; meiosis I halves the count and the failure changes it by one either way.

Why: The haploid count for the insect is 12.
In meiosis I both members of one pair go to one cell, so that cell holds 13 and the other 11.
Meiosis II copies each count into two gametes.
So the four cells hold 13, 13, 11 and 11.

Q14 T52-q14

A lamb has three copies of one chromosome in every body cell. A student says: “The extra copy must be a damaged chromosome, and the damage is what changes the lamb’s development.”

Which of the following statements about the student’s claim is correct?

  1. A. ✓ The student is wrong: the extra copy is an ordinary chromosome, and three copies of each gene make more product than two
  2. B. The student is right: a chromosome that fails to separate is damaged as it is pulled to the wrong pole
    Nondisjunction moves a whole chromosome to the wrong pole and changes nothing along it.
    All three copies are ordinary, undamaged chromosomes.
  3. C. The student is wrong: the extra copy is ordinary, and the lamb develops differently because an odd number of chromosomes cannot be shared evenly in mitosis
    In mitosis each chromosome’s two sister chromatids go one to each cell, so an odd count is shared evenly.
    Three copies of each gene make more product than two.
  4. D. The student is right: the extra copy carries new alleles that neither parent’s chromosomes had
    The alleles on the extra copy are ordinary alleles one parent carried.
    What changed is the number of copies, not any allele.

Why: All three copies of the chromosome are ordinary, undamaged copies.
Every gene on that chromosome is present three times instead of two.
A cell makes a gene’s product from each copy it holds.
So the lamb’s cells make more of each of those genes’ products, which changes development.

Q15 T52-q15

An elephant’s body cells hold 56 chromosomes. An elephant egg with 27 chromosomes is fertilized by a normal sperm with 28.

What does the zygote hold?

  1. A. 54 chromosomes: one chromosome missing from both sets
    The egg is short by one chromosome and the sperm is normal.
    So the zygote is short by exactly one, not by two.
  2. B. ✓ 55 chromosomes: one copy of one of them, a monosomy
  3. C. 56 chromosomes, because the sperm makes up the missing one
    Fertilization joins the two sets as they are.
    The sperm brings its 28 and nothing fills the egg’s gap.
  4. D. 57 chromosomes: three copies of one of them, a trisomy
    A gamete with one chromosome too few brings the zygote down by one, to 55, not up to 57.

Why: The zygote holds the egg’s 27 and the sperm’s 28: 55 in all.
Every chromosome is present twice except one, which is present once.
One copy of one chromosome is a monosomy.

Q16 T52-q16

A calf has three copies of one chromosome in every body cell. Cattle body cells hold 60 chromosomes.

Which gamete could have produced this zygote?

  1. A. A gamete with 29 chromosomes, missing that chromosome
    A gamete missing that chromosome would give a zygote with one copy of it, a monosomy, not three.
  2. B. A sperm with 30 chromosomes, one of them carrying a new allele
    A new allele changes the DNA along one chromosome; it does not add one, and a 30-chromosome sperm with a 30-chromosome egg gives two copies of every chromosome.
  3. C. ✓ An egg or sperm with 31 chromosomes, two copies of that chromosome
  4. D. A sperm carrying all 60 of the bull’s chromosomes
    A 60-chromosome sperm would give a zygote with 90 chromosomes, three copies of every chromosome, not of one.

Why: Three copies of one chromosome come from a gamete that carried two copies of it.
A normal cattle gamete carries 30 chromosomes.
A gamete from a nondisjunction carries 31, with two copies of one chromosome.
Joined to a normal gamete with one copy, it gives three copies: a trisomy.

Q17 T52-q17

In 200 cells at metaphase I, the long pair’s maternal member faces the left pole in about 100 of them.

In about how many of those 100 cells does the short pair’s maternal member also face the left pole?

  1. A. All 100
    Each pair faces either way on its own, so in about half of those 100 cells the short pair’s maternal member faces left and in the other half right.
  2. B. None
    There is no balancing between pairs.
    Each pair lines up independently of the others, so the short pair’s maternal member faces left in about half of the cells.
  3. C. ✓ About 50
  4. D. About 25
    25 would be 25% of the 100 cells; the short pair has two ways to face, each equally likely, so it faces left in about half of the cells.

Why: Each homologous pair lines up at metaphase I facing either way regardless of the other pairs.
So among the 100 cells in which the long pair’s maternal member faces left, the short pair’s maternal member faces left in about half, 50, and right in the other half.

Q18 T52-q18

In a lizard, one homolog carries alleles Q and R at two positions and the other carries q and r. After meiosis with one crossover between the two positions, a gamete carries a chromosome reading q with r.

Which of the following describes the combination q with r?

  1. A. Recombinant
    The parent’s homologs carried Q with R and q with r, so q with r is a combination one homolog already had.
  2. B. ✓ Parental
  3. C. Made by independent orientation
    Independent orientation decides which whole chromosome of each pair a gamete receives.
    It changes no combination along a chromosome.
  4. D. Carried by a sister chromatid of the Q R chromosome
    Both sister chromatids of the Q R chromosome read Q with R.
    q with r comes from the other homolog.

Why: The parent’s two homologs carried Q with R and q with r.
A combination one homolog already carried is parental.
So q with r is parental; the crossover made the recombinant combinations Q with r and q with R.

FRQ 1 T52-frq1 · Scientific Investigation

A gorilla’s body cells hold 48 chromosomes. A researcher counted the chromosomes in each of the four cells produced by one meiosis in a male gorilla; the counts are drawn below.

The four cells from one meiosis in a gorilla, with their chromosome counts: 25, 23, 24 and 24.
The four cells from one meiosis in a gorilla, with their chromosome counts: 25, 23, 24 and 24.

(a) Determine the division in which the failure happened, using the counts. (1 pt)

Model answer The failure was in meiosis II.
A normal cell from this meiosis holds 24 chromosomes.
Two of the four cells hold 24, so meiosis I sent one member of every pair to each of the two cells it made.
The other two cells, 25 and 23, came from one of those two cells.
So that cell failed when it divided, in meiosis II.
Rubric
  • Award 1 point for: the decision (meiosis II) AND the ground (two cells hold the normal count of 24, so meiosis I was normal and only one of its two cells then divided unevenly).

Slip Choosing meiosis I because there is a cell over and a cell under. Both patterns have that; a failure in meiosis I leaves no cell with the normal count, and two cells here hold 24.

(b) Describe what happened at anaphase in the division that failed. (1 pt)

Model answer At anaphase II the two sister chromatids of one chromosome did not separate.
Both chromatids went to the same pole.
So the cell at that pole received both chromatids, 25 chromosomes, and the cell at the other pole received neither chromatid of that chromosome, 23 chromosomes.
Rubric
  • Award 1 point for: at anaphase II both sister chromatids of one chromosome went to the same pole (nondisjunction).

Slip Saying the two members of a homologous pair failed to part. Homologous pairs part in meiosis I; in meiosis II the partners are the two sister chromatids of one chromosome.

(c) Predict the chromosome number of the zygote made when the 23-chromosome sperm fertilizes a normal egg, and identify the zygote as a trisomy or a monosomy. (1 pt)

Model answer 47 chromosomes.
The sperm brings 23 and the normal egg brings 24.
So the zygote holds two copies of every chromosome except one, which it holds once.
One copy of one chromosome is a monosomy.
Rubric
  • Award 1 point for: 47, a monosomy (one copy of one chromosome).

Slip Calling it a trisomy. A gamete with one chromosome too few gives one copy of that chromosome, a monosomy.

(d) In a second gorilla cell, one homologous pair undergoes nondisjunction in meiosis I. Predict the chromosome count of each of the four cells that this meiosis produces. (1 pt)

Model answer 25, 25, 23 and 23.
In meiosis I both members of the failing pair go to one cell.
So that cell holds 25 chromosomes and the other cell holds 23.
Meiosis II parts the sister chromatids of every chromosome, so it changes no count.
So the 25 cell gives two cells of 25 and the 23 cell gives two cells of 23.
Rubric
  • Award 1 point for: 25, 25, 23, 23 (any order).

Slip Giving 25, 23, 24, 24. That is the pattern of a failure in meiosis II; a failure in meiosis I leaves no cell with the normal count of 24.

FRQ 2 T52-frq2 · Conceptual Analysis

In a species of fish, one gene position sits on the long pair of chromosomes and a second gene position sits on the short pair. In the mother, one homolog of the long pair carries A and the other carries a; one homolog of the short pair carries B and the other carries b. The father carries the same alleles the same way.

(a) Explain how one mother can make both eggs carrying A with B and eggs carrying A with b. (1 pt)

Model answer The A position and the B position sit on different chromosome pairs.
Each homologous pair lines up at metaphase I facing either way, regardless of the other pairs.
So which member of each pair an egg receives is decided pair by pair.
Therefore the chromosome carrying A can go into an egg with the chromosome carrying B, or with the one carrying b.
Rubric
  • Award 1 point for: independent orientation, each pair facing either way at metaphase I on its own, so A goes with B or with b.

Slip Naming crossing over. The two genes are on different pairs, so no exchange between chromatids is needed for A to go with b.

(b) Suppose instead that both positions sat on the same chromosome, with A and B on one homolog and a and b on the other. Describe how a gamete could still receive A with b. (1 pt)

Model answer In prophase I the two homologs lie paired.
A chromatid of one homolog and a chromatid of the other break at the same point, between the two gene positions.
The two chromatids exchange the broken pieces.
So one chromatid now carries A with b, a combination neither homolog had.
Rubric
  • Award 1 point for: crossing over between non-sister chromatids in prophase I, breaking at the same point between the genes and exchanging pieces.

Slip Saying sister chromatids swap. Sister chromatids are identical copies; a swap between them changes nothing.

(c) Explain how random fertilization adds variation beyond what meiosis made in each parent. (1 pt)

Model answer Meiosis made every egg and every sperm a different combination.
Any egg can fuse with any sperm.
So each zygote joins one of the mother’s kinds of egg with one of the father’s kinds of sperm.
The kinds of zygote are therefore the kinds of egg multiplied by the kinds of sperm.
Rubric
  • Award 1 point for: any egg can fuse with any sperm, so the combinations multiply beyond the gametes’ own variety.

Slip Saying fertilization makes the gametes vary. The gametes varied already; fertilization multiplies the combinations by pairing them at random.

APBIO-U05-L12 The white flower that came back

Topic 5.3 · Mendelian Genetics · 66 steps

Left: two photographs of real pea flowers, a purple-flowered one and a white-flowered one, the two parents. Right: the cross drawn as three rows of flower symbols: the purple parent crossed with the white parent at the top; a row of six purple-flowered plants grown from their seeds in the middle; and at the bottom the next generation, three purple and one white
Left: two photographs of real pea flowers, a purple-flowered one and a white-flowered one, the two parents. Right: the cross drawn as three rows of flower symbols: the purple parent crossed with the white parent at the top; a row of six purple-flowered plants grown from their seeds in the middle; and at the bottom the next generation, three purple and one white

Photos: Vikiçizer, Wikimedia Commons, CC BY-SA 4.0; net_efekt, Flickr via Wikimedia Commons, CC BY 2.0 (both resized).

Here is a row of pea plants, every one of them purple-flowered. They grew from the seeds of a purple-flowered plant crossed with a white-flowered one.

Their own seeds were sown the next year. Most of the new plants are purple again, but one plant in every four is white. Where did the white come from, and where was it hiding?

Unit 5 · Heredity

1A trait and a true-breeding line

2

Video: Watch: A trait and a true-breeding line

The two pea flowers, purple and white; a line of purple plants giving purple flowers year after year; then a line that gives both colors, which is not true-breeding.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12a.mp4

3

Where did the white flower go for a whole generation, and how did it come back?

4

Gregor Mendel began with two lines of pea plants, in the 1850s and 1860s.

5

One line gave only purple flowers, generation after generation. The other line gave only white flowers, generation after generation.

6

He crossed the two lines. Every offspring plant flowered purple.

7

Then he crossed those purple offspring with each other. Among their offspring, 705 plants flowered purple and 224 flowered white: about three purple to one white.

8

The white did not vanish. Every purple plant in between carried it, unseen.

9

Flower color is a feature of the plant that we can see. A feature we can see, such as flower color or seed shape, is called a .

10

Purple flowers and white flowers are two versions of one trait, flower color. Round seeds and wrinkled seeds are two versions of another trait, seed shape.

11

Mendel's purple line gave only purple flowers, in every plant, year after year.

12

When a line gives only its own version of a trait, generation after generation, we call the line , because every plant it breeds stays true to the line.

13

A line whose plants gave purple flowers one year and some white flowers the next is not true-breeding for flower color.

14

What you are expected to know Say whether a described line of plants is true-breeding for a trait.

15
Check q1

A gardener sowed a new packet of pea seeds this spring. Every plant flowered purple.

Is this line true-breeding for flower color?

  1. A. Yes
    One generation has flowered.
    A true-breeding line gives only its own version of the trait generation after generation.
  2. B. No
    Nothing so far rules the line out.
    The line has flowered for one generation only, so the next generations decide.
  3. C. ✓ Cannot tell yet

Why: A true-breeding line gives only its own version of a trait, generation after generation.
One spring of only purple flowers is one generation.
So the gardener cannot tell yet.

16Quick quiz: trait, true-breeding mixed practice

17
Check q2

A gardener describes one pea plant.

Which of the following describes a trait of the plant?

  1. A. The plant came from a true-breeding line
    Coming from a true-breeding line says where the plant came from.
    A trait is a feature of the plant we can see.
  2. B. ✓ The plant's seeds are round
  3. C. The plant is the offspring of two crossed lines
    Being the offspring of two crossed lines says where the plant came from.
    A trait is a feature of the plant we can see.

Why: A trait is a feature of an organism that we can see.
Round seeds can be seen.
So round seeds is a trait of the plant.

18
Check q3

A breeder keeps several lines of pea plants.

Which of the following describes a true-breeding line?

  1. A. A line that gave only one version of a trait in its first year
    One year is one generation.
    A true-breeding line gives only its own version of the trait generation after generation.
  2. B. A line whose plants show two versions of a trait, some plants each, every year
    A line whose plants show two versions of a trait does not stay true to one version.
  3. C. ✓ A line that gives only its own version of a trait, generation after generation

Why: A true-breeding line gives only its own version of a trait.
It does so generation after generation.

19
Check q4

A pea line has given only round seeds for eight generations.

Is the line true-breeding for seed shape?

  1. A. ✓ Yes
  2. B. No
    Every generation for eight generations gave only round seeds, the line's own version of seed shape.
  3. C. Cannot tell yet
    Eight generations is generation after generation.
    Every one gave only round seeds.

Why: The line gave only round seeds, its own version of seed shape.
It did so for eight generations, generation after generation.
So the line is true-breeding for seed shape.

20
Check q5

A pea line gave only tall plants last year. This year some of its plants are tall and some are short.

Is the line true-breeding for stem height?

  1. A. Yes
    This year the line gave two versions of stem height, tall and short.
    A true-breeding line gives only its own version.
  2. B. ✓ No
  3. C. Cannot tell yet
    The line has already given two versions of the trait.
    Later generations cannot undo that.

Why: This year the line gave tall plants and short plants.
A true-breeding line gives only its own version of the trait.
So the line is not true-breeding for stem height.

21
Check q6

A pea line has given only yellow seeds, generation after generation.

Is the line true-breeding for seed color?

  1. A. ✓ Yes
  2. B. No
    Every generation gave only yellow seeds, the line's own version of seed color.
  3. C. Cannot tell yet
    The line has been followed generation after generation, and every one gave only yellow seeds.

Why: The line gave only yellow seeds, its own version of seed color.
It did so generation after generation.
So the line is true-breeding for seed color.

22
Check q7

Most plants of a pea line have purple flowers, but every year a few of its plants flower white.

Is the line true-breeding for flower color?

  1. A. Yes
    Every year the line gives two versions of flower color, purple and white.
    A true-breeding line gives only its own version.
  2. B. ✓ No
  3. C. Cannot tell yet
    The line has been followed year after year, and every year some plants flower white.

Why: Every year the line gives purple flowers and white flowers.
A true-breeding line gives only its own version of the trait.
So the line is not true-breeding for flower color.

23
Practice writing an answer

A breeder has kept one line of pea plants for ten years. Every year, every seed the line made was wrinkled.

(a) Explain how the ten years of wrinkled seeds show that the line is true-breeding for seed shape. (1 pt)

Frame The line is true-breeding for seed shape because …

Model answer The line is true-breeding for seed shape because it gave only wrinkled seeds every year for ten years.
Wrinkled is the line's own version of seed shape.
So the line gave only its own version of the trait, generation after generation.
Rubric
  • Award 1 point for: the line gave only its own version of the trait (wrinkled seeds) generation after generation (ten years), which is what true-breeding means.

24P, F1, F2: naming the generations

25

Video: Watch: P, F1, F2: naming the generations

Mendel's cross drawn as a tree of flowers; the labels P, F1 and F2 written beside each row as the row appears.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12b.mp4

26

Here is a drawing of Mendel's cross as a tree of flowers, one row per generation.

Mendel's crosses drawn as a tree: a true-breeding purple line crossed with a true-breeding white line, the P generation; all their offspring purple, the F1; the F1 crossed with each other giving 705 purple and 224 white, the F2, about three to one
Mendel's crosses drawn as a tree: a true-breeding purple line crossed with a true-breeding white line, the P generation; all their offspring purple, the F1; the F1 crossed with each other giving 705 purple and 224 white, the F2, about three to one
27

Mendel crossed a plant from the true-breeding purple line with a plant from the true-breeding white line. These two plants are the parents of the cross.

28

Because these two plants are the parents, we call them the , P for parental.

29

The offspring of the P generation are the children of the cross. We call them the F1 generation: F for filial, which means of the children, and 1 because they are the first generation of children.

30

Mendel then crossed the F1 plants with each other. Their offspring are the second generation of children, so we call them the F2 generation.

31

To name a generation, count the generations after the P generation: the first is F1, the second is F2.

32

What you are expected to know Name the generation a plant belongs to when two lines are crossed and their offspring are crossed again: P, F1 or F2.

33
Check q8

Mendel crossed two F1 pea plants with each other.

Which generation are their offspring?

  1. A. P
    The P generation is the two plants crossed first.
    These offspring are two generations after them.
  2. B. F1
    The F1 plants are the parents here.
    Their offspring are the next generation.
  3. C. ✓ F2

Why: The F1 plants are the first generation after the P generation.
Their offspring are the second generation after the P generation.
So their offspring are the F2.

34Quick quiz: P generation, F1, F2 mixed practice

35
Check q9

Two true-breeding pea lines are crossed, and the cross is followed for two more generations.

Which plants are the P generation?

  1. A. ✓ The two plants crossed first
  2. B. The offspring of the two plants crossed first
    The offspring of the two plants crossed first are the first generation of children, the F1.
  3. C. The offspring of the F1 plants
    The offspring of the F1 plants are the second generation of children, the F2.

Why: The P generation is the parents of the cross.
The parents are the two plants crossed first.

36
Check q10

Two true-breeding pea lines are crossed, and the cross is followed for two more generations.

Which plants are the F1 generation?

  1. A. The two true-breeding plants crossed first
    The two plants crossed first are the parents, the P generation.
  2. B. ✓ The offspring of the P generation
  3. C. The offspring of the F1 plants crossed with each other
    The offspring of the F1 plants are the second generation of children, the F2.

Why: F stands for filial, of the children.
The F1 are the first generation of children: the offspring of the P generation.

37
Check q11

A true-breeding tall pea plant is crossed with a true-breeding short pea plant.

Which generation are these two plants?

  1. A. ✓ P
  2. B. F1
    The F1 are the offspring of the two plants crossed first.
    These two plants are the ones crossed first.
  3. C. F2
    The F2 are two generations after the two plants crossed first.

Why: The two plants crossed first are the parents of the cross.
So they are the P generation.

38
Check q12

Two true-breeding pea plants are crossed. Their offspring are all tall.

Which generation are the tall offspring?

  1. A. P
    The P generation is the two plants crossed.
    These plants are their offspring.
  2. B. ✓ F1
  3. C. F2
    The F2 are the offspring of the F1 plants.
    These plants are the offspring of the P generation.

Why: The two plants crossed are the P generation.
Their offspring are the first generation of children.
So the tall offspring are the F1.

39
Check q13

A round-seeded F1 pea plant grew from a seed made by a cross between two true-breeding plants.

Which generation were the two true-breeding plants?

  1. A. ✓ P
  2. B. F1
    The F1 plant is the child.
    The two plants that made its seed are its parents.
  3. C. F2
    The F2 come after the F1, not before.

Why: The two true-breeding plants made the seed the F1 plant grew from.
So they are the parents of the cross, the P generation.

40
Check q14

A grower crosses two true-breeding pea lines and keeps the seeds.

Which generation are the plants that grow from those seeds?

  1. A. P
    The P generation is the two lines crossed.
    The seeds are their offspring.
  2. B. ✓ F1
  3. C. F2
    The F2 come from crossing the F1 plants with each other.
    These seeds come straight from the P generation.

Why: The two crossed plants are the P generation.
The seeds they make grow into their offspring, the first generation of children.
So the plants are the F1.

41
Check q15

F1 pea plants are crossed with each other. Their seeds grow into plants.

Which generation are those plants?

  1. A. P
    The P generation is the two plants crossed first, two generations before these.
  2. B. F1
    The F1 plants are the parents here.
    These plants are their offspring.
  3. C. ✓ F2

Why: The F1 plants are the first generation of children.
Their offspring are the second generation of children.
So those plants are the F2.

42
Practice writing an answer

Mendel crossed two true-breeding pea lines. He then crossed the offspring of that cross with each other and grew their seeds.

(a) Explain why the plants grown from those seeds belong to the F2 generation. (1 pt)

Frame The plants belong to the F2 generation because …

Model answer The plants belong to the F2 generation because the two true-breeding plants Mendel crossed first are the P generation.
Their offspring are the first generation of children after the P generation, the F1.
The plants grown from the F1 plants' seeds are the second generation of children after the P generation.
So they are the F2.
Rubric
  • Award 1 point for: the plants are the second generation of offspring after the P generation (the offspring of the F1), and F counts the generations of children.

43Mendel's two crosses: the white came back

44

Video: Watch: Mendel's two crosses: the white came back

The tree of flowers again: the all-purple F1 row, then the F2 row with 705 purple and 224 white, and the white flower traced back through the purple F1 to the white parent.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12c.mp4

45

Every one of the F1 plants flowered purple: the white had vanished.

Mendel's crosses drawn as a tree: a true-breeding purple line crossed with a true-breeding white line, the P generation; all their offspring purple, the F1; the F1 crossed with each other giving 705 purple and 224 white, the F2, about three to one
Mendel's crosses drawn as a tree: a true-breeding purple line crossed with a true-breeding white line, the P generation; all their offspring purple, the F1; the F1 crossed with each other giving 705 purple and 224 white, the F2, about three to one
46

Mendel crossed the F1 plants with each other. The F2 plants were 705 purple and 224 white: about three purple to one white.

47

The white had come back, unchanged, after a generation in which no plant showed it.

48

So the F1 plants had not lost the white trait. Each F1 plant carried the white trait without showing it.

49

A trait can be present in a plant without showing.

50

What you are expected to know Say what the white F2 plants show about the purple F1 plants.

51
Check q16

Mendel also crossed a true-breeding round-seeded pea line with a true-breeding wrinkled-seeded line. Every F1 seed was round. When the F1 plants were crossed with each other, the F2 seeds were 5,474 round and 1,850 wrinkled.

About what percentage of the F2 seeds were wrinkled?

  1. A. ✓ About 25%
  2. B. About 50%
    The F2 seeds were 5,474 round and 1,850 wrinkled, about three round to one wrinkled.
    One in four is 25%.
  3. C. About 75%
    About 75% of the F2 seeds were round.
    The wrinkled seeds were the other quarter.

Why: The F2 seeds were 5,474 round and 1,850 wrinkled.
1,850 out of 7,324 seeds is about 25%.
So about 25% of the F2 seeds were wrinkled: about three round to one wrinkled.

52
Check q17

Mendel crossed a true-breeding round-seeded pea line with a true-breeding wrinkled-seeded line. Every F1 seed was round. When the F1 plants were crossed with each other, the F2 seeds were 5,474 round and 1,850 wrinkled.

Which of the following happened to the wrinkled trait in the F1 generation?

  1. A. The F1 plants lost the wrinkled trait
    The wrinkled trait came back in the F2 in the same form as before.
    A trait that comes back unchanged was never lost.
  2. B. The F1 plants blended the wrinkled trait with the round trait
    Every F1 seed was fully round, with no blend.
    The F2 seeds were each round or wrinkled, not a blend.
  3. C. ✓ The F1 plants carried the wrinkled trait without showing it
  4. D. The F1 plants changed the wrinkled trait into the round trait
    The wrinkled trait came back in the F2 unchanged.
    A trait that comes back unchanged was not changed into another.

Why: Every F1 seed was round, so no F1 plant showed the wrinkled trait.
In the F2 the wrinkled trait returned unchanged, in about 25% of the seeds.
A trait that returns unchanged was carried through the generation between.
So every F1 plant carried the wrinkled trait without showing it.

53
Practice writing an answer

Mendel crossed a true-breeding round-seeded pea line with a true-breeding wrinkled-seeded line. Every F1 seed was round. When the F1 plants were crossed with each other, the F2 seeds were 5,474 round and 1,850 wrinkled.

(a) Explain how the F2 seeds show that the F1 plants carried the wrinkled trait. (1 pt)

Frame The F1 plants carried the wrinkled trait because …

Model answer The F1 plants carried the wrinkled trait because the wrinkled trait came back in the F2.
About 25% of the F2 seeds were wrinkled.
Those seeds grew on F1 plants, and every F1 plant had itself grown from a round seed.
So the wrinkled trait passed through the F1 plants unseen.
A trait that passes through a generation unseen is carried by that generation.
Rubric
  • Award 1 point for: the wrinkled trait reappeared in the F2 (about 25% of the seeds), so the round-seeded F1 plants must have carried it unseen.
54
Check q18

A student looks at Mendel's flower-color cross and says: “The white trait was destroyed in the F1 plants and made again, from nothing, in the F2.”

Which of the following statements about the student's claim is correct?

  1. A. The student is right
    The white trait came back in the F2 exactly as it was in the white parent.
    A trait that comes back unchanged was carried, not made anew.
  2. B. ✓ The student is wrong

Why: The white trait came back in the F2 exactly as it was in the white parent.
A trait that comes back unchanged was carried through the F1.
So the F1 plants carried the white trait without showing it, and the student is wrong.

55

Back to the row of purple-flowered pea plants grown from a purple-flowered plant crossed with a white-flowered one: they are F1 plants, like Mendel's.

56

Among their offspring, the F2, one plant in every four flowered white.

57

The white came from the white parent. Every purple F1 plant carried the white trait without showing it, and the white F2 plants brought it back into view.

58Mixed practice mixed practice

59
Check q19

A true-breeding tall pea line is crossed with a true-breeding short line. Every F1 plant is tall. The F1 plants are crossed with each other.

Which of the following does Mendel's pattern predict for the F2 plants?

  1. A. All tall
    A trait that vanishes in the F1 is carried, not lost.
    It returns in the F2.
  2. B. ✓ About three tall to one short
  3. C. About half tall and half short
    The F1 × F1 cross gives the vanished trait in about 25% of the offspring.
  4. D. All of medium height
    Mendel's F1 plants all showed one trait, with no blend.
    The pattern predicts each F2 plant tall or short, not in between.

Why: Mendel's crosses give all one trait in the F1 and about three to one in the F2.
The short trait, carried unseen in the F1, comes back in about 25% of the F2.

60
Check q20

A breeder's line of guinea pigs has had only black coats for twelve generations.

Is the line true-breeding for coat color?

  1. A. ✓ Yes
  2. B. No
    Twelve generations gave only black coats, the line's own version of coat color.

Why: The line gave only black coats, its own version of coat color.
It did so for twelve generations, generation after generation.
So the line is true-breeding for coat color.

61
Check q21

A true-breeding purple-kernel maize line is crossed with a true-breeding yellow-kernel line. The kernels from that cross grow into plants.

Which generation are those plants?

  1. A. P
    The P generation is the two lines crossed.
    These plants grew from their kernels.
  2. B. ✓ F1
  3. C. F2
    The F2 come from crossing the F1 plants with each other, one generation later.

Why: The two crossed plants are the P generation.
The kernels they make grow into their offspring, the first generation of children.
So those plants are the F1.

62
Check q22

Mendel crossed a true-breeding yellow-seeded pea line with a true-breeding green-seeded line. Every F1 seed was yellow. When the F1 plants were crossed with each other, the F2 seeds were 6,022 yellow and 2,001 green.

What does the return of green seeds in the F2 show about the F1 plants?

  1. A. The F1 plants had lost the green trait
    A trait that comes back unchanged in the F2 was never lost.
  2. B. The F1 plants blended the yellow trait with the green trait
    Every F1 seed was fully yellow, and the F2 green seeds were fully green: no blend.
  3. C. ✓ The F1 plants carried the green trait without showing it

Why: Every F1 seed was yellow, so no F1 plant showed the green trait.
The green trait came back unchanged in about 25% of the F2 seeds.
A trait that comes back unchanged was carried through the F1.
So the F1 plants carried the green trait without showing it.

63
Check q23

In Mendel's flower-color cross, the white trait vanished for one generation and then came back.

Which generation was the first to show white flowers again?

  1. A. P
    The white parent belongs to the P generation.
    The white vanished after it and came back later.
  2. B. F1
    Every F1 plant flowered purple.
  3. C. ✓ F2

Why: The white parent is in the P generation.
Every F1 plant flowered purple.
The F2 plants were 705 purple and 224 white.
So the F2 was the first generation to show white again.

64
Check q24

A pea line has given only inflated pods for twenty generations. This year a grower crosses one of its plants with a plant from a true-breeding constricted-pod line.

Which of the following describes the inflated-pod line?

  1. A. ✓ A true-breeding line
  2. B. An F1 generation
    The F1 are the offspring of two crossed plants.
    This line existed for twenty generations before the grower crossed it.
  3. C. A trait
    A trait is a feature we can see, such as inflated pods.
    The line is the plants that carry it.

Why: The line gave only inflated pods, its own version of pod shape.
It did so for twenty generations.
So the inflated-pod line is a true-breeding line.

65
Practice writing an answer

A true-breeding purple-kernel maize line is crossed with a true-breeding yellow-kernel line. Every kernel from the cross is purple. The plants grown from those kernels, the F1, are crossed with each other, and their kernels are counted: 300 purple and 100 yellow.

(a) Explain how the counts of purple and yellow kernels show that the F1 plants carried the yellow trait. (1 pt)

Frame The F1 plants carried the yellow trait because …

Model answer The F1 plants carried the yellow trait because the yellow trait came back in the kernels their crosses made.
Every kernel from the first cross was purple, so no F1 plant showed the yellow trait.
The F1 crosses gave 300 purple and 100 yellow kernels, so the yellow trait came back in about 25% of them.
A trait that comes back unchanged after a generation was carried through that generation.
Rubric
  • Award 1 point for: the yellow trait reappeared in about 25% of the F2 kernels (100 of 400) although every F1 kernel was purple, so the F1 plants carried it unseen.

Glossary

trait
A feature of an organism that can be seen, such as flower color or seed shape. Purple flowers and white flowers are two versions of one trait.
true-breeding
A line of plants or animals that gives only its own version of a trait, generation after generation. Mendel's purple line gave only purple flowers.
P generation, F1, F2
The P generation is the parents, the two true-breeding lines Mendel started from. The F1 is their offspring (all purple in the purple × white cross). The F2 is the offspring of the F1 crossed with each other (705 purple and 224 white: about three to one). F stands for filial, of the children.

APBIO-U05-L12A Two alleles, one on each homolog

Topic 5.3 · Mendelian Genetics · 59 steps

One F1 pea plant's homologous pair: two rods side by side, each with a centromere dot, the darker rod carrying P and labelled from the purple parent, the lighter rod carrying p and labelled from the white parent; to the right, the question: how do we write down what this plant carries?
One F1 pea plant's homologous pair: two rods side by side, each with a centromere dot, the darker rod carrying P and labelled from the purple parent, the lighter rod carrying p and labelled from the white parent; to the right, the question: how do we write down what this plant carries?

Now consider one F1 plant from Mendel’s cross. Here is its homologous pair for the flower-color gene, drawn as two rods, each marked with a dot at its centromere, because the cell is not dividing.

The homolog from the purple parent carries the purple allele, P. The homolog from the white parent carries the white allele, p. How do we write down what this plant carries, and what do we call it?

Unit 5 · Heredity

1Write the genotype

2

Video: Watch: Write the genotype

The F1 plant's homologous pair with P and p written on at the same height; the two letters copied out beneath as Pp; then one homolog moving into a gamete, carrying one letter.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Aa.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Aa.mp4

3

How do we write down the two alleles a plant carries?

4

A plant carries two alleles of each gene, one on each homolog. We write the two as a pair of letters: this F1 plant is Pp.

5

A gamete carries one homolog, so it carries one allele of each gene.

6

Every cross that follows is worked out from these pairs of letters.

7
Check q1

A pea plant’s homologous pair carries the flower-color gene at the same position on both homologs, and the two copies of the gene can be different versions.

What do we call two different versions of one gene?

  1. A. ✓ Alleles
  2. B. Homologs
    Homologs are the two chromosomes of a pair.
    The versions of a gene they carry are alleles.

Why: Two different versions of one gene are two alleles of that gene.

8

A homologous pair carries the same genes at the same positions, one homolog from each parent.

A homologous pair drawn as two rods side by side, each with a centromere dot near its upper third, the darker rod from the mother and the lighter rod from the father, with the flower-color gene at the same height on both: the darker rod carries P, the lighter rod carries p
A homologous pair drawn as two rods side by side, each with a centromere dot near its upper third, the darker rod from the mother and the lighter rod from the father, with the flower-color gene at the same height on both: the darker rod carries P, the lighter rod carries p
9

At any one position, the two homologs may carry the same allele or two different alleles.

10

Suppose the purple parent is the mother. In every drawing here, the homolog from the mother is the darker rod and the homolog from the father is the lighter rod. Each homolog is one rod with a centromere dot, because this cell is not dividing.

11

The flower-color gene has two alleles. We write the purple allele P and the white allele p.

12

Every pea plant carries two alleles of the flower-color gene, one on each homolog: P and P, P and p, or p and p.

Three homologous pairs side by side, each rod with a centromere dot: the first pair carries P on both rods, written PP; the second carries P on one rod and p on the other, written Pp; the third carries p on both, written pp
Three homologous pairs side by side, each rod with a centromere dot: the first pair carries P on both rods, written PP; the second carries P on one rod and p on the other, written Pp; the third carries p on both, written pp
13

When we write down the two alleles a plant carries for a gene, one from each homolog, that pair of letters is called the plant's , because it describes the plant by its genes, not its looks.

14

The three genotypes for flower color are PP, Pp and pp.

15

A plant's genotype is always two letters, because a body cell carries two homologs.

16

A gamete carries one homolog, so it carries one allele. So a gamete's genotype is one letter, P or p, and the pair of letters is the plant's.

Left: a body cell's homologous pair carrying P and p, labelled genotype Pp. Right: two gametes drawn as small circles, one holding a rod carrying P and one holding a rod carrying p, labelled one allele each
Left: a body cell's homologous pair carrying P and p, labelled genotype Pp. Right: two gametes drawn as small circles, one holding a rod carrying P and one holding a rod carrying p, labelled one allele each
17

What you are expected to know Write a plant's genotype for one gene as its two alleles, one from each homolog.

18
Check q2

In pea plants, the stem-height gene has two alleles, T and t. The homologous pair of one plant is drawn below.

A homologous pair, two rods each with a centromere dot, carrying the stem-height gene at the same height on both: the darker rod carries T, the lighter rod carries t
A homologous pair, two rods each with a centromere dot, carrying the stem-height gene at the same height on both: the darker rod carries T, the lighter rod carries t

Which of the following is this plant's genotype for stem height?

  1. A. TT
    The lighter homolog, from the father, carries t.
    So the two alleles are T and t, not T and T.
  2. B. ✓ Tt
  3. C. T
    A genotype is both alleles, one from each homolog.
    So a genotype is two letters.

Why: The darker homolog carries T and the lighter homolog carries t.
A genotype is both alleles, one from each homolog.
So the genotype is Tt.

19
Check q3

A student looks at the gamete drawn below, from a Tt pea plant, and says: “This gamete's genotype is Tt.”

One gamete drawn as a circle holding a single rod with a centromere dot; the rod carries T at the stem-height gene
One gamete drawn as a circle holding a single rod with a centromere dot; the rod carries T at the stem-height gene

Is the student correct?

  1. A. Yes
    A gamete carries one homolog of the pair, so one allele.
    Its genotype is one letter, T.
  2. B. ✓ No

Why: A gamete carries one homolog of the pair.
So it carries one allele, T.
Its genotype is T, one letter; Tt is the plant's genotype.
The student is wrong.

20Quick quiz: genotype mixed practice

21
Check q4

A pea plant's flower-color gene has the alleles P and p.

Which of the following is the plant's genotype for flower color?

  1. A. The one allele its gamete carries for the gene
    A gamete carries one allele.
    A genotype is the two alleles the plant carries.
  2. B. The version of the trait it shows, purple or white
    The trait a plant shows is its looks.
    A genotype describes the plant by its genes: the two alleles it carries.
  3. C. ✓ The two alleles it carries for the gene, one on each homolog

Why: A genotype is the two alleles a plant carries for a gene, one on each homolog.
For flower color it is PP, Pp or pp.

22
Check q5

The homologous pair of a pea plant is drawn below. The stem-height gene has the alleles T and t.

A homologous pair, two rods each with a centromere dot, carrying the stem-height gene at the same height on both: both rods carry T
A homologous pair, two rods each with a centromere dot, carrying the stem-height gene at the same height on both: both rods carry T

Which of the following is the plant's genotype?

  1. A. ✓ TT
  2. B. Tt
    Both homologs carry T.
  3. C. tt
    Both homologs carry T.

Why: Both homologs carry T.
A genotype is both alleles.
So the genotype is TT.

23
Check q6

The homologous pair of a pea plant is drawn below. The seed-shape gene has the alleles R and r.

A homologous pair, two rods each with a centromere dot, carrying the seed-shape gene at the same height on both: both rods carry r
A homologous pair, two rods each with a centromere dot, carrying the seed-shape gene at the same height on both: both rods carry r

Which of the following is the plant's genotype?

  1. A. RR
    Both homologs carry r.
  2. B. Rr
    Both homologs carry r.
  3. C. ✓ rr

Why: Both homologs carry r.
A genotype is both alleles.
So the genotype is rr.

24
Check q7

The homologous pair of a pea plant is drawn below. The flower-color gene has the alleles P and p.

A homologous pair, two rods each with a centromere dot, carrying the flower-color gene at the same height on both: the darker rod carries p, the lighter rod carries P
A homologous pair, two rods each with a centromere dot, carrying the flower-color gene at the same height on both: the darker rod carries p, the lighter rod carries P

Which of the following is the plant's genotype?

  1. A. PP
    The lighter homolog carries P, but the darker homolog carries p.
  2. B. ✓ Pp
  3. C. pp
    The darker homolog carries p, but the lighter homolog carries P.

Why: One homolog carries p and the other carries P.
Which homolog carries which does not matter.
So the genotype is Pp.

25
Check q8

One gamete from a pea plant is drawn below. The seed-shape gene has the alleles R and r.

One gamete drawn as a circle holding a single rod with a centromere dot; the rod carries R at the seed-shape gene
One gamete drawn as a circle holding a single rod with a centromere dot; the rod carries R at the seed-shape gene

How many alleles of the seed-shape gene does this gamete carry?

  1. A. ✓ One
  2. B. Two
    A gamete carries one homolog of the pair.
    So a gamete carries one allele.

Why: A gamete carries one homolog of each pair.
One homolog carries one allele of the gene.
So this gamete carries one allele, R.

26
Practice writing an answer

In pea plants, the pod-color gene has the alleles G and g. One plant's homologous pair carries G on the homolog from the mother and g on the homolog from the father.

(a) Explain why this plant's genotype for pod color is written with two letters, Gg. (1 pt)

Frame The genotype is written with two letters because …

Model answer The genotype is written with two letters because the plant's body cells carry two homologs of the pair.
The homolog from the mother carries G and the homolog from the father carries g.
A genotype lists both alleles the plant carries, one from each homolog.
So the genotype is Gg.
Rubric
  • Award 1 point for: a genotype is both alleles the plant carries, one on each of its two homologs (G on one, g on the other), so it takes two letters.

27Homozygous or heterozygous

28

Video: Watch: Homozygous or heterozygous

Four drawn pairs in turn, TT, Tt, tT and tt, each judged the same or different; the table of the three flower-color genotypes filled in row by row.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Ab.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Ab.mp4

29

Look at the three genotypes again: PP, Pp and pp.

Three homologous pairs side by side, each rod with a centromere dot: the first pair carries P on both rods, written PP; the second carries P on one rod and p on the other, written Pp; the third carries p on both, written pp
Three homologous pairs side by side, each rod with a centromere dot: the first pair carries P on both rods, written PP; the second carries P on one rod and p on the other, written Pp; the third carries p on both, written pp
30

When the two alleles are the same, PP or pp, we call the plant for the gene, because homo- means the same.

31

When the two alleles differ, Pp, we call the plant for the gene, because hetero- means different.

32

For example, here is a pair carrying T on both homologs. This plant is homozygous, because its two alleles are the same.

A homologous pair, two rods each with a centromere dot, both carrying T at the same height
A homologous pair, two rods each with a centromere dot, both carrying T at the same height
33

But here is a pair carrying T on the darker homolog and t on the lighter homolog. This plant is heterozygous, because its two alleles differ.

A homologous pair, two rods each with a centromere dot, the darker rod carrying T and the lighter rod carrying t at the same height
A homologous pair, two rods each with a centromere dot, the darker rod carrying T and the lighter rod carrying t at the same height
34

And here is a pair carrying t on the darker homolog and T on the lighter homolog. This plant is still heterozygous, because its two alleles differ: which homolog carries which does not matter.

A homologous pair, two rods each with a centromere dot, the darker rod carrying t and the lighter rod carrying T at the same height
A homologous pair, two rods each with a centromere dot, the darker rod carrying t and the lighter rod carrying T at the same height
35

But here is a pair carrying t on both homologs. This plant is homozygous, because its two alleles are the same.

A homologous pair, two rods each with a centromere dot, both carrying t at the same height
A homologous pair, two rods each with a centromere dot, both carrying t at the same height
36

We write a heterozygous genotype with the capital letter first: Tt, never tT.

37

Here is a table of the three genotypes for the flower-color gene: the two alleles each carries, and whether it is homozygous or heterozygous.

A table with three columns, genotype, the two alleles, and homozygous or heterozygous, and three rows: PP, P and P, homozygous; Pp, P and p, heterozygous; pp, p and p, homozygous
A table with three columns, genotype, the two alleles, and homozygous or heterozygous, and three rows: PP, P and P, homozygous; Pp, P and p, heterozygous; pp, p and p, homozygous
38

Back to the F1 plant from the opening: its homologous pair carries P from the purple parent on one homolog and p from the white parent on the other.

39

So its genotype is Pp, and it is heterozygous for the flower-color gene.

40

What you are expected to know Say whether a plant's genotype is homozygous or heterozygous.

41
Check q9

The homologous pair of a pea plant is drawn below. The pod-color gene has the alleles G and g.

A homologous pair, two rods each with a centromere dot, carrying the pod-color gene at the same height on both: both rods carry G
A homologous pair, two rods each with a centromere dot, carrying the pod-color gene at the same height on both: both rods carry G

Is the plant homozygous or heterozygous for pod color?

  1. A. ✓ Homozygous
  2. B. Heterozygous
    Both homologs carry G.
    Two of the same allele is homozygous.

Why: Both homologs carry G.
The two alleles are the same.
So the plant is homozygous, GG.

42
Practice writing an answer

In pea plants, the stem-height gene has two alleles, T and t. The homologous pair of one plant is drawn below.

A homologous pair, two rods each with a centromere dot, carrying the stem-height gene at the same height on both: the darker rod carries T, the lighter rod carries t
A homologous pair, two rods each with a centromere dot, carrying the stem-height gene at the same height on both: the darker rod carries T, the lighter rod carries t

(a) Explain why this plant is heterozygous for stem height. (1 pt)

Frame The plant is heterozygous because …

Model answer The plant is heterozygous because its two alleles differ.
The darker homolog, from the mother, carries T.
The lighter homolog, from the father, carries t.
So the genotype is Tt, two different alleles of one gene.
Rubric
  • Award 1 point for: the two homologs carry different alleles (T and t), and two different alleles is heterozygous.

43Quick quiz: homozygous, heterozygous mixed practice

44
Check q10

A pea plant carries two alleles of the seed-shape gene.

Which of the following describes a plant that is homozygous for the gene?

  1. A. Its two alleles for the gene differ
    Two alleles that differ is heterozygous.
    Homo- means the same.
  2. B. It carries one allele for the gene
    Every plant carries two alleles of the gene, one on each homolog.
    Homozygous says the two are the same.
  3. C. ✓ Its two alleles for the gene are the same

Why: Homo- means the same.
A homozygous plant carries two of the same allele: RR or rr.

45
Check q11

In pea plants, the pod-color gene has the alleles G and g.

Which of the following genotypes is heterozygous?

  1. A. GG
    GG is two of the same allele: homozygous.
  2. B. ✓ Gg
  3. C. gg
    gg is two of the same allele: homozygous.

Why: Hetero- means different.
Gg carries two different alleles, G and g.
So Gg is heterozygous.

46
Check q12

The homologous pair of a pea plant is drawn below. The pod-shape gene has the alleles I and i.

A homologous pair, two rods each with a centromere dot, carrying the pod-shape gene at the same height on both: the darker rod carries I, the lighter rod carries i
A homologous pair, two rods each with a centromere dot, carrying the pod-shape gene at the same height on both: the darker rod carries I, the lighter rod carries i

Is the plant homozygous or heterozygous for pod shape?

  1. A. Homozygous
    One homolog carries I and the other carries i.
    Two different alleles is heterozygous.
  2. B. ✓ Heterozygous

Why: One homolog carries I and the other carries i.
The two alleles differ.
So the plant is heterozygous, Ii.

47
Check q13

The homologous pair of a pea plant is drawn below. The pod-shape gene has the alleles I and i.

A homologous pair, two rods each with a centromere dot, carrying the pod-shape gene at the same height on both: both rods carry i
A homologous pair, two rods each with a centromere dot, carrying the pod-shape gene at the same height on both: both rods carry i

Is the plant homozygous or heterozygous for pod shape?

  1. A. ✓ Homozygous
  2. B. Heterozygous
    Both homologs carry i.
    Two of the same allele is homozygous.

Why: Both homologs carry i.
The two alleles are the same.
So the plant is homozygous, ii.

48
Check q14

The homologous pair of a pea plant is drawn below. The seed-shape gene has the alleles R and r.

A homologous pair, two rods each with a centromere dot, carrying the seed-shape gene at the same height on both: the darker rod carries R, the lighter rod carries r
A homologous pair, two rods each with a centromere dot, carrying the seed-shape gene at the same height on both: the darker rod carries R, the lighter rod carries r

Is the plant homozygous or heterozygous for seed shape?

  1. A. Homozygous
    One homolog carries R and the other carries r.
    Two different alleles is heterozygous.
  2. B. ✓ Heterozygous

Why: One homolog carries R and the other carries r.
The two alleles differ.
So the plant is heterozygous, Rr.

49
Check q15

A pea plant's genotype for pod color is written gg.

Is the plant homozygous or heterozygous for pod color?

  1. A. ✓ Homozygous
  2. B. Heterozygous
    The genotype gg is two of the same allele.
    Two of the same allele is homozygous.

Why: The genotype gg lists two alleles, g and g.
The two alleles are the same.
So the plant is homozygous.

50
Practice writing an answer

In pea plants, the pod-color gene has the alleles G and g. One plant's genotype for pod color is gg.

(a) Explain why this plant is homozygous for pod color. (1 pt)

Frame The plant is homozygous because …

Model answer The plant is homozygous because its two alleles are the same.
One homolog carries g and the other homolog also carries g.
So the genotype gg is two of the same allele.
Rubric
  • Award 1 point for: the two homologs carry the same allele (g and g), and two of the same allele is homozygous.

51Mixed practice mixed practice

52
Check q16

Four homologous pairs are drawn below, each carrying the pea pod-color gene with alleles G and g.

Four homologous pairs numbered 1 to 4, each rod with a centromere dot: pair 1 carries G on both rods; pair 2 carries G on the darker rod and g on the lighter; pair 3 carries g on both; pair 4 carries g on the darker rod and G on the lighter
Four homologous pairs numbered 1 to 4, each rod with a centromere dot: pair 1 carries G on both rods; pair 2 carries G on the darker rod and g on the lighter; pair 3 carries g on both; pair 4 carries g on the darker rod and G on the lighter

Which of the pairs belong to heterozygous plants?

  1. A. Pair 2 only
    Pair 4 also carries one G and one g.
    Which homolog carries which does not matter.
  2. B. Pairs 1 and 3
    Pairs 1 and 3 each carry the same allele on both homologs, GG and gg.
    Two of the same allele is homozygous.
  3. C. ✓ Pairs 2 and 4
  4. D. Pairs 1, 2 and 4
    Pair 1 carries G on both homologs.
    Two of the same allele is homozygous.

Why: Heterozygous means the two alleles differ.
Pairs 2 and 4 each carry one G and one g, in either order, so both are heterozygous, Gg.
Pairs 1 and 3 are homozygous.

53
Check q17

A Pp pea plant makes a gamete.

Which of the following does the gamete carry for the flower-color gene?

  1. A. ✓ One allele, P or p
  2. B. P only
    A gamete gets one homolog of the pair, and either homolog can be the one it gets.
  3. C. Both P and p
    A gamete carries one homolog of each pair.
    So it carries one allele, not the pair.
  4. D. Two alleles, PP or pp
    A gamete carries one homolog, so one allele.
    Two alleles is the plant's genotype; a gamete's genotype is one allele.

Why: A gamete carries one homolog of each pair.
So for this gene it carries one allele, P or p.
The plant's genotype is two alleles; the gamete's genotype is the one it carries.

54
Check q18

In pea plants, the seed-shape gene has the alleles R and r. A plant is homozygous for seed shape.

Which of the following could be the plant's genotype?

  1. A. Rr only
    Rr is two different alleles: heterozygous.
  2. B. ✓ RR or rr
  3. C. RR, Rr or rr
    Rr is two different alleles: heterozygous.
    A homozygous plant carries two of the same allele.

Why: Homozygous means the two alleles are the same.
The two alleles could both be R or both be r.
So the genotype is RR or rr.

55
Check q19

A gamete from a Gg pea plant carries g.

Which of the following statements about the gamete's genotype for pod color is correct?

  1. A. ✓ Its genotype is g
  2. B. Its genotype is gg
    The gamete carries one homolog, so one g, not two.
  3. C. It has no genotype
    A gamete does have a genotype: the one allele it carries.

Why: A gamete carries one homolog, so one allele, g.
A gamete's genotype is that one letter.
So the gamete's genotype is g; gg is a plant's genotype.

56
Check q20

A pea plant's genotype for pod shape is written as a pair of letters.

How many alleles of the pod-shape gene does the genotype list?

  1. A. One
    A body cell carries two homologs, and each carries one allele of the gene.
  2. B. ✓ Two
  3. C. Four
    A body cell carries two homologs of the pair, not four.

Why: A body cell carries two homologs of the pair.
Each homolog carries one allele of the pod-shape gene.
So the genotype lists two alleles.

57
Check q21

Every F1 plant from Mendel's purple × white cross got a P from its purple parent and a p from its white parent.

Which of the following is each F1 plant's genotype for flower color?

  1. A. PP
    Each F1 plant got a p from its white parent, so its two alleles are not P and P.
  2. B. ✓ Pp
  3. C. pp
    Each F1 plant got a P from its purple parent, so its two alleles are not p and p.

Why: Each F1 plant carries P on one homolog and p on the other.
A genotype is both alleles.
So each F1 plant is Pp.

58
Practice writing an answer

Every F1 plant from Mendel's purple × white cross got a P from its purple parent and a p from its white parent.

(a) Explain why every F1 plant is heterozygous for flower color. (1 pt)

Frame Every F1 plant is heterozygous because …

Model answer Every F1 plant is heterozygous because its two alleles differ.
Each F1 plant carries P on the homolog from its purple parent.
Each F1 plant carries p on the homolog from its white parent.
So its genotype is Pp, two different alleles of one gene.
Rubric
  • Award 1 point for: each F1 plant carries two different alleles (P from one parent, p from the other), so its genotype Pp is heterozygous.

Glossary

genotype
The two alleles an organism carries for a gene, one on each homolog, written as two letters: PP, Pp or pp. A gamete carries one allele, so its genotype is one letter.
homozygous
Carrying two of the same allele for a gene: PP or pp. Homo- means the same.
heterozygous
Carrying two different alleles for a gene: Pp. Hetero- means different.

APBIO-U05-L12B Which allele shows

Topic 5.3 · Mendelian Genetics · 62 steps

One F1 pea plant: on the left its homologous pair, two rods with centromere dots, the darker one carrying P and the lighter one carrying p; on the right a photograph of its flower, a real purple pea flower
One F1 pea plant: on the left its homologous pair, two rods with centromere dots, the darker one carrying P and the lighter one carrying p; on the right a photograph of its flower, a real purple pea flower

Photo: Vikiçizer, Wikimedia Commons, CC BY-SA 4.0 (resized).

Here is one of the F1 plants from Mendel’s cross. Its homologous pair carries P on one chromosome and p on the other, and its flowers are purple.

The plant carries the white allele, yet not a trace of white shows in its flowers. Why does only the purple show?

Unit 5 · Heredity

1Which allele shows: dominant and recessive

2

Video: Watch: Which allele shows: dominant and recessive

The F1 plant's pair, P and p, beside its purple flower; the capital and lowercase letters written on, dominant and recessive labelled.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Ba.mp4

3

Why does a plant carrying two different alleles show one allele's trait and not the other's?

4

One P is enough to make the flower purple. So a purple plant could be PP or Pp, and only a white plant shows its genotype for certain.

5
Check q1

Every F1 plant in Mendel's cross carries P on one homolog and p on the other.

Which word describes an F1 plant's genotype, Pp?

  1. A. Homozygous
    Homozygous means two of the same allele.
    P and p are two different alleles.
  2. B. ✓ Heterozygous

Why: P and p are two different alleles of one gene.
Two different alleles is heterozygous.

6

Every F1 plant is Pp, and every F1 plant is purple: one P is enough to make the flower purple.

One F1 plant: its homologous pair, two rods with centromere dots, carries P and p, and beside it its flower is purple
One F1 plant: its homologous pair, two rods with centromere dots, carries P and p, and beside it its flower is purple
7

When one allele's trait shows in a heterozygous plant, we call that allele , because its trait dominates what you see. Purple, P, is dominant.

8

When an allele's trait shows in a plant whose two alleles are both that allele, and not in a heterozygous plant, we call it , because its trait recedes from view whenever the other allele is present.

9

White, p, is recessive.

10

That is why the dominant allele is written as a capital letter and the recessive allele as the same letter in lowercase: P and p.

Two boxed letters: a capital P labelled dominant, its trait shows in Pp; a lowercase p labelled recessive, its trait shows in pp
Two boxed letters: a capital P labelled dominant, its trait shows in Pp; a lowercase p labelled recessive, its trait shows in pp
11

What you are expected to know Say which allele is dominant from what a heterozygous plant or animal looks like.

12

What you are expected to know Write the dominant allele as the capital letter and the recessive allele as the small letter.

13
Check q2

In pea plants, pod color has two alleles, G and g. A plant with the genotype Gg has green pods; a gg plant has yellow pods.

Which allele is dominant?

  1. A. ✓ G
  2. B. g
    The Gg plant has green pods, so green is the trait that shows in a heterozygous plant.
    Yellow pods show in the gg plant.
  3. C. Neither
    The Gg plant has green pods, not a mix.
    One trait shows and the other does not.

Why: The heterozygous plant, Gg, has green pods.
So green is the trait that shows when the two alleles differ.
The allele whose trait shows in a heterozygous plant is dominant.
So G is dominant, and g is recessive.

14
Practice writing an answer

In pea plants, pod color has two alleles, G and g. A plant with the genotype Gg has green pods; a gg plant has yellow pods.

(a) Explain why G is the dominant allele. (1 pt)

Frame G is dominant because …

Model answer G is dominant because its trait shows in a heterozygous plant.
The Gg plant carries one G and one g.
The Gg plant has green pods.
So green, the trait of G, shows when only one G is present.
Rubric
  • Award 1 point for: the heterozygous plant Gg shows green, G's trait, so G is dominant.
15
Check q3

Imagine a beetle in which one gene sets eye color. The table below gives the eye color for each genotype.

A table with two columns, genotype and eye color, and three rows: EE dark, Ee dark, ee light
A table with two columns, genotype and eye color, and three rows: EE dark, Ee dark, ee light

Which allele is recessive?

  1. A. E
    Ee beetles have dark eyes, E's trait.
    Light eyes show in ee beetles only.
  2. B. ✓ e

Why: Light eyes show in ee beetles, whose two alleles are both e.
Light eyes do not show in Ee beetles.
A trait that shows when both alleles are that allele, and not in a heterozygous animal, is recessive.
So e is recessive.

16Quick quiz: dominant allele, recessive allele mixed practice

17
Check q4

A gene has two alleles.

Which of the following describes a dominant allele?

  1. A. ✓ Its trait shows in a heterozygous plant or animal
  2. B. Its trait recedes from view in a heterozygous plant or animal
    A trait that recedes from view in a heterozygous plant or animal is the recessive allele's trait.
  3. C. It is the more common of the two alleles
    How common an allele is does not decide which is dominant.
    Dominant says which trait shows in a heterozygous plant or animal.

Why: A dominant allele's trait shows in a heterozygous plant or animal.
It is written as a capital letter.

18
Check q5

A gene has two alleles.

Which of the following describes a recessive allele?

  1. A. Its trait shows in a heterozygous plant or animal
    A trait that shows in a heterozygous plant or animal is the dominant allele's trait.
  2. B. ✓ Its trait recedes from view in a heterozygous plant or animal
  3. C. It is the rarer of the two alleles in a population
    How rare an allele is does not decide which is recessive.
    Recessive says the trait recedes from view in a heterozygous plant or animal.

Why: A recessive allele's trait shows when both alleles are that allele.
In a heterozygous plant or animal its trait recedes from view.
It is written as the same letter in lowercase.

19
Check q6

In fruit flies, an Ee fly has a gray body and an ee fly has an ebony body.

Which allele is dominant?

  1. A. ✓ E
  2. B. e
    The Ee fly has a gray body, E's trait.
    The allele whose trait shows in a heterozygous animal is dominant.

Why: The heterozygous fly, Ee, has a gray body.
Gray is E's trait.
The allele whose trait shows in a heterozygous animal is dominant.
So E is dominant.

20
Check q7

In pea plants, an Ii plant has inflated pods and an ii plant has constricted pods.

Which trait is recessive?

  1. A. Inflated pods
    Inflated pods show in the Ii plant, so inflated is the dominant trait.
    Constricted pods show in the ii plant.
  2. B. ✓ Constricted pods

Why: Constricted pods show in the ii plant, whose two alleles are both i.
Constricted pods do not show in the Ii plant.
So constricted pods is the recessive trait.

21
Check q8

In guinea pigs, BB guinea pigs and Bb guinea pigs have black coats, and bb guinea pigs have white coats.

Which allele is dominant?

  1. A. ✓ B
  2. B. b
    The Bb guinea pig has a black coat, B's trait.

Why: The heterozygous guinea pig, Bb, has a black coat.
Black is B's trait.
So B is dominant.

22
Check q9

Imagine a kind of lizard in which a rare tail pattern shows in Ll lizards and in LL lizards. Most lizards are ll and have plain tails.

Which allele is dominant?

  1. A. ✓ L
  2. B. l
    Dominant says only which trait shows in a heterozygous animal.
    The Ll lizard shows the pattern, L's trait.
  3. C. Cannot tell
    The Ll lizard shows the pattern, so L's trait shows in a heterozygous animal.
    That decides it, however rare the pattern is.

Why: The heterozygous lizard, Ll, shows the tail pattern.
The pattern is L's trait.
So L is dominant, even though most lizards are ll.

23
Check q10

Imagine a kind of fish in which most fish are striped. Striped fish are ss. Plain fish are SS or Ss.

Which allele is recessive?

  1. A. S
    Plain shows in Ss fish, so S is dominant.
    Striped shows in ss fish.
  2. B. ✓ s

Why: Striped shows in ss fish, whose two alleles are both s.
Striped does not show in Ss fish.
So s is recessive, even though most fish are striped.

24
Practice writing an answer

In pea plants, an Ii plant has inflated pods and an ii plant has constricted pods.

(a) Explain why i is the recessive allele. (1 pt)

Frame i is recessive because …

Model answer i is recessive because its trait does not show in a heterozygous plant.
The Ii plant carries one i, and its pods are inflated, not constricted.
Constricted pods show in the ii plant, whose two alleles are both i.
Rubric
  • Award 1 point for: constricted pods (i's trait) show in the ii plant and not in the heterozygous Ii plant, so i is recessive.

25Dominant does not mean stronger or more common

26

Video: Watch: Dominant does not mean stronger or more common

A hand with six fingers beside a hand with five: the rare trait labelled dominant, the common trait labelled recessive.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Bb.mp4

27

Dominant does not mean stronger, better or more common. It says only which trait shows in a heterozygous plant or animal.

28

Take a human example. A hand with six fingers comes from a dominant allele, and six fingers is rare.

29

Five fingers is the recessive trait, and five fingers is the common one.

30

So a dominant trait can be rare, and a recessive trait can be the common one.

31

What you are expected to know Say why a rare trait can still be the dominant one.

32
Check q11

Imagine a wild flower in which almost every plant has blue petals and a few plants have white petals. White petals show in Ww plants and blue petals in ww plants. A student says: “White is rare, so the white allele must be recessive.”

Which of the following statements about the student's claim is correct?

  1. A. The student is right
    Dominant says only which trait shows in a heterozygous plant.
    White shows in Ww, so W is dominant, and it is rare at the same time.
  2. B. ✓ The student is wrong

Why: White shows in Ww plants, so W's trait shows in a heterozygous plant.
The allele whose trait shows in a heterozygous plant is dominant.
How common a trait is does not decide which allele is dominant.
So W is dominant and rare at the same time.

33
Practice writing an answer

Imagine a wild flower in which almost every plant has blue petals and a few plants have white petals. White petals show in Ww plants and blue petals in ww plants. A student says: “White is rare, so the white allele must be recessive.”

(a) Explain why the student is wrong. (1 pt)

Frame The student is wrong because …

Model answer The student is wrong because dominant says only which trait shows in a heterozygous plant.
White shows in Ww plants.
So W is dominant.
How common a trait is depends on how many plants carry the allele.
Here few plants carry W, so white petals are rare and W is dominant at the same time.
Rubric
  • Award 1 point for: dominance is about what shows in a heterozygous plant (white shows in Ww, so W is dominant); how common a trait is does not show which allele is dominant.

34From what you see to what could be carried

35

Video: Watch: From what you see to what could be carried

A purple flower with its two possible pairs beneath it, PP and Pp, and a white flower with its one, pp.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L12Bc.mp4

36
Check q12

A pea plant has purple flowers.

Which of the following is part of the plant's phenotype?

  1. A. ✓ The purple color of its flowers
  2. B. The order of bases in its flower-color gene
    The order of bases in a gene is the allele's DNA, not a feature that can be seen.

Why: A phenotype is the set of features that can be seen.
The purple color of the flowers can be seen.

37

A plant's set of features that can be seen, such as its purple flowers, is called its . Its genotype is the two alleles behind those features.

38

Work backward from the flower. A purple plant could be PP or Pp, because P shows in both.

A purple flower on the left with two pairs beneath it, PP and Pp; a white flower on the right with one pair beneath it, pp
A purple flower on the left with two pairs beneath it, PP and Pp; a white flower on the right with one pair beneath it, pp
39

A white plant can only be pp, because white shows when both alleles are p and in no other genotype.

40

So two plants with the same phenotype, purple, can have different genotypes, PP and Pp.

41

The other way round never happens for a gene like this: two plants with the same genotype have the same flower color.

42

A plant whose two alleles are both the dominant one, PP, is called homozygous dominant. A plant whose two alleles are both the recessive one, pp, is called homozygous recessive.

43

What you are expected to know List the genotypes a phenotype allows: a plant showing the dominant trait is homozygous dominant or heterozygous; a plant showing the recessive trait is homozygous recessive and nothing else.

44
Check q13

A seed catalog describes a pea seed as round. For seed shape, the allele R gives round seeds and is dominant; r gives wrinkled seeds and is recessive.

Which of the following genotypes could the round seed have?

  1. A. RR only
    A round seed shows the dominant trait, and one R is enough for that.
    So the seed could also be Rr.
  2. B. Rr only
    A round seed could also carry two copies of R.
  3. C. Rr or rr
    An rr seed has no R, so it is wrinkled, not round.
  4. D. ✓ RR or Rr

Why: Round is the dominant trait.
Round shows whenever at least one R is present.
So the seed is RR or Rr, and looking at it cannot tell which.

45
Check q14

In pea plants, the allele R gives round seeds and is dominant; r gives wrinkled seeds and is recessive. One seed in a catalog is described as wrinkled.

Which of the following genotypes could the wrinkled seed have?

  1. A. Rr only
    An Rr seed carries R, and R shows, so an Rr seed is round.
  2. B. ✓ rr only
  3. C. Rr or rr
    R shows whenever it is present, so an Rr seed is round.
    Wrinkled shows with no R.
  4. D. RR, Rr or rr
    A round seed could be RR or Rr.
    A wrinkled seed shows the recessive trait, which shows when both alleles are r.

Why: Wrinkled is the recessive trait.
A recessive trait shows when both alleles are recessive.
So the seed is rr.

46

Back to the F1 plant from the opening: its homologous pair carries P and p, and its flowers are purple.

47

It shows purple because P is dominant: one P is enough, and the white recedes from view.

48

A purple flower could be PP or Pp. A white flower can only be pp.

49Quick quiz: phenotype mixed practice

50
Check q15

In pea plants, the allele R gives round seeds and r gives wrinkled seeds.

Which of the following is a phenotype?

  1. A. ✓ Round seeds
  2. B. Rr
    Rr is a pair of alleles: a genotype.
  3. C. A gamete carrying R
    A gamete carrying R is a cell with one allele, not a feature that can be seen.

Why: A phenotype is a feature that can be seen.
Round seeds can be seen.

51
Check q16

Suppose that in a breed of dog a dark nose (N) is dominant to a pink nose (n). A dog has a pink nose.

Can you tell the dog's genotype from its nose color?

  1. A. ✓ Yes
  2. B. No
    A pink nose shows with no N.
    So the dog must be nn.

Why: Pink nose is the recessive trait.
It shows when both alleles are n.
So the dog is nn, and nothing else.
Yes, the nose color tells you the genotype.

52
Check q17

Suppose that in a breed of dog a dark nose (N) is dominant to a pink nose (n). A dog has a dark nose.

Can you tell the dog's genotype from its nose color?

  1. A. Yes
    A dark nose shows in NN dogs and in Nn dogs.
    So the nose color leaves two genotypes open.
  2. B. ✓ No

Why: Dark nose is the dominant trait.
It shows in NN dogs and in Nn dogs.
So the dog is NN or Nn, and its nose cannot tell you which.
No, the nose color does not tell you the genotype.

53
Practice writing an answer

In pea plants, the yellow-seed allele Y is dominant and the green-seed allele y is recessive. A plant has yellow seeds.

(a) Explain why the plant's seed color leaves two genotypes open. (1 pt)

Frame The seed color leaves two genotypes open because …

Model answer The seed color leaves two genotypes open because yellow is the dominant trait.
Yellow shows in YY plants and in Yy plants.
So a yellow-seeded plant may be YY or Yy.
Its seed color alone does not tell which.
Rubric
  • Award 1 point for: yellow is dominant, so it shows in both YY and Yy plants, and the seed color leaves both genotypes open.

54Mixed practice mixed practice

55
Check q18

In cattle, one gene decides whether horns grow. Its allele H gives a hornless head and its allele h gives horns. Cattle with the genotype Hh have no horns.

Which of the following statements about the alleles is correct?

  1. A. h is dominant
    Hh cattle have no horns, so h's trait, horns, does not show in the heterozygous animal.
    The allele whose trait shows there is dominant: H.
  2. B. H is recessive
    The trait that shows in Hh, no horns, belongs to H.
    The allele whose trait shows in a heterozygous animal is dominant.
  3. C. ✓ H is dominant
  4. D. Neither is dominant
    One trait does show in Hh: the cattle have no horns, so H's trait dominates.

Why: The heterozygous animal, Hh, has no horns.
So no horns is the trait that shows when the two alleles differ.
H, the allele for that trait, is dominant and h is recessive.

56
Check q19

In pea plants the yellow-seed allele Y is dominant and the green-seed allele y is recessive. Two plants in a field both have yellow seeds.

Which of the following can you say about the two plants' genotypes for seed color?

  1. A. Both plants must be homozygous, YY
    Yellow shows with one Y as well as with two, so a yellow-seeded plant may be Yy.
  2. B. Both plants must be heterozygous, Yy
    A YY plant also has yellow seeds, and nothing about the seeds rules YY out.
  3. C. One plant must be YY and the other Yy
    Both plants could be YY, or both could be Yy, or one could be each.
  4. D. ✓ Each plant is YY or Yy, and the two need not match

Why: Yellow is the dominant trait, so each yellow-seeded plant is YY or Yy.
Two plants with the same phenotype can have different genotypes, or the same.
Looking at the seeds cannot say which.

57
Check q20

In a breed of dog, a gene for coat texture has two alleles. A pup gets a wire-coat allele from one parent and a smooth-coat allele from the other, and grows a wire coat.

Which of the following describes the pup?

  1. A. The pup is heterozygous, and the smooth-coat allele is dominant
    The trait that shows in this heterozygous pup is the wire coat, so the wire-coat allele is dominant.
  2. B. The pup is homozygous, and the wire-coat allele is dominant
    The pup carries one allele for each coat, so it is heterozygous.
  3. C. ✓ The pup is heterozygous, and the wire-coat allele is dominant
  4. D. The pup is heterozygous, and neither allele is dominant
    One trait does show in the pup, the wire coat, so one allele is dominant.

Why: The pup got one allele from each parent, and they differ: heterozygous.
The wire coat is what shows in that heterozygous pup, so the wire-coat allele is dominant.

58
Check q21

In pea plants, purple flowers (P) are dominant to white (p). Two plants have exactly the same genotype for flower color, Pp.

Which of the following can you say about their flower color?

  1. A. ✓ Both plants are purple
  2. B. Both plants are white
    White shows in pp plants, and these plants each carry a P.
  3. C. It cannot be predicted from the genotype
    For a gene like this the genotype fixes the phenotype: Pp is purple every time.
  4. D. One is purple and one is white
    The same genotype for this gene gives the same phenotype; Pp always shows the dominant trait.

Why: The same genotype gives the same flower color for this gene.
Pp carries a P, P is dominant, so both plants are purple.
It is the other direction, phenotype to genotype, that can be uncertain.

59
Check q22

In guinea pigs, a black coat (B) is dominant to a white coat (b). A guinea pig has a white coat.

Which of the following genotypes could the guinea pig have?

  1. A. BB or Bb
    A guinea pig carrying B has a black coat, because B is dominant.
  2. B. ✓ bb only
  3. C. Bb only
    A Bb guinea pig carries B, and B shows, so it has a black coat.

Why: White is the recessive trait.
A recessive trait shows when both alleles are recessive.
So the white guinea pig is bb.

60
Check q23

In tomato plants, a true-breeding tall line crossed with a true-breeding dwarf line gives all tall F1. In the F2 from F1 × F1, a grower counts 302 tall plants and 98 dwarf plants.

Which trait is dominant?

  1. A. Dwarf
    The trait that shows in every F1 plant, tall, is the dominant one.
  2. B. ✓ Tall
  3. C. Neither
    The F2 plants are each tall or dwarf, about three to one, not a mix.
    That is the pattern of one dominant and one recessive allele.

Why: Every F1 plant carries one tall allele and one dwarf allele, and every F1 plant is tall.
The allele whose trait shows in a heterozygous plant is dominant.
So tall is dominant, and about three tall to one dwarf in the F2 (302 : 98) is Mendel's pattern.

61
Practice writing an answer

In tomato plants, a true-breeding tall line crossed with a true-breeding dwarf line gives all tall F1. In the F2 from F1 × F1, a grower counts 302 tall plants and 98 dwarf plants.

(a) Explain how the F2 counts show that every F1 plant carried the dwarf allele. (1 pt)

Frame Every F1 plant carried the dwarf allele because …

Model answer Every F1 plant carried the dwarf allele because the dwarf trait came back in the F2.
The F2 had 98 dwarf plants out of 400, about 25%.
Every F1 plant was tall, so the dwarf allele did not show in the F1.
A dwarf F2 plant gets its two dwarf alleles from its two F1 parents.
So the tall F1 plants carried the dwarf allele without showing it: they were heterozygous.
Rubric
  • Award 1 point for: dwarf plants reappeared in about 25% of the F2 (98 of 400), and a dwarf plant's alleles came from its tall F1 parents, so every F1 plant carried the dwarf allele unseen (heterozygous).

Glossary

dominant allele
The allele whose trait shows in a heterozygous plant or animal; written as a capital letter. Dominant does not mean stronger or more common.
recessive allele
The allele whose trait shows when both alleles are that allele, and not in a heterozygous plant or animal; written as the same letter in lowercase.
phenotype
An organism's set of features that can be seen, such as purple flowers or round seeds. Its genotype is the two alleles behind them.

APBIO-U05-L13 One allele per gamete

Topic 5.3 · Mendelian Genetics · 92 steps

A cell at anaphase I: an oval cell with a pole at each end, a spindle fiber from each pole, and two X-shaped chromosomes being pulled apart, the darker one moving left carrying P on both chromatids and the lighter one moving right carrying p on both chromatids
A cell at anaphase I: an oval cell with a pole at each end, a spindle fiber from each pole, and two X-shaped chromosomes being pulled apart, the darker one moving left carrying P on both chromatids and the lighter one moving right carrying p on both chromatids

Here is a cell from a Pp pea plant, drawn at anaphase I of meiosis. Its two homologs carry the flower-color gene: P on one, p on the other.

The spindle fibers have begun to pull. Where does each allele go, and what will the gametes carry?

Unit 5 · Heredity

1One allele per gamete

2

Video: Watch: One allele per gamete

The two X-shaped homologs parting with their letters, the P chromosome to one pole and the p chromosome to the other; then the two chromatids of each parting in turn, so every gamete carries one allele.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13a.mp4

3

What does meiosis do to a Pp plant’s two alleles?

4

At anaphase I the spindle pulls the two homologs to opposite poles. So the P chromosome goes to one pole and the p chromosome goes to the other.

5

Each gamete receives one homolog. So each gamete carries one allele: half of the gametes carry P and half carry p.

6

Once you can predict each parent’s gametes, a table of the two parents’ gametes lists every offspring they can make.

7
Check q1

A Pp plant’s cell is at anaphase I. Its P chromosome and its p chromosome are the two members of one homologous pair.

Where does the spindle send the P chromosome and the p chromosome?

  1. A. To the same pole
    Anaphase I parts the two members of every homologous pair.
  2. B. ✓ To opposite poles

Why: At anaphase I the spindle pulls the two members of each homologous pair to opposite poles.
So each new cell receives one chromosome of the pair.

8

So each new cell receives one chromosome of every homologous pair.

A cell at anaphase I: the darker P chromosome, an X of two chromatids each carrying P, is pulled toward the left pole, and the lighter p chromosome, an X of two chromatids each carrying p, toward the right pole, by a spindle fiber to each centromere; labels name the pole, the centromere and a spindle fiber
A cell at anaphase I: the darker P chromosome, an X of two chromatids each carrying P, is pulled toward the left pole, and the lighter p chromosome, an X of two chromatids each carrying p, toward the right pole, by a spindle fiber to each centromere; labels name the pole, the centromere and a spindle fiber
9

The P allele sits on one homolog and the p allele sits on the other. So no cell after anaphase I holds both alleles.

10
Check q2

At anaphase I each chromosome is still an X of two chromatids. Every chromosome was copied once, in S phase.

Which alleles do the two chromatids of the P chromosome carry?

  1. A. One chromatid carries P and the other carries p
    The two chromatids are the two copies of one chromosome, made in S phase.
    So both carry the same allele, P.
  2. B. ✓ Both chromatids carry P

Why: S phase copied the P chromosome into two identical sister chromatids.
So both chromatids carry P.

11

So both chromatids of the P chromosome carry P, and both chromatids of the p chromosome carry p.

Two X-shaped chromosomes side by side: the left one, darker, carries P on both of its chromatids; the right one, lighter, carries p on both of its chromatids
Two X-shaped chromosomes side by side: the left one, darker, carries P on both of its chromatids; the right one, lighter, carries p on both of its chromatids
12
Check q3

The cell that received the P chromosome now goes through meiosis II.

What does meiosis II pull apart in that cell?

  1. A. ✓ The two sister chromatids of each chromosome
  2. B. The two members of each homologous pair
    The two members of each pair parted at anaphase I, and this cell holds one of them.
    Meiosis II parts the two sister chromatids of each chromosome.

Why: Meiosis II pulls the two sister chromatids of each chromosome to opposite poles.
So in this cell it parts the two P chromatids.

13

Meiosis II pulls the two sister chromatids of the P chromosome to opposite poles. So the cell that received the P chromosome gives two gametes, and each of them carries P.

On the left one X-shaped chromosome carrying allele P on both chromatids; an arrow points right to two small cells, each holding a single chromosome carrying allele P
On the left one X-shaped chromosome carrying allele P on both chromatids; an arrow points right to two small cells, each holding a single chromosome carrying allele P
14

So for one gene the rule has three steps.
The two alleles sit at the same position on the two homologs.
The homologs part at anaphase I.
So each gamete carries exactly one of the two alleles.

15

Mendel worked out this rule from his counts, long before anyone knew what a chromosome did. When the two alleles of a gene part as gametes form, we call it the , because to segregate means to separate.

16

A Pp plant therefore makes two kinds of gamete: half carry P and half carry p. A PP plant has P on both homologs, so it makes only P gametes.

Left: a Pp plant's pair of homologs, two rods with centromere dots, above four gametes, two carrying P and two carrying p. Right: a PP plant's pair above four gametes, all carrying P
Left: a Pp plant's pair of homologs, two rods with centromere dots, above four gametes, two carrying P and two carrying p. Right: a PP plant's pair above four gametes, all carrying P
17

What you are expected to know Predict a parent’s gametes from its two alleles: each gamete carries one of the two, and a heterozygous plant’s gametes are half one allele and half the other.

18
Check q4

A pea plant is Pp for the flower-color gene.

Which of the following can one of its gametes carry for this gene?

  1. A. ✓ P or p
  2. B. P and p
    A gamete carries one homolog of the pair.
    P sits on one homolog and p sits on the other.

Why: P sits on one homolog and p sits on the other.
The two homologs part at anaphase I.
So a gamete receives one of them and carries P or p, never both.

19
Practice writing an answer

A pea plant is Pp for the flower-color gene.

(a) Explain why no gamete of this plant carries both P and p. (1 pt)

Model answer P sits on one homolog and p sits on the other homolog of the pair.
At anaphase I the spindle pulls the two homologs to opposite poles.
So each gamete receives one homolog.
So each gamete carries one allele, P or p, never both.
Rubric
  • Award 1 point for: the two alleles sit on the two homologs, and the homologs part at anaphase I, so a gamete receives one homolog and so one allele.

Slip Saying the gamete carries Pp because the plant is Pp. Pp is the plant’s genotype; a gamete carries one homolog, so one allele.

(b) Explain why half of the gametes carry P and half carry p. (1 pt)

Model answer Anaphase I sends the P chromosome to one cell and the p chromosome to the other cell.
So every meiosis makes one cell with P and one cell with p.
Meiosis II gives two gametes from each cell, both carrying that cell’s allele.
So half of the gametes carry P and half carry p.
Rubric
  • Award 1 point for: each meiosis sends P to one cell and p to the other in equal number, so the gametes are half P and half p.
20
Check q5

A pea plant is Pp for the flower-color gene. A student says: “Its gametes are Pp, because every gamete comes from a Pp plant.”

Is the student correct?

  1. A. Yes: the gametes are Pp
    Pp names the two alleles on the plant’s two homologs.
    A gamete carries one homolog, so one allele.
  2. B. ✓ No: each gamete is P or p

Why: Pp names the two alleles on the plant’s two homologs.
The homologs part at anaphase I.
So a gamete carries one homolog and one allele, P or p.

21
Check q6

A bull is heterozygous, Bb, for a coat-color gene.

Which of the following does the law of segregation predict for 200 of his sperm cells?

  1. A. ✓ About 100 sperm carry B and about 100 carry b
  2. B. About 150 sperm carry B and about 50 carry b
    Three to one is a ratio of offspring, not of gametes.
    The two homologs part at anaphase I, so half of the sperm get each one.
  3. C. All 200 sperm carry B
    Dominance decides which trait shows in the bull, not which homolog goes into a sperm.
    Half of the sperm get the b homolog.
  4. D. All 200 sperm carry both B and b
    A sperm carries one homolog of the pair.
    So a sperm carries one allele, never both.

Why: The B homolog and the b homolog part at anaphase I.
So each sperm carries one of them.
Half of the sperm get the B homolog and half get the b homolog.
So over 200 sperm, about 100 carry B and about 100 carry b.

22Quick quiz: law of segregation mixed practice

23
Check q7

A plant makes gametes.

What does the law of segregation say?

  1. A. ✓ The two alleles of a gene part when gametes form
  2. B. The two alleles of a gene stay together in every gamete
    The two alleles sit on the two homologs, and the homologs part at anaphase I.
    So a gamete carries one allele, never both.
  3. C. Every gamete carries the dominant allele
    Dominance decides which trait shows in a plant, not which homolog enters a gamete.
    Half of a heterozygous plant’s gametes carry the recessive allele.

Why: The law of segregation says that the two alleles of a gene part when gametes form.
So each gamete carries one of the two alleles.

24
Practice writing an answer

A plant is heterozygous for a gene and makes gametes.

(a) State the law of segregation. (1 pt)

Model answer The two alleles of a gene part when gametes form, so each gamete carries one of the two alleles.
Rubric
  • Award 1 point for: the two alleles of a gene part (separate) when gametes form, so each gamete carries one allele.

(b) Identify the stage of meiosis in which the two alleles part. (1 pt)

Model answer The two alleles part at anaphase I, when the spindle pulls the two homologs to opposite poles.
Rubric
  • Award 1 point for: anaphase I (the two homologs part).
25
Check q8

A dog is Dd for a coat-color gene.

Which of the following can its gametes carry for this gene?

  1. A. D only
    The two homologs part at anaphase I, so half of the gametes get the d homolog.
  2. B. d only
    Half of the gametes get the D homolog.
  3. C. ✓ Half carry D and half carry d

Why: D sits on one homolog and d sits on the other.
The homologs part at anaphase I.
So half of the gametes carry D and half carry d.

26
Check q9

A dog is dd for the same coat-color gene.

Which of the following can its gametes carry for this gene?

  1. A. D only
    Both homologs carry d.
  2. B. ✓ d only
  3. C. Half carry D and half carry d
    Both homologs carry d.

Why: Both homologs carry d.
Each gamete gets one homolog.
So every gamete carries d.

27
Check q10

A dog is DD for the same coat-color gene.

Which of the following can its gametes carry for this gene?

  1. A. ✓ D only
  2. B. d only
    Both homologs carry D.
  3. C. Half carry D and half carry d
    Both homologs carry D.

Why: Both homologs carry D.
Each gamete gets one homolog.
So every gamete carries D.

28
Check q11

A pea plant is Ee for a gene.

How many alleles of that gene does one of its gametes carry?

  1. A. ✓ One
  2. B. Two
    A gamete carries one homolog of the pair, so it carries one allele.

Why: The two alleles sit on the two homologs.
A gamete gets one homolog.
So a gamete carries one allele.

29
Check q12

A pea plant is Ee for a gene.

Can one of its gametes carry both E and e?

  1. A. Yes
    E sits on one homolog and e sits on the other, and the homologs part at anaphase I.
    So a gamete gets one of them.
  2. B. ✓ No

Why: E and e sit on the two homologs of one pair.
The homologs go to opposite poles at anaphase I.
So no gamete carries both.

30
Check q13

The homologous pair of a plant is drawn below.

A homologous pair drawn as two rods with centromere dots, one darker and one lighter, each carrying one boxed letter at the same height
A homologous pair drawn as two rods with centromere dots, one darker and one lighter, each carrying one boxed letter at the same height

Which of the following can the plant’s gametes carry for this gene?

  1. A. A only
    Half of the gametes get the other homolog, which carries a.
  2. B. a only
    Half of the gametes get the other homolog, which carries A.
  3. C. ✓ Half carry A and half carry a

Why: One homolog carries A and the other carries a.
The homologs part at anaphase I.
So half of the gametes carry A and half carry a.

31The egg in the ovule, the sperm in the pollen grain

32

Video: Watch: The egg in the ovule, the sperm in the pollen grain

A pea flower cut open: the ovules in its base, each holding an egg; the anthers dusted with pollen grains; one pollen grain landing and its sperm travelling down to an egg, so that one egg fused with one sperm grows into one seed.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13b.mp4

33
Check q14

In an animal, a new individual begins at fertilization.

Which two cells fuse at fertilization?

  1. A. ✓ A sperm and an egg
  2. B. Two body cells
    Body cells never fuse.
    Fertilization is a sperm fusing with an egg.

Why: At fertilization a sperm fuses with an egg.
The two gametes together make the zygote.

34

Now consider a pea plant. A pea plant makes gametes too: eggs and sperm.

35

Here is a drawing of a pea flower cut open.

A pea flower drawn cut open: two petals behind, and in the middle a pod-shaped ovary holding three small ovules, each with a dot for its egg; a stalk rises from the ovary to the stigma, where one pollen grain sits; two anthers on stalks either side carry pollen grains. Labels on the drawing name a pollen grain (carries the sperm), an anther, an ovule (holds the egg), the egg and a petal
A pea flower drawn cut open: two petals behind, and in the middle a pod-shaped ovary holding three small ovules, each with a dot for its egg; a stalk rises from the ovary to the stigma, where one pollen grain sits; two anthers on stalks either side carry pollen grains. Labels on the drawing name a pollen grain (carries the sperm), an anther, an ovule (holds the egg), the egg and a petal
36

In the base of the flower sit small cases, and each case holds one egg. A small case that holds the plant’s egg is called an .

37

The anthers make tiny grains, and each grain carries the plant’s sperm. A grain that carries the plant’s sperm is called a .

38

A pollen grain lands on the top of the flower. Its sperm travels down to an egg inside an ovule.

39

The sperm fuses with the egg, and the fertilized egg grows into a seed. So each seed is one egg fused with one sperm.

A circle labeled egg carrying P, a plus sign, a small circle labeled pollen grain carrying p, an arrow, and an oval labeled seed reading Pp
A circle labeled egg carrying P, a plus sign, a small circle labeled pollen grain carrying p, an arrow, and an oval labeled seed reading Pp
40

A Pp plant’s eggs are half P and half p. The sperm in its pollen grains are half P and half p too.

41

From here on, a P pollen grain means a pollen grain whose sperm carries P. A P egg means an egg that carries P.

42

What you are expected to know Identify where a flowering plant’s two gametes sit: the egg inside an ovule, the sperm inside a pollen grain.

43
Check q15

A pea flower is pollinated.

Which structure carries the plant’s sperm?

  1. A. The ovule
    The ovule holds the egg.
    The pollen grain carries the sperm.
  2. B. ✓ The pollen grain

Why: The anthers make pollen grains.
Each pollen grain carries the plant’s sperm.
The egg waits inside an ovule.

44
Check q16

A pea plant sets one seed.

Which of the following made the fertilized egg that grew into that seed?

  1. A. ✓ One egg and one sperm
  2. B. Two eggs and two sperm
    One sperm fuses with one egg.
    That one fertilized egg grows into the seed.

Why: A seed grows from one fertilized egg.
One sperm fused with that one egg.
So one egg and one sperm made the fertilized egg.

45Quick quiz: ovule, pollen grain mixed practice

46
Check q17

A pea flower is cut open.

What is an ovule?

  1. A. ✓ The case in the flower’s base that holds the egg
  2. B. The grain the anther makes to carry the sperm
    The grain that carries the sperm is a pollen grain.
    The ovule holds the egg.
  3. C. The seed that grows from the fertilized egg
    A seed grows from an egg after a sperm fuses with it.
    The ovule is the case that holds the egg before that.

Why: An ovule is a small case in the base of the flower.
It holds the plant’s egg.

47
Check q18

A pea flower is cut open.

What is a pollen grain?

  1. A. The case that holds the egg
    The case that holds the egg is the ovule.
    A pollen grain carries the sperm.
  2. B. ✓ The grain that carries the sperm
  3. C. The fertilized egg that grows into a seed
    The fertilized egg grows into the seed.
    A pollen grain carries the sperm that fertilizes the egg.

Why: The anthers make pollen grains.
A pollen grain carries the plant’s sperm.

48
Practice writing an answer

A flowering plant makes eggs and sperm.

(a) State what an ovule holds. (1 pt)

Model answer An ovule holds the plant’s egg.
Rubric
  • Award 1 point for: the egg.

(b) State what a pollen grain carries. (1 pt)

Model answer A pollen grain carries the plant’s sperm.
Rubric
  • Award 1 point for: the sperm.
49
Check q19

A p pollen grain lands on a pea flower.

Which allele does the sperm in that pollen grain carry?

  1. A. ✓ p
  2. B. P
    A p pollen grain is a pollen grain whose sperm carries p.

Why: A p pollen grain means a pollen grain whose sperm carries p.
So the sperm carries p.

50
Check q20

A P egg is fertilized by the sperm from a p pollen grain.

Which genotype does the seed have?

  1. A. PP
    The sperm brought p, not P.
  2. B. ✓ Pp
  3. C. pp
    The egg brought P, not p.

Why: The egg brings P and the sperm brings p.
The seed carries both alleles.
So the seed is Pp.

51Build the square

52

Video: Watch: Build the square

Two Pp plants crossed: the pollen grains’ alleles written along the top edge, the eggs’ alleles down the left edge, then the four cells filling one at a time with the two alleles each pairing brings together: PP, Pp, Pp, pp.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13c.mp4

53
Check q21

A Pp pea plant makes eggs and pollen grains.

Which alleles do its eggs carry?

  1. A. P only
    Half of the eggs receive the p homolog.
  2. B. ✓ Half carry P and half carry p
  3. C. Each egg carries P and p
    An egg carries one homolog, so one allele.

Why: The two homologs part at anaphase I.
So each egg carries one allele.
Half of the eggs carry P and half carry p.

54

Now cross two Pp plants: the eggs of one plant are fertilized by the sperm from the other plant’s pollen grains. What offspring can they give, and in what proportions?

55

Each parent’s gametes are half P and half p: two kinds of egg, and two kinds of pollen grain. So there are four pairings of an egg with a pollen grain.

56

Here is a table that lists the four pairings: the pollen grains’ alleles along the top edge, the eggs’ alleles down the left edge, and one cell for each pairing.

An empty two-by-two grid: the pollen grains' alleles P and p written along the top edge, the eggs' alleles P and p written down the left edge; the four cells are empty
An empty two-by-two grid: the pollen grains' alleles P and p written along the top edge, the eggs' alleles P and p written down the left edge; the four cells are empty
57

Each cell holds the two alleles that egg and that pollen grain bring together. Where the P egg fuses with the P pollen grain, the cell reads PP.

The same grid with one cell filled: where the P egg fuses with the P pollen grain, the cell reads PP
The same grid with one cell filled: where the P egg fuses with the P pollen grain, the cell reads PP
58

Where the P egg fuses with the p pollen grain, the cell reads Pp.

The same grid with two cells filled: PP top left, and Pp top right where the P egg fuses with the p pollen grain
The same grid with two cells filled: PP top left, and Pp top right where the P egg fuses with the p pollen grain
59

The p egg with the P pollen grain gives Pp. The p egg with the p pollen grain gives pp.

The grid with all four cells filled: PP top left, Pp top right, Pp bottom left, pp bottom right
The grid with all four cells filled: PP top left, Pp top right, Pp bottom left, pp bottom right
60

Half of the eggs carry each allele, and so do half of the pollen grains. So the four cells are four equally likely pairings.

61

Pp appears in two of the four cells. So the four cells hold three genotypes: PP, Pp and pp.

62

When a table lists every pairing of one parent’s gametes with the other’s, we call it a , after Reginald Punnett, who first drew one.

63

When a breeding cross follows a single gene, like this one, we call it a : mono- means one, and a hybrid is an offspring that carries two different alleles.

64

The edges of a Punnett square carry gametes, one allele each. A parent’s genotype, two letters, never goes on an edge.

65

Now consider a Pp plant crossed with a pp plant. The pp parent has p on both homologs, so every one of its gametes carries p.

66

So both edge labels down the side read p, and the cells are Pp, Pp, pp, pp.

The square for a Pp plant crossed with a pp plant: P and p along the top, p and p down the side; cells Pp, Pp, pp, pp
The square for a Pp plant crossed with a pp plant: P and p along the top, p and p down the side; cells Pp, Pp, pp, pp
67

What you are expected to know Build a Punnett square for one gene: the two parents’ gametes on the two edges, one allele each, and in every cell the two alleles that egg and that pollen grain bring together.

68
Check q22

In pea plants, green pods (G) are dominant to yellow pods (g). A Gg plant is crossed with a gg plant. The square below has the Gg parent along the top and the gg parent down the side, with the edges still blank.

An empty two-by-two grid labeled for a Gg parent along the top and a gg parent down the left edge; the edge labels and the four cells are all blank
An empty two-by-two grid labeled for a Gg parent along the top and a gg parent down the left edge; the edge labels and the four cells are all blank

Which of the following goes along the top edge for the Gg parent?

  1. A. Gg
    The edges carry gametes, one allele each.
    A Gg parent makes G gametes and g gametes.
  2. B. G only
    Half of a Gg parent’s gametes carry g.
  3. C. ✓ G and g

Why: The edges of a Punnett square carry gametes, one allele each.
A Gg parent makes half G gametes and half g gametes.
So G and g go along the top edge.

69
Check q23

A Gg plant is crossed with a gg plant.

Which of the following goes down the side for the gg parent?

  1. A. ✓ g and g
  2. B. gg
    The edges carry gametes, one allele each.
    A gg parent makes only g gametes.
  3. C. G and g
    A gg parent has no G to give.

Why: A gg parent has g on both homologs.
So every gamete carries g.
The two edge labels down the side are g and g.

70
Check q24

A Gg plant is crossed with a gg plant. In the square for the cross, the top-left cell sits under the G gamete and beside a g gamete.

Which of the following goes in that cell?

  1. A. G
    A cell holds the two alleles the two gametes bring together: the G above it and the g beside it.
  2. B. gg
    The gamete above this cell is G, and the gamete beside it is g.
  3. C. ✓ Gg

Why: A cell holds the allele above it together with the allele beside it.
Above this cell is G, and beside it is g.
So the cell reads Gg.

71
Check q25

In mice, black fur (B) is dominant to brown (b). A Bb mouse is crossed with a bb mouse. Four squares drawn for this cross are numbered 1 to 4 below, with the Bb mouse’s gametes meant to run along the top edge and the bb mouse’s down the left edge.

Four two-by-two squares numbered 1 to 4 for a Bb mouse crossed with a bb mouse, the Bb mouse's gametes along the top edge and the bb mouse's down the left edge. Square 1: Bb and bb on the edges, cells Bbbb, Bbbb, bbBb, bbbb. Square 2: B and b along the top, b and b down the side, cells Bb, bb, Bb, bb. Square 3: B and b along the top, b and b down the side, cells Bb, Bb, bb, bb. Square 4: B and B along the top, b and b down the side, cells Bb, Bb, Bb, Bb
Four two-by-two squares numbered 1 to 4 for a Bb mouse crossed with a bb mouse, the Bb mouse's gametes along the top edge and the bb mouse's down the left edge. Square 1: Bb and bb on the edges, cells Bbbb, Bbbb, bbBb, bbbb. Square 2: B and b along the top, b and b down the side, cells Bb, bb, Bb, bb. Square 3: B and b along the top, b and b down the side, cells Bb, Bb, bb, bb. Square 4: B and B along the top, b and b down the side, cells Bb, Bb, Bb, Bb

Which of the following squares is drawn correctly?

  1. A. Square 1
    Square 1 puts the parents’ genotypes, Bb and bb, on the edges; the edges carry gametes, one allele each.
  2. B. ✓ Square 2
  3. C. Square 3
    Each cell combines the gamete above it with the gamete beside it.
    Square 3’s top-right cell sits under b and beside b, so it should read bb, not Bb.
  4. D. Square 4
    Square 4 puts B and B along the top, but a Bb mouse makes half B gametes and half b gametes.

Why: The Bb parent’s gametes are B and b; the bb parent’s are b and b.
Square 2 has those on its edges.
Each cell combines the allele above it with the allele beside it: Bb, bb in the top row, Bb, bb in the bottom.
So square 2 is correct.

72

Here again is the cell from the Pp pea plant at anaphase I: the P chromosome pulled to one pole, the p chromosome pulled to the other.

A cell at anaphase I: the darker P chromosome, an X of two chromatids each carrying P, is pulled toward the left pole, and the lighter p chromosome, an X of two chromatids each carrying p, toward the right pole, by a spindle fiber to each centromere
A cell at anaphase I: the darker P chromosome, an X of two chromatids each carrying P, is pulled toward the left pole, and the lighter p chromosome, an X of two chromatids each carrying p, toward the right pole, by a spindle fiber to each centromere
73

Each gamete received one homolog, so one allele: half of the gametes carry P and half carry p.

74

Two Pp parents’ gametes on the edges of a Punnett square gave PP, Pp, Pp and pp.

75Quick quiz: Punnett square, monohybrid cross mixed practice

76
Check q26

Two pea plants are crossed.

What is a Punnett square?

  1. A. ✓ A table of every pairing of the two parents’ gametes
  2. B. A table of the two parents’ genotypes
    The parents’ genotypes never go on a Punnett square.
    Its edges carry the parents’ gametes and its cells the offspring.
  3. C. A drawing of the two parents’ homologous pairs
    A homologous pair is drawn as two rods.
    A Punnett square is a table of gamete pairings.

Why: A Punnett square is a table.
One parent’s gametes run along the top edge and the other’s down the left edge.
Each cell holds the offspring genotype that pairing makes.

77
Check q27

Two pea plants are crossed.

What is a monohybrid cross?

  1. A. A breeding cross between two true-breeding lines
    Mendel’s P generation cross was between two true-breeding lines, but a monohybrid cross is named for the one gene it follows.
  2. B. ✓ A breeding cross that follows a single gene
  3. C. A breeding cross that follows two genes
    Mono- means one.
    A monohybrid cross follows one gene.

Why: Mono- means one, and a hybrid carries two different alleles.
A monohybrid cross follows a single gene.

78
Practice writing an answer

Two pea plants are crossed for one gene, and their Punnett square has two edges and four cells.

(a) State what sits on the edges of the Punnett square. (1 pt)

Model answer The parents’ gametes sit on the edges, one allele each: one parent’s gametes along the top edge and the other’s down the left edge.
Rubric
  • Award 1 point for: the parents’ gametes, one allele each (not the parents’ genotypes).

(b) State what sits in each cell of the Punnett square. (1 pt)

Model answer Each cell holds the offspring genotype: the two alleles that the gamete above it and the gamete beside it bring together.
Rubric
  • Award 1 point for: the offspring genotype made by the gamete above and the gamete beside the cell.
79
Check q28

Two Tt pea plants are crossed. In the square below, one cell is shaded.

A two-by-two square with T and t along the top edge and T and t down the left edge; the top-right cell is shaded and marked with a question mark; the other cells are empty
A two-by-two square with T and t along the top edge and T and t down the left edge; the top-right cell is shaded and marked with a question mark; the other cells are empty

Which genotype sits in the shaded cell?

  1. A. TT
    The gamete above the shaded cell is t.
  2. B. ✓ Tt
  3. C. tt
    The gamete beside the shaded cell is T.

Why: The shaded cell sits under t and beside T.
So it holds T and t: Tt.

80
Check q29

An Rr pea plant is crossed with an rr plant. In the square below, one cell is shaded.

A two-by-two square with R and r along the top edge and r and r down the left edge; the bottom-right cell is shaded and marked with a question mark; the other cells are empty
A two-by-two square with R and r along the top edge and r and r down the left edge; the bottom-right cell is shaded and marked with a question mark; the other cells are empty

Which genotype sits in the shaded cell?

  1. A. RR
    The rr parent has no R to give, and the gamete above the shaded cell is r.
  2. B. Rr
    The gamete above the shaded cell is r, and the gamete beside it is r.
  3. C. ✓ rr

Why: The shaded cell sits under r and beside r.
So it holds r and r: rr.

81
Check q30

A YY pea plant is crossed with a yy plant. In the square below, one cell is shaded.

A two-by-two square with Y and Y along the top edge and y and y down the left edge; the top-left cell is shaded and marked with a question mark; the other cells are empty
A two-by-two square with Y and Y along the top edge and y and y down the left edge; the top-left cell is shaded and marked with a question mark; the other cells are empty

Which genotype sits in the shaded cell?

  1. A. YY
    The gamete beside the shaded cell is y.
  2. B. ✓ Yy
  3. C. yy
    The gamete above the shaded cell is Y.

Why: The shaded cell sits under Y and beside y.
So it holds Y and y: Yy.

82
Check q31

Two Gg pea plants are crossed. In the square below, one cell is shaded.

A two-by-two square with G and g along the top edge and G and g down the left edge; the bottom-right cell is shaded and marked with a question mark; the other cells are empty
A two-by-two square with G and g along the top edge and G and g down the left edge; the bottom-right cell is shaded and marked with a question mark; the other cells are empty

Which genotype sits in the shaded cell?

  1. A. GG
    The gamete above the shaded cell is g, and the gamete beside it is g.
  2. B. Gg
    Both gametes at the shaded cell carry g.
  3. C. ✓ gg

Why: The shaded cell sits under g and beside g.
So it holds g and g: gg.

83
Check q32

A Bb mouse is crossed with a BB mouse. In the square below, one cell is shaded.

A two-by-two square with B and b along the top edge and B and B down the left edge; the top-left cell is shaded and marked with a question mark; the other cells are empty
A two-by-two square with B and b along the top edge and B and B down the left edge; the top-left cell is shaded and marked with a question mark; the other cells are empty

Which genotype sits in the shaded cell?

  1. A. ✓ BB
  2. B. Bb
    The gamete above the shaded cell is B, and the gamete beside it is B.
  3. C. bb
    The BB parent has no b to give.

Why: The shaded cell sits under B and beside B.
So it holds B and B: BB.

84Mixed practice mixed practice

85
Check q33

A pea plant is homozygous for the white-flower allele, pp.

Which of the following do its gametes carry for the flower-color gene?

  1. A. Half carry P and half carry p
    Both homologs of a pp plant carry p, so it has no P to pass on.
  2. B. Each gamete carries the pair, pp
    A gamete carries one homolog, so one allele.
  3. C. All of the gametes carry P
    A plant can only pass on alleles it has, and a pp plant carries no P.
  4. D. ✓ Every gamete carries p

Why: Both homologs of a pp plant carry p.
Each gamete receives one homolog.
So every gamete carries a single p.

86
Check q34

A student draws the square below for a Tt pea plant crossed with a tt plant.

A two-by-two square with Tt written over both columns and tt beside both rows, and every cell reading Tttt
A two-by-two square with Tt written over both columns and tt beside both rows, and every cell reading Tttt

Which of the following is wrong with the square?

  1. A. ✓ Genotypes, not gametes, sit on the edges
  2. B. The edges should hold the two phenotypes, tall and short
    The edges of a Punnett square hold gametes, one allele each.
    Tall and short are phenotypes, not gametes.
  3. C. The cells should each hold one allele
    A cell holds the two alleles an egg and a pollen grain bring together, so two letters is right for a cell.
  4. D. Nothing; a square for Tt × tt looks like this
    The edges of a Punnett square carry gametes, one allele each; here they carry Tt and tt, and the cells have four letters.

Why: The edges of a Punnett square are gametes, one allele each.
The Tt parent’s gametes are T and t, and the tt parent’s are t and t.
The student wrote the genotypes instead, so every cell came out with four letters.

87
Check q35

A pea flower is pollinated, and one of its eggs is fertilized.

Where does the sperm fuse with the egg?

  1. A. Inside the pollen grain
    The pollen grain carries the sperm to the flower.
    The egg waits inside the ovule, and the sperm travels down to it.
  2. B. ✓ Inside the ovule

Why: The egg sits inside an ovule.
The sperm leaves the pollen grain and travels down to the egg.
So the sperm fuses with the egg inside the ovule.

88
Check q36

A pea plant is TT for the stem-height gene.

How many kinds of gamete does it make for this gene?

  1. A. ✓ One
  2. B. Two
    Both homologs carry T.
    So every gamete carries T, and there is one kind.

Why: Both homologs carry T.
Each gamete receives one homolog.
So every gamete carries T: one kind of gamete.

89
Check q37

In pea plants, round seeds (R) are dominant to wrinkled seeds (r). An Rr plant is crossed with an rr plant.

Which genotypes appear among the offspring?

  1. A. Rr only
    Half of the Rr parent’s gametes carry r, and every gamete of the rr parent carries r, so some offspring are rr.
  2. B. RR and rr
    The rr parent has no R to give, so no offspring is RR.
  3. C. ✓ Rr and rr
  4. D. RR, Rr and rr
    The rr parent has no R to give, so no offspring is RR.

Why: The Rr parent’s gametes are R and r; the rr parent’s are r and r.
The cells are Rr, Rr, rr, rr.
So the offspring are Rr and rr.

90
Check q38

Two Pp pea plants are crossed, and the Punnett square for the cross has four cells.

How many different genotypes do the four cells hold?

  1. A. Two
    The cells hold PP, Pp and pp.
  2. B. ✓ Three
  3. C. Four
    Two of the four cells hold the same genotype, Pp.

Why: The four cells are PP, Pp, Pp and pp.
Pp appears twice.
So the four cells hold three genotypes.

91
Practice writing an answer

In pea plants, round seeds (R) are dominant to wrinkled seeds (r). Two heterozygous plants, Rr and Rr, are crossed.

(a) Identify the gametes an Rr plant makes for this gene, and explain why no gamete carries both R and r. (1 pt)

Model answer Half of its gametes carry R and half carry r.
R sits on one homolog and r sits on the other.
A gamete carries one homolog of the pair.
So a gamete gets one of the two alleles and never both.
Rubric
  • Award 1 point for: half R and half r, because each gamete receives one homolog and the two alleles sit on different homologs.

Slip Writing Rr as the gamete. Rr is the plant’s genotype; a gamete carries one allele.

(b) State each parent’s gametes and the genotype in each of the four cells of the Punnett square for this cross. (1 pt)

Model answer Each Rr parent makes gametes R and r.
So R and r go along the top edge and down the side.
Top left, R with R: RR.
Top right, R with r: Rr.
Bottom left, r with R: Rr.
Bottom right, r with r: rr.The square for Rr crossed with Rr: R and r along the top and down the side; cells RR, Rr, Rr, rrRRRrRrrrRrRrRr × Rr: RR, Rr, Rr, rr
Rubric
  • Award 1 point for: R and r on each edge (one allele per gamete) and the cells RR, Rr, Rr, rr.

Slip Writing Rr on each edge. The edges carry the gametes, one allele each, not the parents’ genotypes.

(c) A student writes Rr on both edges of the square. Describe what is wrong with that square. (1 pt)

Model answer The edges carry the parents’ genotypes instead of their gametes.
Each edge should carry R and r, one allele each, because every gamete carries one allele.
With Rr on both edges every cell would hold four letters.
Rubric
  • Award 1 point for: the edges should carry the gametes, R and r, one allele each, not the two-letter genotype.

Slip Saying the parents are on the wrong sides. Either parent may take either edge; the fault is what is written on the edges.

(d) Identify the number of cells, out of the four, that hold each genotype. (1 pt)

Model answer One cell holds RR, two cells hold Rr, and one cell holds rr.
Rubric
  • Award 1 point for: one RR, two Rr, one rr.

Slip Counting four different genotypes. Two of the four cells hold the same genotype, Rr.

Glossary

law of segregation
The two alleles of a gene separate when gametes form, because they sit on the two homologs and the homologs part at anaphase I; so each gamete carries exactly one of the two alleles.
ovule
A small case in the base of a flower that holds the plant's egg. After a sperm fuses with the egg, the fertilized egg grows into a seed.
pollen grain
A tiny grain made by a flower's anthers that carries the plant's sperm. A P pollen grain is a pollen grain whose sperm carries the P allele.
Punnett square
A table with one parent's gametes along the top edge and the other's down the left edge; each cell holds the two alleles one egg and one pollen grain (or sperm) bring together.
monohybrid cross
A breeding cross that follows a single gene, such as Pp × Pp for flower color. Mono- means one; a hybrid is an offspring carrying two different alleles.

APBIO-U05-L13B Why three to one, and why not exactly

Topic 5.3 · Mendelian Genetics · 75 steps

The filled Punnett square for Pp crossed with Pp: P and p along the top, P and p down the side, cells PP, Pp, Pp, pp; beside it Mendel's count, 705 purple and 224 white, and the square's three to one, which would be 697 and 232
The filled Punnett square for Pp crossed with Pp: P and p along the top, P and p down the side, cells PP, Pp, Pp, pp; beside it Mendel's count, 705 purple and 224 white, and the square's three to one, which would be 697 and 232

Here is the filled square for two Pp pea plants, PP, Pp, Pp and pp, and beside it Mendel’s count of their offspring: 705 purple plants and 224 white.

The square says three purple to one white, which for 929 plants would be 697 and 232. Mendel counted 705 and 224. Why three to one, and why not exactly?

Unit 5 · Heredity

1Read the square twice

2

Video: Watch: Read the square twice

The same Pp × Pp square read twice: first by genotype, its cells grouped as one PP, two Pp and one pp; then by phenotype, the three purple-giving cells grouped against the one white; the two ratios written beside it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13Ba.mp4

3

What does a filled square tell you, and what does it not promise?

4

Read the square once by genotype: one PP, two Pp, one pp. Read it again by what the plants look like: three purple to one white.

5

To turn a ratio into the count you expect for a batch, you multiply the total by each kind’s share of the ratio.

6

A real batch strays a little from that count, because each seed is its own event with the same chance.

7
Check q1

Two Pp pea plants are crossed.

Which genotypes fill the four cells of their Punnett square?

  1. A. ✓ PP, Pp, Pp, pp
  2. B. PP, PP, pp, pp
    Each parent’s gametes are P and p.
    The P egg with the p pollen grain gives Pp, in two of the four cells.
  3. C. Pp, Pp, Pp, Pp
    The P egg with the P pollen grain gives PP, and the p egg with the p pollen grain gives pp.

Why: Each Pp parent’s gametes are P and p.
The four pairings give PP, Pp, Pp and pp.

8
Check q2

In pea plants, purple flowers (P) are dominant to white flowers (p).

Which flower color does a Pp plant show?

  1. A. White
    P is dominant, so its trait shows in a heterozygous plant.
    A Pp plant is purple.
  2. B. ✓ Purple

Why: P is dominant.
A dominant allele’s trait shows in a heterozygous plant.
So a Pp plant is purple.

9

Read the Pp × Pp square once by genotype. One cell is PP, two are Pp and one is pp: one to two to one.

The Pp × Pp square with its cells grouped by genotype: one PP cell, two Pp cells shaded, one pp cell; beside it the genotypic ratio 1 PP : 2 Pp : 1 pp
The Pp × Pp square with its cells grouped by genotype: one PP cell, two Pp cells shaded, one pp cell; beside it the genotypic ratio 1 PP : 2 Pp : 1 pp
10

When we write the offspring genotypes in this proportion, 1 PP : 2 Pp : 1 pp, we call it the cross’s , because it counts genotypes.

11

Now read the same square by phenotype.

The same square with its three purple-giving cells shaded together, PP, Pp and Pp, and the pp cell unshaded; beside it the phenotypic ratio 3 purple : 1 white
The same square with its three purple-giving cells shaded together, PP, Pp and Pp, and the pp cell unshaded; beside it the phenotypic ratio 3 purple : 1 white
12

P is dominant. So PP and Pp are both purple, and pp is white: three purple to one white.

13

When we write the offspring phenotypes in this proportion, 3 purple : 1 white, we call it the , because it counts phenotypes.

14

The two ratios come from the same square and are not the same thing. The genotypic ratio has three kinds, or classes, of offspring; the phenotypic ratio has two classes.

15

What you are expected to know Read a filled square twice, as a genotypic ratio and as a phenotypic ratio, and say which is which.

16
Check q3

Two Pp pea plants are crossed. Their square gives PP, Pp, Pp, pp.

Which kind of ratio is 1 PP : 2 Pp : 1 pp?

  1. A. ✓ Genotypic
  2. B. Phenotypic
    PP, Pp and pp are genotypes, three classes.
    A ratio that counts genotypes is genotypic.

Why: PP, Pp and pp are genotypes.
A ratio that counts genotypes is the genotypic ratio.
So 1 : 2 : 1 is genotypic.

17
Check q4

Two Pp pea plants are crossed.

Which kind of ratio is 3 purple : 1 white?

  1. A. Genotypic
    Purple and white are phenotypes, two classes, and the three purple cells hold two different genotypes.
    A ratio that counts phenotypes is phenotypic.
  2. B. ✓ Phenotypic

Why: Purple and white are phenotypes.
A ratio that counts phenotypes is the phenotypic ratio.
So 3 : 1 is phenotypic.

18Quick quiz: genotypic ratio, phenotypic ratio mixed practice

19
Check q5

Two pea plants are crossed and their square is filled.

What is a genotypic ratio?

  1. A. ✓ The ratio of offspring genotypes read from the square
  2. B. The ratio of offspring phenotypes read from the square
    A ratio of phenotypes is the phenotypic ratio.
    The genotypic ratio counts genotypes, such as PP, Pp and pp.
  3. C. The number of cells in the square
    A two-by-two square always has four cells.
    The genotypic ratio counts how many cells hold each genotype.

Why: A genotypic ratio counts the offspring genotypes in a square.
For Pp × Pp it is 1 PP : 2 Pp : 1 pp.

20
Check q6

Two pea plants are crossed and their square is filled.

What is a phenotypic ratio?

  1. A. The number of cells in the square
    A two-by-two square always has four cells.
    The phenotypic ratio counts how many cells show each phenotype.
  2. B. The ratio of offspring genotypes read from the square
    A ratio of genotypes is the genotypic ratio.
    The phenotypic ratio groups the genotypes that look the same.
  3. C. ✓ The ratio of offspring phenotypes read from the square

Why: A phenotypic ratio counts the offspring phenotypes in a square, grouping genotypes that look the same.
For Pp × Pp it is 3 purple : 1 white.

21
Practice writing an answer

In pea plants, inflated pods (I) are dominant to constricted pods (i). Two Ii plants are crossed, and their square gives II, Ii, Ii and ii.

(a) Identify the genotypic ratio of the offspring. (1 pt)

Model answer The genotypic ratio is 1 II : 2 Ii : 1 ii.
Rubric
  • Award 1 point for: 1 II : 2 Ii : 1 ii.

(b) Identify the phenotypic ratio of the offspring. (1 pt)

Model answer The phenotypic ratio is 3 inflated : 1 constricted.
Rubric
  • Award 1 point for: 3 inflated : 1 constricted.

(c) Explain why the phenotypic ratio has two classes while the genotypic ratio has three. (1 pt)

Model answer I is dominant, so II and Ii both have inflated pods.
The phenotypic ratio groups II and Ii together as inflated.
So the three genotypes give only two phenotypes.
Rubric
  • Award 1 point for: II and Ii look the same (both inflated, because I is dominant), so two genotype classes make one phenotype class.
22
Check q7

In pea plants, tall (T) is dominant to short (t). The square below is for a Tt plant crossed with a tt plant.

The square for Tt crossed with tt: T and t along the top, t and t down the side; cells Tt, Tt, tt, tt
The square for Tt crossed with tt: T and t along the top, t and t down the side; cells Tt, Tt, tt, tt

Which of the following is the phenotypic ratio, tall to short?

  1. A. ✓ 1 : 1
  2. B. 3 : 1
    Here two cells are Tt, tall, and two are tt, short.
  3. C. 1 : 2 : 1
    Only two phenotypes appear here, tall and short.

Why: Two cells are Tt and two are tt.
Tt is tall and tt is short.
So the phenotypic ratio is 1 tall : 1 short.

23
Check q8

A Tt plant is crossed with a tt plant; the square is drawn below.

The square for Tt crossed with tt: T and t along the top, t and t down the side; cells Tt, Tt, tt, tt
The square for Tt crossed with tt: T and t along the top, t and t down the side; cells Tt, Tt, tt, tt

Which of the following is the genotypic ratio?

  1. A. 3 Tt : 1 tt
    The square has four cells.
    Two cells are Tt and two are tt, so the two genotypes have equal shares.
  2. B. 1 TT : 2 Tt : 1 tt
    No cell here is TT, because the tt parent has no T to give.
  3. C. ✓ 1 Tt : 1 tt

Why: Two cells are Tt and two are tt.
No cell is TT.
So the genotypic ratio is 1 Tt : 1 tt.

24
Check q9

Two Gg pea plants are crossed.

How many different genotypes appear among the offspring?

  1. A. 2
    The offspring show two pod colors, green and yellow.
    But the cells of the square hold GG, Gg and gg.
  2. B. ✓ 3
  3. C. 4
    The square has four cells, but two of them hold the same genotype, Gg.
    The genotypes are GG, Gg and gg.

Why: The Gg × Gg square gives GG, Gg, Gg and gg.
Two cells hold Gg.
So three different genotypes appear: GG, Gg and gg.

25
Check q10

In tomato plants, tall (T) is dominant to dwarf (t). Two tall plants are crossed and give 152 tall and 48 dwarf offspring, close to 3 : 1.

Which of the following genotypic ratios fits these offspring?

  1. A. 3 TT : 1 tt
    The 152 tall plants are of two genotypes, TT and Tt, so 3 : 1 counts phenotypes, not genotypes.
  2. B. 1 TT : 1 Tt : 1 tt
    The Tt × Tt square has four cells and two of them are Tt, so Tt is twice as common as TT.
  3. C. 1 Tt : 1 tt
    A 1 : 1 genotypic ratio comes from Tt × tt and gives half dwarf offspring, not one in four.
  4. D. ✓ 1 TT : 2 Tt : 1 tt

Why: Two tall parents that give one dwarf offspring in four are both Tt.
Their square gives 1 TT : 2 Tt : 1 tt.
Grouping TT with Tt as tall gives the 3 : 1 seen.

26From a ratio to a count

27

Video: Watch: From a ratio to a count

The 3 : 1 ratio cut into its four shares; the equation written once; then 929 plants turned into 697 purple and 232 white, the total multiplied by three quarters and by one quarter.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13Bb.mp4

28
Check q11

A Tt pea plant is crossed with a tt plant. Their square gives Tt, Tt, tt, tt.

Which kind of ratio is 1 tall : 1 short?

  1. A. Genotypic
    Tall and short are phenotypes.
    The genotypic ratio here is 1 Tt : 1 tt.
  2. B. ✓ Phenotypic

Why: Tall and short are phenotypes.
A ratio that counts phenotypes is the phenotypic ratio.

29

Now suppose a grower sows 929 seeds from a Pp × Pp cross. How many purple plants should the grower expect?

30

A ratio is made of shares. The phenotypic ratio 3 : 1 has four shares in all: three of the four shares are purple and one share is white.

31

Here is a table of the two ratios and their shares.

A table of shares: the genotypic ratio 1 : 2 : 1 has four shares, one, two and one of four; the phenotypic ratio 3 : 1 has four shares, three of four and one of four
A table of shares: the genotypic ratio 1 : 2 : 1 has four shares, one, two and one of four; the phenotypic ratio 3 : 1 has four shares, three of four and one of four
32
Check q12 numeric entry

Two Gg pea plants are crossed. Their offspring show the phenotypic ratio 3 green : 1 yellow.

How many shares does this ratio have in all?

Answer: 4  (tolerance ±0)

Working
Write down the values in the question:
phenotypic ratio = 3 green : 1 yellow
Add the parts of the ratio:
shares in all = 3 + 1 = 4
33

To turn a ratio into the expected count for one class of offspring, you multiply the total by that class’s share of the ratio. Here is the equation.

the expected count for one class of offspring
34
Worked example

Two Pp pea plants are crossed and 200 offspring are grown. How many are expected to be PP, Pp and pp, and how many are expected to be purple?

Write down the values in the question:
total offspring = 200
genotypic ratio = 1 PP : 2 Pp : 1 pp, so 4 shares in all
PP share = 1 of 4
Pp share = 2 of 4
pp share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected PP=200×14=50
expected Pp=200×24=100
expected pp=200×14=50
Group the genotypes by phenotype:
expected purple=50+100=150
expected white=50
35
Worked example

A grower sows 929 seeds from a Pp × Pp cross. How many purple plants and how many white plants should the grower expect?

Write down the values in the question:
total offspring = 929
phenotypic ratio = 3 purple : 1 white, so 4 shares in all
purple share = 3 of 4
white share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected purple=929×34=696.75
expected white=929×14=232.25
Round each to the nearest whole plant:
expected purple≈697
expected white≈232
36

What you are expected to know Calculate the expected count in each class from the total and that class’s share of the ratio.

37
Check q13 numeric entry

In pea plants, purple flowers (P) are dominant to white (p). Two Pp pea plants are crossed and 120 offspring are grown.

Calculate how many of the 120 are expected to be purple.

Part 1. PP is 1 share of the 4 in the Pp × Pp genotypic ratio. How many PP offspring are expected?

Answer: 30  (tolerance ±0)

Working
Multiply the total by the PP share of the ratio:
expected PP=120×14=30

Part 2. Pp is 2 shares of the 4. How many Pp offspring are expected?

Answer: 60  (tolerance ±0)

Working
Multiply the total by the Pp share of the ratio:
expected Pp=120×24=60

Part 3. pp is 1 share of the 4. How many pp offspring are expected?

Answer: 30  (tolerance ±0)

Working
Multiply the total by the pp share of the ratio:
expected pp=120×14=30

Answer: 90  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 120
genotypic ratio = 1 PP : 2 Pp : 1 pp, so 4 shares in all
PP share = 1 of 4, Pp share = 2 of 4, pp share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected PP=120×14=30
expected Pp=120×24=60
expected pp=120×14=30
Group the genotypes by phenotype:
expected purple=30+60=90
38
Check q14 numeric entry

In pea plants, tall (T) is dominant to short (t). Two Tt plants are crossed and 80 offspring are grown.

Calculate how many of the 80 are expected to be short.

Answer: 20  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 80
phenotypic ratio = 3 tall : 1 short, so 4 shares in all
short (tt) share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected tt=80×14=20
39
Check q15 numeric entry

In mice, black fur (B) is dominant to brown (b). A Bb mouse is crossed with a bb mouse and 60 pups are born.

Calculate how many of the 60 pups are expected to be brown.

Answer: 30  (tolerance ±0)

Working
Write down the values in the question:
total pups = 60
square = Bb, Bb, bb, bb, so the phenotypic ratio is 1 black : 1 brown, 2 shares in all
brown (bb) share = 1 of 2
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected bb=60×12=30

40Quick quiz: expected counts mixed practice

41
Check q16 numeric entry

In pea plants, purple flowers (P) are dominant to white (p). Two Pp plants are crossed and 160 offspring are grown.

Calculate how many of the 160 are expected to be white.

Answer: 40  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 160
phenotypic ratio = 3 purple : 1 white, so 4 shares in all
white (pp) share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected white=160×14=40
42
Check q17 numeric entry

In mice, black fur (B) is dominant to brown (b). Bb mice are crossed with bb mice and 300 pups are born in all.

Calculate how many of the 300 pups are expected to be brown.

Answer: 150  (tolerance ±0)

Working
Write down the values in the question:
total pups = 300
square = Bb, Bb, bb, bb, so the phenotypic ratio is 1 black : 1 brown, 2 shares in all
brown (bb) share = 1 of 2
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected brown=300×12=150
43
Check q18 numeric entry

In pea plants, tall (T) is dominant to short (t). Two Tt plants are crossed and 96 offspring are grown.

Calculate how many of the 96 are expected to be Tt.

Answer: 48  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 96
genotypic ratio = 1 TT : 2 Tt : 1 tt, so 4 shares in all
Tt share = 2 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected Tt=96×24=48
44
Check q19 numeric entry

In pea plants, round seeds (R) are dominant to wrinkled (r). Two Rr plants are crossed and 72 seeds are collected.

Calculate how many of the 72 seeds are expected to be round.

Answer: 54  (tolerance ±0)

Working
Write down the values in the question:
total seeds = 72
phenotypic ratio = 3 round : 1 wrinkled, so 4 shares in all
round (RR and Rr) share = 3 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected round=72×34=54
45
Check q20 numeric entry

In pea plants, tall (T) is dominant to short (t). Tt plants are crossed with tt plants and 500 offspring are grown in all.

Calculate how many of the 500 are expected to be tall.

Answer: 250  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 500
square = Tt, Tt, tt, tt, so the phenotypic ratio is 1 tall : 1 short, 2 shares in all
tall (Tt) share = 1 of 2
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected tall=500×12=250

46A ratio is a chance, not a promise

47

Video: Watch: A ratio is a chance, not a promise

Single rows of kernels on ears of Pp × Pp corn counted one by one, each straying from three to one; then fifty rows counted together sitting close to it; then a run of purple seeds, and the next seed’s chance staying one in four.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L13Bc.mp4

48
Check q21 numeric entry

Two Pp pea plants are crossed and 48 offspring are grown.

Calculate the expected count of pp offspring.

Answer: 12  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 48
genotypic ratio = 1 PP : 2 Pp : 1 pp, so 4 shares in all
pp share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected pp=48×14=12
49

Each seed on a Pp plant pollinated by another Pp plant comes from its own egg and its own pollen grain. So each seed is a separate event with the same chance: one chance in four of pp.

50

Now consider corn instead of peas. An ear of corn carries its kernels in rows, and one row is a batch of seeds you can count at a glance.

51

In corn, purple kernels (P) are dominant to yellow kernels (p).

52

Here is a table of kernel counts from rows on ears of Pp × Pp corn, with the count three to one would give beside each.

A table of rows of kernels from ears of Pp × Pp corn, with the count three to one would give: a row of 24 kernels with 17 purple and 7 yellow (18 and 6 expected); a row of 20 kernels with 13 purple and 7 yellow (15 and 5 expected); a row of 28 kernels with 24 purple and 4 yellow (21 and 7 expected); and all 1,200 kernels of fifty rows together, 906 purple and 294 yellow (900 and 300 expected)
A table of rows of kernels from ears of Pp × Pp corn, with the count three to one would give: a row of 24 kernels with 17 purple and 7 yellow (18 and 6 expected); a row of 20 kernels with 13 purple and 7 yellow (15 and 5 expected); a row of 28 kernels with 24 purple and 4 yellow (21 and 7 expected); and all 1,200 kernels of fifty rows together, 906 purple and 294 yellow (900 and 300 expected)
53

One row of 24 kernels had 17 purple and 7 yellow; three to one would give 18 and 6.

54

Another row of 28 kernels had 24 purple and 4 yellow; three to one would give 21 and 7. A single row strays from the ratio.

55

Fifty rows counted together held 1,200 kernels: 906 purple and 294 yellow. Three to one would give 900 and 300. The more kernels you count, the closer the split comes to the ratio.

56

Earlier seeds do not change the next seed’s chance. Each seed is its own egg fusing with its own pollen grain.

57

So after four purple seeds in a row, the fifth seed still has one chance in four of being yellow.

58

What you are expected to know Explain why a small batch of offspring need not match the ratio, and judge whether a stated count is consistent with the cross that produced it.

59
Check q22

In corn, purple kernels (P) are dominant to yellow (p). Two Pp plants are crossed. One row of 36 kernels on an ear from the cross has 25 purple and 11 yellow.

Is this row consistent with a Pp × Pp cross?

  1. A. ✓ Yes
  2. B. No
    The square gives a chance for each kernel, not an exact count for a row.
    27 and 9 is the expected count, and a real row strays from it.

Why: Three to one predicts 27 purple and 9 yellow for a row of 36.
25 and 11 is two kernels away from that.
One row of 36 kernels is a small batch, so straying this far is expected.

60
Practice writing an answer

In corn, purple kernels (P) are dominant to yellow (p). Two Pp plants are crossed. One row of 36 kernels on an ear from the cross has 25 purple and 11 yellow.

(a) Explain why 25 purple and 11 yellow in one row of 36 kernels is consistent with a Pp × Pp cross, which predicts three to one. (1 pt)

Model answer Three to one predicts 27 purple and 9 yellow for a row of 36.
Each kernel is its own egg fused with its own sperm, with one chance in four of yellow.
So a batch strays from the expected count by chance.
25 and 11 is two kernels from 27 and 9, and 36 kernels is a small batch.
So the row is consistent with the cross.
Rubric
  • Award 1 point for: each kernel is its own chance, so a small batch strays from the predicted ratio; 25 : 11 is close to the 27 : 9 the square predicts.
61
Check q23

In mice, black fur (B) is dominant to brown (b). Two Bb mice have four pups, all black. A student says: “The next pup must be brown, because the litter is due a brown one.”

Is the student correct?

  1. A. Yes: the next pup will be brown
    Each pup is its own egg fused with its own sperm.
    Earlier pups do not change the next pup’s chance.
  2. B. ✓ No: the next pup has one chance in four of being brown

Why: Each pup comes from its own egg and its own sperm.
So each pup has the same chance, one in four of bb.
Four black pups do not change the chance for the fifth.

62
Check q24

Suppose that in one species of fish, blue (B) is dominant to gold (b). Two Bb fish spawn. Their first four offspring are all blue.

Which of the following is the chance that the next offspring is blue?

  1. A. Certain
    A Bb × Bb pair can produce a gold bb offspring.
    Four blue in a row is what small batches do.
  2. B. Less than three in four
    Each fertilization is a separate event with the same chance.
  3. C. ✓ Three in four
  4. D. One in four
    No offspring is ever due.
    Earlier offspring do not change the chance for the next.

Why: Every fertilization has the same chance, set by the square: three cells of four give blue.
Four blue offspring in a row do not change that for the fifth.

63

Here again is the filled square for two Pp pea plants, PP, Pp, Pp and pp, beside Mendel’s count: 705 purple and 224 white.

The filled Punnett square for Pp crossed with Pp, cells PP, Pp, Pp, pp; beside it three lines: the square says 3 purple : 1 white; for 929 plants, 697 purple and 232 white; Mendel counted 705 purple and 224 white
The filled Punnett square for Pp crossed with Pp, cells PP, Pp, Pp, pp; beside it three lines: the square says 3 purple : 1 white; for 929 plants, 697 purple and 232 white; Mendel counted 705 purple and 224 white
64

Three of the square’s four equally likely cells hold a P. So the square says three purple to one white.

65

Each of Mendel’s 929 plants was its own chance. So his count came close to 697 and 232, and not exactly.

66Mixed practice mixed practice

67
Check q25

Two Yy pea plants (yellow seeds, Y, dominant to green, y) are crossed.

Which of the following is the cross’s phenotypic ratio?

  1. A. 1 YY : 2 Yy : 1 yy
    1 YY : 2 Yy : 1 yy is the genotypic ratio, three classes of genotype; the phenotypic ratio groups YY and Yy together as yellow.
  2. B. ✓ 3 yellow : 1 green
  3. C. 1 yellow : 1 green
    1 : 1 is what Yy × yy gives; two heterozygous plants give one green in four, not half.
  4. D. 2 yellow : 1 green
    The square has four cells, three yellow-giving and one green, not three cells.

Why: The Yy × Yy square gives YY, Yy, Yy, yy.
The first three show yellow and the last green.
So the phenotypic ratio is 3 yellow : 1 green.

68
Check q26

In pea plants, purple flowers (P) are dominant to white (p). Three growers each cross two Pp plants and grow the offspring: one grows 8 plants, one grows 40, and one grows 800.

Which batch is most likely to sit close to three purple to one white?

  1. A. The batch of 8
    A small batch strays furthest from the ratio: two white plants in eight, or none, is common.
  2. B. All three equally
    The chance for each plant is the same, one in four, but a batch of 8 strays from that share far more than a batch of 800 does.
  3. C. ✓ The batch of 800
  4. D. None of them
    Large batches do sit close to the ratio; fifty rows of kernels together sat near three to one even though single rows strayed.

Why: Every plant has the same one-in-four chance of pp.
The more plants you count, the closer the split comes to the ratio.
So the batch of 800 sits closest to three purple to one white.

69
Check q27 numeric entry

In pea plants, green pods (G) are dominant to yellow pods (g). Two Gg plants are crossed and 240 offspring are grown.

Calculate how many of the 240 offspring are expected to be heterozygous, Gg.

Answer: 120  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 240
genotypic ratio = 1 GG : 2 Gg : 1 gg, so 4 shares in all
Gg share = 2 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected Gg=240×24=120
70
Check q28

In mice, black fur (B) is dominant to brown (b). Two Bb mice have a litter of six pups, and all six are black.

Which of the following does this litter show about the cross?

  1. A. The parents cannot both be Bb
    The square gives each pup a three-in-four chance of black, not a fixed count.
  2. B. One parent must be BB
    A Bb × Bb pair can give six black pups in a row.
  3. C. The square does not apply to this pair of mice
    The square is right for every pair of Bb mice: it gives a chance for each pup, and small batches stray from it.
  4. D. ✓ The litter is consistent with Bb × Bb

Why: Each pup has a three-in-four chance of being black.
So six black pups in a row happen often enough from Bb × Bb.
A small batch straying from three to one breaks nothing.

71
Check q29

In pea plants, a heterozygous purple plant, Pp, is crossed with a white plant, pp.

Which of the following describes the offspring?

  1. A. ✓ Half purple and half white
  2. B. Three purple to one white
    Here one parent is pp and contributes only p.
    The square is Pp, Pp, pp, pp.
  3. C. All of the offspring purple
    Half of the Pp parent’s gametes carry p, and every gamete of the pp parent carries p.
    So half of the offspring are pp and white.
  4. D. All of the offspring white
    Half of the offspring receive P from the Pp parent and are purple.

Why: The Pp parent’s gametes are P and p; the pp parent’s are all p.
The square gives Pp, Pp, pp, pp: half purple, half white.

72
Check q30

In mice, black fur (B) is dominant to brown (b). Bb mice are crossed with bb mice, and 50 pups are born: 22 brown and 28 black. The expected count of brown pups is 25.

Is a count of 22 brown pups consistent with the cross?

  1. A. ✓ Yes
  2. B. No
    Each pup is its own chance, one in two of brown.
    A batch of 50 strays from the expected 25, and 22 is close to 25.

Why: Each pup has the same chance, one in two of bb.
So a batch of 50 strays from the expected count of 25.
22 is close to 25, so the count is consistent.

73
Check q31

In pea plants, tall (T) is dominant to short (t). Two Tt plants are crossed.

Which of the following is the genotypic ratio of the offspring?

  1. A. 3 tall : 1 short
    3 tall : 1 short counts phenotypes.
    The genotypic ratio counts TT, Tt and tt.
  2. B. ✓ 1 TT : 2 Tt : 1 tt
  3. C. 1 Tt : 1 tt
    1 Tt : 1 tt is the genotypic ratio of Tt × tt.
    Two Tt parents also give TT.

Why: The Tt × Tt square gives TT, Tt, Tt, tt.
So the genotypic ratio is 1 TT : 2 Tt : 1 tt.

74
Practice writing an answer

In pea plants, round seeds (R) are dominant to wrinkled seeds (r). Two heterozygous plants, Rr and Rr, are crossed and 400 seeds are collected.

(a) Identify the genotypic ratio of the offspring. (1 pt)

Model answer The genotypic ratio is 1 RR : 2 Rr : 1 rr.
Rubric
  • Award 1 point for: 1 RR : 2 Rr : 1 rr, named as the genotypic ratio.

Slip Giving 3 : 1 as the genotypic ratio. The genotypic ratio has three classes; 3 : 1 groups RR and Rr together, which makes it the phenotypic ratio.

(b) Identify the phenotypic ratio of the offspring. (1 pt)

Model answer The phenotypic ratio is 3 round : 1 wrinkled, because RR and Rr are both round.
Rubric
  • Award 1 point for: 3 round : 1 wrinkled, named as the phenotypic ratio.

Slip Writing 1 : 2 : 1 for the phenotypes. Only two phenotypes appear, because RR and Rr look the same.

(c) Calculate the number of wrinkled seeds expected among the 400. (1 pt)

Answer: 100  (tolerance ±0)

Model answer About 100 of the 400 seeds are expected to be wrinkled.
Working
Write down the values in the question:
total seeds = 400
phenotypic ratio = 3 round : 1 wrinkled, so 4 shares in all
wrinkled (rr) share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected rr=400×14=100
Rubric
  • Award 1 point for: 100 wrinkled seeds.

(d) The grower counts 112 wrinkled seeds instead. Explain why a count of 112 is still consistent with the Rr × Rr cross. (1 pt)

Model answer The square gives each seed a one-in-four chance of being rr.
So 100 wrinkled seeds is the expected count for 400.
Each seed is its own chance, so a real batch of 400 seeds strays from the expected count.
112 is close to 100.
So the count is consistent with Rr × Rr.
Rubric
  • Award 1 point for: the ratio is a chance for each seed, so a batch strays from the expected 100, and 112 is close enough to be consistent.

Slip Saying the cross must give exactly 100 wrinkled seeds. The square predicts a chance for each seed, not an exact count for the batch.

Glossary

genotypic ratio
The ratio of offspring genotypes read from a square: for Pp × Pp, 1 PP : 2 Pp : 1 pp.
phenotypic ratio
The ratio of offspring phenotypes read from a square, grouping genotypes that look the same: for Pp × Pp, 3 purple : 1 white.

APBIO-U05-L14 Rules of chance

Topic 5.3 · Mendelian Genetics · 80 steps

The filled Punnett square for Pp crossed with Pp: P and p along the top, P and p down the side, cells PP, Pp, Pp, pp; beside it two questions, the chance of pp and the chance of purple
The filled Punnett square for Pp crossed with Pp: P and p along the top, P and p down the side, cells PP, Pp, Pp, pp; beside it two questions, the chance of pp and the chance of purple

Here is the filled Punnett square for two Pp pea plants: PP, Pp, Pp and pp.

The plants set seed. Two questions about the next seed: what is the chance that it is pp, and what is the chance that it is purple?

Unit 5 · Heredity

1A number for a chance

2

Video: Watch: A number for a chance

One chance in four written as the fraction one quarter on a line from 0 to 1; 0 marked never, 1 marked always; a ratio set aside as not a probability.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L14a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L14a.mp4

3
Check q1

Two Pp pea plants are crossed. The square gives one pp cell in four. The first seed grown is pp.

What is the chance that the next seed is pp?

  1. A. ✓ One chance in four
  2. B. Less than one chance in four
    Each seed comes from its own egg and its own pollen grain.
    The first seed does not change the next seed’s chance.
  3. C. More than one chance in four
    Each seed comes from its own egg and its own pollen grain.
    The first seed does not change the next seed’s chance.

Why: Each seed is its own fertilization.
The parents are still Pp and Pp.
So the next seed has one chance in four of being pp, whatever the first seed was.

4

How do you put a number on the chance of a seed’s genotype? One chance in four is a chance said in words, and now we write it as a number.

A horizontal line from 0 on the left to 1 on the right, with marks at 0, ¼, ½, ¾ and 1; 0 is labeled never happens and 1 is labeled always happens; an arrow points at the ¼ mark, labeled the next seed is pp
A horizontal line from 0 on the left to 1 on the right, with marks at 0, ¼, ½, ¾ and 1; 0 is labeled never happens and 1 is labeled always happens; an arrow points at the ¼ mark, labeled the next seed is pp
5

Two rules then do the rest. When two things must both happen, multiply. When either of two outcomes will do, add.

6

When we write how likely an outcome is as a number between 0 and 1, we call that number its : 0 means the outcome never happens, 1 means it always happens.

7

We write a probability as a fraction. The next seed is pp with probability 14: one cell of the four.

8

The next seed is PP, Pp or pp with probability 1. Every cell of the square is one of those three, so that outcome always happens.

9

The next seed is PP and pp at once with probability 0. No seed has two genotypes, so that outcome never happens.

10

A number greater than 1, or less than 0, is never a probability. Three to one is a ratio, not a probability: it compares two classes of offspring.

11

One more convention: when a check asks you to type a probability, type it as a decimal, so a probability of 14 is typed 0.25.

12

What you are expected to know Say what a probability is: a number between 0 and 1 that says how likely an outcome is, written as a fraction or typed as a decimal.

13
Check q2

Two Pp pea plants are crossed. The next seed is pp with probability 14.

Which of the following does the probability 14 say about the seeds?

  1. A. ✓ Each seed has a probability of ¼ of being pp
  2. B. Exactly one seed in every four is pp
    A probability says how likely each seed is to be pp.
    It fixes no count in every four seeds.
  3. C. The fourth seed is pp
    A probability says how likely each seed is to be pp.
    It names no seed.

Why: Each seed comes from one egg and one pollen grain.
One pairing of the four gives pp.
So each seed has a probability of ¼ of being pp.
The probability says that about every seed, and fixes no count.

14
Check q3

A student writes that a seed is purple with probability 1.5.

Can 1.5 be a probability?

  1. A. Yes
    A probability is a number between 0 and 1.
    1 already means the outcome always happens.
  2. B. ✓ No

Why: A probability is a number between 0 and 1.
1 means the outcome always happens.
No outcome happens more than always, so 1.5 cannot be a probability.

15Quick quiz: probability mixed practice

16
Check q4

A breeder describes how likely the next offspring is to be white.

Which of the following is a probability?

  1. A. ✓ A number between 0 and 1 that says how likely the outcome is
  2. B. The number of white offspring for every one that is not white
    A count of one class against another is a ratio, such as 3 : 1.
  3. C. The number of cells in the Punnett square
    A Punnett square for one gene has four cells whatever the probability of each outcome.

Why: A probability says how likely an outcome is.
It is a number between 0 and 1.
0 means never, and 1 means always.

17
Check q5

A student says a seed is yellow with probability 0.3.

Can 0.3 be a probability?

  1. A. ✓ Yes
  2. B. No
    0.3 lies between 0 and 1, so it can be a probability.

Why: A probability is a number between 0 and 1.
0.3 lies between 0 and 1.
So 0.3 can be a probability.

18
Check q6

A student says the probability of tall offspring is 3 : 1.

Can 3 : 1 be a probability?

  1. A. Yes
    3 : 1 compares two classes of offspring.
    A probability is one number between 0 and 1.
  2. B. ✓ No

Why: 3 : 1 is a ratio: it compares tall offspring with short ones.
A probability is one number between 0 and 1.
So 3 : 1 cannot be a probability; the probability of tall here would be ¾.

19
Check q7

Two pp plants are crossed. A student says a seed is pp with probability 1.

Can 1 be a probability?

  1. A. ✓ Yes
  2. B. No
    1 is the probability of an outcome that always happens.
    Every cell of pp × pp is pp.

Why: 1 means the outcome always happens.
Every cell of the pp × pp square is pp.
So the seed is pp with probability 1.

20
Check q8

A student says a pup is brown with probability 1.25.

Can 1.25 be a probability?

  1. A. Yes
    A probability is a number between 0 and 1.
    1.25 is greater than 1.
  2. B. ✓ No

Why: A probability is a number between 0 and 1.
1.25 is greater than 1.
So 1.25 cannot be a probability.

21
Check q9

Two PP plants are crossed. A student says a seed is pp with probability 0.

Can 0 be a probability?

  1. A. ✓ Yes
  2. B. No
    0 is the probability of an outcome that never happens.
    No cell of PP × PP is pp.

Why: 0 means the outcome never happens.
No cell of the PP × PP square is pp.
So the seed is pp with probability 0, and 0 is a probability.

22
Practice writing an answer

In mice, black fur (B) is dominant to brown (b). Two Bb mice are crossed. The square gives one bb cell in four, so a pup is brown, bb, with probability 14.

(a) Explain what the probability 14 says about the next pup. (1 pt)

Frame The probability 14 says that …

Model answer The probability 14 says that the next pup has a probability of ¼ of being brown.
Each pup comes from one egg and one sperm.
One pairing of the four gives bb.
So each pup has a probability of ¼ of being bb, whatever the pups before it were.
Rubric
  • Award 1 point for: each pup has a probability of ¼ of being bb; the probability says how likely each pup is, not a fixed count of pups.

23Both at once: multiply

24

Video: Watch: Both at once: multiply

Two arrows meeting in the pp cell of the square, one from the p egg and one from the p pollen grain; the two probabilities multiplied; the formula-sheet line typeset as it is on the sheet.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L14b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L14b.mp4

25

For the next seed to be pp, two things must both happen: the egg must carry p, and the pollen grain must carry p.

The Pp × Pp square with the pp cell shaded; an arrow runs from the p on the top edge down into that cell and a second arrow runs from the p on the left edge across into it
The Pp × Pp square with the pp cell shaded; an arrow runs from the p on the top edge down into that cell and a second arrow runs from the p on the left edge across into it
26

A Pp plant’s egg carries p with probability 12, and so does its pollen grain.

27

The egg and the pollen grain come from two different plants’ meioses, so which allele one carries has no effect on which allele the other carries. Two events like that are independent.

28

For two independent events that must both happen, multiply their probabilities.

If A and B are independent, then this holds. The multiplication rule, as the AP formula sheet prints it: for two independent events that must both happen, multiply their probabilities
29

When two independent events must both happen and we multiply their probabilities to find the probability of both, we call it the ; your formula sheet prints it exactly as above.

30
Worked example

Two Pp pea plants are crossed. What is the probability that the next seed is pp?

Write down the values in the question:
P(p from the egg)=12
P(p from the pollen)=12
the two gametes are independent
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(A and B)=P(A)×P(B)
P(pp)=12×12
P(pp)=14=0.25
31

What you are expected to know Calculate the probability that an offspring has a genotype that needs one particular allele from each parent: multiply the two independent gamete probabilities, as the formula sheet’s multiplication rule says.

32
Check q10 numeric entry

Two Pp pea plants are crossed.

Calculate the probability that the next seed is PP, as a decimal.

Part 1. What is the probability that the egg carries P, as a decimal?

Answer: 0.5  (tolerance ±0.005)

Working
Half of a Pp plant's eggs carry P:
P(P from the egg)=12=0.5

Part 2. What is the probability that the pollen grain carries P, as a decimal?

Answer: 0.5  (tolerance ±0.005)

Working
Half of a Pp plant's pollen grains carry P:
P(P from the pollen)=12=0.5

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
P(P from the egg)=12
P(P from the pollen)=12
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(PP)=12×12
P(PP)=14=0.25
33
Check q11 numeric entry

In cats, short hair (L) is dominant to long hair (l). Two Ll cats have a kitten.

Calculate the probability that the kitten is long-haired, as a decimal.

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
long hair = ll
P(l from the mother)=12
P(l from the father)=12
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(ll)=12×12
P(ll)=14=0.25

34Either one: add

35

Video: Watch: Either one: add

The three purple-giving cells of the square shaded together and their probabilities added; then a Pp seed shown as both purple and heterozygous, so those two outcomes cannot be added.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L14c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L14c.mp4

36

For the next seed to be purple, either of two outcomes will do: it can be PP, or it can be Pp. No seed can be both at once, so the two outcomes are mutually exclusive.

The Pp × Pp square with the three purple-giving cells shaded together, PP, Pp and Pp, and the pp cell left unshaded
The Pp × Pp square with the three purple-giving cells shaded together, PP, Pp and Pp, and the pp cell left unshaded
37

When two outcomes cannot both happen and either one will do, we add their probabilities and call it the ; your formula sheet prints it exactly as above.

If A and B are mutually exclusive, then this holds. The addition rule, as the AP formula sheet prints it: for two outcomes that cannot both happen, when either one will do, add their probabilities
38

Which rule applies is decided by the question, not by the numbers:

  1. A p egg and a p pollen grain: both must happen, so multiply.
  2. PP or Pp: either one will do, and no seed is both, so add.
  3. Purple or heterozygous: do not add, because a Pp seed is both purple and heterozygous, so the two outcomes are not mutually exclusive.

39
Worked example

Two Pp pea plants are crossed. What is the probability that the next seed is purple?

Write down the values in the question:
purple = PP or Pp
P(PP)=14
P(Pp)=14+14=12
PP and Pp are mutually exclusive
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(A or B)=P(A)+P(B)
P(purple)=14+12
P(purple)=34=0.75
40

What you are expected to know Calculate the probability that an offspring has one of several genotypes or phenotypes that cannot both be true of it, by adding their probabilities.

41

What you are expected to know Say why adding is wrong when the outcomes overlap.

42
Check q12 numeric entry

Two Pp pea plants are crossed.

Calculate the probability that the next seed is heterozygous, as a decimal.

Part 1. What is the probability of a P egg fusing with a p pollen grain, as a decimal?

Answer: 0.25  (tolerance ±0.005)

Working
Both events must happen, so multiply:
P(P egg and p pollen)=12×12=14=0.25

Part 2. What is the probability of a p egg fusing with a P pollen grain, as a decimal?

Answer: 0.25  (tolerance ±0.005)

Working
Both events must happen, so multiply:
P(p egg and P pollen)=12×12=14=0.25

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(P egg and p pollen)=14
P(p egg and P pollen)=14
the two Pp cells are mutually exclusive
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(Pp)=14+14
P(Pp)=12=0.5
43
Check q13 numeric entry

In cats, short hair (L) is dominant to long hair (l). Two Ll cats have a kitten.

Calculate the probability that the kitten is short-haired, as a decimal.

Answer: 0.75  (tolerance ±0.005)

Working
Write down the values in the question:
short hair = LL or Ll
P(LL)=14
P(Ll)=12
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(short hair)=14+12
P(short hair)=34=0.75
44
Check q14

In cats, short hair (L) is dominant to long hair (l). Two Ll cats have a kitten.

Which of the following gives the probability that the kitten is short-haired or heterozygous?

  1. A. Add the probability of short hair to the probability of Ll
    An Ll kitten is both short-haired and heterozygous, so adding counts every Ll kitten twice.
    The sum, 1.25, is more than 1, and no probability is.
  2. B. Multiply the probability of short hair by the probability of Ll
    Multiplying is for two events that must both happen.
  3. C. Subtract the probability of Ll from the probability of short hair
    Subtracting leaves the short-haired kittens that are not Ll: LL only.
    Either outcome will do, and every Ll kitten is short-haired.
  4. D. ✓ Take the probability of short hair alone

Why: Every Ll kitten is short-haired.
So every heterozygous kitten is already inside the short-haired group.
The two outcomes are not mutually exclusive, so the addition rule does not apply.
The probability of short-haired or heterozygous is just the probability of short hair, 0.75.

45
Practice writing an answer

In cats, short hair (L) is dominant to long hair (l). Two Ll cats have a kitten. A student wants the probability that the kitten is short-haired or heterozygous.

(a) Explain why the addition rule is the wrong rule for short-haired or heterozygous. (1 pt)

Frame The addition rule is the wrong rule here because …

Model answer The addition rule is the wrong rule here because the two outcomes can both be true of one kitten.
Every Ll kitten is heterozygous.
Every Ll kitten is also short-haired, because L is dominant.
So short-haired and heterozygous are not mutually exclusive.
The addition rule needs outcomes that cannot both happen.
Adding here would count every Ll kitten twice.
Rubric
  • Award 1 point for: an Ll kitten is both short-haired and heterozygous, so the outcomes overlap and are not mutually exclusive; the addition rule needs outcomes that cannot both happen.

46Quick quiz: multiplication rule, addition rule mixed practice

47
Check q15

The formula sheet prints two rules of probability.

Which of the following describes the multiplication rule?

  1. A. Multiply the probabilities of two mutually exclusive outcomes
    Outcomes that cannot both happen are the addition rule’s case, when either one will do.
  2. B. ✓ Multiply the probabilities of independent events that must both happen
  3. C. Add the probabilities of independent events that must both happen
    Adding is for one outcome or the other; two events that must both happen are multiplied.

Why: The multiplication rule is for two independent events that must both happen.
Both events must happen, so their probabilities are multiplied.
P(A and B) = P(A) × P(B).

48
Check q16

The formula sheet prints two rules of probability.

Which of the following describes the addition rule?

  1. A. ✓ Add the probabilities of mutually exclusive outcomes when either one will do
  2. B. Add the probabilities of independent events that must both happen
    Two events that must both happen are the multiplication rule’s case.
  3. C. Multiply the probabilities of mutually exclusive outcomes
    Outcomes that cannot both happen, when either will do, are added, not multiplied.

Why: The addition rule is for two outcomes that cannot both happen, when either one will do.
Either one will do, so their probabilities are added.
P(A or B) = P(A) + P(B).

49
Practice writing an answer

The multiplication rule and the addition rule each apply to a pair of events only under a condition.

(a) State the condition under which the multiplication rule applies to two events. (1 pt)

Model answer The two events must be independent: which one happens has no effect on the other, and both must happen.
Rubric
  • Award 1 point for: the two events are independent and both must happen.

(b) State the condition under which the addition rule applies to two outcomes. (1 pt)

Model answer The two outcomes must be mutually exclusive: they cannot both happen, and either one will do.
Rubric
  • Award 1 point for: the two outcomes are mutually exclusive (cannot both happen) and either one will do.
50
Check q17 numeric entry

A Pp plant’s egg carries P with probability 0.5. A pp plant’s pollen grain carries p with probability 1.

Calculate the probability that a seed gets P from the egg and p from the pollen grain, as a decimal.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(P from the egg)=0.5
P(p from the pollen)=1
the two gametes are independent
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(P egg and p pollen)=(0.5)×(1)=0.5
51
Check q18 numeric entry

From Tt × Tt, a seed is TT with probability 0.25 and tt with probability 0.25.

Calculate the probability that a seed is TT or tt, as a decimal.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(TT)=0.25
P(tt)=0.25
TT and tt are mutually exclusive
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(TT or tt)=(0.25)+(0.25)=0.5
52
Check q19 numeric entry

A Gg plant’s egg carries g with probability 0.5. Another Gg plant’s pollen grain carries g with probability 0.5.

Calculate the probability that a seed gets g from the egg and g from the pollen grain, as a decimal.

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
P(g from the egg)=0.5
P(g from the pollen)=0.5
the two gametes are independent
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(gg)=(0.5)×(0.5)=0.25
53
Check q20 numeric entry

From Rr × Rr, a plant is RR with probability 0.25 and Rr with probability 0.5.

Calculate the probability that a plant is RR or Rr, as a decimal.

Answer: 0.75  (tolerance ±0.005)

Working
Write down the values in the question:
P(RR)=0.25
P(Rr)=0.5
RR and Rr are mutually exclusive
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(RR or Rr)=(0.25)+(0.5)=0.75
54
Check q21 numeric entry

A Dd dog’s sperm carries d with probability 0.5. A Dd dog’s egg carries D with probability 0.5.

Calculate the probability that a puppy gets d from the sperm and D from the egg, as a decimal.

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
P(d from the sperm)=0.5
P(D from the egg)=0.5
the two gametes are independent
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(d sperm and D egg)=(0.5)×(0.5)=0.25
55
Check q22

Two Gg pea plants are crossed. For the next seed to be gg, the egg must carry g and the pollen grain must carry g.

Multiply or add?

  1. A. ✓ Multiply
  2. B. Add
    The egg must carry g and the pollen grain must carry g.
    Both events must happen, so multiply.

Why: Two things must both happen: a g egg and a g pollen grain.
The two gametes are independent.
For two independent events that must both happen, multiply.

56
Check q23

Two Gg pea plants are crossed. For the next seed to have green pods, it can be GG or it can be Gg.

Multiply or add?

  1. A. Multiply
    A seed is GG or Gg, never both, and either one gives green pods.
  2. B. ✓ Add

Why: Either outcome will do: GG or Gg.
No seed is both.
For two outcomes that cannot both happen, when either will do, add.

57
Check q24

Two Bb mice have two pups. For both pups to be brown, the first pup must be bb and the second pup must be bb.

Multiply or add?

  1. A. ✓ Multiply
  2. B. Add
    The first pup must be bb and the second pup must be bb.
    Each pup is its own event, and both must happen.

Why: Both pups must be brown.
Each pup is an independent event.
For two independent events that must both happen, multiply.

58
Check q25

Two Tt pea plants are crossed. For the next seed to be homozygous, it can be TT or it can be tt.

Multiply or add?

  1. A. Multiply
    A seed is TT or tt, never both, and either one is homozygous.
  2. B. ✓ Add

Why: Either outcome will do: TT or tt.
No seed is both.
So add the two probabilities, 0.25 each, to get 0.5.

59
Check q26

Two Ll cats have a kitten. A student wants the probability that the kitten is short-haired or heterozygous.

Can the student add the two probabilities?

  1. A. Yes
    Every Ll kitten is both short-haired and heterozygous.
    So the two outcomes are not mutually exclusive, and adding counts every Ll kitten twice.
  2. B. ✓ No

Why: Every Ll kitten is short-haired and heterozygous at the same time.
The two outcomes overlap.
The addition rule needs outcomes that cannot both happen.
So no, the student cannot add.

60
Check q27

Two Ll cats have a kitten. A student wants the probability that the kitten is LL or ll.

Can the student add the two probabilities?

  1. A. ✓ Yes
  2. B. No
    A kitten has one genotype, so it is LL or ll, never both.
    The two outcomes are mutually exclusive, and either will do.

Why: A kitten cannot be both LL and ll.
The two outcomes are mutually exclusive, and either will do.
So yes, the student can add the two probabilities, 0.25 each.

61Several offspring in a row

62

Video: Watch: Several offspring in a row

Four pups born one after another, each with its own probability one half; the four halves multiplied into one sixteenth.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L14d.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L14d.mp4

63

Each offspring is its own event, independent of the ones before it. So a run of offspring is another case for the multiplication rule.

64

Suppose a black Bb guinea pig is crossed with a white bb one. Each pup is black with probability 12.

Four black guinea pig pups in a row, each drawn as a small dark oval and labeled black pup; under the row, the caption says each pup has probability one half and all four together have probability one sixteenth
Four black guinea pig pups in a row, each drawn as a small dark oval and labeled black pup; under the row, the caption says each pup has probability one half and all four together have probability one sixteenth
65

Four black pups in a row from that cross are unlikely but not rare: they happen about one time in sixteen.

66
Worked example

A black Bb guinea pig is crossed with a white bb guinea pig. What is the probability that its first four pups are all black?

Write down the values in the question:
P(a pup is black)=12
number of pups = 4
each pup is independent of the others
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(four black)=12×12×12×12
P(four black)=(12)4=116=0.0625
67

What you are expected to know Calculate the probability of a run of offspring by multiplying each offspring’s probability, one factor for each offspring.

68
Check q28 numeric entry

In pea plants, purple flowers (P) are dominant to white (p). Two Pp pea plants are crossed and their seeds are sown.

Calculate the probability that the first three seeds all grow into purple-flowered plants, as a decimal to two decimal places.

Part 1. What is the probability that one seed grows into a purple-flowered plant, as a decimal?

Answer: 0.75  (tolerance ±0.005)

Working
Purple is PP or Pp, so add:
P(purple)=14+12=34=0.75

Answer: 0.42  (tolerance ±0.005)

Working
Write down the values in the question:
P(a seed grows purple)=34
number of seeds = 3
each seed is independent of the others
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(three purple)=34×34×34
P(three purple)=(34)3=2764≈0.42
69
Check q29 numeric entry

In mice, black fur (B) is dominant to brown (b). Two Bb mice have two pups.

Calculate the probability that both pups are brown, as a decimal.

Answer: 0.0625  (tolerance ±0.0005)

Working
Write down the values in the question:
brown = bb
P(a pup is bb)=12×12=14
number of pups = 2
each pup is independent of the other
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(both brown)=14×14=116=0.0625
70

Back to the opening square, Pp × Pp, here again. pp needs a p from the egg and a p from the pollen grain, so 12×12=14; purple is PP or Pp, so 14+12=34.

The filled Punnett square for Pp crossed with Pp: P and p along the top, P and p down the side; cells PP, Pp, Pp, pp
The filled Punnett square for Pp crossed with Pp: P and p along the top, P and p down the side; cells PP, Pp, Pp, pp
71

Multiply for both, add for either. A run of offspring is just more multiplying: four purple seeds in a row from Pp × pp have probability (12)4=116.

72Mixed practice mixed practice

73
Check q30

Two Ll cats (short hair, L, dominant to long, l) have had three long-haired kittens in a row. A breeder says: “The fourth kitten is almost certain to be short-haired, because four long-haired kittens in a row would be a one-in-256 event.”

Is the breeder right?

  1. A. The breeder is right
    Each fertilization is an independent event: the parents are still Ll and Ll, so the fourth kitten is ll with probability 0.25, as every kitten is.
  2. B. ✓ The breeder is wrong

Why: The breeder is wrong.
One in 256 is the probability of four long-haired kittens in a row, multiplied out before any were born.
Three are already born.
The fourth kitten is its own independent event with the same one-in-four chance of ll as every kitten.

74
Practice writing an answer

Two Ll cats (short hair, L, dominant to long, l) have had three long-haired kittens in a row.

(a) Explain why the fourth kitten’s probability of long hair is 14, whatever the first three were. (1 pt)

Frame The fourth kitten’s probability is 14 because …

Model answer The fourth kitten’s probability is 14 because each kitten comes from its own egg and its own sperm.
So each fertilization is an independent event.
The first three kittens do not change which alleles the parents pass on next.
The parents are still Ll and Ll, so the fourth kitten is ll with probability 0.25.
One in 256 was the probability of four long-haired kittens before any was born; it no longer applies to the fourth.
Rubric
  • Award 1 point for: each fertilization is independent, so the fourth kitten is ll with probability 0.25; one in 256 is for a run of four before any is born.
75
Check q31 numeric entry

Two Ll cats have a litter of three kittens.

Calculate the probability that all three kittens are long-haired, as a decimal to four decimal places.

Answer: 0.0156  (tolerance ±0.0005)

Working
Write down the values in the question:
P(a kitten is long-haired)=14
number of kittens = 3
each kitten is independent of the others
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(three long-haired)=14×14×14
P(three long-haired)=(14)3=164≈0.0156
76
Check q32

In pea plants, tall (T) is dominant to short (t). A Tt pea plant is crossed with a tt plant.

Which of the following is the probability of a tall offspring from Tt × tt?

  1. A. ✓ 0.5
  2. B. 0.25
    The probability comes from the gametes of this cross, not from another square: the Tt parent gives T with probability 0.5 and the tt parent always gives t.
  3. C. 0.75
    0.75 is the probability of tall from two heterozygous parents; here one parent is tt and contributes only t.
  4. D. 1
    Dominance decides what a Tt offspring looks like, not how likely a T is to be passed on; half of the Tt parent’s gametes carry t.

Why: A tall offspring here must be Tt: a T from the Tt parent, probability 0.5, and a t from the tt parent, probability 1.
Multiply: 0.5.

77
Check q33

Two Pp pea plants are crossed. A student wants the probability that a seed is either homozygous dominant, PP, or heterozygous, Pp, and adds 0.25 and 0.5 to get 0.75.

Is the student’s method right?

  1. A. No
    PP and Pp do not overlap: a seed has one genotype, so it is one or the other.
  2. B. ✓ Yes

Why: PP and Pp are two different genotypes.
A seed cannot be both, so the two outcomes are mutually exclusive.
So the addition rule applies, and the total is 0.75.

78
Practice writing an answer

Two Pp pea plants are crossed. A student wants the probability that a seed is either purple or heterozygous, Pp, and adds 0.75 and 0.5 to get 1.25.

(a) Explain why adding the two probabilities is wrong here. (1 pt)

Model answer Every Pp seed is heterozygous.
Every Pp seed is also purple, because P is dominant.
So one seed can be both purple and heterozygous: the two outcomes are not mutually exclusive.
The addition rule needs outcomes that cannot both happen, so adding counts every Pp seed twice.
The probability of purple or heterozygous is the probability of purple alone, 0.75.
Rubric
  • Award 1 point for: a Pp seed is both purple and heterozygous, so the outcomes overlap and are not mutually exclusive; the addition rule needs outcomes that cannot both happen.
79
Practice writing an answer

A trait in people is recessive: only aa children show it. Two parents are both Aa.

(a) Calculate the probability that their first child shows the trait. (1 pt)

Answer: 0.25  (tolerance ±0.005)

Model answer The probability that the first child is aa is 0.25.
Working
Write down the values in the question:
P(a from the mother)=12
P(a from the father)=12
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(aa)=12×12=14=0.25
Rubric
  • Award 1 point for: 0.25.

(b) Calculate the probability that their first two children both show the trait. (1 pt)

Answer: 0.0625  (tolerance ±0.0005)

Model answer Each child is aa with probability 0.25.
The two children are independent events.
So multiply the two probabilities: one sixteenth, 0.0625.
Working
Write down the values in the question:
P(a child is aa)=14
number of children = 2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(both aa)=14×14=116=0.0625
Rubric
  • Award 1 point for: 0.0625 (one sixteenth).

(c) Their first child shows the trait. State the probability that their second child shows it, and explain why. (2 pt)

Model answer The probability is still 0.25.
Each child comes from its own egg and its own sperm.
So each fertilization is an independent event.
The first child’s genotype does not change which alleles the parents pass on to the second child.
The parents are still Aa and Aa, so the second child is aa with probability 0.25.
Rubric
  • Award 1 point for: 0.25, unchanged.
  • Award 1 point for: each fertilization is independent, so an earlier child does not change the probability for the next.

Slip Lowering the second child’s chance because the first one showed the trait. Earlier children do not change the probability for the next; the parents’ genotypes set it every time.

Glossary

probability
A number between 0 and 1 that says how likely an outcome is: 0 means it never happens, 1 means it always happens. Written as a fraction; in a check, typed as a decimal (a probability of 1/4 is 0.25).
multiplication rule
For two independent events that must both happen, multiply their probabilities: P(A and B) = P(A) × P(B). The two gametes of a breeding cross are independent.
addition rule
For two outcomes that cannot both happen, when either one will do, add their probabilities: P(A or B) = P(A) + P(B).

APBIO-U05-L15 PP or Pp?

Topic 5.3 · Mendelian Genetics · 55 steps

Left: a photograph of a black guinea pig standing on sandy ground. Middle: a photograph of a white guinea pig lying on a cloth. Right: the two genotypes the black one could have, BB or Bb, with a question mark, and the words the same coat either way
Left: a photograph of a black guinea pig standing on sandy ground. Middle: a photograph of a white guinea pig lying on a cloth. Right: the two genotypes the black one could have, BB or Bb, with a question mark, and the words the same coat either way

Photos: Thomas Stüven, Wikimedia Commons, CC BY 2.0 (cropped and resized); Popular Science Monthly (1910), Wikimedia Commons, public domain (resized).

Here is a black guinea pig, and beside it a white one. In guinea pigs, black coat, B, is dominant to white coat, b.

So the black guinea pig could be BB or it could be Bb, and you cannot tell which by looking at it. But you can breed it. Which cross would tell you, and what result would settle it?

Unit 5 · Heredity

1Cross it with the recessive

2

Video: Watch: Cross it with the recessive

The two possible test crosses drawn side by side, BB × bb giving all black and Bb × bb giving half and half; the first white pup appearing and settling it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L15a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L15a.mp4

3

How do you find out which genotype hides behind a dominant trait? Cross the black guinea pig with a white one, bb.

4

A white guinea pig gives every pup a b. So each pup’s coat shows what the black parent gave it.

Two Punnett squares side by side for a black guinea pig crossed with a white bb guinea pig. Left, if the black parent is BB: B and B along the top, b and b down the side, every cell Bb, all pups black. Right, if it is Bb: B and b along the top, b and b down the side, cells Bb, Bb, bb, bb, half the pups black and half white
Two Punnett squares side by side for a black guinea pig crossed with a white bb guinea pig. Left, if the black parent is BB: B and B along the top, b and b down the side, every cell Bb, all pups black. Right, if it is Bb: B and b along the top, b and b down the side, cells Bb, Bb, bb, bb, half the pups black and half white
5

If the black parent is BB, every pup gets a B, and every pup is black. If it is Bb, half the pups get B and are black, and half get b and are white: one to one.

6

When we cross an individual that shows the dominant trait with one that is homozygous recessive to find out its genotype, we call it a , because the offspring test which genotype the parent has.

7

A single white pup settles it. That pup needed a b from each parent, so the black parent is Bb.

Five pups in a row from the black × white cross: four black and one white; the white one is ringed
Five pups in a row from the black × white cross: four black and one white; the white one is ringed
8

All black pups make BB likely, but they do not prove it: a Bb parent can give black pups every time by chance.

9

It does not matter which parent is the mother, because each parent gives each pup one allele whatever its sex. A Bb mother with a bb father, or the reverse: both give one black to one white.

10

What you are expected to know Use a test cross to find the genotype of an individual that shows the dominant trait.

11
Check q1

In pea plants, purple flowers (P) are dominant to white (p). A purple plant of unknown genotype is crossed with a white plant. The cross gives 48 purple and 52 white offspring.

Which of the following is the purple parent’s genotype?

  1. A. PP
    A PP parent gives every offspring a P, so none could be white.
    Here 52 are white.
  2. B. pp
    A pp plant has white flowers, and this parent is purple.
  3. C. ✓ Pp
  4. D. It cannot be told from these offspring
    White offspring from a purple parent mean the parent carried p.

Why: The white parent gives only p, so a white offspring got its other p from the purple parent.
A purple parent that carries p is Pp.
Pp × pp gives about half purple and half white, as the 48 : 52 shows.

12
Practice writing an answer

In pea plants, purple flowers (P) are dominant to white (p). A purple plant of unknown genotype is crossed with a white plant. The cross gives 48 purple and 52 white offspring.

(a) Explain how the white offspring show that the purple parent is Pp. (1 pt)

Frame The white offspring show that the purple parent is Pp because …

Model answer The white offspring show that the purple parent is Pp because each white offspring is pp.
A pp offspring needed a p from each parent.
The white parent gave one p.
So the other p came from the purple parent.
A purple parent that carries p is Pp.
Rubric
  • Award 1 point for: a white offspring is pp and received one p from the white parent, so its other p came from the purple parent, which is therefore Pp.
13
Check q2

In bean plants, purple stems (P) are dominant to green (p). In one cross a Pp plant supplied the eggs and a pp plant the pollen, giving 48 purple and 52 green offspring. In the reverse cross a pp plant supplied the eggs and a Pp plant the pollen, giving 51 purple and 49 green.

Which of the following explains why the two crosses give the same result?

  1. A. ✓ Each parent gives every offspring a single allele, whatever the parent’s sex
  2. B. The egg carries both alleles and the pollen carries neither
    Every gamete, egg or pollen grain, carries one allele of the gene; the pollen contributes as much as the egg.
  3. C. Purple is dominant, so the Pp parent always passes on P
    A Pp parent passes on P to half of its gametes and p to the other half; dominance decides what shows, not what is passed on.
  4. D. An egg and a pollen grain always carry the same allele as each other
    The egg and the pollen grain come from two different plants, so they need not carry the same allele; what is the same is that each carries one allele.

Why: A Pp parent gives P or p to each offspring and a pp parent gives p, whether the Pp parent supplies the eggs or the pollen.
So both crosses are Pp × pp and both give about one purple to one green.

14Quick quiz: test cross mixed practice

15
Check q3

In pea plants, tall (T) is dominant to short (t). A breeder wants to know whether a tall plant is TT or Tt.

Which of the following crosses is a test cross?

  1. A. ✓ The tall plant crossed with a short tt plant
  2. B. The tall plant crossed with another tall plant
    The other tall plant can give a T itself, so a tall offspring could have got its T from either parent.
  3. C. Two short tt plants crossed with each other
    Neither short plant is the tall plant whose genotype is in question.

Why: A test cross pairs the individual that shows the dominant trait with one that is homozygous recessive.
The tt plant gives every offspring a t.
So each offspring’s height shows the allele the tall plant gave.

16
Practice writing an answer

In a test cross, the individual that shows the dominant trait is crossed with one that is homozygous recessive.

(a) Explain why the partner in a test cross is homozygous recessive. (1 pt)

Frame The partner is homozygous recessive because …

Model answer The partner is homozygous recessive because it can give each offspring only the recessive allele.
So each offspring’s other allele came from the parent being tested.
An offspring that shows the recessive trait received the recessive allele from that parent.
So the offspring show which alleles the tested parent carries.
Rubric
  • Award 1 point for: the homozygous recessive partner gives only the recessive allele, so each offspring’s phenotype shows the allele the tested parent gave.
17
Check q4

In guinea pigs, black (B) is dominant to white (b). A black guinea pig is crossed with a white bb guinea pig. The pups are 3 black and 2 white.

Which of the following is the black parent’s genotype?

  1. A. BB for certain
    A white pup is bb and needed a b from each parent.
    A BB parent has no b to give.
  2. B. ✓ Bb for certain
  3. C. Probably BB, not certain
    A white pup needed a b from the black parent, so the black parent carries b.

Why: A white pup is bb.
It got one b from the white parent, so its other b came from the black parent.
A black parent that carries b is Bb.
So the black parent is Bb for certain.

18
Check q5

Another black guinea pig is crossed with a white bb guinea pig. The first four pups are all black.

Which of the following is the black parent’s genotype?

  1. A. BB for certain
    A Bb parent can give four black pups in a row by chance.
    So four black pups make BB likely, not certain.
  2. B. Bb for certain
    A white pup is what proves Bb, and none has been born.
  3. C. ✓ Probably BB, not certain

Why: A black pup can come from a BB parent or a Bb parent.
A Bb parent can give black pups four times in a row by chance.
So BB is likely, but nothing proves it.
A white pup would settle it.

19
Check q6

In pea plants, tall (T) is dominant to short (t). A tall plant is crossed with a short tt plant. The bar chart shows the offspring.

A bar chart with two bars, tall offspring and short offspring, from a tall pea plant crossed with a short tt plant; the value is printed above each bar
A bar chart with two bars, tall offspring and short offspring, from a tall pea plant crossed with a short tt plant; the value is printed above each bar

Which of the following is the tall parent’s genotype?

  1. A. TT for certain
    A short offspring is tt and needed a t from each parent.
    A TT parent has no t to give.
  2. B. ✓ Tt for certain
  3. C. Probably TT, not certain
    Each short offspring needed a t from the tall parent.

Why: The chart shows short offspring.
A short offspring is tt and got one t from the tall parent.
So the tall parent carries t and is Tt for certain.
About half tall and half short is what Tt × tt gives.

20
Check q7

A tall pea plant of unknown genotype is crossed with a short tt plant. The bar chart shows the offspring: 60 tall and 0 short.

A bar chart with two bars, tall offspring and short offspring, from a tall pea plant crossed with a short tt plant; the value is printed above each bar
A bar chart with two bars, tall offspring and short offspring, from a tall pea plant crossed with a short tt plant; the value is printed above each bar

Which of the following is the tall parent’s genotype?

  1. A. TT for certain
    A Tt parent gives 60 tall offspring in a row with probability (½)⁶⁰: tiny, but not zero.
    So the run cannot prove TT.
  2. B. Tt for certain
    A short offspring is what proves Tt, and none is in the chart.
  3. C. ✓ Probably TT, not certain

Why: Every offspring is tall, and none is short.
A Tt parent gives 60 tall offspring in a row with probability (½)⁶⁰: tiny, but not zero.
So TT is very likely, but the run never proves TT.
A short offspring would settle it.

21
Check q8

In dogs, a dark coat (D) is dominant to a light coat (d). A dark dog is crossed with a light dd dog, and one light pup is born.

Which of the following is the dark parent’s genotype?

  1. A. DD
    A light pup is dd and needed a d from each parent.
    A DD parent has no d to give.
  2. B. ✓ Dd

Why: A light pup is dd.
One d came from the light parent, so the other d came from the dark parent.
A dark parent that carries d is Dd.

22
Check q9

A breeder wants to find out whether an animal showing the dominant trait is homozygous or heterozygous.

In a test cross, which of the following partners is the unknown animal crossed with?

  1. A. ✓ Homozygous recessive
  2. B. Homozygous dominant
    A homozygous dominant partner gives every offspring the dominant allele, so every offspring shows the dominant trait whatever the first parent is, and nothing is learned.
  3. C. Heterozygous
    A heterozygous partner can give the dominant allele itself, so an offspring showing the dominant trait could have got it from either parent.

Why: A homozygous recessive partner gives every offspring only the recessive allele.
So each offspring’s look shows the allele the unknown parent gave.
That is what a test cross uses.

23A result the model forbids

24

Video: Watch: A result the model forbids

The square for two pp parents filling with pp in every cell and a purple offspring appearing anyway; then four black pups in a row from Bb × bb, unlikely but with a cell of their own; the forbidden cell crossed out.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L15b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L15b.mp4

25

Suppose that in a plant purple flowers, P, are dominant to white, p, and two white plants, pp and pp, are crossed. Every cell of their square is pp.

The square for two white pp plants: p and p along the top, p and p down the side; every cell pp
The square for two white pp plants: p and p along the top, p and p down the side; every cell pp
26

So a purple offspring has probability zero: not small, zero.

27

A probability of zero is a prediction the data can break. If the cross gives 117 white offspring and one purple, the model has been broken.

28

Then one of three things is true: a parent was not pp after all, a stray pollen grain from a purple plant fertilized the egg, or the single-gene model is wrong for this trait. A brand-new mutation is a fourth, rare, route.

29

So the one purple flower sends the breeder back to check the parents’ genotypes, the pollination and the model.

30

Now consider the black guinea pig again, a Bb parent crossed with a white bb one. Each pup is black with probability one half, and four black pups in a row have probability one sixteenth.

Four black guinea pig pups in a row, each drawn as a small dark oval with a head and labeled black pup; under the row, the caption says each pup has probability one half and all four together have probability one sixteenth
Four black guinea pig pups in a row, each drawn as a small dark oval with a head and labeled black pup; under the row, the caption says each pup has probability one half and all four together have probability one sixteenth
31

Four black pups in a row are unlikely, but they break nothing: the square has a cell for a black pup. So four black pups favor BB without ruling out Bb.

32

A white pup rules out BB, because a white pup needs a b from the black parent. A run of black pups, however long, never does.

Two labeled boxes side by side. Left: a purple offspring from pp × pp, probability zero, labeled a prediction the data can break. Right: four black pups in a row from Bb × bb, probability one sixteenth, labeled unlikely, and nothing is broken when it happens
Two labeled boxes side by side. Left: a purple offspring from pp × pp, probability zero, labeled a prediction the data can break. Right: four black pups in a row from Bb × bb, probability one sixteenth, labeled unlikely, and nothing is broken when it happens
33

What you are expected to know Use an offspring the square gives no cell to as the one result that breaks a claim about the parents or about the model.

34
Check q10

In a plant, purple flowers (P) are dominant to white (p). Two white-flowered plants, both pp, are crossed. Their labels are checked. One of their 118 offspring has purple flowers.

Can the single-gene model give this result?

  1. A. Yes
    Two pp parents can give only p alleles, so the square for pp × pp has no purple cell: the probability is zero, not small.
  2. B. ✓ No

Why: Both parents are pp, and their labels were checked.
Two pp parents can give only p alleles.
So the single-gene model gives a purple offspring a probability of zero.
One purple offspring appeared.
So the model cannot give this result.

35
Practice writing an answer

In a plant, purple flowers (P) are dominant to white (p). Two white-flowered plants, both pp, are crossed. Their labels are checked. One of their 118 offspring has purple flowers.

(a) Explain what the purple offspring shows about the single-gene model. (1 pt)

Frame The purple offspring shows that …

Model answer The purple offspring shows that the single-gene model is wrong for this cross.
Under the model a white plant is pp.
Two pp parents can give only p alleles.
So the model gives a purple offspring a probability of zero.
One purple offspring appeared, so the data break that prediction.
The single-gene model cannot give this result, so the model is wrong for this cross.
Rubric
  • Award 1 point for: two pp parents can give only p alleles, so the single-gene model gives a purple offspring a probability of zero; the purple offspring breaks that prediction, so the model is wrong for this cross.
  • Accept with or without a named route to the P allele (stray pollen, a new mutation).
36
Check q11

A black guinea pig is crossed with a white bb guinea pig, and its first four pups are all black. A student says: “Four black pups prove that the black parent is BB.”

Is the student correct?

  1. A. Yes
    A Bb parent gives four black pups in a row with probability one sixteenth: unlikely, not impossible.
  2. B. ✓ No

Why: A Bb parent gives each pup a B with probability one half.
Four black pups in a row from a Bb parent have probability one sixteenth.
That result is unlikely, not impossible.
So four black pups favor BB but do not prove it.

37
Practice writing an answer

A black guinea pig, BB or Bb, is crossed with a white bb guinea pig. The first four pups are all black.

(a) Calculate the probability that a Bb × bb cross gives four black pups in a row. (1 pt)

Answer: 0.0625  (tolerance ±0.0005)

Model answer The probability is one sixteenth, 0.0625.
Working
Write down the values in the question:
P(a pup is black)=12
number of pups = 4
each pup is independent of the others
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(four black)=(12)4=116=0.0625
Rubric
  • Award 1 point for: (1/2)⁴ = 1/16 = 0.0625.

(b) Explain what the four black pups show about the black parent. (1 pt)

Model answer A BB parent gives black pups every time.
A Bb parent gives four black pups in a row with probability one sixteenth: unlikely, not impossible.
So the four black pups favor BB.
They do not rule out Bb; a white pup would.
Rubric
  • Award 1 point for: four black pups from a Bb parent are unlikely but possible, so BB is favored but not proved; a white pup would rule BB out.
38
Check q12

In a breed of chicken, a feather gene has two alleles, C and c. A hen is cc and a rooster is Cc.

Which of the following genotypes can their chicks have for this gene?

  1. A. ✓ Cc or cc
  2. B. CC, Cc or cc
    A CC chick would need a C from each parent, and the cc hen has no C to give.
  3. C. CC or Cc
    A CC chick needs a C from the hen, which has none, and a cc chick is possible when the rooster gives c.
  4. D. cc only
    Half of the rooster’s sperm carry C, and a C sperm with a c egg makes a Cc chick.

Why: The hen’s eggs all carry c; the rooster’s sperm carry C or c.
So the chicks are Cc or cc; CC is a result this cross cannot produce.

39

Back to the black guinea pig and the two crosses it could give, drawn here again. Cross it with a white bb one.

Two Punnett squares side by side for a black guinea pig crossed with a white bb guinea pig. Left, if the black parent is BB: B and B along the top, b and b down the side, every cell Bb, all pups black. Right, if it is Bb: B and b along the top, b and b down the side, cells Bb, Bb, bb, bb, half the pups black and half white
Two Punnett squares side by side for a black guinea pig crossed with a white bb guinea pig. Left, if the black parent is BB: B and B along the top, b and b down the side, every cell Bb, all pups black. Right, if it is Bb: B and b along the top, b and b down the side, cells Bb, Bb, bb, bb, half the pups black and half white
40

One white pup proves the black parent is Bb, because BB gives no cell for a white pup. All black pups make BB likely, but not certain.

41Quick quiz: possible or impossible mixed practice

42
Check q13

In pea plants, tall (T) is dominant to short (t). A Tt plant is crossed with a tt plant.

Is a tt offspring possible or impossible?

  1. A. ✓ Possible
  2. B. Impossible
    The Tt parent gives t to half of its gametes, and the tt parent gives t every time.
    A t with a t is a tt offspring.

Why: The Tt parent gives t with probability one half.
The tt parent gives t every time.
The square has two tt cells of four, so a tt offspring is possible.

43
Check q14

In pea plants, purple flowers (P) are dominant to white (p). A PP plant is crossed with a pp plant.

Is a pp offspring possible or impossible?

  1. A. Possible
    A pp offspring needs a p from each parent.
    The PP parent has no p to give.
  2. B. ✓ Impossible

Why: The PP parent gives every offspring a P.
So every cell of the square is Pp.
The square has no pp cell, so a pp offspring is impossible.

44
Check q15

In mice, black fur (B) is dominant to brown (b). Two Bb mice have four pups, and all four are brown.

Is this litter possible or impossible?

  1. A. ✓ Possible
  2. B. Impossible
    Each pup is bb with probability 14.
    Four bb pups in a row are unlikely, and the square has a cell for each.

Why: Bb × Bb gives each pup a bb cell, probability 14.
Four brown pups in a row have a small probability, not zero.
The square has a cell for every one of them, so the litter is possible.

45
Check q16

In guinea pigs, a rough coat (R) is dominant to a smooth coat (r). Two smooth-coated rr guinea pigs are crossed.

Is a rough-coated pup possible or impossible?

  1. A. Possible
    A rough-coated pup needs an R.
    Neither rr parent has an R to give.
  2. B. ✓ Impossible

Why: Both parents are rr, so every gamete carries r.
Every cell of the square is rr.
The square has no cell with an R, so a rough-coated pup is impossible.

46
Check q17

In a breed of dog, a coat gene has two alleles, D and d. A Dd dog is mated with a dd dog.

Is a DD puppy possible or impossible?

  1. A. Possible
    A DD puppy needs a D from each parent.
    The dd parent has no D to give.
  2. B. ✓ Impossible

Why: The dd parent gives every puppy a d.
So the cells of the square are Dd and dd only.
The square has no DD cell, so a DD puppy is impossible.

47Mixed practice mixed practice

48
Check q18

A rough-coated guinea pig is test-crossed with a smooth-coated rr guinea pig.

How many rough-coated pups in a row would prove that the rough-coated parent is RR?

  1. A. One rough pup
    An Rr parent gives a rough pup or a smooth pup with equal chance each time; there is no fixed order.
  2. B. Four rough pups in a row
    One time in sixteen is unlikely, not impossible; an Rr parent can give four rough pups in a row.
  3. C. Eight rough pups in a row
    One time in 256 is small but not zero; eight rough pups make RR very likely and still do not rule Rr out.
  4. D. ✓ No number of rough pups proves RR

Why: A rough pup can come from an RR or an Rr parent, so a run of rough pups only makes RR more likely.
A smooth pup needs an r from the rough parent, so one smooth pup proves Rr; nothing proves RR.

49
Practice writing an answer

A rough-coated guinea pig is test-crossed with a smooth-coated rr guinea pig. The first eight pups are all rough-coated.

(a) Calculate the probability, as a decimal to four decimal places, that an Rr parent gives eight rough-coated pups in a row in this cross. (1 pt)

Answer: 0.0039  (tolerance ±5e-05)

Model answer The probability is 0.0039, one in 256.
Working
Write down the values in the question:
Rr × rr: rough (Rr) = 2 cells of 4, so P(rough pup) = 1/2
eight pups, each its own fertilization
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(eight rough)=(12)8=1256≈0.0039
Rubric
  • Award 1 point for: (1/2)⁸ = 1/256 ≈ 0.0039.

(b) Explain why eight rough-coated pups in a row leave the rough-coated parent’s genotype, RR or Rr, undecided. (1 pt)

Model answer An Rr parent gives a rough pup with probability ½ each time.
So eight rough pups in a row from an Rr parent has probability 1/256: small, not zero.
A result with a probability above zero is one the data cannot rule out.
So eight rough pups make RR more likely but do not prove it.
One smooth pup would prove Rr, because a smooth pup needs an r from the rough parent.
Rubric
  • Award 1 point for: an Rr parent can give eight rough pups by chance (1/256 is small, not zero), so the run makes RR likely but does not rule Rr out; one smooth pup would prove Rr.
50
Check q19 numeric entry

In sheep, white wool (W) is dominant to black wool (w). A Ww ram is crossed with black ww ewes and 70 lambs are born.

Calculate how many of the 70 lambs are expected to have black wool.

Answer: 35  (tolerance ±0)

Working
Write down the values in the question:
total lambs = 70
square = Ww, Ww, ww, ww, so the phenotypic ratio is 1 white : 1 black, 2 shares in all
black (ww) share = 1 of 2
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected ww=70×12=35
51
Check q20

In a breed of dog, a coat gene has two alleles, D and d. Two dd dogs are mated.

Which of the following genotypes can their puppies have?

  1. A. ✓ dd only
  2. B. Dd only
    A Dd puppy needs a D, which neither parent has; the puppies get d from each parent.
  3. C. Dd or dd
    A Dd puppy needs a D from one parent, and neither dd parent has one.
  4. D. DD, Dd or dd
    Neither parent carries D, so no puppy can receive one.

Why: Both parents are dd, so every egg and every sperm carries d, and every puppy is dd.
Any other genotype is a result this mating cannot produce.

52
Check q21

In a breed of dog, a coat gene has two alleles, and D is dominant to d. A breeder keeps records of four litters.

Which of the following litters is a result the single-gene model forbids?

  1. A. Dd × dd giving six puppies that all show the D trait
    Dd × dd gives each puppy a one-in-two chance of the D trait, so six in a row has a probability of one in 64: unlikely, not forbidden.
  2. B. ✓ dd × dd giving one puppy that shows the D trait
  3. C. Dd × Dd giving four dd puppies in a row
    Dd × Dd gives each puppy a one-in-four chance of dd, so four in a row has a probability of one in 256: unlikely, not forbidden.
  4. D. DD × dd giving no dd puppies at all
    DD × dd gives every puppy a D, so no dd puppy is exactly what the model predicts.

Why: Only a result with no cell in the square is forbidden.
Two dd parents pass on only d, so a D-trait puppy has probability zero; the other three litters are unlikely or expected, and none of them breaks the model.

53
Check q22

In cattle, a hornless head (H) is dominant to horns (h). A hornless bull is test-crossed with horned hh cows.

Which of the following observations would prove that the bull is Hh?

  1. A. The bull’s own parents both being hornless
    Two hornless parents can each be HH or Hh, and a hornless bull from them is HH or Hh either way; his parents’ heads do not show which.
  2. B. ✓ One horned calf
  3. C. Eight hornless calves in a row
    A run of hornless calves favors HH but never rules out Hh; eight in a row from an Hh bull is unlikely, not impossible.
  4. D. One hornless calf
    Both an HH and an Hh bull give hornless calves, so one hornless calf does not tell the two apart.

Why: The horned cows give only h, so a horned calf got its second h from the bull, and a bull that carries h is Hh.
Hornless calves and hornless parents fit HH and Hh alike; one horned calf proves Hh.

54
Practice writing an answer

In sheep, white wool (W) is dominant to black wool (w). A farmer has a white ram and wants to know its genotype. The farmer crosses it with several black ewes.

(a) Identify the genotype of the black ewes, and explain why they are the right partners for finding the ram’s genotype. (1 pt)

Model answer The ewes are ww, because black is the recessive trait: it shows when both alleles are w.
Every egg they make carries w, so a lamb’s wool color shows which allele the ram gave it.
Rubric
  • Award 1 point for: the ewes are ww, so every egg carries w and each lamb’s color shows the ram’s allele.

Slip Choosing white ewes. A white ewe may carry w herself, so a black lamb would not prove the ram gave a w.

(b) Suppose the ram is Ww. State the ram’s gametes, the ewes’ gametes and the genotype in each of the four cells of the Punnett square, and predict the lambs. (1 pt)

Model answer A Ww ram makes two kinds of gamete, W and w.
A ww ewe makes one kind, w.
So W and w go along one edge and w and w down the other.
The cells are Ww, Ww, ww and ww.
Ww lambs are white and ww lambs are black.
So about half of the lambs are white and half are black, 1 : 1.The square for a Ww ram crossed with a ww ewe, with W and w along the top and w and w down the side, and the cells Ww, Ww, ww and wwWwWwwwwwWwwwram gametes W or wewe gametes w onlyWw × ww gives half white and half black
Rubric
  • Award 1 point for: ram gametes W and w, ewe gametes w only, cells Ww, Ww, ww, ww, and about half white to half black (1 : 1).

Slip Predicting three white to one black. Three to one comes from two heterozygous parents; here the ewes are ww.

(c) The third lamb born has black wool. Support the claim that the ram is Ww with evidence from that lamb. (1 pt)

Model answer Evidence: a black lamb is ww.
It received one w from its ww mother, so its other w came from the ram.
A ram that can give a w carries w, and a white ram that carries w is Ww.
So the black lamb supports the claim that the ram is Ww.
Rubric
  • Award 1 point for: the evidence (a black lamb, ww) AND the reasoning (its mother gave one w, so the other w came from the ram, who therefore carries w and is Ww).

Slip Saying the black lamb came from the ewe alone. The ewe supplies one w; a ww lamb needs the second w from the ram.

(d) Suppose instead that the first eight lambs are all white. Explain how sure the farmer can be that the ram is WW. (1 pt)

Model answer The farmer can conclude that the ram is probably WW, but not certainly.
A Ww ram gives a white lamb with probability one half, and eight white lambs in a row have probability (12)8=1256, unlikely but not impossible. Only a black lamb would rule WW out.
Rubric
  • Award 1 point for: eight white lambs favor WW but do not rule out Ww, because a Ww ram gives all white lambs with a small but nonzero probability; only a black lamb would rule WW out.

Slip Concluding the ram is certainly WW. A run of dominant offspring can only make the homozygous genotype likely; a recessive offspring is what would settle it.

Glossary

test cross
A cross between an individual showing the dominant trait, whose genotype is unknown, and one that is homozygous recessive. Any recessive offspring proves the unknown parent heterozygous; all dominant offspring make homozygous likely but not certain.

APBIO-U05-L15B Which allele is dominant?

Topic 5.3 · Mendelian Genetics · 24 steps

Two rows of beetles drawn with heads, antennae and six legs: on the left a true-breeding spotted line, five beetles each with two dark spots on each wing case; on the right a true-breeding unspotted line, five plain beetles; below, a caption that most wild beetles are unspotted
Two rows of beetles drawn with heads, antennae and six legs: on the left a true-breeding spotted line, five beetles each with two dark spots on each wing case; on the right a true-breeding unspotted line, five plain beetles; below, a caption that most wild beetles are unspotted

Imagine two true-breeding lines of a beetle: one line always spotted, the other always unspotted. In the wild, most beetles of this kind are unspotted.

One of the two alleles is dominant. Which is it, spotted or unspotted, and how could crosses tell you?

Unit 5 · Heredity

1Read dominance from the crosses

2

Video: Watch: Read dominance from the crosses

The two true-breeding beetle lines crossed and every F1 coming out spotted; the F1 crossed together and the F2 sorting into about three spotted to one unspotted; the wild population set aside as no evidence.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L15Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L15Ba.mp4

3

How do you decide which allele is dominant when nobody has told you? Breeding tells you.

4

How common a trait is does not tell you: being the common kind does not make unspotted the dominant trait.

5

Cross the two true-breeding lines: every F1 beetle is spotted.

Two true-breeding beetle lines at the top, labelled P generation, spotted and unspotted, crossed; below them a row of six F1 beetles, all spotted; below that the F2 from F1 × F1: three spotted beetles and one unspotted, captioned 152 spotted and 48 unspotted, about three to one
Two true-breeding beetle lines at the top, labelled P generation, spotted and unspotted, crossed; below them a row of six F1 beetles, all spotted; below that the F2 from F1 × F1: three spotted beetles and one unspotted, captioned 152 spotted and 48 unspotted, about three to one
6

Every F1 beetle carries one allele from each line, so every F1 is heterozygous. The trait every F1 shows is the dominant one: spotted.

7

Now cross the F1 beetles with each other: 152 spotted and 48 unspotted, about three to one.

The square for two heterozygous F1 beetles, Ss × Ss: S and s along the top and down the side; cells SS, Ss, Ss, ss, with the three spotted-giving cells shaded
The square for two heterozygous F1 beetles, Ss × Ss: S and s along the top and down the side; cells SS, Ss, Ss, ss, with the three spotted-giving cells shaded
8

Three of the four cells of the F1 × F1 square hold the dominant allele. So the dominant trait is the 75% majority of the F2.

9

So the two readings agree: spotted is dominant, although most wild beetles are unspotted. Dominance says only which trait shows in a heterozygous animal.

10

What tells you which allele is dominant:

  1. The trait every F1 shows when two true-breeding lines are crossed: yes.
  2. The trait in the 75% majority when the F1 are crossed with each other: yes.
  3. The trait that vanishes in the F1 and returns in 25% of the F2: that one is the recessive trait.
  4. The trait most of the wild population shows: no, that does not show which allele is dominant.

11

What you are expected to know Decide which allele is dominant from breeding data: the trait every F1 shows when two true-breeding lines are crossed, or the trait in the 75% majority when the F1 are crossed with each other.

12
Check q1

A true-breeding line of black rabbits is crossed with a true-breeding line of chocolate-brown rabbits. Every F1 rabbit is black.

Which coat color is dominant?

  1. A. Chocolate
    Every F1 rabbit received one allele from each line, so every F1 is heterozygous.
    Every F1 is black, so black is the trait that shows in a heterozygous rabbit.
  2. B. ✓ Black
  3. C. Neither
    The F1 alone settles it: every F1 is heterozygous, and every F1 is black.

Why: Every F1 rabbit has one allele from each line, so every F1 is heterozygous.
Every F1 is black.
So black is the trait that shows in a heterozygous rabbit, and black is dominant.

13
Practice writing an answer

A true-breeding line of black rabbits is crossed with a true-breeding line of chocolate-brown rabbits. All 24 F1 rabbits are black. Crossing the F1 rabbits with each other gives 61 black and 19 chocolate offspring.

(a) Explain how the F1 rabbits show that black is dominant. (1 pt)

Frame The F1 rabbits show that black is dominant because …

Model answer The F1 rabbits show that black is dominant because every F1 rabbit is heterozygous and every F1 rabbit is black.
Each F1 rabbit received a black allele from one true-breeding line and a chocolate allele from the other.
So each F1 carries both alleles.
The trait that shows when both alleles are present is the dominant one.
That trait is black.
So black is dominant.
Rubric
  • Award 1 point for: every F1 rabbit is heterozygous and every one is black, so black shows with one copy.
14
Check q2

In people, five fingers on a hand is the common trait and six fingers is rare. A student says: “Five fingers is the common trait, so five fingers must be the dominant trait.”

Is the student correct?

  1. A. Yes
    How common a trait is does not show which allele is dominant.
    Dominant says only which trait shows in a heterozygous person.
  2. B. ✓ No

Why: Dominant names the trait that shows in a heterozygous person.
How common a trait is does not decide which trait that is.
A dominant trait can be rare, and a recessive trait can be the common one.
So the count of hands cannot tell the student which trait is dominant.

15
Check q3

Two true-breeding lines of a flowering plant are crossed: one with red flowers, one with white. All 40 F1 plants have red flowers.

Which of the following do the F1 plants show about the two alleles?

  1. A. White is dominant
    Every F1 does carry the white allele.
    But the trait that shows in those heterozygous plants is red.
    The allele whose trait shows is dominant.
  2. B. Nothing yet
    The F1 alone settles it: every F1 has one allele from each true-breeding line, so every F1 is heterozygous, and the trait a heterozygous plant shows is the dominant one.
  3. C. ✓ Red is dominant

Why: Each F1 plant received a red allele from one line and a white allele from the other.
So every F1 is heterozygous.
Every F1 is red.
So red is the trait that shows in a heterozygous plant, and red is dominant.

16

Back to the two beetle lines, crossed here again: which pattern is dominant is settled by breeding, not by counting the wild population.

Two true-breeding beetle lines at the top, labelled P generation, spotted and unspotted, crossed; below them a row of six F1 beetles, all spotted; below that the F2 from F1 × F1: three spotted beetles and one unspotted, captioned 152 spotted and 48 unspotted, about three to one
Two true-breeding beetle lines at the top, labelled P generation, spotted and unspotted, crossed; below them a row of six F1 beetles, all spotted; below that the F2 from F1 × F1: three spotted beetles and one unspotted, captioned 152 spotted and 48 unspotted, about three to one
17

Every F1 is spotted, and the F1 × F1 offspring are three spotted to one unspotted. So spotted is dominant and unspotted is recessive.

18Quick quiz: which trait is dominant? mixed practice

19
Check q4

In the peppered moth, a true-breeding dark-winged line is crossed with a true-breeding pale-winged line. All F1 moths are dark.

Which wing color is dominant?

  1. A. Pale
    Every F1 is heterozygous, and every F1 is dark.
    The trait a heterozygous moth shows is dominant.
  2. B. ✓ Dark

Why: Every F1 moth is heterozygous, one allele from each line.
Every F1 is dark.
So dark is the trait that shows in a heterozygous moth, and dark is dominant.

20
Check q5

In fruit flies, a true-breeding line with full-sized wings is crossed with a true-breeding line with tiny, crumpled wings. Every F1 fly has full-sized wings.

Which wing form is recessive?

  1. A. ✓ Tiny, crumpled wings
  2. B. Full-sized wings
    The trait a heterozygous F1 shows is the dominant one, full-sized wings.
    The trait that vanished in the F1 is recessive.

Why: Every F1 is heterozygous.
Every F1 has full-sized wings, so full-sized wings are dominant.
Tiny, crumpled wings vanished in the F1.
The trait that vanishes in the F1 is the recessive one.

21
Check q6

In mice, F1 mice from two true-breeding lines, one black-furred and one brown-furred, are crossed with each other. The bar chart shows their F2 offspring.

A bar chart with two bars, black-furred and brown-furred mice in the F2 of an F1 × F1 cross; the value is printed above each bar
A bar chart with two bars, black-furred and brown-furred mice in the F2 of an F1 × F1 cross; the value is printed above each bar

Which fur color is dominant?

  1. A. ✓ Black
  2. B. Brown
    In the F2 of an F1 × F1 cross, the dominant trait is the 75% majority.
    Black-furred mice are about 75% of the F2.

Why: The F1 × F1 square has three cells of four holding the dominant allele.
So the dominant trait is the 75% majority of the F2.
Black-furred mice are about 75%, 147 of 200.
So black is dominant.

22
Check q7

In one population of the grove snail, most snails have a banded shell.

Does that count tell you which allele is dominant?

  1. A. Yes
    Dominant says only which trait shows in a heterozygous snail.
    A dominant trait can be rare, and a recessive trait can be the common one.
  2. B. ✓ No

Why: How common a trait is does not show which allele is dominant.
Dominant says only which trait shows in a heterozygous snail.
So the wild count does not tell you which allele is dominant.
Only crosses can.

23
Check q8

In the grove snail, a true-breeding banded line is crossed with a true-breeding unbanded line. Every F1 snail is unbanded.

Which shell pattern shows in a heterozygous snail?

  1. A. ✓ Unbanded
  2. B. Banded
    Every F1 is heterozygous, and every F1 is unbanded.
    So unbanded is the trait that shows in a heterozygous snail.

Why: Every F1 snail has one allele from each line, so every F1 is heterozygous.
Every F1 is unbanded.
So unbanded is the trait that shows in a heterozygous snail: unbanded is dominant, even where most wild snails are banded.

APBIO-U05-L15C The twenty-two matching pairs and the one odd pair

Topic 5.3 · Mendelian Genetics · 36 steps

Left: a photograph of a real human karyotype from a male, 46 stained, banded chromosomes laid out in pairs from the longest to the shortest, the last pair unequal in length. Right: 23 pairs of chromosomes drawn as rods in two rows, each rod with a centromere dot, the pairs shrinking in length along the rows; in every pair one rod is darker and one lighter; the last pair at the bottom right is a long rod beside a short rod, labelled X and Y
Left: a photograph of a real human karyotype from a male, 46 stained, banded chromosomes laid out in pairs from the longest to the shortest, the last pair unequal in length. Right: 23 pairs of chromosomes drawn as rods in two rows, each rod with a centromere dot, the pairs shrinking in length along the rows; in every pair one rod is darker and one lighter; the last pair at the bottom right is a long rod beside a short rod, labelled X and Y

Photo: National Human Genome Research Institute, Wikimedia Commons, public domain (resized).

Here is a human karyotype: the chromosomes of one cell, sorted into 23 pairs. Twenty-two of the pairs match, member for member.

The last pair does not. One chromosome is large and one is small. What are the 22 matching pairs called, and what is the odd pair?

Unit 5 · Heredity

1Autosomes and the sex chromosomes

2

Video: Watch: Autosomes and the sex chromosomes

The 22 matching pairs picked out on the karyotype one row at a time and named autosomes; then the odd last pair, a long X and a short Y, named the sex chromosomes; then a female's karyotype, whose two X's sit last.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L15Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L15Ca.mp4

3
Check q1

A human body cell has 46 chromosomes, sorted by size into 23 pairs.

Which of the following describes the two chromosomes of one matching pair?

  1. A. ✓ The same length, with the same genes in the same order
  2. B. The same length, with different genes
    Two chromosomes of one matching pair carry the same genes in the same order.
    One came from the mother and one from the father.
  3. C. Different lengths, with the same genes
    Chromosomes of different lengths carry different genes.
    The two members of a matching pair are the same length.

Why: The two chromosomes of a matching pair are a homologous pair.
They are the same length.
They carry the same genes in the same order, one from each parent.

4

Which chromosomes decide sex, and what do we call the rest? Look first at the 22 pairs that match.

A human karyotype drawn as 23 pairs of rods in two rows, one rod of each pair darker and one lighter, the pairs shrinking in length along the rows; pairs 1, 5, 10, 15 and 22 are numbered; the last pair at the bottom right is a long rod beside a short rod, labelled X and Y
A human karyotype drawn as 23 pairs of rods in two rows, one rod of each pair darker and one lighter, the pairs shrinking in length along the rows; pairs 1, 5, 10, 15 and 22 are numbered; the last pair at the bottom right is a long rod beside a short rod, labelled X and Y
5

The two members of a matching pair are the same length. They carry the same genes in the same order.

6

When a chromosome belongs to one of the 22 pairs whose members match, we call it an ; this cell has 44 autosomes.

7

The last pair is different. In this cell one member is a large chromosome, the X, and the other is a small one, the Y.

An X chromosome drawn as a long rod with a centromere dot beside a Y chromosome drawn as a rod about half as long, labelled X and Y
An X chromosome drawn as a long rod with a centromere dot beside a Y chromosome drawn as a rod about half as long, labelled X and Y
8

Typically a female has two X chromosomes, XX, and a male has one X and one Y, XY. So a cell whose karyotype ends in a long X beside a short Y came from a male.

9

When a pair of chromosomes sets whether a person is female or male, we call them the : two X’s in a female, an X and a Y in a male.

10

The X and the Y differ in size and in genes. Yet they count as a pair: they pair up in meiosis I as the autosome pairs do, and they part at anaphase I.

11

A karyotype always puts the sex chromosomes last, at the bottom right. So in a female’s karyotype the last pair is two X chromosomes of the same length, and their place tells you which pair they are.

12

For example, take pair 1 of this cell: two long chromosomes of the same length, one from each parent. This pair is autosomes.

Four boxed drawings of one pair of chromosomes each. Box 1: two long rods of the same length, one darker and one lighter, labelled pair 1, autosomes. Box 2: two short rods of the same length, labelled pair 21, autosomes. Box 3: a long rod beside a rod half as long, labelled last pair of a male, sex chromosomes. Box 4: two long rods of the same length, labelled last pair of a female, sex chromosomes
Four boxed drawings of one pair of chromosomes each. Box 1: two long rods of the same length, one darker and one lighter, labelled pair 1, autosomes. Box 2: two short rods of the same length, labelled pair 21, autosomes. Box 3: a long rod beside a rod half as long, labelled last pair of a male, sex chromosomes. Box 4: two long rods of the same length, labelled last pair of a female, sex chromosomes
13

Take pair 21: two short chromosomes of the same length, one from each parent. This pair is autosomes too.

14

But take the last pair of this cell: a long X beside a short Y. This pair is the sex chromosomes.

15

And take the last pair of a female's cell: two X chromosomes of the same length, placed last. This pair is the sex chromosomes too.

16

What you are expected to know Tell the one pair of sex chromosomes from the 22 pairs of autosomes on a human karyotype.

17

What you are expected to know Read XX or XY from the last pair of a human karyotype.

18
Check q2

The karyotype below shows the 23 pairs of one human cell.

A karyotype of 23 pairs of rods in two rows, one rod of each pair darker and one lighter, the pairs shrinking in length along the rows; every pair is two rods of the same length, and the last pair at the bottom right is two long rods of the same length with no label under them
A karyotype of 23 pairs of rods in two rows, one rod of each pair darker and one lighter, the pairs shrinking in length along the rows; every pair is two rods of the same length, and the last pair at the bottom right is two long rods of the same length with no label under them

What does the last pair show about this cell?

  1. A. The cell is from a male
    Two X chromosomes are the female pattern.
    A male’s last pair is an X beside a shorter Y.
  2. B. ✓ The cell is from a female

Why: The last pair is two chromosomes of the same length.
Both chromosomes are X’s, because a karyotype puts the sex chromosomes last.
XX is the female pattern, so the cell is from a female.

19
Practice writing an answer

The karyotype below shows the 23 pairs of one human cell.

A karyotype of 23 pairs of rods in two rows, one rod of each pair darker and one lighter, the pairs shrinking in length along the rows; every pair is two rods of the same length, and the last pair at the bottom right is two long rods of the same length with no label under them
A karyotype of 23 pairs of rods in two rows, one rod of each pair darker and one lighter, the pairs shrinking in length along the rows; every pair is two rods of the same length, and the last pair at the bottom right is two long rods of the same length with no label under them

(a) Explain what in the last pair shows that the cell is from a female. (1 pt)

Model answer The last pair is two chromosomes of the same length.
A karyotype puts the sex chromosomes last, so both chromosomes are X’s.
XX is the female pattern.
A male’s last pair would be a long X beside a short Y.
Rubric
  • Award 1 point for: the last pair is two X chromosomes of the same length, the female pattern; a male’s last pair is an X with a shorter Y.
20
Check q3

A human cheek cell has 46 chromosomes.

Which of its chromosomes are autosomes?

  1. A. The 23 chromosomes that came from the mother
    Autosomes are sorted by which pair they belong to, not by which parent they came from.
    Each parent gave 22 autosomes and one sex chromosome.
  2. B. The X and the Y only
    The X and the Y are the sex chromosomes, the one pair that is not autosomes.
  3. C. All 46
    Autosome names the chromosomes of the 22 matching pairs only.
    The two sex chromosomes are left out, so 44 of the 46 are autosomes.
  4. D. ✓ The 44 chromosomes in the 22 matching pairs

Why: An autosome is a chromosome from one of the 22 pairs whose members match.
22 pairs is 44 chromosomes.
The other two, the X and the Y or two X’s, are the sex chromosomes.

21

Back to the karyotype of the male cell, drawn here again: 22 of its pairs match, member for member. Those 44 chromosomes are the autosomes.

A human karyotype drawn as 23 pairs of rods in two rows, one rod of each pair darker and one lighter, the pairs shrinking in length along the rows; pairs 1, 5, 10, 15 and 22 are numbered; the last pair at the bottom right is a long rod beside a short rod, labelled X and Y
A human karyotype drawn as 23 pairs of rods in two rows, one rod of each pair darker and one lighter, the pairs shrinking in length along the rows; pairs 1, 5, 10, 15 and 22 are numbered; the last pair at the bottom right is a long rod beside a short rod, labelled X and Y
22

The odd pair is the sex chromosomes: a long X beside a short Y. So this cell came from a male.

23Quick quiz: autosome, sex chromosomes mixed practice

24
Check q4

A human cell’s chromosomes are sorted by size into pairs.

Which of the following is an autosome?

  1. A. ✓ A chromosome from one of the 22 pairs whose members match
  2. B. The X chromosome of the odd last pair
    The X is one of the two sex chromosomes.
    Autosomes are the chromosomes of the 22 matching pairs.
  3. C. The Y chromosome of the odd last pair
    The Y is one of the two sex chromosomes.
    Autosomes are the chromosomes of the 22 matching pairs.

Why: The 22 pairs whose members match are the autosome pairs.
A chromosome from one of those pairs is an autosome.
The X and the Y are the sex chromosomes.

25
Check q5

The karyotype below shows the 23 pairs of one human cell. One pair is ringed.

A karyotype of 23 pairs of rods in two rows; the first pair, at the top left, is ringed
A karyotype of 23 pairs of rods in two rows; the first pair, at the top left, is ringed

Which of the following is the ringed pair?

  1. A. ✓ A pair of autosomes
  2. B. The pair of sex chromosomes
    The ringed pair is pair 1: two chromosomes of the same length, in the first place, not the last.

Why: The ringed pair is two chromosomes of the same length.
It sits first, not last.
So it is one of the 22 matching pairs: a pair of autosomes.

26
Check q6

A human cell’s chromosomes are sorted by size into pairs.

Which of the following are the sex chromosomes?

  1. A. The 44 chromosomes of the autosome pairs
    The autosome pairs are the 22 pairs whose members match, not the sex chromosomes.
  2. B. The chromosomes that came from the mother
    The mother gave 22 autosomes and one X.
    The sex chromosomes are sorted by which pair they belong to, not by which parent gave them.
  3. C. ✓ The X and the Y

Why: The sex chromosomes are the pair that sets whether a person is female or male.
That pair is the X and the Y.
A female has XX and a male has XY.

27
Check q7

The karyotype below shows the 23 pairs of one human cell. One pair is ringed.

A karyotype of 23 pairs of rods in two rows; the last pair, at the bottom right, one long rod beside one short rod, is ringed
A karyotype of 23 pairs of rods in two rows; the last pair, at the bottom right, one long rod beside one short rod, is ringed

Which of the following is the ringed pair?

  1. A. A pair of autosomes
    The ringed pair is one long chromosome beside one short one, placed last.
    The members of an autosome pair are the same length.
  2. B. ✓ The pair of sex chromosomes

Why: The ringed pair sits last, at the bottom right.
Its two members differ in length: a long X and a short Y.
So it is the pair of sex chromosomes.

28
Check q8

A human skin cell holds 46 chromosomes.

How many of them are autosomes?

  1. A. 23
    23 is the number of pairs, or the number of chromosomes one parent gave.
    Autosomes are counted as chromosomes.
  2. B. ✓ 44
  3. C. 46
    The X and the Y, or the two X’s, are sex chromosomes, not autosomes.

Why: Autosomes are the chromosomes of the 22 matching pairs.
22 pairs is 44 chromosomes.
The last pair is the two sex chromosomes, so 44 of the 46 are autosomes.

29
Check q9

The karyotype below shows the 23 pairs of one human cell. One pair is ringed.

A karyotype of 23 pairs of rods in two rows; the first pair of the bottom row, at the bottom left, is ringed
A karyotype of 23 pairs of rods in two rows; the first pair of the bottom row, at the bottom left, is ringed

Which of the following is the ringed pair?

  1. A. ✓ A pair of autosomes
  2. B. The pair of sex chromosomes
    The ringed pair is pair 13: two chromosomes of the same length, and not in the last place.

Why: The ringed pair is two chromosomes of the same length.
It sits at the start of the bottom row, not at the end.
So it is one of the 22 matching pairs: a pair of autosomes.

30
Practice writing an answer

In a male’s cell the X is a large chromosome and the Y is a small one, and the two carry different genes.

(a) Explain why the X and the Y still count as one pair of chromosomes. (1 pt)

Frame The X and the Y count as one pair because …

Model answer The X and the Y count as one pair because they pair up in meiosis I.
The 22 autosome pairs pair up in meiosis I in the same way.
At anaphase I the X and the Y part, one to each pole, as every pair does.
So the cell treats the X and the Y as a pair.
Rubric
  • Award 1 point for: the X and the Y pair up in meiosis I (and part at anaphase I) as the autosome pairs do.
31
Check q10

The karyotype below shows the 23 pairs of one human cell. One pair is ringed.

A karyotype of 23 pairs of rods in two rows; the second-to-last pair of the bottom row, two short rods of the same length, is ringed
A karyotype of 23 pairs of rods in two rows; the second-to-last pair of the bottom row, two short rods of the same length, is ringed

Which of the following is the ringed pair?

  1. A. ✓ A pair of autosomes
  2. B. The pair of sex chromosomes
    The ringed pair is pair 22: two short chromosomes of the same length, one place before the last.

Why: The ringed pair is two chromosomes of the same length.
It sits one place before the last pair.
So it is one of the 22 matching pairs: a pair of autosomes.

32
Check q11

The karyotype below shows the 23 pairs of one human cell. One pair is ringed.

A karyotype of 23 pairs of rods in two rows; the last pair, at the bottom right, two long rods of the same length, is ringed
A karyotype of 23 pairs of rods in two rows; the last pair, at the bottom right, two long rods of the same length, is ringed

Which of the following is the ringed pair?

  1. A. A pair of autosomes
    The ringed pair sits last, at the bottom right, where a karyotype puts the sex chromosomes.
    Two X’s are the same length.
  2. B. ✓ The pair of sex chromosomes

Why: The ringed pair sits last, at the bottom right.
A karyotype always puts the sex chromosomes last.
Its two members are the same length, so they are two X’s: the sex chromosomes of a female.

33
Check q12

The karyotype below shows the 23 pairs of one human cell.

A karyotype of 23 pairs of rods in two rows, the pairs shrinking in length along the rows; the last pair at the bottom right is one long rod beside one short rod, with no label under them
A karyotype of 23 pairs of rods in two rows, the pairs shrinking in length along the rows; the last pair at the bottom right is one long rod beside one short rod, with no label under them

Which of the following is this cell?

  1. A. A female’s cell, XX
    Two X chromosomes are the same length.
    Here the last pair is a long chromosome beside a short one.
  2. B. ✓ A male’s cell, XY

Why: The last pair is one long chromosome beside one short one.
The X is long and the Y is short.
XY is the male pattern, so the cell is a male’s.

34
Check q13

In a human cell, the last pair of chromosomes is one long chromosome beside one short chromosome.

Which of the following is this cell?

  1. A. ✓ A male’s cell, XY
  2. B. A female’s cell, XX
    Two X chromosomes are the same length.
    A long chromosome beside a short one is an X beside a Y.

Why: The X is a long chromosome and the Y is a short one.
A long chromosome beside a short one in the last pair is an X and a Y.
XY is the male pattern, so the cell is a male’s.

35
Check q14

The karyotype below shows the 23 pairs of one human cell.

A karyotype of 23 pairs of rods in two rows, the pairs shrinking in length along the rows; the last pair at the bottom right is two long rods of the same length, with no label under them
A karyotype of 23 pairs of rods in two rows, the pairs shrinking in length along the rows; the last pair at the bottom right is two long rods of the same length, with no label under them

Which of the following is this cell?

  1. A. A male’s cell, XY
    A male’s last pair is a long X beside a short Y.
    Here the last pair is two chromosomes of the same length.
  2. B. ✓ A female’s cell, XX

Why: The last pair is two chromosomes of the same length.
A karyotype puts the sex chromosomes last, so both chromosomes are X’s.
XX is the female pattern, so the cell is a female’s.

Glossary

autosome
A chromosome from one of the 22 pairs whose members match; a human cell has 44 autosomes and one pair of sex chromosomes.
sex chromosomes
The X and the Y, the pair that sets whether a person is female (XX) or male (XY); they pair in meiosis I although they differ in size and genes.

APBIO-U05-L16 Who is who on a family tree

Topic 5.3 · Mendelian Genetics · 57 steps

Left: a photograph of three women of one family standing together, a grandmother in the middle, her daughter and her granddaughter either side. Right: a family tree of three rows: two unshaded grandparents at the top; four shapes in the middle row, of which the first, a square, is shaded; and two grandchildren at the bottom, of which the first, a circle, is shaded. Each shaded shape's parents are unshaded
Left: a photograph of three women of one family standing together, a grandmother in the middle, her daughter and her granddaughter either side. Right: a family tree of three rows: two unshaded grandparents at the top; four shapes in the middle row, of which the first, a square, is shaded; and two grandchildren at the bottom, of which the first, a circle, is shaded. Each shaded shape's parents are unshaded

Photo: Bill Branson, National Cancer Institute, via Wikimedia Commons, public domain (resized).

Here is a family drawn as squares and circles across three rows: grandparents at the top, their children in the middle, grandchildren at the bottom.

Two of the shapes are shaded. Each shaded person has a trait, and in each case the person’s two parents are unshaded: they do not have it. How can two parents without a trait have a child with it?

Unit 5 · Heredity

1Who is who on a family tree

2

Video: Watch: Who is who on a family tree

A family drawn shape by shape as the viewer watches: squares and circles, the shading, the line between two parents, the line down to their children in birth order, the rows numbered I, II, III and the people numbered from the left.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16a.mp4

3

How do you read a family’s history of one trait from a drawing? Here is a small family drawn as a record of who has a trait, generation by generation.

A two-generation pedigree: I-1, a square, and I-2, a circle, joined by a horizontal line; a vertical line drops to a horizontal line over their three children, II-1 a shaded circle, II-2 an unshaded square and II-3 an unshaded circle; a caption reads: rows are generations; children sit in birth order, left to right
A two-generation pedigree: I-1, a square, and I-2, a circle, joined by a horizontal line; a vertical line drops to a horizontal line over their three children, II-1 a shaded circle, II-2 an unshaded square and II-3 an unshaded circle; a caption reads: rows are generations; children sit in birth order, left to right
4

When a family is drawn as a record of who has a trait, generation by generation, we call the drawing a .

5

Here is a table of the pedigree's key: each symbol, and what it means.

A table with two columns, symbol and meaning, and six rows: a square, a male; a circle, a female; a shaded shape, a person who has the trait; an unshaded shape, a person who does not have the trait; a line between two shapes, the two are parents; a line down to a row of shapes, their children in birth order from the left
A table with two columns, symbol and meaning, and six rows: a square, a male; a circle, a female; a shaded shape, a person who has the trait; an unshaded shape, a person who does not have the trait; a line between two shapes, the two are parents; a line down to a row of shapes, their children in birth order from the left
6

Squares are males and circles are females.

7

A shaded shape is a person who has the trait. An unshaded shape is a person who does not.

8

A horizontal line joins two parents. A vertical line drops from it to their children.

9

Brothers and sisters sit on their line in birth order from left to right, so the oldest is on the left.

10

The rows are generations, numbered I, II, III from the top.

11

The people in a row are numbered from the left. So II-3 is the third person in row II.

12

A pedigree records phenotypes: shaded or unshaded is what each person shows. Nobody's genotype is written on it.

13

What you are expected to know Read who is who on a pedigree: male or female, has the trait or not, whose child, whose partner, and born in which order.

14
Check q1

A family's pedigree for a trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle

Which of the following describes III-1?

  1. A. A female who has the trait
    III-1 is a square.
    A square is a male.
  2. B. A male who does not have the trait
    III-1 is shaded.
    A shaded shape is a person who has the trait.
  3. C. A female who does not have the trait
    III-1 is a square, so a male, and it is shaded, so he has the trait.
  4. D. ✓ A male who has the trait

Why: III-1 is the first shape in row III.
It is a square, so III-1 is a male.
It is shaded, so he has the trait.

15Quick quiz: pedigree mixed practice

16
Check q2

A geneticist records a family.

What is a pedigree?

  1. A. ✓ A family drawn as a record of who has a trait, generation by generation
  2. B. A list of the genotypes of everyone in a family
    A pedigree records what each person shows, not their genotypes.
  3. C. A square showing the offspring two parents can produce
    A square showing the offspring two parents can produce is a Punnett square.

Why: A pedigree is a family drawn as a record of who has a trait, generation by generation: squares, circles, shading and the lines between them.

17
Practice writing an answer

On a pedigree, one shape is shaded and the shapes around it are not.

(a) State what the shaded shape means. (1 pt)

Model answer A shaded shape is a person who has the trait.
Rubric
  • Award 1 point for: a shaded shape is a person who has (shows) the trait.
18
Check q3

A family's pedigree for a trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle

Is II-2 male or female?

  1. A. Male
    II-2 is a circle, and a circle is a female.
  2. B. ✓ Female

Why: II-2 is the second shape in row II.
It is a circle, so II-2 is a female.

19
Check q4

A family's pedigree for a trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle

Does II-2 have the trait?

  1. A. ✓ Yes
  2. B. No
    II-2 is shaded, and a shaded shape is a person who has the trait.

Why: II-2 is shaded.
A shaded shape is a person who has the trait, so II-2 has it.

20
Check q5

A family's pedigree for a trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle

Who are III-1's parents?

  1. A. I-1 and I-2
    I-1 and I-2 are the couple in row I, and the vertical line from their horizontal line drops to row II, not row III.
  2. B. II-1 and II-2
    II-1 and II-2 are brother and sister on one sibling line, and no horizontal line joins them as parents.
  3. C. ✓ II-3 and II-4

Why: A horizontal line joins two parents, and a vertical line drops from it to their children.
The line above III-1 rises to the horizontal line joining II-3 and II-4, so they are the parents.

21
Check q6

A family's pedigree for a trait is drawn below. I-1 and I-2 are joined by a horizontal line.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle

What is I-2 to I-1?

  1. A. A parent
    A parent sits in the row above, and I-1 and I-2 sit in the same row, joined side by side.
  2. B. ✓ A partner
  3. C. A child
    A child hangs from a vertical line in the row below.
    I-2 sits beside I-1 in the same row.

Why: A horizontal line joins two parents.
I-1 and I-2 are joined by one, so I-2 is I-1's partner, and the children below are theirs.

22
Check q7

A family's pedigree for a trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle

Which child of I-1 and I-2 was born first?

  1. A. ✓ II-1
  2. B. II-2
    Children hang from the sibling line in birth order from left to right, and II-2 is second from the left.
  3. C. II-3
    II-3 is the third from the left, so the third born.

Why: Children sit on their sibling line in birth order from left to right.
II-1 is the leftmost child of I-1 and I-2, so II-1 was born first.

23Two unshaded parents, a shaded child

24

Video: Watch: Two unshaded parents, a shaded child

The opening family with its people numbered; one allele traced from each parent into the shaded child; the word carrier written under each unshaded parent of a shaded child.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16b.mp4

25
Check q8

A trait is recessive.

Which people show it?

  1. A. ✓ Only people with two copies of its allele
  2. B. Anyone with at least one copy
    A person with one copy of a recessive allele and one dominant allele shows the dominant allele's trait.

Why: A recessive trait shows when both alleles are the recessive allele.
So only people with two copies of the allele for the trait show it.

26

Now look again at the family of the opening page, with everyone numbered. II-1 is shaded, and his parents I-1 and I-2 are unshaded.

The opening family drawn as a pedigree with its people numbered: I-1 and I-2 unshaded; their children II-1, a shaded square, II-2 an unshaded circle and II-3 an unshaded square; II-3 joined to II-4, an unshaded circle; their children III-1, a shaded circle, and III-2, an unshaded square
The opening family drawn as a pedigree with its people numbered: I-1 and I-2 unshaded; their children II-1, a shaded square, II-2 an unshaded circle and II-3 an unshaded square; II-3 joined to II-4, an unshaded circle; their children III-1, a shaded circle, and III-2, an unshaded square
27

Every person receives one allele of the gene from each parent.

28

II-1 has the trait. So II-1 received the trait's allele from I-1 and from I-2.

29

So I-1 and I-2 each carry the trait's allele. Yet neither of them shows the trait.

30

An allele that a person can carry without showing its trait is a recessive allele. So the trait is recessive.

31

Each parent also carries the dominant allele. The dominant allele is the one that shows.

32

When a person carries an allele without showing its trait, we call that person a , because they carry the allele unseen.

33

On the pedigree a carrier is drawn unshaded, like anyone else without the trait.

34

So two unshaded parents with a shaded child are both carriers. Here is the family again with its carriers named: I-1 and I-2, and II-3 and II-4 above the shaded III-1.

The same pedigree with the word carrier written under I-1, I-2, II-3 and II-4, the four unshaded parents of the two shaded people; a caption reads: each unshaded parent of a shaded child is a carrier
The same pedigree with the word carrier written under I-1, I-2, II-3 and II-4, the four unshaded parents of the two shaded people; a caption reads: each unshaded parent of a shaded child is a carrier
35

Unshaded tells you only that the person does not show the trait. It never tells you on its own that the person carries no allele for it.

36

What you are expected to know Identify the carriers a pedigree fixes: two unshaded parents with a shaded child are both carriers.

37
Check q9

A family's pedigree for a recessive trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 a shaded circle, II-2 an unshaded square, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded circle, III-3 an unshaded square
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 a shaded circle, II-2 an unshaded square, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded circle, III-3 an unshaded square

Which people must be carriers?

  1. A. I-1 and I-2 only
    III-2 is shaded and her parents II-3 and II-4 are unshaded.
    Each gave III-2 the trait's allele without showing the trait, so both are carriers too.
  2. B. ✓ I-1, I-2, II-3 and II-4
  3. C. I-1, I-2, II-2, II-3 and II-4
    II-2 is unshaded and has no shaded child.
    Nothing on the pedigree shows that II-2 carries the allele.

Why: II-1 and III-2 are shaded, so each received the trait's allele from both parents.
II-1's parents are I-1 and I-2; III-2's parents are II-3 and II-4.
All four are unshaded, so each carries the allele without showing the trait.
So I-1, I-2, II-3 and II-4 are carriers.

38
Practice writing an answer

A family's pedigree for a recessive trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 a shaded circle, II-2 an unshaded square, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded circle, III-3 an unshaded square
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 a shaded circle, II-2 an unshaded square, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded circle, III-3 an unshaded square

(a) Explain how the pedigree shows that II-3 and II-4 are carriers. (1 pt)

Model answer III-2 is shaded, so III-2 has the trait.
The trait is recessive, so III-2 has two copies of its allele.
III-2 received one copy from II-3 and one from II-4.
So II-3 and II-4 each carry the trait's allele.
II-3 and II-4 are both unshaded, so neither shows the trait.
A person who carries an allele without showing its trait is a carrier, so II-3 and II-4 are carriers.
Rubric
  • Award 1 point for: their child III-2 has the recessive trait, so III-2 received the trait's allele from each of them; each is unshaded, so each carries the allele without showing the trait.

Slip Saying II-3 and II-4 are carriers because they are unshaded. Unshaded alone fixes nothing; their shaded child fixes them.

39
Check q10

A student looks at the same pedigree for a recessive trait, drawn below, and says: “II-2 is unshaded, so II-2 carries no allele for the trait.”

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 a shaded circle, II-2 an unshaded square, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded circle, III-3 an unshaded square
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 a shaded circle, II-2 an unshaded square, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded circle, III-3 an unshaded square

Which statement about the student's claim is correct?

  1. A. The student is right
    A carrier is unshaded too.
    Unshaded says only that II-2 does not show the trait.
  2. B. ✓ The student is wrong

Why: A pedigree records phenotypes.
Unshaded means II-2 does not show the trait.
A carrier is unshaded too, so the drawing alone cannot say whether II-2 carries the allele.
So the student is wrong: II-2 may carry the allele unseen.

40

Back to the family of the opening page: grandparents at the top, their children in the middle, grandchildren at the bottom, with two shaded shapes. Each shaded person has two unshaded parents.

The same pedigree with the word carrier written under I-1, I-2, II-3 and II-4, the four unshaded parents of the two shaded people; a caption reads: each unshaded parent of a shaded child is a carrier
The same pedigree with the word carrier written under I-1, I-2, II-3 and II-4, the four unshaded parents of the two shaded people; a caption reads: each unshaded parent of a shaded child is a carrier
41

Each of those parents gave the shaded child the trait's allele without showing the trait. So all four are carriers.

42Quick quiz: carrier mixed practice

43
Check q11

A geneticist describes one person in a family.

What is a carrier?

  1. A. ✓ A person who carries an allele without showing its trait
  2. B. A person who has the trait and shows it
    A person who has the trait shows it; a carrier does not.
  3. C. A person drawn as a shaded shape on the pedigree
    A shaded shape is a person who has the trait; a carrier is drawn unshaded.

Why: A carrier carries an allele without showing its trait, so a carrier is drawn unshaded.

44
Practice writing an answer

A pedigree records a recessive trait in a family.

(a) Describe how a carrier appears on the pedigree, and explain why. (1 pt)

Model answer A carrier appears unshaded.
A pedigree records what each person shows.
A carrier does not show the trait.
Rubric
  • Award 1 point for: unshaded, because a pedigree records phenotypes and a carrier does not show the trait.
45
Check q12

A family's pedigree for a recessive trait is drawn below. II-1 is unshaded.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Generation II, their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle; II-3 is joined to II-4, an unshaded square. Generation III, the children of II-3 and II-4: III-1 a shaded square and III-2 an unshaded circle

Does II-1 carry the trait's allele?

  1. A. Yes
    A pedigree records phenotypes.
    Unshaded says only that II-1 does not show the trait.
    A carrier looks the same as anyone else without the trait.
  2. B. No
    An unshaded person can still carry the allele unseen.
    The drawing shows what II-1 looks like, not II-1's alleles.
  3. C. ✓ Cannot tell

Why: A pedigree records phenotypes.
Unshaded means II-1 does not show the trait.
A carrier is unshaded too, so the drawing alone cannot say whether II-1 carries the allele.

46
Check q13

A family's pedigree for a recessive trait is drawn below. I-2 is unshaded.

Two unshaded parents, I-1 and I-2, with three children: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle
Two unshaded parents, I-1 and I-2, with three children: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle

Is I-2 a carrier?

  1. A. ✓ Yes
  2. B. No
    I-2's child II-2 is shaded, so II-2 received the trait's allele from I-2.
    I-2 is unshaded, so I-2 carries the allele without showing the trait.
  3. C. Cannot tell
    The pedigree does fix I-2: her shaded child II-2 received the trait's allele from her.

Why: II-2 is shaded, so II-2 received the trait's allele from each parent.
I-2 is a parent of II-2, so I-2 carries the allele.
I-2 is unshaded, so I-2 does not show the trait.
So I-2 is a carrier.

47
Check q14

A family's pedigree for a recessive trait is drawn below. II-2 is shaded.

Two unshaded parents, I-1 and I-2, with three children: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle
Two unshaded parents, I-1 and I-2, with three children: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle

Is II-2 a carrier?

  1. A. Yes
    A carrier carries the allele without showing the trait.
    II-2 shows the trait.
  2. B. ✓ No
  3. C. Cannot tell
    II-2 is shaded, so II-2 shows the trait, and the pedigree says so.

Why: II-2 is shaded, so II-2 shows the trait.
A carrier carries the allele without showing the trait.
So II-2 is not a carrier: II-2 has the trait.

48
Check q15

A family's pedigree for a recessive trait is drawn below. I-1 is shaded and his children II-1 and II-2 are unshaded.

A shaded father, I-1, and an unshaded mother, I-2, with two children: II-1 an unshaded square and II-2 an unshaded circle
A shaded father, I-1, and an unshaded mother, I-2, with two children: II-1 an unshaded square and II-2 an unshaded circle

Is II-1 a carrier?

  1. A. ✓ Yes
  2. B. No
    I-1 has the trait, so every allele he passes on is the trait's allele.
    II-1 received one allele from I-1.
  3. C. Cannot tell
    The pedigree does fix II-1: his father I-1 has the trait and gave him the trait's allele.

Why: I-1 is shaded, so I-1 has the recessive trait and both of his alleles are the trait's allele.
II-1 received one allele from I-1, so II-1 carries the trait's allele.
II-1 is unshaded, so he does not show the trait.
So II-1 is a carrier.

49Mixed practice mixed practice

50
Check q16

On a pedigree, II-4 is an unshaded circle.

What does that tell you about II-4?

  1. A. ✓ A female in the second generation who does not have the trait
  2. B. A male in the second generation who does not have the trait
    A circle is a female; a square would be a male.
  3. C. A female in the second generation who carries no allele for the trait
    A pedigree records phenotypes: unshaded says only that she does not have the trait, and she could be a carrier.
  4. D. The fourth child of the family's first couple, who has the trait
    Unshaded means she does not have the trait, and fourth from the left is her place in the row, not who her parents are.

Why: A circle is a female, row II is the second generation, and an unshaded shape is a person who does not show the trait.
Whether she carries the allele the drawing cannot say.

51
Check q17

A family's pedigree for a recessive trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle

Who are III-2's parents?

  1. A. I-1 and I-2
    I-1 and I-2 are the couple in row I, and their vertical line drops to row II.
  2. B. ✓ II-3 and II-4
  3. C. II-1 and II-2
    II-1 and II-2 are brother and sister on one sibling line, and no horizontal line joins them as parents.

Why: The line above III-2 rises to the horizontal line joining II-3 and II-4.
A horizontal line joins two parents, so II-3 and II-4 are III-2's parents.

52
Check q18

A family's pedigree for a recessive trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle

Which child of I-1 and I-2 was born last?

  1. A. II-1
    II-1 is the leftmost child, so the first born.
  2. B. II-2
    II-2 is second from the left, so the second born.
  3. C. ✓ II-3

Why: Children sit on their sibling line in birth order from left to right.
II-3 is the rightmost child of I-1 and I-2, so II-3 was born last.

53
Check q19

A family's pedigree for a recessive trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle

Which people must be carriers?

  1. A. I-1 and I-2 only
    III-2 is shaded and his parents II-3 and II-4 are unshaded.
    Each gave III-2 the trait's allele, so both are carriers too.
  2. B. II-1, II-3 and II-4
    II-1 is unshaded and has no shaded child, so nothing fixes II-1 as a carrier.
    I-1 and I-2 gave the shaded II-2 the allele, so they are carriers.
  3. C. ✓ I-1, I-2, II-3 and II-4
  4. D. Everyone who is unshaded
    Some unshaded people carry the allele and some do not.
    The pedigree fixes a carrier only through a shaded child.

Why: II-2 and III-2 are shaded, so each received the trait's allele from both parents.
II-2's parents are I-1 and I-2; III-2's parents are II-3 and II-4.
All four are unshaded, so all four are carriers.

54
Check q20

A family's pedigree for a recessive trait is drawn below. III-2 is shaded.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle

Is III-2 a carrier?

  1. A. Yes
    A carrier carries the allele without showing the trait.
    III-2 shows the trait.
  2. B. ✓ No
  3. C. Cannot tell
    III-2 is shaded, so III-2 shows the trait, and the pedigree says so.

Why: III-2 is shaded, so III-2 shows the trait.
A carrier does not show the trait.
So III-2 is not a carrier.

55
Check q21

A family's pedigree for a recessive trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle

Who is II-3's partner?

  1. A. II-2
    II-2 is II-3's sister: both hang from the sibling line under I-1 and I-2.
  2. B. ✓ II-4
  3. C. I-2
    I-2 is II-3's mother, in the row above.

Why: A horizontal line joins two parents.
II-3 is joined to II-4 by one, and their children hang below, so II-4 is II-3's partner.

56
Practice writing an answer

A family's pedigree for a recessive trait is drawn below.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded square; II-3 is joined to II-4, an unshaded circle. The children of II-3 and II-4: III-1 an unshaded square, III-2 a shaded square, III-3 an unshaded circle

(a) Identify every person the pedigree fixes as a carrier, and explain how it fixes them. (1 pt)

Model answer I-1, I-2, II-3 and II-4 are carriers.
II-2 is shaded, so II-2 received the trait's allele from I-1 and from I-2.
III-2 is shaded, so III-2 received the trait's allele from II-3 and from II-4.
All four parents are unshaded, so each carries the allele without showing the trait.
Rubric
  • Award 1 point for: I-1, I-2, II-3 and II-4, because each is an unshaded parent of a shaded child (II-2, or III-2) and so gave that child the trait's allele without showing the trait.

Slip Naming every unshaded person. Unshaded alone fixes nothing; only a shaded child fixes a parent as a carrier.

(b) III-3 is unshaded. Describe what the pedigree tells you about whether III-3 carries the trait's allele. (1 pt)

Model answer The pedigree cannot say.
III-3 is unshaded, so III-3 does not show the trait.
A carrier is unshaded too.
III-3 has no child on the pedigree, so nothing fixes whether III-3 carries the allele.
Rubric
  • Award 1 point for: the pedigree cannot say; unshaded means only that III-3 does not show the trait, and a carrier is also unshaded.

Slip Saying III-3 carries no allele because she is unshaded. Unshaded rules out having the trait, nothing more.

Glossary

pedigree
A family drawn as a record of who has a trait, generation by generation: squares are males, circles females, a shaded shape has the trait, a horizontal line joins parents, a vertical line drops to their children in birth order, and rows are generations I, II, III.
carrier
A person who carries an allele without showing its trait. On a pedigree a carrier is unshaded.

APBIO-U05-L16A Fix the genotypes; dominant or recessive?

Topic 5.3 · Mendelian Genetics · 73 steps

A family tree of three rows with every person numbered: I-1, a square, and I-2, a circle, both unshaded; their children II-1, a shaded square, II-2, an unshaded circle, and II-3, an unshaded square; II-3 joined to II-4, an unshaded circle; their children III-1, a shaded circle, and III-2, an unshaded square. Text beside it asks which genotypes the drawing fixes
A family tree of three rows with every person numbered: I-1, a square, and I-2, a circle, both unshaded; their children II-1, a shaded square, II-2, an unshaded circle, and II-3, an unshaded square; II-3 joined to II-4, an unshaded circle; their children III-1, a shaded circle, and III-2, an unshaded square. Text beside it asks which genotypes the drawing fixes

Now consider the same family with everyone numbered. The trait shows in II-1 and III-1, and neither has a parent who shows it.

Write r for the trait’s allele and R for the other allele of the gene. Which people’s genotypes does this pedigree fix, and which does it leave open?

Unit 5 · Heredity

1On one of the 22 ordinary pairs

2

Video: Watch: On one of the 22 ordinary pairs

A karyotype with the 22 pairs of autosomes ringed and the X and Y set apart; a gene placed on an autosome and its trait named autosomal recessive, then autosomal dominant; a gene placed on the Y and its trait named not autosomal.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Aa.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Aa.mp4

3
Check q1

A human body cell holds 23 pairs of chromosomes. In 22 of the pairs the two members match; the last pair is the X and the Y.

What is a chromosome from one of the 22 matching pairs called?

  1. A. ✓ An autosome
  2. B. A sex chromosome
    The sex chromosomes are the X and the Y, the one pair whose members need not match.

Why: A chromosome that belongs to one of the 22 pairs whose members match is an autosome.
The X and the Y are the sex chromosomes.

4

Now think about where a trait's gene sits. Most genes sit on one of the 22 pairs of autosomes; a few sit on the X or the Y.

5

When a recessive trait's gene sits on an autosome, we call the trait , because its allele is recessive and its chromosome is an autosome.

6

For example, albinism, the absence of pigment in skin, hair and eyes, is autosomal recessive. Its gene sits on an autosome, and the trait shows only in a person with two copies of its allele.

7

When a dominant trait's gene sits on an autosome, we call the trait .

8

For example, six fingers on a hand, polydactyly, is autosomal dominant. Its gene sits on an autosome, and one copy of its allele is enough to show the trait.

9

But suppose a gene sits on the Y chromosome. Its trait is not autosomal, because the Y is a sex chromosome.

10

Here is a table comparing the three cases: where the gene sits, how many copies of the allele show the trait, and the name.

A table with four columns, trait, where its gene sits, copies of the allele needed to show it, and name, and three rows: albinism, an autosome, two, autosomal recessive; polydactyly (six fingers), an autosome, one, autosomal dominant; a trait whose gene sits on the Y, the Y, a sex chromosome, one, not autosomal
A table with four columns, trait, where its gene sits, copies of the allele needed to show it, and name, and three rows: albinism, an autosome, two, autosomal recessive; polydactyly (six fingers), an autosome, one, autosomal dominant; a trait whose gene sits on the Y, the Y, a sex chromosome, one, not autosomal
11

Every pedigree that follows is for an autosomal trait.

12

What you are expected to know Classify a trait as autosomal recessive or autosomal dominant from where its gene sits and how many copies of its allele show it.

13Quick quiz: autosomal recessive, autosomal dominant mixed practice

14
Check q2

A trait is named autosomal recessive.

What does the word autosomal tell you?

  1. A. ✓ The trait's gene sits on an autosome, one of the 22 matching pairs
  2. B. The trait's gene sits on the X or the Y, the pair of sex chromosomes
    The X and the Y are the sex chromosomes; autosomal means the gene sits on one of the 22 matching pairs.

Why: Autosomal means the trait's gene sits on an autosome, one of the 22 pairs of chromosomes whose members match.

15
Check q3

A trait is named autosomal recessive.

What does the word recessive tell you?

  1. A. One copy of the trait's allele is enough to show the trait
    One copy being enough is what dominant means.
  2. B. ✓ The trait shows only in a person with two copies of its allele

Why: Recessive means the trait shows only in a person with two copies of its allele.
A person with one copy shows the dominant trait instead.

16
Check q4

A trait is autosomal dominant.

How many copies of its allele does a person need to show the trait?

  1. A. ✓ One
  2. B. Two
    Two copies needed is what recessive means; a dominant trait shows with one.

Why: Dominant means one copy of the allele is enough to show the trait.

17
Practice writing an answer

A trait is described as autosomal dominant.

(a) State what the word autosomal tells you about the trait's gene. (1 pt)

Model answer The trait's gene sits on an autosome, one of the 22 pairs of chromosomes whose members match.
Rubric
  • Award 1 point for: the gene sits on an autosome (one of the 22 ordinary or matching pairs), not on the X or the Y.
18
Check q5

A recessive trait's gene sits on chromosome 15, one of the 22 matching pairs.

Which name fits the trait?

  1. A. ✓ Autosomal recessive
  2. B. Autosomal dominant
    The trait is recessive, so it is not a dominant trait.
  3. C. Not autosomal
    Chromosome 15 is one of the 22 matching pairs, so it is an autosome.

Why: Chromosome 15 is one of the 22 matching pairs, so it is an autosome.
The trait is recessive.
So the trait is autosomal recessive.

19
Check q6

A dominant trait's gene sits on the X chromosome.

Which name fits the trait?

  1. A. Autosomal recessive
    The X is a sex chromosome, not an autosome, and the trait is dominant.
  2. B. Autosomal dominant
    The X is a sex chromosome, not an autosome.
  3. C. ✓ Not autosomal

Why: The X is a sex chromosome, not one of the 22 matching pairs.
So the trait is not autosomal.

20
Check q7

A dominant trait's gene sits on chromosome 4, one of the 22 matching pairs.

Which name fits the trait?

  1. A. Autosomal recessive
    The trait is dominant, so it is not a recessive trait.
  2. B. ✓ Autosomal dominant
  3. C. Not autosomal
    Chromosome 4 is one of the 22 matching pairs, so it is an autosome.

Why: Chromosome 4 is one of the 22 matching pairs, so it is an autosome.
The trait is dominant.
So the trait is autosomal dominant.

21
Check q8

A trait shows only in people with two copies of its allele. Its gene sits on chromosome 9, one of the 22 matching pairs.

Which name fits the trait?

  1. A. ✓ Autosomal recessive
  2. B. Autosomal dominant
    A trait that needs two copies of its allele to show is recessive, not dominant.
  3. C. Not autosomal
    Chromosome 9 is one of the 22 matching pairs, so it is an autosome.

Why: A trait that shows only with two copies of its allele is recessive.
Chromosome 9 is an autosome.
So the trait is autosomal recessive.

22Fix the genotypes you can

23

Video: Watch: Fix the genotypes you can

The numbered family; rr written under each shaded person, Rr under each unshaded parent of an rr child, a question mark under each unshaded child of two carriers; then the sheep record worked one animal at a time.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Ab.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Ab.mp4

24
Check q9

Two unshaded parents have a shaded child with a recessive trait.

Which of the following describes the two parents?

  1. A. ✓ Both parents are carriers
  2. B. Neither parent carries the trait's allele
    The shaded child received the trait's allele from each parent, so both carry it.
  3. C. One parent is a carrier and the other carries no allele for the trait
    The shaded child has two copies of the trait's allele, one from each parent, so both parents carry it.

Why: The shaded child received the trait's allele from each parent.
Neither parent shows the trait.
So both parents are carriers.

25

Now consider the family of the opening page, with everyone numbered. The trait shows in II-1 and III-1, and neither has a parent who shows it.

The opening family drawn as a pedigree with its people numbered: I-1 and I-2 unshaded; their children II-1, a shaded square, II-2 an unshaded circle and II-3 an unshaded square; II-3 joined to II-4, an unshaded circle; their children III-1, a shaded circle, and III-2, an unshaded square
The opening family drawn as a pedigree with its people numbered: I-1 and I-2 unshaded; their children II-1, a shaded square, II-2 an unshaded circle and II-3 an unshaded square; II-3 joined to II-4, an unshaded circle; their children III-1, a shaded circle, and III-2, an unshaded square
26

Write r for the trait’s allele and R for the other allele of the gene. The trait is autosomal recessive, so a person shows it when both alleles are r.

27

Here are the deductions, one kind of person at a time:

  1. A shaded person has the recessive trait, so is rr: II-1 and III-1.
  2. An unshaded parent of an rr child gave that child an r, and is unshaded, so is a carrier, Rr: I-1, I-2, II-3 and II-4.
  3. An unshaded child of two carriers could be RR or Rr, and the drawing cannot say which: mark it ?.

28

Unshaded fixes only one thing: the person is not rr. It never tells you on its own whether the person is RR or Rr.

29

Here is the same pedigree with a genotype under each person. II-2 and III-2 are the two people the drawing leaves undecided.

The same pedigree with a genotype under each shape: I-1 Rr, I-2 Rr, II-1 rr, II-2 a question mark, II-3 Rr, II-4 Rr, III-1 rr, III-2 a question mark
The same pedigree with a genotype under each shape: I-1 Rr, I-2 Rr, II-1 rr, II-2 a question mark, II-3 Rr, II-4 Rr, III-1 rr, III-2 a question mark
30

What you are expected to know Deduce the genotypes an autosomal recessive pedigree fixes: shaded is rr, an unshaded parent of an rr child is Rr, and an unshaded child of two carriers is RR or Rr, marked undecided.

31
Check q10

Suppose a trait in a flock of sheep is autosomal recessive: rr sheep show it, RR and Rr sheep do not. The flock's breeding record is drawn below. II-1 is shaded.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children: II-1, a shaded circle, and II-2, an unshaded square; II-2 is joined to II-3, an unshaded circle. Their children: III-1, a shaded square, and III-2, an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children: II-1, a shaded circle, and II-2, an unshaded square; II-2 is joined to II-3, an unshaded circle. Their children: III-1, a shaded square, and III-2, an unshaded circle

Which genotype does II-1 have?

  1. A. RR
    An RR sheep carries no r, so it cannot show a recessive trait.
    II-1 is shaded, so II-1 shows the trait.
  2. B. Rr
    An Rr sheep carries an R, and R shows, so an Rr sheep does not show the trait.
    II-1 is shaded.
  3. C. ✓ rr

Why: II-1 is shaded, so II-1 shows the trait.
The trait is recessive, so it shows when both alleles are r.
So II-1 is rr.

32
Check q11

Suppose a trait in a flock of sheep is autosomal recessive: rr sheep show it, RR and Rr sheep do not. The flock's breeding record is drawn below. I-1 is unshaded, and I-1 is a parent of II-1, who is shaded.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children: II-1, a shaded circle, and II-2, an unshaded square; II-2 is joined to II-3, an unshaded circle. Their children: III-1, a shaded square, and III-2, an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children: II-1, a shaded circle, and II-2, an unshaded square; II-2 is joined to II-3, an unshaded circle. Their children: III-1, a shaded square, and III-2, an unshaded circle

Which genotype does I-1 have? Choose RR or Rr if the record leaves it open.

  1. A. RR
    II-1 is rr and received one r from each parent.
    So I-1 gave II-1 an r, and an RR sheep has no r to give.
  2. B. ✓ Rr
  3. C. RR or Rr
    The record does fix I-1. II-1 is rr, so I-1 gave II-1 an r.
    I-1 is unshaded, so I-1 also carries an R.

Why: II-1 is rr, so II-1 received an r from each parent.
So I-1 gave II-1 an r.
I-1 is unshaded, so I-1 also carries an R.
So I-1 is a carrier, Rr.

33
Check q12

Suppose a trait in a flock of sheep is autosomal recessive: rr sheep show it, RR and Rr sheep do not. The flock's breeding record is drawn below. III-2 is unshaded, and both of III-2's parents are carriers, Rr.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children: II-1, a shaded circle, and II-2, an unshaded square; II-2 is joined to II-3, an unshaded circle. Their children: III-1, a shaded square, and III-2, an unshaded circle
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children: II-1, a shaded circle, and II-2, an unshaded square; II-2 is joined to II-3, an unshaded circle. Their children: III-1, a shaded square, and III-2, an unshaded circle

Which genotype does III-2 have? Choose RR or Rr if the record leaves it open.

  1. A. ✓ RR or Rr
  2. B. RR
    An unshaded child of two carriers could have received R from one parent and r from the other.
    Unshaded rules out rr only.
  3. C. Rr
    An unshaded child of two carriers could have received an R from each parent.
    Unshaded rules out rr only.

Why: III-2 is unshaded, so III-2 is not rr.
Each Rr parent could have given III-2 an R or an r.
So III-2 is RR or Rr, and the record cannot say which.

34
Check q13

A family's pedigree for an autosomal recessive trait is drawn below. R is the other allele and r is the trait's allele.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded circle, II-2 a shaded square, II-3 an unshaded circle; II-3 is joined to II-4, a shaded square. The children of II-3 and II-4: III-1 an unshaded circle and III-2 an unshaded square
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded circle, II-2 a shaded square, II-3 an unshaded circle; II-3 is joined to II-4, a shaded square. The children of II-3 and II-4: III-1 an unshaded circle and III-2 an unshaded square

Which genotype does III-1 have? Choose RR or Rr if the pedigree leaves it open.

  1. A. RR
    III-1's father II-4 is shaded, so he is rr and gave III-1 an r.
    An RR person has no r.
  2. B. ✓ Rr
  3. C. RR or Rr
    The pedigree does fix III-1: her shaded father II-4 gave her an r, and she is unshaded, so she also carries an R.

Why: II-4 is shaded, so II-4 is rr and gives every child an r.
So III-1 received an r from II-4.
III-1 is unshaded, so III-1 also carries an R.
So III-1 is Rr.

35
Check q14

A family's pedigree for an autosomal recessive trait is drawn below. R is the other allele and r is the trait's allele.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded circle, II-2 a shaded square, II-3 an unshaded circle; II-3 is joined to II-4, a shaded square. The children of II-3 and II-4: III-1 an unshaded circle and III-2 an unshaded square
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, an unshaded circle. Their children left to right: II-1 an unshaded circle, II-2 a shaded square, II-3 an unshaded circle; II-3 is joined to II-4, a shaded square. The children of II-3 and II-4: III-1 an unshaded circle and III-2 an unshaded square

Which genotype does II-3 have? Choose RR or Rr if the pedigree leaves it open.

  1. A. RR
    II-3 is an unshaded child of two carriers, I-1 and I-2.
    She could have received an r from one of them.
  2. B. Rr
    II-3 is unshaded, and her children are unshaded too.
    Nothing on the pedigree shows that she received an r.
  3. C. ✓ RR or Rr

Why: II-3 is unshaded, so II-3 is not rr.
Her parents I-1 and I-2 are carriers, so she could be RR or Rr.
Her children III-1 and III-2 are unshaded, so they do not fix her either.
So II-3 is RR or Rr.

36Quick quiz: fix the genotype mixed practice

37
Check q15

A family's pedigree for an autosomal recessive trait is drawn below. R is the other allele and r is the trait's allele.

A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle

Which genotype does I-1 have?

  1. A. RR
    I-1 is shaded, so I-1 shows the recessive trait.
    An RR person does not show it.
  2. B. Rr
    I-1 is shaded, so I-1 shows the recessive trait.
    An Rr person does not show it.
  3. C. ✓ rr
  4. D. RR or Rr
    I-1 is shaded, and a shaded person is fixed as rr.

Why: I-1 is shaded, so I-1 shows the trait.
A recessive trait shows when both alleles are r.
So I-1 is rr.

38
Check q16

A family's pedigree for an autosomal recessive trait is drawn below. R is the other allele and r is the trait's allele.

A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle

Which genotype does II-1 have? Choose RR or Rr if the pedigree leaves it open.

  1. A. RR
    II-1's father I-1 is shaded, so he is rr and gave II-1 an r.
    An RR person has no r.
  2. B. ✓ Rr
  3. C. rr
    II-1 is unshaded, so II-1 does not show the trait and is not rr.
  4. D. RR or Rr
    The pedigree does fix II-1: his shaded father gave him an r, and he is unshaded, so he also carries an R.

Why: I-1 is shaded, so I-1 is rr and gives every child an r.
II-1 received an r from I-1.
II-1 is unshaded, so II-1 also carries an R.
So II-1 is Rr.

39
Check q17

A family's pedigree for an autosomal recessive trait is drawn below. R is the other allele and r is the trait's allele.

A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle

Which genotype does I-2 have? Choose RR or Rr if the pedigree leaves it open.

  1. A. RR
    I-2 could also be Rr: her children are unshaded, so each received an R from her, but that does not rule out an r as her other allele.
  2. B. Rr
    Nothing on the pedigree shows that I-2 carries an r: her children's r came from their shaded father.
  3. C. rr
    I-2 is unshaded, so I-2 is not rr.
  4. D. ✓ RR or Rr

Why: I-2 is unshaded, so I-2 is not rr.
Her children II-1 and II-2 are unshaded, so each received an R from her.
Their r came from their shaded father I-1, so nothing shows whether I-2 also carries an r.
So I-2 is RR or Rr.

40
Check q18

A family's pedigree for an autosomal recessive trait is drawn below. R is the other allele and r is the trait's allele.

A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle

Which genotype does II-3 have? Choose RR or Rr if the pedigree leaves it open.

  1. A. RR
    II-3's child III-1 is shaded, so she is rr and received an r from II-3.
    An RR person has no r.
  2. B. ✓ Rr
  3. C. rr
    II-3 is unshaded, so II-3 does not show the trait and is not rr.
  4. D. RR or Rr
    The pedigree does fix II-3: his shaded child III-1 received an r from him, and he is unshaded, so he also carries an R.

Why: III-1 is shaded, so III-1 is rr and received an r from each parent.
So II-3 gave III-1 an r.
II-3 is unshaded, so II-3 also carries an R.
So II-3 is Rr.

41
Check q19

A family's pedigree for an autosomal recessive trait is drawn below. R is the other allele and r is the trait's allele.

A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle
A three-generation pedigree. Generation I: I-1, a shaded square, and I-2, an unshaded circle. Their children: II-1 an unshaded square and II-2 an unshaded circle; II-2 is joined to II-3, an unshaded square. The children of II-2 and II-3: III-1 a shaded circle, III-2 an unshaded square, III-3 an unshaded circle

Which genotype does III-3 have? Choose RR or Rr if the pedigree leaves it open.

  1. A. RR
    III-3 is an unshaded child of two carriers, II-2 and II-3.
    She could have received an r from one of them.
  2. B. Rr
    III-3 is an unshaded child of two carriers.
    She could have received an R from each of them.
  3. C. rr
    III-3 is unshaded, so III-3 is not rr.
  4. D. ✓ RR or Rr

Why: III-3 is unshaded, so III-3 is not rr.
Her parents II-2 and II-3 are both carriers, Rr, because their child III-1 is rr.
Each parent could have given III-3 an R or an r.
So III-3 is RR or Rr.

42Dominant or recessive?

43

Video: Watch: Dominant or recessive?

The two decisive families side by side, the deciding child ringed in each; then a pedigree with the trait in every generation, and a carrier marrying in at each generation to show that a recessive trait can do the same.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Ac.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Ac.mp4

44

Now suppose you are handed a pedigree for an autosomal trait and nothing more. Is the trait dominant or recessive?

45

Two kinds of family decide it. Here they are side by side.

Two pedigrees side by side. Left: two shaded parents, I-1 and I-2, with three children of whom the middle one, II-2, is unshaded; caption: shaded parents, unshaded child: dominant. Right: two unshaded parents with three children of whom the middle one, II-2, is shaded; caption: unshaded parents, shaded child: recessive
Two pedigrees side by side. Left: two shaded parents, I-1 and I-2, with three children of whom the middle one, II-2, is unshaded; caption: shaded parents, unshaded child: dominant. Right: two unshaded parents with three children of whom the middle one, II-2, is shaded; caption: unshaded parents, shaded child: recessive
46

For example, on the left, two shaded parents have an unshaded child. This trait is dominant, because two parents with a recessive trait would both be rr and every one of their children would be shaded too.

47

But on the right, two unshaded parents have a shaded child. This trait is recessive, because a dominant trait shows in every person who carries its allele, and neither parent shows it.

48

So the two families are a yes/no pair. Shaded parents with an unshaded child means dominant, and unshaded parents with a shaded child means recessive.

49

Here is a pedigree in which the trait appears in every generation. That pattern is consistent with a dominant trait.

A three-generation pedigree in which the trait appears in every generation: I-1 shaded, I-2 unshaded; their children II-1 shaded, II-2 unshaded, II-3 shaded; II-3 joined to II-4, unshaded; their children III-1 shaded and III-2 unshaded
A three-generation pedigree in which the trait appears in every generation: I-1 shaded, I-2 unshaded; their children II-1 shaded, II-2 unshaded, II-3 shaded; II-3 joined to II-4, unshaded; their children III-1 shaded and III-2 unshaded
50

It does not prove one. A recessive trait can appear in every generation too, when a carrier marries into the family at each generation.

51

So "it skips a generation" and "it is in every generation" are hints, not proofs. Look for the family that rules one of the two out: shaded parents with an unshaded child, or unshaded parents with a shaded child.

52

What you are expected to know Judge from a pedigree whether a trait is dominant or recessive by finding the decisive family: shaded parents with an unshaded child means dominant, unshaded parents with a shaded child means recessive.

53
Check q20

The pedigree below records an autosomal trait in one family.

A two-generation pedigree: I-1, a shaded square, and I-2, a shaded circle, have three children: II-1 a shaded circle, II-2 an unshaded square, II-3 a shaded square
A two-generation pedigree: I-1, a shaded square, and I-2, a shaded circle, have three children: II-1 a shaded circle, II-2 an unshaded square, II-3 a shaded square

Which people decide whether the trait is dominant or recessive?

  1. A. No one here
    Two generations are enough when they hold a decisive family, and this one does: two shaded parents with an unshaded child.
  2. B. I-1 and I-2 with their shaded children
    Shaded children of shaded parents fit dominant and recessive alike, so they decide nothing.
  3. C. ✓ I-1 and I-2 with II-2

Why: I-1 and I-2 both show the trait, and their child II-2 does not.
That can happen only if the trait is dominant and both parents are heterozygous, each passing II-2 the recessive allele.

54
Practice writing an answer

The pedigree below records an autosomal trait in one family.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have three children: II-1 an unshaded square, II-2 an unshaded circle, II-3 a shaded circle
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have three children: II-1 an unshaded square, II-2 an unshaded circle, II-3 a shaded circle

(a) Determine whether the trait is dominant or recessive, naming the people whose shading decides it. (1 pt)

Model answer The trait is recessive.
I-1 and I-2 do not show the trait, and their child II-3 does.
A dominant trait shows in every person who carries its allele.
If the trait were dominant, the parent who passed II-3 the allele would show it too, and neither parent does.
So the trait is recessive, and I-1 and I-2 are both carriers who each passed II-3 the recessive allele.
Rubric
  • Award 1 point for: recessive, resting on I-1 and I-2 both unshaded while their child II-3 is shaded; two unshaded parents can have a shaded child only if the trait is recessive and both parents are carriers.
55

Back to the family of the opening page, numbered I-1 to III-2, with II-1 and III-1 shaded and every parent unshaded. Unshaded parents with a shaded child fixed the trait as recessive and the four parents as carriers, Rr.

The same pedigree with a genotype under each shape: I-1 Rr, I-2 Rr, II-1 rr, II-2 a question mark, II-3 Rr, II-4 Rr, III-1 rr, III-2 a question mark
The same pedigree with a genotype under each shape: I-1 Rr, I-2 Rr, II-1 rr, II-2 a question mark, II-3 Rr, II-4 Rr, III-1 rr, III-2 a question mark
56

The two shaded people are rr. II-2 and III-2 stay undecided, RR or Rr: the pedigree fixes what it can and no more.

57Quick quiz: dominant, recessive or cannot tell? mixed practice

58
Check q21

A family's pedigree for an autosomal trait is drawn below.

Two shaded parents, I-1 and I-2, with three children: II-1 an unshaded square, II-2 a shaded circle, II-3 a shaded circle
Two shaded parents, I-1 and I-2, with three children: II-1 an unshaded square, II-2 a shaded circle, II-3 a shaded circle

Is the trait dominant or recessive?

  1. A. ✓ Dominant
  2. B. Recessive
    If the trait were recessive, both shaded parents would be homozygous for it, and every child would be shaded.
    II-1 is unshaded.
  3. C. This family does not decide it
    This family is decisive: two shaded parents, I-1 and I-2, have an unshaded child, II-1.

Why: I-1 and I-2 both show the trait, and their child II-1 does not.
Two shaded parents can have an unshaded child in one case: the trait is dominant and both parents are heterozygous.

59
Check q22

A family's pedigree for an autosomal trait is drawn below.

Two unshaded parents, I-1 and I-2, with three children: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle
Two unshaded parents, I-1 and I-2, with three children: II-1 an unshaded square, II-2 a shaded circle, II-3 an unshaded circle

Is the trait dominant or recessive?

  1. A. Dominant
    A dominant trait shows in everyone who carries the allele.
    I-1 and I-2 are unshaded, so neither carries a dominant allele, yet II-2 has the trait.
  2. B. ✓ Recessive
  3. C. This family does not decide it
    This family is decisive: two unshaded parents, I-1 and I-2, have a shaded child, II-2.

Why: I-1 and I-2 do not show the trait, and their child II-2 does.
Two unshaded parents can have a shaded child in one case: the trait is recessive and both parents are carriers.

60
Check q23

A family's pedigree for an autosomal trait is drawn below.

A shaded father, I-1, and an unshaded mother, I-2, with three children: II-1 a shaded square, II-2 an unshaded circle, II-3 a shaded circle
A shaded father, I-1, and an unshaded mother, I-2, with three children: II-1 a shaded square, II-2 an unshaded circle, II-3 a shaded circle

Is the trait dominant or recessive?

  1. A. Dominant
    A shaded parent with an unshaded partner can have shaded and unshaded children whether the trait is dominant or recessive.
    Nothing here rules recessive out.
  2. B. Recessive
    The same children could come from a heterozygous shaded parent with a dominant allele.
    Nothing here rules dominant out.
  3. C. ✓ This family does not decide it

Why: The parents are one shaded and one unshaded, and the children are mixed.
That fits a dominant trait and a recessive trait alike.
Neither decisive family is here, so this family leaves dominant and recessive both possible.

61
Check q24

A family's pedigree for an autosomal trait is drawn below.

Two unshaded parents, I-1 and I-2, with two children: II-1 a shaded circle and II-2 a shaded square
Two unshaded parents, I-1 and I-2, with two children: II-1 a shaded circle and II-2 a shaded square

Is the trait dominant or recessive?

  1. A. Dominant
    I-1 and I-2 are unshaded.
    If the trait were dominant, neither parent would carry the allele, and no child could have it.
    Both children have it.
  2. B. ✓ Recessive
  3. C. This family does not decide it
    This family is decisive: two unshaded parents have shaded children.

Why: I-1 and I-2 do not show the trait, and both of their children do.
Two unshaded parents can have a shaded child in one case: the trait is recessive and both parents are carriers.

62
Check q25

A family's pedigree for an autosomal trait is drawn below.

Two shaded parents, I-1 and I-2, with three children, all shaded: II-1 a square, II-2 a circle, II-3 a square
Two shaded parents, I-1 and I-2, with three children, all shaded: II-1 a square, II-2 a circle, II-3 a square

Is the trait dominant or recessive?

  1. A. Dominant
    Two shaded parents with only shaded children fits a recessive trait too: two rr parents give every child an r from each.
    Nothing here rules recessive out.
  2. B. Recessive
    Two shaded parents with only shaded children fits a dominant trait too.
    Nothing here rules dominant out.
  3. C. ✓ This family does not decide it

Why: Two shaded parents have three shaded children.
A dominant trait gives that, and so does a recessive trait with two rr parents.
The decisive family would be shaded parents with an unshaded child, and there is none here.
So this family leaves dominant and recessive both possible.

63
Check q26

A family's pedigree for an autosomal trait is drawn below.

Two shaded parents, I-1 and I-2, with two children: II-1 a shaded circle and II-2 an unshaded circle
Two shaded parents, I-1 and I-2, with two children: II-1 a shaded circle and II-2 an unshaded circle

Is the trait dominant or recessive?

  1. A. ✓ Dominant
  2. B. Recessive
    If the trait were recessive, both shaded parents would be rr and every child would receive r from each of them.
    II-2 is unshaded.
  3. C. This family does not decide it
    This family is decisive: two shaded parents, I-1 and I-2, have an unshaded child, II-2.

Why: I-1 and I-2 both show the trait, and their child II-2 does not.
Two shaded parents can have an unshaded child in one case: the trait is dominant and both parents are heterozygous.

64
Check q27

A student looks at a pedigree in which the trait appears in every generation and says: “The trait must be dominant.”

Which statement about the student's claim is correct?

  1. A. ✓ The student is wrong. A trait in every generation can be dominant or recessive
  2. B. The student is right. A trait in every generation is dominant by definition
    A recessive trait can also appear in every generation, when a carrier marries in each time.
    So every generation is a hint, not a definition.
  3. C. The student is wrong. A trait in every generation is always recessive
    A trait in every generation is consistent with dominant.
    It simply does not prove it.
  4. D. The student is wrong. Dominant traits skip generations, so the trait is recessive
    Dominant traits do not skip generations: a shaded person has a shaded parent.

Why: A trait in every generation fits dominant, but a recessive trait can appear in every generation too.
Only a decisive family settles it: shaded parents with an unshaded child, or unshaded parents with a shaded child.

65Mixed practice mixed practice

66
Check q28

Six fingers on a hand, polydactyly, is an autosomal dominant trait. A family's pedigree for it is drawn below. F is the allele for six fingers and f is the other allele.

A two-generation pedigree: I-1, a shaded square, and I-2, a shaded circle, have three children: II-1 a shaded circle, II-2 an unshaded square, II-3 a shaded circle
A two-generation pedigree: I-1, a shaded square, and I-2, a shaded circle, have three children: II-1 a shaded circle, II-2 an unshaded square, II-3 a shaded circle

Which genotypes must I-1 and I-2 have?

  1. A. FF and FF
    Two FF parents can give only F to every child, so every child would have six fingers, and II-2 does not.
  2. B. Ff and ff
    I-2 is shaded and has the dominant trait, so she carries at least one F; an ff person has five fingers.
  3. C. ✓ Ff and Ff
  4. D. FF and Ff
    The unshaded child II-2 is ff and received an f from each parent, so neither parent can be FF.

Why: Both parents show the dominant trait, so each has at least one F.
Their child II-2 does not show it, so II-2 is ff and received an f from each parent.
So both parents are Ff.

67
Check q29

A trait appears in a child whose parents both lack it. That child, grown up, has four children with a partner who lacks the trait, and all four are unshaded. Write R for the other allele and r for the trait's allele.

Which statement about the trait and the four children is correct?

  1. A. The trait is recessive, and the four children may each be RR or Rr
    The shaded parent is rr and gives every child an r; each unshaded child therefore has one r and is Rr, a carrier, with no room for RR.
  2. B. The trait is dominant, and the four children are homozygous for the other allele, RR
    A child whose parents both lack the trait shows a recessive trait, not a dominant one.
  3. C. The trait is recessive, and the four children cannot carry the allele
    The shaded parent, rr, passes an r to every child, so all four carry the allele.
  4. D. ✓ The trait is recessive, and every one of the four children is a carrier

Why: Unshaded parents with a shaded child make the trait recessive and the child rr.
That rr parent gives every child an r, so each of the four unshaded children is Rr: a carrier.

68
Check q30

A trait's gene sits on chromosome 2, one of the 22 matching pairs, and the trait shows only in people with two copies of its allele.

Which name fits the trait?

  1. A. ✓ Autosomal recessive
  2. B. Autosomal dominant
    A trait that needs two copies of its allele to show is recessive, not dominant.
  3. C. Not autosomal
    Chromosome 2 is one of the 22 matching pairs, so it is an autosome.

Why: Chromosome 2 is an autosome.
A trait that shows only with two copies of its allele is recessive.
So the trait is autosomal recessive.

69
Check q31

Suppose the trait in the pedigree below is autosomal dominant, with D the allele for the trait and d the other allele.

A three-generation pedigree in which the trait appears in every generation: I-1 shaded, I-2 unshaded; their children II-1 shaded, II-2 unshaded, II-3 shaded; II-3 joined to II-4, unshaded; their children III-1 shaded and III-2 unshaded
A three-generation pedigree in which the trait appears in every generation: I-1 shaded, I-2 unshaded; their children II-1 shaded, II-2 unshaded, II-3 shaded; II-3 joined to II-4, unshaded; their children III-1 shaded and III-2 unshaded

Which genotype does II-2 have?

  1. A. DD
    A DD person shows a dominant trait, and II-2 is unshaded.
  2. B. Dd
    A Dd person shows a dominant trait, and II-2 is unshaded.
  3. C. ✓ dd

Why: II-2 is unshaded, so II-2 does not show the trait.
One D is enough to show a dominant trait.
So II-2 carries no D: II-2 is dd.

70
Check q32

A family's pedigree for six fingers is drawn below.

A two-generation pedigree: I-1, a shaded square, and I-2, a shaded circle, have three children: II-1 a shaded circle, II-2 an unshaded square, II-3 a shaded circle
A two-generation pedigree: I-1, a shaded square, and I-2, a shaded circle, have three children: II-1 a shaded circle, II-2 an unshaded square, II-3 a shaded circle

Which people decide whether six fingers is dominant or recessive?

  1. A. ✓ I-1 and I-2 with their unshaded child II-2
  2. B. I-1 and I-2 with their shaded children II-1 and II-3
    Shaded children of shaded parents fit dominant and recessive alike.
  3. C. No one here decides it
    Two shaded parents, I-1 and I-2, have an unshaded child, II-2.
    That family decides it.

Why: I-1 and I-2 both show the trait, and their child II-2 does not.
If the trait were recessive, both parents would be homozygous for it and every child would show it.
So II-2 shows the trait is dominant, and those three people decide it.

71
Check q33

On a pedigree for an autosomal recessive trait, an unshaded child has two carrier parents, Rr and Rr.

Which genotypes could the child have?

  1. A. RR only
    Each carrier parent could have given the child an r, so Rr is possible too.
  2. B. Rr only
    Each carrier parent could have given the child an R, so RR is possible too.
  3. C. ✓ RR or Rr
  4. D. rr
    The child is unshaded, so the child is not rr.

Why: The child is unshaded, so the child is not rr.
Each Rr parent could have given an R or an r.
So the child is RR or Rr, and the pedigree cannot say which.

72
Practice writing an answer

Albinism, the absence of pigment in skin, hair and eyes, is an autosomal recessive trait. The pedigree below records it in one family. Write A for the other allele and a for the albinism allele.

A three-generation pedigree. I-1 and I-2 unshaded; their children II-1 an unshaded circle, II-2 a shaded square, II-3 an unshaded circle; II-3 joined to II-4, an unshaded square; their children III-1 an unshaded square and III-2 a shaded circle
A three-generation pedigree. I-1 and I-2 unshaded; their children II-1 an unshaded circle, II-2 a shaded square, II-3 an unshaded circle; II-3 joined to II-4, an unshaded square; their children III-1 an unshaded square and III-2 a shaded circle

(a) Identify a family on the pedigree that shows the trait is recessive rather than dominant, and explain why it does. (1 pt)

Model answer I-1 and I-2 are both unshaded, and their child II-2 is shaded.
Two parents without the trait having a child with it can happen only for a recessive trait, with both parents carriers.
(II-3 and II-4 with their shaded daughter III-2 show the same thing.)
Rubric
  • Award 1 point for: naming unshaded parents (I-1 and I-2, or II-3 and II-4) with a shaded child (II-2, or III-2) as the decisive family, with the reason that a dominant trait would show in a parent who carried it; either family earns the point.

Slip Pointing to the trait skipping generation I. Skipping is a hint; the proof is the unshaded couple with a shaded child.

(b) Determine the genotypes of II-3 and II-4 from the pedigree. (1 pt)

Model answer II-3 and II-4 are both Aa.
Their child III-2 has albinism and is aa, so she received an a from each parent.
Neither parent shows the trait, so each also carries an A.
Rubric
  • Award 1 point for: the decision (II-3 and II-4 both Aa) AND the ground (their shaded child III-2 is aa and received an a from each of them; each is unshaded, so each also carries an A).

Slip Giving II-3 as AA because she is unshaded. Unshaded fixes only that she is not aa; her shaded child fixes her as Aa.

(c) II-2 has a child with a partner who is AA. Predict whether that child could have albinism, and justify your prediction. (1 pt)

Model answer No.
Every gamete from the AA partner carries A.
So every child receives an A and shows the ordinary pigment.
Each child is Aa, a carrier, and none is aa.
Rubric
  • Award 1 point for: no child can have albinism, because the AA parent gives every child an A (each child is Aa).

Slip Predicting a one-in-two chance because II-2 is aa. Every child gets an a from II-2, but also an A from the AA partner, so no child is aa.

(d) III-1 is unshaded. Describe what the pedigree fixes about his genotype and what it leaves open. (1 pt)

Model answer III-1 is unshaded, so he is not aa: he is AA or Aa.
He is the child of two carriers, so both are possible and the pedigree cannot say which.
Rubric
  • Award 1 point for: III-1 is AA or Aa (not aa), and the pedigree cannot decide between them.

Slip Writing III-1 as AA because he is unshaded and his sister has the trait. Unshaded rules out aa only; a child of two carriers may be AA or Aa.

Glossary

autosomal recessive
A recessive trait whose gene sits on an autosome, one of the 22 pairs of chromosomes whose members match. It shows only in a person with two copies of its allele, so two unshaded parents can have a shaded child.
autosomal dominant
A dominant trait whose gene sits on an autosome. One copy of its allele is enough to show it, so two shaded parents can have an unshaded child, but two unshaded parents cannot have a shaded one.

APBIO-U05-L16B The next child

Topic 5.3 · Mendelian Genetics · 36 steps

Two unshaded parents drawn as a square labeled II-3 Rr and a circle labeled II-4 Rr, joined by a line; below them a shaded circle labeled III-1 rr, and beside it a dashed outline with a question mark for the next child
Two unshaded parents drawn as a square labeled II-3 Rr and a circle labeled II-4 Rr, joined by a line; below them a shaded circle labeled III-1 rr, and beside it a dashed outline with a question mark for the next child

Here are II-3 and II-4, two unshaded parents whose daughter III-1 has a recessive trait. Her trait fixed both of them as carriers, Rr.

They plan another child. What is the probability that the child has the trait?

Unit 5 · Heredity

1Both parents fixed as carriers

2

Video: Watch: Both parents fixed as carriers

The two carriers' square drawn cell by cell with its one rr cell shaded; the multiplication rule written out; an r from each parent, one half times one half, one quarter.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Ba.mp4

3
Check q1

Two events must both happen: a child receives r from its mother and r from its father.

Which rule gives the probability that both happen?

  1. A. ✓ Multiply the two probabilities
  2. B. Add the two probabilities
    Adding is for either of two outcomes that cannot both happen; both together is multiplying.

Why: For two events that must both happen, multiply the two probabilities: the multiplication rule.

4

Once a pedigree has fixed both parents' genotypes, the next child's probability comes from their Punnett square and those two rules.

5

Here is the Punnett square for II-3 and II-4, both Rr. One cell of the four is rr.

The square for two carriers, Rr and Rr: R and r along the top and down the side; cells RR, Rr, Rr, rr, with the rr cell shaded
The square for two carriers, Rr and Rr: R and r along the top and down the side; cells RR, Rr, Rr, rr, with the rr cell shaded
6
Worked example

Two parents, II-3 and II-4, are both carriers, Rr. What is the probability that their next child has the trait?

Write down the values in the question:
II-3 = Rr, so P(r from II-3) = 1/2
II-4 = Rr, so P(r from II-4) = 1/2
a child with the trait = rr
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(rr)=P(r from II-3)×P(r from II-4)
P(rr)=12×12=14=0.25
A probability is a number between 0 and 1 and carries no unit.
7

What you are expected to know Calculate the probability that a couple's next child shows a trait, from the genotypes the pedigree fixes and the multiplication rule.

8
Check q2 numeric entry

A woman who is a carrier of an autosomal recessive trait, Rr, has a child with a man who has the trait, rr.

Calculate the probability, as a decimal, that the child has the trait.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(r from the mother, Rr) = 1/2
P(r from the father, rr) = 1
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(rr)=12×1=12=0.5

9One parent only probably a carrier

10

Video: Watch: One parent only probably a carrier

The Rr and Rr square with its rr cell crossed out and two of the three remaining cells marked Rr; the two-in-three chance written as one more factor; the whole product worked for an unshaded sister of a shaded brother.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L16Bb.mp4

11

Now suppose one parent's genotype is not fixed. The pedigree says only that the parent is probably a carrier.

12

The chance that the parent carries the allele is one more factor to multiply.

13

Here is the family tree again with every genotype written in. II-2 is an unshaded child of two carriers, so II-2 is RR or Rr.

The family tree with a genotype under each person: I-1 Rr, I-2 Rr, II-1 rr, II-2 RR or Rr, II-3 Rr, II-4 Rr, III-1 rr, III-2 RR or Rr
The family tree with a genotype under each person: I-1 Rr, I-2 Rr, II-1 rr, II-2 RR or Rr, II-3 Rr, II-4 Rr, III-1 rr, III-2 RR or Rr
14

Here is the Punnett square for II-2's parents, I-1 and I-2, both Rr. II-2 is unshaded, so the rr cell is ruled out.

The Rr and Rr square with the rr cell crossed out and the two Rr cells shaded among the three remaining cells; a caption reads: an unshaded child is one of these three cells, and two of the three are Rr
The Rr and Rr square with the rr cell crossed out and the two Rr cells shaded among the three remaining cells; a caption reads: an unshaded child is one of these three cells, and two of the three are Rr
15

The three cells left hold RR once and Rr twice. So the chance that II-2 is a carrier is 23.

16
Worked example

II-2, an unshaded child of two carriers, has a child with a partner who is a known carrier, Rr. What is the probability that the child has the trait?

Write down the values in the question:
P(II-2 is Rr) = 2/3
P(rr child if II-2 is Rr and the partner is Rr) = 1/2 × 1/2 = 1/4
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(child has the trait)=P(II-2 is Rr)×P(rr if II-2 is Rr)
P(child has the trait)=23×14=212=16≈0.167
17

What you are expected to know Calculate the probability when one parent's genotype is uncertain: multiply in that parent's chance of being a carrier as one more factor.

18
Check q3 numeric entry

In the pedigree below, II-1 has an autosomal recessive trait and his sister II-2 does not. II-2 has a child with II-3, who has the trait, rr.

A pedigree: I-1 and I-2, both unshaded, have two children: II-1, a shaded square, and II-2, an unshaded circle. II-2 is joined to II-3, a shaded square
A pedigree: I-1 and I-2, both unshaded, have two children: II-1, a shaded square, and II-2, an unshaded circle. II-2 is joined to II-3, a shaded square

Calculate the probability, as a decimal to three decimal places, that the first child of II-2 and II-3 has the trait.

Part 1. II-2 is an unshaded child of two carriers. What is the probability, as a decimal to three decimal places, that II-2 is a carrier, Rr?

Answer: 0.667  (tolerance ±0.0015)

Working
Count the unshaded cells of the Rr and Rr square:
unshaded cells = RR, Rr, Rr
P(II-2 is Rr)=23≈0.667

Part 2. If II-2 is Rr, what is the probability, as a decimal, that a child of II-2 and II-3 is rr?

Answer: 0.5  (tolerance ±0.005)

Working
Multiply the two parents' chances of passing r:
P(rr if II-2 is Rr)=12×1=12=0.5

Answer: 0.333  (tolerance ±0.0015)

Working
Write down the values in the question:
P(II-2 is Rr) = 2/3
P(rr child if II-2 is Rr) = 1/2 × 1 = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(child has the trait)=P(II-2 is Rr)×P(rr if II-2 is Rr)
P(child has the trait)=23×12=26=13≈0.333
19
Check q4 numeric entry

In the pedigree below, II-1 has an autosomal recessive trait and her brother II-2 does not. II-2 has a child with II-3, who has the trait, rr.

A pedigree: I-1 and I-2, both unshaded, have two children: II-1, a shaded circle, and II-2, an unshaded square. II-2 is joined to II-3, a shaded circle
A pedigree: I-1 and I-2, both unshaded, have two children: II-1, a shaded circle, and II-2, an unshaded square. II-2 is joined to II-3, a shaded circle

Calculate the probability, as a decimal to three decimal places, that the first child of II-2 and II-3 has the trait.

Answer: 0.333  (tolerance ±0.0015)

Working
Write down the values in the question:
P(II-2 is Rr) = 2/3 (an unshaded child of two carriers)
P(rr child if II-2 is Rr) = 1/2 × 1 = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(child has the trait)=P(II-2 is Rr)×P(rr if II-2 is Rr)
P(child has the trait)=23×12=26=13≈0.333
20

When both parents are only probably carriers, both chances multiply in: P(child has the trait)=P(mother is Rr)×P(father is Rr)×14.

21

Back to II-3 and II-4 of the opening page: two unshaded parents whose daughter III-1 has the trait. Her trait fixed both of them as carriers, Rr.

22

So their next child has the trait with probability 14: an r from each parent, 12×12.

23

When a parent is only probably a carrier, that chance multiplies in too.

24Quick quiz: the next child mixed practice

25
Check q5 numeric entry

Two parents are both carriers of an autosomal recessive trait, Rr and Rr.

Calculate the probability, as a decimal, that their next child is rr.

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
P(r from the mother, Rr) = 1/2
P(r from the father, Rr) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(rr)=12×12=14=0.25
26
Check q6 numeric entry

A man who is a carrier of an autosomal recessive trait, Rr, has a child with a woman who has the trait, rr.

Calculate the probability, as a decimal, that the child is rr.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(r from the father, Rr) = 1/2
P(r from the mother, rr) = 1
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(rr)=12×1=12=0.5
27
Check q7 numeric entry

A woman who is RR for an autosomal recessive trait has a child with a man who is a carrier, Rr.

Calculate the probability, as a decimal, that the child is rr.

Answer: 0  (tolerance ±0.005)

Working
Write down the values in the question:
P(r from the mother, RR) = 0
P(r from the father, Rr) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(rr)=0×12=0
28
Check q8 numeric entry

Two parents both have an autosomal recessive trait, rr and rr.

Calculate the probability, as a decimal, that their next child is rr.

Answer: 1  (tolerance ±0.005)

Working
Write down the values in the question:
P(r from the mother, rr) = 1
P(r from the father, rr) = 1
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(rr)=1×1=1
29
Check q9 numeric entry

Two parents are both carriers of an autosomal recessive trait, Rr and Rr.

Calculate the probability, as a decimal, that their next child does not have the trait.

Answer: 0.75  (tolerance ±0.005)

Working
Write down the values in the question:
square for Rr and Rr = RR, Rr, Rr, rr
a child without the trait = RR or Rr = 3 cells of 4
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(RR or Rr)=14+24=34=0.75

30Mixed practice mixed practice

31
Check q10 numeric entry

Two parents are both carriers of an autosomal recessive trait, Rr and Rr.

Calculate the probability, as a decimal, that their next child is a carrier, Rr.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
square for Rr and Rr = RR, Rr, Rr, rr
Rr = 2 cells of 4
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(Rr)=P(R from mother, r from father)+P(r from mother, R from father)
P(Rr)=14+14=12=0.5
32
Check q11

A woman's brother has an autosomal recessive trait; she and their parents do not. She has a child with a known carrier, Rr. A student calculates the probability that the child has the trait as one in four.

What is wrong with the student's calculation?

  1. A. Nothing is wrong with the calculation
    The woman is not a known carrier: she is an unshaded child of two carriers, RR or Rr.
  2. B. ✓ The woman's two-in-three chance of being a carrier is missing
  3. C. The known carrier passes r to every child
    A carrier, Rr, passes r to half of his children, not to every child.
  4. D. The brother's trait means the woman must be rr
    The woman does not have the trait, so she is not rr; her brother's trait fixes their parents as carriers, and makes her RR or Rr.

Why: The woman is an unshaded child of two carriers, so she is Rr with a two-in-three chance.
That factor multiplies the one-in-four chance for two carriers, giving one in six.

33
Check q12 numeric entry

Two parents are both carriers of an autosomal recessive trait, Rr and Rr. They have two children.

Calculate the probability, as a decimal to four decimal places, that both children have the trait.

Answer: 0.0625  (tolerance ±0.0005)

Working
Write down the values in the question:
P(a child is rr) = 1/4
number of children = 2
each child is independent of the other
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(both rr)=14×14=116=0.0625
34
Check q13 numeric entry

In the pedigree below, II-2 and II-3 are partners. II-2's brother II-1 has an autosomal recessive trait; II-3's sister II-4 has the same trait. None of the four parents in row I has it.

A pedigree of two families. Left: I-1 and I-2, both unshaded, have two children, II-1, a shaded square, and II-2, an unshaded circle. Right: I-3 and I-4, both unshaded, have two children, II-3, an unshaded square, and II-4, a shaded circle. II-2 is joined to II-3
A pedigree of two families. Left: I-1 and I-2, both unshaded, have two children, II-1, a shaded square, and II-2, an unshaded circle. Right: I-3 and I-4, both unshaded, have two children, II-3, an unshaded square, and II-4, a shaded circle. II-2 is joined to II-3

Calculate the probability, as a decimal to three decimal places, that the first child of II-2 and II-3 has the trait.

Answer: 0.111  (tolerance ±0.0015)

Working
Write down the values in the question:
P(II-2 is Rr) = 2/3 (an unshaded child of two carriers)
P(II-3 is Rr) = 2/3 (an unshaded child of two carriers)
P(rr child if both are Rr) = 1/4
Write down the equation:
P(A and B and C)=P(A)×P(B)×P(C)
Substitute the values into the equation:
P(child has the trait)=23×23×14=436=19≈0.111
35
Practice writing an answer

Albinism is an autosomal recessive trait. Write A for the other allele and a for the albinism allele. A couple do not have albinism, but their daughter does, so both parents are Aa. Their son does not have albinism.

(a) Calculate the probability, as a decimal, that the couple's next child has albinism. (1 pt)

Answer: 0.25  (tolerance ±0.005)

Model answer The probability is 0.25: one chance in four.
Working
Write down the values in the question:
P(a from the mother, Aa) = 1/2
P(a from the father, Aa) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(aa)=12×12=14=0.25
Rubric
  • Award 1 point for: 0.25 (one in four).

(b) Calculate the probability, as a decimal, that the couple's next child is a carrier, Aa. (1 pt)

Answer: 0.5  (tolerance ±0.005)

Model answer The probability is 0.5: two of the four cells of the Aa and Aa square are Aa.
Working
Write down the values in the question:
square for Aa and Aa = AA, Aa, Aa, aa
Aa = 2 cells of 4
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(Aa)=14+14=12=0.5
Rubric
  • Award 1 point for: 0.5 (one in two).

(c) The son later has a child with a partner who is a known carrier, Aa. Calculate the probability, as a decimal to three decimal places, that their first child has albinism. (1 pt)

Answer: 0.167  (tolerance ±0.0015)

Model answer The probability is about 0.167, one in six.
Working
Write down the values in the question:
P(the son is Aa) = 2/3 (an unshaded child of two carriers)
P(aa child if the son is Aa and the partner is Aa) = 1/2 × 1/2 = 1/4
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(child is aa)=23×14=212=16≈0.167
Rubric
  • Award 1 point for: 0.167 (one in six): the son’s two-in-three chance of being Aa times ¼.

(d) Explain why the son's genotype enters the calculation in part (c) as a probability rather than as Aa. (1 pt)

Model answer The son does not have albinism.
So he is AA or Aa, and the family record does not say which.
His parents are both Aa.
Of the three cells of their square that are not aa, two are Aa.
So the son is a carrier with a two-in-three chance.
That chance is one more factor, so it multiplies into the calculation.
Rubric
  • Award 1 point for: the son is AA or Aa (undecided), so his chance of being a carrier, two in three, multiplies in as a factor.

Slip Treating the son as a known carrier, Aa. Unshaded rules out aa only; a child of two carriers is Aa with a two-in-three chance.

APBIO-U05-L17 Two genes at once

Topic 5.3 · Mendelian Genetics · 65 steps

A cell at metaphase I: an oval cell with a pole at each end and two homologous pairs of X-shaped chromosomes across its middle. The long pair, above, carries R on its dark left homolog and r on its light right homolog; the short pair, below, carries y on its light left homolog and Y on its dark right homolog. A spindle fiber joins each chromosome to the pole it faces
A cell at metaphase I: an oval cell with a pole at each end and two homologous pairs of X-shaped chromosomes across its middle. The long pair, above, carries R on its dark left homolog and r on its light right homolog; the short pair, below, carries y on its light left homolog and Y on its dark right homolog. A spindle fiber joins each chromosome to the pole it faces

Here is a cell from a pea plant that is heterozygous for two traits, RrYy: round or wrinkled seeds, yellow or green seeds. It is drawn at metaphase I, and its two genes sit on two different chromosome pairs.

The long pair carries R on one homolog and r on the other; the short pair carries Y and y. When this cell makes gametes, do R and Y have to travel together, or can a gamete get R with y?

Unit 5 · Heredity

1Each pair lines up on its own

2

Video: Watch: Each pair lines up on its own

The RrYy cell at metaphase I; the short pair turning round while the long pair stays; the two orientations side by side, R reaching the left pole with y in one and with Y in the other.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17a.mp4

3

Do the alleles of two genes travel into a gamete together, or separately? The answer is decided at metaphase I.

4

At metaphase I each homologous pair faces either way on its own. So which way the pair carrying R and r faces has no effect on which way the pair carrying Y and y faces.

5

A gamete gets R with Y as often as it gets R with y. This holds for genes on different chromosomes.

6
Check q1

A cell with a long homologous pair and a short one is at metaphase I. The long pair faces its maternal member toward the left pole.

Which way does the short pair face?

  1. A. ✓ Either way, regardless of the long pair
  2. B. Its maternal member toward the left pole, like the long pair
    Each pair lines up on its own.
    The long pair’s facing has no hold on the short pair’s.

Why: Each homologous pair faces either way at metaphase I, regardless of the other pairs.
So which member of each pair a gamete receives is decided pair by pair.

7

Now put letters on the pairs of the RrYy cell. The long pair carries R on its maternal homolog and r on its paternal homolog; the short pair carries Y on its maternal homolog and y on its paternal homolog.

8

Here are the two orientations of the cell. In both cells the long pair faces R toward the left pole.

Two cells at metaphase I side by side, each with a pole at each end and two X-shaped homologous pairs across the middle. In both cells the long pair, above, has the dark R homolog on the left and the light r homolog on the right. In the left cell the short pair, below, has the light y homolog on the left and the dark Y homolog on the right. In the right cell the short pair is turned round: the dark Y homolog on the left and the light y homolog on the right
Two cells at metaphase I side by side, each with a pole at each end and two X-shaped homologous pairs across the middle. In both cells the long pair, above, has the dark R homolog on the left and the light r homolog on the right. In the left cell the short pair, below, has the light y homolog on the left and the dark Y homolog on the right. In the right cell the short pair is turned round: the dark Y homolog on the left and the light y homolog on the right
9

In the left cell the short pair faces y toward the left pole. So the left pole receives R with y, and the right pole receives r with Y.

10

In the right cell the short pair faces Y toward the left pole. So the left pole receives R with Y, and the right pole receives r with y.

11

The short pair faces either way regardless of the long pair. So the two orientations are equally common.

12

So R travels with y in half of the cells and with Y in the other half. Which allele of one gene a gamete gets has no effect on which allele of the other gene it gets.

13

When the alleles of two genes on different chromosomes go into gametes independently of each other, we call it the , because to assort means to sort into groups, and each pair sorts on its own.

14

Each homologous pair facing either way on its own at metaphase I is the physical basis of the law of independent assortment.

15

What you are expected to know Explain the law of independent assortment from metaphase I: each homologous pair faces either way on its own, so the alleles of two genes on different chromosomes go into gametes independently of each other.

16
Check q2

Suppose a pea plant is RrYy, and its two genes sit on different chromosomes.

Can one of its gametes carry R with y?

  1. A. ✓ Yes
  2. B. No
    The pair carrying R and r faces either way regardless of the pair carrying Y and y.
    In half of the cells R faces the same pole as y.

Why: Each homologous pair faces either way on its own at metaphase I.
So in half of the cells R and y face the same pole.
The pole’s gametes then carry R with y.

17
Practice writing an answer

A pea plant is RrYy, and its two genes sit on different chromosomes. Some of its gametes carry R with y.

(a) Explain why some of the plant’s gametes carry R with y. (1 pt)

Frame Some gametes carry R with y because …

Model answer Some gametes carry R with y because each homologous pair faces either way on its own at metaphase I.
So in half of the cells the pair carrying R and r faces R toward the same pole as the pair carrying Y and y faces y.
Anaphase I then pulls R and y to that pole.
So the gametes made from that pole carry R with y.
Rubric
  • Award 1 point for: each homologous pair faces either way independently at metaphase I, so in some cells R and y face the same pole and go into the same gamete.
18
Check q3

A pea plant is RrYy, and its two genes sit on different chromosomes. A student says: “This plant got R and Y from the same parent, so its gametes carry R and Y together.”

Is the student correct?

  1. A. Yes: R and Y stay together in every gamete
    Where an allele came from has no hold on which pole it faces.
    The pair carrying R and r faces either way regardless of the pair carrying Y and y.
  2. B. ✓ No: a gamete carries R with y as often as R with Y

Why: R and r sit on one pair; Y and y sit on another pair.
Each pair faces either way on its own at metaphase I.
So R faces the same pole as y in half of the cells.
A gamete carries R with y as often as R with Y.

19
Check q4

In maize, purple kernels (P) are dominant to yellow (p) and smooth kernels (S) are dominant to wrinkled (s); the two genes sit on different chromosomes. A plant is PpSs.

Which gamete types does this plant make?

  1. A. PS only
    Half of the gametes get p and half get s, in every combination.
  2. B. PS and ps only
    The pair carrying P and p faces either way regardless of the pair carrying S and s.
    So P goes with s as often as with S.
  3. C. Pp and Ss
    A gamete carries one allele of each gene, one homolog from each pair, never both alleles of one gene.
  4. D. ✓ PS, Ps, pS and ps

Why: Each pair faces either way regardless of the other.
So a gamete gets P or p and, independently, S or s.
That gives four types: PS, Ps, pS and ps.

20Quick quiz: law of independent assortment mixed practice

21
Check q5

Two genes sit on different chromosomes.

What does the law of independent assortment say about their alleles?

  1. A. The two alleles of one gene go into different gametes
    The two alleles of one gene parting is the law of segregation.
    The law of independent assortment is about the alleles of two different genes.
  2. B. ✓ The alleles of the two genes go into gametes independently of each other
  3. C. The dominant allele of each gene goes into every gamete
    A gamete gets one allele of each gene, dominant or recessive.
    Half of the gametes get the recessive allele.

Why: The law of independent assortment says that the alleles of two genes on different chromosomes go into gametes independently of each other.

22
Check q6

The law of independent assortment has a physical basis in meiosis.

Which event is that physical basis?

  1. A. The two sister chromatids of each chromosome part at anaphase II
    Sister chromatids are identical copies, so parting them changes nothing about which alleles travel together.
  2. B. The two homologs of a pair swap segments in prophase I
    Swapping segments moves alleles between the two homologs of one pair.
    The law of independent assortment is about two different pairs.
  3. C. ✓ Each homologous pair faces either way on its own at metaphase I

Why: Each homologous pair faces either way on its own at metaphase I.
So which member of one pair a gamete gets has no effect on which member of another pair it gets.
That is the law of independent assortment.

23
Practice writing an answer

Two genes sit on different chromosomes.

(a) State the law of independent assortment. (1 pt)

Model answer The alleles of two genes on different chromosomes go into gametes independently of each other.
Rubric
  • Award 1 point for: the alleles of two genes (on different chromosomes) go into gametes independently of each other.

24Four kinds of gamete

25

Video: Watch: Four kinds of gamete

The four gamete types of the RrYy plant appearing one at a time, RY, Ry, rY and ry, each from one pole of one orientation; the four counted as equally common, one quarter each.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17b.mp4

26

Now count the gamete types of the RrYy plant.

27

From the long pair a gamete gets R or r. From the short pair it gets Y or y.

28

Each allele of the first gene goes with each allele of the second. So the gamete types are RY, Ry, rY and ry.

Four gametes in a row, each a circle carrying two alleles: RY, Ry, rY and ry; under each is written ¼
Four gametes in a row, each a circle carrying two alleles: RY, Ry, rY and ry; under each is written ¼
29

The two orientations are equally common, and each orientation gives two of the four gamete types. So the four gamete types are equally common.

30
Worked example

An RrYy pea plant makes gametes; its two genes sit on different chromosomes. How many gamete types does it make, and what fraction of its gametes is each type?

Write down the values in the question:
seed-shape gene: R or r, 2 alleles
seed-color gene: Y or y, 2 alleles
the four types are equally common
Each allele of one gene goes with each allele of the other, so multiply the two counts:
gamete types=2×2=4
RY, Ry, rY, ry
fraction of each type=14
A fraction of the gametes is a number between 0 and 1 and carries no unit.
31

When a breeding cross follows two genes at once, we call it a : di- means two, and the parents are hybrids, heterozygous, for both genes.

32

So the dominant alleles do not travel together. RY and ry are two of the four gamete types, each as common as Ry and rY.

33

What you are expected to know List the four gamete types of a plant heterozygous for two genes on different chromosomes, RY, Ry, rY and ry, each ¼ of its gametes.

34Quick quiz: dihybrid cross mixed practice

35
Check q7

A breeder crosses two plants.

Which of the following is a dihybrid cross?

  1. A. Rr × Rr
    Rr × Rr follows one gene, seed shape.
    A dihybrid cross follows two genes at once, with both parents heterozygous for both.
  2. B. ✓ RrYy × RrYy
  3. C. Yy × yy
    Yy × yy follows one gene, and the yy parent is homozygous.
    A dihybrid cross follows two genes at once, with both parents heterozygous for both.

Why: A dihybrid cross follows two genes at once, and each parent is heterozygous for both.
RrYy × RrYy is heterozygous for seed shape and for seed color in both parents.

36
Practice writing an answer

Two RrYy pea plants are crossed.

(a) State what makes this cross a dihybrid cross. (1 pt)

Model answer The cross follows two genes at once, seed shape and seed color, and each parent is heterozygous for both genes.
Rubric
  • Award 1 point for: two genes followed at once, with each parent heterozygous for both.
37
Check q8

A guinea pig has the genotype CcLl. The two genes sit on different chromosomes.

Is cL one of the gamete types this guinea pig makes?

  1. A. ✓ Yes
  2. B. No
    The pair carrying C and c faces either way regardless of the pair carrying L and l.
    So c goes with L as often as with l.

Why: Each pair lines up at metaphase I facing either way on its own.
So a gamete gets c or C and, independently, L or l.
cL is one of the four gamete types.

38
Check q9

A CcLl guinea pig makes gametes.

Is Cc one of the gamete types it makes?

  1. A. Yes
    C and c sit on the two homologs of one pair, and those two homologs go to opposite poles.
    So a gamete carries C or c, never both.
  2. B. ✓ No

Why: C and c are the two alleles of one gene, one on each homolog of one pair.
The two homologs part at anaphase I, so a gamete gets one of them, never both.
A gamete carries one allele of each gene: CL, Cl, cL or cl.

39
Check q10 numeric entry

A guinea pig is CcLl: the two genes sit on different chromosomes.

Calculate the number of gamete types it makes.

Answer: 4  (tolerance ±0)

Working
Write down the values in the question:
gene 1: C or c, 2 alleles
gene 2: L or l, 2 alleles
Each allele of one gene goes with each allele of the other, so multiply the two counts:
gamete types=2×2=4
CL, Cl, cL, cl
40
Check q11 numeric entry

The CcLl guinea pig makes gametes.

Calculate the fraction of its gametes that carry C with l, as a decimal.

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
number of gamete types = 4 (CL, Cl, cL, cl)
each type equally common
Write down the equation:
fraction of one type=1number of types
Substitute the values into the equation:
fraction carrying Cl=14=0.25
41
Check q12 numeric entry

Another guinea pig has the genotype CCLl. The two genes sit on different chromosomes.

Calculate the number of gamete types it makes.

Answer: 2  (tolerance ±0)

Working
Write down the values in the question:
gene 1: C or C, 1 allele
gene 2: L or l, 2 alleles
Each allele of one gene goes with each allele of the other, so multiply the two counts:
gamete types=1×2=2
CL, Cl
42
Check q13

In a CcLl guinea pig, C and L are the dominant alleles.

Does C have to go into the same gamete as L?

  1. A. ✓ No
  2. B. Yes
    Dominance decides what shows in the animal, not which homolog a gamete receives.
    The pair carrying C and c faces either way regardless of the pair carrying L and l.

Why: The two pairs line up at metaphase I independently.
So C goes with L in half of the gametes that carry C, and with l in the other half.
The dominant alleles do not have to travel together.

43Where the laws hold

44

Video: Watch: Where the laws hold

Two pairs of rods, R and r on one pair and Y and y on the other, each pair turning on its own; then one pair of rods carrying R above Y on one homolog and r above y on the other, one whole homolog sliding into a gamete with both its alleles.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17c.mp4

45

The law of independent assortment describes genes on different chromosomes, like seed shape and seed color in peas. Their two pairs face either way on their own at metaphase I.

Left: two homologous pairs drawn as rods with centromere dots, one dark and one light in each pair; a long pair carrying R on the dark rod and r on the light rod, and a short pair carrying Y on the dark rod and y on the light rod, captioned two genes on different chromosomes. Right: one homologous pair, the dark rod carrying R above Y and the light rod carrying r above y, captioned two genes on the same chromosome
Left: two homologous pairs drawn as rods with centromere dots, one dark and one light in each pair; a long pair carrying R on the dark rod and r on the light rod, and a short pair carrying Y on the dark rod and y on the light rod, captioned two genes on different chromosomes. Right: one homologous pair, the dark rod carrying R above Y and the light rod carrying r above y, captioned two genes on the same chromosome
46

The law of segregation holds for every gene. The two alleles of one gene sit on the two homologs of one pair, and those two homologs always part at anaphase I.

47

Now consider two genes close together on the same chromosome. Both genes sit on each homolog, so one whole homolog carries an allele of each into a gamete.

48

So the two alleles on one homolog tend to travel together into a gamete. Two genes on the same chromosome need not follow the law of independent assortment.

49

That is a boundary of the law of independent assortment, not a failure of meiosis. Meiosis parts the two homologs of a pair, not the two genes on one homolog.

50

What you are expected to know State where each of Mendel’s two laws holds: the law of independent assortment for genes on different chromosomes, and the law of segregation for every gene.

51
Check q14

Suppose two genes sit close together on the same chromosome.

Do the alleles of the two genes have to go into gametes independently of each other?

  1. A. Yes
    The law of independent assortment describes genes on different chromosomes.
    Two genes on one homolog ride into a gamete together.
  2. B. ✓ No

Why: Both genes sit on each homolog of the pair.
A gamete receives one whole homolog.
So the alleles on that homolog travel together, and the two genes need not assort independently.

52
Practice writing an answer

In a plant, two genes A and B sit close together on the same chromosome. In one AaBb plant, A and B sit on one homolog and a and b on the other. The plant makes mostly AB and ab gametes.

(a) Explain why the gametes are mostly AB and ab. (1 pt)

Model answer A gamete receives one whole homolog of the pair.
One homolog carries A with B, and the other carries a with b.
So a gamete that receives the first homolog carries A and B, and a gamete that receives the second carries a and b.
So the two genes travel together, and the gametes are mostly AB and ab.
Rubric
  • Award 1 point for: a gamete receives one whole homolog, and A sits with B on one homolog and a with b on the other, so the two genes travel together.
53
Check q15

In one plant, two genes sit close together on the same chromosome, and the plant makes mostly two kinds of gamete for those genes instead of four. A student says: “Meiosis must have gone wrong in this plant.”

Is the student correct?

  1. A. The student is right
    The law of independent assortment describes genes on different chromosomes.
    Two genes on one chromosome sit on the same homolog, so a normal meiosis carries them into a gamete together.
  2. B. ✓ The student is wrong

Why: The law of independent assortment holds for genes on different chromosomes.
These two genes sit on the same chromosome, so each homolog carries both into a gamete together.
That is a boundary of the law, not a failure of meiosis.
So the student is wrong.

54
Check q16

Suppose that in one kind of plant, white fruit (W) is dominant to yellow (w) and flat fruit (D) is dominant to round (d). A WwDd plant is crossed with a wwdd plant, and the offspring appear in four equal classes: white flat, white round, yellow flat and yellow round.

What do the four equal classes show about the two genes?

  1. A. The two genes sit close together on the same chromosome
    Two genes close together on one chromosome would give mostly the two parental combinations, WD and wd, not four equal classes.
  2. B. The wwdd parent gave no alleles to the offspring
    The wwdd parent gave a w and a d to every offspring.
    The four classes come from the WwDd parent’s four gamete types.
  3. C. ✓ The two genes assort independently of each other
  4. D. White fruit and flat fruit are both recessive traits
    White and flat appear in offspring that got W and D from only one parent.
    So each shows with one copy, which is what dominant means.

Why: The wwdd parent gave w and d to every offspring.
So each offspring’s look shows the gamete it got from the WwDd parent.
Four equal classes mean four equally common gamete types, WD, Wd, wD and wd: the two genes assorted independently.

55

Back to the RrYy cell at metaphase I, its long pair carrying R and r and its short pair carrying Y and y, both pairs at the equator. Here are its two orientations again.

Two cells at metaphase I side by side, each with a pole at each end and two X-shaped homologous pairs across the middle. In both cells the long pair, above, has the dark R homolog on the left and the light r homolog on the right. In the left cell the short pair, below, has the light y homolog on the left and the dark Y homolog on the right. In the right cell the short pair is turned round: the dark Y homolog on the left and the light y homolog on the right
Two cells at metaphase I side by side, each with a pole at each end and two X-shaped homologous pairs across the middle. In both cells the long pair, above, has the dark R homolog on the left and the light r homolog on the right. In the left cell the short pair, below, has the light y homolog on the left and the dark Y homolog on the right. In the right cell the short pair is turned round: the dark Y homolog on the left and the light y homolog on the right
56

Because the two pairs face either way independently, the plant makes RY, Ry, rY and ry in equal numbers: R goes with y as often as with Y. This holds for genes on different chromosomes.

57Mixed practice mixed practice

58
Check q17

A fruit fly is EeVv, and the two genes sit on different chromosomes.

Which fraction of its gametes carries e with V?

  1. A. ✓ 14
  2. B. 12
    Half of the gametes carry e, and half of those carry V.
    Half of a half is ¼.
  3. C. 1
    Half of the gametes carry E, not e.
    eV is one of four equally common types.

Why: The fly makes four gamete types, EV, Ev, eV and ev.
The four gamete types are equally common.
So eV is ¼ of the gametes.

59
Check q18

In a plant, two genes A and B sit close together on the same chromosome. In one AaBb plant, A and B sit on one homolog and a and b on the other.

Which gametes does this plant make?

  1. A. AB only
    Dominance decides what shows in a plant, not which homolog a gamete receives.
    Half of the gametes get the homolog carrying a and b.
  2. B. ✓ Mostly AB and ab
  3. C. Mostly Ab and aB
    Meiosis parts the two homologs of a pair from each other, not the two genes on one homolog.
    A and B ride into a gamete together on their homolog.
  4. D. AB, Ab, aB and ab in equal numbers
    The law of independent assortment holds for genes on different chromosomes.
    Two genes close together on one chromosome ride on the same homolog.

Why: A gamete receives one homolog of the pair.
One homolog carries A with B and the other a with b.
So the two genes travel together, and the gametes are mostly AB and ab.

60
Check q19

A cell is at metaphase I. Its long pair faces R toward the left pole; its short pair carries Y and y.

Which way can the short pair face Y?

  1. A. Toward the left pole only
    Which way the long pair faces has no effect on the short pair.
    The short pair faces either way on its own.
  2. B. Toward the right pole only
    Nothing turns the short pair the opposite way to the long pair.
    The short pair faces either way on its own.
  3. C. ✓ Toward either pole

Why: Each homologous pair lines up at metaphase I on its own.
The long pair’s facing does not set the short pair’s facing.
So the short pair can face Y toward either pole.

61
Check q20

A pea plant is RRYy, and the two genes sit on different chromosomes.

Which gamete types does this plant make?

  1. A. RY only
    Half of the gametes get Y and half get y.
    Every gamete gets R, so the types are RY and Ry.
  2. B. ✓ RY and Ry
  3. C. RY, Ry, rY and ry
    The plant has no r allele to give.
    Every gamete gets R, so only two types are possible.

Why: Every gamete gets R, because both homologs of the long pair carry R.
A gamete gets Y or y.
So the plant makes two gamete types, RY and Ry.

62
Check q21

Mendel found two laws.

Which law holds for every gene, whichever chromosome the gene sits on?

  1. A. ✓ The law of segregation
  2. B. The law of independent assortment
    The law of independent assortment holds for genes on different chromosomes.
    Two genes on one chromosome need not follow it.

Why: The two alleles of any gene sit on the two homologs of one pair.
Those two homologs always part at anaphase I.
So the law of segregation holds for every gene.

63
Check q22

In maize, purple kernels (P) are dominant to yellow (p) and smooth kernels (S) are dominant to wrinkled (s); the two genes sit on different chromosomes. A student says: “A PpSs plant makes only PS and ps gametes, because P and S are the dominant alleles.”

Is the student correct?

  1. A. Yes: the dominant alleles travel together
    Dominance decides what shows in the plant, not which pole an allele faces.
    The pair carrying P and p faces either way regardless of the pair carrying S and s.
  2. B. ✓ No: the plant makes PS, Ps, pS and ps in equal numbers

Why: Each pair faces either way on its own at metaphase I.
So P goes with s as often as with S.
The plant makes four gamete types, PS, Ps, pS and ps, in equal numbers.

64
Practice writing an answer

In maize, purple kernels (P) are dominant to yellow (p) and smooth kernels (S) are dominant to wrinkled (s); the two genes sit on different chromosomes. A PpSs plant is crossed with a ppss plant. The ppss parent gives p and s to every kernel, so each kernel’s color and shape show which gamete came from the PpSs parent. Of 400 kernels, 98 are purple smooth, 104 purple wrinkled, 101 yellow smooth and 97 yellow wrinkled.

(a) Explain how the four counts demonstrate the law of independent assortment. (2 pt)

Model answer Each kernel’s color and shape show the gamete from the PpSs parent.
So the four kernel classes count the four gamete types PS, Ps, pS and ps.
The four counts are about equal, so the four gamete types are about equally common.
So P went with s as often as with S: which allele of one gene a gamete got had no effect on which allele of the other gene it got.
That is the law of independent assortment.
Rubric
  • Award 1 point for: the four kernel classes count the PpSs parent’s four gamete types, which are about equally common.
  • Award 1 point for: P went with s as often as with S, so the alleles of the two genes went into gametes independently of each other.

Glossary

law of independent assortment
The alleles of two genes on different chromosomes go into gametes independently of each other, because each homologous pair lines up at metaphase I facing either way regardless of the other pairs. An RrYy plant makes RY, Ry, rY and ry gametes, each ¼ of its gametes.
dihybrid cross
A breeding cross that follows two genes at once, such as RrYy × RrYy for seed shape and seed color. Di- means two; the parents are heterozygous for both genes.

APBIO-U05-L17B The sixteen-cell square

Topic 5.3 · Mendelian Genetics · 52 steps

An empty four-by-four grid: RY, Ry, rY and ry written along the top edge and the same four gametes down the left edge; every one of the sixteen cells is empty
An empty four-by-four grid: RY, Ry, rY and ry written along the top edge and the same four gametes down the left edge; every one of the sixteen cells is empty

Here is a grid with four gametes along the top and four down the side: RY, Ry, rY and ry, the four gamete types an RrYy pea plant makes.

Two RrYy plants are crossed. What offspring can they give, and in what proportions?

Unit 5 · Heredity

1Build the sixteen-cell Punnett square

2

Video: Watch: Build the sixteen-cell Punnett square

The four-by-four square filling row by row, each cell taking one allele of each gene from its row’s gamete and one from its column’s gamete; the nine different genotypes counted at the end.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17Ba.mp4

3

What does a cross of two RrYy plants give? Each plant makes four gamete types, so the Punnett square needs four gametes along each edge: sixteen cells.

4

Fill each cell with the alleles its egg and its pollen grain bring together. Then the sixteen cells show every offspring the cross can give.

5
Check q1

A pea plant is RrYy, and its two genes sit on different chromosomes.

Which gamete types does this plant make?

  1. A. RY and ry
    The pair carrying R and r faces either way regardless of the pair carrying Y and y.
    So R goes with y as often as with Y.
  2. B. Rr and Yy
    A gamete carries one allele of each gene, never both alleles of one gene.
  3. C. ✓ RY, Ry, rY and ry

Why: Each pair faces either way regardless of the other.
So a gamete gets R or r and, independently, Y or y.
That gives four types: RY, Ry, rY and ry.

6
Check q2

A Punnett square has gametes along its edges and offspring in its cells.

What is written along the top edge and the left edge?

  1. A. ✓ The gamete types each parent makes
  2. B. The genotypes of the offspring
    The offspring genotypes go in the cells.
    Each edge carries one parent’s gamete types.

Why: Each edge of a Punnett square carries one parent’s gamete types.
A cell brings one gamete from each edge together, and that is an offspring genotype.

7

Now consider a cross of two RrYy plants. Each plant makes the four gamete types RY, Ry, rY and ry.

An empty four-by-four grid with the gametes RY, Ry, rY and ry written along the top edge and the same four down the left edge
An empty four-by-four grid with the gametes RY, Ry, rY and ry written along the top edge and the same four down the left edge
8

So the Punnett square carries four gametes along the top and four down the side. That gives sixteen cells.

9

Fill each cell with the two alleles of each gene its egg and its pollen grain bring together. The RY egg with the four pollen types gives RRYY, RRYy, RrYY and RrYy.

The same grid with its first row filled: the RY egg with each pollen type gives RRYY, RRYy, RrYY, RrYy
The same grid with its first row filled: the RY egg with each pollen type gives RRYY, RRYy, RrYY, RrYy
10

Fill the other three rows the same way, cell by cell. Every cell holds one allele of each gene from its row’s gamete and one from its column’s gamete.

The full sixteen-cell square, no cell shaded: each cell holds a four-letter genotype made of one allele of each gene from its row's gamete and one from its column's gamete; RrYy appears in four cells
The full sixteen-cell square, no cell shaded: each cell holds a four-letter genotype made of one allele of each gene from its row's gamete and one from its column's gamete; RrYy appears in four cells
11

The sixteen cells hold nine different genotypes. Several genotypes appear in more than one cell: RrYy appears four times.

12

What you are expected to know Build the sixteen-cell Punnett square for two dihybrids: the four gamete types along each edge, and in each cell the two alleles of each gene that its row’s gamete and its column’s gamete bring together.

13
Check q3

In the Punnett square for RrYy × RrYy, one cell brings together an Ry gamete from one parent and an Ry gamete from the other.

Which genotype does that cell hold?

  1. A. RRYY
    Each Ry gamete carries y, not Y, so the cell holds two y alleles.
  2. B. ✓ RRyy
  3. C. Ry
    A cell holds two alleles of each gene, one from each gamete.
    Ry is one gamete’s alleles, not the cell’s four.

Why: Each gamete gives one allele of each gene.
Ry gives R and y, and the other Ry gives R and y.
So the cell holds RRyy.

14
Check q4

In the Punnett square for RrYy × RrYy, one cell brings together an rY gamete from one parent and an ry gamete from the other.

Which genotype does that cell hold?

  1. A. ✓ rrYy
  2. B. RrYy
    Neither gamete carries R.
    rY gives r and ry gives r, so the cell holds two r alleles.
  3. C. rrYY
    The ry gamete carries y, not Y.
    So the cell holds one Y and one y.

Why: Each gamete gives one allele of each gene.
rY gives r and Y; ry gives r and y.
So the cell holds rrYy.

15
Check q5

In the Punnett square for RrYy × RrYy, one cell brings together an RY gamete from one parent and an ry gamete from the other.

Which genotype does that cell hold?

  1. A. RRYY
    The ry gamete gives r and y, so the cell holds one r and one y.
  2. B. rryy
    The RY gamete gives R and Y, so the cell holds one R and one Y.
  3. C. ✓ RrYy

Why: Each gamete gives one allele of each gene.
RY gives R and Y; ry gives r and y.
So the cell holds RrYy.

16
Check q6

A student’s Punnett square for RrYy × RrYy is drawn below. One of its sixteen cells is filled wrongly. Rows are named by the gamete on the left edge and columns by the gamete on the top edge.

A four-by-four square for RrYy × RrYy with RY, Ry, rY, ry along the top edge and down the left edge; all sixteen cells are filled with four-letter genotypes
A four-by-four square for RrYy × RrYy with RY, Ry, rY, ry along the top edge and down the left edge; all sixteen cells are filled with four-letter genotypes

Which cell is wrong?

  1. A. Row RY, column RY
    An RY egg with RY pollen gives RRYY, and that is what the cell reads.
  2. B. ✓ Row rY, column Ry
  3. C. Row ry, column ry
    Row ry, column ry reads rryy, which is right.
  4. D. Row Ry, column rY
    Row Ry, column rY reads RrYy, which is right.
    RRyy would need an R and a y from both gametes.

Why: The cell in row rY, column Ry brings together an rY gamete, r and Y, with an Ry gamete, R and y.
Together they give RrYy.
That cell reads RRyy, so it is the wrong one.

17
Practice writing an answer

A student’s Punnett square for RrYy × RrYy is drawn below. Rows are named by the gamete on the left edge and columns by the gamete on the top edge. One of its sixteen cells is filled wrongly.

A four-by-four square for RrYy × RrYy with RY, Ry, rY, ry along the top edge and down the left edge; all sixteen cells are filled with four-letter genotypes
A four-by-four square for RrYy × RrYy with RY, Ry, rY, ry along the top edge and down the left edge; all sixteen cells are filled with four-letter genotypes

(a) Identify the cell that is filled wrongly, state what it should read, and explain how the two gametes give it. (1 pt)

Model answer The wrong cell is in row rY, column Ry: it reads RRyy.
The row’s gamete, rY, gives r and Y.
The column’s gamete, Ry, gives R and y.
Together they give one R and one r, and one Y and one y.
So the cell should read RrYy.
Rubric
  • Award 1 point for: the cell in row rY, column Ry should read RrYy, because the gamete rY gives r and Y and the gamete Ry gives R and y.

18Quick quiz: which genotype sits in this cell? mixed practice

19
Check q7

In maize, purple kernels (P) are dominant to yellow (p) and smooth kernels (S) are dominant to wrinkled (s); the two genes sit on different chromosomes. Two PpSs plants are crossed. Rows are named by the gamete on the left edge and columns by the gamete on the top edge. The shaded cell is in row Ps, column pS.

An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row Ps, column pS is shaded
An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row Ps, column pS is shaded

Which genotype sits in the shaded cell?

  1. A. ✓ PpSs
  2. B. PPSs
    The pS gamete gives p, not P.
    So the cell holds one P and one p.
  3. C. ppSs
    The Ps gamete gives P, not p.
    So the cell holds one P and one p.

Why: Ps gives P and s; pS gives p and S.
So the cell holds PpSs.

20
Check q8

Two PpSs maize plants are crossed. Rows are named by the gamete on the left edge and columns by the gamete on the top edge. The shaded cell is in row ps, column ps.

An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row ps, column ps is shaded
An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row ps, column ps is shaded

Which genotype sits in the shaded cell?

  1. A. PpSs
    Neither gamete carries P or S.
    The cell holds two p alleles and two s alleles.
  2. B. ps
    A cell holds two alleles of each gene, one from each gamete.
    ps is one gamete’s alleles, not the cell’s four.
  3. C. ✓ ppss

Why: ps gives p and s; the other ps gives p and s.
So the cell holds ppss.

21
Check q9

Two PpSs maize plants are crossed. Rows are named by the gamete on the left edge and columns by the gamete on the top edge. The shaded cell is in row PS, column Ps.

An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row PS, column Ps is shaded
An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row PS, column Ps is shaded

Which genotype sits in the shaded cell?

  1. A. PpSs
    Both gametes carry P.
    So the cell holds two P alleles.
  2. B. ✓ PPSs
  3. C. PPSS
    The Ps gamete carries s, not S.
    So the cell holds one S and one s.

Why: PS gives P and S; Ps gives P and s.
So the cell holds PPSs.

22
Check q10

Two PpSs maize plants are crossed. Rows are named by the gamete on the left edge and columns by the gamete on the top edge. The shaded cell is in row pS, column pS.

An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row pS, column pS is shaded
An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row pS, column pS is shaded

Which genotype sits in the shaded cell?

  1. A. ppSs
    Both gametes carry S.
    So the cell holds two S alleles.
  2. B. PpSS
    Neither gamete carries P.
    So the cell holds two p alleles.
  3. C. ✓ ppSS

Why: pS gives p and S; the other pS gives p and S.
So the cell holds ppSS.

23
Check q11

Two PpSs maize plants are crossed. Rows are named by the gamete on the left edge and columns by the gamete on the top edge. The shaded cell is in row Ps, column ps.

An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row Ps, column ps is shaded
An empty four-by-four grid with the gametes PS, Ps, pS and ps written along the top edge and the same four down the left edge; the cell in row Ps, column ps is shaded

Which genotype sits in the shaded cell?

  1. A. ✓ Ppss
  2. B. PpSs
    Both gametes carry s.
    So the cell holds two s alleles.
  3. C. ppss
    The Ps gamete carries P.
    So the cell holds one P and one p.

Why: Ps gives P and s; ps gives p and s.
So the cell holds Ppss.

24Read the square: 9 : 3 : 3 : 1

25

Video: Watch: Read the square: 9 : 3 : 3 : 1

The nine round-yellow cells shaded together, then the three, the three and the one; the ratio written beside the square with its nine genotypes counted; 160 seeds shared out as 90, 30, 30 and 10.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17Bb.mp4

26

Now read the filled Punnett square by what the seeds look like. A seed is round if its genotype has at least one R, and yellow if it has at least one Y.

27

Nine cells have at least one R and at least one Y. Those nine seeds are round and yellow.

The full sixteen-cell square: the nine cells that hold at least one R and at least one Y are shaded; the other seven are RRyy, Rryy, Rryy, rrYY, rrYy, rrYy and rryy
The full sixteen-cell square: the nine cells that hold at least one R and at least one Y are shaded; the other seven are RRyy, Rryy, Rryy, rrYY, rrYy, rrYy and rryy
28

Three cells have an R and two y alleles: round and green. Three cells have two r alleles and a Y: wrinkled and yellow.

29

One cell, rryy, has no R and no Y. That seed is wrinkled and green.

30

Here is a table of the four phenotype classes, the cells in each and the genotypes in each.

A table of the four phenotype classes of RrYy × RrYy: round yellow, 9 cells of 16; round green, 3 cells; wrinkled yellow, 3 cells; wrinkled green, 1 cell
A table of the four phenotype classes of RrYy × RrYy: round yellow, 9 cells of 16; round green, 3 cells; wrinkled yellow, 3 cells; wrinkled green, 1 cell
31

So 9 : 3 : 3 : 1 is a phenotypic ratio. It sits over nine genotypes.

32
Check q12

A Pp × Pp cross gives a phenotypic ratio of 3 purple : 1 white, and 120 seeds are collected.

To find the expected count of white seeds, what do you multiply the 120 by?

  1. A. ✓ White’s share of the ratio, 1 of 4
  2. B. The number of classes, 2
    The number of classes says how many kinds of seed there are, not what fraction of the seeds each kind is.
  3. C. The ratio’s first number, 3
    The 3 is purple’s share; white’s share is 1 of the 4 shares.

Why: Expected count = total × that class’s share of the ratio.
A 3 : 1 ratio has four shares, and white has one of them.
So the expected white count is 120 × ¼.

33

A 9 : 3 : 3 : 1 ratio has sixteen shares in all: nine for round yellow, three for round green, three for wrinkled yellow and one for wrinkled green.

34
Worked example

Two RrYy pea plants are crossed and 160 seeds are collected. How many seeds are expected in each of the four phenotype classes?

Write down the values in the question:
total seeds = 160
phenotypic ratio = 9 : 3 : 3 : 1, so 16 shares in all
round yellow share = 9 of 16
round green share = 3 of 16
wrinkled yellow share = 3 of 16
wrinkled green share = 1 of 16
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected round yellow=160×916=90
expected round green=160×316=30
expected wrinkled yellow=160×316=30
expected wrinkled green=160×116=10
An expected count is a number of seeds.
35

What you are expected to know Read the phenotypic ratio 9 : 3 : 3 : 1 from the sixteen-cell Punnett square.

36

What you are expected to know Say that the four phenotype classes sit over nine genotypes.

37

What you are expected to know Turn the 9 : 3 : 3 : 1 ratio into expected counts for a stated total.

38
Check q13

In maize, purple kernels (P) are dominant to yellow (p) and smooth kernels (S) are dominant to wrinkled (s); the two genes sit on different chromosomes. Two PpSs plants are crossed and the sixteen-cell Punnett square is filled.

How many of the sixteen cells are purple and smooth?

  1. A. 1
    One cell of sixteen is ppss, yellow and wrinkled.
    Purple and smooth needs at least one P and at least one S.
  2. B. 3
    Three cells of sixteen are purple and wrinkled.
    Purple and smooth needs at least one P and at least one S.
  3. C. ✓ 9

Why: A kernel is purple and smooth if its genotype has at least one P and at least one S.
Nine of the sixteen cells do.
So nine cells are purple and smooth.

39
Check q14 numeric entry

In pea plants, round seeds (R) are dominant to wrinkled (r) and yellow seeds (Y) are dominant to green (y); the two genes sit on different chromosomes. Two RrYy pea plants are crossed and 240 seeds are collected.

Calculate how many of the 240 seeds are expected to be wrinkled and green.

Answer: 15  (tolerance ±0)

Working
Write down the values in the question:
total seeds = 240
phenotypic ratio = 9 : 3 : 3 : 1, so 16 shares in all
wrinkled green (rryy) share = 1 of 16
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected wrinkled green=240×116=15
40
Check q15 numeric entry

In maize, purple kernels (P) are dominant to yellow (p) and smooth kernels (S) are dominant to wrinkled (s); the two genes sit on different chromosomes. Two PpSs plants are crossed and 320 kernels are collected.

Calculate how many of the 320 kernels are expected to be purple and wrinkled.

Answer: 60  (tolerance ±0)

Working
Write down the values in the question:
total kernels = 320
phenotypic ratio = 9 : 3 : 3 : 1, so 16 shares in all
purple wrinkled (PPss, Ppss) share = 3 of 16
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected purple wrinkled=320×316=60
41
Check q16

In fruit flies, gray body (E) is dominant to ebony (e) and normal wings (V) are dominant to vestigial (v); the two genes sit on different chromosomes. A student fills the sixteen-cell Punnett square for two EeVv flies and counts the cells: 9 gray normal, 3 gray vestigial, 3 ebony normal, 1 ebony vestigial. The student says: “So the genotypic ratio is 9 : 3 : 3 : 1.”

Is the student correct?

  1. A. ✓ No: 9 : 3 : 3 : 1 is the phenotypic ratio
  2. B. Yes: 9 : 3 : 3 : 1 is the genotypic ratio
    Gray or ebony and normal or vestigial are what the flies look like.
    A ratio of looks is a phenotypic ratio.

Why: Gray or ebony and normal or vestigial are phenotypes.
So 9 : 3 : 3 : 1 groups the cells by phenotype: it is the phenotypic ratio.
The same sixteen cells hold nine different genotypes.

42
Check q17

Two EeVv fruit flies are crossed and the sixteen-cell Punnett square is filled.

How many different genotypes do the sixteen cells hold?

  1. A. 4
    Four is the number of phenotype classes.
    Each class holds one or more genotypes.
  2. B. ✓ 9
  3. C. 16
    Several genotypes appear in more than one cell; EeVv appears four times.
    So there are fewer genotypes than cells.

Why: Each of the sixteen cells holds a genotype, and several genotypes repeat.
Counting each genotype once gives nine.

43

Back to the grid with RY, Ry, rY and ry along both edges. Two RrYy plants fill its sixteen cells with nine genotypes.

The full sixteen-cell square: the nine cells that hold at least one R and at least one Y are shaded; the other seven are RRyy, Rryy, Rryy, rrYY, rrYy, rrYy and rryy
The full sixteen-cell square: the nine cells that hold at least one R and at least one Y are shaded; the other seven are RRyy, Rryy, Rryy, rrYY, rrYy, rrYy and rryy
44

By phenotype the cells sort into nine round yellow, three round green, three wrinkled yellow and one wrinkled green: 9 : 3 : 3 : 1.

45Mixed practice mixed practice

46
Check q18

In maize, purple kernels (P) are dominant to yellow (p) and smooth kernels (S) are dominant to wrinkled (s). In the Punnett square for two PpSs plants, one cell brings together a pS gamete from one parent and a Ps gamete from the other.

Which genotype does that cell hold?

  1. A. PPSS
    Each gamete carries one P or one p, and the two gametes here carry one of each.
  2. B. ✓ PpSs
  3. C. ppss
    The Ps gamete carries P and the pS gamete carries S.

Why: pS gives p and S; Ps gives P and s.
So the cell holds PpSs.

47
Check q19 numeric entry

In fruit flies, gray body (E) is dominant to ebony (e) and normal wings (V) are dominant to vestigial (v); the two genes sit on different chromosomes. Two EeVv flies are crossed and 480 offspring are counted.

Calculate how many of the 480 offspring are expected to be ebony with vestigial wings.

Answer: 30  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 480
phenotypic ratio = 9 : 3 : 3 : 1, so 16 shares in all
ebony vestigial (eevv) share = 1 of 16
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected ebony vestigial=480×116=30
48
Check q20

Two RrYy pea plants are crossed and the sixteen-cell Punnett square is filled.

How many of the sixteen cells hold the genotype RrYy?

  1. A. 1
    RrYy needs an R from one gamete and an r from the other, and a Y from one and a y from the other.
    Four pairings of gametes do that.
  2. B. 2
    RY × ry and ry × RY give RrYy.
    So do Ry × rY and rY × Ry: four pairings in all.
  3. C. ✓ 4

Why: RrYy comes from RY × ry, ry × RY, Ry × rY and rY × Ry.
So four of the sixteen cells hold RrYy.

49
Check q21 numeric entry

In pea plants, round seeds (R) are dominant to wrinkled (r) and yellow seeds (Y) are dominant to green (y); the two genes sit on different chromosomes. Two RrYy pea plants are crossed and 96 seeds are collected.

Calculate how many of the 96 seeds are expected to be round and green.

Answer: 18  (tolerance ±0)

Working
Write down the values in the question:
total seeds = 96
phenotypic ratio = 9 : 3 : 3 : 1, so 16 shares in all
round green (RRyy, Rryy) share = 3 of 16
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected round green=96×316=18
50
Check q22

In pea plants, round seeds (R) are dominant to wrinkled (r) and yellow seeds (Y) are dominant to green (y).

Which of the following genotypes gives a round green seed?

  1. A. rrYY
    rrYY has no R, so the seed is wrinkled, and it has Y, so the seed is yellow.
  2. B. RrYy
    RrYy has a Y, so the seed is yellow, not green.
  3. C. ✓ RRyy

Why: Round needs at least one R; green needs two y alleles.
RRyy has two R alleles and two y alleles.
So RRyy gives a round green seed.

51
Practice writing an answer

In pea plants, round seeds (R) are dominant to wrinkled (r) and yellow seeds (Y) are dominant to green (y); the two genes sit on different chromosomes. A plant heterozygous for both genes, RrYy, is crossed with a wrinkled green plant, rryy, and 200 seeds are collected.

(a) State each parent’s gametes and the genotype in each of the four cells of the Punnett square for this cross. (1 pt)

Model answer The RrYy parent’s gametes are RY, Ry, rY and ry.
The rryy parent makes only one gamete type, ry.
So the Punnett square has one row of four cells.
The four cells are RrYy, Rryy, rrYy and rryy.The RrYy parent’s four gametes along the top edge and the rryy parent’s one gamete down the left edge. The four cells read RrYy, Rryy, rrYy and rryyRrYyRryyrrYyrryyRYRyrYryrygametes of the RrYy parentrryy parentRrYy × rryy: four classes, one cell each
Rubric
  • Award 1 point for: RY, Ry, rY and ry as the RrYy parent's gametes, ry as the rryy parent's only gamete, and the cells RrYy, Rryy, rrYy, rryy.

Slip Giving the rryy parent four gamete types. A homozygous parent makes one gamete type, so its edge carries ry alone.

(b) Identify the phenotypic ratio of the offspring. (1 pt)

Model answer 1 round yellow : 1 round green : 1 wrinkled yellow : 1 wrinkled green.
Each of the four cells is one phenotype class, so round yellow, round green, wrinkled yellow and wrinkled green are equally common.
Rubric
  • Award 1 point for: 1 : 1 : 1 : 1 across the four phenotypes, named.

Slip Writing 9 : 3 : 3 : 1. That ratio needs both parents to be RrYy; here one parent gives only recessive alleles.

(c) Calculate the number of wrinkled green seeds expected among the 200. (1 pt)

Answer: 50  (tolerance ±0)

Model answer About 50 of the 200 seeds are expected to be wrinkled and green.
Working
Write down the values in the question:
total seeds = 200
phenotypic ratio = 1 : 1 : 1 : 1, so 4 shares in all
wrinkled green (rryy) share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected rryy=200×14=50
Rubric
  • Award 1 point for: 50 wrinkled green seeds.

(d) A student predicts 9 : 3 : 3 : 1 for this cross. Explain why that prediction is wrong here. (1 pt)

Model answer 9 : 3 : 3 : 1 comes from two parents that each make four gamete types, RrYy × RrYy, so that sixteen pairings fill the Punnett square.
Here the rryy parent makes only ry gametes.
So there are four pairings, one cell each.
So round yellow, round green, wrinkled yellow and wrinkled green are equally common.
Rubric
  • Award 1 point for: 9 : 3 : 3 : 1 needs both parents heterozygous for both genes (sixteen cells); the rryy parent gives only ry, so the square has four cells and the classes are 1 : 1 : 1 : 1.

Slip Saying 9 : 3 : 3 : 1 is the ratio for any two-gene cross. It is the ratio for two dihybrid parents only.

APBIO-U05-L17C Multiply the one-gene answers

Topic 5.3 · Mendelian Genetics · 29 steps

Two small Punnett squares side by side: on the left Rr crossed with Rr, cells RR, Rr, Rr, rr with the three round cells shaded; on the right Yy crossed with Yy, cells YY, Yy, Yy, yy with the three yellow cells shaded; between them a multiplication sign and beside them the question, where does nine of sixteen come from
Two small Punnett squares side by side: on the left Rr crossed with Rr, cells RR, Rr, Rr, rr with the three round cells shaded; on the right Yy crossed with Yy, cells YY, Yy, Yy, yy with the three yellow cells shaded; between them a multiplication sign and beside them the question, where does nine of sixteen come from

Here are two small squares, one for seed shape, Rr × Rr, and one for seed color, Yy × Yy: the two genes of an RrYy × RrYy cross, taken one at a time.

Round fills three cells of four in the first; yellow fills three of four in the second. The sixteen-cell square gave round and yellow together in nine cells of sixteen. Where does that nine come from?

Unit 5 · Heredity

1Multiply the one-gene answers

2

Video: Watch: Multiply the one-gene answers

The two small squares side by side with their shaded cells; the two shares multiplied and the product matched against the sixteen-cell square; then a third small square added for a third gene.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L17Ca.mp4

3

Is there a shorter way than sixteen cells? Yes, because the two genes are independent.

4

The shape allele a seed gets has no effect on the color allele it gets. So treat each gene as its own monohybrid cross, and multiply the two probabilities.

5
Check q1

Two independent events must both happen.

How do you find the probability of both happening?

  1. A. Add the two probabilities
    Adding is for one event or the other, when the two cannot both happen.
    Both happening is the multiplication rule.
  2. B. ✓ Multiply the two probabilities

Why: For two independent events that must both happen, multiply their probabilities.
That is the multiplication rule.

6

Round is three cells of four in the shape square. Yellow is three cells of four in the color square.

Two small squares side by side: Rr × Rr with cells RR, Rr, Rr, rr, three of the four shaded for round; and Yy × Yy with cells YY, Yy, Yy, yy, three of the four shaded for yellow; between them a multiplication sign, and to the right: 9 of 16
Two small squares side by side: Rr × Rr with cells RR, Rr, Rr, rr, three of the four shaded for round; and Yy × Yy with cells YY, Yy, Yy, yy, three of the four shaded for yellow; between them a multiplication sign, and to the right: 9 of 16
7

A round yellow seed needs both: round from the shape square and yellow from the color square. The two are independent, so the multiplication rule applies.

If A and B are independent, then this holds. The multiplication rule, as the AP formula sheet prints it: for two independent events that must both happen, multiply their probabilities
8

So any two-gene question is two one-gene questions multiplied. Find each gene’s probability from its own small square, then multiply the two probabilities.

9
Worked example

In fruit flies, gray body (E) is dominant to ebony (e), and normal wings (V) are dominant to vestigial wings (v); the two genes sit on different chromosomes. An EeVv fly is crossed with an Eevv fly. What is the probability that an offspring has an ebony body and vestigial wings?

Write down the values in the question:
Ee × Ee: ebony (ee) = 1 cell of 4, so P(ee) = 1/4
Vv × vv: vestigial (vv) = 2 cells of 4, so P(vv) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(ebony and vestigial)=P(ee)×P(vv)
P(ebony and vestigial)=14×12=18=0.125
A probability is a number between 0 and 1 and carries no unit.
10

A full Punnett square for that cross has eight cells, and one of the eight is eevv: the same one in eight. Here are the two small squares for it.

Two small squares side by side for the fruit-fly cross: Ee × Ee with cells EE, Ee, Ee, ee and the one ebony cell, ee, shaded; and Vv × vv with cells Vv, Vv, vv, vv and the two vestigial cells, vv, shaded; between them a multiplication sign
Two small squares side by side for the fruit-fly cross: Ee × Ee with cells EE, Ee, Ee, ee and the one ebony cell, ee, shaded; and Vv × vv with cells Vv, Vv, vv, vv and the two vestigial cells, vv, shaded; between them a multiplication sign
11

Now consider three genes on three different chromosomes. A full Punnett square would need sixty-four cells.

Three small squares in a row labelled gene A, gene B and gene C, each with one cell shaded, joined by multiplication signs; beside them the words: a full square would need 64 cells
Three small squares in a row labelled gene A, gene B and gene C, each with one cell shaded, joined by multiplication signs; beside them the words: a full square would need 64 cells
12

The product is three one-gene answers multiplied. So the product method scales to three genes; the 64-cell square is impractical.

13

What you are expected to know Calculate the probability of a two-gene genotype or phenotype by treating each gene as its own monohybrid cross and multiplying the two probabilities.

14
Check q2 numeric entry

In pea plants, round seeds (R) are dominant to wrinkled (r) and yellow seeds (Y) are dominant to green (y); the two genes sit on different chromosomes. Two RrYy pea plants are crossed.

Calculate the probability, as a decimal to four decimal places, that a seed is round and green.

Part 1. From Rr × Rr, what is the probability, as a decimal, that a seed is round (RR or Rr)?

Answer: 0.75  (tolerance ±0.005)

Working
Count the round cells of the Rr × Rr square:
round (RR, Rr, Rr) = 3 cells of 4
P(round)=34=0.75

Part 2. From Yy × Yy, what is the probability, as a decimal, that a seed is green (yy)?

Answer: 0.25  (tolerance ±0.005)

Working
Count the green cells of the Yy × Yy square:
green (yy) = 1 cell of 4
P(green)=14=0.25

Answer: 0.1875  (tolerance ±0.0006)

Working
Write down the values in the question:
P(round) = 3/4
P(green) = 1/4
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(round and green)=P(round)×P(green)
P(round and green)=34×14=316=0.1875
15
Check q3 numeric entry

In rabbits, black coat (B) is dominant to brown (b), and short hair (L) is dominant to long hair (l); the two genes sit on different chromosomes. A BbLl rabbit is crossed with a bbLl rabbit.

Calculate the probability, as a decimal to four decimal places, that a young rabbit, a kit, is brown with long hair.

Answer: 0.125  (tolerance ±0.0005)

Working
Write down the values in the question:
Bb × bb: brown (bb) = 2 cells of 4, so P(bb) = 1/2
Ll × Ll: long hair (ll) = 1 cell of 4, so P(ll) = 1/4
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(brown and long)=12×14=18=0.125
16

Back to the two small squares, Rr × Rr and Yy × Yy, with round in three cells of four and yellow in three cells of four.

Two small squares side by side: Rr × Rr with cells RR, Rr, Rr, rr, three of the four shaded for round; and Yy × Yy with cells YY, Yy, Yy, yy, three of the four shaded for yellow; between them a multiplication sign, and to the right: 9 of 16
Two small squares side by side: Rr × Rr with cells RR, Rr, Rr, rr, three of the four shaded for round; and Yy × Yy with cells YY, Yy, Yy, yy, three of the four shaded for yellow; between them a multiplication sign, and to the right: 9 of 16
17

The nine of sixteen round yellow is 34×34: round from the shape square, yellow from the color square, multiplied because the two genes are independent.

18Quick quiz: multiply the one-gene answers mixed practice

19
Check q4 numeric entry

In pea plants, from Tt × Tt the probability that a plant is tall is 3/4. From Pp × Pp the probability that a plant has purple flowers is 3/4. The two genes sit on different chromosomes.

Calculate the probability, as a decimal to four decimal places, that a plant is tall with purple flowers.

Answer: 0.5625  (tolerance ±0.0005)

Working
Write down the values in the question:
P(tall) = 3/4
P(purple) = 3/4
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(tall and purple)=34×34=916=0.5625
20
Check q5 numeric entry

In pea plants, from Rr × rr the probability that a seed is round is 1/2. From Yy × yy the probability that a seed is yellow is 1/2. The two genes sit on different chromosomes.

Calculate the probability, as a decimal, that a seed is round and yellow.

Answer: 0.25  (tolerance ±0.0005)

Working
Write down the values in the question:
P(round) = 1/2
P(yellow) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(round and yellow)=12×12=14=0.25
21
Check q6 numeric entry

In fruit flies, from Ee × Ee the probability that a fly has a gray body is 3/4. From Vv × vv the probability that a fly has vestigial wings is 1/2. The two genes sit on different chromosomes.

Calculate the probability, as a decimal, that a fly is gray with vestigial wings.

Answer: 0.375  (tolerance ±0.0005)

Working
Write down the values in the question:
P(gray) = 3/4
P(vestigial) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(gray and vestigial)=34×12=38=0.375
22
Check q7 numeric entry

In maize, from Pp × Pp the probability that a kernel is yellow is 1/4. From Ss × ss the probability that a kernel is wrinkled is 1/2. The two genes sit on different chromosomes.

Calculate the probability, as a decimal, that a kernel is yellow and wrinkled.

Answer: 0.125  (tolerance ±0.0005)

Working
Write down the values in the question:
P(yellow) = 1/4
P(wrinkled) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(yellow and wrinkled)=14×12=18=0.125
23
Check q8 numeric entry

In fruit flies, from Ee × Ee the probability that a fly has a gray body is 3/4. From Vv × Vv the probability that a fly has vestigial wings is 1/4. The two genes sit on different chromosomes.

Calculate the probability, as a decimal to four decimal places, that a fly is gray with vestigial wings.

Answer: 0.1875  (tolerance ±0.0006)

Working
Write down the values in the question:
P(gray) = 3/4
P(vestigial) = 1/4
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(gray and vestigial)=34×14=316=0.1875

24Mixed practice mixed practice

25
Check q9 numeric entry

In pea plants, round seeds (R) are dominant to wrinkled (r) and yellow seeds (Y) are dominant to green (y); the two genes sit on different chromosomes. Two RrYy pea plants are crossed.

Calculate the probability, as a decimal to four decimal places, that a seed has the genotype RRYY.

Answer: 0.0625  (tolerance ±0.0005)

Working
Write down the values in the question:
Rr × Rr: RR = 1 cell of 4, so P(RR) = 1/4
Yy × Yy: YY = 1 cell of 4, so P(YY) = 1/4
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(RRYY)=14×14=116=0.0625
26
Check q10

In fruit flies, gray body (E) is dominant to ebony (e), and normal wings (V) are dominant to vestigial wings (v); the two genes sit on different chromosomes. An EeVv fruit fly is crossed with another EeVv fly.

Which calculation gives the probability of a gray-bodied fly with normal wings?

  1. A. 34+34
    Gray and normal must both happen in one fly.
    Both together is the multiplication rule, not addition.
  2. B. 14×14
    One cell of four is the recessive share in each cross.
    Gray (EE or Ee) and normal (VV or Vv) are each three cells of four.
  3. C. ✓ 34×34
  4. D. 916−116
    Both together is the product of the two one-gene probabilities, not a difference of cells.

Why: Ee × Ee gives gray in three cells of four.
Vv × Vv gives normal wings in three cells of four.
Both together is the product, 34×34=916.

27
Check q11 numeric entry

Two pea plants heterozygous for three genes on three different chromosomes, AaBbCc and AaBbCc, are crossed.

Calculate the probability, as a decimal to four decimal places, that an offspring is aabbcc.

Answer: 0.0156  (tolerance ±5e-05)

Working
Write down the values in the question:
P(aa) = 1/4
P(bb) = 1/4
P(cc) = 1/4
Write down the equation:
P(A and B and C)=P(A)×P(B)×P(C)
Substitute the values into the equation:
P(aabbcc)=14×14×14=164≈0.0156
28
Check q12

In pea plants, tall (T) is dominant to dwarf (t) and purple flowers (P) are dominant to white (p); the two genes sit on different chromosomes. A TtPp plant is crossed with a ttpp plant.

Which calculation gives the probability that an offspring is tall with purple flowers?

  1. A. ✓ 12×12
  2. B. 12+12
    Tall and purple must both be true of one plant, so the two one-gene probabilities are multiplied, not added.
    Adding would give one, a certainty.
  3. C. 34×34
    Three cells of four is the dominant share from two heterozygous parents.
    Here each gene is Tt × tt or Pp × pp, and the dominant trait fills two cells.
  4. D. 14×14
    One cell of four is the recessive share from two heterozygous parents.
    Here tall is Tt in two cells of four, and so is purple.

Why: Tt × tt gives tall (Tt) in two cells of four.
Pp × pp gives purple (Pp) in two cells of four.
Both together is the product, 12×12=14.

APBIO-U05-L18 Does the data fit? Stating the null, finding chi-square

Topic 5.3 · Mendelian Genetics · 74 steps

Two bars side by side: the left bar, 88 gray-bodied flies, stops below a dashed line marked 100; the right bar, 112 ebony-bodied flies, rises above it
Two bars side by side: the left bar, 88 gray-bodied flies, stops below a dashed line marked 100; the right bar, 112 ebony-bodied flies, rises above it

Here are the offspring of one fruit-fly cross, sorted into two piles: 88 gray-bodied flies and 112 ebony-bodied flies.

The cross was a heterozygous gray-bodied fly with an ebony-bodied one, a test cross. So the model predicts one gray to one ebony: 100 and 100. The piles are not 100 and 100. Is that chance, or is the model wrong?

Unit 5 · Heredity

1The null hypothesis for a breeding cross

2

Video: Watch: The null hypothesis for a breeding cross

The 88 and 112 flies beside the 100 and 100 the model predicts; the null hypothesis written on screen, then three statements sorted beside it: null, alternative, neither.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L18a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L18a.mp4

3
Check q1

A heterozygous plant is crossed with a homozygous recessive plant: a test cross.

Which ratio of dominant-phenotype offspring to recessive offspring does this test cross predict?

  1. A. 3 : 1
    3 : 1 comes from two heterozygous parents.
    Here one parent is homozygous recessive.
  2. B. ✓ 1 : 1

Why: The homozygous recessive parent gives every offspring the recessive allele.
The heterozygous parent gives the dominant allele to half the offspring and the recessive allele to the other half.
So half the offspring show the dominant trait and half show the recessive trait: 1 : 1.

4
Check q2

In tomato plants, tall (T) is dominant to dwarf (t). A grower crosses two Tt plants, predicts 3 tall : 1 dwarf, and counts 130 tall and 30 dwarf plants.

Which of the following is the null hypothesis for this cross?

  1. A. ✓ The plants fit 3 : 1, and the gap from 120 : 40 is due to chance
  2. B. Tall plants are more common than 3 : 1 predicts, so the gap from 120 : 40 is real
    That statement claims a real difference from the model.
    The null hypothesis claims no real difference.

Why: The null hypothesis is the statement that there is no real difference between the counts and the model.
The model predicts 120 tall and 40 dwarf.
So the null hypothesis says the plants fit 3 : 1, and the gap from 120 : 40 is chance.

5

How do you decide whether a batch of counts fits a model, or breaks it?

6

First you say what “fits” would mean. Then you put one number on the gap between the counts and the model.

7

For these flies the test cross predicts 100 gray-bodied and 100 ebony-bodied. The breeder counted 88 and 112.

A table with two rows, gray-bodied and ebony-bodied: counted 88 and 112, predicted by the 1 : 1 model 100 and 100
A table with two rows, gray-bodied and ebony-bodied: counted 88 and 112, predicted by the 1 : 1 model 100 and 100
8

Two things could explain the gap. The first is chance: each fertilization is its own chance event, so a batch of 200 flies strays from the ratio by chance.

9

The second is that the model is wrong.

10

Gray and ebony might not be one gene with one dominant allele, or the parents might not be what the breeder thinks.

11

The statement that there is no real difference, that any difference is due to chance, is called the . It is the statement the counts are tested against.

12

For a breeding cross the null hypothesis reads: the offspring occur in the ratio the model predicts, and any difference between the counts and that ratio is due to chance.

13

Which of these is the null hypothesis for the flies?

  1. “The offspring occur in a 1 : 1 ratio, and the difference between 88 : 112 and 100 : 100 is due to chance.” This is the null hypothesis: it claims no real difference.
  2. “Gray-bodied flies are rarer than a 1 : 1 model predicts.” This is not the null hypothesis: it claims a real difference.
  3. “The cross gives exactly 100 gray and 100 ebony.” This is not the null hypothesis: the null hypothesis expects chance to make the counts stray.
  4. “The one-gene model is wrong.” This is not the null hypothesis: it claims a real difference, in other words.

14

A statement that the counts really differ from the model is called the alternative hypothesis. Naming the reason, a wrong model or a wrong parent, does not stop a statement being the alternative hypothesis.

15

The null hypothesis is always the no-difference statement, whatever the breeder hopes or expects. In a breeding cross the ratio the breeder predicted is the null hypothesis: the counts fit it, apart from chance.

16

What you are expected to know State the null hypothesis for a breeding cross, and tell it from the alternative hypothesis.

17
Check q3

In pea plants, purple flowers (P) are dominant to white (p). A Pp plant is crossed with a pp plant and the grower predicts 1 purple : 1 white. The 120 plants are 71 purple-flowered and 49 white-flowered.

Which of the following statements is the null hypothesis for this cross?

  1. A. Purple-flowered plants are more common than a 1 : 1 model predicts
    The statement claims a real difference from the model.
    So the statement is the alternative hypothesis.
  2. B. The cross gives exactly 60 purple-flowered and 60 white-flowered plants
    The null hypothesis expects the counts to stray from 60 : 60 by chance.
    So the null hypothesis never predicts an exact result.
  3. C. One parent is PP rather than Pp, so the plants do not fit 1 : 1
    A reason for a real difference is an alternative hypothesis, not the null hypothesis.
  4. D. ✓ The plants occur in a 1 : 1 ratio, and the difference from 60 : 60 is due to chance

Why: The null hypothesis says there is no real difference.
So the null hypothesis says the plants fit the 1 : 1 model, and the gap between 71 : 49 and 60 : 60 is chance.
Every other option claims a real difference or an exact count.

18
Check q4

A grower crosses two heterozygous pea plants and predicts 3 tall : 1 short. A student writes the null hypothesis for the cross as: “The 3 : 1 model is wrong.”

Which of the following is correct about the student's statement?

  1. A. ✓ The student has written the alternative hypothesis
  2. B. The student has written the null hypothesis correctly
    The null hypothesis is the no-difference statement.
    “The model is wrong” claims a real difference.
  3. C. The student has written a statement that is neither hypothesis
    A claim that the model is wrong is a claim of a real difference.
    That is the alternative hypothesis.

Why: “The 3 : 1 model is wrong” claims a real difference between the counts and the model.
A claim of a real difference is the alternative hypothesis.
The null hypothesis says the plants occur in a 3 : 1 ratio and any difference is due to chance.

19Quick quiz: null hypothesis mixed practice

20
Check q5

A breeder predicts a ratio from a Punnett square and counts the offspring.

Which of the following is the null hypothesis for the cross?

  1. A. The offspring differ from the ratio the model predicts, for a reason the breeder can name
    A claim that the counts really differ from the model is the alternative hypothesis.
  2. B. The offspring occur in exactly the ratio the model predicts, with no straying at all
    The null hypothesis expects chance to make the counts stray from the ratio.
    So the null hypothesis never promises exact counts.
  3. C. ✓ The offspring occur in the ratio the model predicts, and any difference from it is due to chance

Why: The null hypothesis is the no-difference statement.
For a breeding cross, the no-difference statement is that the offspring fit the predicted ratio and any gap is chance.

21
Check q6

A grower crosses two Pp plants, predicts 3 purple : 1 white, and counts 47 purple and 13 white plants. A student writes: “The plants occur in a 3 : 1 ratio, and the difference between 47 : 13 and 45 : 15 is due to chance.”

Which of the following is the student's statement?

  1. A. ✓ The null hypothesis
  2. B. The alternative hypothesis
    The statement claims no real difference.
    The alternative hypothesis claims a real difference.
  3. C. Neither
    The statement says the counts fit the model and the gap is chance.
    That is exactly the no-difference statement.

Why: The statement says the counts fit the model and the gap is chance.
That is the no-difference statement, so the statement is the null hypothesis.

22
Check q7

A grower crosses two Pp plants, predicts 3 purple : 1 white, and counts 47 purple and 13 white plants. A student writes: “White plants are rarer than the 3 : 1 model predicts.”

Which of the following is the student's statement?

  1. A. The null hypothesis
    The statement claims a real difference from the model.
    The null hypothesis claims no real difference.
  2. B. ✓ The alternative hypothesis
  3. C. Neither
    A claim that the counts really differ from the model is the alternative hypothesis.

Why: The statement claims that white plants really are rarer than the model predicts.
A claim of a real difference is the alternative hypothesis.

23
Check q8

In mice, black fur (B) is dominant to brown (b). A breeder crosses a Bb mouse with a bb mouse, predicts 1 black : 1 brown, and counts 28 black and 32 brown pups. A student writes: “The cross gives exactly 30 black and 30 brown pups.”

Which of the following is the student's statement?

  1. A. The null hypothesis
    The null hypothesis expects chance to make the counts stray from 30 : 30.
    The null hypothesis never promises exact counts.
  2. B. The alternative hypothesis
    The statement does not claim that the counts really differ from the model.
    The statement promises exact counts, which neither hypothesis does.
  3. C. ✓ Neither

Why: The null hypothesis says the pups fit 1 : 1 and stray from 30 : 30 by chance.
The alternative hypothesis says the pups really differ from 1 : 1.
A promise of exactly 30 and 30 is neither.

24
Check q9

In mice, black fur (B) is dominant to brown (b). A breeder crosses a Bb mouse with a bb mouse, predicts 1 black : 1 brown, and counts 28 black and 32 brown pups. A student writes: “The pups occur in a 1 : 1 ratio, and the gap between the counts and that ratio is due to chance.”

Which of the following is the student's statement?

  1. A. ✓ The null hypothesis
  2. B. The alternative hypothesis
    The statement claims no real difference, only chance.
    The alternative hypothesis claims a real difference.
  3. C. Neither
    The statement names the model's ratio and puts the gap down to chance.
    That is the null hypothesis.

Why: The statement says the pups fit the 1 : 1 model and the gap is chance.
That is the no-difference statement, so the statement is the null hypothesis.

25
Check q10

In mice, black fur (B) is dominant to brown (b). A breeder crosses a Bb mouse with a bb mouse, predicts 1 black : 1 brown, and counts 28 black and 32 brown pups. A student writes: “The black parent is really BB, so the 1 : 1 model is wrong.”

Which of the following is the student's statement?

  1. A. The null hypothesis
    The statement says the model is wrong, which is a claim of a real difference.
    The null hypothesis says the model fits.
  2. B. ✓ The alternative hypothesis
  3. C. Neither
    A reason why the counts really differ from the model is one form of the alternative hypothesis.

Why: The statement claims that the model is wrong: the counts really differ from 1 : 1.
An alternative hypothesis is any statement that the model's prediction is wrong.
Naming the reason, a BB parent, does not stop the statement being one.

26
Check q11

A breeder crosses two Bb mice, predicts 3 black : 1 brown, and counts 61 black and 19 brown pups. A student writes: “The pups occur in a 3 : 1 ratio, and the gap between the counts and that ratio is due to chance.”

Which of the following is the student's statement?

  1. A. ✓ The null hypothesis
  2. B. The alternative hypothesis
    The statement claims no real difference from the model.
    The alternative hypothesis claims a real difference.
  3. C. Neither
    The statement names the model's ratio and puts the gap down to chance.
    That is exactly the null hypothesis.

Why: The statement says the pups fit the 3 : 1 model and the gap is chance.
That is the no-difference statement, so the statement is the null hypothesis.

27
Practice writing an answer

In guinea pigs, black coat (B) is dominant to white (b). A breeder crosses a Bb guinea pig with a bb guinea pig and predicts 1 black : 1 white. The 40 pups are 17 black and 23 white.

(a) State the null hypothesis for this cross. (1 pt)

Model answer The pups occur in a 1 : 1 ratio of black to white, and any difference between 17 : 23 and 20 : 20 is due to chance.
Rubric
  • Award 1 point for: the pups fit the predicted 1 : 1 ratio and the difference from it is due to chance (a no-difference statement).
  • Do not award: a statement that white pups are more common than predicted, or that the model is wrong (those are alternative hypotheses).

Slip Writing that white pups are more common than the model predicts. That claims a real difference, so it is the alternative hypothesis, not the null hypothesis.

28Observed and expected: o and e

29

Video: Watch: Observed and expected: o and e

The two numbers for one class labeled on screen: the count the breeder made, o, and the count the ratio predicts for the total, e; the expected-count equation written once with e on its left.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L18b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L18b.mp4

30
Check q12

Two Pp pea plants are crossed, and their Punnett square predicts 3 purple : 1 white. There are 120 offspring.

Which of the following is the expected count of white-flowered plants?

  1. A. 1 plant
    The digits of a ratio are shares of the total, not counts of offspring.
    25% of 120 plants is 30 plants.
  2. B. ✓ 30 plants
  3. C. 90 plants
    90 plants is 3 of the 4 shares: the purple-flowered class.
    The white-flowered class is 1 of the 4 shares.

Why: A 3 : 1 ratio has four shares.
The white-flowered class is 1 of the 4 shares.
25% of 120 plants is 30 plants.

31

To measure the gap you need two numbers for each class: the number you counted, and the number the model predicts for the total you grew.

32

The number you counted in a class is called the . Your formula sheet writes it o.

33

The number the model's ratio predicts for that class is called the . Your formula sheet writes it e, and the equation is the one you know.

The expected count for a class, e: the total number of offspring times that class's share of the predicted ratio
34

What you are expected to know Tell the observed count, o, from the expected count, e, for each class of a breeding cross.

35
Check q13 numeric entry

In tobacco plants, green seedlings (G) are dominant to albino seedlings (g). A Gg plant is crossed with a gg plant, and the test cross predicts 1 green : 1 albino. There are 250 seedlings.

Calculate the expected count of albino seedlings, e.

Answer: 125  (tolerance ±0)

Working
Write down the values in the question:
total = 250
ratio = 1 : 1, so 2 shares in all
albino share = 1 of 2
Write down the equation:
e=total×that class's share of the ratio
Substitute the values into the equation:
ealbino=250×12=125

36Quick quiz: observed count (o), expected count (e) mixed practice

37
Check q14

A breeder has two numbers for one class of a breeding cross.

Which of the following is the expected count, e, of the class?

  1. A. ✓ The number the ratio predicts for that class, for the total grown
  2. B. The number of offspring the breeder counted in that class
    The number counted is the observed count, o.
  3. C. The digit that class has in the model's ratio, 3 or 1
    A ratio's digit is a share of the total, not a count.
    The expected count is the total times that share.

Why: The expected count is the model's prediction for that class.
It is the total times that class's share of the ratio, written e.

38
Check q15

A grower sorts 180 pea plants and finds 131 purple-flowered.

Is 131 the observed count or the expected count?

  1. A. ✓ The observed count, o
  2. B. The expected count, e
    131 is the number the grower found by counting.
    A count made from the plants is the observed count, o.

Why: 131 is the number the grower counted.
The number counted in a class is the observed count, o.

39
Check q16

For 180 pea plants from a Pp × Pp cross, the 3 : 1 model predicts 135 purple-flowered.

Is 135 the observed count or the expected count?

  1. A. The observed count, o
    135 is the number the model predicts, not a number anyone counted.
    A predicted number is the expected count, e.
  2. B. ✓ The expected count, e

Why: 135 is the number the ratio predicts for the total of 180.
The number the model predicts for a class is the expected count, e.

40
Check q17

A breeder counts 41 black pups in a litter of 96.

Is 41 the observed count or the expected count?

  1. A. ✓ The observed count, o
  2. B. The expected count, e
    41 is the number the breeder found by counting the pups.
    A count made from the litter is the observed count, o.

Why: 41 is the number the breeder counted.
The number counted in a class is the observed count, o.

41
Check q18

For a litter of 96 pups, the 1 : 1 model predicts 48 black pups.

Is 48 the observed count or the expected count?

  1. A. The observed count, o
    48 comes from the ratio and the total, not from counting pups.
    A predicted number is the expected count, e.
  2. B. ✓ The expected count, e

Why: 48 is the number the 1 : 1 model predicts for 96 pups.
The number the model predicts is the expected count, e.

42
Check q19

Of 240 corn kernels on one ear, 19 are yellow.

Is 19 the observed count or the expected count?

  1. A. ✓ The observed count, o
  2. B. The expected count, e
    19 is a number found by counting the kernels.
    A count made from the ear is the observed count, o.

Why: 19 is the number of yellow kernels counted on the ear.
The number counted in a class is the observed count, o.

43
Practice writing an answer

For one class of a breeding cross, a student has two numbers: 57, the number of striped fish she counted, and 60, the number the 1 : 1 model predicts for her 120 fish.

(a) Identify which number is the observed count, o, and which is the expected count, e. (1 pt)

Model answer 57 is the observed count, o, because the student counted 57 striped fish.
60 is the expected count, e, because the 1 : 1 model predicts 60 striped fish for 120 fish.
Rubric
  • Award 1 point for: 57 is the observed count (o) AND 60 is the expected count (e).

Slip Swapping the two. The observed count is the number counted; the expected count is the number the model predicts.

44How far from the model: chi-square

45

Video: Watch: How far from the model: chi-square

The 88 and 112 bars against the dashed line at 100; the chi-square equation typeset as the formula sheet prints it; one class at a time squared, divided by its own e, and added to the sum until 2.88 stands on screen.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L18c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L18c.mp4

46

Here are the fly counts against the expected line. Gray-bodied fell short of 100 by twelve flies, and ebony-bodied overshot by twelve.

Two bars, 88 gray-bodied and 112 ebony-bodied flies, with a dashed line at the expected count of 100 crossing both; gridlines every 20 flies
Two bars, 88 gray-bodied and 112 ebony-bodied flies, with a dashed line at the expected count of 100 crossing both; gridlines every 20 flies
47

One number should say how far, in all, the counts sit from the model. That number is built in three steps:

  1. For each class, take the gap between its observed count and its expected count, and square it.
  2. Divide that squared gap by the class's own expected count.
  3. Add the classes.

48

That sum is called , written χ2 and said “kye-square”. Here is its equation, exactly as your formula sheet prints it.

Chi-square, as the AP formula sheet writes it: for every class, the observed count o minus the expected count e, squared, divided by e; then the classes added up
49
Worked example

The test cross gave 88 gray-bodied and 112 ebony-bodied flies; the 1 : 1 model predicts 100 and 100. Calculate chi-square.

Write down the values in the question:
o = 88 (gray), 112 (ebony)
e = 100 (gray), 100 (ebony)
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
χ2=∑(o−e)2e
gray: (88−100)2100=144100=1.44
ebony: (112−100)2100=144100=1.44
Add the classes:
χ2=1.44+1.44=2.88(no unit)
50

Every class is squared and divided by its own expected count, never by another class's expected count.

51

The sum has no unit: chi-square is a plain number. The bigger chi-square is, the further the counts sit from the model.

52

What you are expected to know Calculate chi-square for the classes of a breeding cross with the formula-sheet equation, one class per line, and add the classes to one number with no unit.

53
Check q20 numeric entry

In tomato plants, tall (T) is dominant to dwarf (t). Two Tt plants are crossed; their square predicts 3 tall : 1 dwarf, so for 160 offspring the expected counts are 120 tall and 40 dwarf. The observed counts are 123 tall and 37 dwarf.

Calculate chi-square for this cross, to three significant figures.

Part 1. Work out the tall class's term, the squared gap divided by its expected count. What is the tall class's term, to three decimal places?

Answer: 0.075  (tolerance ±0.0015)

Working
Square the gap and divide by the class's own expected count:
tall: (123−120)2120=9120=0.075

Part 2. Work out the dwarf class's term. What is the dwarf class's term, to three decimal places?

Answer: 0.225  (tolerance ±0.0015)

Working
Square the gap and divide by the class's own expected count:
dwarf: (37−40)240=940=0.225

Answer: 0.3  (tolerance ±0.005)

Working
Write down the values in the question:
o = 123 (tall), 37 (dwarf)
e = 120 (tall), 40 (dwarf)
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
tall: (123−120)2120=0.075
dwarf: (37−40)240=0.225
Add the classes:
χ2=0.075+0.225=0.300
54
Check q21 numeric entry

In tobacco plants, green seedlings (G) are dominant to albino seedlings (g). Two Gg plants are crossed; their square predicts 3 green : 1 albino, so for 200 seedlings the expected counts are 150 green and 50 albino. The tray holds 160 green and 40 albino seedlings.

Calculate chi-square for this cross, to three significant figures.

Answer: 2.67  (tolerance ±0.005)

Working
Write down the values in the question:
o = 160 (green), 40 (albino)
e = 150 (green), 50 (albino)
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
green: (160−150)2150=100150=0.67
albino: (40−50)250=10050=2.00
Add the classes:
χ2=0.67+2.00=2.67
55

Back to the 200 flies from the test cross: 88 gray-bodied and 112 ebony-bodied, against the 100 and 100 the 1 : 1 model predicts.

A table with two rows, gray-bodied and ebony-bodied: counted 88 and 112, predicted by the 1 : 1 model 100 and 100
A table with two rows, gray-bodied and ebony-bodied: counted 88 and 112, predicted by the 1 : 1 model 100 and 100
56

The null hypothesis says the flies fit 1 : 1 and the gap is chance. The expected counts, 100 and 100, came from the ratio and the total.

57

Chi-square, χ2=2.88, put one number on the gap.

58Quick quiz: chi-square (χ²) mixed practice

59
Check q22

A breeder has the observed count and the expected count for every class of a breeding cross.

Which of the following is chi-square?

  1. A. The number of offspring whose class did not fit the model's ratio
    Chi-square is built from squared gaps divided by expected counts, not by counting offspring.
  2. B. ✓ For every class, the squared gap between o and e divided by e, added over the classes
  3. C. The gap between o and e in the largest class, as a percentage of the total
    Every class enters the sum, and nothing is divided by the total or multiplied by 100.

Why: For each class the equation takes o minus e, squares it, and divides by that class's own e.
Then the classes are added.
That sum is chi-square.

60
Practice writing an answer

A breeder calculates chi-square for the classes of a breeding cross.

(a) State what chi-square measures. (1 pt)

Model answer Chi-square measures how far the observed counts sit from the counts the model predicts, over all the classes, as one number with no unit.
Rubric
  • Award 1 point for: chi-square measures the gap between the observed counts and the expected counts (how far the counts sit from the model), over all the classes.

Slip Saying chi-square counts the offspring that did not fit. It is a sum of squared gaps divided by expected counts, not a count.

61
Check q23 numeric entry

For one class of a breeding cross, o = 52 offspring and e = 40 offspring.

Calculate that class's term in the chi-square sum, (o − e)²/e.

Answer: 3.6  (tolerance ±0.005)

Working
Write down the values in the question:
o = 52
e = 40
Write down the class's term:
(o−e)2e
Substitute the values:
(52−40)240=14440=3.60
62
Check q24 numeric entry

For one class of a breeding cross, o = 18 offspring and e = 25 offspring.

Calculate that class's term in the chi-square sum, (o − e)²/e.

Answer: 1.96  (tolerance ±0.005)

Working
Write down the values in the question:
o = 18
e = 25
Write down the class's term:
(o−e)2e
Substitute the values:
(18−25)225=4925=1.96
63
Check q25 numeric entry

For one class of a breeding cross, o = 66 offspring and e = 60 offspring.

Calculate that class's term in the chi-square sum, (o − e)²/e.

Answer: 0.6  (tolerance ±0.005)

Working
Write down the values in the question:
o = 66
e = 60
Write down the class's term:
(o−e)2e
Substitute the values:
(66−60)260=3660=0.600
64
Check q26 numeric entry

For one class of a breeding cross, o = 12 offspring and e = 20 offspring.

Calculate that class's term in the chi-square sum, (o − e)²/e.

Answer: 3.2  (tolerance ±0.005)

Working
Write down the values in the question:
o = 12
e = 20
Write down the class's term:
(o−e)2e
Substitute the values:
(12−20)220=6420=3.20
65
Check q27 numeric entry

For one class of a breeding cross, o = 231 offspring and e = 220 offspring.

Calculate that class's term in the chi-square sum, (o − e)²/e.

Answer: 0.55  (tolerance ±0.005)

Working
Write down the values in the question:
o = 231
e = 220
Write down the class's term:
(o−e)2e
Substitute the values:
(231−220)2220=121220=0.550

66Mixed practice mixed practice

67
Check q28

For one class of a breeding cross, the observed count is 30 and the expected count is 40. Another class has an expected count of 60.

Which of the following is the first class's term in the chi-square sum?

  1. A. The squared gap, 100, divided by 60
    Each class is divided by its own expected count, never by another class's expected count.
  2. B. The gap, 10, divided by 40
    The gap is squared before dividing, so that a shortfall and an overshoot both count as positive.
  3. C. ✓ The squared gap, 100, divided by 40
  4. D. The squared gap, 100, divided by 30
    The divisor is the expected count, not the observed count.

Why: For each class the equation takes observed minus expected, squares it, and divides by that class's own expected count.
Here that is the squared gap, 100, over this class's expected count of 40.

68
Check q29 numeric entry

In fruit flies, normal wings (V) are dominant to vestigial wings (v). A Vv fly is test-crossed with a vv fly and the model predicts 1 normal : 1 vestigial. The table below gives the counts for 100 offspring.

A table with two rows, normal wings and vestigial wings: observed 45 and 55; the 1 : 1 model predicts 50 and 50
A table with two rows, normal wings and vestigial wings: observed 45 and 55; the 1 : 1 model predicts 50 and 50

Calculate chi-square for this cross.

Answer: 1  (tolerance ±0.005)

Working
Write down the values in the question:
o = 45 (normal), 55 (vestigial)
e = 50 (normal), 50 (vestigial)
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
normal: (45−50)250=2550=0.50
vestigial: (55−50)250=2550=0.50
Add the classes:
χ2=0.50+0.50=1.00
69
Check q30

In corn, purple kernels (P) are dominant to yellow (p). A breeder crosses two Pp plants, predicts 3 purple : 1 yellow, and counts 77 purple and 19 yellow kernels. A student writes: “Yellow kernels are rarer than the 3 : 1 model predicts.”

Which of the following is the student's statement?

  1. A. The null hypothesis
    The statement claims a real difference from the model.
    The null hypothesis claims no real difference.
  2. B. ✓ The alternative hypothesis
  3. C. Neither
    A claim that the counts really differ from the model is the alternative hypothesis.

Why: The statement claims that yellow kernels really are rarer than the model predicts.
A claim of a real difference is the alternative hypothesis.

70
Check q31 numeric entry

Two Gg tobacco plants are crossed; their square predicts 3 green : 1 albino, and 80 seedlings come up.

Calculate the expected count of albino seedlings.

Answer: 20  (tolerance ±0)

Working
Write down the values in the question:
total = 80
ratio = 3 : 1, so 4 shares in all
albino share = 1 of 4
Write down the equation:
e=total×that class's share of the ratio
Substitute the values into the equation:
ealbino=80×14=20
71
Check q32

A student calculates chi-square for a breeding cross and gets 3.0.

Which statement about the 3.0 is right?

  1. A. ✓ It is a plain number with no unit, measuring how far the counts sit from the model
  2. B. It is a count of the offspring whose class did not fit the model's ratio
    Chi-square is a sum of squared gaps divided by expected counts, not a count of any offspring.
  3. C. It is a percentage of the offspring that fell outside the model's predicted ratio
    Nothing in the equation is divided by the total or multiplied by 100.
  4. D. It is the number of classes the offspring were sorted into for the test
    The number of classes is what the counts are sorted into.
    Chi-square adds a term for each class.

Why: Chi-square adds up each class's squared gap divided by its expected count.
The units cancel.
What is left is one plain number for how far the counts sit from the model.

72
Check q33 numeric entry

Two Gg tobacco plants are crossed and 400 seedlings come up: 285 green and 115 albino. Their square predicts 3 green : 1 albino, so the expected counts are 300 green and 100 albino.

Calculate chi-square for this cross.

Answer: 3  (tolerance ±0.005)

Working
Write down the values in the question:
o = 285 (green), 115 (albino)
e = 300 (green), 100 (albino)
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
green: (285−300)2300=225300=0.75
albino: (115−100)2100=225100=2.25
Add the classes:
χ2=0.75+2.25=3.00
73
Practice writing an answer

In sheep, white wool (W) is dominant to black wool (w). A shepherd crosses a Ww ram with a ww ewe flock and predicts 1 white lamb : 1 black lamb. Over a season the flock bears 60 lambs: 36 white and 24 black.

(a) State the null hypothesis for this cross. (1 pt)

Model answer The lambs occur in a 1 : 1 ratio of white to black, and any difference between the observed counts and that ratio is due to chance.
Rubric
  • Award 1 point for: the lambs fit the predicted 1 : 1 ratio and the difference from it is due to chance (a no-difference statement).
  • Do not award: a statement that white lambs are more common than predicted, or that the model is wrong (those are alternative hypotheses).

Slip Writing that white lambs are more common than the model predicts. That claims a real difference, so it is the alternative hypothesis, not the null hypothesis.

(b) Calculate the expected count of black lambs. (1 pt)

Answer: 30  (tolerance ±0)

Model answer The expected count of black lambs is 30.
Working
Write down the values in the question:
total = 60
ratio = 1 : 1, so 2 shares in all
black share = 1 of 2
Write down the equation:
e=total×that class's share of the ratio
Substitute the values into the equation:
eblack=60×12=30
Rubric
  • Award 1 point for: 30 black lambs expected (and 30 white).

(c) Calculate chi-square for the flock's counts. (1 pt)

Answer: 2.4  (tolerance ±0.005)

Model answer Chi-square for the flock is 2.40.
Working
Write down the values in the question:
o = 36 (white), 24 (black)
e = 30 (white), 30 (black)
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
white: (36−30)230=3630=1.20
black: (24−30)230=3630=1.20
Add the classes:
χ2=1.20+1.20=2.40
Rubric
  • Award 1 point for: chi-square 2.40 (accept 2.4).

(d) Describe, from the size of the chi-square value alone, what 2.40 says about how far the flock's counts sit from the counts the 1 : 1 model predicts. (1 pt)

Model answer Chi-square measures how far the observed counts sit from the counts the 1 : 1 model predicts.
For each class it takes the gap between the observed count and the expected count, squares that gap, and divides by the expected count.
Then it adds the classes.
The result, 2.40, is a plain number with no unit.
A larger value would mean the counts sat further from the model.
Rubric
  • Award 1 point for: chi-square measures the gap between the observed counts and the model's expected counts, with a larger value meaning a larger gap (no unit). The description alone earns the point; no verdict is asked for.
  • Award also for: a comparison of 2.40 with a critical value (2.40 is below 3.84, the value for one degree of freedom, so the gap is small enough to be chance), with or without the size-only description.

Slip Saying that 2.40 is the number of lambs that did not fit, or a percentage. Chi-square is a plain unitless number built from squared gaps divided by expected counts.

Glossary

null hypothesis
The statement that there is no real difference and any difference is due to chance. For a breeding cross: the offspring occur in the ratio the model predicts, and the counts differ from it only by chance.
observed count (o)
The number of offspring actually counted in one class, written o on the formula sheet.
expected count (e)
The number of offspring the model's ratio predicts for one class, for the total actually grown: the total times that class's share of the ratio. Written e on the formula sheet.
chi-square (χ²)
One number for how far the observed counts sit from the expected counts: for each class, observed minus expected, squared, divided by that class's expected count; then the classes added. It has no unit.

APBIO-U05-L19 Does the data fit? Reading the verdict

Topic 5.3 · Mendelian Genetics · 69 steps

On the left, chi-square equals 2.88 in large type; on the right, a small copy of the formula sheet's chi-square table with eight columns for degrees of freedom 1 to 8 and two rows for p values 0.05 and 0.01
On the left, chi-square equals 2.88 in large type; on the right, a small copy of the formula sheet's chi-square table with eight columns for degrees of freedom 1 to 8 and two rows for p values 0.05 and 0.01

Here is the number from the fly cross, chi-square equals 2.88, and beside it the chi-square table printed on your formula sheet: eight columns and two rows.

The flies were sorted into two classes, gray-bodied and ebony-bodied. Which column of the table is theirs, which row, and what does the number in that cell mean for the 1 : 1 model?

Unit 5 · Heredity

1Degrees of freedom

2

Video: Watch: Degrees of freedom

The 200 flies in two boxes: the first count chosen, the second fixed by the total; the count of free classes written as classes minus one, then the same for four classes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L19a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L19a.mp4

3
Check q1

A breeder calculates chi-square for the two classes of a test cross and gets 2.88.

Which of the following does 2.88 measure?

  1. A. ✓ How far the observed counts sit from the counts the model predicts
  2. B. How many offspring did not fit the model's ratio
    Chi-square is a sum of squared gaps divided by expected counts.
    It is not a count of offspring.

Why: For each class, chi-square takes the gap between o and e, squares it, and divides by e.
Then the classes are added.
So 2.88 measures how far the counts sit from the model.

4

How do you read a chi-square against the table and reach a verdict?

5

The table needs two things from you: a column and a row. Then it gives one number back, and the verdict comes from comparing chi-square with that number.

6

The column depends on the number of classes, not on the number of flies. Once the total, 200, and the gray-bodied count, 88, are fixed, the ebony-bodied count is fixed too: 112.

Two boxes side by side labeled gray-bodied and ebony-bodied, under a total of 200; the first box reads 88, free to vary, and the second reads 112, fixed by the total
Two boxes side by side labeled gray-bodied and ebony-bodied, under a total of 200; the first box reads 88, free to vary, and the second reads 112, fixed by the total
7

Only one class, the gray-bodied, is free to vary. The number of classes free to vary is called the : the number of classes minus one, so two classes give one degree of freedom.

8

Degrees of freedom count classes, never offspring. Two hundred flies in two classes give one degree of freedom, and so would two thousand flies.

9

Three flower classes from a breeding cross, red, pink and white, give two degrees of freedom: once two counts and the total are fixed, the third count is fixed too.

10

What you are expected to know Identify the degrees of freedom for a set of counts: the number of classes minus one.

11
Check q2 numeric entry

A breeder sorts the offspring of a breeding cross into five classes.

Calculate the degrees of freedom for a chi-square test of the five classes.

Answer: 4  (tolerance ±0)

Working
Write down the values in the question:
classes = 5
Subtract one from the number of classes:
degrees of freedom=5−1=4

12Quick quiz: degrees of freedom mixed practice

13
Check q3

A breeder is about to read the chi-square table for a set of counts.

Which of the following is the degrees of freedom for the test?

  1. A. ✓ The number of classes minus one
  2. B. The number of offspring minus one
    Degrees of freedom count classes, never offspring.
  3. C. The number of classes
    One class is fixed by the total, so one class is not free to vary.

Why: Once the total and all but one class are fixed, the last class is fixed too.
So the classes free to vary are the number of classes minus one.

14
Check q4 numeric entry

A breeder sorts the kernels on an ear of corn into three color classes.

Calculate the degrees of freedom for a chi-square test of the three classes.

Answer: 2  (tolerance ±0)

Working
Write down the values in the question:
classes = 3
Subtract one from the number of classes:
degrees of freedom=3−1=2
15
Check q5 numeric entry

A grower counts 800 seedlings and sorts them into four classes.

Calculate the degrees of freedom for a chi-square test of the four classes.

Answer: 3  (tolerance ±0)

Working
Write down the values in the question:
classes = 4
seedlings counted = 800 (this number plays no part)
Subtract one from the number of classes:
degrees of freedom=4−1=3
16
Check q6 numeric entry

A grower counts 500 seedlings from a test cross and sorts them into two classes, tall and dwarf.

Calculate the degrees of freedom for a chi-square test of these counts.

Answer: 1  (tolerance ±0)

Working
Write down the values in the question:
classes = 2
seedlings counted = 500 (this number plays no part)
Subtract one from the number of classes:
degrees of freedom=2−1=1
17
Check q7 numeric entry

A breeder sorts the fruits of a breeding cross into six classes.

Calculate the degrees of freedom for a chi-square test of the six classes.

Answer: 5  (tolerance ±0)

Working
Write down the values in the question:
classes = 6
Subtract one from the number of classes:
degrees of freedom=6−1=5
18
Check q8 numeric entry

A breeder counts 60 pups from a test cross and sorts them into two classes, black and brown.

Calculate the degrees of freedom for a chi-square test of these counts.

Answer: 1  (tolerance ±0)

Working
Write down the values in the question:
classes = 2
pups counted = 60 (this number plays no part)
Subtract one from the number of classes:
degrees of freedom=2−1=1
19
Practice writing an answer

A grower counts 300 plants from a breeding cross and sorts them into six classes.

(a) State the degrees of freedom for a chi-square test of these counts, and justify the number. (1 pt)

Model answer Five degrees of freedom.
Six classes minus one is five, because once five counts and the total are fixed, the sixth count is fixed too.
The 300 plants play no part.
Rubric
  • Award 1 point for: five degrees of freedom, because degrees of freedom are the number of classes minus one (6 − 1 = 5).

Slip Using the 300 plants: 299 degrees of freedom. Degrees of freedom count classes, never offspring.

20The critical value from the table

21

Video: Watch: The critical value from the table

A finger tracing the formula sheet's table: down the column for one degree of freedom, along the p = 0.05 row, stopping on 3.84; the 0.01 row named once as the stricter row.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L19b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L19b.mp4

22

Here is the chi-square table your formula sheet prints. It is a table with a column for each number of degrees of freedom, 1 to 8, and two rows.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
23

Each row is labeled by a number, 0.05 or 0.01. That row label is called a . It is a label you read, not a number you calculate here.

24

The 0.05 means this: if the model were right, a gap this large or larger would turn up by chance in fewer than 5 batches in 100. The 0.01 row asks for a gap that chance would give in fewer than 1 batch in 100, which is why it is stricter.

25

The 0.05 row is the row to read. Your exam questions use the 0.05 row unless they say otherwise.

26

The 0.01 row is a stricter test: its values are larger, so a gap has to be bigger before the model is rejected.

27

For the flies, go down the column for one degree of freedom and along the 0.05 row: 3.84.

The same chi-square table with the cell for one degree of freedom in the p = 0.05 row ringed: 3.84
The same chi-square table with the cell for one degree of freedom in the p = 0.05 row ringed: 3.84
28

The number in that cell is called the , because chi-square has to exceed it before the model is rejected.

29

What you are expected to know Read the critical value for a set of counts from the formula sheet's table, at the degrees of freedom and the p = 0.05 row.

30
Check q9 numeric entry

Two pea plants heterozygous for two genes are crossed, and the seeds are sorted into four classes: round yellow, round green, wrinkled yellow and wrinkled green. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Read the critical value for the four seed classes from the p = 0.05 row.

Part 1. Count the classes and subtract one. How many degrees of freedom do four classes give?

Answer: 3  (tolerance ±0)

Working
Subtract one from the number of classes:
degrees of freedom=4−1=3

Answer: 7.81  (tolerance ±0.005)

Working
Write down the values in the question:
classes = 4
degrees of freedom = 3
row = p value 0.05
Read the table at the column for 3 degrees of freedom and the 0.05 row:
critical value = 7.81

31Quick quiz: critical value, p value mixed practice

32
Check q10

A breeder has chi-square for a set of counts and the formula sheet's table.

Which of the following is the critical value?

  1. A. The chi-square value the breeder calculated from the observed and expected counts
    Chi-square is calculated from the counts.
    The critical value is read from the table.
  2. B. ✓ The number in the table at the degrees of freedom and the p = 0.05 row
  3. C. The number of classes minus one, which picks the table's column
    The number of classes minus one is the degrees of freedom.
    It picks the column; the critical value is the number in the cell.

Why: The critical value is read from the table, not calculated.
Go down the column for the degrees of freedom and along the 0.05 row.
The number in that cell is the critical value.

33
Check q11

The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Which of the following is a p value in the table?

  1. A. The number in a cell, such as 3.84
    The number in a cell is a critical value.
  2. B. The number of degrees of freedom at the top of a column
    The column heading is the degrees of freedom.
  3. C. ✓ The label of a row, 0.05 or 0.01

Why: The table has two rows.
Each row is labeled 0.05 or 0.01, and that label is the p value.
The 0.05 row is the one to read unless the question says otherwise.

34
Practice writing an answer

The formula sheet's chi-square table has two rows, labeled p = 0.05 and p = 0.01.

(a) Describe how the critical values in the 0.01 row differ from those in the 0.05 row. (1 pt)

Model answer In every column the 0.01 row's critical value is larger than the 0.05 row's.
So chi-square has to be larger before the null hypothesis is rejected: the 0.01 row is a stricter test.
Rubric
  • Award 1 point for: the 0.01 row's critical values are larger, so the test is stricter (chi-square must be larger to reject).

Slip Saying the 0.01 row is easier to reject on because 0.01 is a smaller number. The row label is smaller, but the critical values in it are larger.

35
Check q12 numeric entry

A chi-square test has two degrees of freedom. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Read the critical value at p = 0.05.

Answer: 5.99  (tolerance ±0.005)

Working
Write down the values in the question:
degrees of freedom = 2
row = p value 0.05
Read the table at the column for 2 degrees of freedom and the 0.05 row:
critical value = 5.99
36
Check q13 numeric entry

A breeder crosses two plants and sorts the offspring into five classes. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Read the critical value for a chi-square test of those counts at p = 0.05.

Answer: 9.49  (tolerance ±0.005)

Working
Write down the values in the question:
classes = 5
row = p value 0.05
Subtract one from the number of classes:
degrees of freedom=5−1=4
Read the table at the column for 4 degrees of freedom and the 0.05 row:
critical value = 9.49
37
Check q14 numeric entry

A chi-square test compares six classes at p = 0.05. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Read the critical value for the test.

Answer: 11.07  (tolerance ±0.005)

Working
Write down the values in the question:
classes = 6
row = p value 0.05
Subtract one from the number of classes:
degrees of freedom=6−1=5
Read the table at the column for 5 degrees of freedom and the 0.05 row:
critical value = 11.07
38
Check q15 numeric entry

A chi-square test has four degrees of freedom, and the stricter row is to be read. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Read the critical value at p = 0.01.

Answer: 13.28  (tolerance ±0.005)

Working
Write down the values in the question:
degrees of freedom = 4
row = p value 0.01
Read the table at the column for 4 degrees of freedom and the 0.01 row:
critical value = 13.28

39Reject, or fail to reject

40

Video: Watch: Reject, or fail to reject

A number line with 3.84 marked; 2.88 placed before it and 9.6 beyond it; the verdict written in its two allowed wordings, and the two banned wordings crossed out.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L19c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L19c.mp4

41
Check q16

A breeder predicts 1 black : 1 brown pups from a test cross of mice.

Which of the following is the null hypothesis for the cross?

  1. A. The pups differ from 1 : 1 because one parent is not the genotype the breeder thinks
    A claim of a real difference, with its reason, is the alternative hypothesis.
  2. B. ✓ The pups occur in a 1 : 1 ratio, and any difference from it is due to chance

Why: The null hypothesis is the no-difference statement.
For a breeding cross, the no-difference statement is that the pups fit the predicted ratio and any gap is chance.

42

If chi-square is larger than the critical value, reject the null hypothesis: the counts do not fit the model.

A number line from 0 to 10 with the critical value 3.84 marked by a dashed line; the region beyond it is shaded and labeled reject, the region before it labeled fail to reject; the fly value 2.88 sits before the line and the value 9.6 sits well beyond it
A number line from 0 to 10 with the critical value 3.84 marked by a dashed line; the region beyond it is shaded and labeled reject, the region before it labeled fail to reject; the fly value 2.88 sits before the line and the value 9.6 sits well beyond it
43

If chi-square is not larger than the critical value, fail to reject the null hypothesis: the counts are consistent with the model.

44

The verdict is always written in one of those two wordings.

  1. The flies: χ2=2.88, critical value 3.84, so 2.88<3.84. Fail to reject the null hypothesis: the counts are consistent with 1 : 1.
  2. Another test cross, 144 tall to 96 dwarf of 240: χ2=9.6, so 9.6>3.84. Reject the null hypothesis: the counts do not fit 1 : 1.
  3. “The test proves the model is right” is a wrong wording: a test never proves a model.
  4. “We accept the null hypothesis” is a wrong wording too: the counts are consistent with the model, and other models might fit them as well.

45

So failing to reject is not accepting or proving. It says only that chance can account for the gap.

46

Rejecting says the model does not fit the counts. It does not say what is true instead.

47

What you are expected to know State the verdict of a chi-square test in one of its two wordings: reject, or fail to reject, the null hypothesis.

48

What you are expected to know Say what the verdict means for the model.

49
Check q17

A grower predicted 1 tall : 1 dwarf from a test cross of tomato plants. Of 300 seedlings, 176 were tall and 124 dwarf, and chi-square for the counts is 9.01. The critical value for one degree of freedom at p = 0.05 is 3.84.

Which of the following is the verdict?

  1. A. Fail to reject the null hypothesis
    9.01 is larger than the critical value 3.84.
    So the gap is too large to be chance.
  2. B. Accept the null hypothesis
    “Accept” is never one of the two wordings, and 9.01 exceeds 3.84 in any case.
  3. C. The test proves the 1 : 1 model wrong
    A chi-square test never proves anything.
    It says the counts do not fit the model.
  4. D. ✓ Reject the null hypothesis

Why: Chi-square, 9.01, is larger than the critical value, 3.84.
So the null hypothesis is rejected: the counts do not fit the 1 : 1 model.
Why they do not fit is a separate question the test does not answer.

50
Check q18

In guinea pigs, black coat (B) is dominant to white (b). A test cross of a Bb guinea pig with a bb one was predicted to give 1 black : 1 white. Of 80 pups, 44 were black and 36 white; chi-square for the counts is 0.80, and the critical value is 3.84. A student says: “0.80 is below 3.84, so the test proves the 1 : 1 model is right.”

Which of the following is correct about the student's claim?

  1. A. The student is right
    A chi-square test never proves a model.
    It only fails to find a gap too large for chance.
  2. B. ✓ The student is wrong

Why: 0.80 is not larger than 3.84, so fail to reject the null hypothesis.
Failing to reject means chance can account for the gap.
Other models might fit the counts too.
So the student is wrong: the counts are consistent with the model, but the model is not proved.

51
Practice writing an answer

In guinea pigs, black coat (B) is dominant to white (b). A test cross of a Bb guinea pig with a bb one was predicted to give 1 black : 1 white. Of 80 pups, 44 were black and 36 white; chi-square for the counts is 0.80, and the critical value for one degree of freedom at p = 0.05 is 3.84. The verdict is: fail to reject the null hypothesis.

(a) Explain what this verdict tells the breeder about the 1 : 1 model. (1 pt)

Model answer 0.80 is not larger than 3.84, so the null hypothesis is not rejected.
Failing to reject means the gap between 44 : 36 and 40 : 40 is small enough to be chance.
Other models could also give counts this close to 40 : 40.
So the counts are consistent with the 1 : 1 model, but they do not prove it.
Rubric
  • Award 1 point for: failing to reject means only that chance can account for the gap (the counts are consistent with the model); other models could fit the same counts, so the model is not proved.

Slip Writing that the model is accepted or proved. Failing to reject says only that the gap is small enough to be chance.

52

Back to the fly cross: 88 gray-bodied and 112 ebony-bodied flies, chi-square 2.88, and the formula sheet's table beside it.

The same chi-square table with the cell for one degree of freedom in the p = 0.05 row ringed: 3.84
The same chi-square table with the cell for one degree of freedom in the p = 0.05 row ringed: 3.84
53

Two classes give one degree of freedom. The 0.05 row of the table gives a critical value of 3.84.

54

2.88 is not larger than 3.84. So fail to reject the null hypothesis: the counts are consistent with the 1 : 1 model, and the model stands.

55Quick quiz: reject, or fail to reject? mixed practice

56
Check q19

A test cross gives two classes of seedlings, and chi-square for the counts is 5.20. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Which of the following is the verdict on the null hypothesis at p = 0.05?

  1. A. ✓ Reject
  2. B. Fail to reject
    Chi-square is larger than the critical value, so the gap between the counts and the model is too large to be chance.

Why: Two classes give one degree of freedom, and the 0.05 row reads 3.84.
5.20 is larger than 3.84.
So reject the null hypothesis: the counts do not fit the model.

57
Check q20

A test cross gives two classes of pups, and chi-square for the counts is 1.10. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Which of the following is the verdict on the null hypothesis at p = 0.05?

  1. A. Reject
    Chi-square is not larger than the critical value, so the gap between the counts and the model is small enough to be chance.
  2. B. ✓ Fail to reject

Why: Two classes give one degree of freedom, and the 0.05 row reads 3.84.
1.10 is not larger than 3.84.
So fail to reject the null hypothesis: the counts are consistent with the model.

58
Check q21

A dihybrid cross gives four classes of kernels, and chi-square for the counts is 7.05. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Which of the following is the verdict on the null hypothesis at p = 0.05?

  1. A. Reject
    Chi-square is not larger than the critical value, so the gap between the counts and the model is small enough to be chance.
  2. B. ✓ Fail to reject

Why: Four classes give three degrees of freedom, and the 0.05 row reads 7.81.
7.05 is not larger than 7.81.
So fail to reject the null hypothesis: the counts are consistent with the model.

59
Check q22

A dihybrid cross gives four classes of fruits, and chi-square for the counts is 8.30. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Which of the following is the verdict on the null hypothesis at p = 0.05?

  1. A. ✓ Reject
  2. B. Fail to reject
    Chi-square is larger than the critical value, so the gap between the counts and the model is too large to be chance.

Why: Four classes give three degrees of freedom, and the 0.05 row reads 7.81.
8.30 is larger than 7.81.
So reject the null hypothesis: the counts do not fit the model.

60
Check q23

A cross between two heterozygous plants gives three classes of flowers, and chi-square for the counts is 12.00. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Which of the following is the verdict on the null hypothesis at p = 0.05?

  1. A. ✓ Reject
  2. B. Fail to reject
    Chi-square is larger than the critical value, so the gap between the counts and the model is too large to be chance.

Why: Three classes give two degrees of freedom, and the 0.05 row reads 5.99.
12.00 is larger than 5.99.
So reject the null hypothesis: the counts do not fit the model.

61
Check q24

A cross between two heterozygous plants gives three classes of flowers, and chi-square for the counts is 0.45. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Which of the following is the verdict on the null hypothesis at p = 0.05?

  1. A. Reject
    Chi-square is not larger than the critical value, so the gap between the counts and the model is small enough to be chance.
  2. B. ✓ Fail to reject

Why: Three classes give two degrees of freedom, and the 0.05 row reads 5.99.
0.45 is not larger than 5.99.
So fail to reject the null hypothesis: the counts are consistent with the model.

62Mixed practice mixed practice

63
Check q25 numeric entry

A chi-square test compares two classes at p = 0.05. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

Read the critical value for the test from the p = 0.05 row.

Answer: 3.84  (tolerance ±0.005)

Working
Write down the values in the question:
classes = 2
row = p value 0.05
Subtract one from the number of classes:
degrees of freedom=2−1=1
Read the table at the column for 1 degree of freedom and the 0.05 row:
critical value = 3.84
64
Check q26

Two tomato plants heterozygous for height are crossed and give 123 tall and 37 dwarf offspring. Chi-square against 3 : 1 is 0.300, and against the p = 0.05 critical value of 3.84 the null hypothesis was not rejected. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

If the stricter row, p = 0.01, is read instead, what is the critical value?

  1. A. 0.01
    0.01 is the row's label, its p value, not a critical value.
    The critical value is the number in that row's cell for one degree of freedom.
  2. B. 3.84
    The two rows hold different values.
    For one degree of freedom the 0.01 row reads 6.63, larger than the 0.05 row's 3.84.
  3. C. ✓ 6.63

Why: In the 0.01 row the cell for one degree of freedom reads 6.63, larger than 3.84.
Chi-square, 0.300, is smaller than 6.63 as well, so the verdict is the same: fail to reject the null hypothesis.
A model that survives the 0.05 row survives the stricter row too.

65
Practice writing an answer

Two tomato plants heterozygous for height are crossed and give 123 tall and 37 dwarf offspring. Chi-square against 3 : 1 is 0.300, and against the p = 0.05 critical value of 3.84 the null hypothesis was not rejected. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

(a) Determine whether the null hypothesis is rejected at the p = 0.01 row of the table. (1 pt)

Model answer In the 0.01 row the cell for one degree of freedom reads 6.63, larger than 3.84.
Chi-square, 0.300, is smaller than 6.63 as well.
So the null hypothesis is not rejected at this row either.
A model that survives the 0.05 row survives the stricter row too.
Rubric
  • Award 1 point for: the decision (fail to reject the null hypothesis) AND the ground (0.300 is smaller than 6.63, the table's value for one degree of freedom at p = 0.01).
66
Check q27 numeric entry

Squash fruits from a dihybrid cross are sorted into four classes: white disk, white sphere, yellow disk and yellow sphere.

Calculate the degrees of freedom for a chi-square test of the four classes.

Answer: 3  (tolerance ±0)

Working
Write down the values in the question:
classes = 4
Subtract one from the number of classes:
degrees of freedom=4−1=3
67
Check q28

A breeder's chi-square value for a four-class cross is 9.2, and the critical value for three degrees of freedom at p = 0.05 is 7.81.

What does the verdict say about the breeder's model?

  1. A. ✓ The model does not fit the counts
  2. B. The model is proved wrong
    A test never proves anything.
    Rejecting says the model does not fit these counts, and does not name the true explanation.
  3. C. The counts are consistent with the model
    9.2 is larger than 7.81, so the null hypothesis is rejected: the counts do not fit the model.
  4. D. The model is accepted
    “Accept” is never one of the two verdict wordings, and 9.2 exceeds 7.81 in any case.

Why: 9.2 is larger than 7.81, so the null hypothesis is rejected: the model does not fit these counts.
The test does not show which other model would fit.

68
Practice writing an answer

In fruit flies, normal wings (V) are dominant to vestigial wings (v). A Vv fly is crossed with a vv fly, so the model predicts 1 normal : 1 vestigial. Of 200 offspring, 110 have normal wings and 90 have vestigial wings; chi-square for these counts against the model is 2.00. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01; the 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the 0.01 row 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09

(a) Identify the critical value at p = 0.05 for this test. (1 pt)

Answer: 3.84  (tolerance ±0.005)

Model answer Two classes give one degree of freedom, and the 0.05 row of the table gives a critical value of 3.84.
Working
Write down the values in the question:
classes = 2 (normal, vestigial)
Subtract one from the number of classes:
degrees of freedom=2−1=1
Read the table at the column for 1 degree of freedom and the 0.05 row:
critical value = 3.84
Rubric
  • Award 1 point for: a critical value of 3.84.

(b) State the degrees of freedom you read the critical value at. Then state the verdict of the test, in the correct wording, and justify it with the two numbers. (1 pt)

Model answer One degree of freedom, because two classes give one.
Fail to reject the null hypothesis, because chi-square, 2.00, is smaller than the critical value, 3.84.
Rubric
  • Award 1 point for: fail to reject the null hypothesis, because 2.00 is smaller than 3.84; the degrees of freedom stated (one, from two classes).
  • Do not award: “accept the null hypothesis” or “the model is proved”.

Slip Writing that the counts prove the 1 : 1 model or that the null hypothesis is accepted. The test can only fail to find a gap too large for chance.

(c) Explain what the verdict means for the 1 : 1 model. (1 pt)

Model answer The counts, 110 and 90, are consistent with the 1 : 1 model: the gap from 100 and 100 is small enough to be chance.
Rubric
  • Award 1 point for: the counts are consistent with 1 : 1, because chance can account for the gap.

Slip Treating the verdict as proof that wing shape is one gene with normal dominant.

(d) In a second cross of the same kind, 130 offspring had normal wings and 70 had vestigial wings, and chi-square against 1 : 1 is 18.0. State the verdict for that cross in the correct wording, and describe what it tells the breeder about the 1 : 1 model. (1 pt)

Model answer Reject the null hypothesis, because 18.0 is larger than the critical value, 3.84: the counts do not fit the 1 : 1 model.
The test does not say what is true instead.
Rubric
  • Award 1 point for: reject the null hypothesis, because 18.0 is larger than 3.84, so the counts do not fit 1 : 1.
  • Award also if the answer adds that the test does not say which other model fits.
  • Do not award: “accept” or “prove” as the verdict wording. A guess at which model is right instead neither earns nor loses the point.

Slip Writing that the test proves wing shape is not one gene with normal dominant. Rejecting says the 1 : 1 model does not fit these counts; it does not say why.

Glossary

p value
The label of a row in the chi-square table, 0.05 or 0.01: the share of batches in which chance alone would give a gap this large if the model were right. The course reads the 0.05 row; the 0.01 row is a stricter test with larger values.
degrees of freedom
The number of classes minus one: two classes give one degree of freedom, four classes give three. Degrees of freedom count classes, never offspring.
critical value
The number read from the chi-square table for the degrees of freedom and the p = 0.05 row. If chi-square is larger than it, reject the null hypothesis; if not, fail to reject.

APBIO-U05-L19B When chi-square is the wrong tool

Topic 5.3 · Mendelian Genetics · 36 steps

Two bars side by side: mean stem height 18.2 centimeters in sandy soil and 19.0 centimeters in clay soil, each with an error bar; beside them the question, will chi-square work here
Two bars side by side: mean stem height 18.2 centimeters in sandy soil and 19.0 centimeters in clay soil, each with an error bar; beside them the question, will chi-square work here

Here are two bars: the mean stem height of ten pea plants grown in sandy soil and ten grown in clay soil, in centimeters, each with an error bar.

A student asks whether the two soils differ and reaches for chi-square, because it worked for the fly counts. Will chi-square work here?

Unit 5 · Heredity

1What chi-square needs

2

Video: Watch: What chi-square needs

Five data sets sorted one at a time into the ones chi-square can test and the ones it cannot, a yes or no appearing beside each; the two mean bars with their overlapping error bars as the case it cannot.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L19Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L19Ba.mp4

3
Check q1

A breeder calculates chi-square for the classes of a breeding cross.

Which numbers does the chi-square equation use?

  1. A. ✓ The observed count and the expected count of each class
  2. B. The mean of each class and its standard error
    Chi-square takes o minus e for each class, squared, divided by e.
    No mean and no standard error appears in the equation.

Why: For each class the equation takes the observed count o minus the expected count e, squares it, and divides by e.
So the equation needs an o and an e for every class.

4
Check q2

Two means are compared, each with a ±2SE error bar. Imagine the two bars overlap.

What do the data show?

  1. A. ✓ No difference is shown between the two means
  2. B. The two means are the same
    Overlapping bars mean no difference shown, not the same.
    The true means may still differ.
  3. C. The two means differ
    When two ±2SE bars overlap, the two true means could both lie in the shared range.

Why: Each ±2SE bar is the range its true mean is likely to lie in.
When two ±2SE bars overlap, the two true means could both lie in the shared range.
So the data do not show a difference.

5

When is chi-square the right test, and when is it the wrong one?

6

Chi-square needs counts of individuals sorted into classes, and a ratio a model predicts for each class. Numbers of purple-flowered and white-flowered plants against 3 : 1 are that kind of data.

A table with two rows, purple-flowered and white-flowered plants: counted 71 and 29; the 3 : 1 model predicts 75 and 25
A table with two rows, purple-flowered and white-flowered plants: counted 71 and 29; the 3 : 1 model predicts 75 and 25
7

Here are the two bars again: mean stem heights, measured in centimeters. Nothing is counted into classes, and no model predicts a ratio.

Two bars of mean stem height in centimeters, 18.2 cm in sandy soil and 19.0 cm in clay soil, each with an error bar from 16.8 to 19.6 cm and from 17.6 to 20.4 cm, which overlap; gridlines every 5 cm; the legend reads: error bars represent ±2SE
Two bars of mean stem height in centimeters, 18.2 cm in sandy soil and 19.0 cm in clay soil, each with an error bar from 16.8 to 19.6 cm and from 17.6 to 20.4 cm, which overlap; gridlines every 5 cm; the legend reads: error bars represent ±2SE
8

So chi-square has no observed counts and no expected counts to work on. For two means, the tool is Unit 3's: the ±2SE error bars and the overlap rule.

9

Each ±2SE bar is the range its true mean is likely to lie in.

10

Read from the ends of the bars, the sandy-soil bar runs from 16.8 to 19.6 cm and the clay-soil bar from 17.6 to 20.4 cm.

11

The two bars overlap. So the data do not show a difference between the two soils.

12

Percentages and means are not counts. A percentage of purple-flowered plants can be turned back into a count if the total is known; a mean height never can.

13

Here is a table comparing counts in classes with means: what is recorded, what the model gives, and which test to use.

A table with two columns, counts in classes and means, and three rows: what is recorded, the number in each class against a measured value averaged; what the model gives, a ratio and so expected counts against no ratio but a ±2SE error bar; which test, chi-square against the ±2SE overlap rule
A table with two columns, counts in classes and means, and three rows: what is recorded, the number in each class against a measured value averaged; what the model gives, a ratio and so expected counts against no ratio but a ±2SE error bar; which test, chi-square against the ±2SE overlap rule
14

For example, take counts of purple-flowered and white-flowered plants against 3 : 1. Chi-square can test this, because the plants are counted into classes and a model predicts a ratio.

15

But take the same plants recorded as 72% purple and 28% white, with no total. Chi-square cannot test this, because percentages are not counts.

16

And take the same percentages with the total, 200 plants. Chi-square can test this, because the total turns the percentages back into counts: 144 purple and 56 white.

17

But take the mean stem height of the purple-flowered plants and of the white-flowered plants, in centimeters. Chi-square cannot test this, because a mean is a measured value, not a count, and no model predicts a ratio.

18

And take the mean mass of the seeds from each kind of plant, in grams. Chi-square cannot test this, because a mean is a measured value, not a count, and no model predicts a ratio.

A table of five data sets with a yes or no beside each: counts of purple and white plants against 3 : 1, yes; 72% purple and 28% white with no total, no; 72% and 28% of 200 plants, yes; mean stem height in centimeters, no; mean seed mass in grams, no
A table of five data sets with a yes or no beside each: counts of purple and white plants against 3 : 1, yes; 72% purple and 28% white with no total, no; 72% and 28% of 200 plants, yes; mean stem height in centimeters, no; mean seed mass in grams, no
19

So chi-square tests counts of individuals in classes against a predicted ratio. Means, measured values and percentages without a total are not that kind of data.

20

What you are expected to know Judge whether chi-square fits a data set: counts of individuals in classes against a predicted ratio, or not.

21
Check q3

Four sets of data are collected from a pea field.

Which set can a chi-square test be used on?

  1. A. The mean mass of ten seeds from tall plants and ten from short plants, in grams
    A mean mass is a measured value, not a count of individuals in classes, and no ratio is predicted for it.
  2. B. The mean height of the plants in two beds, with ±2SE error bars
    Means with error bars are compared with the ±2SE overlap rule, not with chi-square.
    Nothing is counted into classes.
  3. C. The percentage of a field that flowered by June, with no plant counts
    A percentage without the total cannot be turned back into counts, and chi-square needs counts.
  4. D. ✓ Counts of purple-flowered and white-flowered plants from a Pp × Pp cross, against 3 : 1

Why: Chi-square compares counts of individuals in classes with the counts a ratio predicts.
Only the numbers of purple-flowered and white-flowered plants against 3 : 1 are that kind of data.
The others are means or a percentage without a total.

22
Check q4

A student reports that 78% of the seedlings from a breeding cross were green and 22% were albino, and wants to test the counts against 3 : 1 with chi-square.

What does the student need before calculating chi-square?

  1. A. ✓ The total number of seedlings, to turn the percentages back into counts
  2. B. The mean height of the seedlings in each class, measured in centimeters
    Chi-square works on counts of individuals in classes.
    A mean height is a measured value and plays no part.
  3. C. The ±2SE error bars on the two percentages, to apply the overlap rule
    Error bars belong to means, where the overlap rule is the tool.
    Chi-square needs counts.
  4. D. The percentages rounded to 75% and 25% so that they match the model's ratio
    Rounding the observed percentages toward the model hides the very gap the test measures.

Why: Chi-square needs observed counts.
A percentage can be turned back into a count with the total.
So the total number of seedlings is what the student needs.

23

Back to the two bars: the mean stem height of ten pea plants in sandy soil, 18.2 cm, and of ten in clay soil, 19.0 cm, each with a ±2SE error bar.

Two bars of mean stem height in centimeters, 18.2 cm in sandy soil and 19.0 cm in clay soil, each with an error bar from 16.8 to 19.6 cm and from 17.6 to 20.4 cm, which overlap; gridlines every 5 cm; the legend reads: error bars represent ±2SE
Two bars of mean stem height in centimeters, 18.2 cm in sandy soil and 19.0 cm in clay soil, each with an error bar from 16.8 to 19.6 cm and from 17.6 to 20.4 cm, which overlap; gridlines every 5 cm; the legend reads: error bars represent ±2SE
24

The two heights are means, not counts, and no model predicts a ratio for them. So chi-square is the wrong tool.

25

Read from the ends of the bars, 16.8 to 19.6 cm overlaps 17.6 to 20.4 cm. So the data do not show that the two soils differ.

26Quick quiz: can chi-square test it? mixed practice

27
Check q5

A student has this data set: counts of tall and dwarf pea plants from a Tt × tt cross, against 1 : 1.

Can chi-square test it?

  1. A. ✓ Yes
  2. B. No
    Tall and dwarf are counts of individuals in two classes, and 1 : 1 is a predicted ratio: chi-square has what it needs.
  3. C. Only if the total is known
    The counts are given, so the total is already known: add the tall and dwarf counts.

Why: These are counts of individuals in two classes, and the model predicts a ratio for them, 1 : 1.
That is exactly the data chi-square needs.

28
Check q6

A student has this data set: the mean root length of ten seedlings grown in wet soil and ten grown in dry soil, with ±2SE error bars.

Can chi-square test it?

  1. A. Yes
    A mean root length is a measured value, not a count of seedlings in classes, and no model predicts a ratio.
    The overlap rule is the tool for two means.
  2. B. ✓ No
  3. C. Only if the total is known
    A mean root length is a measured value.
    Knowing the total would not turn it into counts in classes against a predicted ratio.

Why: A mean root length is a measured value.
No seedlings are counted into classes, and no model predicts a ratio, so chi-square has nothing to work on.
The overlap rule is the tool for two means.

29
Check q7

A student has this data set: 65% striped and 35% plain zebrafish from one breeding pair, with no total given.

Can chi-square test it?

  1. A. Yes
    Percentages are not counts.
    Only with the total do 65% and 35% become observed counts of striped and plain fish.
  2. B. No
    Percentages are not counts.
    With the total, 65% and 35% become observed counts of striped and plain fish.
    Chi-square can then compare those counts with the predicted ratio.
  3. C. ✓ Only if the total is known

Why: Percentages are not counts.
With the total, 65% and 35% become observed counts of striped and plain fish.
Chi-square can then compare those counts with the predicted ratio.

30
Check q8

A student has this data set: counts of the four coat classes of guinea pigs from a dihybrid cross, against 9 : 3 : 3 : 1.

Can chi-square test it?

  1. A. ✓ Yes
  2. B. No
    The four coat classes are counts of individuals, and 9 : 3 : 3 : 1 is a predicted ratio: chi-square has what it needs.
  3. C. Only if the total is known
    The four counts are given, so the total is already known: add them.

Why: These are counts of individuals in four classes, and the model predicts a ratio for them, 9 : 3 : 3 : 1.
That is the data chi-square needs.

31
Check q9

A student has this data set: the mean mass, in grams, of the eggs a hen laid at two temperatures.

Can chi-square test it?

  1. A. Yes
    A mean mass is a measured value, not a count of eggs in classes, and no model predicts a ratio.
  2. B. ✓ No
  3. C. Only if the total is known
    A mean mass is a measured value.
    No eggs are counted into classes, and no model predicts a ratio, so chi-square has nothing to work on.

Why: A mean mass is a measured value.
No eggs are counted into classes, and no model predicts a ratio, so chi-square has nothing to work on.

32
Check q10

A student has this data set: 52% red-eyed and 48% white-eyed flies from a test cross, with no total given.

Can chi-square test it?

  1. A. Yes
    Percentages are not counts.
    Only with the total do 52% and 48% become observed counts of red-eyed and white-eyed flies.
  2. B. No
    Percentages are not counts.
    With the total, 52% and 48% become observed counts of red-eyed and white-eyed flies.
    Chi-square can then compare those counts with 1 : 1.
  3. C. ✓ Only if the total is known

Why: Percentages are not counts.
With the total, 52% and 48% become observed counts of red-eyed and white-eyed flies.
Chi-square can then compare those counts with 1 : 1.

33Mixed practice mixed practice

34
Check q11

A class measures the mean mass of twenty seeds from each of two pea lines and wants to know whether the lines differ.

Which tool fits this question?

  1. A. Chi-square, with the two lines as the two classes and their seed counts as the observed counts
    The data are measured masses, not counts of individuals sorted into classes by a trait.
  2. B. ✓ The ±2SE error bars on the two means, and the overlap rule
  3. C. Chi-square, with each seed's mass as one class and its own expected count
    A mass is a measurement, not a class, and chi-square has no ratio to compare it with.
  4. D. Chi-square against a 1 : 1 ratio of seeds between the two pea lines
    No model predicts a ratio of seeds between two lines; the question is whether two means differ.

Why: Two means from measured values are compared with their ±2SE error bars: if the bars overlap the data do not show a difference.
Chi-square needs counts in classes against a predicted ratio, which these data are not.

35
Practice writing an answer

A breeder raises fruit flies at two temperatures. The breeder measures the mean wing length of the normal-winged flies at each temperature, with ±2SE error bars. The breeder also records the normal-winged and vestigial-winged flies from a Vv × vv cross at each temperature; the cross predicts 1 normal : 1 vestigial. At the cooler temperature the flies are recorded as counts, at the warmer one only as percentages.

(a) Identify the tool for deciding whether the two mean wing lengths differ. (1 pt)

Model answer The ±2SE error bars on the two mean values and the overlap rule: if the bars overlap, the data do not show a difference.
Rubric
  • Award 1 point for: the ±2SE overlap rule (comparing the two means' error bars).

Slip Naming chi-square. A mean wing length is a measurement, not a count of individuals in classes.

(b) Explain how the cooler temperature's record gives chi-square what it needs. (1 pt)

Model answer At the cooler temperature the flies are counted into two classes, normal-winged and vestigial-winged, so each class has an observed count.
The cross predicts 1 normal : 1 vestigial, so the ratio and the total give each class an expected count.
Chi-square needs an observed count and an expected count for every class, and this record supplies both.
Rubric
  • Award 1 point for: the record has counts of individuals in classes (observed counts) AND a predicted ratio that gives expected counts, which is what chi-square needs.

Slip Naming means or percentages as what chi-square needs. It needs counts of individuals in classes and a ratio a model predicts.

(c) The breeder wants to test the warmer temperature's record, 47% vestigial-winged, against 1 : 1. Describe what the breeder needs to add to that record first. (1 pt)

Model answer The breeder needs the total number of flies counted at that temperature.
A percentage is not a count.
With the total, 47% becomes an observed count of vestigial-winged flies and 53% an observed count of normal-winged flies.
The total and the 1 : 1 ratio then give the expected counts.
Rubric
  • Award 1 point for: the total number of flies at that temperature, so that the percentages become observed counts (chi-square needs counts, not percentages).

Slip Calculating chi-square from 47 and 53 as if they were counts. They are percentages of an unknown total, and the gap chi-square measures depends on how many flies were counted.

(d) Evaluate the claim that chi-square can be used on the two mean wing lengths. (1 pt)

Model answer The claim is not supported.
Chi-square needs counts of individuals sorted into classes and a ratio a model predicts for each class.
A mean wing length is a measured value: no flies are counted into classes and no ratio is predicted.
So the equation has no observed or expected counts to use, and chi-square cannot be used on the two mean wing lengths.
Rubric
  • Award 1 point for: the judgement (the claim is not supported) AND the ground (chi-square needs counts in classes against a predicted ratio, and a mean length is a measured value with neither).

Slip Proposing chi-square with the two temperatures as two classes. The temperatures are conditions, not classes of counted individuals, and no ratio is predicted between them.

APBIO-U05-L20 From the cross to the verdict

Topic 5.3 · Mendelian Genetics · 27 steps

Left: a photograph of a real ear of corn lying on straw, its kernels purple, red, yellow and white. Right: an ear of corn drawn as a long oval full of small kernels of four kinds, filled smooth, filled wrinkled, pale smooth and pale wrinkled; beside it the four counts, 180 purple smooth, 54 purple wrinkled, 60 yellow smooth and 26 yellow wrinkled
Left: a photograph of a real ear of corn lying on straw, its kernels purple, red, yellow and white. Right: an ear of corn drawn as a long oval full of small kernels of four kinds, filled smooth, filled wrinkled, pale smooth and pale wrinkled; beside it the four counts, 180 purple smooth, 54 purple wrinkled, 60 yellow smooth and 26 yellow wrinkled

Photo: Jim, the Photographer, Flickr via Wikimedia Commons, CC BY 2.0 (cropped and resized).

Here is a breeder’s ear of corn, drawn, beside a photograph of a real ear with kernels of several colors. The breeder’s ear carries 320 kernels of four kinds: purple and smooth, purple and wrinkled, yellow and smooth, yellow and wrinkled.

The breeder counted them: 180, 54, 60 and 26. She had predicted 9 : 3 : 3 : 1. Do the kernels fit her prediction, or has something other than chance been at work?

Unit 5 · Heredity

1The whole chain, once

2

Video: Watch: The whole chain, once

The chain drawn as one left-to-right strip, from the parents' square to the verdict, each box appearing when the one before it is done; the corn's numbers written into each box as it appears.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L20a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L20a.mp4

3
Check q1

Two pea plants RrYy are crossed, and the two genes are on different chromosomes.

Which phenotypic ratio does the sixteen-cell square predict?

  1. A. ✓ 9 : 3 : 3 : 1
  2. B. 3 : 1
    3 : 1 is one gene's ratio from two heterozygous parents.
    Two genes give four phenotype classes.
  3. C. 1 : 1 : 1 : 1
    1 : 1 : 1 : 1 comes from a dihybrid test cross, RrYy × rryy.

Why: Each parent makes four kinds of gamete in equal numbers.
The square has sixteen cells: nine round yellow, three round green, three wrinkled yellow and one wrinkled green.

4
Check q2

A breeder predicts 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green for the seeds of that cross.

Which of the following is the null hypothesis for the cross?

  1. A. The seeds differ from 9 : 3 : 3 : 1 because the two genes sit on one chromosome
    A claim of a real difference, with its reason, is the alternative hypothesis.
  2. B. ✓ The seeds occur in a 9 : 3 : 3 : 1 ratio, and any difference from it is due to chance

Why: The null hypothesis is the no-difference statement.
For a breeding cross, the no-difference statement is that the seeds fit the predicted ratio and any gap is chance.

5
Check q3

The breeder counts 160 seeds in all. Wrinkled green is the class with 1 of the 16 shares.

Which of the following is the expected count of wrinkled green seeds?

  1. A. 1 seed
    The digits of a ratio are shares of the total, not counts.
    160 seeds split into 16 equal shares is 10 seeds a share.
  2. B. ✓ 10 seeds
  3. C. 16 seeds
    16 is the number of shares in all, not a count of seeds.

Why: The expected count is the total times that class's share of the ratio.
160 seeds split into 16 equal shares is 10 seeds a share.
Wrinkled green has one share, so its expected count is 10 seeds.

6
Check q4

For the wrinkled green class, the observed count is 14 seeds and the expected count is 10 seeds.

Which of the following is that class's term in the chi-square sum?

  1. A. ✓ The squared gap, 16, divided by 10
  2. B. The squared gap, 16, divided by 14
    Each class is divided by its own expected count, not by its observed count.
  3. C. The gap, 4, divided by 10
    The gap is squared before dividing.

Why: For each class the equation takes o minus e, squares it, and divides by that class's own e.
Here that is the squared gap, 16, over the expected count, 10.

7
Check q5

The seeds are sorted into four classes. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01

Which of the following is the critical value at p = 0.05?

  1. A. 3.84
    3.84 is the critical value for one degree of freedom, two classes.
  2. B. ✓ 7.81
  3. C. 9.49
    9.49 is the critical value for four degrees of freedom, five classes.

Why: Four classes give three degrees of freedom.
Down the column for 3 and along the 0.05 row, the cell reads 7.81.

8
Check q6

Chi-square for the 160 seeds is 1.90, and the critical value is 7.81.

Which of the following is the verdict on the null hypothesis?

  1. A. ✓ Fail to reject
  2. B. Reject
    1.90 is not larger than 7.81, so the gap between the counts and the model is small enough to be chance.

Why: 1.90 is not larger than the critical value, 7.81.
So fail to reject the null hypothesis: the seed counts are consistent with 9 : 3 : 3 : 1.

9

How do you take a breeding cross all the way from the parents to a verdict? Testing a model against counts follows a fixed order:

  1. The model gives a Punnett square, and the square gives a ratio.
  2. The null hypothesis says the counts fit that ratio apart from chance.
  3. The ratio and the total give the expected counts.
  4. Chi-square puts one number on the gap.
  5. The number of classes gives the degrees of freedom.
  6. The table gives the critical value.
  7. Comparing chi-square with the critical value gives the verdict, and the verdict says something about the model.

Seven boxes joined by arrows: the square and its ratio; the null hypothesis; expected counts; chi-square; degrees of freedom; the critical value; the verdict and what it means
Seven boxes joined by arrows: the square and its ratio; the null hypothesis; expected counts; chi-square; degrees of freedom; the critical value; the verdict and what it means
10

The breeder's model: kernel color and kernel texture are two genes on different chromosomes. Purple (P) is dominant to yellow (p), smooth (S) is dominant to wrinkled (s), and both parents were PpSs.

11

That model gives a sixteen-cell square: 9 : 3 : 3 : 1.

The sixteen-cell Punnett square for two parents PpSs: gametes PS, Ps, pS and ps along the top and down the side; the cells hold the sixteen combinations, of which nine are purple smooth, three purple wrinkled, three yellow smooth and one yellow wrinkled
The sixteen-cell Punnett square for two parents PpSs: gametes PS, Ps, pS and ps along the top and down the side; the cells hold the sixteen combinations, of which nine are purple smooth, three purple wrinkled, three yellow smooth and one yellow wrinkled
12

Here is a table of the breeder's counts against the shares of her model. The table your formula sheet prints gives the critical value for four classes.

A table of the four kernel classes with observed counts 180, 54, 60 and 26 and the shares the 9 : 3 : 3 : 1 model gives them, 9, 3, 3 and 1 of 16
A table of the four kernel classes with observed counts 180, 54, 60 and 26 and the shares the 9 : 3 : 3 : 1 model gives them, 9, 3, 3 and 1 of 16
13
Worked example

An ear of corn from a PpSs × PpSs cross carries 320 kernels: 180 purple smooth, 54 purple wrinkled, 60 yellow smooth and 26 yellow wrinkled. The model predicts 9 : 3 : 3 : 1. Test the counts against the model and state the verdict.

Write down the values in the question:
total = 320
ratio = 9 : 3 : 3 : 1, so 16 shares in all
o = 180, 54, 60, 26
State the null hypothesis:
The kernels occur in a 9 : 3 : 3 : 1 ratio, and any difference between the counts and that ratio is due to chance.
Write down the equation for the expected counts:
e=total×that class's share of the ratio
Substitute the values into the equation:
epurple smooth=320×916=180
epurple wrinkled=320×316=60
eyellow smooth=320×316=60
eyellow wrinkled=320×116=20
Write down the chi-square equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
purple smooth: (180−180)2180=0
purple wrinkled: (54−60)260=3660=0.60
yellow smooth: (60−60)260=0
yellow wrinkled: (26−20)220=3620=1.80
Add the classes:
χ2=0+0.60+0+1.80=2.40(no unit)
Find the degrees of freedom and read the critical value from the table:
degrees of freedom=4−1=3
critical value (3 degrees of freedom, p = 0.05 row) = 7.81
Compare and state the verdict:
2.40<7.81
Fail to reject the null hypothesis: the counts are consistent with 9 : 3 : 3 : 1.
14

The verdict's meaning for the model: the kernel counts are consistent with two genes assorting independently. The verdict does not prove the model; it says the gap from 180 : 60 : 60 : 20 is small enough to be chance.

The chi-square table from the formula sheet with the cell for three degrees of freedom in the p = 0.05 row ringed: 7.81
The chi-square table from the formula sheet with the cell for three degrees of freedom in the p = 0.05 row ringed: 7.81
15

What you are expected to know Test a breeding cross with the whole chain: the ratio from the square, the null hypothesis, expected counts and chi-square, the degrees of freedom and critical value, and the verdict with what it means for the model.

16
Check q7 numeric entry

In squash, white fruit (W) is dominant to yellow (w) and disk shape (D) is dominant to sphere (d); the two genes are on different chromosomes. Two WwDd plants are crossed, so the model predicts 9 white disk : 3 white sphere : 3 yellow disk : 1 yellow sphere. The 240 fruits are counted in the table below.

A table of four squash classes with observed counts 126 white disk, 48 white sphere, 42 yellow disk and 24 yellow sphere, and their shares 9, 3, 3 and 1 of 16
A table of four squash classes with observed counts 126 white disk, 48 white sphere, 42 yellow disk and 24 yellow sphere, and their shares 9, 3, 3 and 1 of 16

Calculate chi-square for the four classes.

Part 1. White disk is 9 of the 16 shares. What is the expected count of white disk fruits?

Answer: 135  (tolerance ±0)

Working
Multiply the total by the class's share:
ewhite disk=240×916=135

Part 2. Yellow sphere is 1 of the 16 shares. What is the expected count of yellow sphere fruits?

Answer: 15  (tolerance ±0)

Working
Multiply the total by the class's share:
eyellow sphere=240×116=15

Part 3. Work out the white disk class's term, the squared gap divided by its expected count. What is the white disk term?

Answer: 0.6  (tolerance ±0.005)

Working
Square the gap and divide by the class's own expected count:
white disk: (126−135)2135=81135=0.60

Part 4. Work out the yellow sphere class's term. What is the yellow sphere term?

Answer: 5.4  (tolerance ±0.005)

Working
Square the gap and divide by the class's own expected count:
yellow sphere: (24−15)215=8115=5.40

Part 5. Count the classes and subtract one. How many degrees of freedom does this test have?

Answer: 3  (tolerance ±0)

Working
Subtract one from the number of classes:
degrees of freedom=4−1=3

Answer: 6.4  (tolerance ±0.005)

Working
Write down the values in the question:
total = 240
ratio = 9 : 3 : 3 : 1, so 16 shares in all
o = 126, 48, 42, 24
Write down the equation for the expected counts:
e=total×that class's share of the ratio
Substitute the values into the equation:
ewhite disk=240×916=135
ewhite sphere=240×316=45
eyellow disk=240×316=45
eyellow sphere=240×116=15
Write down the chi-square equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
white disk: (126−135)2135=0.60
white sphere: (48−45)245=0.20
yellow disk: (42−45)245=0.20
yellow sphere: (24−15)215=5.40
Add the classes:
χ2=0.60+0.20+0.20+5.40=6.40
17
Check q8

For the squash fruits, chi-square is 6.40 and there are three degrees of freedom. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01

Which of the following is the verdict on the null hypothesis?

  1. A. Reject
    The critical value for three degrees of freedom at p = 0.05 is 7.81.
    6.40 is smaller than 7.81, so the null hypothesis is not rejected.
  2. B. ✓ Fail to reject

Why: Four classes give three degrees of freedom, and the 0.05 row's critical value is 7.81.
Chi-square, 6.40, is smaller than 7.81.
So the null hypothesis is not rejected: the fruit counts are consistent with two genes assorting independently.

18
Practice writing an answer

For the squash fruits, chi-square is 6.40 and there are three degrees of freedom; the yellow sphere class alone contributed 5.40 of the 6.40. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01

(a) Determine the verdict on the null hypothesis at p = 0.05, and explain what the verdict means for the breeder’s model. (1 pt)

Model answer Four classes give three degrees of freedom, and the 0.05 row's critical value is 7.81.
Chi-square, 6.40, is smaller than 7.81, so the verdict is fail to reject the null hypothesis.
The gap between the counts and 9 : 3 : 3 : 1 is small enough to be chance.
So the fruit counts are consistent with two genes assorting independently, even though the yellow sphere class strayed a long way.
The verdict does not prove the model.
Rubric
  • Award 1 point for: the verdict fail to reject (6.40 is smaller than the critical value 7.81 for three degrees of freedom at p = 0.05) AND its meaning: the counts are consistent with 9 : 3 : 3 : 1; the model is not rejected, though it is not proved.
19

Back to the ear of corn: 320 kernels, 180 purple smooth, 54 purple wrinkled, 60 yellow smooth and 26 yellow wrinkled, against the breeder's 9 : 3 : 3 : 1.

A table of the four kernel classes with observed counts 180, 54, 60 and 26 and the shares the 9 : 3 : 3 : 1 model gives them, 9, 3, 3 and 1 of 16
A table of the four kernel classes with observed counts 180, 54, 60 and 26 and the shares the 9 : 3 : 3 : 1 model gives them, 9, 3, 3 and 1 of 16
20

The parents' square predicts 9 : 3 : 3 : 1, and the ratio and the total, 320, make the expected counts 180 : 60 : 60 : 20.

21

Chi-square for the counts is 2.40, and three degrees of freedom give a critical value of 7.81.

22

2.40 is not larger than 7.81, so the verdict is fail to reject the null hypothesis: the kernels behaved as two independently assorting genes predict.

23Mixed practice mixed practice

24
Check q9

A breeder tests the kernel counts from a different ear of corn, from another PpSs × PpSs cross, against 9 : 3 : 3 : 1 and gets a chi-square of 12.1, above the critical value of 7.81.

What can the breeder conclude?

  1. A. ✓ The counts do not fit the model, and the test does not say why
  2. B. The two genes must sit together on the same chromosome
    Rejecting the null hypothesis says the model does not fit.
    It does not name the reason, and several reasons could produce the gap.
  3. C. The counts fit the model, because 12.1 is a large number
    A chi-square larger than the critical value means the gap is too big for chance, so the model is rejected, not supported.
  4. D. The parents were not PpSs, which the test has now proved
    A test never proves a cause.
    It rejects the model as a fit for these counts.

Why: 12.1 is larger than 7.81, so the null hypothesis is rejected: the counts do not fit 9 : 3 : 3 : 1.
Which explanation is right, wrong parents, a shared chromosome or something else, is a separate question.

25
Check q10

A student tests 480 kernels from a PpSs × PpSs cross, sorted into four classes, against 9 : 3 : 3 : 1. The student states the null hypothesis as the counts fitting the ratio, any difference due to chance. The student finds each expected count as 480 times that class's share of the ratio. The student divides each squared gap by its own expected count and adds the four. The student reads the critical value at 479 degrees of freedom.

Which step of the student's chain is wrong?

  1. A. Stating the null hypothesis as the counts fitting the ratio with differences due to chance
    That is the null hypothesis in its correct form: the no-difference statement, with the gap due to chance.
  2. B. Finding each expected count as 480 times that class's share of the ratio
    The expected counts are the total times each class's share of the ratio, exactly as the student did: 270, 90, 90 and 30.
  3. C. Dividing each squared gap by its own expected count before adding
    Each class is divided by its own expected count, which is the formula-sheet equation.
  4. D. ✓ Reading the critical value at 479 degrees of freedom

Why: Degrees of freedom count classes, never offspring: four classes give three degrees of freedom, and the critical value is read there, 7.81.
Every other step of the chain is right.

26
Practice writing an answer

In rabbits, black fur (B) is dominant to white fur (b). A breeder crosses two black rabbits believed to be both Bb and predicts 3 black : 1 white. The 100 young rabbits, the kits, are 66 black and 34 white. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01

(a) State the null hypothesis for this cross. (1 pt)

Model answer The kits occur in a 3 : 1 ratio of black to white, and any difference between the observed counts and that ratio is due to chance.
Rubric
  • Award 1 point for: the kits fit the predicted 3 : 1 ratio and the difference from it is due to chance.
  • Do not award: a statement that white kits are more common than predicted, or that the parents are not both Bb.

Slip Writing that there are more white kits than the model predicts. That is the alternative hypothesis; the null hypothesis is the no-difference statement.

(b) Calculate the expected count of white kits. (1 pt)

Answer: 25  (tolerance ±0)

Model answer The expected count of white kits is 25.
Working
Write down the values in the question:
total = 100
ratio = 3 : 1, so 4 shares in all
white share = 1 of 4
Write down the equation:
e=total×that class's share of the ratio
Substitute the values into the equation:
ewhite=100×14=25
Rubric
  • Award 1 point for: 25 white kits expected (and 75 black).

(c) Calculate chi-square for the breeder's counts. (1 pt)

Answer: 4.32  (tolerance ±0.005)

Model answer Chi-square for the counts is 4.32.
Working
Write down the values in the question:
o = 66 (black), 34 (white)
e = 75 (black), 25 (white)
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
black: (66−75)275=8175=1.08
white: (34−25)225=8125=3.24
Add the classes:
χ2=1.08+3.24=4.32
Rubric
  • Award 1 point for: chi-square 4.32.

(d) Identify the degrees of freedom and the critical value from the table, state the verdict in the correct wording, and explain what the verdict means for the breeder's belief that both parents are Bb. (2 pt)

Model answer Two classes give one degree of freedom, and the 0.05 row gives a critical value of 3.84.
Chi-square, 4.32, is larger than 3.84, so reject the null hypothesis: the counts do not fit the 3 : 1 model.
The gap between 66 : 34 and 75 : 25 is too large to be chance, so the breeder's model does not fit these kits.
The test does not say why.
One possibility to check is that a parent is not Bb.
Rubric
  • Award 1 point for: one degree of freedom, critical value 3.84, and the verdict reject the null hypothesis because 4.32 is larger than 3.84.
  • Award 1 point for: the counts do not fit the 3 : 1 model (the gap is too large to be chance), and the test does not say which explanation is right, so the belief that both parents are Bb is in doubt but not disproved.

Slip Writing that the test proves one parent is BB or bb. Rejecting the null hypothesis says the model does not fit; it does not name the cause.

APBIO-U05-P53 Practice questions: Topic 5.3

Topic 5.3 · Mendelian Genetics · 10 MCQ · 2 FRQ · for APBIO-U05-T53

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation. The first free-response question walks you through a chi-square test one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a question needs a critical value, the chi-square table is drawn with it; read the p = 0.05 row.

Video: Watch first: Mendelian genetics, summed up

Alleles, genotype and phenotype; one allele per gamete and the square read twice; multiply for both, add for either; the test cross; reading a pedigree; two genes and 9 : 3 : 3 : 1; the null hypothesis, chi-square, the table and the verdict.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T53-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T53-summary.mp4

Q1 P53-q01

In goats, the coat-color gene has two alleles, D and d. One goat's homologous pair is drawn below.

One goat's homologous pair carrying the coat-color gene; the allele each chromosome carries is written on it.
One goat's homologous pair carrying the coat-color gene; the allele each chromosome carries is written on it.

What is this goat's genotype, and is it homozygous or heterozygous?

  1. A. D, homozygous
    A genotype is both alleles, one on each homolog.
    So a genotype is two letters.
  2. B. ✓ DD, homozygous
  3. C. Dd, heterozygous
    Both chromosomes carry D.
    So the two alleles do not differ.
  4. D. DD, heterozygous
    Heterozygous means the two alleles are different.
    Here both alleles are D, so they are the same.

Why: Both chromosomes of the pair carry D, so the genotype is DD, and two of the same allele make the goat homozygous.

Q2 P53-q02

In goats, a Dd goat has a dark coat and a dd goat has a light coat.

Which allele is dominant, and why?

  1. A. d, because a light coat needs two copies to show
    An allele whose trait needs two copies to show is the recessive one.
  2. B. D, because dark is the common coat color in goats
    How common a color is does not show which allele is dominant; a dominant trait can be rare.
  3. C. ✓ D, because a dark coat shows in the heterozygote Dd
  4. D. Neither, because Dd goats show a mix of both colors
    A Dd goat is dark, not a mix; one trait shows and the other does not.

Why: The heterozygote Dd is dark, so dark is the trait that shows when the two alleles differ, and D is dominant; a light coat shows only in dd, so d is recessive.

Q3 P53-q03

A breeder has grown a line of marigolds from its own seed for eight years, and every plant in every year had orange flowers. This spring she also bought one marigold plant in flower; its flowers are orange.

Which of the following can the breeder call true-breeding for orange flowers?

  1. A. ✓ The eight-year line only
  2. B. The bought plant only
    The bought plant has flowered for one generation.
    A true-breeding line gives only its own version of a trait generation after generation, and the eight-year line has.
  3. C. Both the line and the bought plant
    The bought plant has flowered once, so the breeder cannot tell yet whether its line stays orange.
    One generation is not enough to call a line true-breeding.
  4. D. Neither the line nor the bought plant
    The line gave only orange flowers for eight generations.
    That is what true-breeding means.

Why: A true-breeding line gives only its own version of a trait, generation after generation.
The line gave only orange flowers for eight years, so the breeder can call it true-breeding.
The bought plant has flowered once, and one generation is not enough to tell.

Q4 P53-q04

Suppose a breeder has two true-breeding lines of a plant, one with smooth leaves and one with hairy leaves. She crosses a plant from each line, and every offspring has smooth leaves. She then crosses those offspring with each other and counts 148 smooth-leaved and 52 hairy-leaved plants.

Which of the following names the 148 smooth-leaved and 52 hairy-leaved plants?

  1. A. The P generation
    The P generation is the two true-breeding parent lines the breeder started with.
  2. B. The F1 generation
    The F1 are the offspring of the two parent lines: the plants that all had smooth leaves.
  3. C. ✓ The F2 generation
  4. D. The F3 generation
    The offspring of F1 crossed with F1 are the F2.
    Counting begins at the first cross, not at the parent lines.

Why: The two true-breeding lines are the P generation.
Their offspring, all smooth-leaved, are the F1.
The F1 crossed with each other give the F2.
So the 148 smooth-leaved and 52 hairy-leaved plants are the F2.

Q5 P53-q05

A breeder crosses two Dd goats (dark coat, D, dominant to light, d).

What is the probability that an offspring has a light coat?

  1. A. ✓ 0.25
  2. B. 0.5
    0.5 is the chance that one gamete carries d.
    A light coat needs d from both parents, one cell of the four.
  3. C. 0.75
    0.75 is the probability of a dark coat: DD, Dd and dD, three of the four cells.
  4. D. 1 : 3
    1 : 3 is a ratio; it compares the light class with the dark class.
    A probability is one number between 0 and 1.

Why: A probability is one number between 0 and 1.
The Dd × Dd square has four cells: DD, Dd, dD and dd.
Only dd gives a light coat.
So the probability of a light coat is 1 cell in 4, which is 0.25.

Q6 P53-q06

A dark-coated goat could be DD or Dd. It is crossed with a light-coated goat, dd, and all eight offspring are dark.

What can be concluded about the dark parent, and why?

  1. A. It is certainly DD, because a Dd parent would have given at least one light offspring among eight
    A Dd parent gives dark offspring half the time, and eight dark in a row is unlikely but possible.
  2. B. Nothing at all, because crossing with a light dd goat reveals nothing about a dark goat's genotype
    Crossing with the homozygous recessive is exactly the cross that reveals a hidden d: a light offspring would prove Dd.
  3. C. It is certainly Dd, because a dark goat crossed with a light one is always heterozygous
    The cross partner's genotype does not show the dark parent's genotype; only the offspring do, and here none is light.
  4. D. ✓ Probably DD, but Dd is still possible: eight dark from Dd × dd has a probability of about 1 in 256

Why: Only a light offspring can prove Dd.
All dark favors DD, but a Dd parent gives eight dark in a row with probability (12)8=1256, so Dd is unlikely, not ruled out.

Q7 P53-q07

A breeder takes pollen grains from one sunflower plant and puts them on the flowers of a second sunflower plant. Seeds form in the base of those flowers.

Which two things fused to make each seed?

  1. A. A pollen grain and an ovule
    A pollen grain carries the sperm; it is not the sperm.
    An ovule holds the egg; it is not the egg.
  2. B. A pollen grain and an egg
    A pollen grain carries the sperm; it is not the sperm.
    The sperm inside the grain fuses with the egg.
  3. C. A sperm and an ovule
    An ovule holds the egg; it is not the egg.
    The sperm fuses with the egg inside the ovule.
  4. D. ✓ A sperm and an egg

Why: The anthers of the first plant make pollen grains, and each grain carries a sperm.
An ovule in the base of the second plant's flower holds an egg.
The sperm travels down to the egg and fuses with it.
The fused cell grows into the seed.

Q8 P53-q08

Tay-Sachs disease comes from one gene on chromosome 15, one of the 22 pairs of chromosomes whose members match. A child shows the disease only when both of the child's alleles are the disease allele.

Which of the following describes Tay-Sachs disease?

  1. A. Autosomal dominant
    One copy of the allele does not show the disease, so the allele is recessive, not dominant.
  2. B. ✓ Autosomal recessive
  3. C. Dominant, not autosomal
    Chromosome 15 is one of the 22 matching pairs, an autosome, so the trait is autosomal.
    One copy does not show the disease, so the allele is recessive.
  4. D. Recessive, not autosomal
    Autosomal names where the gene sits, not whether the allele is dominant.
    Chromosome 15 is an autosome, so the trait is autosomal.

Why: A chromosome from one of the 22 matching pairs is an autosome, so the gene sits on an autosome.
The disease shows only in a child with two copies of the allele, so the allele is recessive.
A recessive trait whose gene sits on an autosome is autosomal recessive.

Q9 P53-q09

Suppose a breeder has a plant that is JjQq for two genes on different chromosomes, and crosses it with another JjQq plant. Her sixteen-cell Punnett square is drawn below with one cell still to fill, marked ?. The four kinds of pollen grain are written along the top and the four kinds of egg down the side.

A sixteen-cell Punnett square for JjQq × JjQq: the four kinds of pollen grain along the top, the four kinds of egg down the side; one cell still to fill, marked ?.
A sixteen-cell Punnett square for JjQq × JjQq: the four kinds of pollen grain along the top, the four kinds of egg down the side; one cell still to fill, marked ?.

Which genotype belongs in the marked cell?

  1. A. JjQQ
    The marked cell's row is headed jQ and its column Jq, so the egg brings Q and the pollen grain brings q: Qq, not QQ.
  2. B. ✓ JjQq
  3. C. jjQq
    The marked cell's column is headed Jq, so the pollen grain brings J.
    With the egg's j the cell holds Jj, not jj.
  4. D. jQ
    jQ is the egg's two alleles, one of each gene.
    A cell holds four letters: two alleles of each gene, one from the egg and one from the pollen grain.

Why: The marked cell is in the row headed jQ and the column headed Jq.
The egg brings j and Q; the pollen grain brings J and q.
So the cell holds Jj and Qq: JjQq.

Q10 P53-q10

The chromosomes of one human cell are sorted by size into pairs, with the sex chromosomes placed last. The last pair is a long chromosome beside a much shorter chromosome.

Which of the following describes the cell's chromosomes?

  1. A. No sex chromosomes and 46 autosomes
    The X and the Y are the sex chromosomes, the one pair whose members do not match.
    The autosomes are the 44 chromosomes of the 22 matching pairs.
  2. B. One sex chromosome, the Y, and 45 autosomes
    The X is a sex chromosome too: the X and the Y are one pair, though they differ in length.
    The autosomes are the 44 chromosomes of the matching pairs.
  3. C. ✓ XY and 44 autosomes
  4. D. XX and 44 autosomes
    A long chromosome beside a much shorter one is an X beside a Y, the male pattern.
    XX would be two chromosomes of the same length.

Why: Twenty-two of the pairs match, member for member: those 44 chromosomes are the autosomes.
The last pair is the sex chromosomes.
A long chromosome beside a much shorter one is an X beside a Y.
So the cell has 44 autosomes and the sex chromosomes XY.

FRQ 1 P53-frq1 · Scientific Investigation scaffolded

In zebrafish, striped (T) is dominant to plain (t). A breeder crosses two Tt fish and predicts striped and plain offspring in a 3 : 1 ratio. She raises 280 offspring and counts 200 striped and 80 plain. Her counts are in the table below, with the chi-square table from the formula sheet beneath it.

Top: the counts of striped and plain fish among 280 offspring of two Tt zebrafish. Bottom: the chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.
Top: the counts of striped and plain fish among 280 offspring of two Tt zebrafish. Bottom: the chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.

(a) State the null hypothesis for her chi-square test. (1 pt)

Frame The null hypothesis is that the offspring occur in a … ratio, and that any difference between the observed counts and the expected counts is due to …

Hint Which statement says there is no real difference between the counts and the model?

Model answer The null hypothesis is that the offspring occur in a 3 : 1 ratio of striped to plain, and that any difference between the observed counts and the expected counts is due to chance alone.
Rubric
  • Award 1 point for: the null hypothesis stated as no real difference between the observed counts and the 3 : 1 model, any difference being due to chance.

Slip Stating that the counts differ from 3 : 1. That is the alternative hypothesis.

(b) Calculate the expected number of striped fish among the 280. (1 pt)

Frame Expected striped = total × the striped share of the ratio = …

Hint What share of a 3 : 1 ratio is the striped class, and what is that share of 280?

Answer: 210  (tolerance ±0)

Model answer Expected striped = total × the striped share of the ratio = 280 × 34 = 210 (and 70 plain are expected).
Working
Write down the values in the question:
total offspring = 280
predicted ratio = 3 striped : 1 plain
Write down the equation:
tex:\text{expected count} = \text{total} \times \text{that class's share of the ratio}
Substitute the values into the equation:
tex:e_{\text{striped}} = 280 \times \frac{3}{4} = 210
tex:e_{\text{plain}} = 280 \times \frac{1}{4} = 70
Rubric
  • Award 1 point for: 210 striped fish expected.

(c) Calculate the chi-square value for her counts. (1 pt)

Frame χ² = (o − e)²/e for striped + (o − e)²/e for plain = …

Hint Use the equation on the formula sheet, one class per line, each class divided by its own expected count.

Answer: 1.9  (tolerance ±0.01)

Model answer χ² = 0.476 + 1.429 = 1.90.
Working
Write down the values in the question:
o = 200 striped, 80 plain
e = 210 striped, 70 plain
Write down the equation:
tex:\chi^2 = \sum \frac{(o - e)^2}{e}
Substitute the values into the equation, one class per line:
tex:\frac{(200 - 210)^2}{210} = \frac{100}{210} = 0.476
tex:\frac{(80 - 70)^2}{70} = \frac{100}{70} = 1.429
tex:\chi^2 = 0.476 + 1.429 = 1.90
Rubric
  • Award 1 point for: χ² = 1.90 (accept 1.89 to 1.91).

(d) Identify the degrees of freedom and the critical value at p = 0.05. (1 pt)

Frame There are … classes, so … degrees of freedom; the critical value at p = 0.05 is …

Hint How many classes of fish are there, and which row and column of the table do you read?

Model answer There are two classes, striped and plain, so there is 1 degree of freedom; the critical value at p = 0.05 is 3.84.
Rubric
  • Award 1 point for: 1 degree of freedom and a critical value of 3.84.

Slip Reading the degrees of freedom as the number of fish or the number of classes.

(e) Determine whether the null hypothesis is rejected, and state what that means for her 3 : 1 model. (1 pt)

Frame χ² = … is … than the critical value …, so the null hypothesis is …; the counts are … with the 3 : 1 model.

Hint Compare the calculated value with the critical value: which is larger, and what does that comparison decide?

Model answer χ² = 1.90 is smaller than the critical value 3.84.
So the null hypothesis is not rejected.
The counts are consistent with the 3 : 1 model.
That does not prove the model; it says the counts give no reason to doubt it.
Rubric
  • Award 1 point for: the decision (fail to reject the null hypothesis) AND the ground (1.90 is less than 3.84), so the counts are consistent with 3 : 1 (not ‘accept’ or ‘prove’).

Slip Writing ‘accept the null hypothesis’. Failing to reject is not accepting.

FRQ 2 P53-frq2 · Scientific Investigation

In goats, a dark coat (D) is dominant to a light coat (d) and long ears (E) to short ears (e). A breeder believes the two genes sit on different chromosomes and predicts that crossing two DdEe goats will give the four classes in a 9 : 3 : 3 : 1 ratio. She counts 160 offspring; her counts are in the table below, with the chi-square table beneath it.

Top: the counts of the four coat-and-ear classes among 160 offspring of two DdEe goats. Bottom: the chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.
Top: the counts of the four coat-and-ear classes among 160 offspring of two DdEe goats. Bottom: the chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.

(a) Identify the four kinds of gamete a DdEe goat makes if the two genes are on different chromosomes, and explain why they are equally likely. (1 pt)

Model answer DE, De, dE and de, each with probability 14.
The homologous pair carrying D and d and the pair carrying E and e line up at metaphase I facing either way regardless of each other.
So which allele of each gene a gamete receives is decided independently, and the four combinations are equally likely.
Rubric
  • Award 1 point for: DE, De, dE, de, each 25%, because the two homologous pairs line up independently at metaphase I.

Slip Listing only DE and de. The dominant alleles do not have to travel together when the genes are on different chromosomes.

(b) Calculate the expected number of light-coated, short-eared offspring among the 160. (1 pt)

Answer: 10  (tolerance ±0)

Model answer 10 light-coated, short-eared goats are expected (the 1-in-16 class).
Working
Write down the values in the question:
total offspring = 160
predicted ratio = 9 : 3 : 3 : 1
Write down the equation:
tex:\text{expected count} = \text{total} \times \text{that class's share of the ratio}
Substitute the values into the equation:
tex:e_{\text{light, short}} = 160 \times \frac{1}{16} = 10
tex:e_{\text{dark, long}} = 160 \times \frac{9}{16} = 90
tex:e_{\text{dark, short}} = e_{\text{light, long}} = 160 \times \frac{3}{16} = 30
Rubric
  • Award 1 point for: 10 light-coated, short-eared goats expected.

(c) Calculate the chi-square value for the breeder's counts against the 9 : 3 : 3 : 1 prediction. (1 pt)

Answer: 4.18  (tolerance ±0.02)

Model answer χ² = 4.18.
Working
Write down the values in the question:
o = 88, 26, 30, 16
e = 90, 30, 30, 10
Write down the equation:
tex:\chi^2 = \sum \frac{(o - e)^2}{e}
Substitute the values into the equation, one class per line:
tex:\frac{(88 - 90)^2}{90} = 0.044
tex:\frac{(26 - 30)^2}{30} = 0.533
tex:\frac{(30 - 30)^2}{30} = 0
tex:\frac{(16 - 10)^2}{10} = 3.6
tex:\chi^2 = 0.044 + 0.533 + 0 + 3.6 = 4.18
Rubric
  • Award 1 point for: χ² = 4.18 (accept 4.16 to 4.20).

(d) Identify the degrees of freedom and the critical value at p = 0.05, and evaluate the breeder's belief that the two genes sit on different chromosomes. (1 pt)

Model answer Four classes give 3 degrees of freedom.
The critical value at p = 0.05 is 7.81.
χ² = 4.18 is smaller than 7.81, so the null hypothesis is not rejected.
So the counts are consistent with 9 : 3 : 3 : 1, and with the two genes assorting independently.
The belief is supported by the counts but not proved: the test finds no reason to doubt it.
Rubric
  • Award 1 point for: 3 degrees of freedom and a critical value of 7.81, AND the judgement (the belief is consistent with the counts, not proved) with its ground (4.18 is less than 7.81, so the null hypothesis is not rejected).

Slip Saying the test proves the genes are on different chromosomes. Failing to reject leaves the belief standing; it does not prove it.

APBIO-U05-T53 End-of-topic test: Mendelian Genetics

Topic 5.3 · Mendelian Genetics · 19 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation, then open the scoring guide and mark your own work against it. Where a question needs a critical value, the chi-square table is drawn with it; read the p = 0.05 row.

Q1 T53-q01

In peppers, a true-breeding red-fruited line was crossed with a true-breeding yellow-fruited line. Every F1 plant had red fruit. When the F1 plants were crossed with each other, the F2 plants were 316 red-fruited and 104 yellow-fruited. A student says the F1 plants lost the yellow allele.

Which statement about the student's claim is correct?

  1. A. The student is right: the yellow allele was lost in the F1, and a new mutation brought yellow back in the F2 plants
    An allele that returns unchanged in about 25% of the F2 was never lost; a new mutation is rare and would not strike 104 of 420 plants.
  2. B. ✓ The student is wrong: every F1 plant carried the yellow allele without showing it, and yellow came back in the F2
  3. C. The student is wrong: the F1 plants kept yellow by being yellow inside their fruit and red on the outside
    Every F1 fruit was red through and through.
    The yellow allele was present in every F1 plant, but red is dominant, so the allele did not show at all.
  4. D. The student is right: the yellow allele was lost in the F1, and the F2 yellow plants came from red pollen only
    Red pollen carries a red allele or a yellow allele, never a yellow fruit.
    The 104 yellow F2 plants each received a yellow allele from both F1 parents.

Why: Yellow vanished in the F1 and returned unchanged in about 25% of the F2, 104 of 420.
A yellow F2 plant needs a yellow allele from each parent.
So every F1 plant carried the yellow allele.
Red is dominant, so the allele did not show: carried, not lost.

Q2 T53-q02

In horses, the coat-color gene has two alleles, K and k. One horse's homologous pair is drawn below.

One horse's homologous pair carrying the coat-color gene; the allele each chromosome carries is written on it.
One horse's homologous pair carrying the coat-color gene; the allele each chromosome carries is written on it.

What is this horse's genotype for coat color, and is the horse homozygous or heterozygous?

  1. A. KK, homozygous
    One chromosome carries K and the other carries k.
    So the two alleles are not both K. Homozygous means the two alleles are the same, and these two differ.
  2. B. kk, homozygous
    One chromosome carries K, so the two alleles are not both k.
    Homozygous means the two alleles are the same, and these two differ.
  3. C. ✓ Kk, heterozygous
  4. D. Kk, homozygous
    Kk is one K and one k.
    Two different alleles is what heterozygous means; homozygous is two of the same.

Why: One chromosome carries K and the other k, so the genotype is Kk, and two different alleles make the horse heterozygous.

Q3 T53-q03

In peppers, red fruit (R) is dominant to yellow fruit (r). Two pepper plants both have yellow fruit.

Which genotypes do the two plants have?

  1. A. Rr and Rr
    An Rr plant carries R.
    R is dominant, so an Rr plant has red fruit, and neither plant is red.
  2. B. RR and rr
    An RR plant has red fruit.
    Both plants have yellow fruit, so neither plant is RR.
  3. C. Rr and rr
    An Rr plant has red fruit, and both of these plants have yellow fruit.
  4. D. ✓ rr and rr

Why: Yellow is the recessive trait.
A recessive trait shows only when both alleles are recessive.
Each plant has yellow fruit, so each plant is rr.

Q4 T53-q04

Two Rr pepper plants are crossed. The Punnett square below has one cell still to fill, marked ?.

A Punnett square for two Rr pepper plants, with one cell still to fill, marked ?.
A Punnett square for two Rr pepper plants, with one cell still to fill, marked ?.

Which genotype belongs in the marked cell?

  1. A. ✓ Rr
  2. B. rr
    The marked cell is where the r egg down the side fuses with the R pollen along the top, so it holds one of each.
  3. C. RR
    The marked cell's column is headed R but its row is headed r, so the cell brings together R and r.
  4. D. R
    A cell holds the two alleles the egg and the pollen bring together, so it is two letters.

Why: The marked cell is in the column headed R and the row headed r, so the egg brings r and the pollen brings R: the cell reads Rr.

Q5 T53-q05

Two Rr pepper plants (red fruit dominant to yellow) are crossed.

Which of these is the genotypic ratio of their offspring?

  1. A. 3 red : 1 yellow
    3 red : 1 yellow groups RR and Rr together by what shows, which makes it the phenotypic ratio.
  2. B. ✓ 1 RR : 2 Rr : 1 rr
  3. C. 1 RR : 1 rr
    It leaves out the two Rr cells of the square, which are half of the offspring.
  4. D. 2 red : 1 yellow
    The square has four cells, not three, and this ratio is by phenotype rather than genotype.

Why: The Rr × Rr square gives RR, Rr, Rr, rr: read by genotype that is 1 RR : 2 Rr : 1 rr; read by phenotype it is 3 red : 1 yellow.

Q6 T53-q06

In horses, a black coat (K) is dominant to a chestnut coat (k). Two Kk horses have six foals over the years: five black and one chestnut.

Are these foals consistent with the parents both being Kk, and why?

  1. A. No; two Kk horses give three black foals for every chestnut one, so six foals cannot be five and one
    A Punnett square gives a chance for each foal, not a fixed count for six foals; three to one is what large numbers approach.
  2. B. No; five to one shows that one parent must be KK
    A KK parent would give no chestnut foals at all, and there is one.
  3. C. ✓ Yes; each foal has the same one-in-four chance of chestnut, and six foals is a small batch
  4. D. No; six foals from two Kk horses always include at least two chestnut ones
    Each foal is its own event; six foals can include any number of chestnut ones from none to six.

Why: Each foal is a separate event with a one-in-four chance of chestnut.
Six foals is a small batch, so the count strays from three to one.
For six foals the square expects four or five black and one or two chestnut.
Five black and one chestnut sits inside that range.

Q7 T53-q07

In horses, a black coat (K) is dominant to a chestnut coat (k). Two Kk horses are bred.

What is the probability that a foal is homozygous, KK or kk?

  1. A. 34
    34 is the probability of a black coat, KK or Kk; the question asks for the two homozygous genotypes, KK or kk.
  2. B. 14
    14 is the probability of KK alone, or of kk alone; the question allows either.
  3. C. 116
    KK and kk are two outcomes that cannot both be true of one foal, and either will do, so their probabilities are added, not multiplied.
  4. D. ✓ 12

Why: KK and kk are two outcomes that cannot both be true of one foal.
Either one satisfies the question.
So add the two probabilities: P(KK or kk)=P(KK)+P(kk)=14+14=12.

Q8 T53-q08

In peppers, red fruit (R) is dominant to yellow (r). Two Rr plants are crossed.

What is the probability that an offspring plant has the genotype RR?

  1. A. ✓ 14
  2. B. 12
    12 is the chance that one gamete carries R; an RR plant needs R from the egg and R from the pollen, two independent events.
  3. C. 34
    34 is the probability of red fruit, the three cells of the square that carry an R; RR is one of those cells.
  4. D. 1
    The egg's R and the pollen's R must both happen, so the two probabilities are multiplied, not added; adding gives a probability of one, which would make every offspring RR.

Why: An RR plant needs R from the egg and R from the pollen.
The two gametes are independent, so P(RR)=P(R from the egg)×P(R from the pollen)=12×12=14.

Q9 T53-q09

Two Kk horses (black coat, K, dominant to chestnut, k) have three foals, one after another.

What is the probability that all three foals are chestnut?

  1. A. 112
    112 is one foal’s 14 divided by three; three chestnut foals in a row needs the three 14 chances multiplied.
  2. B. 34
    34 is the probability of one foal being black; three chestnut foals in a row multiplies three one-in-four chances.
  3. C. ✓ 164
  4. D. 14
    14 is the chance for one foal; each further foal multiplies it by 14 again.

Why: Each foal independently has P(kk)=14; for all three, multiply: (14)3=164.

Q10 T53-q10

A red-fruited pepper plant could be RR or Rr. It is crossed with a yellow-fruited plant, rr, and the offspring are 22 red-fruited and 20 yellow-fruited.

What is the red-fruited parent's genotype, and how do the offspring show it?

  1. A. RR, because more than half of the offspring are red
    An RR parent gives every offspring an R, so none could be yellow; 20 yellow offspring rule RR out.
  2. B. ✓ Rr, because yellow offspring received an r allele from it
  3. C. RR, because an Rr parent would give only yellow offspring with rr
    Rr × rr gives half red and half yellow, which is what 22 : 20 shows.
  4. D. It stays undecided, because a cross with a yellow-fruited plant shows nothing
    Crossing with the homozygous recessive is exactly how the genotype is decided: any yellow offspring means the red parent carried r.

Why: A yellow (rr) offspring needs an r from each parent.
The red parent gave r to 20 offspring, so it carried r: Rr.
The 22 : 20 split is the 1 : 1 that Rr × rr predicts.

Q11 T53-q11

A breeder keeps zebrafish. She crosses a true-breeding striped line with a true-breeding plain line. Every F1 fish is striped. Crossing the F1 fish with each other gives 149 striped and 51 plain F2 fish. In her tanks, most of the fish are plain.

Which allele is dominant, and what shows it?

  1. A. ✓ The striped allele: all the F1 and about 75% of the F2 are striped
  2. B. The plain allele, because plain is by far the more common pattern in her tanks
    How common a trait is in the tanks does not show which allele is dominant; a dominant trait can be rare.
  3. C. The plain allele, because the plain fish vanished and came back in the F2
    A trait that vanishes in the F1 and returns in 25% of the F2 is the recessive one.
  4. D. Neither allele, because striped and plain fish are both present in the F2
    The F1 all showed one trait and the F2 split about three to one, which is the pattern of one dominant and one recessive allele.

Why: The trait every F1 shows, and that about 75% of the F2 show, is the dominant one: striped.
How common plain fish are in the tanks does not show which allele is dominant.

Q12 T53-q12

In horses, a black coat (K) is dominant to a chestnut coat (k). A Kk black horse is mated with a kk chestnut horse.

Which foal is impossible from this mating?

  1. A. A black foal with genotype Kk
    The black parent gives K to half of its gametes and the chestnut parent gives k to every gamete, so half of the foals are Kk, and Kk is black.
  2. B. A chestnut foal with genotype kk
    The black parent gives k to half of its gametes and the chestnut parent gives k to every gamete, so half of the foals are kk, and kk is chestnut.
  3. C. A chestnut foal whose black parent gave it k
    The black parent is Kk, so it gives k to half of its gametes; a foal that receives that k and the chestnut parent's k is kk, chestnut.
  4. D. ✓ A black foal with genotype KK

Why: The chestnut parent is kk, so every one of its gametes carries k.
So every foal receives a k from the chestnut parent.
A KK foal needs a K from each parent.
So a KK foal has a probability of zero from this mating.

Q13 T53-q13

A rare condition is recorded in the family below.

A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.
A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.

Is the condition dominant or recessive, and which family decides it?

  1. A. Dominant: II-2 has the condition although neither of her parents has it
    A dominant trait shows in anyone who carries the allele, so two unaffected parents have none to pass on and cannot have an affected child.
  2. B. Dominant: the condition appears in generation II and again in generation III
    Appearing in two generations is consistent with either mode and decides nothing.
  3. C. ✓ Recessive: an affected child, II-2, of two unaffected parents, I-1 and I-2
  4. D. Recessive: the condition skips generation I
    Skipping a generation is a hint, not a proof; the decisive family is the one that rules a mode out: I-1 and I-2, unaffected, with their affected daughter II-2.

Why: Two unaffected parents with an affected child rule dominance out: a dominant trait would show in a parent who carried it.
So the condition is recessive, and I-1, I-2, II-3 and II-4 are all carriers.

Q14 T53-q14

The pedigree below records a rare condition in one family. Write A for the ordinary allele and a for the allele that causes the condition.

A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.
A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.

Which genotype must II-4 have?

  1. A. AA
    II-4's child III-1 has the condition, so III-1 is aa and received an a from each parent.
    An AA parent has no a to pass on.
  2. B. ✓ Aa
  3. C. aa
    II-4 does not have the condition, so he is not aa.
    He carries the a allele without showing it.
  4. D. AA or Aa; the pedigree cannot decide
    An unaffected child of two carriers is AA or Aa.
    II-4 is an unaffected parent of an affected child, so he must carry a: Aa.

Why: III-1 has the condition, so III-1 is aa.
III-1 received one allele from each parent, so II-4 passed an a.
II-4 does not have the condition, so he also has an A.
So II-4 is Aa, a carrier.

Q15 T53-q15

In the pedigree below, II-1 has a child with a man who has the condition.

A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.
A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.

What is the probability that the child shows the condition?

  1. A. 14
    14 is the probability when both parents are carriers.
    Here the father has the condition, so he passes its allele to every child.
  2. B. ✓ 13
  3. C. 12
    12 treats II-1 as a certain carrier.
    II-1 is an unshaded child of two carriers, so she is a carrier with probability 23, not 1.
  4. D. 23
    23 is the probability that II-1 is a carrier.
    The child must also receive the allele from her, which has probability 12.

Why: II-2 has the condition and her parents do not, so the condition is recessive and both parents are carriers.
II-1, their unshaded daughter, is a carrier with probability 23.
A carrier passes the allele with probability 12; the affected father always passes it.
So 23×12=13.

Q16 T53-q16

In peppers, fruit pungency and fruit color are controlled by two genes on different chromosomes: hot fruit (M) is dominant to mild (m), and red (R) to yellow (r). A plant is MMRr.

Which gametes does this plant make, and in what proportions?

  1. A. MR only
    The plant is Rr for color.
    The homologs carrying R and r part at anaphase I, so half of the gametes receive R and half receive r.
  2. B. MM and Rr, half each
    A gamete carries one allele of each gene, not both alleles of one gene.
    MM and Rr are the plant's genotypes for the two genes, not gametes.
  3. C. ✓ MR and Mr, half each
  4. D. MR, Mr, mR and mr, each 25%
    The plant is MM for pungency.
    Every gamete receives one M, so no gamete carries m.
    Four kinds of gamete need a plant heterozygous for both genes.

Why: The plant is MM, so every gamete receives M. The plant is Rr, so half of the gametes receive R and half receive r.
The two genes sit on different chromosomes, so each gamete carries one allele of each gene: MR or Mr, half each.

Q17 T53-q17

Two MmRr pepper plants are crossed (hot, M, dominant to mild, m; red, R, dominant to yellow, r; the genes are on different chromosomes) and 480 offspring are grown.

How many offspring are expected to have mild, yellow fruit?

  1. A. ✓ 30
  2. B. 90
    90 is the expected count of one of the 3-in-16 classes (hot yellow, or mild red); mild yellow is the 1-in-16 class.
  3. C. 120
    120 is 25% of 480, the share for one gene's recessive class alone; two recessive traits together are one sixteenth.
  4. D. 270
    270 is the 9-in-16 class, hot red, the two dominant traits together.

Why: In the 9 : 3 : 3 : 1 ratio, mild yellow (mmrr) is the 1-in-16 class: 480×116=30.

Q18 T53-q18

An MmRr pepper plant is crossed with an Mmrr plant (hot, M, dominant to mild, m; red, R, dominant to yellow, r; genes on different chromosomes).

What is the probability that an offspring has mild, yellow fruit?

  1. A. 116
    116 is the answer for two heterozygous parents; here the second parent is rr, so yellow has probability 12, not 14.
  2. B. ✓ 18
  3. C. 12
    12 is the probability of yellow fruit alone, from Rr × rr; the fruit must be mild as well, which has probability 14.
  4. D. 34
    34 is 14+12.
    Mild and yellow must both happen, so the two probabilities are multiplied, not added.

Why: Treat each gene alone: P(mm)=14 from Mm × Mm and P(rr)=12 from Rr × rr.
Both events must happen, so multiply: 14×12=18.

Q19 T53-q19

In peppers, the fruit-pungency gene and the fruit-color gene sit on different chromosomes. In another plant, two genes sit close together on the same chromosome.

Which statement about Mendel's law of independent assortment is correct?

  1. A. The law applies to neither case, because it describes only the seven traits of Mendel's peas
    The law holds for any two genes on different chromosomes in any organism, not only in peas.
  2. B. The law applies to both cases, because every gene assorts independently of every other gene
    Independent assortment comes from two homologous pairs facing either way at metaphase I; two genes on one chromosome sit on the same pair and so travel together.
  3. C. ✓ The law holds for the pepper genes on different chromosomes; two genes on one chromosome need not follow it
  4. D. The law applies to the two genes on one chromosome, because alleles inherited together assort independently
    Being inherited together is the opposite of assorting independently.

Why: Independent assortment rests on separate homologous pairs lining up independently at metaphase I.
Genes on different chromosomes follow it; two genes on the same chromosome need not, because they ride the same pair.

FRQ 1 T53-frq1 · Scientific Investigation

A pepper breeder crosses two red-fruited plants that she knows to be Rr (red, R, dominant to yellow, r) and predicts that the offspring will show red and yellow fruit in a 3 : 1 ratio. She grows 300 offspring and records the fruit color of each. Her counts are in the table below; the chi-square table from the formula sheet is drawn beneath it. She also weighs the fruit of every plant and records the mean fruit mass of the red-fruited plants and of the yellow-fruited plants.

Top: the breeder's counts of fruit color among 300 offspring of Rr × Rr pepper plants. Bottom: the chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.
Top: the breeder's counts of fruit color among 300 offspring of Rr × Rr pepper plants. Bottom: the chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.

(a) State the null hypothesis for the breeder's chi-square test. (1 pt)

Model answer The null hypothesis is that the offspring occur in a 3 : 1 ratio of red to yellow, and that any difference between the observed counts and 225 : 75 is due to chance alone; there is no real difference between the counts and the model.
Rubric
  • Award 1 point for: the null hypothesis stated as no real difference between the observed counts and the 3 : 1 model (any difference is due to chance).
  • Do not award: a statement that the counts differ from 3 : 1 (the alternative hypothesis), or a bare ‘3 : 1’ with no mention of chance.

Slip Stating that the counts will differ from 3 : 1. That is the alternative hypothesis; the null is the ‘only chance’ statement.

(b) Calculate the expected number of red-fruited plants among the 300 offspring. (1 pt)

Answer: 225  (tolerance ±0)

Model answer 225 red-fruited plants are expected (and 75 yellow-fruited).
Working
Write down the values in the question:
total offspring = 300
predicted ratio = 3 red : 1 yellow
Write down the equation:
tex:\text{expected count} = \text{total} \times \text{that class's share of the ratio}
Substitute the values into the equation:
tex:e_{\text{red}} = 300 \times \frac{3}{4} = 225
tex:e_{\text{yellow}} = 300 \times \frac{1}{4} = 75
Rubric
  • Award 1 point for: 225 red-fruited plants expected.

(c) Calculate the chi-square value for the breeder's counts. (1 pt)

Answer: 2.15  (tolerance ±0.01)

Model answer χ² = 2.15.
Working
Write down the values in the question:
o = 214 red, 86 yellow
e = 225 red, 75 yellow
Write down the equation:
tex:\chi^2 = \sum \frac{(o - e)^2}{e}
Substitute the values into the equation, one class per line:
tex:\frac{(214 - 225)^2}{225} = \frac{121}{225} = 0.538
tex:\frac{(86 - 75)^2}{75} = \frac{121}{75} = 1.613
tex:\chi^2 = 0.538 + 1.613 = 2.15
Rubric
  • Award 1 point for: χ² = 2.15 (accept 2.14 to 2.16).

(d) Identify the degrees of freedom and the critical value at p = 0.05, and determine whether the null hypothesis is rejected. (1 pt)

Model answer There are two classes, red and yellow, so there is 1 degree of freedom.
The critical value at p = 0.05 is 3.84.
The calculated χ² of 2.15 is smaller than 3.84.
So the null hypothesis is not rejected: the counts are consistent with the 3 : 1 model.
Rubric
  • Award 1 point for: 1 degree of freedom and a critical value of 3.84, AND the decision (fail to reject the null hypothesis) with its ground (2.15 is less than 3.84).
  • Do not award: ‘accept the null hypothesis’ or ‘the model is proved’.

Slip Taking the degrees of freedom as 300 or as 2. Degrees of freedom count the classes minus one, and there are two classes.

(e) The breeder wants to test whether the mean fruit mass of red-fruited plants differs from that of yellow-fruited plants. Explain why chi-square is the wrong test for that comparison, and identify the tool she should use instead. (1 pt)

Model answer Chi-square compares counts of individuals in categories with the counts a ratio predicts.
A mean fruit mass is a measured value, not a count, and there is no predicted ratio to compare it with.
She should compare the two mean values using their standard errors: plot each mean with ±2SE error bars and apply the overlap rule.
Rubric
  • Award 1 point for: chi-square needs counts in categories against a predicted ratio, and a mean mass is a measured value, not a count; AND a tool for comparing two means named: ±2SE (95% confidence interval) error bars with the overlap rule, or a t-test.
  • Accept: either tool; the point rests on the reason and one named tool.

Slip Converting the mean masses into a ratio and calculating chi-square from them. Means are not counts, and no model predicts a ratio of masses.

FRQ 2 T53-frq2 · Conceptual Analysis

Cystic fibrosis is a condition caused by one gene on an autosome, one of the 22 pairs of chromosomes that are not sex chromosomes. The pedigree below records it in one family: filled shapes show the condition.

A pedigree for cystic fibrosis across three generations. Squares are males, circles females; a filled shape shows the condition.
A pedigree for cystic fibrosis across three generations. Squares are males, circles females; a filled shape shows the condition.

(a) Support the claim that the cystic fibrosis allele is recessive with evidence from one set of parents and their children in the pedigree. (1 pt)

Model answer Evidence: I-1 and I-2 do not have the condition, yet their daughter II-1 does.
If the allele were dominant, anyone carrying it would show the condition.
So unaffected parents could not carry a dominant allele, and could not pass it on.
Two unaffected parents with an affected child are possible only if both carry a recessive allele without showing it.
So these parents and their child support the claim that the allele is recessive.
Rubric
  • Award 1 point for: the evidence (two unaffected parents, I-1 and I-2 or II-3 and II-4, have an affected child) AND the reasoning that links it to the claim (a dominant allele would show in a carrier, so unaffected parents could not pass one on).

Slip Naming the family without the link, or arguing from the condition skipping a generation. Support a claim needs the evidence and the reasoning that ties it to the claim.

(b) Write F for the ordinary allele and f for the allele that causes the condition. Identify the genotypes of I-1, I-2 and II-1. (1 pt)

Model answer II-1 has the condition, so she is ff.
She received one f from each parent.
I-1 and I-2 do not have the condition, so each also has an F.
So I-1 is Ff, I-2 is Ff and II-1 is ff.
Rubric
  • Award 1 point for: I-1 Ff, I-2 Ff and II-1 ff (all three).

Slip Writing I-1 or I-2 as FF. An FF parent has no f to pass on, so an ff child rules FF out.

(c) III-1 has cystic fibrosis. Calculate the probability that the next child of II-3 and II-4 also has it. (1 pt)

Answer: 0.25  (tolerance ±0.005)

Model answer III-1 is ff, so each of II-3 and II-4 passed an f.
Neither II-3 nor II-4 has the condition, so each also has an F: both are Ff.
The next child has the condition only if it receives f from both: 12×12=14, which is 0.25.
Working
Write down the values:
II-3 and II-4 are both Ff (unaffected parents of an ff child)
tex:P(f \text{ from II-3}) = \frac{1}{2}
tex:P(f \text{ from II-4}) = \frac{1}{2}
Write down the equation:
tex:P(A \text{ and } B) = P(A) \times P(B)
Substitute the values into the equation:
tex:P(ff) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25
Rubric
  • Award 1 point for: 14 (0.25), from II-3 and II-4 both Ff.

Slip Writing 12. II-3 and II-4 are each Ff, not ff; the child needs an f from each, so the two halves are multiplied.

(d) Explain why II-2's genotype stays undecided from the pedigree, and give the two genotypes it could be. (1 pt)

Model answer II-2 does not have the condition, so he is not ff.
An unaffected person can be FF or a carrier, Ff, and the two look the same.
His parents are both Ff.
So he could have received F from both (FF), or F from one parent and f from the other (Ff).
Nothing in the pedigree tells which, so his genotype is FF or Ff and cannot be fixed.
Rubric
  • Award 1 point for: unaffected means not ff, so II-2 is FF or Ff, and the two genotypes look the same, so the pedigree cannot decide between them.

Slip Fixing II-2 as Ff because his parents are carriers. Being a carrier's child makes Ff more likely (two chances in three among the unaffected children) but does not fix it.

APBIO-U05-L21 Pink between red and white

Topic 5.4 · Non-Mendelian Genetics · 76 steps

Left: three photographs of real snapdragon flowers side by side, labeled red, pink and white. Right: the cross drawn as flower symbols: a red parent and a white parent at the top, joined by two short lines to one pink offspring below, and the question: why is the pink one pink?
Left: three photographs of real snapdragon flowers side by side, labeled red, pink and white. Right: the cross drawn as flower symbols: a red parent and a white parent at the top, joined by two short lines to one pink offspring below, and the question: why is the pink one pink?

Photos: Suresh Prasad, Wikimedia Commons, CC BY-SA 4.0; Sabina Bajracharya, Wikimedia Commons, CC BY-SA 4.0; Robert Scarth, Wikimedia Commons, CC BY-SA 2.0 (resized).

Here are three snapdragons in a row: red, pink, white. The pink one’s parents were the red one and the white one, drawn beside them.

Mendel’s rule said one trait should hide the other. Why is the pink one pink?

Unit 5 · Heredity

1Neither allele hides the other

2

Video: Watch: Neither allele hides the other

The red and white parents with their alleles written on, the pink offspring between them, and the two pinks crossed to bring red and white back.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21a.mp4

3

What happens when neither allele hides the other?

4

In a pink snapdragon the heterozygote is a blend of the two homozygotes. In a roan calf the heterozygote shows both traits side by side, red hairs and white hairs.

5

In Mendel’s peas the heterozygote looks like one parent. So one look at the heterozygote sorts a trait into one of three kinds.

6
Check q1

A Pp pea plant has purple flowers, and a PP pea plant has purple flowers too.

Which allele is dominant?

  1. A. ✓ P
  2. B. p
    A pp plant is white.
    The Pp plant is purple, like the PP plant, so P hides p.
  3. C. Neither
    The Pp plant looks like the PP plant.
    So one allele, P, hides the other.

Why: The Pp plant is purple, like the PP plant.
So P hides p in the heterozygote.
The allele whose trait shows in the heterozygote is dominant, so P is dominant.

7

Now consider snapdragons. Cross a true-breeding red snapdragon with a true-breeding white one.

8

Every offspring flowers pink: neither the red nor the white shows on its own.

A red snapdragon on the left and a white snapdragon on the right, each labeled; a downward arrow from each to a row of four offspring flowers, all pink, labeled: every offspring is pink
A red snapdragon on the left and a white snapdragon on the right, each labeled; a downward arrow from each to a row of four offspring flowers, all pink, labeled: every offspring is pink
9

In these snapdragons neither allele hides the other. So neither allele is dominant.

10

The two alleles are written as one letter with a superscript, Cᴿ for red and Cᵂ for white. The one letter shows that they are alleles of one gene.

11

The pink plant is CᴿCᵂ: one red allele and one white allele.

A homologous pair drawn as two rods with centromere dots, one darker and one lighter; the darker rod carries C with a superscript R in a paper box and the lighter rod C with a superscript W at the same height; caption: the pink plant is C-R C-W
A homologous pair drawn as two rods with centromere dots, one darker and one lighter; the darker rod carries C with a superscript R in a paper box and the lighter rod C with a superscript W at the same height; caption: the pink plant is C-R C-W
12

In every drawing, the chromosome from the mother is drawn dark and the chromosome from the father is drawn light.

13

A CᴿCᴿ plant has two red alleles. Its petal cells make full red pigment.

14

A CᴿCᵂ plant has one red allele. So its petal cells make about half as much red pigment.

Three flowers with their genotypes: C-R C-R red, labeled two red alleles, full pigment; C-R C-W pink, labeled one red allele, about half the pigment; C-W C-W white, labeled no red allele, no pigment
Three flowers with their genotypes: C-R C-R red, labeled two red alleles, full pigment; C-R C-W pink, labeled one red allele, about half the pigment; C-W C-W white, labeled no red allele, no pigment
15

Less red pigment looks pink.

16

When neither allele hides the other, the heterozygote is a blend. We call this , because the red allele’s dominance is incomplete: it shows in the heterozygote only in part.

17

Pink does not mean the alleles blended into one. Each pink plant still carries one Cᴿ and one Cᵂ.

18

Cross two pink plants, and red and white flowers come back among the offspring.

Two pink snapdragons crossed, above a row of four offspring flowers labeled red, pink, pink, white; caption: red and white come back, so the alleles never blended
Two pink snapdragons crossed, above a row of four offspring flowers labeled red, pink, pink, white; caption: red and white come back, so the alleles never blended
19

What you are expected to know Describe incomplete dominance: the heterozygote is a blend of the two homozygotes, and the two alleles come back unchanged in the next generation.

20
Check q2

A four o’clock plant carries one red allele, Cᴿ, and one white allele, Cᵂ. Its flowers are pink.

Which allele is dominant?

  1. A. Cᴿ
    If Cᴿ hid Cᵂ, the CᴿCᵂ plant would be red like a CᴿCᴿ plant.
    It is pink, so Cᴿ does not hide Cᵂ.
  2. B. Cᵂ
    If Cᵂ hid Cᴿ, the CᴿCᵂ plant would be white like a CᵂCᵂ plant.
    It is pink, so Cᵂ does not hide Cᴿ.
  3. C. ✓ Neither

Why: The CᴿCᵂ plant is pink.
Pink is a blend of red and white.
So neither allele hides the other.
Therefore neither allele is dominant.

21
Practice writing an answer

A four o’clock plant carries one red allele, Cᴿ, and one white allele, Cᵂ. Its flowers are pink, and neither allele is dominant.

(a) Explain why the CᴿCᵂ plant is pink rather than red. (1 pt)

Frame The CᴿCᵂ plant has …

Model answer The CᴿCᵂ plant has one red allele.
A CᴿCᴿ plant has two red alleles, and its petal cells make full red pigment.
One red allele makes about half as much red pigment as two.
So the CᴿCᵂ plant’s petal cells make less red pigment.
Less red pigment looks pink.
Rubric
  • Award 1 point for: one red allele makes less red pigment than two, and less red pigment looks pink.
22
Check q3

A student looks at a pink four o’clock plant, CᴿCᵂ, and says: “The red allele and the white allele blended into one pink allele.”

Which statement about the student’s claim is correct?

  1. A. The student is right: the plant now carries one pink allele
    Cross two pink plants and red and white flowers come back among the offspring.
    So each pink parent still carried one Cᴿ and one Cᵂ.
  2. B. ✓ The student is wrong: the plant still carries one Cᴿ and one Cᵂ

Why: Cross two pink plants and red and white flowers come back among the offspring.
A red flower needs two Cᴿ alleles and a white flower two Cᵂ alleles.
So each pink plant still carried one Cᴿ and one Cᵂ.
Pink is what the two look like together.

23
Check q4

In horses, a chestnut horse crossed with a cremello horse gives palomino foals, a golden coat between the two colors. A breeder crosses two palomino horses.

Which coat colors are possible among the foals?

  1. A. Palomino only
    A palomino is a heterozygote, one chestnut allele and one cremello allele.
    So a foal can receive two chestnut alleles, two cremello alleles, or one of each.
  2. B. Chestnut and cremello only
    A foal that receives one chestnut allele and one cremello allele is palomino, so palomino foals appear too.
  3. C. Palomino and chestnut only
    A foal that receives the cremello allele from both parents is cremello.
    Each palomino parent passes its cremello allele to half of its gametes.
  4. D. ✓ Chestnut, palomino and cremello

Why: A palomino horse is the heterozygote: one chestnut allele and one cremello allele.
Each palomino parent passes one of the two to each gamete.
Two chestnut alleles give chestnut, two cremello alleles give cremello, one of each gives palomino.
So all three colors are possible.

24Quick quiz: incomplete dominance mixed practice

25
Check q5

A plant carries two different alleles of one flower-color gene.

What is incomplete dominance?

  1. A. Each allele’s trait shows whole and separate in the heterozygote
    A blend has no separate red part and white part.
    In incomplete dominance every part of the heterozygote is in between.
  2. B. One allele hides the other, so the heterozygote looks like one homozygote
    One allele hiding the other is complete dominance.
    In incomplete dominance the heterozygote looks like neither homozygote.
  3. C. ✓ Neither allele hides the other, so the heterozygote is a blend of the two homozygotes

Why: In incomplete dominance neither allele hides the other.
So the heterozygote is a blend of the two homozygotes.

26
Practice writing an answer

In one plant, a CᴿCᵂ heterozygote has pink flowers, between the red flowers of CᴿCᴿ plants and the white flowers of CᵂCᵂ plants.

(a) State what incomplete dominance is. (1 pt)

Model answer Neither allele hides the other, so the heterozygote is a blend of the two homozygotes.
Rubric
  • Award 1 point for: neither allele hides the other; the heterozygote is a blend (in between) of the two homozygotes.

(b) Explain how the pink CᴿCᵂ plant demonstrates incomplete dominance. (1 pt)

Model answer The CᴿCᵂ plant is the heterozygote.
Its flowers are pink, between the red of CᴿCᴿ and the white of CᵂCᵂ.
So neither allele hides the other.
A heterozygote that is a blend of the two homozygotes shows incomplete dominance.
Rubric
  • Award 1 point for: the heterozygote is a blend of the two homozygotes (pink between red and white), so neither allele hides the other.
27
Check q6

In a plant, a red-flowered parent crossed with a white-flowered parent gives pink-flowered offspring.

Is a pink offspring a heterozygote?

  1. A. ✓ Yes
  2. B. No
    The pink offspring received a red allele from one parent and a white allele from the other.
    Two different alleles make a heterozygote.

Why: The red parent gave the offspring a red allele and the white parent gave it a white allele.
An individual with two different alleles is a heterozygote.

28
Check q7

In a plant, a red-flowered parent crossed with a white-flowered parent gives pink-flowered offspring.

Does a pink offspring carry a pink allele?

  1. A. Yes
    The pink offspring carries one red allele and one white allele.
    Pink is what the two look like together.
  2. B. ✓ No

Why: The pink offspring carries one red allele and one white allele.
Neither allele changed.
Pink is what the two alleles look like together, not a third allele.

29Both alleles show, side by side

30

Video: Watch: Both alleles show, side by side

A patch of roan coat under a magnifier, red hairs beside white hairs and no hair pink; then the red cow and the white bull the roan calf came from, with their alleles written on.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21b.mp4

31

Now consider cattle instead of snapdragons. A coat is made of many hairs, and you can look at the hairs one by one.

32

Here is a photograph of a roan Shorthorn heifer, a young cow.

A photograph of a roan Shorthorn heifer standing in a field, seen from the side: a red-and-white coat with red hairs and white hairs mixed over the shoulders and flanks and larger white patches on the belly and back
A photograph of a roan Shorthorn heifer standing in a field, seen from the side: a red-and-white coat with red hairs and white hairs mixed over the shoulders and flanks and larger white patches on the belly and back
33

In one coat-color gene of these cattle, Cᴿ gives red hairs and Cᵂ gives white hairs.

34

A red cow, CᴿCᴿ, crossed with a white bull, CᵂCᵂ, gives a roan calf, CᴿCᵂ.

A red cow and a white bull labeled, an arrow down to a patch of their roan calf's coat: short red hairs and short white hairs side by side, with the caption: red hairs and white hairs, no hair is pink
A red cow and a white bull labeled, an arrow down to a patch of their roan calf's coat: short red hairs and short white hairs side by side, with the caption: red hairs and white hairs, no hair is pink
35

Look closely at the roan calf’s coat. Some hairs are red and some are white, and no hair is pink.

36

So the heterozygote shows both parents’ traits, each one whole and separate.

37

When both alleles’ traits show separately in the heterozygote, we call it , because the two alleles are dominant together: co- means together.

38

In incomplete dominance the heterozygote blends the two traits. In codominance the heterozygote shows the two traits side by side.

39

What you are expected to know Describe codominance: both homozygotes’ traits show separately in the heterozygote, red hairs and white hairs on one roan coat.

40
Check q8

In one cattle breed, Cᴿ gives red hairs and Cᵂ gives white hairs. A CᴿCᵂ calf is roan.

What does its coat show?

  1. A. Red hairs only
    The white allele’s trait shows too.
    A roan coat has white hairs beside the red hairs.
  2. B. Pink hairs, every hair a blend of red and white
    No hair on a roan coat is pink.
    Each hair is fully red or fully white.
  3. C. ✓ Red hairs and white hairs, each hair one color

Why: The roan calf carries Cᴿ and Cᵂ.
Cᴿ gives red hairs and Cᵂ gives white hairs.
Both alleles’ traits show separately, so the coat has red hairs and white hairs.

41
Check q9

Suppose in one plant a heterozygote for a petal-color gene has petals with red patches and white patches, and no part of any petal is pink.

What does this heterozygote show?

  1. A. The color of one parent only
    Both red and white appear on the petals.
    So both alleles’ traits show.
  2. B. A blend of red and white
    A blend would be pink all through.
    Here each patch is fully red or fully white.
  3. C. ✓ Red and white, each separate

Why: The petals have red patches and white patches.
Each patch is one color, and no patch is pink.
So both parents’ traits show, each one separate.

42Quick quiz: codominance mixed practice

43
Check q10

An animal carries two different alleles of one coat-color gene.

What is codominance?

  1. A. ✓ Both alleles’ traits show separately in the heterozygote
  2. B. The heterozygote is a blend of the two homozygotes
    A blend is incomplete dominance.
    In codominance each trait shows whole and separate.
  3. C. One allele hides the other in the heterozygote
    One allele hiding the other is complete dominance.
    In codominance both alleles’ traits show.

Why: In codominance both alleles are dominant together.
So both alleles’ traits show separately in the heterozygote.

44
Practice writing an answer

In people, an LᴹLᴺ person’s red blood cells carry the M marker and the N marker, each as its own molecule, and no marker of an in-between kind.

(a) State what codominance is. (1 pt)

Model answer Both alleles’ traits show separately in the heterozygote.
Rubric
  • Award 1 point for: both alleles’ traits (products) show separately in the heterozygote.

(b) Explain how the LᴹLᴺ person demonstrates codominance. (1 pt)

Model answer The LᴹLᴺ person is the heterozygote.
That person’s red cells carry the M marker and the N marker, each whole and separate.
No marker is in between M and N.
So both alleles’ traits show separately, which is codominance.
Rubric
  • Award 1 point for: the heterozygote carries both markers, each whole and separate (none in between), so both alleles’ traits show: codominance.
45
Check q11

A roan mammal’s coat has red hairs and white hairs.

Is any hair on the coat a blend of red and white?

  1. A. Yes
    Each hair on a roan coat is fully red or fully white.
  2. B. ✓ No

Why: In codominance each allele’s trait shows whole and separate.
So each hair is red or white, and no hair is a blend.

46
Check q12

A gene shows codominance.

How many of the two alleles’ traits show in the heterozygote?

  1. A. ✓ Both
  2. B. One
    One trait showing is complete dominance.
    In codominance both alleles are dominant together.

Why: In codominance both alleles are dominant together.
So both alleles’ traits show in the heterozygote.

47Which kind of dominance?

48

Video: Watch: Which kind of dominance?

A Pp pea, a pink snapdragon and a roan calf in a row, each heterozygote sorted into its kind of dominance with the same sentence; then the table of the three kinds filling.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21c.mp4

49

So there are three kinds of dominance. You sort a trait into one of them by looking at the heterozygote.

50

Look at a Pp pea plant, the heterozygote: it looks like one of the two homozygotes, purple like PP. So this is complete dominance.

51

Look at a CᴿCᵂ snapdragon, the heterozygote: it is a blend of the two homozygotes, pink between red and white. So this is incomplete dominance.

52

Look at a CᴿCᵂ calf, the heterozygote: it shows both homozygotes’ traits separately, red hairs and white hairs. So this is codominance.

Three panels in a row. Left: a purple pea flower labeled Pp pea, like one homozygote, and the verdict complete dominance. Middle: a pink snapdragon labeled C-R C-W snapdragon, a blend, and the verdict incomplete dominance. Right: a patch of roan coat labeled C-R C-W calf, both traits separate, and the verdict codominance
Three panels in a row. Left: a purple pea flower labeled Pp pea, like one homozygote, and the verdict complete dominance. Middle: a pink snapdragon labeled C-R C-W snapdragon, a blend, and the verdict incomplete dominance. Right: a patch of roan coat labeled C-R C-W calf, both traits separate, and the verdict codominance
53

To tell incomplete dominance from codominance, look at one small part of the heterozygote.

54

A pink petal is pink all through: a blend. A roan coat is red hairs beside white hairs: both traits shown separately.

Left: a single pink petal, pink all through, labeled incomplete dominance, a blend. Right: a patch of roan coat with red hairs beside white hairs, labeled codominance, both shown separately
Left: a single pink petal, pink all through, labeled incomplete dominance, a blend. Right: a patch of roan coat with red hairs beside white hairs, labeled codominance, both shown separately
55

Here is a table of the three kinds of dominance, what the heterozygote looks like in each, and one example of each.

A table with three rows, one per kind of dominance, and three columns: the kind, what the heterozygote looks like, and one example. Complete dominance: like one homozygote; a Pp pea, purple. Incomplete dominance: a blend of the two; a C-R C-W snapdragon, pink. Codominance: both traits, side by side; a C-R C-W calf, red and white hairs
A table with three rows, one per kind of dominance, and three columns: the kind, what the heterozygote looks like, and one example. Complete dominance: like one homozygote; a Pp pea, purple. Incomplete dominance: a blend of the two; a C-R C-W snapdragon, pink. Codominance: both traits, side by side; a C-R C-W calf, red and white hairs
56

Neither Cᴿ nor Cᵂ hides the other, so the snapdragon and the calf both break Mendel’s rule.

57

What you are expected to know Classify a trait as complete dominance, incomplete dominance or codominance from what the heterozygote looks like.

58

Here again are the three snapdragons: red, pink and white, the pink one the offspring of the red one and the white one.

Three snapdragon flowers in a row labeled red, pink and white, the pink one in the middle
Three snapdragon flowers in a row labeled red, pink and white, the pink one in the middle
59

Neither Cᴿ nor Cᵂ hid the other in the pink flower. One red allele gave half the pigment, so the flower is a blend: pink.

60

A roan calf shows both alleles side by side instead. Mendel’s rule, one allele hides the other, holds only for complete dominance.

61Quick quiz: which kind of dominance? mixed practice

62
Check q13

In one mammal’s coat-color gene, an animal with one allele of each has red hairs mixed with white hairs, and every hair is either red or white.

Which kind of dominance does this gene show?

  1. A. Complete dominance
    Under complete dominance the heterozygote looks like one of the homozygotes, all red or all white.
    This heterozygote shows both colors.
  2. B. Incomplete dominance
    Under incomplete dominance the heterozygote is a blend, and every part of it is in between.
    Here each hair is fully red or fully white.
  3. C. ✓ Codominance

Why: The animal with one allele of each is the heterozygote.
Its coat has red hairs and white hairs, and every hair is one color or the other.
So both traits show separately in the heterozygote.
Two alleles whose traits both show separately are codominant.

63
Check q14

In one animal, a BB individual has a black coat, a bb individual has a white coat, and a Bb individual has a black coat.

Which kind of dominance does coat color show in this animal?

  1. A. ✓ Complete dominance
  2. B. Incomplete dominance
    Under incomplete dominance the Bb heterozygote would be a blend, a gray between black and white.
    The Bb individual is as black as a BB individual.
  3. C. Codominance
    Under codominance the Bb heterozygote would show black and white separately, black patches beside white patches.
    The Bb individual is all black.

Why: The Bb individual is the heterozygote.
It looks like the BB homozygote, black.
So one allele, B, hides the other.
One allele hiding the other in the heterozygote is complete dominance.

64
Check q15

In radishes, a long-rooted plant crossed with a round-rooted plant gives offspring with oval roots.

Which kind of dominance does root shape show in these radishes?

  1. A. Complete dominance
    Under complete dominance the offspring would be long-rooted or round-rooted, like one parent.
    Their roots are oval, in between.
  2. B. ✓ Incomplete dominance
  3. C. Codominance
    Under codominance each root would show both shapes separately, a long part beside a round part.
    An oval root is one shape in between the two.

Why: The offspring of a long-rooted plant and a round-rooted plant are the heterozygotes.
Their roots are oval, a shape between long and round.
So the heterozygote is a blend of the two homozygotes.
A heterozygote that is a blend shows incomplete dominance.

65
Check q16

In one bird, an individual with one black-feather allele and one white-feather allele has black feathers and white feathers, and no feather is gray.

Which kind of dominance does feather color show in this bird?

  1. A. Complete dominance
    Under complete dominance the heterozygote would be all black or all white, like one homozygote.
    This bird shows both colors.
  2. B. Incomplete dominance
    Under incomplete dominance every feather would be gray, in between.
    Here each feather is fully black or fully white.
  3. C. ✓ Codominance

Why: The bird with one allele of each is the heterozygote.
It has black feathers and white feathers, and no feather is in between.
So both traits show separately in the heterozygote.
Two alleles whose traits both show separately are codominant.

66
Check q17

In one cattle breed, a true-breeding black bull crossed with a true-breeding red cow gives calves that are all black.

Which kind of dominance does coat color show in these cattle?

  1. A. ✓ Complete dominance
  2. B. Incomplete dominance
    Under incomplete dominance the calves would be a blend, a shade between black and red.
    The calves are as black as the bull.
  3. C. Codominance
    Under codominance the calves would show black and red separately, black hairs beside red hairs.
    The calves are all black.

Why: The calves of a true-breeding black bull and a true-breeding red cow are the heterozygotes.
They are black, like the black parent.
So the black allele hides the red allele.
One allele hiding the other in the heterozygote is complete dominance.

67
Check q18

In one flowering plant, an individual with one red allele and one yellow allele has orange petals, orange all through.

Which kind of dominance does petal color show in this plant?

  1. A. Complete dominance
    Under complete dominance the heterozygote would be red or yellow, like one of the homozygotes.
    Its petals are orange, in between.
  2. B. ✓ Incomplete dominance
  3. C. Codominance
    Under codominance the petals would show red and yellow separately, red patches beside yellow patches.
    Orange all through is one color in between the two.

Why: The plant with one red allele and one yellow allele is the heterozygote.
Its petals are orange, a color between red and yellow, and orange all through.
So the heterozygote is a blend of the two homozygotes.
A heterozygote that is a blend shows incomplete dominance.

68Mixed practice mixed practice

69
Check q19

A plant with one red allele and one white allele of a flower-color gene has pink petals.

Which of the following is the pink plant?

  1. A. ✓ A heterozygote
  2. B. A homozygote
    A homozygote carries two copies of one allele.
    The pink plant carries two different alleles, red and white.

Why: The pink plant carries one red allele and one white allele.
Two different alleles of a gene make a heterozygote.

70
Check q20

In one plant, the offspring of a red-flowered line and a white-flowered line have white petals with red spots, and each spot is fully red.

Which kind of dominance does petal color show in this plant?

  1. A. Complete dominance
    Under complete dominance the petals would be all red or all white, like one parent.
    These petals show both colors.
  2. B. Incomplete dominance
    Under incomplete dominance the petals would be pink all through.
    Here each spot is fully red on a fully white petal.
  3. C. ✓ Codominance

Why: The offspring of the two lines are the heterozygotes.
Their petals show red and white separately: red spots on white.
Both traits showing separately in the heterozygote is codominance.

71
Check q21

Two pink CᴿCᵂ plants are crossed. A student predicts: “Every offspring will be pink, because each parent passes on a pink allele.”

Is the student correct?

  1. A. Yes: every offspring is pink
    There is no pink allele.
    Each parent passes on Cᴿ or Cᵂ, so some offspring receive two Cᴿ alleles and are red.
  2. B. ✓ No: red and white offspring appear as well as pink

Why: Each pink parent carries Cᴿ and Cᵂ, and passes one of them to each gamete.
An offspring with two Cᴿ alleles is red and one with two Cᵂ alleles is white.
So red and white offspring appear as well as pink.

72
Check q22

In snapdragons, red and white flower color show incomplete dominance, and the alleles are written Cᴿ and Cᵂ.

Why are the alleles written as one letter with superscripts instead of a capital C and a small c?

  1. A. ✓ Neither allele is dominant, so a capital and a small letter would mislead
  2. B. Cᴿ is dominant to Cᵂ, and the superscript R marks the dominant allele
    A CᴿCᵂ plant is pink, not red.
    So Cᴿ does not hide Cᵂ.
  3. C. Pink is a third allele, Cᴾ, and the superscripts leave room for it
    There is no pink allele.
    A pink plant carries one Cᴿ and one Cᵂ.

Why: A capital letter and a small letter say that one allele hides the other.
In snapdragons neither allele hides the other.
So the two alleles take one letter with a superscript each, which says they are two alleles of one gene and neither is dominant.

73
Check q23

In one flowering plant, a red CᴿCᴿ plant crossed with a yellow CʸCʸ plant gives orange CᴿCʸ offspring. Two of the orange plants are crossed.

Which petal colors can appear among their offspring?

  1. A. Orange only
    Each orange parent passes on Cᴿ or Cʸ.
    An offspring with two Cᴿ alleles is red and one with two Cʸ alleles is yellow.
  2. B. Red and yellow only
    An offspring that receives Cᴿ from one parent and Cʸ from the other is orange again.
    That is the most common outcome.
  3. C. ✓ Red, orange and yellow

Why: Each orange parent carries Cᴿ and Cʸ, and passes one of them to each gamete.
Two Cᴿ alleles give red, two Cʸ alleles give yellow, and one of each gives orange.
So all three colors can appear.

74
Check q24

A CᴿCᵂ calf is roan, and a CᴿCᵂ snapdragon is pink.

In which of the two is the heterozygote a blend of the two homozygotes?

  1. A. The calf
    The roan calf has red hairs and white hairs, each hair one color.
    Both traits show separately, so the calf is not a blend.
  2. B. ✓ The snapdragon
  3. C. Both
    Only the snapdragon is a blend.
    The calf shows red and white separately, hair by hair.

Why: The pink snapdragon is pink all through, between red and white: a blend.
The roan calf has red hairs beside white hairs: both traits shown separately.
So only the snapdragon is a blend.

75
Practice writing an answer

In Andalusian chickens, a true-breeding black-feathered bird crossed with a true-breeding white-feathered bird gives offspring that are all blue-gray, a color between black and white. A breeder then crosses two of the blue-gray birds. Among their chicks she finds black chicks and white chicks as well as blue-gray ones.

(a) Explain how the blue-gray offspring of the first cross demonstrate incomplete dominance. (1 pt)

Model answer Each blue-gray bird received a black allele from one parent and a white allele from the other.
So each blue-gray bird is the heterozygote.
Its feathers are blue-gray, between black and white, so neither allele hides the other.
A heterozygote that is a blend of the two homozygotes shows incomplete dominance.
Rubric
  • Award 1 point for: the blue-gray birds are heterozygotes whose color is a blend (between) of the two homozygotes, so neither allele hides the other: incomplete dominance.

Slip Calling it codominance. Codominance would give black feathers and white feathers on one bird; blue-gray all through is a blend.

(b) Explain how the black chicks and the white chicks of the second cross show that the black and white alleles were unchanged in their blue-gray parents. (1 pt)

Model answer A black chick needs two black alleles, one from each parent.
A white chick needs two white alleles, one from each parent.
So each blue-gray parent still carried a black allele and a white allele to pass on.
The two alleles were unchanged; blue-gray is what the two look like together.
Rubric
  • Award 1 point for: black and white chicks each need two of one allele, so each blue-gray parent still carried one black allele and one white allele, unchanged.

Slip Saying the alleles blended into a blue-gray allele. A blended allele could not give black or white chicks back.

Glossary

incomplete dominance
Neither allele hides the other, so the heterozygote is a blend of the two homozygotes: a CᴿCᵂ snapdragon is pink between red and white. The alleles are unchanged and come back in the next generation.
codominance
Both alleles’ traits show separately in the heterozygote: a CᴿCᵂ roan calf has red hairs and white hairs, and no hair is pink.

APBIO-U05-L21B Two pinks crossed

Topic 5.4 · Non-Mendelian Genetics · 28 steps

The filled Punnett square for two pink snapdragons: C-R and C-W along the top, C-R and C-W down the side, cells C-R C-R, C-R C-W, C-R C-W, C-W C-W; beside it two pink flowers over a tray drawn as a grid of small sprouts labeled 120 seedlings, and the question: what colors, in what numbers?
The filled Punnett square for two pink snapdragons: C-R and C-W along the top, C-R and C-W down the side, cells C-R C-R, C-R C-W, C-R C-W, C-W C-W; beside it two pink flowers over a tray drawn as a grid of small sprouts labeled 120 seedlings, and the question: what colors, in what numbers?

Here is the square for two pink snapdragons, CᴿCᵂ crossed with CᴿCᵂ, and beside it a tray of 120 seedlings from that cross, waiting to flower.

What colors will they show, and in what numbers?

Unit 5 · Heredity

1Read the square when pink shows

2

Video: Watch: Read the square when pink shows

The pink × pink square filling, then read twice: by genotype and by phenotype, with a red, a pink and a white flower appearing over the cells and the same 1 : 2 : 1 written both times; then the total of 120 shared out as 30, 60 and 30.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Ba.mp4

3

What does a cross of two heterozygotes give when the heterozygote shows?

4

The square for two pink snapdragons is the same as the square for Pp × Pp: one CᴿCᴿ, two CᴿCᵂ, one CᵂCᵂ. But now the pink flowers can be told apart from the red ones.

5

So the phenotypic ratio carries the same numbers as the genotypic ratio, 1 red : 2 pink : 1 white. The expected counts then follow from the total, as for any ratio.

6
Check q1

The filled square for Pp × Pp in pea plants has the cells PP, Pp, Pp and pp.

Which ratio does the square give when it is read by genotype?

  1. A. ✓ 1 PP : 2 Pp : 1 pp
  2. B. 3 purple : 1 white
    3 purple : 1 white is the square read by phenotype.
    Read by genotype, the four cells hold three genotypes: one PP, two Pp, one pp.

Why: Read by genotype, one cell is PP, two cells are Pp and one cell is pp.
So the genotypic ratio is 1 PP : 2 Pp : 1 pp.

7

A filled square is read twice: by genotype for the genotypic ratio, and by phenotype for the phenotypic ratio.

8

Read the square for two pink snapdragons by genotype first. One cell is CᴿCᴿ, two are CᴿCᵂ and one is CᵂCᵂ, so the genotypic ratio is 1 : 2 : 1.

The C-R C-W by C-R C-W square with its two heterozygote cells shaded; beside it the genotypic ratio: 1 C-R C-R, 2 C-R C-W (the shaded cells), 1 C-W C-W, so 1 : 2 : 1
The C-R C-W by C-R C-W square with its two heterozygote cells shaded; beside it the genotypic ratio: 1 C-R C-R, 2 C-R C-W (the shaded cells), 1 C-W C-W, so 1 : 2 : 1
9

Now read the same square by phenotype. CᴿCᴿ is red, CᴿCᵂ is pink and CᵂCᵂ is white, so the phenotypic ratio is 1 red : 2 pink : 1 white.

The same square, and beside it three flowers with their counts: a red flower labeled 1 red (C-R C-R), a pink flower labeled 2 pink (C-R C-W), a white flower labeled 1 white (C-W C-W); phenotypic ratio 1 : 2 : 1
The same square, and beside it three flowers with their counts: a red flower labeled 1 red (C-R C-R), a pink flower labeled 2 pink (C-R C-W), a white flower labeled 1 white (C-W C-W); phenotypic ratio 1 : 2 : 1
10

Here is a table of the two readings, one row each. The two ratios carry the same numbers, because a pink flower shows the heterozygote.

A table with two rows and three columns: the reading, its classes, its ratio. Read by genotype: C-R C-R, C-R C-W, C-W C-W; 1 : 2 : 1. Read by phenotype: red, pink, white; 1 : 2 : 1
A table with two rows and three columns: the reading, its classes, its ratio. Read by genotype: C-R C-R, C-R C-W, C-W C-W; 1 : 2 : 1. Read by phenotype: red, pink, white; 1 : 2 : 1
11

Under complete dominance the heterozygote looks like one homozygote. So two genotype classes merge into one phenotype class.

12

That merge is where 3 : 1 came from.

13
Check q2

A cross of pea plants is predicted to give a 3 purple : 1 white ratio, four shares in all, and 200 offspring are grown.

How is the expected count of white offspring found?

  1. A. The ratio’s digit for white, on its own
    The digits of a ratio are shares of the total, not counts of offspring.
  2. B. ✓ The total times white’s share of the ratio

Why: An expected count is the total times that class’s share of the ratio.
White’s share is 1 of the 4 shares, so the expected count of white offspring is 25% of the 200.

14

The expected count of a class is expected count=total×that class's share of the ratio. A 1 : 2 : 1 ratio has four shares in all: one, two and one.

15
Worked example

Two pink snapdragons, CᴿCᵂ and CᴿCᵂ, are crossed and 120 seedlings are grown to flowering. How many are expected to be red, pink and white?

Write down the values in the question:
total offspring = 120
phenotypic ratio = 1 red : 2 pink : 1 white, so 4 shares in all
red (CᴿCᴿ) share = 1 of 4
pink (CᴿCᵂ) share = 2 of 4
white (CᵂCᵂ) share = 1 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected red=120×14=30
expected pink=120×24=60
expected white=120×14=30
16

What you are expected to know Predict the offspring of two heterozygotes when the heterozygote shows: 1 : 2 : 1 by phenotype as well as by genotype, and the expected count of each class from the total and its share.

17
Check q3 numeric entry

In roan cattle, Cᴿ gives red hairs, Cᵂ gives white hairs, and a CᴿCᵂ calf is roan, with red hairs and white hairs together. On one farm, roan cows are mated with a roan bull, and 80 calves are born.

Calculate how many of the 80 calves are expected to be roan.

Part 1. Red, CᴿCᴿ, is 1 share of the 4 in the ratio. How many red calves are expected?

Answer: 20  (tolerance ±0)

Working
Multiply the total by the red share of the ratio:
expected red=80×14=20

Part 2. White, CᵂCᵂ, is also 1 share of the 4. How many white calves are expected?

Answer: 20  (tolerance ±0)

Working
Multiply the total by the white share of the ratio:
expected white=80×14=20

Answer: 40  (tolerance ±0)

Working
Write down the values in the question:
total calves = 80
phenotypic ratio = 1 red : 2 roan : 1 white, so 4 shares in all
roan (CᴿCᵂ) share = 2 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected roan=80×24=40
18
Check q4 numeric entry

In four o’clock plants, red (Cᴿ) and white (Cᵂ) show incomplete dominance, and a CᴿCᵂ plant is pink. Two pink plants are crossed and 200 offspring are grown to flowering.

Calculate how many of the 200 offspring are expected to be pink.

Answer: 100  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 200
phenotypic ratio = 1 red : 2 pink : 1 white, so 4 shares in all
pink (CᴿCᵂ) share = 2 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected pink=200×24=100
19
Check q5 numeric entry

In snapdragons, a pink plant, CᴿCᵂ, is crossed with a white plant, CᵂCᵂ, and 72 offspring are grown to flowering.

Calculate how many of the 72 offspring are expected to be white.

Answer: 36  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 72
square = CᴿCᵂ, CᴿCᵂ, CᵂCᵂ, CᵂCᵂ, so the phenotypic ratio is 1 pink : 1 white, 2 shares in all
white (CᵂCᵂ) share = 1 of 2
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected white=72×12=36
20
Check q6

A student predicts the offspring of two pink snapdragons, CᴿCᵂ × CᴿCᵂ, and writes: “phenotypic ratio 3 pink : 1 white”.

What is wrong with the student’s prediction?

  1. A. Nothing is wrong with it
    3 : 1 needs one allele to hide the other.
    Here the heterozygote has its own color, pink, so the two CᴿCᵂ cells do not merge with the CᴿCᴿ cell.
  2. B. ✓ It should be 1 red : 2 pink : 1 white
  3. C. It should be 1 pink : 1 white
    The cross also gives red offspring.
    One cell of the four is CᴿCᴿ, and CᴿCᴿ is red.
    So 25% of the offspring are red.
  4. D. It should be 3 red : 1 white
    Red is not dominant here.
    The heterozygote, CᴿCᵂ, is pink, not red.
    So the pink cells do not join the red class.

Why: The square for CᴿCᵂ × CᴿCᵂ gives one CᴿCᴿ cell, two CᴿCᵂ cells and one CᵂCᵂ cell.
CᴿCᴿ is red, CᴿCᵂ is pink and CᵂCᵂ is white.
The heterozygote looks like neither homozygote, so no two classes merge.
So the ratio is 1 red : 2 pink : 1 white.

21
Practice writing an answer

A student predicts the offspring of two pink snapdragons, CᴿCᵂ × CᴿCᵂ, and writes: “phenotypic ratio 3 pink : 1 white”. The prediction should be 1 red : 2 pink : 1 white.

(a) Explain why this cross gives three phenotypes rather than two. (1 pt)

Model answer The square for CᴿCᵂ × CᴿCᵂ gives one CᴿCᴿ cell, two CᴿCᵂ cells and one CᵂCᵂ cell.
CᴿCᴿ is red, CᴿCᵂ is pink and CᵂCᵂ is white.
The heterozygote, CᴿCᵂ, looks like neither homozygote, so no two genotype classes merge.
Therefore every genotype class is its own color class, and the phenotypic ratio is 1 red : 2 pink : 1 white.
Rubric
  • Award 1 point for: the heterozygote CᴿCᵂ has its own phenotype, pink (incomplete dominance), so each genotype shows a different phenotype: 1 : 2 : 1.
22

Here again are the square for two pink snapdragons, CᴿCᵂ × CᴿCᵂ, and the tray of 120 seedlings beside it.

The C-R C-W by C-R C-W square on the left; on the right a tray of small sprouts labeled 120 seedlings, and beneath it three flowers with their expected counts: about 30 red, about 60 pink, about 30 white
The C-R C-W by C-R C-W square on the left; on the right a tray of small sprouts labeled 120 seedlings, and beneath it three flowers with their expected counts: about 30 red, about 60 pink, about 30 white
23

The 120 seedlings should flower as about 30 red, 60 pink and 30 white. The colors sit in the same 1 : 2 : 1 as the genotypes, because pink shows the heterozygote.

24Mixed practice mixed practice

25
Check q7

A grower wants a bed of snapdragons that flower pink, every one of them, grown from the seeds of one cross between two of her plants.

Which cross gives only pink offspring?

  1. A. ✓ A red plant crossed with a white plant
  2. B. Two pink plants crossed
    Two CᴿCᵂ plants give red and white offspring as well as pink.
    One offspring in every four is CᴿCᴿ, red, and one is CᵂCᵂ, white.
  3. C. A pink plant crossed with a white plant
    The white parent gives every offspring a Cᵂ and the pink parent gives Cᴿ to half.
    So half of the offspring are CᴿCᵂ, pink, and half CᵂCᵂ, white.
  4. D. A pink plant crossed with a red plant
    The red parent gives every offspring a Cᴿ and the pink parent gives Cᵂ to half.
    So half of the offspring are CᴿCᵂ, pink, and half CᴿCᴿ, red.

Why: A red plant is CᴿCᴿ, so every gamete it makes carries Cᴿ.
A white plant is CᵂCᵂ, so every gamete it makes carries Cᵂ.
Every offspring receives one Cᴿ and one Cᵂ, so every offspring is CᴿCᵂ.
CᴿCᵂ is pink, so every offspring is pink.

26
Check q8

In snapdragons, a pink plant, CᴿCᵂ, is crossed with a red plant, CᴿCᴿ.

Which phenotypic ratio does this cross predict?

  1. A. 1 red : 2 pink : 1 white
    1 : 2 : 1 comes from two heterozygotes.
    Here the red parent gives only Cᴿ.
    So no offspring can be CᵂCᵂ, and no offspring is white.
  2. B. 3 red : 1 pink
    The red parent’s gametes all carry Cᴿ and the pink parent’s are half Cᴿ, half Cᵂ.
    So the four cells are CᴿCᴿ, CᴿCᵂ, CᴿCᴿ, CᴿCᵂ: two red, two pink.
  3. C. ✓ 1 red : 1 pink
  4. D. 1 pink : 1 white
    A white offspring needs a Cᵂ from each parent.
    The red parent, CᴿCᴿ, has no Cᵂ to give.
    So no offspring is white.

Why: The red parent, CᴿCᴿ, makes only Cᴿ gametes.
The pink parent, CᴿCᵂ, makes half Cᴿ gametes and half Cᵂ gametes.
So the square’s four cells are CᴿCᴿ, CᴿCᵂ, CᴿCᴿ and CᴿCᵂ.
Two cells are red and two are pink, so the phenotypic ratio is 1 red : 1 pink.

27
Practice writing an answer

A grower crosses a true-breeding red snapdragon with a true-breeding white one, and every F1 plant flowers pink. She then crosses two of the pink F1 plants and grows 244 F2 plants to flowering. She counts 63 red, 119 pink and 62 white.

(a) Make a claim about the kind of dominance flower color shows in these snapdragons, and support your claim with the F1 plants. (1 pt)

Model answer Claim: flower color shows incomplete dominance.
Evidence: every F1 plant received Cᴿ from the red parent and Cᵂ from the white parent, so every F1 plant is a heterozygote, CᴿCᵂ.
Every F1 plant is pink.
Pink is a blend of red and white, so neither allele hides the other in the heterozygote.
So the F1 plants support the claim.
Rubric
  • Award 1 point for: the claim (incomplete dominance) AND the support (the F1 heterozygotes are pink, a blend of the two parents’ colors, so neither allele hides the other).

Slip Calling it codominance. Codominance would show red and white separately in one flower; a pink flower is a blend.

(b) Calculate the number of pink plants expected among the 244 F2 plants. (1 pt)

Answer: 122  (tolerance ±0)

Model answer About 122 of the 244 F2 plants are expected to be pink.
Working
Write down the values in the question:
total F2 plants = 244
phenotypic ratio = 1 red : 2 pink : 1 white, so 4 shares in all
pink (CᴿCᵂ) share = 2 of 4
Write down the equation:
expected count=total×that class's share of the ratio
Substitute the values into the equation:
expected pink=244×24=122
Rubric
  • Award 1 point for: 122 pink plants.

(c) Explain why a count of 63 red, 119 pink and 62 white is consistent with a CᴿCᵂ × CᴿCᵂ cross. (1 pt)

Model answer The CᴿCᵂ × CᴿCᵂ square gives each F2 seed one chance in four of red, two in four of pink and one in four of white.
The expected counts are 61 red, 122 pink and 61 white.
The square gives a chance for each seed, not an exact count.
So a real batch strays a little from the expected counts.
63, 119 and 62 sit close to 61, 122 and 61, so they are consistent with the cross.
Rubric
  • Award 1 point for: the ratio gives a chance for each seed, so a real batch strays a little from the expected 61 : 122 : 61, and the counts are close to it.

Slip Saying the counts must be exactly 61, 122 and 61. The square predicts a chance for each seed, not an exact count for the batch.

(d) Predict the flower colors of the offspring of one pink F2 plant crossed with one red F2 plant, and their proportions. (1 pt)

Model answer The red F2 plant, CᴿCᴿ, gives every offspring a Cᴿ.
The pink F2 plant, CᴿCᵂ, gives half of them Cᴿ and half Cᵂ.
So the cells are CᴿCᴿ, CᴿCᵂ, CᴿCᴿ and CᴿCᵂ: two red, two pink, none white.
So the offspring are half red and half pink, with no white.The pink × red square, its four cells C-R C-R, C-R C-W, C-R C-R and C-R C-WCᴿCᴿCᴿCᵂCᴿCᴿCᴿCᵂCᴿCᵂCᴿCᴿpink parent, Cᴿ or Cᵂred parent, Cᴿ onlypink × red, half red and half pink
Rubric
  • Award 1 point for: half red and half pink (1 red : 1 pink), with no white offspring.

Slip Predicting 1 : 2 : 1 again. That ratio needs two heterozygotes; a red parent passes on Cᴿ only, so no offspring can be white.

APBIO-U05-L21C Three alleles, two per person

Topic 5.4 · Non-Mendelian Genetics · 43 steps

A family drawn as a small pedigree: a circle for the mother labeled type A and a square for the father labeled type B, joined by a line; below them a daughter labeled type AB and a son labeled type O
A family drawn as a small pedigree: a circle for the mother labeled type A and a square for the father labeled type B, joined by a line; below them a daughter labeled type AB and a son labeled type O

Here is a family and their blood types: the mother is type A, the father type B, one child is type AB and the other is type O.

The type O child shows neither parent’s type. How many alleles is this gene carrying, and how can one gene give four blood types?

Unit 5 · Heredity

1Three alleles, two in each person

2

Video: Watch: Three alleles, two in each person

Four red blood cells gaining their markers as the three alleles are named; then three people’s homologous pairs, each carrying two of the three alleles.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Ca.mp4

3

How can one gene have more than two alleles, and why do three alleles give four blood types?

4

The ABO gene puts a marker on the surface of red blood cells. It has three alleles among people, and any one person carries two of the three.

5

Two of the alleles each make a marker, and both show together. The third makes no marker, and shows only when a person carries two of it.

6

Type A cells carry the A marker, type B cells the B marker, type AB cells both, and type O cells neither.

Four red blood cells in a row: a type A cell with triangle markers on its surface, a type B cell with square markers, a type AB cell with both kinds, and a type O cell with none
Four red blood cells in a row: a type A cell with triangle markers on its surface, a type B cell with square markers, a type AB cell with both kinds, and a type O cell with none
7

Here is a table of the gene’s three alleles and what each one makes: Iᴬ makes the A marker, Iᴮ makes the B marker, and i makes no marker at all.

A table with three rows: the allele I-A makes the A marker; the allele I-B makes the B marker; the allele i makes no marker
A table with three rows: the allele I-A makes the A marker; the allele I-B makes the B marker; the allele i makes no marker
8
Check q1

A pea plant carries the flower-color gene at the same position on the two homologs of one pair.

How many alleles of that gene does the plant carry?

  1. A. ✓ Two
  2. B. One
    The gene sits on both homologs of the pair, so the plant carries two alleles.
  3. C. Four
    One position on each of two homologs gives two alleles, not four.

Why: The gene has one position on each homolog of the pair.
The plant has two homologs.
So the plant carries two alleles of the gene.

9
Check q2

The ABO gene has three alleles among people: Iᴬ, Iᴮ and i.

How many alleles of the ABO gene does one person carry?

  1. A. Three, one of each
    A person has two homologs of that chromosome, so two positions for the gene.
  2. B. ✓ Two, one on each homolog

Why: The two alleles of a gene sit at the same position on the two homologs.
So any one person carries two of the three.

10

So three alleles exist among people, but any one person carries two of them, one on each homolog.

Three homologous pairs side by side, each two rods with centromere dots, one darker and one lighter: the first pair carries I-A and I-A, the second I-A and i, the third I-B and i; caption: three alleles exist, and each person carries two
Three homologous pairs side by side, each two rods with centromere dots, one darker and one lighter: the first pair carries I-A and I-A, the second I-A and i, the third I-B and i; caption: three alleles exist, and each person carries two
11

When people are sorted by these markers into the types A, B, AB and O, we call the system the . A and B name the markers; O means neither.

12

The four blood types are four phenotypes, not four alleles. Three alleles exist, and the two a person carries set which of the four types shows.

13

What you are expected to know Name the three ABO alleles and the marker each one makes, and state that any one person carries two of the three.

14
Check q3

A student says: “The ABO gene has three alleles, so every person carries all three of them.”

Is the student correct?

  1. A. Yes: a person carries Iᴬ, Iᴮ and i
    A gene has one position on each of the two homologs.
    Two positions hold two alleles, not three.
  2. B. ✓ No: a person carries two of the three, one on each homolog

Why: The ABO gene has one position on each of the two homologs.
So a person carries two alleles of it.
Three alleles exist among people, but each person carries two of the three.

15Quick quiz: ABO blood groups mixed practice

16
Check q4

People are sorted by the markers on their red blood cells.

What are the ABO blood groups?

  1. A. Three genes on three chromosomes, one gene for each marker
    One gene sets the marker.
    Its three alleles make the A marker, the B marker or no marker.
  2. B. Four alleles of one gene, one allele for each of the four blood types
    Type A, B, AB and O are phenotypes.
    The gene has three alleles, and two of them together set the type.
  3. C. ✓ Blood types A, B, AB and O, set by one gene with three alleles

Why: One gene puts the marker on red blood cells, and it has three alleles.
The two alleles a person carries set one of four types: A, B, AB or O.
Those four types are the ABO blood groups.

17
Practice writing an answer

The ABO gene puts a marker on the surface of red blood cells.

(a) State the three alleles of the ABO gene and what each one makes. (1 pt)

Model answer Iᴬ makes the A marker.
Iᴮ makes the B marker.
i makes no marker.
Rubric
  • Award 1 point for: Iᴬ (A marker), Iᴮ (B marker), i (no marker).

(b) Explain why one person carries only two of the gene’s three alleles. (1 pt)

Model answer The ABO gene sits at one position on each of the two homologs of a pair.
A person has two homologs, so two positions.
So a person carries two alleles, one on each homolog.
Rubric
  • Award 1 point for: the gene has one position on each of the two homologs, so a person carries two alleles.
18
Check q5

A person’s red blood cells carry the B marker.

Which allele makes the B marker?

  1. A. Iᴬ
    Iᴬ makes the A marker.
  2. B. ✓ Iᴮ
  3. C. i
    The i allele makes no marker.

Why: Iᴮ makes the B marker.

19
Check q6

A person’s red blood cells carry no marker at all.

Which allele does that person carry two copies of?

  1. A. Iᴬ
    Iᴬ makes the A marker, so a person with an Iᴬ has A markers.
  2. B. Iᴮ
    Iᴮ makes the B marker, so a person with an Iᴮ has B markers.
  3. C. ✓ i

Why: The i allele makes no marker.
Cells with no marker at all come from two i alleles.

20
Check q7

Suppose a gene has four alleles among the people of a population.

How many alleles of that gene does one person carry?

  1. A. ✓ Two
  2. B. Four
    The gene has one position on each of the two homologs.
    Two positions hold two alleles, whatever the number in the population.

Why: A person has two homologs of each pair, with one position for the gene on each.
So a person carries two alleles, however many exist in the population.

21Six genotypes, four blood types

22

Video: Watch: Six genotypes, four blood types

The six pairs of alleles written out one by one, each red cell gaining its markers: I-A I-B carrying both, i i carrying none, and the six pairs sorting into four blood types in the table.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Cb.mp4

23
Check q8

In one gene, both alleles’ traits show separately in the heterozygote.

Which kind of dominance is this?

  1. A. Complete dominance
    Under complete dominance one allele hides the other, so one trait shows.
  2. B. Incomplete dominance
    Under incomplete dominance the heterozygote is a blend, so neither trait shows whole.
  3. C. ✓ Codominance

Why: Both alleles’ traits showing separately in the heterozygote is codominance.

24

Iᴬ and Iᴮ are codominant. So an IᴬIᴮ person’s red cells carry the A marker and the B marker, each whole and separate: type AB.

One red blood cell carrying triangle markers and square markers on its surface, labeled I-A I-B, type AB: the A marker and the B marker, both whole
One red blood cell carrying triangle markers and square markers on its surface, labeled I-A I-B, type AB: the A marker and the B marker, both whole
25

The i allele makes no marker. So i is recessive to both of the others: it shows only when a person carries two of it.

26

One Iᴬ is enough for the A marker: Iᴬi is type A, like IᴬIᴬ. One Iᴮ is enough for the B marker: Iᴮi is type B, like IᴮIᴮ.

27

Only ii is type O, because neither i makes a marker.

28

Here is a table of the six genotypes and the blood type each one gives.

A table of the six genotypes and the four blood types: I-A I-A and I-A i give type A; I-B I-B and I-B i give type B; I-A I-B gives type AB; i i gives type O
A table of the six genotypes and the four blood types: I-A I-A and I-A i give type A; I-B I-B and I-B i give type B; I-A I-B gives type AB; i i gives type O
29

Three alleles make six pairs. Two of the pairs look like type A and two look like type B, so six genotypes give four blood types.

30

What you are expected to know Match the six ABO genotypes to the four blood types, and explain why three alleles give four types.

31
Check q9

A person’s ABO genotype is IᴬIᴮ.

Which markers do that person’s red blood cells carry?

  1. A. The A marker only
    Iᴮ makes the B marker too.
    Iᴬ and Iᴮ are codominant, so both markers show.
  2. B. The B marker only
    Iᴬ makes the A marker too.
    Iᴬ and Iᴮ are codominant, so both markers show.
  3. C. ✓ The A marker and the B marker
  4. D. No marker
    Both alleles make a marker.
    Only ii cells carry no marker.

Why: Iᴬ makes the A marker and Iᴮ makes the B marker.
Iᴬ and Iᴮ are codominant, so both markers show.
The red cells carry the A marker and the B marker.

32
Practice writing an answer

The ABO gene has three alleles: Iᴬ makes the A marker, Iᴮ makes the B marker, and i makes no marker. Iᴬ and Iᴮ are codominant, and i is recessive to both.

(a) Explain how the six genotypes of this gene give only four blood types. (1 pt)

Model answer Three alleles make six pairs: IᴬIᴬ, Iᴬi, IᴮIᴮ, Iᴮi, IᴬIᴮ and ii.
One Iᴬ is enough for the A marker, so IᴬIᴬ and Iᴬi both show type A.
One Iᴮ is enough for the B marker, so IᴮIᴮ and Iᴮi both show type B.
IᴬIᴮ shows both markers, type AB, and ii shows none, type O.
So six genotypes give four types.
Rubric
  • Award 1 point for: i is recessive, so Iᴬi looks like IᴬIᴬ (type A) and Iᴮi looks like IᴮIᴮ (type B); with IᴬIᴮ (AB) and ii (O) that is four types from six genotypes.
33

Here again is the family: the mother type A, the father type B, one child type AB and the other type O.

A family drawn as a pedigree: a circle for the mother labeled type A and a square for the father labeled type B, joined by a line; below them a daughter labeled type AB and a son labeled type O
A family drawn as a pedigree: a circle for the mother labeled type A and a square for the father labeled type B, joined by a line; below them a daughter labeled type AB and a son labeled type O
34

The ABO gene has three alleles among people: Iᴬ, Iᴮ and i. Each of the four people carries two of them, one on each homolog.

35

The mother’s two alleles give type A, the father’s give type B, one child’s give type AB and the other child’s give type O. Three alleles, two per person, are enough for all four blood types in one family.

36Quick quiz: which blood type? mixed practice

37
Check q10

A person’s ABO genotype is IᴬIᴬ.

Which blood type does the person have?

  1. A. ✓ Type A
  2. B. Type B
    The B marker needs an Iᴮ allele, and IᴬIᴬ carries none.
    Both alleles make the A marker.
  3. C. Type AB
    Type AB needs an Iᴬ and an Iᴮ.
    IᴬIᴬ carries no Iᴮ, so the cells carry the A marker only.
  4. D. Type O
    Type O needs two i alleles, which make no marker.
    Iᴬ makes the A marker, so the cells carry it.

Why: Both alleles are Iᴬ.
Iᴬ makes the A marker.
So the red cells carry the A marker and no other.
That is type A.

38
Check q11

A person’s ABO genotype is Iᴮi.

Which blood type does the person have?

  1. A. Type A
    The A marker needs an Iᴬ allele, and Iᴮi carries none.
  2. B. ✓ Type B
  3. C. Type AB
    Type AB needs an Iᴬ as well as an Iᴮ.
    The i allele makes no marker.
  4. D. Type O
    Type O needs two i alleles.
    This person has one Iᴮ, and one Iᴮ is enough for the B marker.

Why: Iᴮ makes the B marker.
The i allele makes no marker.
So the red cells carry the B marker only.
That is type B.

39
Check q12

A person’s ABO genotype is IᴬIᴮ.

Which blood type does the person have?

  1. A. Type A
    The person also carries Iᴮ, which makes the B marker.
    Iᴬ and Iᴮ are codominant, so both markers show.
  2. B. Type B
    The person also carries Iᴬ, which makes the A marker.
    Iᴬ and Iᴮ are codominant, so both markers show.
  3. C. ✓ Type AB
  4. D. Type O
    Type O needs two i alleles, which make no marker.
    Here both alleles make a marker.

Why: Iᴬ makes the A marker and Iᴮ makes the B marker.
Iᴬ and Iᴮ are codominant, so both markers show.
The red cells carry the A marker and the B marker.
That is type AB.

40
Check q13

A person’s ABO genotype is ii.

Which blood type does the person have?

  1. A. Type A
    The A marker needs an Iᴬ allele, and ii carries none.
  2. B. Type B
    The B marker needs an Iᴮ allele, and ii carries none.
  3. C. Type AB
    Type AB needs an Iᴬ and an Iᴮ.
    Neither i allele makes a marker.
  4. D. ✓ Type O

Why: Both alleles are i.
The i allele makes no marker.
So the red cells carry neither marker.
That is type O.

41
Check q14

A woman is type B.

Which genotypes could she have for the ABO gene?

  1. A. IᴮIᴮ only
    One Iᴮ is enough for the B marker, and i makes no marker.
    So an Iᴮi person is also type B.
  2. B. Iᴮi only
    A person with two Iᴮ alleles also carries the B marker and no A marker.
    Nothing in type B blood rules IᴮIᴮ out.
  3. C. IᴬIᴮ or Iᴮi
    An IᴬIᴮ person’s cells carry the A marker as well.
    That person is type AB, not type B.
  4. D. ✓ IᴮIᴮ or Iᴮi

Why: Her cells carry the B marker, so she carries at least one Iᴮ.
Her cells carry no A marker, so she carries no Iᴬ.
Her second allele is therefore Iᴮ or i.
So a type B person is IᴮIᴮ or Iᴮi.

42
Check q15

A man is type AB.

Which genotype does he have for the ABO gene?

  1. A. ✓ IᴬIᴮ
  2. B. Iᴬi
    The i allele makes no marker.
    An Iᴬi person carries the A marker only, and is type A.
  3. C. IᴮIᴮ
    IᴮIᴮ makes the B marker only.
    A type AB person carries the A marker as well, so he carries an Iᴬ.

Why: His cells carry the A marker, so he carries Iᴬ.
His cells carry the B marker, so he carries Iᴮ.
A person carries two alleles, one on each homolog.
So the man is IᴬIᴮ.

Glossary

ABO blood groups
The four blood types A, B, AB and O, set by one gene with three alleles: Iᴬ (the A marker), Iᴮ (the B marker) and i (no marker). Iᴬ and Iᴮ are codominant, i is recessive to both, and each person carries two of the three alleles.

APBIO-U05-L21D Whose child is this?

Topic 5.4 · Non-Mendelian Genetics · 53 steps

Two small pedigrees side by side. Left: a mother labeled I-A i and a father labeled I-B i, joined by a line, with two children below each marked with a question mark; caption: which blood types can their children have? Right: a mother and a father each marked with a question mark, with two children below labeled type AB and type O; caption: which alleles must these parents carry?
Two small pedigrees side by side. Left: a mother labeled I-A i and a father labeled I-B i, joined by a line, with two children below each marked with a question mark; caption: which blood types can their children have? Right: a mother and a father each marked with a question mark, with two children below labeled type AB and type O; caption: which alleles must these parents carry?

Now consider the same couple: the mother is Iᴬi and the father is Iᴮi. They want to know which blood types their children could have.

A second couple already have two children, one type AB and one type O, and they want to know what alleles they themselves must carry. How does a square give the children’s blood types, and how does the same square work backwards from the children to the parents’ alleles?

Unit 5 · Heredity

1Predict the children

2

Video: Watch: Predict the children

The mother’s two alleles along the top edge and the father’s down the side, the four cells filling one at a time to give types AB, B, A and O, and one cell of the four shaded for the probability of a type AB child.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Da.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Da.mp4

3

How do you predict children’s blood types from their parents’ alleles?

4

A Punnett square with the mother’s two alleles along the top and the father’s two down the side gives one cell for each pairing. Each cell is one blood type a child can have, and each cell is equally likely.

5
Check q1

A student builds a Punnett square for a breeding cross.

What sits on the edges of the square?

  1. A. ✓ The parents’ gametes, one allele each
  2. B. The parents’ genotypes, two alleles each
    A parent’s genotype never goes on an edge.
    Each edge label is one gamete, so one allele.

Why: The edges of a Punnett square carry the parents’ gametes.
A gamete carries one allele of the gene.
So each edge label is one allele.

6
Check q2

A person’s ABO genotype is Iᴬi.

Which blood type does the person have?

  1. A. Type AB
    Type AB needs an Iᴮ as well as an Iᴬ.
    The i allele makes no marker.
  2. B. ✓ Type A
  3. C. Type O
    One Iᴬ is enough for the A marker.
    Only ii is type O.

Why: Iᴬ makes the A marker and i makes no marker.
So the red cells carry the A marker only: type A.

7

Now consider an Iᴬi mother and an Iᴮi father. What blood types can their children have?

8

Her eggs carry Iᴬ or i, and his sperm carry Iᴮ or i. So the square has her two gametes along the top and his two down the side.

An empty two-by-two grid: the mother’s gametes I-A and i along the top edge, the father’s gametes I-B and i down the left edge; the four cells are empty
An empty two-by-two grid: the mother’s gametes I-A and i along the top edge, the father’s gametes I-B and i down the left edge; the four cells are empty
9

Where the Iᴬ egg fuses with the Iᴮ sperm, the cell reads IᴬIᴮ: a type AB child.

The same grid with one cell filled: where the I-A egg fuses with the I-B sperm, the cell reads I-A I-B, type AB
The same grid with one cell filled: where the I-A egg fuses with the I-B sperm, the cell reads I-A I-B, type AB
10

Where the i egg fuses with the Iᴮ sperm, the cell reads Iᴮi: a type B child.

The grid with two cells filled: I-A I-B top left, and I-B i top right where the i egg fuses with the I-B sperm
The grid with two cells filled: I-A I-B top left, and I-B i top right where the i egg fuses with the I-B sperm
11

Fill the other two cells: the Iᴬ egg with the i sperm gives Iᴬi, type A, and the i egg with the i sperm gives ii, type O. One couple can have children of all four types.

The grid with all four cells filled: I-A I-B (type AB), I-B i (type B), I-A i (type A), i i (type O)
The grid with all four cells filled: I-A I-B (type AB), I-B i (type B), I-A i (type A), i i (type O)
12

Half of the eggs carry each of the mother’s alleles, and half of the sperm carry each of the father’s. So the four cells are four equally likely pairings.

13

The probability that a child has one blood type is that type’s share of the four cells. One cell of the four is type AB.

The filled grid for I-A i by I-B i with one cell shaded, the I-A I-B cell; caption: one cell of four is type AB
The filled grid for I-A i by I-B i with one cell shaded, the I-A I-B cell; caption: one cell of four is type AB
14
Worked example

An Iᴬi mother and an Iᴮi father have a child. What is the probability that the child is type AB?

Write down the values in the question:
cells in the square = 4, all equally likely
cells that are type AB (IᴬIᴮ) = 1
Write down the equation:
probability of a type=cells of that typecells in all
Substitute the values into the equation:
probability of type AB=14
15

What you are expected to know Predict the blood types a couple’s children can have, and the probability of each, from a Punnett square of the parents’ ABO genotypes.

16
Check q3

Suppose a mother is type AB, IᴬIᴮ, and a father is type A, Iᴬi. The square below has the mother along the top and the father down the side, with the edges still blank.

An empty two-by-two grid labeled for a type AB mother along the top and a type A father down the left edge; the edge labels and the four cells are all blank
An empty two-by-two grid labeled for a type AB mother along the top and a type A father down the left edge; the edge labels and the four cells are all blank

Which alleles go along the top edge for the mother?

  1. A. Iᴬ only
    Half of her eggs carry Iᴮ.
  2. B. ✓ Iᴬ and Iᴮ
  3. C. IᴬIᴮ
    The edges carry gametes, one allele each.
    Her eggs carry Iᴬ or Iᴮ, never both.

Why: The mother carries Iᴬ on one homolog and Iᴮ on the other.
The two homologs part at anaphase I, so half of her eggs carry Iᴬ and half carry Iᴮ.
So Iᴬ and Iᴮ go along the top edge.

17
Check q4

The mother is IᴬIᴮ and the father is Iᴬi.

Which alleles go down the side for the father?

  1. A. ✓ Iᴬ and i
  2. B. Iᴬ only
    Half of his sperm carry i.
    The i allele makes no marker, but it is still passed on.

Why: The father carries Iᴬ on one homolog and i on the other.
Half of his sperm carry Iᴬ and half carry i.
So Iᴬ and i go down the side.

18
Check q5

The mother is IᴬIᴮ and the father is Iᴬi. In the square, one cell sits under the Iᴮ egg and beside the i sperm.

Which genotype and blood type does that cell hold?

  1. A. Iᴬi, type A
    The egg above this cell carries Iᴮ, not Iᴬ.
  2. B. ii, type O
    The mother has no i to give.
    The egg above this cell carries Iᴮ.
  3. C. ✓ Iᴮi, type B

Why: A cell holds the allele above it together with the allele beside it.
Above this cell is Iᴮ, and beside it is i.
So the cell reads Iᴮi, and one Iᴮ is enough for the B marker: type B.

19
Check q6

The mother is IᴬIᴮ and the father is Iᴬi.

Which blood types are possible for their children?

  1. A. Type AB only
    Only one cell of the four is IᴬIᴮ.
    The other three cells are type A or type B.
  2. B. ✓ Type A, B or AB
  3. C. Type A, B, AB or O
    A type O child needs an i from each parent.
    The mother has no i to give, so no cell is ii.

Why: IᴬIᴬ and Iᴬi are type A, IᴬIᴮ is type AB and Iᴮi is type B.
No cell is ii, because the mother has no i to give.
So the children can be type A, B or AB.

20
Check q7

A father is type AB and a mother is type O.

Which blood types are possible for their children?

  1. A. Type AB or type O
    The mother, ii, passes an i to every child and the father passes Iᴬ or Iᴮ.
    So every child is Iᴬi or Iᴮi, and neither IᴬIᴮ nor ii can form.
  2. B. ✓ Type A or type B
  3. C. Type AB only
    A type AB child needs an Iᴮ or an Iᴬ from the mother.
    An ii mother has neither.
  4. D. Type A, B, AB or O
    The mother can pass only i.
    So the square has two kinds of cell, Iᴬi and Iᴮi, not four types.

Why: The father is IᴬIᴮ, so each of his gametes carries Iᴬ or Iᴮ.
The mother is ii, so every gamete carries i.
So each child is Iᴬi or Iᴮi.
Iᴬi is type A and Iᴮi is type B.

21
Check q8

A mother is Iᴮi and a father is Iᴮi.

Which blood types are possible for their children?

  1. A. ✓ Type B or type O
  2. B. Type B only
    One cell of the four is ii.
    An ii child is type O.
  3. C. Type A, B, AB or O
    Neither parent carries Iᴬ.
    So no child can be type A or type AB.

Why: Each parent’s gametes carry Iᴮ or i.
The four cells are IᴮIᴮ, Iᴮi, Iᴮi and ii.
IᴮIᴮ and Iᴮi are type B, and ii is type O.
So the children can be type B or type O.

22Work back to the parents

23

Video: Watch: Work back to the parents

A type O child’s two i alleles traced back, one to each parent; then a type AB parent’s two gametes, I-A and I-B, with no i among them, and the empty cell for a type O child.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Db.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L21Db.mp4

24
Check q9

In an orchid, purple flowers (P) are dominant to white (p). Two white-flowered orchids, both pp, are crossed.

Can the single-gene model give a purple-flowered offspring from this cross?

  1. A. Yes
    Two pp parents can give only p alleles.
    The square for pp × pp has no cell with a P.
  2. B. ✓ No

Why: Both parents are pp, so every gamete carries p.
Every cell of the square is pp, white.
The model gives a purple offspring a probability of zero.

25

Now work backward, from a child to the parents. A type O child is ii, and received one i from each parent.

26

So each parent of a type O child carried an i.

27

A type A parent is IᴬIᴬ or Iᴬi. IᴬIᴬ has no i to give, so a type A parent of a type O child must be Iᴬi.

A type O child, i i, drawn as a homologous pair at the bottom; above it the two genotypes a type A parent could have, I-A I-A and I-A i; only the I-A i pair has an arrow down to the child, labeled: an i to pass
A type O child, i i, drawn as a homologous pair at the bottom; above it the two genotypes a type A parent could have, I-A I-A and I-A i; only the I-A i pair has an arrow down to the child, labeled: an i to pass
28

A breeding cross cannot give an offspring for which its square has no cell.

29

A type AB parent, IᴬIᴮ, has no i to pass. So none of that parent’s children can be type O.

A type AB parent drawn as a homologous pair carrying I-A and I-B; beside it the two gametes that parent can make, one carrying I-A and one carrying I-B, and an empty space labeled: no gamete carries i, so no type O child
A type AB parent drawn as a homologous pair carrying I-A and I-B; beside it the two gametes that parent can make, one carrying I-A and one carrying I-B, and an empty space labeled: no gamete carries i, so no type O child
30

What you are expected to know Work back from a child’s blood type to the alleles each parent must carry, and say which parent type a child rules out.

31
Check q10

A mother is type A. One of her children is type O.

Which genotype does the mother have for the ABO gene?

  1. A. ✓ Iᴬi
  2. B. IᴬIᴬ
    A type O child is ii and received an i from each parent.
    An IᴬIᴬ mother has no i to pass.
  3. C. ii
    An ii person is type O.
    The mother is type A, so she carries an Iᴬ.
  4. D. IᴬIᴮ
    An IᴬIᴮ person is type AB, not type A.
    An IᴬIᴮ person also has no i to pass to a type O child.

Why: The mother is type A, so she carries at least one Iᴬ.
Her child is type O, so the child is ii.
A type O child received one i from each parent, so the mother passed an i.
Therefore the mother’s two alleles are Iᴬ and i: she is Iᴬi.

32
Practice writing an answer

A father is type B. One of his children is type O.

(a) Explain how the child’s blood type shows the father’s genotype. (1 pt)

Model answer The child is type O, so the child is ii.
A type O child received one i from each parent, so the father passed an i.
The father is type B, so he carries at least one Iᴮ.
Therefore the father’s two alleles are Iᴮ and i: he is Iᴮi.
Rubric
  • Award 1 point for: the type O child is ii and received an i from each parent, so the type B father carries an i as well as an Iᴮ: Iᴮi.
33
Check q11

A student says: “A type AB mother could have a type O child, as long as the father is type O.”

Is the student correct?

  1. A. Yes: the father’s i alleles make the child type O
    A type O child needs an i from each parent.
    An IᴬIᴮ mother has no i to give, whatever the father carries.
  2. B. ✓ No: the mother has no i to give, so none of her children is type O

Why: A type O child is ii and needs an i from each parent.
A type AB mother is IᴬIᴮ, and neither of her alleles is i.
So the square for her cross has no ii cell, whatever the father carries.

34
Check q12

A couple has two children. One child is type AB and the other is type O.

Which blood types could the two parents have?

  1. A. A type A parent and a type O parent
    A type AB child needs an Iᴮ from one parent.
    A type O parent (ii) has none, and a type A parent (IᴬIᴬ or Iᴬi) has none either.
  2. B. Two type O parents
    Two ii parents pass only i.
    So every child is type O and none can be type AB.
  3. C. ✓ A type A parent and a type B parent
  4. D. A type AB parent and a type A parent
    The type AB parent has no i.
    So a type O child, ii, is impossible for this couple.

Why: The type AB child is IᴬIᴮ: one parent passed Iᴬ, the other Iᴮ.
The type O child is ii: each parent also passed an i.
So one parent carries Iᴬ and i, the other Iᴮ and i.
Iᴬi is type A and Iᴮi is type B.

35

Here again are the two couples. The first couple, the mother Iᴬi and the father Iᴮi, can have children of all four types: AB, A, B or O, one cell each.

Two pedigrees side by side. Left: a mother labeled I-A i and a father labeled I-B i with two children labeled AB, A, B or O. Right: a mother labeled I-A i and a father labeled I-B i with two children labeled type AB and type O
Two pedigrees side by side. Left: a mother labeled I-A i and a father labeled I-B i with two children labeled AB, A, B or O. Right: a mother labeled I-A i and a father labeled I-B i with two children labeled type AB and type O
36

The second couple’s children gave their parents away. The type AB child needed an Iᴬ from one parent and an Iᴮ from the other, and the type O child needed an i from each.

37

So the second couple are Iᴬi and Iᴮi as well: a type A parent and a type B parent.

38Quick quiz: can this couple have a child of this type? mixed practice

39
Check q13

A mother is Iᴬi and a father is Iᴮi.

Can they have a type O child?

  1. A. ✓ Yes
  2. B. No
    Each parent carries an i to give.
    The i egg with the i sperm gives ii, type O.

Why: Each parent carries an i.
An i egg fused with an i sperm gives ii, type O.
So the square has a cell for a type O child.

40
Check q14

A mother is IᴬIᴮ and a father is ii.

Can they have a type O child?

  1. A. Yes
    A type O child needs an i from each parent.
    The IᴬIᴮ mother has no i to give.
  2. B. ✓ No

Why: A type O child is ii and needs an i from each parent.
The mother is IᴬIᴮ and has no i.
So the square has no cell for a type O child.

41
Check q15

A mother is ii and a father is Iᴮi.

Can they have a type A child?

  1. A. Yes
    A type A child needs an Iᴬ from one parent.
    Neither ii nor Iᴮi carries an Iᴬ.
  2. B. ✓ No

Why: A type A child needs an Iᴬ.
Neither parent carries an Iᴬ.
So the square has no cell for a type A child.

42
Check q16

A mother is IᴬIᴬ and a father is Iᴮi.

Can they have a type AB child?

  1. A. ✓ Yes
  2. B. No
    Every egg carries Iᴬ, and half of the sperm carry Iᴮ.
    An Iᴬ egg with an Iᴮ sperm gives IᴬIᴮ, type AB.

Why: Every egg carries Iᴬ and half of the sperm carry Iᴮ.
An Iᴬ egg fused with an Iᴮ sperm gives IᴬIᴮ, type AB.
So the square has a cell for a type AB child.

43
Check q17

A mother is Iᴬi and a father is Iᴬi.

Can they have a type O child?

  1. A. ✓ Yes
  2. B. No
    Each parent carries an i to give.
    The i egg with the i sperm gives ii, type O.

Why: Each parent carries an i.
An i egg fused with an i sperm gives ii, type O.
So one cell of the four is a type O child.

44
Check q18

A mother is IᴬIᴮ and a father is IᴬIᴮ.

Can they have a type O child?

  1. A. Yes
    A type O child needs an i from each parent.
    Neither IᴬIᴮ parent carries an i.
  2. B. ✓ No

Why: A type O child is ii.
Neither parent carries an i.
So the square has no cell for a type O child.

45Mixed practice mixed practice

46
Check q19

A person’s ABO genotype is Iᴬi.

How many different alleles of the ABO gene exist among people?

  1. A. Two
    Any one person carries two, but three exist among people: Iᴬ, Iᴮ and i.
  2. B. ✓ Three
  3. C. Four
    The four blood types are phenotypes.
    The gene has three alleles.

Why: The ABO gene has three alleles among people: Iᴬ, Iᴮ and i.
Any one person carries two of the three.

47
Check q20

A mother is type O and a father is type B.

Which of the following blood types could their child have?

  1. A. Type AB
    A type AB child needs an Iᴬ and an Iᴮ.
    The ii mother has neither, and the father has no Iᴬ.
  2. B. Type A
    A type A child needs an Iᴬ.
    Neither an ii mother nor a type B father carries an Iᴬ.
  3. C. ✓ Type B

Why: The mother is ii and passes an i to every child.
The father carries an Iᴮ and can pass it on.
An i egg fused with an Iᴮ sperm gives Iᴮi, type B.

48
Check q21

A man is type A.

Which genotypes could he have for the ABO gene?

  1. A. IᴬIᴬ only
    One Iᴬ is enough for the A marker, and i makes no marker.
    So an Iᴬi man is also type A.
  2. B. ✓ IᴬIᴬ or Iᴬi
  3. C. IᴬIᴮ or Iᴬi
    An IᴬIᴮ man’s cells carry the B marker as well.
    He is type AB, not type A.

Why: His cells carry the A marker, so he carries at least one Iᴬ.
His cells carry no B marker, so he carries no Iᴮ.
His second allele is Iᴬ or i, so he is IᴬIᴬ or Iᴬi.

49
Check q22

A woman is type AB.

Which alleles can she pass to a child?

  1. A. ✓ Iᴬ or Iᴮ
  2. B. Iᴬ, Iᴮ or i
    A type AB woman is IᴬIᴮ.
    She carries no i to pass.
  3. C. i only
    A type AB woman is IᴬIᴮ.
    Neither of her alleles is i.

Why: A type AB woman is IᴬIᴮ.
Each of her eggs carries one of her two alleles.
So she passes Iᴬ or Iᴮ to a child, never i.

50
Check q23

An IᴬIᴮ person’s red blood cells carry the A marker and the B marker.

Which kind of dominance do Iᴬ and Iᴮ show?

  1. A. Complete dominance
    Under complete dominance one allele would hide the other, and the cells would carry one marker.
  2. B. Incomplete dominance
    Under incomplete dominance the cells would carry a marker in between A and B.
    They carry the A marker and the B marker, each whole.
  3. C. ✓ Codominance

Why: The IᴬIᴮ person is the heterozygote.
That person’s cells carry the A marker and the B marker, each whole and separate.
Both alleles’ traits showing separately in the heterozygote is codominance.

51
Check q24

A mother is IᴮIᴮ and a father is ii.

Which blood types are possible for their children?

  1. A. ✓ Type B only
  2. B. Type B or type O
    A type O child needs an i from each parent.
    The IᴮIᴮ mother has no i to give.
  3. C. Type A, B, AB or O
    Neither parent carries an Iᴬ.
    So no child can be type A or type AB.

Why: Every egg carries Iᴮ and every sperm carries i.
So every child is Iᴮi.
Iᴮi is type B, so every child is type B.

52
Practice writing an answer

A woman is type A and her husband is type B. Their first child is type O.

(a) Identify the ABO genotype of each parent. (1 pt)

Model answer The mother is Iᴬi and the father is Iᴮi.
Rubric
  • Award 1 point for: mother Iᴬi, father Iᴮi.

Slip Writing the mother as IᴬIᴬ. An IᴬIᴬ mother has no i to give a type O child.

(b) Calculate the probability that the couple’s next child is type AB. Enter the probability as a decimal. (1 pt)

Answer: 0.25  (tolerance ±0.001)

Model answer The probability that the next child is type AB is 1 in 4, or 0.25.
Working
Write down the values in the question:
mother Iᴬi, father Iᴮi: cells in the square = 4, all equally likely
cells that are type AB (IᴬIᴮ) = 1
Write down the equation:
probability of a type=cells of that typecells in all
Substitute the values into the equation:
probability of type AB=14=0.25
Rubric
  • Award 1 point for: 1/4 (0.25), from the one IᴬIᴮ cell of the four.

(c) Explain how the first child’s blood type shows the mother’s genotype. (1 pt)

Model answer The child is type O, so the child is ii.
A type O child received one i from each parent, so the mother passed an i.
The mother is type A, so she carries an Iᴬ.
Therefore the mother’s two alleles are Iᴬ and i: she is Iᴬi.
Rubric
  • Award 1 point for: the type O child is ii and received an i from each parent, so the type A mother carries an i as well as an Iᴬ: Iᴬi.

(d) The couple’s second child is type A. Explain how this child demonstrates that Iᴬ is dominant to i. (1 pt)

Model answer The second child received Iᴬ from the mother and i from the father, so the child is Iᴬi.
Iᴬ makes the A marker and i makes no marker.
The child’s cells carry the A marker, so the child is type A: one Iᴬ shows and the i does not.
An allele that shows in the heterozygote over the other is dominant, so Iᴬ is dominant to i.
Rubric
  • Award 1 point for: the child is the heterozygote Iᴬi and shows type A (the Iᴬ shows, the i does not), so Iᴬ is dominant to i.

APBIO-U05-L22 Genes that travel together

Topic 5.4 · Non-Mendelian Genetics · 91 steps

Four bars side by side on a baseline: 421 gray normal and 407 black vestigial rise high; 46 gray vestigial and 42 black normal are short; each bar carries its count
Four bars side by side on a baseline: 421 gray normal and 407 black vestigial rise high; 46 gray vestigial and 42 black normal are short; each bar carries its count

Photos: André Karwath, Wikimedia Commons, CC BY-SA 2.5; Salem.slima, Wikimedia Commons, CC BY-SA 4.0 (resized).

Suppose a test cross of fruit flies gave these offspring, sorted into four classes: a fly with a gray body and normal wings, heterozygous for both genes, GgWw, was crossed with a black-bodied, vestigial-winged fly, ggww.

Independent assortment predicts four equal classes. The counts: 421 gray normal, 407 black vestigial, 46 gray vestigial, 42 black normal. Why two big classes and two small ones?

Unit 5 · Heredity

1Which gamete made this offspring?

2

Video: Watch: Which gamete made this offspring?

The ggww parent handing every offspring a g and a w; each of the four offspring classes read back as the one gamete the GgWw parent gave it; the four counts as bars.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22a.mp4

3

Why does a dihybrid test cross give two big classes and two small ones, instead of four equal ones?

4

In a test cross the ggww parent gives every offspring a g and a w. So each offspring’s look shows which gamete the GgWw parent gave it.

5

The body-color gene and the wing gene sit on the same chromosome. So G rides into a gamete with W, and g rides into a gamete with w.

6

The combinations the parent’s own chromosomes carried make the two big classes. The two small classes, 46 and 42, carry combinations neither homolog had.

7

Those two small classes came from a crossover between the two genes in prophase I.

8
Check q1

A breeder wants to know whether an animal showing the dominant trait is homozygous or heterozygous, so she sets up a test cross.

In a test cross, which partner is the animal crossed with?

  1. A. ✓ A homozygous recessive individual
  2. B. A homozygous dominant individual
    A homozygous dominant partner gives every offspring a dominant allele, so every offspring shows the dominant trait whatever the tested parent gave.
  3. C. Another individual showing the dominant trait
    A partner that shows the dominant trait may carry a hidden dominant allele, so an offspring’s look would not show what the tested parent gave.

Why: A homozygous recessive partner gives every offspring only recessive alleles.
So each offspring’s look shows the allele the tested parent gave.
That is what a test cross uses.

9

Now consider fruit flies. Vestigial wings are small stubs.

10

In these flies a gray body (G) is dominant to a black body (g). Normal wings (W) are dominant to vestigial wings (w).

11

Here are two photographs: a fly with normal wings, and five flies with vestigial wings.

Two photographs side by side. Left: one fruit fly seen from the side, its two wings lying flat along its back and reaching past the end of its body. Right: five fruit flies on a pale background, each with two short stubs where the wings would be
Two photographs side by side. Left: one fruit fly seen from the side, its two wings lying flat along its back and reaching past the end of its body. Right: five fruit flies on a pale background, each with two short stubs where the wings would be
12

Now consider the test cross itself: a GgWw fly crossed with a ggww fly.

13

The ggww parent has g on both homologs and w on both homologs. So every one of its gametes carries g and w.

14

So every offspring received a g and a w from the ggww parent. The offspring’s other two alleles came in the GgWw parent’s gamete.

15

So each offspring’s look shows which gamete the GgWw parent gave it.

16

Here is a table of the four offspring classes: the gamete the GgWw parent gave, the offspring’s genotype, what the offspring looks like, and how many there were.

A table with four rows: the gamete from the GgWw parent, the offspring genotype after a gw gamete joins it, the offspring phenotype and the count: GW gives GgWw, gray normal, 421; gw gives ggww, black vestigial, 407; Gw gives Ggww, gray vestigial, 46; gW gives ggWw, black normal, 42
A table with four rows: the gamete from the GgWw parent, the offspring genotype after a gw gamete joins it, the offspring phenotype and the count: GW gives GgWw, gray normal, 421; gw gives ggww, black vestigial, 407; Gw gives Ggww, gray vestigial, 46; gW gives ggWw, black normal, 42
17

A gray normal offspring is GgWw. It got a GW gamete from the GgWw parent.

18

A black vestigial offspring is ggww. It got a gw gamete.

19

A gray vestigial offspring is Ggww. It got a Gw gamete.

20

A black normal offspring is ggWw. It got a gW gamete.

21

Here are the four counts as a bar graph: 421 gray normal, 407 black vestigial, 46 gray vestigial and 42 black normal.

A bar graph of the four offspring classes of the GgWw × ggww test cross: gray normal 421, black vestigial 407, gray vestigial 46, black normal 42; gridlines every 100 offspring
A bar graph of the four offspring classes of the GgWw × ggww test cross: gray normal 421, black vestigial 407, gray vestigial 46, black normal 42; gridlines every 100 offspring
22

So the GgWw parent made many GW gametes and many gw gametes, but few Gw gametes and few gW gametes.

23

What you are expected to know Read a test-cross offspring’s look as the gamete the heterozygous parent gave it: the homozygous recessive parent gave only recessive alleles, so every other allele came in that one gamete.

24Quick quiz: which gamete did the parent give? mixed practice

25
Check q2

In a plant, purple flowers (P) are dominant to white (p) and tall stems (T) are dominant to short (t). A PpTt plant is crossed with a pptt plant. One offspring has white flowers and a tall stem.

Which gamete did the PpTt parent give that offspring?

  1. A. ✓ pT
  2. B. PT
    The offspring’s flowers are white, so it carries no P.
    The PpTt parent gave it p.
  3. C. pt
    The offspring’s stem is tall, so it carries a T.
    The pptt parent gave only t, so the T came from the PpTt parent.
  4. D. Pt
    The offspring’s flowers are white, so the PpTt parent gave it p, not P.

Why: The pptt parent gave the offspring p and t.
The offspring has white flowers, so its other allele for that gene is p.
The offspring has a tall stem, so its other allele for that gene is T.
So the PpTt parent gave it pT.

26
Check q3

In an insect, long wings (L) are dominant to short wings (l) and red eyes (E) are dominant to brown eyes (e). An LlEe insect is crossed with an llee insect. One offspring has long wings and brown eyes.

Which gamete did the LlEe parent give that offspring?

  1. A. LE
    The offspring’s eyes are brown, so it carries no E.
    The LlEe parent gave it e.
  2. B. ✓ Le
  3. C. lE
    The offspring’s wings are long, so it carries an L.
    The llee parent gave only l, so the L came from the LlEe parent.
  4. D. le
    The offspring’s wings are long, so the LlEe parent gave it L, not l.

Why: The llee parent gave the offspring l and e.
The offspring has long wings, so its other allele for that gene is L.
The offspring has brown eyes, so its other allele for that gene is e.
So the LlEe parent gave it Le.

27
Check q4

In a mammal, black fur (B) is dominant to brown fur (b) and a short tail (S) is dominant to a long tail (s). A BbSs animal is crossed with a bbss animal. One offspring has brown fur and a long tail.

Which gamete did the BbSs parent give that offspring?

  1. A. BS
    The offspring’s fur is brown and its tail is long, so it carries no B and no S.
  2. B. Bs
    The offspring’s fur is brown, so it carries no B.
    The BbSs parent gave it b.
  3. C. bS
    The offspring’s tail is long, so it carries no S.
    The BbSs parent gave it s.
  4. D. ✓ bs

Why: The bbss parent gave the offspring b and s.
The offspring has brown fur, so its other allele for that gene is b.
The offspring has a long tail, so its other allele for that gene is s.
So the BbSs parent gave it bs.

28
Check q5

In a plant, round seeds (R) are dominant to wrinkled seeds (r) and yellow seeds (Y) are dominant to green seeds (y). An RrYy plant is crossed with an rryy plant. One offspring has round yellow seeds.

Which gamete did the RrYy parent give that offspring?

  1. A. rY
    The offspring’s seeds are round, so it carries an R.
    The rryy parent gave only r, so the R came from the RrYy parent.
  2. B. Ry
    The offspring’s seeds are yellow, so it carries a Y.
    The rryy parent gave only y, so the Y came from the RrYy parent.
  3. C. ✓ RY
  4. D. ry
    The offspring’s seeds are round and yellow, so it carries an R and a Y, and both came from the RrYy parent.

Why: The rryy parent gave the offspring r and y.
The offspring has round seeds, so its other allele for that gene is R.
The offspring has yellow seeds, so its other allele for that gene is Y.
So the RrYy parent gave it RY.

29
Check q6

In a fish, a striped body (D) is dominant to a plain body (d) and a forked tail (F) is dominant to a rounded tail (f). A DdFf fish is crossed with a ddff fish. One offspring has a plain body and a forked tail.

Which gamete did the DdFf parent give that offspring?

  1. A. DF
    The offspring’s body is plain, so it carries no D.
    The DdFf parent gave it d.
  2. B. ✓ dF
  3. C. Df
    The offspring’s body is plain and its tail is forked, so it carries no D and one F.
  4. D. df
    The offspring’s tail is forked, so it carries an F.
    The ddff parent gave only f, so the F came from the DdFf parent.

Why: The ddff parent gave the offspring d and f.
The offspring has a plain body, so its other allele for that gene is d.
The offspring has a forked tail, so its other allele for that gene is F.
So the DdFf parent gave it dF.

30Genes that travel together

31

Video: Watch: Genes that travel together

The body-color gene and the wing gene placed on one homologous pair, G with W on the dark homolog and g with w on the light one; the homologs parting at anaphase I, each carrying both of its alleles into a gamete; the two big bars lit as the parental combinations.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22b.mp4

32
Check q7

A plant is heterozygous for two genes.

Which pairs of genes obey the law of independent assortment?

  1. A. ✓ Genes on different chromosomes
  2. B. Genes on the same chromosome
    Two genes on the same chromosome need not follow independent assortment.
  3. C. Every pair of genes
    Independent assortment comes from each homologous pair facing either way on its own at metaphase I, so it applies to genes on different pairs.

Why: At metaphase I each homologous pair faces either way on its own.
So the alleles of two genes on different chromosomes go into gametes independently of each other.
Two genes on the same chromosome need not follow independent assortment.

33

Independent assortment predicts four equal classes from a dihybrid test cross. The fly gave two big classes and two small ones.

34

Independent assortment applies to genes on different chromosomes. So the fly’s two genes may sit on the same chromosome.

35

Now put the body-color gene and the wing gene on one homologous pair. One homolog carries G and W; the other homolog carries g and w.

A homologous pair drawn as two rods with centromere dots, the left rod darker and the right rod lighter: the left rod carries G in a box near its middle and W in a box lower down; the right rod carries g and w at the same two heights; labels name the body-color gene and the wing gene
A homologous pair drawn as two rods with centromere dots, the left rod darker and the right rod lighter: the left rod carries G in a box near its middle and W in a box lower down; the right rod carries g and w at the same two heights; labels name the body-color gene and the wing gene
36

At anaphase I the spindle pulls the two homologs of the pair to opposite poles.

37

G and W sit on one homolog, so they move together. g and w sit on the other homolog, so they move together.

38

So one gamete receives G with W, and the other gamete receives g with w.

The same homologous pair above two gametes: the gamete on the left holds the darker rod carrying G and W, the gamete on the right holds the lighter rod carrying g and w
The same homologous pair above two gametes: the gamete on the left holds the darker rod carrying G and W, the gamete on the right holds the lighter rod carrying g and w
39

Two genes on the same chromosome tend to pass into a gamete together. We call them , because the chromosome they share links them.

40

The combinations the parent’s own chromosomes carried are the parental combinations. The GgWw parent’s chromosomes carried GW and gw, so GW and gw are its parental combinations.

41
Check q8

A parent’s two homologs carried A with B and a with b. One of its gametes carries A with b.

Which kind of combination does that gamete carry?

  1. A. A parental combination
    Neither of the parent’s homologs carried A with b.
  2. B. ✓ A recombinant combination

Why: The parent’s homologs carried AB and ab.
A with b is a combination neither homolog carried.
So the gamete carries a recombinant combination.

42

Most of the GgWw parent’s gametes carried GW or gw. So the two big classes, 421 gray normal and 407 black vestigial, are the parental combinations.

43

Two genes on different chromosomes are never linked, however alike their traits look. Only a shared chromosome links two genes.

44

So a dihybrid test cross of two linked genes gives mostly the two parental combinations, instead of the four equal classes independent assortment predicts.

45

What you are expected to know Explain why two linked genes give mostly the two parental combinations in a test cross: the two genes ride into a gamete on one chromosome, so the big classes carry the combinations the parent’s chromosomes carried.

46
Check q9

Suppose a cross of plants gave the counts graphed below. An AaBb plant was crossed with an aabb plant, and the 800 offspring were sorted into four classes.

A bar graph of four offspring classes from an AaBb × aabb test cross: AB 356, ab 348, Ab 49, aB 47; gridlines every 100 offspring
A bar graph of four offspring classes from an AaBb × aabb test cross: AB 356, ab 348, Ab 49, aB 47; gridlines every 100 offspring

Which two classes are the parental classes?

  1. A. ✓ AB and ab
  2. B. AB and Ab
    Ab is a small class, 49 offspring; a parental combination fills a big class.
  3. C. Ab and aB
    Ab and aB are the two small classes, 49 and 47; a parental combination fills a big class.
  4. D. ab and aB
    aB is a small class, 47 offspring; a parental combination fills a big class.

Why: The two big classes are AB, 356 offspring, and ab, 348 offspring.
Linked genes pass into a gamete together in most meioses.
So the big classes carry the combinations the parent’s chromosomes carried.
So AB and ab are the parental classes.

47
Practice writing an answer

Suppose a cross of plants gave these counts. An AaBb plant whose chromosomes carried A with B and a with b was crossed with an aabb plant. Of the 800 offspring, 356 were AB, 348 were ab, 49 were Ab and 47 were aB.

(a) Explain why AB and ab are the two largest classes. (1 pt)

Model answer The two genes sit on one chromosome.
At anaphase I the whole chromosome moves to one pole.
So A moves with B into a gamete, and a moves with b, in most meioses.
The aabb parent adds a and b to every offspring.
So most offspring are AB or ab.
Rubric
  • Award 1 point for: the two genes sit on one chromosome, so A passes into a gamete with B and a with b (the parental combinations) in most meioses; the aabb parent adds only a and b.

Slip Saying AB and ab are largest because A and B are dominant. Dominance decides what an offspring looks like, not how often a gamete type forms.

48
Check q10

In a mammal, the fur-color gene sits on chromosome 1 and the fur-length gene sits on chromosome 4. A student says: “Both genes change how the coat looks, so they are linked genes.”

Is the student correct?

  1. A. Yes: the two genes are linked
    Only a shared chromosome links two genes.
    Chromosome 1 and chromosome 4 are different chromosomes.
  2. B. ✓ No: the two genes are not linked

Why: Linked genes sit on the same chromosome.
The fur-color gene sits on chromosome 1 and the fur-length gene sits on chromosome 4.
So the two genes are not linked, whatever their traits do.

49
Check q11

In an insect, the eye-color gene and the bristle gene sit close together on one chromosome. An insect heterozygous for both genes is test-crossed.

Which pattern do the four offspring classes show?

  1. A. ✓ Two large classes and two small classes
  2. B. Four classes of about equal size
    Four equal classes come from independent assortment, which applies to genes on different chromosomes.

Why: The two genes sit on one chromosome.
So they pass into a gamete together in most meioses.
So the two parental combinations fill two large classes, and the other two classes are small.

50Quick quiz: linked genes mixed practice

51
Check q12

A plant carries two genes.

What are linked genes?

  1. A. Two genes that change the same trait
    What the traits do has nothing to do with linkage.
    Only a shared chromosome links two genes.
  2. B. ✓ Two genes on the same chromosome
  3. C. Two genes on different chromosomes
    Genes on different chromosomes assort independently, so they are not linked.

Why: Linked genes are two genes on the same chromosome.
So they tend to pass into a gamete together.

52
Practice writing an answer

Two genes in a plant are linked.

(a) State what linked genes are. (1 pt)

Model answer Linked genes are two genes on the same chromosome, so they tend to pass into a gamete together.
Rubric
  • Award 1 point for: two genes on the same chromosome (so they tend to be inherited together).

(b) State what the two large offspring classes of a test cross for two linked genes carry. (1 pt)

Model answer The two large classes carry the parental combinations, the allele combinations the heterozygous parent’s chromosomes carried.
Rubric
  • Award 1 point for: the parental combinations (the combinations the heterozygous parent’s chromosomes carried).
53
Check q13

Suppose a KkLl plant is crossed with a kkll plant. The four offspring classes are graphed below.

A bar graph of the four offspring classes of a KkLl × kkll test cross: KL 238, kl 230, Kl 16, kL 16; gridlines every 100 offspring
A bar graph of the four offspring classes of a KkLl × kkll test cross: KL 238, kl 230, Kl 16, kL 16; gridlines every 100 offspring

Are the two genes linked?

  1. A. ✓ Linked
  2. B. Not linked
    Four equal classes are what independent assortment gives.
    Two large and two small classes mean two combinations went into most gametes together, so the genes sit on one chromosome.

Why: Two classes are large and two are small.
So two combinations of alleles went into most gametes together.
Two genes pass into a gamete together when they sit on one chromosome.
So the two genes are linked.

54
Check q14

Suppose an EeFf insect is crossed with an eeff insect. The four offspring classes are graphed below.

A bar graph of the four offspring classes of an EeFf × eeff test cross: EF 124, ef 121, Ef 128, eF 127; gridlines every 100 offspring
A bar graph of the four offspring classes of an EeFf × eeff test cross: EF 124, ef 121, Ef 128, eF 127; gridlines every 100 offspring

Are the two genes linked?

  1. A. Linked
    Linked genes give two large classes, the parental combinations, and two small ones.
    Four equal classes are what genes on different chromosomes give by independent assortment.
  2. B. ✓ Not linked

Why: The four classes are about equal in size.
So each combination of alleles went into a gamete as often as every other.
That is independent assortment, which genes on different chromosomes show.
So the two genes are not linked.

55
Check q15

Suppose a PpQq plant is crossed with a ppqq plant. The four offspring classes are graphed below.

A bar graph of the four offspring classes of a PpQq × ppqq test cross: PQ 355, pq 349, Pq 148, pQ 148; gridlines every 100 offspring
A bar graph of the four offspring classes of a PpQq × ppqq test cross: PQ 355, pq 349, Pq 148, pQ 148; gridlines every 100 offspring

Are the two genes linked?

  1. A. Not linked
    Four equal classes are what independent assortment gives.
    Two large and two small classes mean two combinations went into most gametes together, so the genes sit on one chromosome.
  2. B. ✓ Linked

Why: Two classes are large and two are small.
So two combinations of alleles went into most gametes together.
Two genes pass into a gamete together when they sit on one chromosome.
So the two genes are linked.

56
Check q16

Suppose a JjKk snail is crossed with a jjkk snail. The four offspring classes are graphed below.

A bar graph of the four offspring classes of a JjKk × jjkk test cross: JK 201, jk 199, Jk 202, jK 198; gridlines every 100 offspring
A bar graph of the four offspring classes of a JjKk × jjkk test cross: JK 201, jk 199, Jk 202, jK 198; gridlines every 100 offspring

Are the two genes linked?

  1. A. ✓ Not linked
  2. B. Linked
    Linked genes give two large classes, the parental combinations, and two small ones.
    Four equal classes are what genes on different chromosomes give by independent assortment.

Why: The four classes are about equal in size.
So each combination of alleles went into a gamete as often as every other.
That is independent assortment, which genes on different chromosomes show.
So the two genes are not linked.

57
Check q17

Suppose an AaCc plant is crossed with an aacc plant. The four offspring classes are graphed below.

A bar graph of the four offspring classes of an AaCc × aacc test cross: AC 461, ac 455, Ac 42, aC 42; gridlines every 100 offspring
A bar graph of the four offspring classes of an AaCc × aacc test cross: AC 461, ac 455, Ac 42, aC 42; gridlines every 100 offspring

Are the two genes linked?

  1. A. ✓ Linked
  2. B. Not linked
    Four equal classes are what independent assortment gives.
    Two large and two small classes mean two combinations went into most gametes together, so the genes sit on one chromosome.

Why: Two classes are large and two are small.
So two combinations of alleles went into most gametes together.
Two genes pass into a gamete together when they sit on one chromosome.
So the two genes are linked.

58
Check q18

In a mammal, the fur-length gene and the tail-length gene sit close together on one chromosome.

Are the two genes linked?

  1. A. Not linked
    The two genes sit on the same chromosome.
    A shared chromosome is what links two genes, whatever their traits do.
  2. B. ✓ Linked

Why: The two genes sit on one chromosome.
Genes on one chromosome pass into a gamete together.
So the two genes are linked.

59Where the small classes come from

60

Video: Watch: Where the small classes come from

The fly’s pair in prophase I with G and W on the dark chromatids and g and w on the light ones; one crossover between the two genes swapping the tip pieces; the four chromatids read out as GW, Gw, gW and gw; the two small bars lit as the recombinant classes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22c.mp4

61

Suppose G always traveled with W. Then the GgWw fly could make no Gw gamete and no gW gamete.

62

Yet 46 gray vestigial and 42 black normal offspring appeared. So some gametes did carry Gw or gW.

63
Check q19

In prophase I the two homologs of a pair lie side by side.

What happens in crossing over?

  1. A. ✓ Chromatids of the two homologs break at the same point and swap pieces
  2. B. The two homologs of the pair are pulled apart to opposite poles of the cell
    The homologs part at anaphase I.
    Crossing over is an exchange of pieces in prophase I.

Why: In crossing over a chromatid of each homolog breaks at the same point.
The two chromatids exchange the pieces.

64

Here is the fly’s chromosome pair in prophase I, drawn apart so the letters can be read. Each homolog is two sister chromatids: both dark chromatids read GW, and both light chromatids read gw.

The fly’s chromosome pair drawn apart: a dark X whose two upper arms each carry a box reading G and, nearer the tip, a box reading W; a light X whose two upper arms each carry g and, nearer the tip, w
The fly’s chromosome pair drawn apart: a dark X whose two upper arms each carry a box reading G and, nearer the tip, a box reading W; a light X whose two upper arms each carry g and, nearer the tip, w
65

Now suppose one crossover falls between the body-color position and the wing position. A dark chromatid and a light chromatid break at the same point and exchange the pieces beyond the break.

66

The piece of the dark chromatid beyond the break carries W. The piece of the light chromatid beyond the break carries w.

67

So the exchange carries W and w across.

The same pair after one crossover between the two positions: the right upper arm of the dark X now ends in a light piece carrying w, and the left upper arm of the light X ends in a dark piece carrying W; the other two chromatids are unchanged
The same pair after one crossover between the two positions: the right upper arm of the dark X now ends in a light piece carrying w, and the left upper arm of the light X ends in a dark piece carrying W; the other two chromatids are unchanged
68

The dark chromatid that took part now reads Gw. The light chromatid that took part now reads gW.

69

The other two chromatids took no part in the crossover. They still read GW and gw.

70

So the two chromatids that took part carry combinations of alleles that neither homolog had. Gw and gW are recombinant; GW and gw are parental.

Four gametes in a row, each holding one chromosome drawn as a rod with two allele boxes: GW, Gw, gW and gw; GW and gw are labeled parental, Gw and gW recombinant
Four gametes in a row, each holding one chromosome drawn as a rod with two allele boxes: GW, Gw, gW and gw; GW and gw are labeled parental, Gw and gW recombinant
71

A gamete built from the Gw chromatid gives a gray vestigial offspring. A gamete built from the gW chromatid gives a black normal offspring.

72

So the 46 gray vestigial and the 42 black normal are the recombinant classes.

73

A crossover falls between two linked positions in only some of the cells going through meiosis. So the recombinant classes are always the small classes, and the parental classes are always the big classes.

74

What you are expected to know Identify the two frequent classes of a linked test cross as parental and the two rare classes as recombinant, and explain that the recombinants come from a crossover between the two genes in prophase I.

75
Check q20

Suppose a cross of tomato plants gave the counts graphed below. An RrSs plant was crossed with an rrss plant, and the 1,000 offspring were sorted into four classes.

A bar graph of four offspring classes from an RrSs × rrss test cross: RS 412, rs 398, Rs 96, rS 94; gridlines every 100
A bar graph of four offspring classes from an RrSs × rrss test cross: RS 412, rs 398, Rs 96, rS 94; gridlines every 100

Which two classes are the recombinant classes?

  1. A. RS and rs
    RS and rs are the two big classes, 412 and 398; a recombinant combination fills a small class.
  2. B. RS and Rs
    RS is a big class, 412 offspring; a recombinant combination fills a small class.
  3. C. ✓ Rs and rS
  4. D. rs and rS
    rs is a big class, 398 offspring; a recombinant combination fills a small class.

Why: The two small classes are Rs, 96 offspring, and rS, 94 offspring.
A crossover between two linked genes falls there in only some meioses.
So the recombinant combinations fill the small classes.
So Rs and rS are the recombinant classes.

76
Practice writing an answer

Suppose a cross of tomato plants gave these counts. An RrSs plant whose chromosomes carried R with S and r with s was crossed with an rrss plant. Of the 1,000 offspring, 412 were RS, 398 were rs, 96 were Rs and 94 were rS.

(a) Explain how the 96 Rs offspring came to be. (1 pt)

Model answer The RrSs parent’s chromosomes carried R with S and r with s.
In prophase I a chromatid of one homolog and a chromatid of the other broke at the same point, between the two genes, and exchanged the pieces.
One chromatid then carried R with s.
A gamete built from that chromatid joined an rs gamete from the rrss parent.
So the offspring carried R with s: an Rs offspring.
Rubric
  • Award 1 point for: a crossover between the two genes in prophase I (chromatids of the two homologs break at the same point and exchange pieces) made a chromatid carrying R with s, and a gamete built from it made the Rs offspring.

Slip Saying a mutation changed S to s in a few gametes. A crossover between the two genes made the new combination; no allele changed.

(b) Explain why the Rs class is much smaller than the RS class. (1 pt)

Model answer A crossover falls between the two genes in only some of the cells going through meiosis.
So only a few gametes carry R with s.
Most gametes carry R with S, the combination the parent’s chromosome carried.
So the Rs class is much smaller than the RS class.
Rubric
  • Award 1 point for: a crossover falls between the two genes in only some meioses, so most gametes keep the parental combination R with S and few carry R with s.

Slip Saying the Rs offspring survived less well. The classes differ in how often each gamete formed, not in survival.

77
Check q21

Suppose a cross of plants gave 318 PS, 306 ps, 41 Ps and 35 pS offspring, from a PpSs plant crossed with a ppss plant. A student says: “The two large classes, PS and ps, are the recombinant classes.”

Is the student correct?

  1. A. Yes: PS and ps are the recombinant classes
    A crossover between two linked genes falls there in only some meioses, so the recombinant combinations fill the small classes.
  2. B. ✓ No: PS and ps are the parental classes

Why: The two large classes carry the combinations the PpSs parent’s chromosomes carried.
So PS and ps are the parental classes.
The small classes, Ps and pS, are the recombinant classes.

78
Check q22

Suppose a cross of insects gave 288 LM, 296 lm, 58 Lm and 58 lM offspring, from an LlMm insect whose chromosomes carried L with M and l with m, crossed with an llmm insect. A student says: “The Lm and lM offspring came from a mutation that changed an allele in a few gametes.”

Is the student correct?

  1. A. Yes: only a mutation can make a combination the parent did not carry
    A crossover between the two genes in prophase I makes a new combination of the same alleles, and no allele changes.
  2. B. ✓ No: a crossover between the two genes made the Lm and lM gametes

Why: In prophase I a chromatid of each homolog breaks at the same point between the two genes, and the two exchange pieces.
One chromatid then carries L with m, and the other l with M.
No allele changed.
So a crossover, not a mutation, made the Lm and lM gametes.

79
Check q23

Suppose a cross of maize plants gave the kernel counts graphed below. An RrSs plant whose chromosomes carried R with S and r with s was crossed with an rrss plant.

A bar graph of four kernel classes from an RrSs × rrss test cross: RS 423, rs 401, Rs 91, rS 85; gridlines every 100
A bar graph of four kernel classes from an RrSs × rrss test cross: RS 423, rs 401, Rs 91, rS 85; gridlines every 100

Which event made the gametes behind the Rs and rS kernels?

  1. A. ✓ A crossover between the two gene positions in prophase I
  2. B. Independent orientation of two chromosome pairs at metaphase I
    The two genes sit on one pair, which orients as a unit; independent orientation shuffles different pairs.
  3. C. Fertilization by the rs gametes of the rrss parent
    The rrss parent gives every kernel an r and an s, so it cannot decide which class a kernel joins.

Why: In prophase I a chromatid of each homolog breaks at the same point and the two exchange pieces.
A break between the two genes leaves one chromatid with R and s and the other with r and S.
Gametes built from those chromatids give the Rs and rS kernels.

80

Back to the test cross: a GgWw fly crossed with a ggww fly gave 421 gray normal, 407 black vestigial, 46 gray vestigial and 42 black normal offspring.

A bar graph of the four offspring classes of the GgWw × ggww test cross: gray normal 421, black vestigial 407, gray vestigial 46, black normal 42; gridlines every 100 offspring
A bar graph of the four offspring classes of the GgWw × ggww test cross: gray normal 421, black vestigial 407, gray vestigial 46, black normal 42; gridlines every 100 offspring
81

The body-color gene and the wing gene sit on one chromosome. So G traveled with W, and g with w, into most gametes, and the two big classes are the parental combinations.

82

A crossover between the two genes in prophase I made a few Gw and gW gametes. So the 46 and 42 are the recombinant classes.

83Mixed practice mixed practice

84
Check q24

In an insect, the wing-color gene and the antenna-length gene both sit on chromosome 7.

Are the two genes linked?

  1. A. ✓ Linked
  2. B. Not linked
    The two genes sit on one chromosome, and a shared chromosome is what links two genes.

Why: Linked genes sit on the same chromosome.
Both genes sit on chromosome 7.
So the two genes are linked.

85
Check q25

Suppose an AaBb beetle whose chromosomes carried A with B and a with b was test-crossed, and the offspring came in four classes: AB 388, ab 402, Ab 61 and aB 49.

Which two classes are the parental classes?

  1. A. Ab and aB
    Ab and aB are combinations neither parental chromosome carried, so they are the recombinant classes.
  2. B. AB and Ab
    Ab is a new combination; only AB is parental.
  3. C. ✓ AB and ab
  4. D. ab and aB
    aB is a new combination; only ab is parental.

Why: The parent’s chromosomes carried A with B and a with b.
So AB and ab are the parental combinations.
They are also the two large classes, because linked genes pass into a gamete together in most meioses.

86
Check q26

Suppose a DdEe moth whose chromosomes carried D with E and d with e was test-crossed, and 58 De and 52 dE offspring appeared.

Which event made the gametes behind those offspring?

  1. A. Independent orientation of two chromosome pairs at metaphase I
    The two genes sit on one pair, which orients as a unit; independent orientation shuffles different pairs.
  2. B. ✓ A crossover between the two gene positions in prophase I
  3. C. Fertilization by the homozygous parent’s gametes
    The homozygous parent gives every offspring a d and an e, so it cannot decide which class an offspring joins.

Why: In prophase I a chromatid of each homolog breaks at the same point and the two exchange pieces.
A break between the two genes leaves one chromatid with D and e and the other with d and E.
Gametes built from those chromatids give the De and dE offspring.

87
Check q27

In a plant, the flower-color gene sits on chromosome 2 and the leaf-shape gene sits on chromosome 5. A plant heterozygous for both genes is test-crossed.

Which pattern do the four offspring classes show?

  1. A. ✓ Four classes of about equal size
  2. B. Two large classes and two small classes
    Two large and two small classes is the pattern of linked genes; these two genes sit on different chromosomes.

Why: The two genes sit on different chromosome pairs.
Each pair faces either way at metaphase I on its own, so every combination of alleles is equally common in the gametes.
So the four offspring classes are about equal in size.

88
Check q28

A student says: “Linked genes always pass into a gamete together, so a test cross of a heterozygote gives only two classes.”

Is the student correct?

  1. A. Yes: the offspring come in two classes
    A crossover between the two genes in prophase I makes new combinations, so a few offspring of two other classes appear.
  2. B. ✓ No: two small recombinant classes appear as well

Why: Linked genes pass into a gamete together in most meioses, not all.
In some meioses a crossover falls between the two genes and makes recombinant chromatids.
So a test cross gives two large parental classes and two small recombinant classes, four classes in all.

89
Check q29

Suppose a DdEe fish whose chromosomes carried D with e and d with E was test-crossed, and the offspring came in four classes: De 310, dE 296, DE 47 and de 43.

Which two classes are the recombinant classes?

  1. A. De and dE
    De and dE are the combinations the parent’s chromosomes carried, so they are parental, and their size says so.
  2. B. ✓ DE and de
  3. C. De and DE
    De is a parental combination; only DE is new.

Why: The parent’s chromosomes carried D with e and d with E.
DE and de are combinations neither chromosome carried, so they are recombinant.
A crossover between the two genes made them, in only some meioses, so the two classes are small.

90
Practice writing an answer

Suppose a GgHh lizard whose chromosomes carried G with h and g with H was test-crossed, and the offspring came in four classes: 305 Gh, 299 gH, 51 GH and 45 gh.

(a) Identify the two recombinant classes. (1 pt)

Model answer The recombinant classes are GH and gh.
Rubric
  • Award 1 point for: GH and gh.

(b) Explain why two of the four classes are so much smaller than the other two. (1 pt)

Model answer The two genes sit on one chromosome, so G rides into a gamete with h, and g with H, unless something separates them.
Only a crossover between the two gene positions in prophase I separates them.
A crossover falls between the two positions in only some of the cells going through meiosis.
So only a few gametes carry GH or gh, and the two recombinant classes are small.
Rubric
  • Award 1 point for: the genes are linked, so the parental combinations pass into most gametes; only a crossover between the two genes in prophase I makes GH or gh, and it falls there in only some meioses.

Slip Saying the recombinant offspring survived less well. The classes differ in how often the gamete formed, not in survival.

Glossary

linked genes
Two genes on the same chromosome, which therefore tend to pass into a gamete together: a test cross of a heterozygote gives mostly the two parental combinations.

APBIO-U05-L22B How far apart are the genes?

Topic 5.4 · Non-Mendelian Genetics · 61 steps

A chromosome drawn as a horizontal ruler with tick marks; two genes are marked on it, body color and wing, with a question mark over the gap between them; to the right four small bars, 421, 407, 46 and 42
A chromosome drawn as a horizontal ruler with tick marks; two genes are marked on it, body color and wing, with a question mark over the gap between them; to the right four small bars, 421, 407, 46 and 42

Here is the fly’s chromosome drawn as a ruler, with the body-color gene and the wing gene marked on it, and beside it the four classes from the test cross: 421, 407, 46 and 42.

The 46 and 42 came from a crossover between the two genes. How far apart are the genes?

Unit 5 · Heredity

1From recombinants to a frequency

2

Video: Watch: From recombinants to a frequency

The two recombinant bars picked out of the four; the division written out on screen, 88 over 916 times 100; the answer, 9.6 %, given to one decimal place.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22Ba.mp4

3

How do you measure the distance between two genes on a chromosome? You count how often a crossover fell between them.

4

The recombinant offspring, as a share of all the offspring, show how often a crossover fell between the two genes.

5

One percent of recombination counts as one unit of distance along the chromosome. So the share of recombinants gives the distance between the two genes.

6
Check q1

In a test cross for two linked genes, the offspring come in two large classes and two small classes.

Which classes are the recombinant classes?

  1. A. ✓ The two small classes
  2. B. The two large classes
    The large classes carry the parental combinations, the ones the heterozygous parent’s chromosomes carried.

Why: A crossover between two linked genes falls there in only some meioses.
So few gametes carry a recombinant combination.
So the recombinant classes are the two small classes.

7

Here is a table of the fly’s four classes, each marked parental or recombinant, with its count.

A table of the four classes with a column saying parental or recombinant: 421 gray normal parental, 407 black vestigial parental, 46 gray vestigial recombinant, 42 black normal recombinant; footer: recombinant offspring 46 and 42, out of 916 in all
A table of the four classes with a column saying parental or recombinant: 421 gray normal parental, 407 black vestigial parental, 46 gray vestigial recombinant, 42 black normal recombinant; footer: recombinant offspring 46 and 42, out of 916 in all
8

The 46 and 42 came from a crossover between the body-color gene and the wing gene. Each of those 88 offspring got a Gw gamete or a gW gamete.

9

The more often a crossover falls between the two genes, the more recombinant gametes the parent makes. So the share of recombinant offspring shows how often a crossover fell between the two genes.

10

The recombinant offspring written as a percentage of all the offspring is called the , because it says how frequently a gamete carried a recombinant chromosome.

The recombination frequency: the recombinant offspring as a percentage of all the offspring
11
Worked example

Suppose the test cross gave 421 gray normal, 407 black vestigial, 46 gray vestigial and 42 black normal offspring, 916 in all. The gray vestigial and black normal classes are the recombinants. Calculate the recombination frequency between the body-color gene and the wing gene.

Write down the values in the question:
recombinant offspring = 46 + 42 = 88
total offspring = 916
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=88916×100%=9.6%
12

Give a recombination frequency to one decimal place: 9.6 %, not 9.607 %. A few hundred offspring cannot fix the frequency more finely than 0.1 %.

13

What you are expected to know Calculate a recombination frequency from test-cross counts: the recombinant offspring divided by all the offspring, multiplied by 100, to one decimal place, with the unit %.

14
Check q2 numeric entry

Suppose a cross of tomato plants gave these counts. An RrSs plant whose chromosomes carried R with S and r with s was crossed with an rrss plant. The 1,000 offspring were 412 RS, 398 rs, 96 Rs and 94 rS.

Calculate the recombination frequency between the two genes.

Part 1. Add the two recombinant classes, Rs and rS. How many recombinant offspring are there?

Answer: 190  (tolerance ±0)

Working
Add the two recombinant classes:
recombinant offspring=96+94=190

Part 2. Divide the recombinant offspring by the total and multiply by 100. What is the recombination frequency?

Answer: 19 %  (tolerance ±0.05)

Working
Substitute into the equation:
recombination frequency=1901000×100%=19.0%

Answer: 19 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 96 + 94 = 190
total offspring = 1,000
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=1901000×100%=19.0%
15
Check q3 numeric entry

Suppose a cross of maize plants gave these counts. An RrSs plant whose chromosomes carried R with S and r with s was crossed with an rrss plant. The 1,000 kernels were 423 RS, 401 rs, 91 Rs and 85 rS.

Calculate the recombination frequency between the two genes.

Answer: 17.6 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 91 + 85 = 176
total offspring = 1,000
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=1761000×100%=17.6%
16
Check q4 numeric entry

Suppose a cross of insects gave these counts. A BbVv insect whose chromosomes carried B with V and b with v was crossed with a bbvv insect. The 1,000 offspring were 430 BV, 420 bv, 75 Bv and 75 bV.

Calculate the recombination frequency between the two genes.

Answer: 15 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 75 + 75 = 150
total offspring = 1,000
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=1501000×100%=15.0%

17Quick quiz: recombination frequency mixed practice

18
Check q5

A test cross for two linked genes gives four offspring classes.

What is the recombination frequency?

  1. A. ✓ The recombinant offspring as a percentage of all the offspring
  2. B. The parental offspring as a percentage of all the offspring
    The recombinant offspring, not the parental offspring, go on top of the fraction.
  3. C. The recombinant offspring as a percentage of the parental offspring
    All the offspring, not the parental offspring, go underneath the fraction.

Why: The recombination frequency is the recombinant offspring divided by all the offspring, multiplied by 100.
So it is the recombinant offspring as a percentage of all the offspring.

19
Practice writing an answer

A test cross for two linked genes gives four offspring classes, two large and two small.

(a) State what the recombination frequency is. (1 pt)

Model answer The recombination frequency is the recombinant offspring as a percentage of all the offspring.
Rubric
  • Award 1 point for: recombinant offspring ÷ total offspring × 100, or the recombinant offspring as a percentage of all the offspring.

(b) State which offspring go on top of the fraction. (1 pt)

Model answer The recombinant offspring, the two small classes added together, go on top of the fraction.
Rubric
  • Award 1 point for: the recombinant offspring (the two small classes).
20
Check q6 numeric entry

Suppose a test cross for two linked genes in a plant gave 600 offspring, 30 of them recombinants.

Calculate the recombination frequency between the two genes.

Answer: 5 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 30
total offspring = 600
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=30600×100%=5.0%
21
Check q7 numeric entry

Suppose a test cross for two linked genes in an insect gave 800 offspring, 96 of them recombinants.

Calculate the recombination frequency between the two genes.

Answer: 12 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 96
total offspring = 800
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=96800×100%=12.0%
22
Check q8 numeric entry

Suppose a test cross for two linked genes in a plant gave 400 offspring, 44 of them recombinants.

Calculate the recombination frequency between the two genes.

Answer: 11 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 44
total offspring = 400
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=44400×100%=11.0%
23
Check q9 numeric entry

Suppose a test cross for two linked genes in a fish gave 700 offspring, 63 of them recombinants.

Calculate the recombination frequency between the two genes.

Answer: 9 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 63
total offspring = 700
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=63700×100%=9.0%
24
Check q10 numeric entry

Suppose a test cross for two linked genes in a snail gave 1,000 offspring, 210 of them recombinants.

Calculate the recombination frequency between the two genes.

Answer: 21 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 210
total offspring = 1,000
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=2101000×100%=21.0%

25From a frequency to a distance

26

Video: Watch: From a frequency to a distance

The chromosome drawn as a ruler with a scale in map units; 9.6 % becoming 9.6 map units and the wing gene placed on the scale; two chromosomes of equal length with different numbers of crossover marks between their genes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L22Bb.mp4

27

The recombination frequency between the body-color gene and the wing gene is 9.6 %. So how far apart are the two genes?

28

Biologists use the recombination frequency itself as the distance.

29

One percent of recombination is used as one unit of distance along the chromosome. We call it a , because with it each gene can be put at a place on a map of the chromosome.

30

So 1 map unit = 1 % recombination. A recombination frequency of 9.6 % puts the two genes 9.6 map units apart.

The chromosome as a ruler with a scale from 0 to 10 map units; the body-color gene sits at 0 and the wing gene at 9.6, and a dashed line between them is labeled 9.6 map units
The chromosome as a ruler with a scale from 0 to 10 map units; the body-color gene sits at 0 and the wing gene at 9.6, and a dashed line between them is labeled 9.6 map units
31

The frequency carries the unit %. The distance carries the unit map units.

32

Working out the distances between genes from their recombination frequencies is called , because the distances build a map of the genes along the chromosome.

33

A map unit measures how often a crossover falls between two genes, not a count of nucleotides.

34

Some stretches of a chromosome cross over more often than others. So two stretches of equal length in nucleotides can differ in map distance.

Two chromosomes drawn as rulers of equal length, one above the other. On the upper one, two genes A and B sit at the two ends and eight small marks between them show crossover points, labeled 20 map units. On the lower one, two genes C and D sit at the two ends with two marks between them, labeled 5 map units
Two chromosomes drawn as rulers of equal length, one above the other. On the upper one, two genes A and B sit at the two ends and eight small marks between them show crossover points, labeled 20 map units. On the lower one, two genes C and D sit at the two ends with two marks between them, labeled 5 map units
35

A recombination frequency never rises above 50 %.

36

One crossover between two genes swaps pieces on two of the four chromatids. The other two chromatids keep the parental combinations.

37

So even a crossover in every cell leaves half the gametes parental. The frequency stops at 50 %.

38

At 50 % the two parental classes and the two recombinant classes are equal in size. Four equal classes are what two unlinked genes give, so the two genes look unlinked.

39

Here is one simplification. Geneticists treat every recombination frequency below 50 % as a distance in map units.

40

What you are expected to know State the distance between two genes in map units from their recombination frequency, using 1 map unit = 1 % recombination, and say what a map unit measures.

41
Check q11 numeric entry

Suppose a test cross for two linked genes in a plant gave a recombination frequency of 14.0 %.

State the distance between the two genes, in map units.

Answer: 14 map units  (tolerance ±0.05)

Working
Write down the values in the question:
recombination frequency = 14.0 %
Write down the equation:
1map unit=1% recombination
Substitute the values into the equation:
map distance=14.0map units
42
Check q12

Two genes on one chromosome are 8 map units apart. A student says: “So the two genes sit 8 thousand nucleotides apart.”

Is the student correct?

  1. A. Yes: one map unit is one thousand nucleotides
    A map unit is one percent recombination, a count of how often a crossover fell between the two genes.
  2. B. ✓ No: a map unit is not a count of nucleotides

Why: One map unit is one percent recombination.
So 8 map units means 8 of every 100 gametes carried the two genes in a new combination.
Crossovers fall more often on some stretches than others.
So the distance does not count the nucleotides between the genes.

43
Practice writing an answer

Suppose a test cross for two genes in a plant gave a recombination frequency of 50 %.

(a) Explain why the two genes look unlinked in these counts. (1 pt)

Model answer At 50 % recombination, half the offspring carry parental combinations and half carry recombinant combinations.
So the two parental classes and the two recombinant classes are equal in size.
Two genes on different chromosomes also give four equal classes, by independent assortment.
So the counts look the same as the counts from two unlinked genes.
Rubric
  • Award 1 point for: at 50 % the four classes are equal, which is what two unlinked genes give by independent assortment, so the counts look like an unlinked cross.
44

Back to the fly’s chromosome, drawn as a ruler: the body-color gene and the wing gene sit on it, and the test cross gave the four classes 421, 407, 46 and 42.

The chromosome as a ruler with a scale from 0 to 10 map units; the body-color gene sits at 0 and the wing gene at 9.6, and a dashed line between them is labeled 9.6 map units
The chromosome as a ruler with a scale from 0 to 10 map units; the body-color gene sits at 0 and the wing gene at 9.6, and a dashed line between them is labeled 9.6 map units
45

88 of the 916 offspring came from a crossover between the two genes: a recombination frequency of 9.6 %.

46

One map unit is one percent recombination. So the two genes are 9.6 map units apart.

47Quick quiz: map unit, gene mapping mixed practice

48
Check q13

Two genes sit on one chromosome.

What is a map unit?

  1. A. ✓ One percent of recombination between two genes
  2. B. One thousand nucleotides between the two genes
    A map unit is not a count of nucleotides.
    It is one percent recombination.
  3. C. One crossover between the two genes in one meiosis
    A map unit counts how often a crossover fell between the two genes across all the offspring, not one crossover in one cell.

Why: One map unit is one percent of recombination between two genes.
Biologists use it as the unit of distance along the chromosome.

49
Check q14

Two genes sit on one chromosome.

What is gene mapping?

  1. A. Reading the order of nucleotides along one gene
    Gene mapping places genes along a chromosome by their map distances, not by reading their nucleotides.
  2. B. ✓ Finding the distances between genes from recombination frequencies
  3. C. Counting how many genes sit on one chromosome
    Gene mapping gives the distances between genes, not the number of genes.

Why: Gene mapping works out the distances between genes from their recombination frequencies.
The distances build a map of the genes along the chromosome.

50
Practice writing an answer

Two genes sit on one chromosome, and a test cross gives their recombination frequency.

(a) State what a map unit is. (1 pt)

Model answer A map unit is one percent of recombination between two genes, used as a unit of distance along the chromosome.
Rubric
  • Award 1 point for: one percent recombination (1 map unit = 1 % recombination).

(b) State what gene mapping is. (1 pt)

Model answer Gene mapping is working out the distances between genes from their recombination frequencies.
Rubric
  • Award 1 point for: finding the distances between genes (in map units) from recombination frequencies.
51
Check q15 numeric entry

Suppose a test cross for two linked genes in an insect gave a recombination frequency of 6.5 %.

State the distance between the two genes, in map units.

Answer: 6.5 map units  (tolerance ±0.05)

Working
Write down the values in the question:
recombination frequency = 6.5 %
Write down the equation:
1map unit=1% recombination
Substitute the values into the equation:
map distance=6.5map units
52
Check q16 numeric entry

Suppose a test cross for two linked genes in a plant gave a recombination frequency of 23.0 %.

State the distance between the two genes, in map units.

Answer: 23 map units  (tolerance ±0.05)

Working
Write down the values in the question:
recombination frequency = 23.0 %
Write down the equation:
1map unit=1% recombination
Substitute the values into the equation:
map distance=23.0map units
53
Check q17

A test cross for two linked genes gives a recombination frequency of 12.5 %, and a student writes down the distance between the two genes.

Which unit does the distance carry?

  1. A. ✓ map units
  2. B. %
    The frequency carries %.
    The distance carries map units.
  3. C. nucleotides
    A map unit is not a count of nucleotides.

Why: The recombination frequency carries the unit %.
One percent recombination is one map unit.
So the distance carries the unit map units.

54Mixed practice mixed practice

55
Check q18 numeric entry

Suppose a test cross for two linked genes in a plant gave 800 offspring, 64 of them recombinants.

Calculate the map distance between the two genes.

Answer: 8 map units  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring = 64
total offspring = 800
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=64800×100%=8.0%
map distance=8.0map units
56
Check q19

Suppose a test cross for two linked genes in a moth gave 500 offspring: 415 parental combinations and 85 recombinants. A student divides 415 by 500, multiplies by 100 and writes: the map distance is 83.0 map units.

What is wrong with the student’s working?

  1. A. The student should have divided 85 by 415, the recombinants by the parental offspring
    415 is the parental count, not the total.
    With 415 underneath, the fraction is the recombinants as a share of parental offspring only.
  2. B. ✓ The student used the parental count; the recombinant count, 85, goes on top
  3. C. The student should have divided 415 by 85, not by the total
    The recombination frequency is a share of all the offspring, so the total, 500, stays underneath.
    The slip is on top, where 415 sits in place of 85.
  4. D. Nothing; 83.0 map units is the distance between the genes
    83.0 % is the share of parental offspring.
    A map distance measures how often a crossover fell between the genes, and the 85 recombinants are the offspring that show one.

Why: The recombination frequency is the recombinant offspring as a share of all the offspring.
So 85 goes on top and the total, 500, underneath.
The student put the parental count, 415, on top, so 83.0 is the share of parental offspring.
The recombination frequency is 17.0 %: 17.0 map units.

57
Check q20

Two genes on one chromosome are 12 map units apart.

What does the distance of 12 map units measure?

  1. A. The number of nucleotides between the two genes, in thousands
    A map unit is not a count of nucleotides.
    Crossovers fall more often on some stretches than others, so two stretches of equal length can have different map distances.
  2. B. The number of genes that sit between the two
    A map distance does not show how many other genes lie between the two.
  3. C. ✓ The share of offspring in which the two genes came apart: 12 in 100
  4. D. The number of crossovers each chromosome has in one meiosis
    The distance does not count a chromosome’s crossovers.
    It counts how often a crossover fell between these two genes.

Why: One map unit is one percent recombination.
So 12 map units means a recombination frequency of 12 %.
So in 12 of every 100 offspring the two genes came apart.
They came apart because a crossover fell between them in the parent’s meiosis.

58
Check q21 numeric entry

Suppose a cross of beetles gave these counts. An AaBb beetle whose chromosomes carried A with B and a with b was crossed with an aabb beetle. The 1,000 offspring were 251 AB, 255 ab, 250 Ab and 244 aB.

Calculate the recombination frequency between the two genes.

Answer: 49.4 %  (tolerance ±0.05)

Working
Write down the values in the question:
recombinant offspring (Ab and aB) = 250 + 244 = 494
total offspring = 1,000
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=4941000×100%=49.4%
59
Check q22

Suppose a test cross for two genes in a beetle gave a recombination frequency of 49.4 %.

What does a recombination frequency this close to 50 % show about the two genes?

  1. A. ✓ The counts cannot tell these genes from unlinked ones
  2. B. The two genes sit very close together on one chromosome
    Close genes are rarely separated, so they give a small recombination frequency, not one near 50 %.
  3. C. The recombinant offspring outnumber the parental offspring
    49.4 % of the offspring are recombinant, so 50.6 % are parental.
    A crossover makes two recombinant chromatids and leaves two parental ones, so recombinants never outnumber parentals.
  4. D. The two genes sit about 50 nucleotides apart on the chromosome
    A map unit is not a nucleotide.
    One map unit is one percent recombination.

Why: At 50 % recombination, half the offspring are parental and half are recombinant.
So the two parental classes and the two recombinant classes are equal in size.
Four equal classes are what two unlinked genes give by independent assortment.
So the counts cannot say whether the genes are linked.

60
Practice writing an answer

Suppose a cross of maize plants gave these counts. A PpSs plant whose chromosomes carried P with S and p with s was crossed with a ppss plant. The 900 kernels were 402 PS, 388 ps, 58 Ps and 52 pS.

(a) Identify the two recombinant classes, and justify the choice. (1 pt)

Model answer The PpSs parent’s chromosomes carried P with S and p with s.
Neither parental chromosome carried P with s or p with S, so those two combinations are new.
A new combination is recombinant.
Therefore the recombinant classes are Ps and pS, the two small classes, 58 and 52, which a crossover between the two genes made.
Rubric
  • Award 1 point for: Ps and pS named as recombinant, because they carry combinations that neither parental chromosome had.
  • Accept: the same two classes justified as the two small classes, because a crossover between two linked genes is the rarer event.

Slip Naming the two large classes as recombinant. The large classes carry the parental combinations; a crossover between two linked genes is the rarer event.

(b) Calculate the recombination frequency between the two genes. (1 pt)

Answer: 12.2 %  (tolerance ±0.05)

Model answer The recombination frequency is 12.2 %.
Working
Write down the values in the question:
recombinant offspring = 58 + 52 = 110
total offspring = 900
Write down the equation:
recombination frequency=recombinant offspringtotal offspring×100%
Substitute the values into the equation:
recombination frequency=110900×100%=12.2%
Rubric
  • Award 1 point for: 12.2 % (110 recombinants of 900).

(c) Identify the map distance between the two genes, with its unit. (1 pt)

Model answer The two genes are 12.2 map units apart, because one map unit is one percent recombination.
Rubric
  • Award 1 point for: 12.2 map units.

Slip Giving the distance in percent, or as 12.2 with no unit. The frequency carries %; the distance carries map units.

(d) Explain why the two recombinant classes are so much smaller than the two parental classes. (1 pt)

Model answer A recombinant kernel needs a crossover to fall between the two genes in the parent’s prophase I.
A crossover falls on that stretch of the chromosome in only some of the cells going through meiosis.
So most gametes carry the parental combinations, P with S or p with s.
Only a few gametes carry a recombinant combination.
Therefore the two recombinant classes are much smaller than the two parental classes.
Rubric
  • Award 1 point for: recombinants need a crossover between the two genes, which happens in only some meioses, so most gametes keep the parental combinations.

Slip Saying the recombinant kernels survive less well. The classes differ in how often each gamete forms, not in survival.

Glossary

recombination frequency
The recombinant offspring of a test cross as a percentage of all the offspring: recombinants ÷ total × 100 %.
map unit
One unit of distance along a chromosome, equal to one percent recombination between two genes; 9.6 % recombination puts two genes 9.6 map units apart.
gene mapping
Working out the distances between genes on a chromosome, in map units, from their recombination frequencies.

APBIO-U05-L23 Putting genes in order

Topic 5.4 · Non-Mendelian Genetics · 52 steps

A chromosome drawn as a horizontal ruler with tick marks and no genes on it; above it three boxes labeled flower color, pod color and seed shape wait to be placed; to the right the three distances: flower color to seed shape 36 %, flower color to pod color 24 %, pod color to seed shape 12 %
A chromosome drawn as a horizontal ruler with tick marks and no genes on it; above it three boxes labeled flower color, pod color and seed shape wait to be placed; to the right the three distances: flower color to seed shape 36 %, flower color to pod color 24 %, pod color to seed shape 12 %

Here is a drawing of a chromosome as a ruler. Beside it wait three genes that sit somewhere on it: flower color, pod color and seed shape. They are drawn off the ruler because you do not yet know their order.

Suppose three crosses told you how often a crossover separates each pair of genes: flower color and seed shape in 36 % of gametes, flower color and pod color in 24 %, pod color and seed shape in 12 %. Which gene sits where?

Unit 5 · Heredity

1Farther apart, more often separated

2

Video: Watch: Farther apart, more often separated

A paired homolog pair with dashed marks along it where a crossover might fall; two genes close together catch one mark, two genes far apart catch five.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L23a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L23a.mp4

3

Why does one pair of genes come apart in 36 % of gametes and another pair in only 12 %? The answer lies in where a crossover can fall.

4
Check q1

In prophase I, the two homologs of a pair lie side by side.

What happens in a crossover?

  1. A. ✓ A chromatid of each homolog breaks at the same point, and the two swap pieces
  2. B. The two homologs of the pair separate and move to opposite poles of the cell
    The homologs move apart in anaphase I.
    A crossover happens earlier, while the homologs are still paired.
  3. C. The two sister chromatids of one homolog separate and move to opposite poles
    Sister chromatids separate in anaphase II.
    A crossover exchanges pieces between the two homologs, not between sisters.

Why: In a crossover, a chromatid of one homolog and a chromatid of the other break at the same point.
The two chromatids exchange the broken pieces.
So each of those chromatids now carries alleles from both homologs.

5
Check q2

Two genes on one chromosome are 12 map units apart.

In what percentage of gametes does a crossover separate the two genes?

  1. A. 6 %
    1 map unit is 1 % recombination.
    So 12 map units is 12 % of gametes, not 6 %.
  2. B. ✓ 12 %
  3. C. 88 %
    88 % is the share of gametes that keep the two genes together.
    The map distance counts the gametes in which they were separated.

Why: 1 map unit is 1 % recombination.
So two genes 12 map units apart are separated by a crossover in 12 % of gametes.

6

Here is a homologous pair in prophase I, lying side by side. The break of a crossover can fall at any point along it.

A homologous pair drawn as two rods lying side by side, with eight dashed marks spaced along them: the points where a crossover might fall
A homologous pair drawn as two rods lying side by side, with eight dashed marks spaced along them: the points where a crossover might fall
7

Two genes come apart only when the break falls between them.

8

Here is a drawing of two rulers, each with two genes on it.

Two rulers. On the upper ruler two genes, A and B, sit close together and one dashed crossover mark lies between them out of eight along the ruler. On the lower ruler two genes, C and D, sit far apart and five of the eight marks lie between them
Two rulers. On the upper ruler two genes, A and B, sit close together and one dashed crossover mark lies between them out of eight along the ruler. On the lower ruler two genes, C and D, sit far apart and five of the eight marks lie between them
9

On the upper ruler, genes A and B sit close together. On the lower ruler, genes C and D sit far apart.

10

Between A and B lies 1 of the 8 possible break points. Between C and D lie 5 of the 8.

11

So a break falls between C and D five times as often as between A and B.

12

The longer the stretch between two genes, the more of the possible break points lie between them.

13

So the farther apart two genes sit, the more often a crossover separates them. So their recombination frequency is higher.

14

Two genes 2 map units apart are separated in 2 % of gametes: they are almost always inherited together. Two genes 40 map units apart are separated in 40 % of gametes.

15

Close genes are not stuck to each other. Little room lies between them, so a break rarely falls there.

16

What you are expected to know Explain why two genes farther apart on a chromosome show a higher recombination frequency: a crossover can fall anywhere along the paired homologs, and a longer stretch between two genes catches more of those breaks.

17
Check q3

On one chromosome of a plant, genes D and E sit 5 map units apart and genes M and N sit 30 map units apart.

Which pair of genes stays together in a gamete more often?

  1. A. ✓ D and E
  2. B. M and N
    A crossover separates two genes only when its break falls between them.
    A break falls in 30 map units far more often than in 5.
  3. C. The two pairs equally often
    A longer stretch between two genes catches more breaks.
    So the two pairs do not come apart equally often.

Why: A crossover separates two genes only when the break falls between them.
D and E have 5 map units between them; M and N have 30.
So a break lands between M and N six times as often.
So D and E stay together far more often.

18
Practice writing an answer

On one chromosome of a plant, genes D and E sit 5 map units apart and genes M and N sit 30 map units apart. D and E stay together in a gamete more often than M and N.

(a) Explain why D and E stay together in a gamete more often than M and N. (1 pt)

Frame A crossover separates two genes only when …

Model answer A crossover separates two genes only when its break falls between them.
The break can fall at any point along the paired homologs.
Only 5 map units lie between D and E, and 30 map units lie between M and N.
So a break falls between M and N about six times as often as between D and E.
So D and E are separated less often, and they stay together in a gamete more often.
Rubric
  • Award 1 point for: a crossover separates two genes only when its break falls between them, and a shorter stretch between two genes catches a break less often.
19
Check q4

Two genes on a chromosome are 3 map units apart and are inherited together in 97 % of gametes. A student says: “The two genes stay together so often because they are fastened to each other and always travel together.”

Is the student correct?

  1. A. Yes
    The 3 % of gametes that carry the two genes apart show that a break can fall between them.
  2. B. ✓ No

Why: Close genes are not fastened together.
Only 3 map units lie between these two genes.
So a break lands there in only 3 gametes of 100.
The genes are close, not stuck.

20
Check q5

Two genes on a chromosome are 3 map units apart and are inherited together in 97 % of gametes.

Why do the two genes stay together so often?

  1. A. Crossing over does not happen on that chromosome
    Crossing over happens along the whole chromosome.
    The short stretch between these two genes rarely catches a break.
  2. B. The two genes are copies of one gene
    Two genes 3 map units apart are two different genes at two positions.
    The distance between them is small, not zero.
  3. C. ✓ Little room lies between them, so a break rarely falls there

Why: A crossover separates two genes only when its break falls between them.
Only 3 map units lie between these two genes.
So a break lands there in only 3 gametes of 100.

21
Check q6

Suppose two genes sit at opposite ends of a long chromosome of a fish.

Among the fish's gametes, how common are recombinant combinations of the two genes?

  1. A. Rare
    A long stretch between two genes catches many of the possible breaks.
    So recombinant gametes are common, not rare.
  2. B. ✓ Common

Why: A crossover separates two genes only when its break falls between them.
The whole length of the chromosome lies between these two genes.
So a break falls between them often, and recombinant combinations are common: up to half of the gametes.

22Quick quiz: which pair stays together? mixed practice

23
Check q7

Here is a table of two gene pairs on one chromosome of a moth and the map distance within each pair.

A table of two gene pairs and their map distances: P and Q 6 map units apart, R and S 25 map units apart
A table of two gene pairs and their map distances: P and Q 6 map units apart, R and S 25 map units apart

Which pair stays together in a gamete more often?

  1. A. ✓ P and Q
  2. B. R and S
    A break falls in 25 map units more often than in 6.
    So R and S come apart more often, and P and Q stay together more often.

Why: The fewer map units between two genes, the less often a break falls between them.
P and Q are 6 map units apart, R and S 25.
So P and Q stay together more often.

24
Check q8

Here is a table of two gene pairs on one chromosome of a fungus and the map distance within each pair.

A table of two gene pairs and their map distances: J and K 30 map units apart, L and M 3 map units apart
A table of two gene pairs and their map distances: J and K 30 map units apart, L and M 3 map units apart

Which pair stays together in a gamete more often?

  1. A. J and K
    A break falls in 30 map units more often than in 3.
    So J and K come apart more often, and L and M stay together more often.
  2. B. ✓ L and M

Why: The fewer map units between two genes, the less often a break falls between them.
L and M are 3 map units apart, J and K 30.
So L and M stay together more often.

25
Check q9

Here is a table of two gene pairs on one chromosome of a plant and the map distance within each pair.

A table of two gene pairs and their map distances: A and B 18 map units apart, C and D 12 map units apart
A table of two gene pairs and their map distances: A and B 18 map units apart, C and D 12 map units apart

Which pair stays together in a gamete more often?

  1. A. A and B
    A break falls in 18 map units more often than in 12.
    So A and B come apart more often, and C and D stay together more often.
  2. B. ✓ C and D

Why: The fewer map units between two genes, the less often a break falls between them.
C and D are 12 map units apart, A and B 18.
So C and D stay together more often.

26
Check q10

Here is a table of two gene pairs on one chromosome of an insect and the map distance within each pair.

A table of two gene pairs and their map distances: T and U 1 map unit apart, V and W 45 map units apart
A table of two gene pairs and their map distances: T and U 1 map unit apart, V and W 45 map units apart

Which pair stays together in a gamete more often?

  1. A. ✓ T and U
  2. B. V and W
    A break falls in 45 map units far more often than in 1.
    So V and W come apart often, and T and U stay together almost always.

Why: The fewer map units between two genes, the less often a break falls between them.
T and U are 1 map unit apart, V and W 45.
So T and U stay together more often.

27
Check q11

Here is a table of two gene pairs on one chromosome of a fish and the map distance within each pair.

A table of two gene pairs and their map distances: G and H 27 map units apart, K and L 22 map units apart
A table of two gene pairs and their map distances: G and H 27 map units apart, K and L 22 map units apart

Which pair stays together in a gamete more often?

  1. A. G and H
    A break falls in 27 map units more often than in 22.
    So G and H come apart more often, and K and L stay together more often.
  2. B. ✓ K and L

Why: The fewer map units between two genes, the less often a break falls between them.
K and L are 22 map units apart, G and H 27.
So K and L stay together more often.

28The largest distance spans the other two

29

Video: Watch: The largest distance spans the other two

The chromosome as a ruler; the two genes with the largest distance placed at the ends; the third gene slid in until both of its distances fit; the two smaller distances added to check.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L23b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L23b.mp4

30

Now consider three genes on one chromosome: flower color, pod color and seed shape, with three distances between them. Which two are the ends, and which sits between?

A ruler with no genes placed on it; beneath it three distances: flower color to seed shape 36 map units, flower color to pod color 24 map units, pod color to seed shape 12 map units
A ruler with no genes placed on it; beneath it three distances: flower color to seed shape 36 map units, flower color to pod color 24 map units, pod color to seed shape 12 map units
31

The two genes farthest apart come apart most often. So the largest of the three distances belongs to the two genes at the ends.

32

Flower color and seed shape are 36 map units apart, the largest of the three distances. So flower color and seed shape go at the two ends of the ruler.

A ruler with the flower-color gene placed at the left end and the seed-shape gene at the right end; a dashed bar above them is labeled 36 map units; the middle of the ruler is empty
A ruler with the flower-color gene placed at the left end and the seed-shape gene at the right end; a dashed bar above them is labeled 36 map units; the middle of the ruler is empty
33

Pod color must sit 24 map units from flower color and 12 map units from seed shape.

34

Only one place does both: between them, nearer to seed shape.

The same ruler with the pod-color gene placed between the other two, 24 map units from flower color and 12 map units from seed shape; three dashed bars above show 36, 24 and 12
The same ruler with the pod-color gene placed between the other two, 24 map units from flower color and 12 map units from seed shape; three dashed bars above show 36, 24 and 12
35

The check is that the two smaller distances add up to the largest: 24+12=36map units. When they do, the middle gene is in place.

36

So the rule for three linked genes: the two genes with the largest distance are the ends, and the third gene sits between them.

37
Worked example

Three linked genes are written A, B and C. Their map distances are A to B 12 map units, B to C 8 map units, and A to C 20 map units. Put the three genes in order.

Write down the three distances:
A to B = 12 map units
B to C = 8 map units
A to C = 20 map units
Find the largest distance; its two genes are the ends:
largest = A to C, 20 map units
so A and C are the two ends
Place the third gene between the ends:
B sits between A and C: A, B, C
Check that the two smaller distances add up to the largest:
12+8=20map units
38

Here is the drawing of A, B and C in place: A and C at the ends, B between them.

A ruler with gene A at the left end, gene C at the right end and gene B between them; dashed bars read 12 map units from A to B, 8 from B to C and 20 from A to C
A ruler with gene A at the left end, gene C at the right end and gene B between them; dashed bars read 12 map units from A to B, 8 from B to C and 20 from A to C
39

What you are expected to know Order three linked genes from their three pairwise map distances: the two genes with the largest distance are the ends, and the third gene sits between them.

40
Check q12

Three linked genes in a moth are written P, Q and R. Their map distances are P to Q 15 map units, Q to R 9 map units, and P to R 24 map units.

Which two genes are the two ends?

  1. A. P and Q
    The two ends are the two genes farthest apart.
    P to Q, 15 map units, is not the largest distance.
  2. B. Q and R
    The two ends are the two genes farthest apart.
    Q to R, 9 map units, is the smallest distance.
  3. C. ✓ P and R

Why: The two ends are the two genes farthest apart.
The largest of the three distances is P to R, 24 map units.
So P and R are the two ends.

41
Check q13

Three linked genes in a moth are written P, Q and R: P to Q 15 map units, Q to R 9 map units, and P to R 24 map units.

Which gene sits between the other two?

  1. A. P
    P is one of the two ends.
    The gene between the ends is the third gene.
  2. B. ✓ Q
  3. C. R
    R is one of the two ends.
    The gene between the ends is the third gene.

Why: The largest distance, P to R at 24 map units, marks P and R as the two ends.
So the third gene, Q, sits between them.
The order is P, Q, R.

42
Check q14

Three linked genes in a fungus are written T, G and Z. Their map distances are T to G 7 map units, G to Z 18 map units, and T to Z 25 map units.

In what order do the three genes sit along the chromosome?

  1. A. G, T, Z
    T and Z have the largest distance, 25 map units.
    So T and Z are the two ends, and T cannot sit between.
  2. B. T, Z, G
    Z is one of the two ends.
    T to Z, 25 map units, is the largest distance.
  3. C. The order cannot be found from three distances
    Three pairwise distances fix the order.
    The largest distance marks the two ends, and the third gene sits between them.
  4. D. ✓ T, G, Z

Why: The largest distance, T to Z at 25 map units, marks T and Z as the two ends.
G sits 7 map units from T and 18 from Z.
So G is between the ends: T, G, Z.

43
Check q15

Three linked genes in a plant: flower color to seed shape is 36 map units, flower color to pod color is 24, and pod color to seed shape is 12. A student draws the order below, with flower color and seed shape side by side.

A ruler with the flower-color gene and the seed-shape gene placed side by side near the left end, and the pod-color gene far to the right
A ruler with the flower-color gene and the seed-shape gene placed side by side near the left end, and the pod-color gene far to the right

What is wrong with the student's drawing?

  1. A. ✓ Flower color and seed shape belong at opposite ends
  2. B. Nothing is wrong with the student’s drawing
    A large map distance means the two genes came apart often.
    Two genes come apart often when they are far apart, not close.
  3. C. The ruler should be read from right to left
    The direction of the ruler does not matter, but the positions do.
    Flower color and seed shape are drawn side by side, and they should be farthest apart.
  4. D. Pod color should sit at the left end of the ruler
    The smallest distance, 12 map units, marks the two genes closest together, pod color and seed shape.
    A small distance does not put a gene at an end.

Why: The largest distance, 36 map units, belongs to the two genes farthest apart.
So flower color and seed shape are the two ends.
Pod color sits between them, 24 from flower color and 12 from seed shape.

44

Here again is the ruler with its three genes and three distances: 36, 24 and 12 map units.

The same ruler with the pod-color gene placed between the other two, 24 map units from flower color and 12 map units from seed shape; three dashed bars above show 36, 24 and 12
The same ruler with the pod-color gene placed between the other two, 24 map units from flower color and 12 map units from seed shape; three dashed bars above show 36, 24 and 12
45

The largest distance, 36 map units, puts flower color and seed shape at the two ends. Pod color sits between them, 24 map units from flower color and 12 from seed shape.

46Quick quiz: which gene sits between? mixed practice

47
Check q16

Here is a table of the map distances between three linked genes of a fungus.

A table of three gene pairs and their map distances: J to K 14, K to L 6, J to L 20
A table of three gene pairs and their map distances: J to K 14, K to L 6, J to L 20

Which gene sits between the other two?

  1. A. J
    J is one of the two ends.
    The largest distance, J to L, marks J and L as the ends.
  2. B. ✓ K
  3. C. L
    L is one of the two ends.
    The largest distance, J to L, marks J and L as the ends.

Why: The largest distance is J to L, 20 map units, so J and L are the ends.
K sits between them.

48
Check q17

Here is a table of the map distances between three linked genes of a moth.

A table of three gene pairs and their map distances: A to B 27, A to C 5, B to C 22
A table of three gene pairs and their map distances: A to B 27, A to C 5, B to C 22

Which gene sits between the other two?

  1. A. A
    A is one of the two ends.
    The largest distance, A to B, marks A and B as the ends.
  2. B. B
    B is one of the two ends.
    The largest distance, A to B, marks A and B as the ends.
  3. C. ✓ C

Why: The largest distance is A to B, 27 map units, so A and B are the ends.
C sits between them.

49
Check q18

Here is a table of the map distances between three linked genes of a plant.

A table of three gene pairs and their map distances: M to N 11, M to O 30, N to O 19
A table of three gene pairs and their map distances: M to N 11, M to O 30, N to O 19

Which gene sits between the other two?

  1. A. M
    M is one of the two ends.
    The largest distance, M to O, marks M and O as the ends.
  2. B. ✓ N
  3. C. O
    O is one of the two ends.
    The largest distance, M to O, marks M and O as the ends.

Why: The largest distance is M to O, 30 map units, so M and O are the ends.
N sits between them.

50
Check q19

Here is a table of the map distances between three linked genes of an insect.

A table of three gene pairs and their map distances: X to Y 13, X to Z 9, Y to Z 4
A table of three gene pairs and their map distances: X to Y 13, X to Z 9, Y to Z 4

Which gene sits between the other two?

  1. A. X
    X is one of the two ends.
    The largest distance, X to Y, marks X and Y as the ends.
  2. B. Y
    Y is one of the two ends.
    The largest distance, X to Y, marks X and Y as the ends.
  3. C. ✓ Z

Why: The largest distance is X to Y, 13 map units, so X and Y are the ends.
Z sits between them.

51
Check q20

Here is a table of the map distances between three linked genes of a fish.

A table of three gene pairs and their map distances: D to E 16, E to F 17, D to F 33
A table of three gene pairs and their map distances: D to E 16, E to F 17, D to F 33

Which gene sits between the other two?

  1. A. D
    D is one of the two ends.
    The largest distance, D to F, marks D and F as the ends.
  2. B. ✓ E
  3. C. F
    F is one of the two ends.
    The largest distance, D to F, marks D and F as the ends.

Why: The largest distance is D to F, 33 map units, so D and F are the ends.
E sits between them.

APBIO-U05-L23B Linked, or just chance?

Topic 5.4 · Non-Mendelian Genetics · 37 steps

Four bars on a baseline, 421, 407, 46 and 42, with a dashed line drawn across all four at the height of 229, labeled 229 predicted for every class
Four bars on a baseline, 421, 407, 46 and 42, with a dashed line drawn across all four at the height of 229, labeled 229 predicted for every class

Suppose a cross of fruit flies gave these four classes of offspring: 421 gray-bodied with normal wings, 407 black-bodied with vestigial wings, 46 gray with vestigial wings and 42 black with normal wings. One parent was heterozygous for both genes; the other was black-bodied and vestigial-winged.

Here are the four counts as bars. The dashed line across them sits at 229, where independent assortment would put every class. A small batch strays from its ratio by chance. Is this linkage, or could chance stray this far from 1 : 1 : 1 : 1?

Unit 5 · Heredity

1Put a number on the gap

2

Video: Watch: Linked, or just chance?

The chi-square test on the four fly counts: the null hypothesis written, the expected count 229 for every class, one class at a time added to the sum, 598 read against 7.81 in the table, the verdict written.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L23Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L23Ba.mp4

3

How do you tell a chance gap from real linkage? You test the counts against 1 : 1 : 1 : 1 with the chi-square test.

4

Here is a table comparing each class's count with the count independent assortment predicts.

A table of the four fly classes with the observed count and the count independent assortment predicts: gray normal 421 against 229, black vestigial 407 against 229, gray vestigial 46 against 229, black normal 42 against 229
A table of the four fly classes with the observed count and the count independent assortment predicts: gray normal 421 against 229, black vestigial 407 against 229, gray vestigial 46 against 229, black normal 42 against 229
5

Two things could explain the gap. The first: each fertilization is its own event, so a batch strays from its ratio by chance.

6

The second: the two genes are linked, so the parental combinations really are more common.

7
Check q1

A breeder tests offspring counts against the ratio that independent assortment predicts.

What does the null hypothesis say?

  1. A. The offspring differ from the predicted ratio for a real reason, not by chance
    A claim of a real difference from the predicted ratio is the alternative hypothesis.
    The null hypothesis says there is no real difference.
  2. B. The offspring occur in exactly the counts the predicted ratio gives, with no gap at all
    The null hypothesis expects chance to make the counts stray from the ratio.
    It never promises exact counts.
  3. C. ✓ The offspring fit the predicted ratio, and any difference from it is due to chance

Why: The null hypothesis is the no-difference statement.
For a breeding cross it says the offspring fit the predicted ratio, and any gap between the counts and that ratio is due to chance.

8

Here the model is independent assortment, 1 : 1 : 1 : 1. So the null hypothesis reads: the offspring occur in a 1 : 1 : 1 : 1 ratio, and any difference between the counts and that ratio is due to chance.

9
Check q2

Independent assortment predicts a ratio for the four offspring classes, and the offspring are counted.

How is the expected count of one class found?

  1. A. The observed count of that class times its share of the ratio
    The expected count comes from the predicted ratio, not from what was counted.
    The class's share of the ratio is applied to the total.
  2. B. ✓ The total times that class's share of the ratio
  3. C. The total divided by the number of offspring counted
    The total divided by itself is 1.
    The expected count is the total times the class's share of the ratio.

Why: The expected count, e, is the count the ratio predicts for the total.
So e is the total times that class's share of the ratio.

10

Each class's share of 1 : 1 : 1 : 1 is 1 of 4, so every class expects the same count: e=916×14=229.

11
Check q3

Chi-square puts one number on the gap between the observed counts, o, and the expected counts, e.

Which of the following is the chi-square equation on your formula sheet?

  1. A. χ2=∑(o−e)e
    Without the square, the gaps above and below the expected count cancel each other.
    Each gap is squared first.
  2. B. χ2=∑(o−e)2o
    Each squared gap is divided by that class's expected count, e, not by its observed count.
  3. C. ✓ χ2=∑(o−e)2e

Why: For each class the gap between the observed and the expected count is squared and divided by the expected count.
The classes are added: χ2=∑(o−e)2e.

12
Check q4

The offspring of one breeding cross are sorted into four classes.

How many degrees of freedom does the chi-square test have?

  1. A. ✓ 3
  2. B. 4
    Degrees of freedom are the number of classes minus one.
    Four classes give 3.
  3. C. 5
    Degrees of freedom are the number of classes minus one, not plus one.
    Four classes give 3.

Why: Degrees of freedom are the number of classes minus one.
Four classes give 3.

13

Here is a table: the chi-square table your formula sheet prints. Each column is a number of degrees of freedom; each row is labeled by a p value.

The chi-square table from the formula sheet with the cell for three degrees of freedom in the p = 0.05 row ringed: 7.81
The chi-square table from the formula sheet with the cell for three degrees of freedom in the p = 0.05 row ringed: 7.81
14

Down the column for 3 degrees of freedom and along the p = 0.05 row sits 7.81, the critical value.

15
Check q5

For one set of counts, chi-square comes out larger than the critical value from the formula sheet.

What is the verdict on the null hypothesis?

  1. A. ✓ Reject
  2. B. Fail to reject
    A chi-square below the critical value would leave the gap small enough to be chance.
    This chi-square is above it.

Why: A chi-square larger than the critical value means the gap between the counts and the predicted ratio is too large to be chance.
So the null hypothesis is rejected.

16
Worked example

The cross gave 421 gray normal, 407 black vestigial, 46 gray vestigial and 42 black normal offspring, 916 in all. Independent assortment predicts 1 : 1 : 1 : 1. Test the counts against that model and state the verdict.

Write down the values in the question:
total = 916
ratio = 1 : 1 : 1 : 1, so 4 shares in all, each class 1 of 4
o = 421, 407, 46, 42
State the null hypothesis:
The offspring occur in a 1 : 1 : 1 : 1 ratio, and any difference between the counts and that ratio is due to chance.
Write down the equation for the expected counts:
e=total×that class's share of the ratio
Substitute the values into the equation:
e=916×14=229 for every class
Write down the chi-square equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
gray normal: (421−229)2229=36864229=160.98
black vestigial: (407−229)2229=31684229=138.36
gray vestigial: (46−229)2229=33489229=146.24
black normal: (42−229)2229=34969229=152.70
Add the classes:
χ2=160.98+138.36+146.24+152.70=598.28≈598(no unit)
Find the degrees of freedom and read the critical value from the table:
degrees of freedom=4−1=3
critical value (3 degrees of freedom, p = 0.05) = 7.81
Compare and state the verdict:
598>7.81
Reject the null hypothesis: the counts do not fit 1 : 1 : 1 : 1.
17

Chi-square came out at 598, far above the critical value of 7.81. So the null hypothesis is rejected: the counts do not fit independent assortment.

18

Rejecting 1 : 1 : 1 : 1 says one thing: the two genes are not assorting independently.

19

The two large classes are the parental combinations, and the two small classes are the recombinants. That pattern is linkage.

20

Chi-square does not say how tightly the two genes are linked. The map distance says that: 9.6 map units here.

21

When two classes are ten times the size of the other two, the verdict is plain before the arithmetic. The test puts a number on it.

22

What you are expected to know Judge from dihybrid test-cross counts whether two genes assort independently or are linked, by testing the counts against 1 : 1 : 1 : 1 with chi-square and stating the verdict.

23
Check q6 numeric entry

Suppose a tomato plant heterozygous for a fruit-color gene and a skin gene is crossed with a plant carrying only the recessive alleles of both genes. If the two genes assort independently, the offspring should come 1 : 1 : 1 : 1. The 1,000 offspring are counted in the table below.

A table of four tomato classes with observed counts: red smooth 412, yellow wrinkled 398, red wrinkled 96, yellow smooth 94; total 1,000
A table of four tomato classes with observed counts: red smooth 412, yellow wrinkled 398, red wrinkled 96, yellow smooth 94; total 1,000

Calculate chi-square for the four classes, to three significant figures.

Part 1. Each class is 1 of the 4 shares of 1 : 1 : 1 : 1. What is the expected count of each class?

Answer: 250  (tolerance ±0)

Working
Multiply the total by the class's share:
e=1000×14=250 for every class

Part 2. Work out the red smooth class's term, the squared gap divided by its expected count, to two decimal places. What is the red smooth term?

Answer: 104.98  (tolerance ±0.006)

Working
Square the gap and divide by the class's own expected count:
red smooth: (412−250)2250=26244250=104.98

Part 3. Work out the yellow wrinkled class's term, to two decimal places. What is the yellow wrinkled term?

Answer: 87.62  (tolerance ±0.006)

Working
Square the gap and divide by the class's own expected count:
yellow wrinkled: (398−250)2250=21904250=87.62

Part 4. Work out the red wrinkled class's term, to two decimal places. What is the red wrinkled term?

Answer: 94.86  (tolerance ±0.006)

Working
Square the gap and divide by the class's own expected count:
red wrinkled: (96−250)2250=23716250=94.86

Answer: 385  (tolerance ±0.5)

Working
Write down the values in the question:
total = 1,000
ratio = 1 : 1 : 1 : 1, each class 1 of 4
o = 412, 398, 96, 94
Write down the equation for the expected counts:
e=total×that class's share of the ratio
Substitute the values into the equation:
e=1000×14=250 for every class
Write down the chi-square equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
red smooth: (412−250)2250=104.98
yellow wrinkled: (398−250)2250=87.62
red wrinkled: (96−250)2250=94.86
yellow smooth: (94−250)2250=97.34
Add the classes:
χ2=104.98+87.62+94.86+97.34=384.80≈385
24
Check q7 numeric entry

Suppose an AaBb beetle is crossed with an aabb beetle. If the two genes assort independently, the offspring should come 1 : 1 : 1 : 1. The 1,000 offspring are 250 AB, 244 ab, 251 Ab and 255 aB.

Calculate chi-square for the four classes, to three significant figures.

Answer: 0.248  (tolerance ±0.0006)

Working
Write down the values in the question:
total = 1,000
ratio = 1 : 1 : 1 : 1, each class 1 of 4
o = 250, 244, 251, 255
Write down the equation for the expected counts:
e=total×that class's share of the ratio
Substitute the values into the equation:
e=1000×14=250 for every class
Write down the chi-square equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
AB: (250−250)2250=0
ab: (244−250)2250=36250=0.144
Ab: (251−250)2250=1250=0.004
aB: (255−250)2250=25250=0.100
Add the classes:
χ2=0+0.144+0.004+0.100=0.248
25
Check q8

Suppose a second beetle cross, an AaBb beetle with an aabb beetle, gave four classes of about 250 each among 1,000 offspring, and chi-square against 1 : 1 : 1 : 1 came out at 0.42. The table below is the formula sheet's.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01

What is the verdict on the null hypothesis?

  1. A. Reject
    The null hypothesis is rejected only when chi-square is larger than the critical value.
    0.42 is below 7.81.
  2. B. ✓ Fail to reject

Why: Four classes give three degrees of freedom, and the 0.05 row's critical value is 7.81.
Chi-square, 0.42, is far below it.
So the null hypothesis is not rejected: the counts are consistent with the two genes assorting independently.

26
Practice writing an answer

Suppose a second beetle cross, an AaBb beetle with an aabb beetle, gave four classes of about 250 each among 1,000 offspring, and chi-square against 1 : 1 : 1 : 1 came out at 0.42. The table below is the formula sheet's. The verdict is to fail to reject the null hypothesis.

The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01
The chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01

(a) Describe what the verdict says about the two genes. (1 pt)

Model answer The counts are consistent with 1 : 1 : 1 : 1, so they are consistent with the two genes assorting independently.
The data give no evidence that the two genes are linked.
Rubric
  • Award 1 point for: the counts are consistent with independent assortment (no evidence of linkage).
  • Accept with or without: the verdict does not prove the genes are on different chromosomes.
  • Do not award: the genes are proved unlinked, or the null hypothesis is accepted.

(b) Explain how comparing 0.42 with the critical value at p = 0.05 gives this verdict. (1 pt)

Model answer Four classes give three degrees of freedom.
The 0.05 row at three degrees of freedom gives a critical value of 7.81.
Chi-square, 0.42, is smaller than 7.81.
So the gap between the counts and 1 : 1 : 1 : 1 is small enough to be chance, and the null hypothesis is not rejected.
Rubric
  • Award 1 point for: 0.42 is below the critical value 7.81 (three degrees of freedom, p = 0.05), so the gap could be chance and the null hypothesis is not rejected.
27
Check q9

Suppose a dihybrid test cross of beetles gave four classes whose chi-square against 1 : 1 : 1 : 1 came out at 312, far above 7.81. A student says: “312 is enormous, so the two genes must sit very close together on the chromosome.”

Is the student correct?

  1. A. Yes
    Chi-square measures the gap between the counts and 1 : 1 : 1 : 1.
    It does not measure the distance between the genes.
  2. B. ✓ No

Why: Rejecting 1 : 1 : 1 : 1 says the two genes are not assorting independently.
How close together they sit is the map distance, worked out from the recombinant share.
Chi-square does not give that distance.

28
Check q10

Suppose a dihybrid test cross of beetles gave four classes whose chi-square against 1 : 1 : 1 : 1 came out at 312, and the null hypothesis was rejected.

Which result says how tightly the two genes are linked?

  1. A. The chi-square value, 312
    Chi-square measures the gap between the counts and the predicted ratio.
    A larger chi-square means a larger gap, not a shorter distance.
  2. B. The critical value, 7.81
    The critical value comes from the table.
    It sets the line chi-square must pass; it describes no property of the genes.
  3. C. ✓ The map distance, in map units

Why: The map distance is the recombinant share of the offspring, in map units.
So the map distance says how tightly the two genes are linked.
Chi-square says only that they are not assorting independently.

29

Here again is the table of the four counts against the 229 that independent assortment predicts.

A table of the four fly classes with the observed count and the count independent assortment predicts: gray normal 421 against 229, black vestigial 407 against 229, gray vestigial 46 against 229, black normal 42 against 229
A table of the four fly classes with the observed count and the count independent assortment predicts: gray normal 421 against 229, black vestigial 407 against 229, gray vestigial 46 against 229, black normal 42 against 229
30

Tested against 1 : 1 : 1 : 1, the fly counts gave a chi-square of 598, far above the critical value of 7.81. So “linked” is a verdict, not a guess.

31Mixed practice mixed practice

32
Check q11 numeric entry

Suppose a plant heterozygous for two genes, TtGg, is crossed with a ttgg plant. If the two genes assort independently, the offspring should come 1 : 1 : 1 : 1. The 1,000 offspring are 300 tall green, 290 short yellow, 210 tall yellow and 200 short green.

Calculate chi-square for the four classes, to three significant figures.

Answer: 32.8  (tolerance ±0.05)

Working
Write down the values in the question:
total = 1,000
ratio = 1 : 1 : 1 : 1, each class 1 of 4
o = 300, 290, 210, 200
Write down the equation for the expected counts:
e=total×that class's share of the ratio
Substitute the values into the equation:
e=1000×14=250 for every class
Write down the chi-square equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
tall green: (300−250)2250=10.0
short yellow: (290−250)2250=6.4
tall yellow: (210−250)2250=6.4
short green: (200−250)2250=10.0
Add the classes:
χ2=10.0+6.4+6.4+10.0=32.8
33
Check q12

A breeder tests 1,000 offspring of a dihybrid test cross, sorted into four classes, against 1 : 1 : 1 : 1.

How many degrees of freedom does the test have?

  1. A. 1
    The number of genes does not set the degrees of freedom.
    The number of classes does, and there are four.
  2. B. ✓ 3
  3. C. 4
    Degrees of freedom are the number of classes minus one.
    Once three counts and the total are fixed, the fourth count is fixed too.
  4. D. 999
    Degrees of freedom count classes, never offspring.
    A thousand offspring in four classes give 3, and so would ten thousand.

Why: Degrees of freedom are the number of classes minus one.
Four classes give 3, whatever the number of offspring or genes.

34
Check q13

For a dihybrid test cross, chi-square against 1 : 1 : 1 : 1 comes out at 2.9, smaller than the critical value for the test. A student writes: “We accept the null hypothesis, so the two genes are proved to be on different chromosomes.”

What is wrong with the student's sentence?

  1. A. The student should have compared 2.9 with 3.84, the critical value for one degree of freedom
    Four classes give three degrees of freedom, whose critical value is 7.81.
    3.84 belongs to one degree of freedom.
  2. B. Nothing; a chi-square smaller than the critical value means the null hypothesis is accepted
    The verdict is worded fail to reject, never accept.
    Other models might fit the counts as well.
  3. C. The verdict should be to reject, because 2.9 is a small gap between the counts and the model
    A chi-square below the critical value means the gap is small enough to be chance.
    So the null hypothesis is not rejected.
  4. D. ✓ Failing to reject is not accepting or proving; the counts are only consistent with the model

Why: A chi-square below the critical value gives the verdict fail to reject: the counts are consistent with 1 : 1 : 1 : 1.
Failing to reject does not accept or prove the model.
So the genes are not proved to be on different chromosomes.

35
Check q14

A student has two results from one linked test cross: the four class counts, and a map distance of 12.2 map units worked out from them. She wants to calculate chi-square.

Which result can she test with chi-square, and why?

  1. A. The map distance, because it is a percentage and chi-square works on percentages
    Chi-square is for counts in categories, never for a percentage or a mean.
    The map distance is a single percentage with no predicted ratio to test it against.
  2. B. Either one, because both results come from the same cross and the same offspring
    A map distance is one percentage, not counts in classes.
    So no expected counts exist to compare it with.
  3. C. ✓ The four counts, because chi-square compares counts in classes with what a ratio predicts
  4. D. Neither, because chi-square works only on the 3 : 1 ratio of a monohybrid cross
    Chi-square tests counts against any predicted ratio, 1 : 1 : 1 : 1 included.

Why: Chi-square compares observed counts in classes with the counts a ratio predicts.
The four class counts can be tested against 1 : 1 : 1 : 1.
The map distance is a single percentage, and no ratio predicts a value for it.

36
Practice writing an answer

In a plant, tall (T) is dominant to short (t) and green pods (G) are dominant to yellow pods (g). A breeder crosses a TtGg plant with a ttgg plant. If the two genes assort independently, the offspring should come 1 : 1 : 1 : 1. Her 1,000 offspring are counted in the table below; the chi-square table from the formula sheet is drawn beneath it.

A table of four plant classes with observed counts: tall green 480, short yellow 470, tall yellow 30, short green 20; total 1,000. Beneath it, the chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01
A table of four plant classes with observed counts: tall green 480, short yellow 470, tall yellow 30, short green 20; total 1,000. Beneath it, the chi-square table from the formula sheet: a column for each number of degrees of freedom, 1 to 8, and a row for each p value, 0.05 and 0.01

(a) State the null hypothesis for the breeder's test. (1 pt)

Model answer The offspring occur in a 1 : 1 : 1 : 1 ratio, as independent assortment predicts, and any difference between the counts and 250 in each class is due to chance.
Rubric
  • Award 1 point for: the counts fit 1 : 1 : 1 : 1 and any difference from it is due to chance (no real difference).
  • Do not award: a statement that the two genes are linked, or that the parental classes are more common than predicted.

Slip Writing that the two genes are linked. That is the alternative hypothesis; the null is the no-difference statement.

(b) Calculate the expected count of each class. (1 pt)

Answer: 250  (tolerance ±0)

Model answer Each class has an expected count of 250.
Working
Write down the values in the question:
total = 1,000
ratio = 1 : 1 : 1 : 1, each class 1 of 4
Write down the equation:
e=total×that class's share of the ratio
Substitute the values into the equation:
e=1000×14=250 for every class
Rubric
  • Award 1 point for: 250 in each class.

(c) Calculate chi-square for the breeder's counts, to three significant figures. (1 pt)

Answer: 810  (tolerance ±0.5)

Model answer Chi-square is 810.
Working
Write down the values in the question:
o = 480, 470, 30, 20
e = 250 for every class
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
tall green: (480−250)2250=211.6
short yellow: (470−250)2250=193.6
tall yellow: (30−250)2250=193.6
short green: (20−250)2250=211.6
Add the classes:
χ2=211.6+193.6+193.6+211.6=810.4≈810
Rubric
  • Award 1 point for: chi-square 810 (accept 810 to 811).

(d) Determine whether the two genes are assorting independently. Justify your determination using the critical value at p = 0.05. (2 pt)

Model answer Four classes give 3 degrees of freedom.
The 0.05 row at 3 degrees of freedom gives a critical value of 7.81.
Chi-square is 810, larger than 7.81, so the null hypothesis is rejected.
So the counts do not fit 1 : 1 : 1 : 1, and the two genes are not assorting independently.
The two large classes are the parental combinations and the two small ones the recombinants: the two genes are linked.
Rubric
  • Award 1 point for: the determination, that the two genes are not assorting independently (accept: linked, or consistent with linkage).
  • Award 1 point for: the justification, that 810 is larger than the critical value 7.81 at 3 degrees of freedom, so the null hypothesis is rejected.
  • Do not award: ‘accept’ or ‘prove’, or degrees of freedom given as 4 or 999.

Slip Reading the critical value at one degree of freedom, 3.84. Four classes give three degrees of freedom, and the value is 7.81.

APBIO-U05-L24 Where the X and the Y go

Topic 5.4 · Non-Mendelian Genetics · 37 steps

A human karyotype drawn as 23 pairs of chromosomes in two rows, each chromosome a rod with a centromere dot, the pairs shrinking in length along the rows; the last pair at the bottom right is a long X beside a short Y, labeled X and Y
A human karyotype drawn as 23 pairs of chromosomes in two rows, each chromosome a rod with a centromere dot, the pairs shrinking in length along the rows; the last pair at the bottom right is a long X beside a short Y, labeled X and Y

Here is the odd pair from one man’s cells: a large X chromosome beside a small Y chromosome.

One of the two came from his mother and one from his father. Which parent gave him which, and what rides along with each?

Unit 5 · Heredity

1Where the X and the Y go

2

Video: Watch: Where the X and the Y go

The X and the Y parting in meiosis I; the eggs all carrying an X, the sperm half X and half Y; then the arrows: the mother’s X to every child, the father’s X to his daughters and his Y to his sons.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L24a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L24a.mp4

3
Check q1

A human cell’s chromosomes are sorted into 23 pairs. In 22 of the pairs the two members match. In the last pair one chromosome is a large X and the other is a small Y.

Which name fits the X and the Y?

  1. A. Autosomes
    The autosomes are the other 22 pairs, whose members match.
  2. B. ✓ Sex chromosomes

Why: The X and the Y are the pair that sets whether a person is female or male: the sex chromosomes.
The other 22 pairs, whose members match, are the autosomes.

4

Which parent gives a child which sex chromosome? The X and the Y part in meiosis I like any other pair.

5

A mother is XX, so every egg she makes carries an X.

6

A father is XY, so half of his sperm carry his X and half carry his Y.

Left: a mother's two X chromosomes above four eggs, each carrying an X. Right: a father's X and Y above four sperm, two carrying an X and two carrying a Y
Left: a mother's two X chromosomes above four eggs, each carrying an X. Right: a father's X and Y above four sperm, two carrying an X and two carrying a Y
7

Put the two parents’ gametes on a square: the eggs, X and X, along the top edge and the sperm, X and Y, down the side.

An empty two-by-two grid: the eggs, X and X, along the top edge; the sperm, X and Y, down the left edge
An empty two-by-two grid: the eggs, X and X, along the top edge; the sperm, X and Y, down the left edge
8

An X sperm fusing with either egg gives XX: a daughter.

The same grid with the top row filled: the X sperm with either egg gives XX
The same grid with the top row filled: the X sperm with either egg gives XX
9

A Y sperm fusing with either egg gives XY: a son.

The grid with all four cells filled: XX, XX in the top row and XY, XY in the bottom row
The grid with all four cells filled: XX, XX in the top row and XY, XY in the bottom row
10

Two cells of the four are XX and two are XY, so half of the children are daughters and half are sons.

11

Read the square by parent: the mother gives an X to every child.

12

The father gives his X to each daughter and his Y to each son.

A mother's XX and a father's XY at the top; a daughter's XX and a son's XY at the bottom. Arrows run from the mother's X to the daughter and to the son, from the father's X to the daughter, and from the father's Y to the son
A mother's XX and a father's XY at the top; a daughter's XX and a son's XY at the bottom. Arrows run from the mother's X to the daughter and to the son, from the father's X to the daughter, and from the father's Y to the son
13

The chromosome a son’s father gave him was the Y. So a son’s X came from his mother.

14

Alleles ride on the X as on any chromosome. Take a human example: a man who is red-green color-blind carries the color-blindness allele, written r, on his one X.

15

His X, with its r, goes to every daughter. Each son receives his Y instead, so his X, with its r, reaches none of his sons.

A color-blind father's X carrying the allele r beside his Y at the top; at the bottom a daughter whose second X carries r, and a son whose X carries no r and whose Y came from the father. Arrows: the father's X with r to the daughter; his Y to the son
A color-blind father's X carrying the allele r beside his Y at the top; at the bottom a daughter whose second X carries r, and a son whose X carries no r and whose Y came from the father. Arrows: the father's X with r to the daughter; his Y to the son
16

What you are expected to know Say which sex chromosome each parent passes to a child: a mother gives an X to every child; a father gives his X to each daughter and his Y to each son.

17
Check q2

A woman and a man have a son.

Which parent gave the son his X chromosome?

  1. A. ✓ His mother
  2. B. His father
    A father gives each son his Y.
    So a son’s X came from his mother.

Why: A father gives his X to each daughter and his Y to each son.
This boy received his father’s Y.
So his one X came from his mother.

18
Practice writing an answer

A woman, XX, and a man, XY, have a son. The son’s X chromosome came from his mother.

(a) Explain why the son’s X chromosome came from his mother. (1 pt)

Model answer The father is XY.
The chromosome that makes a child a son is the Y, so the father gave the son his Y.
The father’s X did not reach the son.
The mother is XX, so every egg she makes carries an X.
So the son’s one X came from his mother.
Rubric
  • Award 1 point for: the father gave the son his Y, so the father’s X did not reach him; the mother is XX and every egg carries an X, so the son’s X is hers.
19
Check q3

A student says: “A son can receive his X chromosome from his father.”

Is the student correct?

  1. A. Yes: a son can carry his father’s X
    A father gives a son his Y.
    So the father’s X never reaches a son.
  2. B. ✓ No: a son’s X came from his mother

Why: The chromosome a father gives a son is the Y.
The son’s X came from his mother’s egg.
So a son cannot receive his X from his father.

20
Check q4

A man is red-green color-blind and carries the color-blindness allele on his X chromosome. His daughter sees colors as most people do. Her son is color-blind. A son’s X came from his mother.

By which route did the color-blindness allele reach the boy?

  1. A. In the man’s sperm straight to the boy, on the man’s X
    A man gives no X to his grandson directly.
    His sperm made his daughter, and her egg made the boy.
  2. B. On the man’s Y to the woman, then in her egg to the boy
    A father gives his Y only to his sons.
    A daughter receives his X.
  3. C. ✓ On the man’s X to the woman, then in her egg to the boy
  4. D. On the boy’s father’s X, in the sperm that made the boy
    A father gives his son a Y, not an X, so the boy’s father gave no X to the boy.

Why: The man’s X, carrying the color-blindness allele, went to his daughter.
Her son’s one X came from her, and it was the X with that allele.
So the allele reached him through her.

21

Here again is the man’s odd pair: a large X beside a small Y. His father’s sperm carried the Y, so the Y came from his father.

A man's two sex chromosomes drawn as rods: a long X with a centromere dot beside a Y about half as long, labeled X and Y
A man's two sex chromosomes drawn as rods: a long X with a centromere dot beside a Y about half as long, labeled X and Y
22

His mother’s egg carried the X, so the X came from his mother. Every allele on that X came from her.

23Quick quiz: which parent gave it? mixed practice

24
Check q5

A boy’s X chromosome came from his mother. Her two X chromosomes came one from each of her parents.

Which of the mother’s parents could the boy’s X have come from?

  1. A. ✓ Either of her parents
  2. B. Her mother only
    The mother’s two X’s came one from each of her parents.
    Either X could have gone into the egg that made the boy.
  3. C. Her father only
    The mother’s two X’s came one from each of her parents.
    Either X could have gone into the egg that made the boy.

Why: The mother carries one X from her mother and one X from her father.
Her two X’s part at anaphase I, so each egg carries one of the two.
So the boy’s X could be either.

25
Check q6

A boy has one Y chromosome.

Which parent gave him his Y?

  1. A. His mother
    A mother is XX and has no Y to give.
    Every egg carries an X.
  2. B. ✓ His father

Why: A mother is XX, so every egg she makes carries an X.
A father is XY, and half of his sperm carry his Y.
So the Y came from his father.

26
Check q7

A girl has two X chromosomes.

Which parent gave her an X?

  1. A. Her mother only
    A father gives his X to each daughter.
    So one of her two X’s came from her father.
  2. B. Her father only
    A mother gives an X to every child.
    So one of her two X’s came from her mother.
  3. C. ✓ Both parents

Why: A mother gives an X to every child.
A father gives his X to each daughter.
So a daughter’s two X’s came one from each parent.

27
Check q8

A father carries an allele on his X chromosome.

Which of his children can receive that allele from him?

  1. A. His sons only
    A son receives his father’s Y, not his X.
    So an allele on the father’s X never reaches a son from him.
  2. B. ✓ His daughters only
  3. C. All of his children
    A father gives his X only to his daughters.
    Each son receives his Y.

Why: A father gives his X to each daughter and his Y to each son.
An allele on his X rides with the X.
So the allele can reach his daughters and none of his sons.

28
Check q9

A mother carries an allele on one of her two X chromosomes.

Which of her children can receive that allele from her?

  1. A. ✓ Any of her children
  2. B. Her daughters only
    A mother gives an X to every child, sons included.
    The Y that makes a son comes from the father.
  3. C. Her sons only
    A daughter also receives one X from her mother.

Why: A mother gives an X to every child, son or daughter.
Her two X’s part at anaphase I, so half of her eggs carry the X with the allele.
So any of her children can receive it.

29Mixed practice mixed practice

30
Check q10

A father’s sperm carrying his Y chromosome fuses with an egg.

Which child grows from the cell they make?

  1. A. A daughter, XX
    Every egg carries an X, so the cell holds the egg’s X and the sperm’s Y: XY.
  2. B. ✓ A son, XY

Why: Every egg carries an X.
The sperm brought the Y.
So the cell is XY, and a son grows from it.

31
Check q11

A mother is XX. One of her eggs is about to fuse with a sperm.

Which sex chromosome does the egg carry?

  1. A. ✓ An X
  2. B. A Y
    A mother is XX and has no Y.
    Every egg she makes carries an X.
  3. C. Either an X or a Y
    A mother’s two sex chromosomes are both X’s, so every egg carries an X and none carries a Y.

Why: The mother is XX.
Her two X’s part at anaphase I, and each egg receives one of them.
So the egg carries an X.

32
Check q12

A man carries an allele on his Y chromosome.

Which of his children receive that allele?

  1. A. His daughters only
    A father gives his daughters his X, never his Y.
  2. B. ✓ His sons only
  3. C. All of his children, sons and daughters
    Only his sons receive his Y; each daughter receives his X.

Why: A father gives his X to each daughter and his Y to each son.
An allele on the Y rides with the Y.
So every son receives it and no daughter does.

33
Check q13

A mother carries an allele on one of her two X chromosomes and has a son.

What is the probability that the son received that allele?

  1. A. 0
    A son’s one X comes from his mother, so he can receive the allele.
  2. B. ✓ 1/2
  3. C. 1
    Her two X’s part at anaphase I, so only half of her eggs carry the X with the allele.

Why: A son’s one X comes from his mother.
Her two X’s part at anaphase I, so half of her eggs carry the X with the allele.
So the probability that the son received it is 1/2.

34
Check q14

A woman’s father is red-green color-blind and carries the color-blindness allele on his X chromosome. She sees colors as most people do.

Which of his sex chromosomes did she receive from her father?

  1. A. ✓ The man’s X
  2. B. The man’s Y
    A father gives his Y only to his sons; every daughter receives his X.
  3. C. Neither one
    Every child receives one sex chromosome from each parent; a daughter receives her father’s X.

Why: A father gives his X to each daughter.
The man has one X, and it carries the color-blindness allele.
So his daughter received that X, with the allele on it.

35
Check q15

A father carries an allele on his X chromosome. He has two daughters and two sons.

How many of his four children received that allele from him?

  1. A. 0
    A father gives his X to each daughter, and the allele rides on that X.
  2. B. ✓ 2
  3. C. 4
    Each son received his father’s Y, not his X, so the two sons did not receive the allele from him.

Why: A father gives his X to each daughter and his Y to each son.
The allele rides on his X, so both daughters received it and neither son did.
So 2 of his 4 children received it.

36
Practice writing an answer

In a mammal, a gene sits on the X chromosome. A male carries the allele a on his X. He fathers 12 daughters and 10 sons.

(a) Predict which of his 22 offspring received the allele a from him. (1 pt)

Model answer All 12 daughters received the allele a from him.
None of the 10 sons did.
Rubric
  • Award 1 point for: all 12 daughters and none of the 10 sons.

(b) Explain how the way the X and the Y part in meiosis I produces this pattern. (1 pt)

Model answer At anaphase I the X and the Y are pulled to opposite poles, so half of his sperm carry his X and half carry his Y.
A sperm carrying his X makes a daughter, and that X carries the allele a.
A sperm carrying his Y makes a son, and the Y carries no allele of this gene.
So every daughter received a and no son did.
Rubric
  • Award 1 point for: the X and the Y part at anaphase I, so each sperm carries one or the other; the X sperm (with a) make daughters and the Y sperm (without a) make sons.

Slip Saying half of the sons received a. A son receives his father’s Y, so no son receives an allele on his father’s X.

APBIO-U05-L24B One X is enough

Topic 5.4 · Non-Mendelian Genetics · 58 steps

On the left a man's sex chromosomes, a long X carrying the allele r beside a short Y, labeled color-blind; on the right a woman's two X's, one carrying the ordinary allele R and the other r, labeled sees red and green
On the left a man's sex chromosomes, a long X carrying the allele r beside a short Y, labeled color-blind; on the right a woman's two X's, one carrying the ordinary allele R and the other r, labeled sees red and green

Here are the sex chromosomes of a man and of a woman. The man and the woman each carry the red-green color-blindness allele, r, on one X: his only X, and one of her two.

He is color-blind. She sees red and green as most people do. Why does the same allele show in him and hide in her?

Unit 5 · Heredity

1One X, nothing opposite it

2

Video: Watch: One X, nothing opposite it

A single X with the color-blindness allele on it and nothing opposite it, beside two X’s where the ordinary allele hides the same allele.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L24Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L24Ba.mp4

3
Check q1

In peas, purple flower color, P, is dominant to white, p. A pea plant carries one P and one p.

Which color are its flowers?

  1. A. ✓ Purple
  2. B. White
    One dominant P is enough for purple to show.
    White, p, is recessive: P hides it whenever P is present.

Why: P is dominant, so its trait shows in a heterozygous plant.
This plant carries one P.
So its flowers are purple.

4

Why does one allele show in one person and hide in another? A recessive trait shows only when no dominant allele is present to hide it.

5

Write the ordinary allele R and the color-blindness allele r. Red-green color blindness is recessive: one R hides r.

6

The woman carries r on one X and R on the other. Her R hides her r, so she sees red and green as most people do.

A woman's two X chromosomes side by side: one carries the ordinary allele R and the other the color-blindness allele r
A woman's two X chromosomes side by side: one carries the ordinary allele R and the other the color-blindness allele r
7

She is a carrier: she carries r without showing it.

8

The man carries r on his one X, and his other sex chromosome is the Y.

9

The Y carries no allele of this gene, so nothing sits opposite his r.

A man's X chromosome carrying the allele r beside his short Y, which carries no allele of this gene; the space opposite the r box on the Y is empty
A man's X chromosome carrying the allele r beside his short Y, which carries no allele of this gene; the space opposite the r box on the Y is empty
10

So the rule for any recessive allele on the X: an XY individual has one X, so a recessive allele on it has no second X carrying a masking allele, and its trait shows.

11

What you are expected to know Explain why a recessive allele on the X shows in a male with one copy and hides in a female who also carries the ordinary allele.

12
Check q2

A man carries the red-green color-blindness allele, r, on his X chromosome. Red-green color blindness is recessive.

Which of the following describes the man?

  1. A. The man sees red and green
    The man has one X, so no second X carries an R to hide his r.
  2. B. ✓ The man is color-blind

Why: The man has one X, and his r sits on it.
His Y carries no allele of this gene, so nothing hides his r.
So his r shows, and he is color-blind.

13
Check q3

A woman carries the red-green color-blindness allele, r, on one X chromosome and the ordinary allele, R, on the other. Red-green color blindness is recessive.

Which of the following describes the woman?

  1. A. ✓ The woman sees red and green
  2. B. The woman is color-blind
    Her other X carries R.
    One R is enough to hide r.

Why: The woman carries r on one X and R on the other.
R is dominant, so her R hides her r.
So she sees red and green.

14
Practice writing an answer

A man carries the red-green color-blindness allele, r, on his X chromosome. His Y chromosome carries no allele of this gene. Red-green color blindness is recessive. The man is color-blind.

(a) Explain why the man is color-blind although he carries only one copy of r. (1 pt)

Model answer The man is XY, so he has one X.
His r sits on that X.
His Y carries no allele of this gene, so no second X carries a masking allele R.
A recessive allele’s trait shows when no dominant allele hides it.
So his single r shows, and he is color-blind.
Rubric
  • Award 1 point for: he has one X, so no second X carries the masking allele R; nothing hides his r, so the recessive trait shows.
15
Check q4

A student says: “A man who carries r on his X sees red and green, because his Y carries a hidden R that masks the r.”

Is the student correct?

  1. A. Yes: the man’s Y carries an R
    The Y carries no allele of this gene.
    Nothing on it masks the r.
  2. B. ✓ No: the man’s Y carries no allele of this gene

Why: The color-blindness gene sits on the X.
The Y carries no allele of this gene.
So nothing sits opposite the man’s r, and his r shows.

16Common in men, rare in women

17

Video: Watch: Common in men, rare in women

One r showing in a man; one r hidden in a woman; the table of about 1 in 12 men and about 1 in 200 women; then a woman with r on both X’s.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L24Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L24Bb.mp4

18

How common is red-green color blindness in men, and in women?

19

One copy of r is enough for the trait to show in a man. A woman needs two copies.

20

Here is a table comparing how common red-green color blindness is in men and in women.

A table: about 1 in 12 men and about 1 in 200 women are red-green color-blind
A table: about 1 in 12 men and about 1 in 200 women are red-green color-blind
21

About 1 in 12 men are red-green color-blind. About 1 in 200 women are.

22

So red-green color blindness is common in men and rare in women.

23

Rare is not never. A woman whose two X chromosomes both carry r has no R to hide it, so she is color-blind too.

A woman's two X chromosomes, each carrying the allele r; no R is present
A woman's two X chromosomes, each carrying the allele r; no R is present
24

What you are expected to know Predict that a recessive trait carried on the X is common in males and rare in females, and that rare is not never.

25
Check q5

In a mammal, a recessive allele on the X chromosome causes a trait.

Which sex shows the trait more often?

  1. A. ✓ Males
  2. B. Females
    A female with the allele on one X usually carries the ordinary allele on her other X, and it hides the trait.
  3. C. Both sexes equally
    A male has one X, so one copy shows; a female needs the allele on both X’s.

Why: A male has one X, so a recessive allele on it has no second X carrying a masking allele, and the trait shows.
A female needs the allele on both X’s.
So males show the trait more often.

26
Practice writing an answer

In a mammal, a recessive allele on the X chromosome causes a pale coat. In one large group of these animals, pale coats are common among males and rare among females.

(a) Explain why a male with one copy of the allele is pale while a female needs two copies, and how that makes pale coats rarer among females. (1 pt)

Model answer A male has one X, so a pale-coat allele on it has no second X carrying a masking allele, and his coat is pale.
So every male who carries the allele is pale.
A female has two X’s, so the allele on one X is hidden when the other X carries the ordinary allele.
A female is pale only when both X’s carry the allele, which is far rarer.
So far fewer females than males are pale.
Rubric
  • Award 1 point for: a male’s single X means one copy shows (no masking allele), so every male who carries it is pale; a female needs the allele on both X’s because the ordinary allele on the other X hides it, so far fewer females are pale.
27
Check q6

A student says: “Red-green color blindness is carried on the X, so only men are color-blind.”

Is the student correct?

  1. A. Yes: only men are color-blind
    A woman whose two X’s both carry r has no R to hide it.
    She is color-blind.
  2. B. ✓ No: a woman can be color-blind too

Why: A woman is color-blind when both of her X’s carry r and no R hides it.
Two copies of r in one woman are rare.
So a woman can be color-blind: rare is not never.

28
Check q7

A woman’s two X chromosomes both carry the red-green color-blindness allele, r. Red-green color blindness is recessive.

Which of the following describes the woman?

  1. A. The woman sees red and green
    Neither of her X’s carries R, so nothing hides her r.
  2. B. ✓ The woman is color-blind

Why: A recessive trait shows when no dominant allele is present to hide it.
Both of her X’s carry r and neither carries R.
So she is color-blind.

29The names

30

Video: Watch: The names

An X and a Y side by side, each with an allele box; the words sex-linked above both, X-linked under the X and Y-linked under the Y.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L24Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L24Bc.mp4

31

What do we call traits like this one? When a trait’s gene sits on a sex chromosome, we call it a , because its gene rides on a sex chromosome and travels, linked, with it.

An X chromosome and a Y chromosome side by side, each with an allele box; the X is labeled X-linked and the Y labeled Y-linked; above both, the words sex-linked
An X chromosome and a Y chromosome side by side, each with an allele box; the X is labeled X-linked and the Y labeled Y-linked; above both, the words sex-linked
32

A sex-linked gene on the X is called .

33

A sex-linked gene on the Y is called .

34

The red-green color-blindness gene sits on the X, so red-green color blindness is X-linked.

35

Its allele r is recessive. So red-green color blindness is an X-linked recessive trait.

36

What you are expected to know Name a trait whose gene sits on a sex chromosome as sex-linked, and say whether it is X-linked or Y-linked from the chromosome that carries the gene.

37

Back to the man and the woman: each carries r on one X. The man’s only X carries r, and his Y carries no allele of this gene.

38

Nothing hides his r, so he is color-blind.

39

The woman’s other X carries R. Her R hides her r, so she sees red and green as most people do.

40Quick quiz: sex-linked trait, X-linked and Y-linked mixed practice

41
Check q8

Which of the following is a sex-linked trait?

  1. A. A trait whose gene sits on one of the 22 matching pairs
    A gene on one of the 22 matching pairs, the autosomes, gives an autosomal trait.
  2. B. A trait that shows in one sex only
    A sex-linked trait is named for where its gene sits, not for which sex shows it.
    A woman with r on both X’s shows red-green color blindness.
  3. C. ✓ A trait whose gene sits on the X or the Y chromosome

Why: The X and the Y are the sex chromosomes.
A trait whose gene sits on a sex chromosome is sex-linked.

42
Check q9

Which of the following describes an X-linked gene?

  1. A. ✓ A gene that sits on the X chromosome
  2. B. A gene that sits on the Y chromosome
    A gene on the Y is Y-linked.
  3. C. A gene that sits on one of the autosomes
    A gene on an autosome gives an autosomal trait.

Why: A sex-linked gene on the X is called X-linked.

43
Check q10

In a mammal, a gene sits on the Y chromosome.

Which name fits a trait of this gene?

  1. A. X-linked
    An X-linked gene sits on the X.
    This gene sits on the Y.
  2. B. ✓ Y-linked
  3. C. Autosomal
    An autosomal gene sits on one of the matching pairs.
    The Y is a sex chromosome.

Why: The gene sits on the Y, a sex chromosome.
A sex-linked gene on the Y is called Y-linked.

44
Check q11

In cats, a gene sits on the X chromosome. A female cat with one copy of an allele and one ordinary allele looks the same as a cat with two ordinary alleles.

Which name fits the allele’s trait?

  1. A. Autosomal recessive
    The gene sits on the X, not on an autosome.
    A trait whose gene is on the X is X-linked.
  2. B. X-linked dominant
    A dominant allele shows in a female who carries one copy.
    This female carries one copy and does not show the trait, so the allele is recessive.
  3. C. ✓ X-linked recessive
  4. D. Y-linked
    A Y-linked gene sits on the Y.
    This gene sits on the X.

Why: The gene sits on the X, so the trait is X-linked.
A female with one copy and one ordinary allele does not show the trait, so the ordinary allele hides it: the allele is recessive.
The trait is X-linked recessive.

45
Check q12

In a mammal, a gene sits on the Y chromosome, and one of its alleles causes a trait.

Which name fits the trait?

  1. A. Autosomal recessive
    The gene sits on the Y, a sex chromosome, not on an autosome.
  2. B. X-linked dominant
    An X-linked gene sits on the X.
    This gene sits on the Y.
  3. C. X-linked recessive
    An X-linked gene sits on the X.
    This gene sits on the Y.
  4. D. ✓ Y-linked

Why: The gene sits on the Y.
A sex-linked gene on the Y is called Y-linked.
So the trait is Y-linked.

46
Check q13

In mice, a gene sits on chromosome 7, one of the matching pairs. A mouse with one copy of an allele looks the same as a mouse with two ordinary alleles, and a mouse with two copies shows the allele’s trait.

Which name fits the trait?

  1. A. ✓ Autosomal recessive
  2. B. X-linked dominant
    An X-linked gene sits on the X.
    Chromosome 7 is an autosome.
  3. C. X-linked recessive
    An X-linked gene sits on the X.
    Chromosome 7 is an autosome.
  4. D. Y-linked
    A Y-linked gene sits on the Y.
    Chromosome 7 is an autosome.

Why: Chromosome 7 is one of the matching pairs, an autosome, so the trait is autosomal.
One copy does not show and two copies do, so the allele is recessive.
The trait is autosomal recessive.

47
Check q14

In people, a gene sits on chromosome 11, one of the 22 matching pairs. A woman with one copy of an allele and one ordinary allele shows the allele’s trait.

Which name fits the trait?

  1. A. Autosomal recessive
    A recessive allele hides in a person who also carries the ordinary allele.
    This woman carries the ordinary allele and still shows the trait, so the allele is dominant.
  2. B. ✓ Autosomal dominant
  3. C. X-linked recessive
    An X-linked gene sits on the X.
    Chromosome 11 is an autosome.
  4. D. Y-linked
    A Y-linked gene sits on the Y.
    Chromosome 11 is an autosome.

Why: Chromosome 11 is one of the matching pairs, an autosome, so the trait is autosomal.
A woman with one copy and one ordinary allele shows the trait, so one copy is enough: the allele is dominant.
The trait is autosomal dominant.

48
Check q15

In fruit flies, a gene on the X has an allele for red eyes and an allele for white eyes. A female with one allele of each has red eyes.

Which name fits the white-eye trait?

  1. A. Autosomal recessive
    An autosomal trait’s gene sits on an autosome, one of the matching pairs.
    This gene sits on the X.
  2. B. X-linked dominant
    A dominant white-eye allele would show in a female carrying one copy.
    This female carries one white-eye allele and has red eyes, so the red-eye allele hides the white-eye allele.
  3. C. ✓ X-linked recessive
  4. D. Y-linked
    A Y-linked gene sits on the Y.
    This gene sits on the X.

Why: The gene sits on the X, so the trait is X-linked.
A female with one white-eye allele and one red-eye allele has red eyes, so one red-eye allele hides the white-eye allele: white eyes are recessive.
The trait is X-linked recessive.

49
Practice writing an answer

In fruit flies, the gene for eye color sits on the X chromosome. The white-eye allele is recessive to the red-eye allele.

(a) Explain why the white-eye trait is called X-linked recessive. (1 pt)

Model answer The eye-color gene sits on the X, a sex chromosome, so the trait is sex-linked.
The sex chromosome that carries the gene is the X, so the trait is X-linked.
The white-eye allele is recessive.
So the white-eye trait is X-linked recessive.
Rubric
  • Award 1 point for: the gene sits on the X, a sex chromosome (so X-linked), and the white-eye allele is recessive.

50Mixed practice mixed practice

51
Check q16

In a mammal, a recessive allele on the X chromosome causes white fur.

Which sex has white fur more often?

  1. A. ✓ Males
  2. B. Females
    A female with the allele on one X usually carries the ordinary allele on her other X, and it hides the white fur.
  3. C. Both sexes equally
    A male has one X, so one copy shows; a female needs the allele on both X’s.

Why: A male has one X, so a recessive allele on it has no second X carrying a masking allele, and white fur shows.
A female needs the allele on both X’s.
So males have white fur more often.

52
Check q17

In a mammal, a gene sits on the X chromosome. A male with one copy of an allele shows its trait. A female with one copy and one ordinary allele looks the same as a female with two ordinary alleles.

Which name fits the trait?

  1. A. Autosomal recessive
    The gene sits on the X, a sex chromosome, not on an autosome.
  2. B. ✓ X-linked recessive
  3. C. X-linked dominant
    A dominant allele shows in a female who carries one copy.
    This female carries one copy and does not show the trait, so the allele is recessive.
  4. D. Y-linked
    A Y-linked gene sits on the Y.
    This gene sits on the X.

Why: The gene sits on the X, so the trait is X-linked.
A female with one copy and one ordinary allele does not show the trait, so the ordinary allele hides it: the allele is recessive.
The trait is X-linked recessive.

53
Check q18

In a mammal, a recessive allele on the X chromosome causes short whiskers. A female carries that allele on one X and the ordinary allele on the other.

Which of the following describes the female?

  1. A. The female has short whiskers
    Her other X carries the ordinary allele, and one ordinary allele is enough to hide the recessive allele.
  2. B. ✓ The female has ordinary whiskers

Why: The female carries the ordinary allele on her other X.
The short-whisker allele is recessive, so the ordinary allele hides it.
So the female has ordinary whiskers.

54
Check q19

In a mammal, a recessive allele on the X chromosome causes a curled tail. A male carries that allele on his X.

Which of the following describes the male?

  1. A. ✓ The male has a curled tail
  2. B. The male has an ordinary tail
    The male has one X, so no second X carries an ordinary allele to hide the curled-tail allele.

Why: The male has one X, and the curled-tail allele sits on it.
His Y carries no allele of this gene, so nothing hides the allele.
So the male has a curled tail.

55
Check q20

In a mammal, a gene sits on chromosome 4, one of the matching pairs.

Which name fits a trait of this gene?

  1. A. ✓ Autosomal
  2. B. X-linked
    An X-linked gene sits on the X.
    Chromosome 4 is one of the matching pairs.
  3. C. Y-linked
    A Y-linked gene sits on the Y.
    Chromosome 4 is one of the matching pairs.

Why: Chromosome 4 is one of the matching pairs, an autosome.
A trait whose gene sits on an autosome is autosomal.

56
Check q21

In a population, a recessive allele on the X chromosome causes a trait.

Why does the trait appear in more males than females?

  1. A. The Y chromosome carries a second copy of the allele in males
    The Y carries no allele of a gene on the X.
    That is why nothing sits opposite the allele in a male.
  2. B. Males inherit more copies of the allele than females do
    A male has one X and so at most one copy of the allele; a female can have two.
  3. C. The allele is dominant in males and recessive in females
    The allele is recessive in both sexes.
    A female can carry a masking allele on her second X and a male cannot.
  4. D. ✓ A male’s one X has no second X carrying a masking allele

Why: A female who carries the allele on one X usually carries the ordinary allele on her other X, and it hides the trait.
A male has one X, so a recessive allele on it has no second X carrying a masking allele, and the trait shows.

57
Practice writing an answer

In a mammal, a gene on the X chromosome has a recessive allele that causes a short tail. A female carries the short-tail allele on one X and the ordinary allele on the other. Her brother carries the short-tail allele on his X.

(a) Predict which of the two animals has a short tail. (1 pt)

Model answer The brother has a short tail.
The female has an ordinary tail.
Rubric
  • Award 1 point for: the brother (male) has the short tail; the female does not.

(b) Explain how the sex chromosomes each animal carries produce this difference. (1 pt)

Model answer The brother is XY, so he has one X, and the short-tail allele sits on it.
His Y carries no allele of this gene, so nothing hides the allele, and his tail is short.
The female is XX, and her other X carries the ordinary allele.
The short-tail allele is recessive, so the ordinary allele hides it, and her tail is ordinary.
Rubric
  • Award 1 point for: the male’s one X carries the allele and his Y carries no masking allele, so it shows; the female’s other X carries the ordinary allele, which hides the recessive allele.

Slip Saying the Y carries a hidden ordinary allele. The Y carries no allele of a gene on the X.

Glossary

sex-linked trait
A trait whose gene sits on a sex chromosome, the X or the Y, so the trait travels with that chromosome.
X-linked
Carried on the X chromosome; a recessive X-linked trait shows in a male with one copy, because his Y carries no masking allele.
Y-linked
Carried on the Y chromosome, so passed from a father to his sons only.

APBIO-U05-L25 Predict an X-linked cross

Topic 5.4 · Non-Mendelian Genetics · 60 steps

An empty two-by-two grid: the mother's gametes X-R and X-r written along the top edge, the father's gametes X-R and Y written down the left edge; all four cells are empty
An empty two-by-two grid: the mother's gametes X-R and X-r written along the top edge, the father's gametes X-R and Y written down the left edge; all four cells are empty

Here is an empty square for a woman who carries the color-blindness allele on one of her X chromosomes, XᴿXʳ, and a man with ordinary vision, XᴿY.

It is a square with a difference: the alleles sit on the X, and the Y carries none. What will their sons and daughters get?

Unit 5 · Heredity

1Build the square

2

Video: Watch: Build the square

The X-linked square filling one cell at a time: each allele written on its X, the Y bare, the mother’s eggs along the top and the father’s sperm down the side.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25a.mp4

3

How do you build a Punnett square when the gene sits on the X? Write each allele on the chromosome that carries it, and write the Y bare.

4

Then put each parent’s two gametes along one edge of the square, as for any other gene.

5
Check q1

A Punnett square predicts the offspring two parents can have.

Which of the following goes along the edges of the square?

  1. A. ✓ Each parent’s gametes
  2. B. Each parent’s genotype
    A parent’s genotype holds two alleles, but a gamete carries one.
    The edges hold the gametes, one allele each.
  3. C. The offspring’s genotypes
    The offspring’s genotypes go in the cells, where one egg’s allele is written with one sperm’s allele.

Why: Each parent makes gametes, and each gamete carries one allele.
The gametes go along the edges.
Each cell joins one gamete from the top edge with one from the side edge: one offspring’s genotype.

6

First the notation. We write the ordinary allele on its X as Xᴿ and the color-blindness allele on its X as Xʳ, so each allele shows the chromosome that carries it.

7

The Y is written bare, Y, because it carries no allele of this gene. So the mother is XᴿXʳ and the father is XᴿY.

8

Now the cross: what children can XᴿXʳ and XᴿY have, and in what proportions? The mother’s eggs, Xᴿ or Xʳ, go along the top edge and the father’s sperm, Xᴿ or Y, down the side.

An empty two-by-two grid: the mother's eggs, X-R and X-r, along the top edge; the father's sperm, X-R and Y, down the left edge
An empty two-by-two grid: the mother's eggs, X-R and X-r, along the top edge; the father's sperm, X-R and Y, down the left edge
9

Where the Xᴿ egg fuses with the Xᴿ sperm, the cell reads XᴿXᴿ: a daughter with two ordinary alleles.

The same grid with one cell filled: where the X-R egg fuses with the X-R sperm, the cell reads X-R X-R
The same grid with one cell filled: where the X-R egg fuses with the X-R sperm, the cell reads X-R X-R
10

Where the Xʳ egg fuses with the Xᴿ sperm, the cell reads XᴿXʳ: a daughter who carries r.

The grid with two cells filled: X-R X-R top left and X-R X-r top right, where the X-r egg fuses with the X-R sperm
The grid with two cells filled: X-R X-R top left and X-R X-r top right, where the X-r egg fuses with the X-R sperm
11

The Y row gives the sons: the Xᴿ egg with the Y sperm gives XᴿY, and the Xʳ egg with the Y sperm gives XʳY.

The grid with all four cells filled: X-R X-R and X-R X-r in the top row, X-R Y and X-r Y in the bottom row
The grid with all four cells filled: X-R X-R and X-R X-r in the top row, X-R Y and X-r Y in the bottom row
12

Now consider a color-blind man, XʳY, and a woman with two ordinary alleles, XᴿXᴿ. Here is their square.

The square for a color-blind man, X-r Y, and a woman X-R X-R: X-R and X-R along the top, X-r and Y down the side; cells X-R X-r, X-R X-r, X-R Y, X-R Y
The square for a color-blind man, X-r Y, and a woman X-R X-R: X-R and X-R along the top, X-r and Y down the side; cells X-R X-r, X-R X-r, X-R Y, X-R Y
13

Every daughter is XᴿXʳ, because each got the father’s Xʳ and one of the mother’s Xᴿ. Every son is XᴿY.

14

What you are expected to know Build a Punnett square for an X-linked gene: alleles written on the X, the Y bare, each parent’s gametes on an edge.

15
Check q2

In fruit flies, red eyes, Xᴿ, are dominant to white eyes, Xʳ, and the gene sits on the X. A carrier female is XᴿXʳ.

Which sex chromosomes can her eggs carry?

  1. A. ✓ Xᴿ or Xʳ
  2. B. Xᴿ only
    The female carries Xᴿ on one X and Xʳ on the other.
    The two X’s part at anaphase I. So half of her eggs carry Xᴿ and half carry Xʳ.
  3. C. XᴿXʳ
    An egg carries one sex chromosome, not both.
    The two X’s part at anaphase I, so each egg receives one of them.

Why: The female carries Xᴿ on one X and Xʳ on the other.
The two X’s part at anaphase I. So each egg receives one of the two: half of her eggs carry Xᴿ and half carry Xʳ.

16
Check q3

A red-eyed male fruit fly is XᴿY.

Which sex chromosomes can his sperm carry?

  1. A. Xᴿ only
    The male carries an X and a Y, and the two part at anaphase I.
    So half of his sperm carry the Y.
  2. B. Xᴿ or Xʳ
    The male has one X, and it carries Xᴿ.
    He has no Xʳ to give.
    His other sex chromosome is the Y.
  3. C. ✓ Xᴿ or Y

Why: The male carries one X, with Xᴿ on it, and one Y.
The X and the Y part at anaphase I.
So half of his sperm carry Xᴿ and half carry Y.

17
Check q4

In fruit flies, red eyes, Xᴿ, are dominant to white eyes, Xʳ, and the gene sits on the X. A carrier female, XᴿXʳ, is crossed with a red-eyed male, XᴿY. Four squares drawn for this cross are numbered 1 to 4 below, with the female’s gametes meant to sit along the top edge and the male’s down the left edge.

Four two-by-two squares numbered 1 to 4 for a carrier female fruit fly, X-R X-r, crossed with a red-eyed male, X-R Y, the female's gametes along the top edge and the male's down the left edge. Square 1: X-R and X-r along the top, X-r and Y down the side, cells X-R X-r, X-r X-r, X-R Y, X-r Y. Square 2: X-R and X-r along the top, X-R and Y down the side, cells X-R X-R, X-R X-r, X-R Y, X-r Y. Square 3: X-R and X-r along the top, X-R and Y down the side, cells X-R X-R, X-R X-r, X-R X-R, X-r X-r. Square 4: X-R and X-R along the top, X-R and Y down the side, cells X-R X-R, X-R X-R, X-R Y, X-R Y
Four two-by-two squares numbered 1 to 4 for a carrier female fruit fly, X-R X-r, crossed with a red-eyed male, X-R Y, the female's gametes along the top edge and the male's down the left edge. Square 1: X-R and X-r along the top, X-r and Y down the side, cells X-R X-r, X-r X-r, X-R Y, X-r Y. Square 2: X-R and X-r along the top, X-R and Y down the side, cells X-R X-R, X-R X-r, X-R Y, X-r Y. Square 3: X-R and X-r along the top, X-R and Y down the side, cells X-R X-R, X-R X-r, X-R X-R, X-r X-r. Square 4: X-R and X-R along the top, X-R and Y down the side, cells X-R X-R, X-R X-R, X-R Y, X-R Y

Which square is drawn correctly?

  1. A. Square 1
    Square 1 gives the red-eyed male an Xʳ sperm; a red-eyed male, XᴿY, makes Xᴿ sperm and Y sperm, so his edge reads Xᴿ and Y.
  2. B. ✓ Square 2
  3. C. Square 3
    Square 3’s edges are right but its Y row is not: a Y sperm with an egg gives XᴿY or XʳY, not a second pair of X’s.
  4. D. Square 4
    Square 4 puts Xᴿ and Xᴿ along the top, but a carrier female makes half Xᴿ eggs and half Xʳ eggs.

Why: The female’s gametes are Xᴿ and Xʳ, the male’s are Xᴿ and Y, and each cell holds the chromosome above it with the chromosome beside it: XᴿXᴿ, XᴿXʳ, XᴿY, XʳY.
Only square 2 has those edges and those cells.

18Read it by sex

19

Video: Watch: Read it by sex

The filled square with its top row, the daughters, shaded and read; then its bottom row, the sons, shaded and read.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25b.mp4

20

Who gets what? Read the square by sex: the cells with two X’s are the daughters, and the cells with a Y are the sons.

21
Check q5

A father is XY. He has sons and daughters.

Which of his children receive his X?

  1. A. His sons only
    A son receives his father’s Y; his one X comes from his mother.
  2. B. ✓ His daughters only

Why: A father gives his X to each daughter and his Y to each son, so only his daughters receive his X.

22

The two cells with two X’s are the daughters, XᴿXᴿ and XᴿXʳ.

The filled square with the top row shaded: the two cells with two X's, X-R X-R and X-R X-r, are the daughters
The filled square with the top row shaded: the two cells with two X's, X-R X-R and X-R X-r, are the daughters
23

Neither daughter is color-blind, because each has an Xᴿ. Half of the daughters carry r.

24

The two cells with a Y are the sons, XᴿY and XʳY.

The filled square with the bottom row shaded: the two cells with a Y, X-R Y and X-r Y, are the sons
The filled square with the bottom row shaded: the two cells with a Y, X-R Y and X-r Y, are the sons
25

Half of the sons are color-blind, because a son’s only X came from his mother, and half of her eggs carry Xʳ.

26

What you are expected to know Read an X-linked square by sex: the two cells with two X’s are the daughters, the two cells with a Y are the sons.

27Quick quiz: read a cell by sex mixed practice

28
Check q6

Xᴿ is the ordinary allele and Xʳ the color-blindness allele. One cell of a square reads XʳY.

Which of the following is this child?

  1. A. ✓ A son
  2. B. A daughter
    The cell holds a Y.
    Only a sperm carries a Y, and a child with a Y is a son.

Why: The cell reads XʳY: one X and one Y.
A child with a Y is a son.

29
Check q7

Xᴿ is the ordinary allele and Xʳ the color-blindness allele. One cell of a square reads XᴿXʳ.

Which of the following is this child?

  1. A. A son
    The cell holds two X’s and no Y.
    A child with two X’s is a daughter.
  2. B. ✓ A daughter

Why: The cell reads XᴿXʳ: two X’s and no Y.
A child with two X’s is a daughter.

30
Check q8

Xᴿ is the ordinary allele and Xʳ the color-blindness allele. One cell of a square reads XʳY.

Which of the following describes this child’s vision?

  1. A. ✓ Color-blind
  2. B. A carrier with ordinary vision
    This child has one X, and it carries r.
    No second X carries an R to hide it, so the trait shows.
  3. C. Ordinary vision, not a carrier
    The one X carries r, so the child does carry r.
    Nothing hides it, so the trait shows.

Why: The cell reads XʳY: one X, carrying r, and a Y.
The Y carries no allele of this gene, so nothing hides the r.
The child is color-blind.

31
Check q9

Xᴿ is the ordinary allele and Xʳ the color-blindness allele. One cell of a square reads XᴿXʳ.

Which of the following describes this child’s vision?

  1. A. Color-blind
    One X carries R.
    One R is enough to hide the r on the other X.
  2. B. ✓ A carrier with ordinary vision
  3. C. Ordinary vision, not a carrier
    The second X carries r.
    The child carries r without showing it.

Why: The cell reads XᴿXʳ: one X carrying R and one carrying r.
The R hides the r, so the child has ordinary vision.
The r is still there, so the child carries it.

32
Check q10

Xᴿ is the ordinary allele and Xʳ the color-blindness allele. One cell of a square reads XʳXʳ.

Which of the following describes this child’s vision?

  1. A. ✓ Color-blind
  2. B. A carrier with ordinary vision
    Both X’s carry r and neither carries R.
    No R hides the r, so the trait shows.
  3. C. Ordinary vision, not a carrier
    Both X’s carry r.
    No R hides the r, so the trait shows.

Why: The cell reads XʳXʳ: both X’s carry r.
Neither carries an R to hide it.
The child is color-blind.

33
Check q11

Xᴿ is the ordinary allele and Xʳ the color-blindness allele. One cell of a square reads XᴿY.

Which of the following describes this child’s vision?

  1. A. Color-blind
    The one X carries R, the ordinary allele.
    No r is present, so the trait cannot show.
  2. B. A carrier with ordinary vision
    The one X carries R.
    The Y carries no allele of this gene, so no r is present anywhere.
  3. C. ✓ Ordinary vision, not a carrier

Why: The cell reads XᴿY: one X, carrying R, and a Y.
The Y carries no allele of this gene, so the child carries no r.
The child has ordinary vision and does not carry r.

34The probability of a named child

35

Video: Watch: The probability of a named child

The probability of a color-blind son laid out on screen as values, equation, substitution and result: one half times one half.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25c.mp4

36

How likely is one named child, say a color-blind son? A child is a color-blind son when two things both happen: the sperm carried the Y, and the egg carried Xʳ.

37
Check q12

Two independent events must both happen.

Which of the following gives the probability that both happen?

  1. A. Add the two probabilities
    Adding is for either event happening: P(A or B) = P(A) + P(B).
    For both, multiply.
  2. B. ✓ Multiply the two probabilities
  3. C. Take the larger of the two probabilities
    Both events happening is less likely than either alone.
    So the larger probability cannot be the answer; multiply the two.

Why: For two independent events that must both happen, multiply their probabilities: P(A and B) = P(A) × P(B).

38

The multiplication rule: multiply for both, P(A and B)=P(A)×P(B).

39

The sperm carrying the Y and the egg carrying Xʳ are independent events, so their two probabilities multiply.

40
Worked example

A carrier woman, XᴿXʳ, and a man with ordinary vision, XᴿY, have a child. What is the probability that the child is a color-blind son?

Write down the values in the question:
P(the sperm carries Y, so a son) = 1/2
P(the egg carries Xʳ) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(A and B)=P(A)×P(B)
P(color-blind son)=P(Y)×P(Xr)=12×12=14=0.25
A probability is a number between 0 and 1 and carries no unit.
41

What you are expected to know Calculate the probability of a named child: multiply the probability of the sex by the probability of the allele.

42
Check q13 numeric entry

A carrier woman, XᴿXʳ, and a color-blind man, XʳY, have a child.

Calculate the probability, as a decimal, that the child is a color-blind daughter.

Part 1. The father’s sperm carries Xʳ or Y. What is the probability, as a decimal, that the child is a daughter?

Answer: 0.5  (tolerance ±0.005)

Working
Half of the sperm carry the X:
P(daughter)=P(Xr sperm)=12=0.5

Part 2. The mother’s eggs carry Xᴿ or Xʳ. What is the probability, as a decimal, that the egg carries Xʳ?

Answer: 0.5  (tolerance ±0.005)

Working
Half of the eggs carry each allele:
P(Xr egg)=12=0.5

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
P(the sperm carries Xʳ, so a daughter) = 1/2
P(the egg carries Xʳ) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(A and B)=P(A)×P(B)
P(color-blind daughter)=12×12=14=0.25
43
Check q14 numeric entry

A carrier woman, XᴿXʳ, and a color-blind man, XʳY, have a child. The child is color-blind as a daughter or as a son; when either outcome will do, the two probabilities add.

Calculate the probability, as a decimal, that the child is color-blind.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(color-blind daughter, XʳXʳ) = 1/4
P(color-blind son, XʳY) = 1/4
Write down the equation:
P(A or B)=P(A)+P(B)
Substitute the values into the equation:
P(A or B)=P(A)+P(B)
P(color-blind)=14+14=12=0.5
44

Back to the woman who carries the color-blindness allele on one X, XᴿXʳ, and the man with ordinary vision, XᴿY. Here is their square filled in.

The grid with all four cells filled: X-R X-R and X-R X-r in the top row, X-R Y and X-r Y in the bottom row
The grid with all four cells filled: X-R X-R and X-R X-r in the top row, X-R Y and X-r Y in the bottom row
45

The top row holds the daughters, XᴿXᴿ and XᴿXʳ: none is color-blind, and half carry r.

46

The bottom row holds the sons, XᴿY and XʳY: half are color-blind, because a son’s only X came from his mother.

47Quick quiz: the probability of a named child mixed practice

48
Check q15 numeric entry

In an insect, a gene on the X has a dominant allele for dark eyes, Xᴰ, and a recessive allele for pale eyes, Xᵈ. A pale-eyed female, XᵈXᵈ, is crossed with a dark-eyed male, XᴰY.

Calculate the probability, as a decimal, that an offspring is a pale-eyed son.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(the sperm carries Y, so a son) = 1/2
P(the egg carries Xᵈ) = 1
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(pale-eyed son)=12×1=12=0.5
49
Check q16 numeric entry

A color-blind woman, XʳXʳ, and a color-blind man, XʳY, have a child.

Calculate the probability, as a decimal, that the child is a color-blind daughter.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(the sperm carries Xʳ, so a daughter) = 1/2
P(the egg carries Xʳ) = 1
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(color-blind daughter)=12×1=12=0.5
50
Check q17 numeric entry

In fruit flies, red eyes, Xᴿ, are dominant to white eyes, Xʳ, and the gene sits on the X. A carrier female, XᴿXʳ, is crossed with a white-eyed male, XʳY.

Calculate the probability, as a decimal, that an offspring is a white-eyed daughter.

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
P(the sperm carries Xʳ, so a daughter) = 1/2
P(the egg carries Xʳ) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(white-eyed daughter)=12×12=14=0.25
51
Check q18 numeric entry

In a mammal, a gene on the X has a dominant allele, Xᴬ, and a recessive allele, Xᵃ, whose trait shows in a male with one copy. A carrier female, XᴬXᵃ, is crossed with a male that shows the trait, XᵃY.

Calculate the probability, as a decimal, that an offspring is a son that shows the trait.

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
P(the sperm carries Y, so a son) = 1/2
P(the egg carries Xᵃ) = 1/2
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(son showing the trait)=12×12=14=0.25
52
Check q19 numeric entry

A color-blind woman, XʳXʳ, and a man with ordinary vision, XᴿY, have a child.

Calculate the probability, as a decimal, that the child is a daughter with ordinary vision.

Answer: 0.5  (tolerance ±0.005)

Working
Write down the values in the question:
P(the sperm carries Xᴿ, so a daughter) = 1/2
P(the daughter has ordinary vision, given the father’s Xᴿ) = 1
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(daughter with ordinary vision)=12×1=12=0.5

53Mixed practice mixed practice

54
Check q20 numeric entry

In fruit flies, red eyes, Xᴿ, are dominant to white eyes, Xʳ, and the gene sits on the X. A carrier female, XᴿXʳ, is crossed with a red-eyed male, XᴿY, and 200 offspring are counted.

Calculate how many of the 200 offspring are expected to have white eyes.

Answer: 50  (tolerance ±0)

Working
Write down the values in the question:
total offspring = 200
square = XᴿXᴿ, XᴿXʳ, XᴿY, XʳY: one cell of four is white-eyed, XʳY
white-eyed share = 1 of 4
Write down the equation:
e=total×that class's share of the ratio
Substitute the values into the equation:
e=total×that class's share of the ratio
ewhite-eyed=200×14=50
55
Check q21

A student draws the square below for a carrier woman, XᴿXʳ, and a color-blind man, XʳY.

A two-by-two square with X-R and X-r along the top edge and X-r and Y-r down the left edge; cells X-R X-r, X-r X-r, X-R Y-r, X-r Y-r
A two-by-two square with X-R and X-r along the top edge and X-r and Y-r down the left edge; cells X-R X-r, X-r X-r, X-R Y-r, X-r Y-r

What is wrong with the square?

  1. A. The mother and the father sit on the wrong edges
    Either parent may take either edge; the fault is what is written on the father’s Y.
  2. B. The cells should each hold one chromosome
    A cell holds the two sex chromosomes an egg and a sperm bring together, so two symbols is right for a cell.
  3. C. The mother’s edge should read Xʳ and Xʳ
    A carrier woman makes half Xᴿ eggs and half Xʳ eggs, so Xᴿ and Xʳ is right for her edge.
  4. D. ✓ An allele is written on the Y

Why: The Y carries no allele of this gene, so it is written bare, Y.
Writing Yʳ puts an allele where there is none, and the two son cells come out wrong.

56
Check q22

In fruit flies, a gene on the X has a dominant allele for normal wings and a recessive allele for miniature wings. A normal-winged female, XᴹXᵐ, who carries the miniature allele is crossed with a normal-winged male, XᴹY.

Which offspring can have miniature wings?

  1. A. None
    The mother carries Xᵐ without showing it, and a son who receives it has no second X to hide it.
  2. B. ✓ Half of the sons only
  3. C. Half of the daughters only
    Every daughter receives the father’s Xᴹ, which hides an Xᵐ from the mother.
  4. D. Half of the daughters and half of the sons
    The daughters all get the father’s Xᴹ and so have normal wings; only sons can show the recessive allele here.

Why: The square is XᴹXᴹ, XᴹXᵐ, XᴹY, XᵐY.
Every daughter has the father’s Xᴹ, so no daughter has miniature wings.
The XᵐY son has one X, and it carries the miniature allele.
Nothing hides it, so half of the sons have miniature wings.

57
Check q23

A color-blind boy has a father with ordinary vision.

Which parent gave the boy the X that carries the color-blindness allele?

  1. A. His father
    A father gives his son a Y, not an X, so no allele on the father’s X reaches a son from him.
  2. B. Either parent
    Each parent did give one sex chromosome, but the father’s was the Y; the X came from the mother.
  3. C. ✓ His mother
  4. D. Neither
    The allele is on an X, and the boy’s one X came from his mother, who carries it.

Why: A son receives his father’s Y and his mother’s X, so the X carrying the color-blindness allele came from his mother, who is a carrier.

58
Practice writing an answer

In fruit flies, white eyes are X-linked recessive. A white-eyed male fly has a red-eyed father.

(a) Determine which parent gave the male fly the X that carries the white-eye allele. (1 pt)

Model answer A male fly is XY, and he received his Y from his father.
So his father gave him no X.
His one X came from his mother.
So the X carrying the white-eye allele came from his mother, who carries the allele without showing it.
Rubric
  • Award 1 point for: his mother.
59
Practice writing an answer

Red-green color blindness is X-linked recessive. Write Xᴿ for the ordinary allele and Xʳ for the color-blindness allele. A color-blind woman, XʳXʳ, and a man with ordinary vision, XᴿY, plan a family.

(a) State each parent’s gametes and the children’s sex chromosomes in each of the four cells of the square for this cross. (1 pt)

Model answer The mother’s eggs are Xʳ and Xʳ.
The father’s sperm are Xᴿ and Y.
The four cells are XᴿXʳ, XᴿXʳ, XʳY and XʳY.The square for X-r X-r crossed with X-R Y, with the cells X-R X-r, X-R X-r, X-r Y and X-r YXᴿXʳXᴿXʳXʳYXʳYXʳXʳXᴿYeggs along the top, sperm down the sideevery daughter XᴿXʳ and every son XʳY
Rubric
  • Award 1 point for: the gametes Xʳ, Xʳ and Xᴿ, Y, with the cells XᴿXʳ, XᴿXʳ, XʳY, XʳY.

Slip Writing an allele on the Y. The Y carries no allele of this gene and is written bare.

(b) Calculate the probability that a child of this couple is color-blind. (1 pt)

Answer: 0.5  (tolerance ±0.005)

Model answer The probability is 0.5, because every son is color-blind and no daughter is; as a fraction, 12.
Working
Write down the values in the question:
P(son) = 1/2
P(color-blind, given a son) = 1
P(color-blind, given a daughter) = 0
Write down the equation:
P(color-blind)=P(son)×P(color-blind given a son)
Substitute the values into the equation:
P(color-blind)=12×1=12=0.5
Rubric
  • Award 1 point for: 1/2 (0.5).

(c) Predict the vision of the couple’s sons and of their daughters. (1 pt)

Model answer Every son is XʳY, so every son is color-blind.
Every daughter is XᴿXʳ.
Her father’s Xᴿ hides her r, so she sees red and green as most people do.
Rubric
  • Award 1 point for: all sons color-blind; all daughters with ordinary vision.

Slip Predicting half of the sons color-blind. Every egg of an XʳXʳ mother carries Xʳ, so every son receives it.

(d) One of the couple’s daughters later has a son with a man who has ordinary vision. Predict the probability that this boy is color-blind, and justify the value. (1 pt)

Model answer One half.
The daughter is XᴿXʳ: she received her mother’s Xʳ and her father’s Xᴿ, so she is a carrier and half of her eggs carry Xʳ.
Her son’s only X comes from her, because his father gave him a Y, which carries no allele of this gene.
So one son in two receives Xʳ, nothing hides it, and he is color-blind.
Rubric
  • Award 1 point for: one half (0.5), because the daughter is a carrier, XᴿXʳ, so half of her eggs carry Xʳ, and her son’s only X comes from her, with no allele on the Y to hide it.

Slip Saying the boy cannot be color-blind because both of his parents see red and green. His mother carries Xʳ without showing it, and his one X came from her.

APBIO-U05-L25A Swap the parents

Topic 5.4 · Non-Mendelian Genetics · 29 steps

Left: a photograph of a fruit fly from the side with a bright red eye. Right: a photograph of a fruit fly seen through a microscope, its eye pale. Between them the words: cross 1, red-eyed female with white-eyed male; cross 2, white-eyed female with red-eyed male; the same offspring both ways?
Left: a photograph of a fruit fly from the side with a bright red eye. Right: a photograph of a fruit fly seen through a microscope, its eye pale. Between them the words: cross 1, red-eyed female with white-eyed male; cross 2, white-eyed female with red-eyed male; the same offspring both ways?

Photos: André Karwath, Wikimedia Commons, CC BY-SA 2.5; Paul Reynolds, Wikimedia Commons, CC BY 2.5 (cropped).

Imagine two crosses of fruit flies set up both ways round. In the first, a red-eyed female is crossed with a white-eyed male. In the second, a white-eyed female is crossed with a red-eyed male.

For a gene on an autosome the two crosses would give the same offspring. Here they do not: in the second cross every son has white eyes. What does that tell you about where the eye-color gene sits?

Unit 5 · Heredity

1Swap the parents

2

Video: Watch: Swap the parents

The two fly crosses drawn side by side, cross 1 filling with red-eyed offspring and cross 2 with red-eyed daughters and white-eyed sons; then a son’s X traced back to his mother.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25Aa.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25Aa.mp4

3

Does it matter which parent carries the allele?

4

For a gene on an autosome it does not. For a gene on the X it does.

5
Check q1

In guinea pigs, black coat, B, is dominant to white coat, b, and the gene sits on an autosome. A bb mother and a Bb father have pups.

Which of the following are their pups?

  1. A. ✓ Half black and half white
  2. B. All of the pups black
    The Bb father gives b to half of his pups, and the bb mother gives every pup a b.
    Those pups are bb: white.
  3. C. All of the pups white
    The Bb father gives B to half of his pups.
    One B is enough for a black coat.

Why: Each parent gives each pup one allele, whatever its sex.
The bb mother gives every pup a b; the Bb father gives half of them B and half b.
So half the pups are Bb, black, and half bb, white.

6

Suppose you cross a red-eyed female fruit fly with a white-eyed male. Then you set up the same cross the other way round: a white-eyed female with a red-eyed male.

7

When the same two types are crossed again with the mother’s type and the father’s type swapped, we call the second cross a , because reciprocal means the other way round.

8

For a gene on an autosome, a reciprocal cross gives the same offspring as the first cross, because each parent gives one allele whichever sex it is.

9

In fruit flies, red eyes, Xᴿ, are dominant to white eyes, Xʳ, and the gene sits on the X. Here the two crosses do not match.

10

Cross 1: a red-eyed female, XᴿXᴿ, with a white-eyed male, XʳY. Every daughter is XᴿXʳ and every son is XᴿY, so every offspring has red eyes.

Two squares side by side. Left: a red-eyed female X-R X-R crossed with a white-eyed male X-r Y; X-R and X-R along the top, X-r and Y down the side; cells X-R X-r, X-R X-r, X-R Y, X-R Y, all red-eyed. Right: a white-eyed female X-r X-r crossed with a red-eyed male X-R Y; X-r and X-r along the top, X-R and Y down the side; cells X-R X-r, X-R X-r red-eyed daughters, X-r Y, X-r Y white-eyed sons
Two squares side by side. Left: a red-eyed female X-R X-R crossed with a white-eyed male X-r Y; X-R and X-R along the top, X-r and Y down the side; cells X-R X-r, X-R X-r, X-R Y, X-R Y, all red-eyed. Right: a white-eyed female X-r X-r crossed with a red-eyed male X-R Y; X-r and X-r along the top, X-R and Y down the side; cells X-R X-r, X-R X-r red-eyed daughters, X-r Y, X-r Y white-eyed sons
11

Cross 2, the reciprocal cross: a white-eyed female, XʳXʳ, with a red-eyed male, XᴿY.

12

Every daughter is XᴿXʳ, red-eyed, and every son is XʳY, white-eyed.

13

Why do the sons follow their mother? A son’s only X came from his mother, and his Y carries no allele of this gene.

A son's two sex chromosomes: a long X, darker, with an allele box reading r, and a shorter Y, lighter, with no allele box; caption: his X came from his mother, his Y from his father
A son's two sex chromosomes: a long X, darker, with an allele box reading r, and a shorter Y, lighter, with no allele box; caption: his X came from his mother, his Y from his father
14

So in cross 2 every son received an Xʳ from his white-eyed mother, and nothing hides it: every son has white eyes.

15

A daughter received an X from each parent. In both crosses one of her X’s carries Xᴿ, so every daughter has red eyes.

16

For an X-linked gene, then, the reciprocal cross gives a different result from the first cross, and the sons take after their mother. That difference is evidence that the gene sits on the X.

17

What you are expected to know Predict both crosses of a reciprocal cross for an X-linked gene, and use the difference between them as evidence that the gene sits on the X.

18
Check q2

In an insect, a gene on the X has a dominant allele for dark wings, Xᴰ, and a recessive allele for pale wings, Xᵈ. Cross 1: a dark-winged female, XᴰXᴰ, with a pale-winged male, XᵈY. Cross 2: a pale-winged female, XᵈXᵈ, with a dark-winged male, XᴰY.

Which cross gives pale-winged sons?

  1. A. Cross 1 only
    In cross 1 the mother is XᴰXᴰ, so every egg carries Xᴰ.
    Every son receives an Xᴰ, so every son has dark wings.
  2. B. ✓ Cross 2 only
  3. C. Both crosses
    In cross 1 the mother is XᴰXᴰ, so every son receives an Xᴰ and has dark wings.
    Only cross 2 has an XᵈXᵈ mother.

Why: A son’s only X comes from his mother.
In cross 2 the mother is XᵈXᵈ, so every son receives an Xᵈ, and nothing hides it.
In cross 1 the mother is XᴰXᴰ, so every son receives an Xᴰ.

19
Practice writing an answer

In an insect, a gene on the X has a dominant allele for dark wings, Xᴰ, and a recessive allele for pale wings, Xᵈ. A pale-winged female, XᵈXᵈ, is crossed with a dark-winged male, XᴰY. Every son has pale wings.

(a) Explain why every son has pale wings. (1 pt)

Frame A son’s only X …

Model answer A son’s only X came from his mother.
His mother is XᵈXᵈ, so every egg she makes carries Xᵈ.
His Y came from his father and carries no allele of this gene.
So nothing hides the Xᵈ, and every son has pale wings.
Rubric
  • Award 1 point for: a son’s only X comes from his mother, who is XᵈXᵈ, and his Y carries no allele to hide the Xᵈ.
20
Check q3

A student says: “For any gene, swapping which parent is the mother and which is the father leaves the offspring the same.”

Which of the following is correct?

  1. A. The claim is true for every gene
    For a gene on the X, a son’s only X comes from his mother.
    Swapping the parents changes which mother gives the sons their X, so the sons change.
  2. B. ✓ The claim is true for a gene on an autosome only
  3. C. The claim is false for every gene
    For a gene on an autosome, each parent gives one allele whichever sex it is.
    Swapping the parents changes nothing there.

Why: For a gene on an autosome, each parent gives one allele whichever sex it is, so the reciprocal cross gives the same offspring.
For a gene on the X, a son’s only X comes from his mother, so the reciprocal cross gives different sons.

21
Check q4

In a mammal, a gene on the X has a dominant allele for a banded coat, Xᴮ, and a recessive allele for a plain coat, Xᵇ. A plain female, XᵇXᵇ, is crossed with a banded male, XᴮY.

Which coats do the offspring have?

  1. A. ✓ Daughters banded and sons plain
  2. B. Daughters plain and sons banded
    A daughter receives the father’s Xᴮ, so she is banded.
    A son’s only X comes from the plain mother, so he is plain.
  3. C. Every offspring has a banded coat
    Every son’s only X comes from the XᵇXᵇ mother.
    So every son has a plain coat.
  4. D. Every offspring has a plain coat
    Every daughter receives the father’s Xᴮ.
    One Xᴮ is enough for a banded coat.

Why: The mother carries Xᵇ on both X’s.
Every son’s only X comes from her, so every son is plain.
Every daughter also receives the father’s Xᴮ, so every daughter is banded.

22

Back to the two crosses of fruit flies set up both ways round. A red-eyed female with a white-eyed male gave every offspring red eyes.

23

The reciprocal cross, a white-eyed female with a red-eyed male, gave red-eyed daughters and white-eyed sons.

24

The sons followed their mother, because a son’s only X came from her. So the eye-color gene sits on the X.

25Quick quiz: reciprocal cross mixed practice

26
Check q5

A breeder’s first cross has a long-haired mother and a short-haired father. Then the breeder sets up a second cross.

Which of the following second crosses is the reciprocal cross of the first?

  1. A. A long-haired female with a short-haired male, again
    Repeating the same cross keeps the mother’s type and the father’s type where they were.
    A reciprocal cross swaps them.
  2. B. Two long-haired animals crossed with each other
    Crossing two long-haired animals changes the types crossed.
    A reciprocal cross keeps the two types and swaps which is the mother.
  3. C. ✓ A short-haired female with a long-haired male

Why: A reciprocal cross keeps the same two types and swaps the mother’s type with the father’s.
So the second cross is a short-haired female with a long-haired male.

27
Check q6

A breeder crosses two types of pea plant.

Which of the following describes the reciprocal cross?

  1. A. ✓ The same two types crossed again, with the parents’ types swapped
  2. B. The offspring of the first cross crossed with each other to give an F2
    Crossing the offspring with each other is the F1 × F1 cross that gives the F2.
    A reciprocal cross uses the two parent types.
  3. C. One of the two types crossed with a homozygous recessive individual
    Crossing with a homozygous recessive individual is a test cross.
    A reciprocal cross swaps the parents’ types.

Why: A reciprocal cross is the same two types crossed again, with the mother’s type and the father’s type swapped.

28
Practice writing an answer

In fruit flies, red eyes are dominant to white eyes. A red-eyed female crossed with a white-eyed male gives all red-eyed offspring. The reciprocal cross, a white-eyed female with a red-eyed male, gives red-eyed daughters and white-eyed sons.

(a) Explain how the two results demonstrate that the eye-color gene sits on the X. (2 pt)

Model answer For a gene on an autosome, each parent gives one allele whichever sex it is, so the reciprocal cross would give the same offspring.
Here the reciprocal cross gives different offspring: the sons follow their mother.
A son’s only X comes from his mother, and his Y carries no allele of the gene.
So a white-eyed mother gives every son white eyes only if the gene sits on the X.
Rubric
  • Award 1 point for: a gene on an autosome would give the same offspring in both crosses, and here the two crosses differ.
  • Award 1 point for: the sons take after their mother because a son’s only X comes from her, so the gene sits on the X.

Glossary

reciprocal cross
The same two types crossed again with the mother's type and the father's type swapped. For a gene on an autosome it gives the same offspring as the first cross; for a gene on the X it gives different sons, who take after their mother.

APBIO-U05-L25B Birds, bees and the Y

Topic 5.4 · Non-Mendelian Genetics · 52 steps

Left: a photograph of a brown rooster with a tall red comb standing between two white hens. Right: an empty two-by-two grid with question marks in its four cells, labelled the hen's eggs along the top and the rooster's sperm down the side; a caption reads in birds the egg sets the sex: how?
Left: a photograph of a brown rooster with a tall red comb standing between two white hens. Right: an empty two-by-two grid with question marks in its four cells, labelled the hen's eggs along the top and the rooster's sperm down the side; a caption reads in birds the egg sets the sex: how?

Photo: Simon Waldherr, Wikimedia Commons, CC BY-SA 4.0 (resized).

Bee photos: USGS Bee Inventory and Monitoring Lab, Wikimedia Commons, CC BY 2.0 and public domain (resized).

Here is a rooster between two hens. In people the sperm sets the sex: an X sperm makes a daughter and a Y sperm makes a son.

In birds it is the egg that sets the sex. How does the egg do it?

Unit 5 · Heredity

1The egg sets the sex

2

Video: Watch: The egg sets the sex

The hen’s Z and W beside the rooster’s two Z’s, then the square filling: every sperm a Z, the egg’s Z or W setting each chick’s sex.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25Ba.mp4

3

Do all species set sex the way people do? They do not, but following the chromosomes works the same way in every species.

4
Check q1

A mother is XX and a father is XY.

Which gamete sets whether a child is a son or a daughter?

  1. A. The egg
    Every egg carries an X.
    The eggs are all alike, so the egg cannot set the sex.
  2. B. ✓ The sperm

Why: Every egg carries an X.
A sperm carries an X or a Y.
An X sperm makes a daughter and a Y sperm makes a son, so the sperm sets the sex.

5

In birds the sex chromosomes are written Z and W, not X and Y. A hen is ZW and a rooster is ZZ.

Left: a hen's sex chromosomes, a long Z beside a shorter W. Right: a rooster's, two long Z's
Left: a hen's sex chromosomes, a long Z beside a shorter W. Right: a rooster's, two long Z's
6

So every sperm a rooster makes carries a Z. Each egg a hen makes carries either a Z or a W.

7

Here is the square: the hen’s eggs, Z and W, along the top edge and the rooster’s sperm, Z and Z, down the side.

A two-by-two square: the hen's eggs, Z and W, along the top edge; the rooster's sperm, Z and Z, down the left edge; cells ZZ, ZW in the top row and ZZ, ZW in the bottom row, the ZW cells shaded
A two-by-two square: the hen's eggs, Z and W, along the top edge; the rooster's sperm, Z and Z, down the left edge; cells ZZ, ZW in the top row and ZZ, ZW in the bottom row, the ZW cells shaded
8

A Z egg with a Z sperm makes ZZ, a male chick. A W egg with a Z sperm makes ZW, a female chick.

9

So in birds the egg sets the chick’s sex, because the sperm are all alike and the eggs differ.

10

What you are expected to know Explain how a hen’s egg sets a chick’s sex: a hen is ZW and a rooster is ZZ.

11
Check q2

In birds, females are ZW and males are ZZ.

Which gamete sets the sex of a chick?

  1. A. ✓ The egg
  2. B. The sperm
    A rooster is ZZ.
    So every sperm carries a Z, and none carries a W. The sperm is the same for every chick.
  3. C. Both gametes
    The rooster gives a Z to every chick, the same for all of them.
    The difference between a ZZ chick and a ZW chick comes from the egg.

Why: A ZZ rooster gives every chick a Z; a ZW hen gives an egg carrying Z or W.
A Z egg makes a ZZ male and a W egg a ZW female, so the egg sets the sex.

12
Check q3

In birds, females are ZW and males are ZZ. A chick is ZW.

Which parent gave the chick its W?

  1. A. The rooster
    A rooster is ZZ.
    He has no W to give.
  2. B. ✓ The hen
  3. C. Either parent
    Only the hen carries a W.
    The rooster is ZZ.

Why: A rooster is ZZ, so every sperm carries a Z.
The W in a ZW chick can only have come from the hen’s egg.

13A drone from an unfertilized egg

14

Video: Watch: A drone from an unfertilized egg

Two eggs side by side: one that no sperm fertilized growing into a drone with one chromosome set, one that a sperm fertilized growing into a worker with two.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25Bb.mp4

15

Now consider honeybees, which set the sex in a third way: by the number of chromosome sets, not by one pair of chromosomes.

Left: a photograph of a drone honeybee from the side, stout, with very large eyes that cover the top of its head. Right: a photograph of a worker honeybee from the side, slimmer, with smaller eyes at the sides of its head
Left: a photograph of a drone honeybee from the side, stout, with very large eyes that cover the top of its head. Right: a photograph of a worker honeybee from the side, slimmer, with smaller eyes at the sides of its head
16
Check q4

An egg is a gamete.

How many chromosome sets does an egg carry?

  1. A. ✓ One set
  2. B. Two sets
    Meiosis halves the number of sets.
    A gamete carries one set: it is haploid.

Why: Meiosis makes the gametes and halves the number of chromosome sets.
So an egg carries one set: it is haploid.

17

A queen lays two kinds of egg: eggs that a sperm fertilized, and eggs that no sperm fertilized.

18

A male bee, a drone, develops from an egg that no sperm fertilized. The egg carries one set, and no sperm added a second, so a drone is haploid.

Left: an egg cell alone, labeled n, with an arrow to a drone labeled male, haploid. Right: an egg cell fusing with a sperm cell, labeled 2n, with an arrow to a worker labeled female, diploid
Left: an egg cell alone, labeled n, with an arrow to a drone labeled male, haploid. Right: an egg cell fusing with a sperm cell, labeled 2n, with an arrow to a worker labeled female, diploid
19

A female bee, a worker or a queen, develops from a fertilized egg. The egg’s set and the sperm’s set make two, so a female bee is diploid.

20

What you are expected to know Explain why a drone bee is haploid: it develops from an egg that no sperm fertilized.

21
Check q5

In honeybees, a drone, a male bee, has a mother but no father.

How many chromosome sets does a drone carry?

  1. A. ✓ One set
  2. B. Two sets
    A drone develops from an egg that no sperm fertilized.
    An egg carries one set.
    No sperm added a second set.
  3. C. Three sets
    An egg carries one set, and an unfertilized egg stays at one set.

Why: An egg is haploid: it carries one set.
A drone develops from an egg that no sperm fertilized.
So no sperm added a second set, and the drone stays haploid, with one set.

22Father to every son

23

Video: Watch: Father to every son

A father’s Y with its allele traced to each son and to no daughter; then the table of the three species filling row by row.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L25Bc.mp4

24

Now consider people again, and a trait whose gene sits on the Y.

25
Check q6

A trait’s gene sits on the Y chromosome.

Which of the following names the trait?

  1. A. X-linked
    An X-linked gene sits on the X.
    This gene sits on the Y.
  2. B. ✓ Y-linked
  3. C. Autosomal
    An autosomal gene sits on one of the ordinary pairs, not on a sex chromosome.

Why: The gene sits on the Y, so the trait is Y-linked.

26

A father gives his Y to every son and to no daughter. So a Y-linked allele reaches all of his sons and none of his daughters.

A father's X and Y at the top, the Y carrying an allele box labeled a; at the bottom a daughter's two X's with no allele box, and a son's X and Y with the a box on the Y. An arrow runs from the father's Y to the son only
A father's X and Y at the top, the Y carrying an allele box labeled a; at the bottom a daughter's two X's with no allele box, and a son's X and Y with the a box on the Y. An arrow runs from the father's Y to the son only
27

A daughter has no Y, so she cannot carry a Y-linked allele, not even hidden.

28

Here is a table comparing people, birds and honeybees: the female’s chromosomes, the male’s chromosomes, and what sets the sex of the young.

A table with four columns, species, female, male and what sets the sex of the young, and three rows: people, XX, XY, the sperm's X or Y; birds, ZW, ZZ, the egg's Z or W; honeybees, diploid 2n, haploid n, whether a sperm fertilized the egg
A table with four columns, species, female, male and what sets the sex of the young, and three rows: people, XX, XY, the sperm's X or Y; birds, ZW, ZZ, the egg's Z or W; honeybees, diploid 2n, haploid n, whether a sperm fertilized the egg
29

The rule differs from species to species, but following the chromosomes works the same way: find which gamete differs, and that gamete sets the sex.

30

What you are expected to know Predict the route of a Y-linked trait: from a father to every son and to no daughter.

31
Check q7

A gene sits on the Y chromosome, and a man carries an allele of it that shows as a trait.

Which of his children receive that allele?

  1. A. None of his children
    A father does pass his Y on: every son receives it.
  2. B. His daughters only
    A father gives his daughters his X, never his Y.
    So an allele on the Y cannot reach a daughter.
  3. C. ✓ His sons only
  4. D. Half of his daughters and half of his sons
    Every son receives his father’s Y, and no daughter does.
    So there is no half and half.

Why: A father gives his Y to every son and to no daughter, so an allele on the Y reaches all of his sons and none of his daughters.

32
Check q8

A man carries an allele on his Y that shows as a trait. A student says: “His daughters carry the allele hidden.”

Which of the following is correct?

  1. A. Yes, each daughter carries the allele hidden
    A father gives each daughter his X, never his Y.
    An allele on the Y cannot reach her, hidden or not.
  2. B. No, each daughter shows the trait
    A daughter receives no Y from her father.
    So she carries no allele from it and cannot show it.
  3. C. ✓ No, a daughter receives no Y from him

Why: A father gives his X to each daughter and his Y to each son.
A daughter receives no Y, so no allele on the Y reaches her, hidden or shown.

33

Here again are the hen and the rooster, with their sex chromosomes.

Left: a hen's sex chromosomes, a long Z beside a shorter W. Right: a rooster's, two long Z's
Left: a hen's sex chromosomes, a long Z beside a shorter W. Right: a rooster's, two long Z's
34

A hen is ZW, so her eggs carry a Z or a W. A rooster is ZZ, so every sperm carries a Z.

35

A W egg makes a ZW chick, a female, and a Z egg makes a ZZ chick, a male. So the hen’s egg sets each chick’s sex.

36

Bees set sex by the number of chromosome sets, and a Y-linked trait passes from a father to his sons. The rules differ, but following the chromosomes works the same way.

37Quick quiz: which gamete sets the sex mixed practice

38
Check q9

In people, a mother is XX and a father is XY. A sperm carrying a Y fertilizes an egg.

Which does the child become?

  1. A. ✓ A son
  2. B. A daughter
    The egg carries an X and the sperm a Y.
    XY is a son.

Why: The egg carries an X and the sperm a Y, so the child is XY: a son.

39
Check q10

In birds, a hen is ZW and a rooster is ZZ. A hen’s egg carries a W, and a sperm fertilizes it.

Which does the chick become?

  1. A. A male, ZZ
    The egg carries a W and the sperm a Z.
    ZW is a female.
  2. B. ✓ A female, ZW

Why: Every sperm carries a Z.
The egg carries a W, so the chick is ZW: a female.

40
Check q11

In birds, a hen is ZW and a rooster is ZZ.

Which sex chromosome does every sperm of a rooster carry?

  1. A. ✓ Z
  2. B. W
    A rooster is ZZ and has no W.
  3. C. Z or W
    A rooster is ZZ.
    Both of his sex chromosomes are Z, so every sperm carries a Z.

Why: A rooster is ZZ.
Each sperm receives one of his two sex chromosomes, and both are Z.

41
Check q12

In honeybees, a drone develops from an egg that no sperm fertilized, and a worker from a fertilized egg. A queen lays an egg that no sperm fertilizes.

Which does the egg become?

  1. A. ✓ A drone
  2. B. A worker
    A worker develops from a fertilized egg.
    No sperm fertilized this egg.

Why: No sperm fertilized the egg.
An unfertilized egg develops into a drone.

42
Check q13

In honeybees, a drone is haploid and a worker is diploid.

How many chromosome sets does a worker carry?

  1. A. One set
    One set is haploid: a drone.
    A worker is diploid.
  2. B. ✓ Two sets

Why: Diploid means two chromosome sets.
A worker is diploid, so she carries two sets.

43Mixed practice mixed practice

44
Check q14

In birds, a hen is ZW and a rooster is ZZ. A hen’s egg carries a Z, and a sperm fertilizes it.

Which does the chick become?

  1. A. ✓ A male
  2. B. A female
    The egg carries a Z and the sperm a Z.
    ZZ is a male.

Why: Every sperm carries a Z.
The egg carries a Z, so the chick is ZZ: a male.

45
Check q15

In birds, a hen is ZW and a rooster is ZZ. A chick is female.

Which sex chromosome did the egg carry?

  1. A. Z
    A Z egg with a Z sperm makes ZZ: a male.
    A female chick is ZW, and her W came from the egg.
  2. B. ✓ W

Why: A female chick is ZW.
Every sperm carries a Z, so the W came from the egg.

46
Check q16

In honeybees, a queen lays an egg and a sperm fertilizes it.

Which does the egg become?

  1. A. ✓ A worker or a queen
  2. B. A drone
    A drone develops from an egg that no sperm fertilized.
    This egg was fertilized.

Why: A sperm fertilized the egg, so the egg carries two sets.
A fertilized egg develops into a female: a worker or a queen.

47
Check q17

A grandfather carries an allele on his Y. His daughter has a son.

Can the grandson have received that allele from his grandfather?

  1. A. Yes
    The grandfather gave his daughter an X, not his Y.
    She has no Y to pass on, so the allele stopped with her generation.
  2. B. ✓ No

Why: The grandfather gave his daughter his X, never his Y.
She has no Y, so her son’s Y came from his own father.
The grandfather’s allele cannot reach the grandson.

48
Check q18

A man carries an allele on his Y that shows as a trait. He has three sons and two daughters.

How many of his five children receive the allele?

  1. A. Two
    His two daughters receive his X, never his Y.
    The allele reaches his sons instead.
  2. B. ✓ Three
  3. C. Five
    His daughters receive no Y.
    Only his three sons receive the allele.

Why: A father gives his Y to every son and to no daughter.
He has three sons, so three children receive the allele.

49
Check q19

In people the sperm sets the sex. In birds the egg sets the sex.

Which of the following is the reason the egg sets the sex in birds?

  1. A. ✓ Every sperm carries a Z, and an egg carries a Z or a W
  2. B. Every egg carries a Z, and a sperm carries a Z or a W
    A rooster is ZZ, so his sperm are all alike.
    The hen is ZW, so her eggs differ.
  3. C. A hen carries no sex chromosomes
    A hen carries two sex chromosomes, a Z and a W.

Why: The gamete that differs sets the sex.
A ZZ rooster’s sperm all carry a Z; a ZW hen’s eggs carry a Z or a W.
So the egg sets the sex.

50
Practice writing an answer

In birds, females are ZW and males are ZZ. The egg sets the sex of the chick.

(a) Explain why the egg, rather than the sperm, sets the sex in birds. (1 pt)

Model answer A ZZ rooster gives every chick a Z; a ZW hen gives an egg carrying Z or W.
A Z egg makes a ZZ male and a W egg a ZW female, so the egg sets the sex.
Rubric
  • Award 1 point for: the hen (ZW) makes Z eggs and W eggs, while every sperm from the rooster (ZZ) carries Z.
51
Practice writing an answer

In honeybees, a drone, a male bee, has a mother but no father. A drone carries one chromosome set.

(a) Explain why a drone carries one chromosome set rather than two. (1 pt)

Frame A drone develops from …

Model answer A drone develops from an egg that no sperm fertilized.
An egg is haploid, so it carries one chromosome set.
Fertilization is what adds a second set, the sperm’s.
No sperm fertilized this egg, so no sperm added a second set.
Therefore the drone carries one set.
Rubric
  • Award 1 point for: the drone develops from an unfertilized egg, and an egg carries one set, so no sperm added a second.

APBIO-U05-L26 One gene, many effects

Topic 5.4 · Non-Mendelian Genetics · 36 steps

Left: a photograph of a stained blood film, one red blood cell bent into a sickle in the middle of a field of round red blood cells. Right: a drawn round red blood cell and a drawn red blood cell bent into a sickle beside it; three arrows run from the sickled cell to three labels: anemia, episodes of pain, damage to the kidneys
Left: a photograph of a stained blood film, one red blood cell bent into a sickle in the middle of a field of round red blood cells. Right: a drawn round red blood cell and a drawn red blood cell bent into a sickle beside it; three arrows run from the sickled cell to three labels: anemia, episodes of pain, damage to the kidneys

Here is a red blood cell from a patient with sickle cell disease, bent into a sickle, beside a round one. The disease brings the patient three problems: anemia, episodes of pain, and damage to the kidneys.

Three problems, and behind them one changed amino acid in one protein. How does one change do three things?

Photo: Gregory Kato, Wikimedia Commons, CC BY-SA 4.0 (resized).

Unit 5 · Heredity

1One protein, three problems

2

Video: Watch: One protein, three problems

The Unit 1 hemoglobin drawing recalled, the red cell bending into a sickle, then the rigid cell breaking down, wedging in a small vessel, and starving a kidney.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L26a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L26a.mp4

3

How can one gene change several traits at once? Sickle cell disease is the case to follow.

4

The sickle cell allele changes one amino acid in hemoglobin. That one change makes the red cells rigid, and the rigid cells cause all three problems.

5
Check q1

In sickle cell disease one hemoglobin chain carries a swapped amino acid near its start.

What do the changed hemoglobin molecules do at low oxygen concentration?

  1. A. ✓ They stick together into stiff fibers
  2. B. They break apart into single amino acids
    The swapped amino acid still forms peptide bonds on both sides, so the chain stays whole.
  3. C. They leave the red cell
    Hemoglobin stays inside the red cell.
    The changed molecules stick to each other there.

Why: The swapped amino acid gives the hemoglobin molecule a sticky patch.
At low oxygen concentration the molecules stick together into stiff fibers.
The fibers bend the red cell into a sickle.

6

Here is the chain from the swapped amino acid to the sickled cell.

Left: a chain of nine amino-acid beads with the sixth bead marked as the swapped amino acid, labeled one hemoglobin chain. An arrow leads to a stack of four straight rods labeled stiff fibers at low oxygen concentration. A second arrow leads to a red cell bent into a sickle, labeled the red cell sickles
Left: a chain of nine amino-acid beads with the sixth bead marked as the swapped amino acid, labeled one hemoglobin chain. An arrow leads to a stack of four straight rods labeled stiff fibers at low oxygen concentration. A second arrow leads to a red cell bent into a sickle, labeled the red cell sickles
7

A sickled cell is rigid. A round red cell folds to squeeze through the narrowest vessels.

8

A rigid cell cannot fold. So it wedges.

A small blood vessel drawn as two lines that narrow toward the right. On the left a round red cell squeezes through the narrow part by folding. On the right a sickled red cell is wedged across the narrow part, and behind it the label reads: no blood gets past
A small blood vessel drawn as two lines that narrow toward the right. On the left a round red cell squeezes through the narrow part by folding. On the right a sickled red cell is wedged across the narrow part, and behind it the label reads: no blood gets past
9

Rigid cells break down early, so the blood holds too few red cells. Too few red cells is the anemia.

10

Rigid cells also block small vessels, so the tissue beyond the block gets too little oxygen. That shortage is the pain.

11

Blockages happen again and again, and each one starves an organ of oxygen for a while. In the kidneys the starved tissue is damaged a little more each time.

A chain of boxes from left to right: one changed gene, then changed hemoglobin, then rigid red cells; from the last box three arrows lead to three boxes: anemia, from cells breaking down early; pain, from small vessels blocked; kidney damage, from repeated blockages
A chain of boxes from left to right: one changed gene, then changed hemoglobin, then rigid red cells; from the last box three arrows lead to three boxes: anemia, from cells breaking down early; pain, from small vessels blocked; kidney damage, from repeated blockages
12

Here is a table of the three problems and what the rigid cells do to cause each one.

A table with three rows: anemia is caused by rigid cells breaking down early, so the blood holds too few red cells; episodes of pain are caused by rigid cells blocking small vessels, so the tissue beyond gets too little oxygen; kidney damage is caused by blockages that happen again and again, each starving the kidney of oxygen
A table with three rows: anemia is caused by rigid cells breaking down early, so the blood holds too few red cells; episodes of pain are caused by rigid cells blocking small vessels, so the tissue beyond gets too little oxygen; kidney damage is caused by blockages that happen again and again, each starving the kidney of oxygen
13

In one study, researchers grew red blood cells from patients' cells and held them at low oxygen concentration: 68 % became rigid. In the same kind of cells with the one changed gene repaired, only 9 % did.

A table with two rows: red blood cells grown from patients' cells and held at low oxygen concentration, 68 % became rigid; the same kind of cells with the one changed gene repaired, 9 % became rigid
A table with two rows: red blood cells grown from patients' cells and held at low oxygen concentration, 68 % became rigid; the same kind of cells with the one changed gene repaired, 9 % became rigid
14

So the one gene causes the rigid cells, and the rigid cells cause all three problems. One gene, one changed protein, three effects.

15

What you are expected to know Explain how one changed amino acid in hemoglobin leads to anemia, episodes of pain and kidney damage.

16
Check q2

A patient with sickle cell disease has red blood cells that stay stiff instead of folding.

Which of the following makes the red cells rigid?

  1. A. Too few red cells left in the blood
    Too few red cells is the anemia.
    The rigid cells cause it; it does not cause them.
  2. B. ✓ Hemoglobin molecules stuck into stiff fibers
  3. C. Small vessels blocked by other red cells
    A blocked vessel is what a rigid cell causes, not what makes it rigid.
  4. D. A second gene active in the red cell
    One changed gene makes the changed hemoglobin.
    No second gene is needed.

Why: The changed hemoglobin molecules stick together into stiff fibers at low oxygen concentration.
The fibers bend the red cell and keep it stiff.
So the red cell cannot fold.

17
Practice writing an answer

A patient with sickle cell disease has anemia: the blood holds too few red cells.

(a) Explain how the changed hemoglobin leads to the anemia. (1 pt)

Model answer At low oxygen concentration the changed hemoglobin molecules stick together into stiff fibers.
The fibers make the red cell rigid.
Rigid red cells break down early.
So the blood holds too few red cells, which is the anemia.
Rubric
  • Award 1 point for: the changed hemoglobin makes the red cells rigid, and rigid cells break down early, so the blood holds too few red cells.

18One gene, several traits

19

Video: Watch: One gene, several traits

The chain from one gene to three problems drawn as one line, then a child inheriting the allele and every arrow lighting up together.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L26b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L26b.mp4

20

The anemia, the pain and the kidney damage look like three separate traits. All three come from one gene.

21

When one gene's expression changes several traits at once, we call it : pleio- means more and -tropy means turning, so one gene turns several ways.

22

Traits that come from one such gene travel as a set.

23

A child who inherits the sickle cell allele from both parents inherits every effect of it, because the anemia, the pain and the kidney damage all come from that one allele.

24

So the three problems are never inherited separately: no child gets the pain without the anemia, and none gets the anemia without the risk to the kidneys.

25

What you are expected to know Identify pleiotropy, one gene whose expression changes several traits, and predict that its traits are inherited together.

26
Check q3

A student says: “The anemia of sickle cell disease and the risk of kidney damage are two different traits, so a child could inherit the anemia alone.”

Is the student correct?

  1. A. Yes
    Both problems come from the one sickle cell allele, through the rigid red cells it produces, so a child who inherits the allele inherits every effect of it.
  2. B. ✓ No

Why: The anemia and the kidney damage both come from one allele of one gene.
That allele makes the red cells rigid.
Rigid cells break down early, the anemia, and block small vessels, damaging the kidneys.
A child inherits the allele whole, so cannot inherit one effect without the other.

27
Check q4

A child inherits the sickle cell allele from both parents.

Which of the following does inheriting the allele predict for the child?

  1. A. One of the three problems, chosen by chance
    The three problems all come from the rigid red cells the allele produces.
    So they arrive together, not one at a time by chance.
  2. B. Anemia only
    The pain and the kidney damage come from the same rigid cells that cause the anemia.
    No other gene is needed.
  3. C. ✓ Anemia, pain and kidney damage together
  4. D. Problems that depend on which parent gave which allele
    The two alleles are the same allele.
    Either parent's copy makes the same changed hemoglobin.

Why: The sickle cell allele makes every red cell rigid.
Rigid cells break down early, so the child has anemia.
Rigid cells block small vessels, so the child has episodes of pain.
Repeated blockages damage the kidneys.
So the child faces all three problems together.

28
Check q5

Suppose that in one patient with sickle cell disease, 72 % of the red cells are rigid. Researchers repair the one faulty gene in a sample of that patient's blood-making cells. Of the red cells those repaired cells make, 8 % are rigid.

What does the drop from 72 % to 8 % show?

  1. A. A low oxygen concentration changes the gene in most of the cells
    The gene was the same in every cell before repair; the low oxygen concentration made the cells rigid because of the changed hemoglobin the gene already coded for.
  2. B. ✓ The one changed gene is behind the rigid cells
  3. C. A second gene makes the remaining cells rigid
    Repairing one gene removed almost all of the rigidity, which one gene could not do if a second gene were the cause.
  4. D. Repairing the gene works in 8 % of the cells
    8 % is the share of cells still rigid after repair, not the share repaired; repair left 92 % flexible.

Why: The only difference between the two batches of cells is the one repaired gene.
Repairing that gene dropped the share of rigid cells from 72 % to 8 %.
So that one gene is behind the rigid cells.

29

Here again is the red blood cell bent into a sickle, and the patient with anemia, episodes of pain and damage to the kidneys.

Left: a chain of nine amino-acid beads with the sixth bead marked as the swapped amino acid, labeled one hemoglobin chain. An arrow leads to a stack of four straight rods labeled stiff fibers at low oxygen concentration. A second arrow leads to a red cell bent into a sickle, labeled the red cell sickles
Left: a chain of nine amino-acid beads with the sixth bead marked as the swapped amino acid, labeled one hemoglobin chain. An arrow leads to a stack of four straight rods labeled stiff fibers at low oxygen concentration. A second arrow leads to a red cell bent into a sickle, labeled the red cell sickles
30

One changed amino acid made the hemoglobin stick into stiff fibers, and the fibers made the red cells rigid.

31

Rigid cells broke down, blocked vessels and starved the kidneys, so the three problems were one gene's work. One gene's effects cannot be inherited separately.

32Quick quiz: pleiotropy mixed practice

33
Check q6

What is pleiotropy?

  1. A. Two genes on one chromosome inherited together
    Two genes on one chromosome inherited together are linked genes.
    Pleiotropy is one gene with several effects.
  2. B. ✓ One gene whose expression changes several traits at once
  3. C. Two alleles of one gene both showing in one trait
    Two alleles both showing in one trait is codominance.
    Pleiotropy is one gene changing several traits.

Why: Pleiotropy is one gene whose expression changes several traits at once.

34
Check q7

In cats, some kittens inherit one particular allele. Every kitten that inherits it has a white coat, many of them have blue eyes, and many of those are deaf.

Which pattern of inheritance do the three traits show?

  1. A. Linked genes
    Linked genes are two or more genes on one chromosome, one gene per trait; here one allele of one gene brings all three traits.
  2. B. Codominance
    Codominance is two alleles of one gene both showing in one trait.
    Here one allele shows in three different traits.
  3. C. Three separate mutations
    Three separate mutations would be three alleles, and they would not be inherited as one.
    Here one inherited allele brings all three traits.
  4. D. ✓ Pleiotropy

Why: The kittens inherit one allele of one gene.
That one allele brings the white coat and, in many kittens, blue eyes and deafness.
One gene changing several traits at once is called pleiotropy.

35
Practice writing an answer

In cats, some kittens inherit one particular allele. Every kitten that inherits it has a white coat, many of them have blue eyes, and many of those are deaf. This is pleiotropy.

(a) Explain how these kittens demonstrate pleiotropy. (1 pt)

Model answer The white coat and, in many kittens, the blue eyes and the deafness all come from one allele of one gene.
A kitten inherits the allele whole, not part of it.
So a kitten that inherits the allele inherits every effect of that one gene.
One gene changing several traits at once is pleiotropy.
Rubric
  • Award 1 point for: one allele of one gene brings all three traits, and a kitten inherits the allele whole, so the traits appear together.

Glossary

pleiotropy
One gene whose expression changes several traits at once, so the traits are inherited together; sickle cell disease, where one changed hemoglobin brings anemia, pain and kidney damage.

APBIO-U05-L27 Genes outside the nucleus

Topic 5.4 · Non-Mendelian Genetics · 38 steps

Left: a photograph of a potted geranium plant, each leaf green in the middle with a broad white edge. Middle: a leaf drawn as a pointed oval with a midrib, green over most of its blade and with three white patches. Right: a single leaf cell drawn as a rounded box holding a nucleus and eight small chloroplasts
Left: a photograph of a potted geranium plant, each leaf green in the middle with a broad white edge. Middle: a leaf drawn as a pointed oval with a midrib, green over most of its blade and with three white patches. Right: a single leaf cell drawn as a rounded box holding a nucleus and eight small chloroplasts

Here is a geranium whose leaves are green in the middle and white at the edge, and beside it a drawing of a leaf where the white fell in patches, and one of its cells with its nucleus and its chloroplasts.

Every cell of this leaf grew from one zygote by mitosis, so every cell carries the same chromosomes. How can some patches be white?

Photo: Yercaud-elango, Wikimedia Commons, CC BY-SA 4.0 (resized).

Unit 5 · Heredity

1Genes outside the nucleus

2

Video: Watch: Genes outside the nucleus

A leaf cell with its nucleus and its chloroplasts, each chloroplast with its own small ring of DNA, and a mitochondrion with its own ring too.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L27a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L27a.mp4

3

How can cells with the same chromosomes look different? Not all of a cell's genes sit in the nucleus.

4

Mitochondria and chloroplasts carry small DNA of their own, with genes of their own. That DNA is the start of the answer.

5
Check q1

Which parts of a plant cell hold DNA of their own, besides the nucleus?

  1. A. The ribosomes
    A ribosome builds proteins from instructions.
    A ribosome carries no DNA of its own.
  2. B. ✓ The mitochondria and the chloroplasts
  3. C. The cell wall
    The cell wall is a layer of cellulose outside the cell membrane.
    The cell wall holds no DNA.

Why: A mitochondrion has its own DNA in one closed loop.
A chloroplast has its own DNA in one closed loop too.
Both descended from bacteria, and each kept a bacterium's loop of DNA.

6

Here is a cell with its nucleus, a mitochondrion and a chloroplast, each organelle with its own small ring of DNA.

A cell drawn as a rounded box. In its center a nucleus holding four short chromosomes, labeled: the nucleus, chromosomes, most of the cell's genes. To the left a mitochondrion and to the right a chloroplast, each drawn with a small ring of DNA inside and labeled beneath the cell: its own ring of DNA
A cell drawn as a rounded box. In its center a nucleus holding four short chromosomes, labeled: the nucleus, chromosomes, most of the cell's genes. To the left a mitochondrion and to the right a chloroplast, each drawn with a small ring of DNA inside and labeled beneath the cell: its own ring of DNA
7

That DNA carries genes of its own. A chloroplast's DNA holds a gene for one of the proteins of photosynthesis, and a mitochondrion's DNA holds a gene for one of the proteins of the electron transport chain.

8

When we speak of the DNA inside a mitochondrion, we call it , because it belongs to the mitochondrion, not to the nucleus.

9

In the same way, when we speak of the DNA inside a chloroplast, we call it , because it belongs to the chloroplast.

10

Most of a cell's genes sit on the chromosomes in the nucleus. The organelle DNA carries a small set: a few dozen genes in a mitochondrion against about twenty thousand in the nucleus.

11

What you are expected to know State where a cell's genes sit: most on the chromosomes in the nucleus, a small set in mitochondrial DNA and in chloroplast DNA.

12
Check q2

In one leaf cell of a plant, some of the chloroplasts make a normal form of one photosynthesis protein and the others make a faulty form of it.

Where does the gene for that protein sit?

  1. A. On a chromosome in the nucleus
    A gene on a chromosome in the nucleus is one set of instructions for the whole cell, so every chloroplast would make the same form of the protein.
  2. B. In a ribosome
    A ribosome builds proteins from instructions; it carries no genes of its own.
  3. C. ✓ In the chloroplast's own DNA
  4. D. In the cell wall
    The cell wall is a layer of cellulose outside the cell membrane and holds no DNA.

Why: Each chloroplast carries its own small ring of DNA, with genes of its own.
So two chloroplasts in one cell can hold different copies of a photosynthesis gene.
Different copies make different forms of the protein.
A gene in the nucleus would give every chloroplast the same instructions.

13Quick quiz: mitochondrial DNA, chloroplast DNA mixed practice

14
Check q3

What is mitochondrial DNA?

  1. A. The chromosomes in the nucleus that code for the mitochondrion's proteins
    The chromosomes in the nucleus are nuclear DNA.
    Mitochondrial DNA is the ring inside the mitochondrion itself.
  2. B. The DNA a mitochondrion takes in from the cytosol
    A mitochondrion takes in no DNA from the cytosol.
    Mitochondrial DNA is the mitochondrion's own ring, copied when the mitochondrion divides.
  3. C. ✓ The small ring of DNA inside a mitochondrion, with genes of its own

Why: Mitochondrial DNA is the small ring of DNA inside a mitochondrion.
It carries a few dozen genes of its own.

15
Check q4

What is chloroplast DNA?

  1. A. ✓ The small ring of DNA inside a chloroplast, with genes of its own
  2. B. The genes for photosynthesis on the chromosomes in the nucleus
    Genes on the chromosomes in the nucleus are nuclear DNA.
    Chloroplast DNA is the ring inside the chloroplast itself.
  3. C. The DNA of the bacteria that live on a leaf
    Bacteria on a leaf are separate organisms.
    Chloroplast DNA is the chloroplast's own ring, inside the plant cell.

Why: Chloroplast DNA is the small ring of DNA inside a chloroplast.
It carries genes of its own, separate from the nucleus.

16
Practice writing an answer

A leaf cell holds a nucleus, mitochondria and chloroplasts.

(a) State the two places in this cell, besides the nucleus, that hold DNA. (1 pt)

Model answer The mitochondria hold mitochondrial DNA.
The chloroplasts hold chloroplast DNA.
Rubric
  • Award 1 point for: the mitochondria and the chloroplasts.

17Dealt out at random

18

Video: Watch: Dealt out at random

A leaf cell dividing and its chloroplasts falling unevenly into the two halves, one line of cells losing its green while the chromosomes in every nucleus stay the same.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L27b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L27b.mp4

19

Now consider the white patches. Every cell of the leaf has the same chromosomes, so what differs between a green cell and a white one?

20

Suppose one chloroplast in a leaf cell carries a faulty copy of a pigment gene in its chloroplast DNA. Chloroplasts divide inside the cell, so the cell comes to hold a mixture of normal and faulty chloroplasts.

A leaf cell drawn as a rounded box holding twelve chloroplasts: nine filled, labeled normal, and three drawn as hollow outlines, labeled faulty pigment gene in their chloroplast DNA
A leaf cell drawn as a rounded box holding twelve chloroplasts: nine filled, labeled normal, and three drawn as hollow outlines, labeled faulty pigment gene in their chloroplast DNA
21
Check q5

A gene sits on a chromosome, and a plant is Aa for that gene.

How many alleles of the gene does each of its gametes carry?

  1. A. Both, A and a
    The homologs part at anaphase I, so a gamete receives one homolog and so one allele.
  2. B. ✓ Exactly one, A or a

Why: The two alleles sit on the two homologs, and the homologs part at anaphase I, so each gamete carries exactly one of the two alleles.

22

Chromosomes are counted into daughter cells and gametes, one of every pair. Organelles are not counted at all: when a cell divides, its chloroplasts fall into the two halves however they happen to lie.

On the left a leaf cell with nine normal chloroplasts and three faulty ones. An arrow leads to two daughter cells: the upper daughter holds seven normal chloroplasts and no faulty one; the lower daughter holds two normal and three faulty ones. A caption reads: the chloroplasts fall into the two halves however they lie
On the left a leaf cell with nine normal chloroplasts and three faulty ones. An arrow leads to two daughter cells: the upper daughter holds seven normal chloroplasts and no faulty one; the lower daughter holds two normal and three faulty ones. A caption reads: the chloroplasts fall into the two halves however they lie
23

So one daughter cell can receive mostly faulty chloroplasts, another mostly normal ones, and a third a mixture.

24

A cell dealt mostly faulty chloroplasts makes no green pigment. Its descendants, made by mitosis, inherit its chloroplasts, and together they form a white patch.

Three leaf cells side by side. The first holds mostly normal chloroplasts and is labeled: makes green pigment, a green patch. The second holds a mixture and is labeled: still green. The third holds mostly faulty chloroplasts and is labeled: no green pigment, a white patch
Three leaf cells side by side. The first holds mostly normal chloroplasts and is labeled: makes green pigment, a green patch. The second holds a mixture and is labeled: still green. The third holds mostly faulty chloroplasts and is labeled: no green pigment, a white patch
25

No square counts organelles into gametes, so an organelle trait never gives 3 : 1 or 1 : 1. A gamete receives whatever mixture of organelles happened to lie in its half.

26

When a trait is passed on by genes outside the nucleus, in mitochondrial DNA or chloroplast DNA, we call it , because the genes are not in the nucleus.

27

What you are expected to know Explain why a trait in chloroplast DNA or mitochondrial DNA gives none of Mendel's ratios: chromosomes are counted into gametes one per pair, but organelles fall into daughter cells at random.

28
Check q6

A cell in a growing leaf holds sixteen chloroplasts: twelve normal and four with a faulty pigment gene in their chloroplast DNA, as drawn below. The cell divides.

A leaf cell drawn as a rounded box holding sixteen chloroplasts: twelve filled, labeled normal, and four drawn as hollow outlines, labeled faulty pigment gene
A leaf cell drawn as a rounded box holding sixteen chloroplasts: twelve filled, labeled normal, and four drawn as hollow outlines, labeled faulty pigment gene

How are the four faulty chloroplasts shared between the two daughter cells?

  1. A. Two to each daughter cell, every time
    Nothing deals the faulty chloroplasts out two to each daughter.
    Alleles are counted into gametes by the parting of the homologs; chloroplasts are not counted at all.
  2. B. Six normal and two faulty to each daughter cell, every time
    Nothing counts the chloroplasts into equal shares.
    Each half gets whatever lies in it.
  3. C. None to either daughter cell, every time
    Division does not destroy a faulty chloroplast.
    A chloroplast with a faulty gene divides and is passed on like any other.
  4. D. ✓ Any number to each daughter cell, from none to all four

Why: Chloroplasts are not counted into daughter cells.
When the cell divides, each chloroplast falls into whichever half it happens to lie in.
So one daughter can get most of the faulty ones, and the other can get few or none.

29
Practice writing an answer

A cell in a growing leaf holds sixteen chloroplasts: twelve normal and four with a faulty pigment gene in their chloroplast DNA, as drawn below. The cell divides. The two daughter cells can receive any split of the faulty chloroplasts.

A leaf cell drawn as a rounded box holding sixteen chloroplasts: twelve filled, labeled normal, and four drawn as hollow outlines, labeled faulty pigment gene
A leaf cell drawn as a rounded box holding sixteen chloroplasts: twelve filled, labeled normal, and four drawn as hollow outlines, labeled faulty pigment gene

(a) Explain why the two daughter cells can receive any split of the faulty chloroplasts, when the two alleles of a chromosome gene are always dealt out one per gamete. (1 pt)

Model answer The two alleles of a chromosome gene sit on the two homologs.
The homologs part at anaphase I, so each gamete receives exactly one of the two alleles.
Chloroplasts are not on chromosomes.
Nothing parts them or counts them.
When the cell divides, each chloroplast falls into whichever half it happens to lie in.
So one daughter cell can receive most of the faulty chloroplasts and the other few or none.
Rubric
  • Award 1 point for: chromosome alleles are counted into gametes by the homologs parting, but chloroplasts are not counted; they fall into the two halves however they lie.
30
Check q7

A leaf has green patches and white patches. A student says: “The cells in the white patches lost the pigment gene from their chromosomes.”

Is the student correct?

  1. A. Yes
    Every cell of the leaf grew from one zygote by mitosis, so every cell carries the same chromosomes; the pigment gene at fault sits in the chloroplast DNA.
  2. B. ✓ No

Why: Every cell of the leaf grew from one zygote by mitosis.
So every cell, green or white, carries the same chromosomes.
The faulty pigment gene sits in the chloroplast DNA.
A white cell was dealt mostly faulty chloroplasts when its parent cell divided.

31
Check q8

In a plant, a pale-leaf trait comes from a gene in chloroplast DNA. A student treats that gene like a gene on a chromosome, writes the two parents as heterozygous, and predicts 3 : 1 pale to green among their seedlings.

What is wrong with the prediction?

  1. A. Nothing is wrong; a gene in chloroplast DNA follows the rules for chromosomes
    The rules for chromosomes rest on the two homologs parting, one to each gamete, and chloroplasts are not dealt out one per gamete at all.
  2. B. The trait is not inherited at all, because only the chromosomes reach the zygote
    A gamete carries organelles as well as chromosomes, whatever mixture lay in its half, so a chloroplast trait is inherited; it simply gives none of Mendel's ratios.
  3. C. ✓ Chloroplasts are not dealt out one per gamete, so none of Mendel's ratios applies
  4. D. The ratio should be 1 : 1 pale to green, as in a test cross of a heterozygote
    1 : 1, like 3 : 1, comes from a square that counts one allele into each gamete, and chloroplasts are not dealt out one per gamete.

Why: Every one of Mendel's ratios comes from a square that counts one allele into each gamete, because the homologs part at anaphase I.
Chloroplasts are not counted at all: a gamete receives whatever chloroplasts lie in its half.
So none of Mendel's ratios applies to a chloroplast-DNA trait.

32

Here again is a leaf cell with its chloroplasts, three of the twelve carrying the faulty pigment gene.

A leaf cell drawn as a rounded box holding twelve chloroplasts: nine filled, labeled normal, and three drawn as hollow outlines, labeled faulty pigment gene in their chloroplast DNA
A leaf cell drawn as a rounded box holding twelve chloroplasts: nine filled, labeled normal, and three drawn as hollow outlines, labeled faulty pigment gene in their chloroplast DNA
33

The white patches are the descendants of cells that were dealt mostly faulty chloroplasts when their parent cells divided. The chromosomes in the nucleus are the same in every cell of the leaf.

34Quick quiz: non-nuclear inheritance mixed practice

35
Check q9

What is non-nuclear inheritance?

  1. A. Inheritance of a trait through genes on the sex chromosomes
    The sex chromosomes are chromosomes in the nucleus.
    Non-nuclear inheritance is through genes outside the nucleus.
  2. B. ✓ Inheritance of a trait through genes outside the nucleus
  3. C. Inheritance of a trait through two genes on one chromosome
    Two genes on one chromosome are linked genes, in the nucleus.
    Non-nuclear inheritance is through genes outside the nucleus.

Why: Non-nuclear inheritance is inheritance of a trait through genes outside the nucleus.
Those genes sit in mitochondrial DNA or chloroplast DNA.

36
Check q10

Which of the following traits is passed on by non-nuclear inheritance?

  1. A. ✓ A trait coded by a gene in mitochondrial DNA
  2. B. A recessive trait carried on one of the 22 ordinary pairs
    The 22 ordinary pairs are chromosomes in the nucleus, so a trait on them follows the rules for chromosomes.
  3. C. A trait coded by a gene on the X chromosome
    The X chromosome is one of the chromosomes in the nucleus.
  4. D. A trait that appears in every generation of a family
    Appearing in every generation does not show where the gene sits.
    A dominant allele on a nuclear chromosome also appears in every generation.

Why: Non-nuclear inheritance is inheritance through genes outside the nucleus.
Mitochondrial DNA is the mitochondrion's own DNA, outside the nucleus.

37
Practice writing an answer

In a plant, a pale-leaf trait comes from a gene in chloroplast DNA.

(a) Explain how this trait demonstrates non-nuclear inheritance. (1 pt)

Model answer The gene sits in chloroplast DNA, outside the nucleus.
Chloroplasts are not counted into gametes one per pair.
A gamete receives whatever chloroplasts lie in its half.
So the trait passes on outside the rules for chromosomes, which is non-nuclear inheritance.
Rubric
  • Award 1 point for: the gene is outside the nucleus, in chloroplast DNA, and chloroplasts are not dealt out one per gamete, so the trait follows the rules for organelles, not for chromosomes.

Glossary

mitochondrial DNA
The small ring of DNA inside a mitochondrion, with a few dozen genes of its own, separate from the chromosomes in the nucleus.
chloroplast DNA
The small ring of DNA inside a chloroplast, with genes of its own, separate from the chromosomes in the nucleus.
non-nuclear inheritance
Inheritance of a trait through genes outside the nucleus, in mitochondrial DNA or chloroplast DNA; because organelles are not dealt out one per gamete, such traits give none of Mendel's ratios.

APBIO-U05-L27B Traits that come only from the mother

Topic 5.4 · Non-Mendelian Genetics · 48 steps

A large round egg cell on the left, packed with twenty small mitochondria; on the right, not to the same scale, a sperm cell with a small head, a short midpiece holding three mitochondria, and a long thin tail
A large round egg cell on the left, packed with twenty small mitochondria; on the right, not to the same scale, a sperm cell with a small head, a short midpiece holding three mitochondria, and a long thin tail

Here is an egg cell and a sperm cell, not drawn to scale. The egg is packed with mitochondria. The sperm carries a few in its midpiece, just behind the head, and they do not last after fertilization.

Whose mitochondria does the zygote end up with, and what follows for a trait written in mitochondrial DNA?

Unit 5 · Heredity

1The egg brings the organelles

2

Video: Watch: The egg brings the organelles

The egg and the sperm at fertilization with the sperm's mitochondria fading, then the plant's egg inside its ovule and the sperm in its pollen grain.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L27Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L27Ba.mp4

3

Why does a trait in mitochondrial DNA come only from the mother? The egg's mitochondria all stay, and the sperm's few break down.

4

So every child's mitochondria came from the mother's egg. In plants the ovule brings the organelles in the same way.

5

At fertilization the sperm and the egg fuse into one cell, the zygote.

On the left an egg cell holding twenty mitochondria and, beside it, a sperm cell with three mitochondria in its midpiece. An arrow leads to the zygote on the right: a cell holding the egg's twenty mitochondria; the sperm's three are drawn faded beside it, labeled: the sperm's few break down
On the left an egg cell holding twenty mitochondria and, beside it, a sperm cell with three mitochondria in its midpiece. An arrow leads to the zygote on the right: a cell holding the egg's twenty mitochondria; the sperm's three are drawn faded beside it, labeled: the sperm's few break down
6

The egg's mitochondria, hundreds of thousands of them, all stay in the zygote. The sperm's few mitochondria usually break down soon after fertilization.

7
Check q1

A flowering plant makes eggs and sperm.

Which structure holds the plant’s egg?

  1. A. ✓ The ovule
  2. B. The pollen grain
    The pollen grain carries the sperm.
    The ovule holds the egg.

Why: An ovule is a small case in the base of the flower.
It holds the plant’s egg.
A pollen grain carries the plant’s sperm.

8

Here is the ovule holding the egg with its organelles, and a pollen grain above it.

The base of a flower: a swollen ovule on its stalk, labeled the ovule, holding a small circle labeled the egg with its organelles; above it a pollen grain, a small circle labeled: pollen grain, carries the sperm
The base of a flower: a swollen ovule on its stalk, labeled the ovule, holding a small circle labeled the egg with its organelles; above it a pollen grain, a small circle labeled: pollen grain, carries the sperm
9

So the rule is this: the egg (in plants, the ovule) supplies the zygote's mitochondria and chloroplasts and the sperm's (the pollen's) usually do not persist.

10

So an organelle trait passes from a mother to all her children and never from a father.

11

When a trait passes from the mother to every child and from the father to none, we call it , because it comes down the maternal line.

12

What you are expected to know Explain why a trait written in mitochondrial DNA or chloroplast DNA comes from the mother and never from the father.

13
Check q2

A sperm cell carries a few mitochondria whose DNA has a faulty gene. It fertilizes an egg whose mitochondria are all normal.

Which mitochondria does the body that grows from the zygote carry?

  1. A. The sperm's
    The sperm's few mitochondria usually break down soon after fertilization.
    So none of them is passed on.
  2. B. A mixture of both
    The sperm's few mitochondria usually break down soon after fertilization, so they are not there to be passed on.
  3. C. Half of each
    The two sets of chromosomes in the nucleus are shared half and half; the mitochondria are not.
    The sperm's few break down, so the egg supplies all of them.
  4. D. ✓ The egg's

Why: The egg's mitochondria all stay in the zygote.
The sperm's few mitochondria usually break down.
So the body carries the egg's mitochondria, and none with the faulty gene.

14
Practice writing an answer

A sperm cell carries a few mitochondria whose DNA has a faulty gene. It fertilizes an egg whose mitochondria are all normal. The body that grows from the zygote carries the egg’s mitochondria.

(a) Explain why only the egg’s mitochondria are passed on. (1 pt)

Model answer The egg's mitochondria all stay in the zygote.
The sperm's few mitochondria usually break down soon after fertilization.
So the body carries the egg's mitochondria and none with the faulty gene.
Rubric
  • Award 1 point for: the sperm’s few mitochondria break down after fertilization, so the zygote’s mitochondria all come from the egg.
15
Check q3

In one plant species a pale-leaf trait is written in chloroplast DNA. A green plant supplies the ovules and a pale plant supplies the pollen, and seeds are grown.

What leaf color do the seedlings show?

  1. A. ✓ All green
  2. B. All pale
    The pollen brings the sperm, not the chloroplasts.
    The zygote's chloroplasts come from the egg in the ovule.
  3. C. Half green and half pale
    1 : 1 comes from a heterozygote's two kinds of gamete at a chromosome gene.
    Every seedling here received its chloroplasts from the same green ovule parent.
  4. D. Three green to one pale
    3 : 1 comes from a square that counts a chromosome gene into gametes.
    Chloroplasts are not counted into gametes; they come from one parent.

Why: The ovule supplies the zygote's chloroplasts.
The pollen's chloroplasts do not persist.
The ovule parent is green, so every seedling carries green chloroplasts.
So every seedling is green.

16Quick quiz: maternal inheritance mixed practice

17
Check q4

What is maternal inheritance?

  1. A. A trait that passes from a mother to her sons only, through her X
    A trait that reaches sons through their mother's X is X-linked.
    Maternal inheritance reaches every child, sons and daughters alike.
  2. B. ✓ A trait that passes from a mother to all her children and from a father to none
  3. C. A trait that passes to a child from whichever parent is affected, mother or father
    A father passes none of his mitochondria on.
    Maternal inheritance comes from the mother only.

Why: Maternal inheritance is a trait that passes from a mother to all her children and from a father to none.
The egg supplies the zygote's organelles, and the sperm's do not persist.

18
Check q5

A trait is written in mitochondrial DNA.

Which parent passes the trait to a child?

  1. A. ✓ The mother
  2. B. The father
    The sperm's few mitochondria break down after fertilization, so a father passes none on.
  3. C. Either parent
    Only the egg's mitochondria stay in the zygote, so only the mother passes the trait on.

Why: The egg supplies the zygote's mitochondria.
The sperm's few mitochondria break down.
So the mother passes a trait in mitochondrial DNA to the child, and the father does not.

19
Practice writing an answer

A man has a trait written in mitochondrial DNA. He and an unaffected woman have three children.

(a) Predict how many of the three children have the trait, and explain how this demonstrates maternal inheritance. (1 pt)

Model answer None of the three children has the trait.
The sperm's few mitochondria break down after fertilization.
So each child's mitochondria came from the unaffected mother's egg.
A trait in mitochondrial DNA passes from a mother and never from a father, which is maternal inheritance.
Rubric
  • Award 1 point for: none, because the father's mitochondria do not persist in the zygote, so each child's mitochondria came from the mother's egg.

20The pattern in a family

21

Video: Watch: The pattern in a family

A three-generation family with the trait following the mothers, then one child of the affected father imagined affected and the pattern breaking.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L27Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L27Bb.mp4

22

So what does the family of a person with a trait in mitochondrial DNA look like?

23

Almost every child of an affected mother is affected, sons and daughters alike, because every child got its mitochondria from her egg. A child escapes only if the egg it grew from happened to carry few of the faulty mitochondria.

24

No child of an affected father is affected, because none of his mitochondria persisted in the zygote.

25

Suppose a hearing loss is written in mitochondrial DNA. In one family, three affected women had 7 of their 8 children affected, and three affected men had 0 of their 9.

26

Here is such a family drawn across three generations: the affected grandmother's children are affected; her affected daughter's children are all affected; her affected son's children are not.

A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, a shaded circle. Their children: II-2, a shaded circle, joined to II-1, an unshaded square; and II-3, a shaded square, joined to II-4, an unshaded circle. The children of II-1 and II-2, III-1 a circle and III-2 a square, are both shaded. The children of II-3 and II-4, III-3 a circle and III-4 a square, are both unshaded
A three-generation pedigree. Generation I: I-1, an unshaded square, and I-2, a shaded circle. Their children: II-2, a shaded circle, joined to II-1, an unshaded square; and II-3, a shaded square, joined to II-4, an unshaded circle. The children of II-1 and II-2, III-1 a circle and III-2 a square, are both shaded. The children of II-3 and II-4, III-3 a circle and III-4 a square, are both unshaded
27

Now read the family one parent at a time.

28

I-2 is an affected mother, and both of her children, II-2 and II-3, are affected. This fits maternal inheritance, because every child of the affected mother is affected.

29

II-3 is an affected father, and neither of his children, III-3 and III-4, is affected. This fits maternal inheritance, because no child of the affected father is affected.

30

Now imagine III-3 were affected. This does not fit maternal inheritance, because a child of the affected father is affected.

The same three-generation pedigree, with one change: III-3, the daughter of the shaded square II-3 and the unshaded circle II-4, is now shaded
The same three-generation pedigree, with one change: III-3, the daughter of the shaded square II-3 and the unshaded circle II-4, is now shaded
31
Check q6

A man has one X chromosome and one Y chromosome.

Which of his children receive his X chromosome?

  1. A. Each son
    A son receives his father's Y.
    A son's X came from his mother.
  2. B. ✓ Each daughter
  3. C. Every child
    A father gives his X to each daughter and his Y to each son.

Why: A father gives his X to each daughter.
A father gives his Y to each son, so a son's X came from his mother.

32

Two questions tell a mitochondrial trait from an X-linked one, as a yes/no pair:

  1. Does an affected father pass it to his daughters? A father gives his X to each daughter, so for an X-linked trait, yes. For a mitochondrial trait, no.
  2. Does an affected mother pass it to every child? For a mitochondrial trait, yes. For an X-linked recessive trait, only to the children who receive that X.

33

Here is one simplification. A few plant species do pass chloroplasts through pollen, but for a breeding cross, treat the pollen as bringing none.

34

What you are expected to know Identify maternal inheritance in a family: every child of an affected mother affected, no child of an affected father, and an affected father's daughters as the family that tells it from an X-linked trait.

35
Check q7

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have four children: II-1, a shaded square, II-2, a shaded circle, II-3, a shaded square, and II-4, a shaded circle
A two-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have four children: II-1, a shaded square, II-2, a shaded circle, II-3, a shaded square, and II-4, a shaded circle

Does this family fit maternal inheritance?

  1. A. ✓ Yes
  2. B. No
    I-1 is an affected mother and all four of her children are affected, sons and daughters alike: every child’s mitochondria came from her egg.

Why: I-1 is an affected mother.
All four of her children are affected.
This fits maternal inheritance, because every child of the affected mother is affected.

36
Check q8

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, a shaded circle, and II-2, an unshaded square
A two-generation pedigree: I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, a shaded circle, and II-2, an unshaded square

Does this family fit maternal inheritance?

  1. A. Yes
    I-1 is an affected father and his daughter II-1 is affected; none of a father’s mitochondria persist in the zygote, so no child of his should be affected.
  2. B. ✓ No

Why: I-1 is an affected father.
His daughter II-1 is affected.
This does not fit maternal inheritance, because a child of the affected father is affected.

37
Check q9

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded square, and I-2, an unshaded circle, have three children: II-1, an unshaded circle, II-2, an unshaded square, and II-3, an unshaded circle
A two-generation pedigree: I-1, a shaded square, and I-2, an unshaded circle, have three children: II-1, an unshaded circle, II-2, an unshaded square, and II-3, an unshaded circle

Does this family fit maternal inheritance?

  1. A. ✓ Yes
  2. B. No
    I-1 is an affected father and none of his three children is affected.
    That is exactly what maternal inheritance predicts, because his mitochondria did not persist in any zygote.

Why: I-1 is an affected father.
None of his three children is affected.
This fits maternal inheritance, because no child of the affected father is affected.

38
Check q10

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have four children: II-1, an unshaded square, II-2, an unshaded circle, II-3, an unshaded square, and II-4, an unshaded circle
A two-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have four children: II-1, an unshaded square, II-2, an unshaded circle, II-3, an unshaded square, and II-4, an unshaded circle

Does this family fit maternal inheritance?

  1. A. Yes
    I-1 is an affected mother and none of her four children is affected; every child’s mitochondria came from her egg, so four unaffected children break the pattern.
  2. B. ✓ No

Why: I-1 is an affected mother.
None of her four children is affected.
This does not fit maternal inheritance, because almost every child of an affected mother is affected.

39
Check q11

A trait is recorded in the family below. II-2 married into the family.

A three-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have two children: II-1, a shaded square, and II-3, a shaded circle. II-1 is joined to II-2, an unshaded circle; their children III-1, a circle, and III-2, a square, are both unshaded
A three-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have two children: II-1, a shaded square, and II-3, a shaded circle. II-1 is joined to II-2, an unshaded circle; their children III-1, a circle, and III-2, a square, are both unshaded

Does this family fit maternal inheritance?

  1. A. ✓ Yes
  2. B. No
    The affected mother I-1 passed the trait to both of her children, and the affected father II-1 passed it to neither of his.
    Both results are what maternal inheritance predicts.

Why: I-1 is an affected mother, and both of her children, II-1 and II-3, are affected.
II-1 is an affected father, and neither of his children is affected.
This fits maternal inheritance, because every child of the affected mother is affected and no child of the affected father is.

40
Check q12

A trait is recorded in the family below.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have two children: II-1, a shaded square, and II-2, an unshaded circle
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have two children: II-1, a shaded square, and II-2, an unshaded circle

Does this family fit maternal inheritance?

  1. A. Yes
    II-1 is affected and his mother I-2 is not; an affected child’s mitochondria came from the mother’s egg, so an affected son of an unaffected mother breaks the pattern.
  2. B. ✓ No

Why: II-1 is an affected son.
His mother I-2 is unaffected.
This does not fit maternal inheritance, because an affected child's mitochondria came from its mother's egg, so the mother of an affected child is affected.

41
Check q13

A woman has a trait. Her two sons and her two daughters all have it, and their father does not. Her affected brother has a son and two daughters, and none of the three has it. A student says: “The trait is X-linked recessive: mothers pass it to their sons.”

Is the student correct?

  1. A. Yes
    The woman's two daughters have the trait, and their father does not; an X-linked recessive daughter needs the allele on her father's only X, which would then show it.
  2. B. ✓ No

Why: The woman's two daughters have the trait.
Their father does not.
An X-linked recessive daughter needs the allele on her father's X, so he would show it.
Every child of the affected mother has it and no child of the affected father does, which is maternal inheritance.

42

Here again is the egg cell packed with mitochondria, and the sperm cell with a few in its midpiece that do not last after fertilization.

On the left an egg cell holding twenty mitochondria and, beside it, a sperm cell with three mitochondria in its midpiece. An arrow leads to the zygote on the right: a cell holding the egg's twenty mitochondria; the sperm's three are drawn faded beside it, labeled: the sperm's few break down
On the left an egg cell holding twenty mitochondria and, beside it, a sperm cell with three mitochondria in its midpiece. An arrow leads to the zygote on the right: a cell holding the egg's twenty mitochondria; the sperm's three are drawn faded beside it, labeled: the sperm's few break down
43

The zygote's mitochondria all came from the egg, so a trait written in mitochondrial DNA reached every child of an affected mother and none of an affected father.

44Quick quiz: read a whole family mixed practice

45
Check q14

The pedigree below records a trait across three generations of one family.

A three-generation pedigree. Generation I: I-1, a shaded circle, and I-2, an unshaded square. Their children: II-2, a shaded square, joined to II-1, an unshaded circle; and II-3, a shaded circle, joined to II-4, an unshaded square. The four children of II-1 and II-2, III-1 to III-4, are all unshaded. The four children of II-3 and II-4, III-5 to III-8, are all shaded
A three-generation pedigree. Generation I: I-1, a shaded circle, and I-2, an unshaded square. Their children: II-2, a shaded square, joined to II-1, an unshaded circle; and II-3, a shaded circle, joined to II-4, an unshaded square. The four children of II-1 and II-2, III-1 to III-4, are all unshaded. The four children of II-3 and II-4, III-5 to III-8, are all shaded

Which mode of inheritance fits this family best?

  1. A. Autosomal dominant
    A dominant allele on a chromosome would pass from the affected father II-2 to about half of his four children, and none of them has it.
  2. B. X-linked recessive
    II-3 is an affected daughter of the unaffected father I-2; an X-linked recessive daughter needs the allele on her father’s only X, which would then show it.
  3. C. Autosomal recessive
    A recessive chromosome trait does not appear in every child of an affected parent unless the other parent carries it too; here every child of each affected mother has it.
  4. D. ✓ Maternal

Why: I-1 and II-3 are affected mothers, and every one of their children has the trait.
II-2 is an affected father, and none of his four children has it.
A trait in mitochondrial DNA follows the mothers, because the egg supplies the zygote's mitochondria.
Maternal inheritance fits best.

46
Check q15

A woman has a trait written in mitochondrial DNA. She has four children, two sons and two daughters, with an unaffected man. Her affected brother has three children, a son and two daughters, with an unaffected woman.

Predict which of the seven children have the trait.

  1. A. About half of her children and about half of his
    Half is the share a dominant chromosome allele gives from either parent; mitochondria come from the egg, so all of hers and none of his receive the trait.
  2. B. Her two sons only, and his two daughters only
    Sons from mothers and daughters from fathers is the route of an X-linked allele; mitochondria travel in the egg to every child and never in the sperm.
  3. C. ✓ All four of her children and none of his
  4. D. All seven, because both parents pass organelles on
    The sperm's few mitochondria break down after fertilization, so a father passes none on.

Why: Every child's mitochondria came from its mother's egg.
The affected woman's eggs carried the faulty mitochondria to all four of her children, sons and daughters alike.
The affected man's sperm carried none of his mitochondria on.
So none of his three children has the trait.

47
Check q16

A man has a trait, and one of his daughters has the same trait. The daughter’s mother is unaffected.

Which mode of inheritance does this father and daughter rule out?

  1. A. Autosomal dominant
    A dominant allele on a chromosome passes from a father to a daughter as easily as to a son.
  2. B. Autosomal recessive
    An affected father can pass a recessive allele on a chromosome to a daughter.
    She shows the trait if her mother passes one too.
  3. C. ✓ Maternal inheritance
  4. D. X-linked recessive inheritance
    A father gives his X to every daughter, so an X-linked allele reaches her.

Why: A father passes none of his mitochondria to any child.
So a trait in mitochondrial DNA never comes from a father.
A father and daughter who share a trait therefore rule maternal inheritance out.

Glossary

maternal inheritance
A trait that passes from a mother to all her children and from a father to none, because the egg (in plants, the ovule) supplies the zygote's mitochondria and chloroplasts and the sperm's usually do not persist.

APBIO-U05-L28 Which pattern is this?

Topic 5.4 · Non-Mendelian Genetics · 79 steps

Four small pedigrees in a row, none labeled. First: an affected mother and unaffected father with three affected children, and the affected son's two children both unaffected. Second: unaffected parents with an affected son, an unaffected son and an unaffected daughter, whose own son is affected. Third: two unaffected parents with an affected daughter and an unaffected son. Fourth: two affected parents with two affected daughters and an unaffected son
Four small pedigrees in a row, none labeled. First: an affected mother and unaffected father with three affected children, and the affected son's two children both unaffected. Second: unaffected parents with an affected son, an unaffected son and an unaffected daughter, whose own son is affected. Third: two unaffected parents with an affected daughter and an unaffected son. Fourth: two affected parents with two affected daughters and an unaffected son

Here are four pedigrees, one trait each, and no labels. In one the trait follows the mothers only; in one it appears mostly in boys; in one two unaffected parents have an affected daughter; in one two affected parents have an unaffected child.

Which pattern is each, and which family gives it away?

Unit 5 · Heredity

1The X-linked recessive route

2

Video: Watch: The X-linked recessive route

The second opening pedigree with the allele drawn travelling on the X: from a carrier mother to her affected son, and from an affected father to his carrier daughter and never to his son; the three signs written beside the family as each appears.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L28a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L28a.mp4

3

How do you name the mode of inheritance from a pedigree? Find the one family that decides it.

4

Start with the second pedigree, the one in which the trait appears mostly in boys.

5

Here is how to read a pedigree: a square is a male, a circle is a female, and a shaded shape has the trait.

6

A line between two shapes joins two parents. The line down from it leads to their children, in birth order.

7
Check q1

A mother and a father have a son.

Which parent gave the son his X chromosome?

  1. A. ✓ His mother
  2. B. His father
    A father gives each son his Y, not his X.
  3. C. Either parent, at random
    A father gives his X to each daughter and his Y to each son, so a son’s X is never his father’s.

Why: A mother gives an X to every child.
A father gives his X to each daughter and his Y to each son.
So a son’s X always came from his mother.

8

So a son’s X always came from his mother, and a daughter has one X from each parent.

9
Check q2

A man carries a recessive allele on his one X chromosome.

Why does the allele’s trait show in him?

  1. A. The allele is dominant in males
    The allele is recessive in both sexes; a female hides it with the ordinary allele on her second X, and a male has no second X.
  2. B. His Y chromosome carries a second copy of the allele
    The Y carries no allele of this gene, so nothing sits opposite the allele on his X.
  3. C. ✓ His one X has no second X carrying a masking allele

Why: A recessive allele is hidden only by a masking allele on a second copy of the gene.
A male has one X, so a recessive allele on it has no second X carrying a masking allele, and the trait shows.

10

So one X is enough for the trait to show.

11

Put the two facts together on a family, and an X-linked recessive trait leaves three signs. Here is a table of them: each sign, what you see on the pedigree, and why.

A table of the signs of an X-linked recessive trait on a pedigree. Mostly males affected: shaded squares and few shaded circles, because a son shows one r allele and a daughter needs two. An affected son: his mother is unaffected and carries the allele, because his only X came from her. An affected father: his daughters are unaffected carriers and no son gets the allele from him, because each daughter gets his X and each son his Y
A table of the signs of an X-linked recessive trait on a pedigree. Mostly males affected: shaded squares and few shaded circles, because a son shows one r allele and a daughter needs two. An affected son: his mother is unaffected and carries the allele, because his only X came from her. An affected father: his daughters are unaffected carriers and no son gets the allele from him, because each daughter gets his X and each son his Y
12

Here is the route in one family: I-1 is affected, his daughter II-1 is unaffected, and her son III-1 is affected.

A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, an unshaded circle, and II-2, an unshaded square. II-1 is joined to II-3, an unshaded square; their children are III-1, a shaded square, III-2, an unshaded circle, and III-3, an unshaded square
A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, an unshaded circle, and II-2, an unshaded square. II-1 is joined to II-3, an unshaded square; their children are III-1, a shaded square, III-2, an unshaded circle, and III-3, an unshaded square
13

The allele travels from I-1 to his daughter II-1 on his X, and from II-1 to her son III-1 on hers.

14

II-2 is unaffected because his father gave him a Y. II-1 is unaffected because her mother gave her an ordinary X, and that Xᴿ masks the allele.

15

Now consider an affected mother. Both of her X chromosomes carry Xʳ, so every son receives an Xʳ from her, and every son is affected.

16

So an unaffected son of an affected mother breaks the route.

17

What you are expected to know Recognize the X-linked recessive route on a pedigree from its signs: affected sons of unaffected carrier mothers, and an affected father’s unaffected carrier daughters.

18Quick quiz: the X-linked recessive route mixed practice

19
Check q3

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have three children: II-1, a shaded square, II-2, a shaded square, and II-3, an unshaded circle
A two-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have three children: II-1, a shaded square, II-2, a shaded square, and II-3, an unshaded circle

Does this family fit the X-linked recessive route?

  1. A. ✓ Yes
  2. B. No
    II-1 and II-2 are the affected sons of the affected I-1; an affected mother gives an Xʳ to every son.

Why: I-1 is affected, so both of her X chromosomes carry Xʳ.
Each son’s only X came from her, so II-1 and II-2 are affected.
II-3 also received I-2’s Xᴿ, which masks her Xʳ.
Every person fits the route.

20
Check q4

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have two children: II-1, an unshaded square, and II-2, a shaded square
A two-generation pedigree: I-1, a shaded circle, and I-2, an unshaded square, have two children: II-1, an unshaded square, and II-2, a shaded square

Does this family fit the X-linked recessive route?

  1. A. Yes
    II-1 is an unaffected son of the affected I-1, and an affected mother gives an Xʳ to every son.
  2. B. ✓ No

Why: I-1 is affected, so both of her X chromosomes carry Xʳ.
A son’s only X came from his mother, so every son of I-1 receives an Xʳ and is affected.
II-1 is her son and is unaffected.
So the family breaks the route.

21
Check q5

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded square, and I-2, an unshaded circle, have three children: II-1, an unshaded circle, II-2, an unshaded circle, and II-3, an unshaded square
A two-generation pedigree: I-1, a shaded square, and I-2, an unshaded circle, have three children: II-1, an unshaded circle, II-2, an unshaded circle, and II-3, an unshaded square

Does this family fit the X-linked recessive route?

  1. A. ✓ Yes
  2. B. No
    I-1 gives his X to each daughter and his Y to his son, so unaffected daughters and an unaffected son fit an affected father.

Why: I-1 is affected, so his one X carries Xʳ.
II-1 and II-2 each received that Xʳ, and each also received I-2’s Xᴿ, which masks it: unaffected carriers.
II-3 received his father’s Y, so he did not receive the allele from him.
Every person fits the route.

22
Check q6

A trait is recorded in the family below.

A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, a shaded circle, and II-2, an unshaded square. II-1 is joined to II-3, an unshaded square; their children are III-1, an unshaded square, and III-2, a shaded square
A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, a shaded circle, and II-2, an unshaded square. II-1 is joined to II-3, an unshaded square; their children are III-1, an unshaded square, and III-2, a shaded square

Does this family fit the X-linked recessive route?

  1. A. Yes
    III-1 is an unaffected son of the affected II-1, and an affected mother gives an Xʳ to every son.
  2. B. ✓ No

Why: II-1 is affected, so both of her X chromosomes carry Xʳ.
Each of her sons received his only X from her, so each would be affected.
III-1 is her son and is unaffected.
So the family breaks the route.

23
Check q7

A trait is recorded in the family below.

A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, an unshaded circle, and II-2, an unshaded circle. II-1 is joined to II-3, an unshaded square; their children are III-1, an unshaded circle, III-2, a shaded square, and III-3, a shaded square
A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, an unshaded circle, and II-2, an unshaded circle. II-1 is joined to II-3, an unshaded square; their children are III-1, an unshaded circle, III-2, a shaded square, and III-3, a shaded square

Does this family fit the X-linked recessive route?

  1. A. ✓ Yes
  2. B. No
    I-1 passes his Xʳ to his daughter II-1, who is unaffected because her mother gave her an Xᴿ; II-1 passes the Xʳ to her sons III-2 and III-3.

Why: I-1 is affected, so his one X carries Xʳ, and each daughter received it.
II-1 and II-2 also received I-2’s Xᴿ, which masks it, so both are unaffected.
III-2 and III-3 each received their only X from II-1, and each is affected.
Every person fits the route.

24The family that rules it out

25

Video: Watch: The family that rules it out

An unaffected father’s X and Y beside his affected daughter’s two X chromosomes; the allele written on each of hers, and the arrow from his X stopping short because it carries the ordinary allele; the family ringed and the words “not X-linked recessive” written.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L28b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L28b.mp4

26

Only-males-affected is a hint, not a proof: a small family can show only affected sons by chance under an autosomal trait too. The decisive evidence is the transmission route, or the one family that breaks it.

27

The family that breaks it is an affected daughter whose father is unaffected. She would need an Xʳ from each parent, and an unaffected father’s only X carries the ordinary allele, Xᴿ.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have two children: II-1, a shaded circle, and II-2, an unshaded square; a dashed ring surrounds I-1 and II-1
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have two children: II-1, a shaded circle, and II-2, an unshaded square; a dashed ring surrounds I-1 and II-1
28

Here are the chromosomes behind that family: the father’s Xᴿ is the only X he can give, so the daughter’s second Xʳ has nowhere to come from. One such daughter rules X-linked recessive inheritance out for the whole pedigree.

Left: an unaffected father's X and Y, the X carrying the ordinary allele R. Right: an affected daughter's two X chromosomes, each carrying the allele r; the one from her father is marked. A caption reads: her father's X carries R, so she cannot have got r from him
Left: an unaffected father's X and Y, the X carrying the ordinary allele R. Right: an affected daughter's two X chromosomes, each carrying the allele r; the one from her father is marked. A caption reads: her father's X carries R, so she cannot have got r from him
29

What you are expected to know Rule X-linked recessive inheritance out with one family: an affected daughter of an unaffected father.

30
Check q8

A trait is recorded in the family below.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have three children: II-1, a shaded square, II-2, a shaded circle, and II-3, an unshaded square
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have three children: II-1, a shaded square, II-2, a shaded circle, and II-3, an unshaded square

Which feature of this pedigree rules X-linked recessive inheritance out?

  1. A. II-1, an affected son of two unaffected parents
    An affected son of two unaffected parents is exactly what an X-linked recessive allele produces: his carrier mother gave him her Xʳ.
  2. B. ✓ II-2, an affected daughter of an unaffected father
  3. C. I-2, an unaffected mother with an affected son
    An unaffected mother with an affected son is the carrier-mother sign of an X-linked recessive trait; it fits the pattern rather than breaking it.
  4. D. II-3, an unaffected son of the same parents
    An unaffected son is possible under every mode: under X-linked recessive inheritance he simply received his mother’s Xᴿ.

Why: II-2 is a daughter who has the trait, so she would need an Xʳ on both of her X chromosomes, one from each parent.
Her father I-1 is unaffected, so his only X carries Xᴿ, and he cannot have given her an Xʳ.
The trait therefore cannot be X-linked recessive.

31
Practice writing an answer

A trait is recorded in the family below. II-2, an affected daughter of the unaffected I-1, rules X-linked recessive inheritance out.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have three children: II-1, a shaded square, II-2, a shaded circle, and II-3, an unshaded square
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have three children: II-1, a shaded square, II-2, a shaded circle, and II-3, an unshaded square

(a) Explain how II-2 rules X-linked recessive inheritance out. (1 pt)

Model answer II-2 has the trait, so under X-linked recessive inheritance she would carry Xʳ on both of her X chromosomes.
One of her X chromosomes came from her father, I-1.
I-1 is unaffected, so his only X carries Xᴿ.
So I-1 could not have given II-2 an Xʳ, and the trait cannot be X-linked recessive.
Rubric
  • Award 1 point for: II-2 would need an Xʳ from her father, and an unaffected father’s only X carries Xᴿ, so he could not have given her one.
32
Check q9

A trait is recorded in the family below. A student says: “Mostly sons are affected, so the trait is X-linked recessive.”

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have four children: II-1, a shaded square, II-2, a shaded square, II-3, a shaded circle, and II-4, an unshaded circle
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have four children: II-1, a shaded square, II-2, a shaded square, II-3, a shaded circle, and II-4, an unshaded circle

Is the student’s conclusion correct?

  1. A. Yes, the trait is X-linked recessive
    II-3 is an affected daughter of the unaffected I-1, and an unaffected father’s only X carries Xᴿ.
  2. B. ✓ No, the trait cannot be X-linked recessive

Why: Mostly affected sons is a hint, not a proof.
II-3 is a daughter who has the trait, so she would need an Xʳ from each parent.
Her father I-1 is unaffected, so his only X carries Xᴿ.
So the trait cannot be X-linked recessive.

33
Check q10

A trait is recorded in the family below. It is consistent with X-linked recessive inheritance.

A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, an unshaded circle, and II-2, an unshaded square. II-1 is joined to II-3, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded square
A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, an unshaded circle, and II-2, an unshaded square. II-1 is joined to II-3, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded square

Which family in the pedigree shows the transmission route of an X-linked recessive allele?

  1. A. I-1 and II-2
    An affected father and an unaffected son show only that the son got his father’s Y; that is consistent with the route but does not show the allele moving.
  2. B. II-1 and II-3
    Two unaffected parents with an unaffected son show nothing on their own; the unaffected mother matters only because her son III-1 is affected.
  3. C. III-1 and III-2
    Two brothers, one affected, fit any mode; the brothers alone do not show where the allele came from.
  4. D. ✓ I-1, II-1 and III-1

Why: An X-linked recessive allele travels on the X. I-1 gives his X, carrying the allele, to his daughter II-1, and she is unaffected because her mother gave her an Xᴿ.
II-1 gives that X to her son III-1, whose only X it is, so he is affected.

34
Practice writing an answer

A trait is recorded in the family below. It is consistent with X-linked recessive inheritance. I-1, II-1 and III-1 show the transmission route.

A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, an unshaded circle, and II-2, an unshaded square. II-1 is joined to II-3, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded square
A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have two children: II-1, an unshaded circle, and II-2, an unshaded square. II-1 is joined to II-3, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded square

(a) Describe the route the X-linked recessive allele takes through those three people. (1 pt)

Model answer An X-linked recessive allele travels on the X.
I-1 gives his X, carrying the allele, to his daughter II-1, and she is unaffected because her mother gave her an Xᴿ.
II-1 gives that X to her son III-1, whose only X it is, so he is affected.
Rubric
  • Award 1 point for: affected man to unaffected carrier daughter to affected grandson; the allele passes on the X from father to daughter, and then to her son.

35Deciding the mode

36

Video: Watch: Deciding the mode

Each of the four opening pedigrees in turn, its decisive family ringed and the mode written beside it: fathers who never pass the trait on; affected sons of unaffected mothers; an affected daughter of an unaffected father; affected parents with an unaffected child.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L28c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L28c.mp4

37

Now take all four pedigrees together, and ask of each which family decides its mode.

38
Check q11

Two unaffected parents have an affected child.

Is the trait dominant or recessive?

  1. A. Dominant
    A parent who carries a dominant allele shows its trait, and both parents are unaffected.
  2. B. ✓ Recessive

Why: Each unaffected parent gave the child one allele for the trait without showing the trait.
An allele that hides in a carrier is recessive.

39

For a trait on an autosome, one of the 22 ordinary pairs, two families decide.

40

Two shaded parents with an unshaded child mean a dominant trait. Two unshaded parents with a shaded child mean a recessive trait.

41
Check q12

A trait is carried in mitochondrial DNA. An affected father has four children.

How many of the four inherit the trait from him?

  1. A. All four
    Every child’s mitochondria came from the egg, not from the sperm.
  2. B. About half
    The sperm’s few mitochondria break down after fertilization, so a father passes none of his to any child.
  3. C. ✓ None

Why: The egg supplies the zygote’s mitochondria, and the sperm’s few break down.
So a father passes none of his mitochondria on, and none of his children inherits the trait from him.

42

So the maternal pattern is this: a trait in mitochondrial DNA or chloroplast DNA passes from an affected mother to every child, and from an affected father to none.

43

Here is a table of the four modes: the family that decides each, and the family that rules it out.

A table of the four modes of inheritance. Autosomal dominant: decided by two affected parents with an unaffected child; ruled out by an affected child of two unaffected parents. Autosomal recessive: decided by two unaffected parents with an affected child; ruled out by two affected parents with an unaffected child. X-linked recessive: no family proves it, affected sons of unaffected mothers are consistent with it; ruled out by an affected daughter of an unaffected father. Maternal: decided by an affected mother whose every child is affected and an affected father whose no child is; ruled out by an affected child of an affected father
A table of the four modes of inheritance. Autosomal dominant: decided by two affected parents with an unaffected child; ruled out by an affected child of two unaffected parents. Autosomal recessive: decided by two unaffected parents with an affected child; ruled out by two affected parents with an unaffected child. X-linked recessive: no family proves it, affected sons of unaffected mothers are consistent with it; ruled out by an affected daughter of an unaffected father. Maternal: decided by an affected mother whose every child is affected and an affected father whose no child is; ruled out by an affected child of an affected father
44

The first opening pedigree: I-1 is affected and all three of her children are affected; II-3 is affected and neither of his children is.

The first opening pedigree, numbered: I-1, a shaded circle, and I-2, an unshaded square, have three shaded children, II-1 a square, II-2 a circle and II-3 a square. II-3 is joined to II-4, an unshaded circle; their children III-1, a circle, and III-2, a square, are unshaded. A dashed ring surrounds II-3 and his two children
The first opening pedigree, numbered: I-1, a shaded circle, and I-2, an unshaded square, have three shaded children, II-1 a square, II-2 a circle and II-3 a square. II-3 is joined to II-4, an unshaded circle; their children III-1, a circle, and III-2, a square, are unshaded. A dashed ring surrounds II-3 and his two children
45

Fathers never pass it on, so the trait is maternal.

46

The third opening pedigree: I-1 and I-2 are unaffected, and their daughter II-1 is affected.

The third opening pedigree, numbered: I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded circle, and II-2, an unshaded square. A dashed ring surrounds the whole family
The third opening pedigree, numbered: I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded circle, and II-2, an unshaded square. A dashed ring surrounds the whole family
47

Two unaffected parents with an affected child mean a recessive trait. An affected daughter of an unaffected father rules the X out, so the trait is autosomal recessive.

48

The fourth opening pedigree: I-1 and I-2 are both affected, and their son II-2 is not.

The fourth opening pedigree, numbered: I-1, a shaded square, and I-2, a shaded circle, have II-1, a shaded circle, II-2, an unshaded square, and II-3, a shaded circle. A dashed ring surrounds the two parents and II-2
The fourth opening pedigree, numbered: I-1, a shaded square, and I-2, a shaded circle, have II-1, a shaded circle, II-2, an unshaded square, and II-3, a shaded circle. A dashed ring surrounds the two parents and II-2
49

Two affected parents with an unaffected child mean a dominant trait, because each parent passed II-2 the recessive allele: autosomal dominant.

50

The second opening pedigree: the affected people are II-1 and III-1, both sons of unaffected mothers, and no affected daughter appears anywhere.

The second opening pedigree, numbered: I-1, an unshaded circle, and I-2, an unshaded square, have II-1, a shaded square, II-2, an unshaded square, and II-3, an unshaded circle. II-3 is joined to II-4, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded circle. Dashed rings surround I-1 with II-1, and II-3 with III-1
The second opening pedigree, numbered: I-1, an unshaded circle, and I-2, an unshaded square, have II-1, a shaded square, II-2, an unshaded square, and II-3, an unshaded circle. II-3 is joined to II-4, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded circle. Dashed rings surround I-1 with II-1, and II-3 with III-1
51

That is the X-linked recessive route, and nothing in the family breaks it.

52

Other patterns exist, and a pedigree can fit more than one mode when its decisive family is missing. Then you write “consistent with”, never “is”, and you name the family that would settle it.

53

A whole pedigree holds several families. Find the one family that decides the mode, and name the mode from it.

54

What you are expected to know Name the mode of inheritance a pedigree shows from its decisive family, or say “consistent with” when no family decides.

55

Back to the four pedigrees of the opening, one trait each and no labels: each gave itself away at one family. Here is a table of the four: the family that decided each, and its mode.

A table of the four opening pedigrees. First: II-3 passed the trait to none of his children and I-1 to all of hers; maternal. Second: affected sons of unaffected mothers and no affected daughter; consistent with X-linked recessive. Third: two unaffected parents with the affected daughter II-1; autosomal recessive. Fourth: two affected parents with the unaffected son II-2; autosomal dominant
A table of the four opening pedigrees. First: II-3 passed the trait to none of his children and I-1 to all of hers; maternal. Second: affected sons of unaffected mothers and no affected daughter; consistent with X-linked recessive. Third: two unaffected parents with the affected daughter II-1; autosomal recessive. Fourth: two affected parents with the unaffected son II-2; autosomal dominant

56Quick quiz: mode of inheritance mixed practice

57
Check q13

A trait is recorded in the family below. II-4 married into the family.

A three-generation pedigree. I-1, an unshaded circle, and I-2, an unshaded square, have three children: II-1, a shaded square, II-2, an unshaded circle, and II-3, an unshaded square. II-2 is joined to II-4, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded circle
A three-generation pedigree. I-1, an unshaded circle, and I-2, an unshaded square, have three children: II-1, a shaded square, II-2, an unshaded circle, and II-3, an unshaded square. II-2 is joined to II-4, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded circle

Which mode of inheritance is this family consistent with?

  1. A. Autosomal dominant
    II-1 and III-1 are affected sons of unaffected parents.
    Under a dominant trait a parent who carries the allele shows it, so an affected child has an affected parent.
  2. B. ✓ X-linked recessive
  3. C. Maternal
    I-1 and II-2 are the mothers of the affected sons, and both are unaffected.
    Under maternal inheritance the mother of every affected child is affected.

Why: II-1 and III-1 are affected sons of unaffected mothers; no affected daughter of an unaffected father appears.
A dominant trait would show in a parent; a maternal trait would show in I-1 and II-2.
Of the three offered, only X-linked recessive fits: consistent with it, not proof of it.

58
Check q14

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded square, and I-2, a shaded circle, have two children: II-1, a shaded circle, and II-2, an unshaded square
A two-generation pedigree: I-1, a shaded square, and I-2, a shaded circle, have two children: II-1, a shaded circle, and II-2, an unshaded square

Which mode of inheritance does this family show?

  1. A. ✓ Autosomal dominant
  2. B. Autosomal recessive
    I-1 and I-2 both have the trait and their son II-2 does not.
    Under a recessive trait two affected parents are both homozygous, so every child would be affected.
  3. C. X-linked recessive
    Under X-linked recessive inheritance an affected mother is XʳXʳ, so every son receives an Xʳ and is affected; II-2 is an unaffected son.
  4. D. Maternal
    Under maternal inheritance every child of the affected mother I-1 would be affected, and II-2 is not.

Why: I-1 and I-2 are both affected, and their son II-2 is not.
Two affected parents can have an unaffected child only if each parent is heterozygous for a dominant allele and each passed II-2 the recessive allele.
So the trait is autosomal dominant.

59
Check q15

A trait is recorded in the family below.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have three children: II-1, an unshaded square, II-2, an unshaded circle, and II-3, a shaded circle
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have three children: II-1, an unshaded square, II-2, an unshaded circle, and II-3, a shaded circle

Which mode of inheritance does this family show?

  1. A. Autosomal dominant
    I-1 and I-2 are unaffected and their daughter II-3 is affected.
    Under a dominant trait an affected child has an affected parent.
  2. B. ✓ Autosomal recessive
  3. C. X-linked recessive
    II-3 is an affected daughter of an unaffected father.
    Under X-linked recessive inheritance she would need an Xʳ from her father, and an unaffected father's only X carries Xᴿ.
  4. D. Maternal
    Under maternal inheritance an affected child's mother is affected, and I-2 is not.

Why: I-1 and I-2 are unaffected, and their daughter II-3 is affected.
Two unaffected parents with an affected child mean the trait is recessive.
The affected child is a daughter of an unaffected father, so the allele cannot be on the X.
So the trait is autosomal recessive.

60
Check q16

A rare trait is recorded in the family below. II-5 married into the family.

A three-generation pedigree. I-1, a shaded circle, and I-2, an unshaded square, have four children, all shaded: II-1, a square, II-2, a circle, II-3, a square, and II-4, a circle. II-1 is joined to II-5, an unshaded circle; their four children III-1, a circle, III-2, a square, III-3, a circle, and III-4, a square, are all unshaded
A three-generation pedigree. I-1, a shaded circle, and I-2, an unshaded square, have four children, all shaded: II-1, a square, II-2, a circle, II-3, a square, and II-4, a circle. II-1 is joined to II-5, an unshaded circle; their four children III-1, a circle, III-2, a square, III-3, a circle, and III-4, a square, are all unshaded

Which mode of inheritance best fits this family?

  1. A. Autosomal dominant
    A dominant trait passes to about half of each affected parent’s children; here I-1 passed it to all four of hers and II-1 to none of his four.
  2. B. Autosomal recessive
    The trait is rare, so the married-in I-2 and II-5 are unlikely to carry a recessive allele, and even two carriers would give it to about half of their children.
  3. C. X-linked recessive
    II-2 and II-4 are affected daughters of the unaffected father I-2, which rules X-linked recessive inheritance out.
  4. D. ✓ Maternal

Why: The affected mother I-1 has four affected children; the affected father II-1 has four unaffected children.
II-2 and II-4, affected daughters of the unaffected I-2, rule X-linked recessive out.
Every child of an affected mother and no child of an affected father: the maternal pattern.

61
Check q17

A trait is recorded in the family below.

A two-generation pedigree: I-1, a shaded circle, and I-2, a shaded square, have three children: II-1, an unshaded square, II-2, a shaded circle, and II-3, a shaded square
A two-generation pedigree: I-1, a shaded circle, and I-2, a shaded square, have three children: II-1, an unshaded square, II-2, a shaded circle, and II-3, a shaded square

Which mode of inheritance does this family show?

  1. A. ✓ Autosomal dominant
  2. B. Autosomal recessive
    I-1 and I-2 both have the trait and their son II-1 does not.
    Under a recessive trait two affected parents are both homozygous and every child would be affected.
  3. C. X-linked recessive
    Under X-linked recessive inheritance the affected mother I-1 is XʳXʳ, so every son receives an Xʳ and is affected; II-1 is an unaffected son.
  4. D. Maternal
    Under maternal inheritance every child of the affected mother I-1 would be affected, and II-1 is not.

Why: I-1 and I-2 are both affected, and their son II-1 is unaffected.
Two affected parents can have an unaffected child only if the trait is dominant and both parents are heterozygous.
So the trait is autosomal dominant.

62
Check q18

A trait is recorded in the family below.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have four children: II-1, a shaded circle, II-2, an unshaded square, II-3, a shaded circle, and II-4, an unshaded square
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have four children: II-1, a shaded circle, II-2, an unshaded square, II-3, a shaded circle, and II-4, an unshaded square

Which mode of inheritance does this family show?

  1. A. Autosomal dominant
    I-1 and I-2 are unaffected and two of their children are affected.
    Under a dominant trait an affected child has an affected parent.
  2. B. ✓ Autosomal recessive
  3. C. X-linked recessive
    II-1 and II-3 are affected daughters of an unaffected father.
    Each daughter would need an Xʳ from I-1, and an unaffected father's only X carries Xᴿ.
  4. D. Maternal
    Under maternal inheritance the mother of an affected child is affected, and I-2 is not.

Why: I-1 and I-2 are unaffected, and their daughters II-1 and II-3 are affected.
Two unaffected parents with an affected child mean the trait is recessive.
II-1 and II-3 are affected daughters of an unaffected father, so the allele cannot be on the X.
So the trait is autosomal recessive.

63Whole pedigrees mixed practice

64
Check q19

A trait is recorded in the family below.

A three-generation pedigree. I-1, a shaded circle, and I-2, a shaded square, have II-1, a shaded square, II-2, an unshaded circle, and II-3, a shaded square. II-3 is joined to II-4, an unshaded circle; their children are III-1, a shaded circle, and III-2, an unshaded square
A three-generation pedigree. I-1, a shaded circle, and I-2, a shaded square, have II-1, a shaded square, II-2, an unshaded circle, and II-3, a shaded square. II-3 is joined to II-4, an unshaded circle; their children are III-1, a shaded circle, and III-2, an unshaded square

Which mode of inheritance does this pedigree show?

  1. A. Autosomal recessive
    I-1 and I-2 both have the trait and their daughter II-2 does not; under a recessive trait two affected parents are both homozygous and every child would be affected.
  2. B. X-linked recessive
    I-1 and I-2 are both affected; under X-linked recessive inheritance every daughter of theirs would receive an Xʳ from each and be affected, yet II-2 is unaffected.
  3. C. ✓ Autosomal dominant
  4. D. Maternal
    II-3, an affected father, has an affected daughter III-1, and a maternal trait never passes from a father.

Why: I-1 and I-2 are both affected, and their daughter II-2 is unaffected.
Two affected parents have an unaffected child only if both are heterozygous, so the trait is dominant.
The affected father II-3 passes it to his daughter III-1, so it is not maternal.
So the trait is autosomal dominant.

65
Check q20

A rare trait is recorded in the family below. II-6 married into the family.

A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have five children, all unshaded: II-1, a circle, II-2, a square, II-3, a circle, II-4, a square, and II-5, a square. II-5 is joined to II-6, a shaded circle who married in; their five children are all shaded: III-1, a circle, III-2, a square, III-3, a circle, III-4, a square, and III-5, a circle
A three-generation pedigree. I-1, a shaded square, and I-2, an unshaded circle, have five children, all unshaded: II-1, a circle, II-2, a square, II-3, a circle, II-4, a square, and II-5, a square. II-5 is joined to II-6, a shaded circle who married in; their five children are all shaded: III-1, a circle, III-2, a square, III-3, a circle, III-4, a square, and III-5, a circle

Which mode of inheritance best fits this pedigree?

  1. A. Autosomal dominant
    A dominant allele passes to about half of an affected parent’s children; here I-1 passed it to none of his five and II-6 to all five of hers.
  2. B. X-linked recessive
    III-1, III-3 and III-5 are affected daughters of an unaffected father, II-5, which rules X-linked recessive inheritance out.
  3. C. ✓ Maternal
  4. D. Autosomal recessive
    A rare recessive allele passes through fathers as readily as through mothers and would need the unaffected II-5 to carry it and pass it to all five of his children.

Why: The affected father I-1 has five unaffected children, and the affected mother II-6 has five affected children, daughters and sons alike.
A trait that every child of an affected mother shows and no child of an affected father shows is maternal.
So maternal inheritance fits best.

66
Check q21

A trait is recorded in the family below.

A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, an unshaded square, II-2, an unshaded circle, and II-3, an unshaded square. II-3 is joined to II-4, an unshaded circle; their children are III-1, a shaded circle, and III-2, an unshaded square
A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, an unshaded square, II-2, an unshaded circle, and II-3, an unshaded square. II-3 is joined to II-4, an unshaded circle; their children are III-1, a shaded circle, and III-2, an unshaded square

Which single family in the pedigree decides the mode?

  1. A. I-1 and I-2 with their three unaffected children
    Two unaffected parents with unaffected children fit every mode and decide nothing.
  2. B. ✓ II-3 and II-4 with III-1
  3. C. II-3 and II-4 with III-2
    III-2 is an unaffected son, and an unaffected child fits every mode; the deciding child is the affected daughter III-1, who rules X-linked recessive out.
  4. D. No family decides it
    One family here does decide: II-3 and II-4 are unaffected and their daughter III-1 is affected.

Why: II-3 and II-4 are unaffected and their daughter III-1 is affected.
Two unaffected parents with an affected child mean the trait is recessive, and because the affected child is a daughter of an unaffected father, the allele cannot be on the X: autosomal recessive.

67
Practice writing an answer

A trait is recorded in the family below. II-3 and II-4 with III-1 decide the mode.

A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, an unshaded square, II-2, an unshaded circle, and II-3, an unshaded square. II-3 is joined to II-4, an unshaded circle; their children are III-1, a shaded circle, and III-2, an unshaded square
A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, an unshaded square, II-2, an unshaded circle, and II-3, an unshaded square. II-3 is joined to II-4, an unshaded circle; their children are III-1, a shaded circle, and III-2, an unshaded square

(a) State the mode of inheritance that family decides, and explain how. (1 pt)

Model answer II-3 and II-4 are unaffected and their daughter III-1 is affected.
Two unaffected parents with an affected child mean the trait is recessive, and because the affected child is a daughter of an unaffected father, the allele cannot be on the X: autosomal recessive.
Rubric
  • Award 1 point for: autosomal recessive, because two unaffected parents have an affected child (recessive), and the affected child is a daughter of an unaffected father, which rules the X out.

68Mixed practice mixed practice

69
Check q22

Red–green color blindness is an X-linked recessive trait. The family below is recorded for it.

A two-generation pedigree for red–green color blindness: I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded square, II-2, an unshaded square, and II-3, an unshaded circle
A two-generation pedigree for red–green color blindness: I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded square, II-2, an unshaded square, and II-3, an unshaded circle

Which person must be a carrier of the color-blindness allele?

  1. A. I-1, the father
    A father gives his son a Y, so II-1’s allele did not come from I-1; and a man with the allele on his one X would be color-blind himself.
  2. B. II-2, the unaffected son
    II-2 has one X, and if it carried the allele he would be color-blind; he received his mother’s Xᴿ.
  3. C. ✓ I-2, the mother
  4. D. II-3, the unaffected daughter
    II-3 may be a carrier or may have received two Xᴿ, and the pedigree cannot say which.

Why: II-1 is color-blind, so his one X carries Xʳ, and a son’s X always comes from his mother.
I-2 is unaffected, so she carries Xʳ on one X and Xᴿ on the other: a carrier.

70
Check q23

A trait is recorded in the family below.

A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, an unshaded circle, II-2, a shaded square, and II-3, an unshaded square. II-4, an unshaded square, is joined to II-1; their children are III-1, a shaded square, III-2, a shaded square, and III-3, an unshaded circle
A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, an unshaded circle, II-2, a shaded square, and II-3, an unshaded square. II-4, an unshaded square, is joined to II-1; their children are III-1, a shaded square, III-2, a shaded square, and III-3, an unshaded circle

Which mode of inheritance is ruled out by this pedigree?

  1. A. X-linked recessive
    II-2, III-1 and III-2 are affected sons of unaffected mothers, the route an X-linked recessive allele takes; no affected daughter of an unaffected father appears to break it.
  2. B. Autosomal recessive
    An autosomal recessive trait can also give affected sons of unaffected parents, when both parents are carriers; only an affected daughter of an unaffected father would rule it out.
  3. C. ✓ Maternal

Why: Under maternal inheritance every affected child has an affected mother.
I-2 is the mother of the affected II-2, and II-1 is the mother of the affected III-1 and III-2.
I-2 and II-1 are both unaffected.
So maternal inheritance is ruled out.

71
Check q24

A trait is recorded in the family below.

A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, an unshaded circle, II-2, a shaded square, and II-3, an unshaded square. II-4, an unshaded square, is joined to II-1; their children are III-1, a shaded square, III-2, a shaded square, and III-3, an unshaded circle
A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, an unshaded circle, II-2, a shaded square, and II-3, an unshaded square. II-4, an unshaded square, is joined to II-1; their children are III-1, a shaded square, III-2, a shaded square, and III-3, an unshaded circle

Which mode of inheritance is consistent with this pedigree?

  1. A. ✓ X-linked recessive
  2. B. Autosomal dominant
    A dominant allele shows in everyone who carries it.
    So an affected child of a dominant trait has an affected parent.
    No affected person here has an affected parent.

Why: II-2, III-1 and III-2 are affected sons of unaffected mothers.
An X-linked recessive allele travels on the X from a carrier mother to a son, whose only X it is, so he shows it.
No affected daughter of an unaffected father breaks the pattern.
So X-linked recessive is consistent.

72
Check q25

A trait is recorded in the family below.

A three-generation pedigree. I-1, a shaded circle, and I-2, a shaded square, have II-1, a shaded circle, II-2, a shaded square, and II-3, an unshaded circle. II-3 is joined to II-4, an unshaded square; their child III-1 is an unshaded square
A three-generation pedigree. I-1, a shaded circle, and I-2, a shaded square, have II-1, a shaded circle, II-2, a shaded square, and II-3, an unshaded circle. II-3 is joined to II-4, an unshaded square; their child III-1 is an unshaded square

Which family decides the mode of inheritance?

  1. A. II-3 and II-4 with III-1
    Two unaffected parents with an unaffected child fit every mode and decide nothing.
  2. B. I-1 and I-2 with II-1 and II-2
    Affected children of affected parents fit dominant and recessive alike.
  3. C. ✓ I-1 and I-2 with II-3

Why: I-1 and I-2 both have the trait and their daughter II-3 does not.
Two affected parents can have an unaffected child only if the trait is dominant and both parents are heterozygous, each passing II-3 the recessive allele.

73
Practice writing an answer

A trait is recorded in the family below. I-1 and I-2 with II-3 decide the mode.

A three-generation pedigree. I-1, a shaded circle, and I-2, a shaded square, have II-1, a shaded circle, II-2, a shaded square, and II-3, an unshaded circle. II-3 is joined to II-4, an unshaded square; their child III-1 is an unshaded square
A three-generation pedigree. I-1, a shaded circle, and I-2, a shaded square, have II-1, a shaded circle, II-2, a shaded square, and II-3, an unshaded circle. II-3 is joined to II-4, an unshaded square; their child III-1 is an unshaded square

(a) State whether the trait is dominant or recessive, and explain how that family shows it. (1 pt)

Model answer I-1 and I-2 both have the trait and their daughter II-3 does not.
Two affected parents can have an unaffected child only if the trait is dominant and both parents are heterozygous, each passing II-3 the recessive allele.
Rubric
  • Award 1 point for: dominant, because I-1 and I-2 both have the trait and their daughter II-3 does not; two affected parents can have an unaffected child only if the trait is dominant and both are heterozygous.
74
Check q26 numeric entry

A woman is a carrier for red–green color blindness, XᴿXʳ. Her partner has normal color vision, XᴿY.

Calculate the probability, as a decimal, that their next child is a color-blind son.

Answer: 0.25  (tolerance ±0.005)

Working
Write down the values in the question:
P(son) = 1/2, because half of the father's sperm carry his Y
P(Xʳ from the mother) = 1/2, because half of her eggs carry Xʳ
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(A and B)=P(A)×P(B)
P(color-blind son)=12×12=14=0.25
75
Check q27

A trait is recorded in the family below.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded square, and II-2, an unshaded square
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded square, and II-2, an unshaded square

What can be decided about the mode of inheritance from this pedigree?

  1. A. Autosomal dominant
    An affected child of two unaffected parents rules a dominant trait out: a parent carrying a dominant allele would show it.
  2. B. Autosomal recessive
    Two unaffected parents with an affected son fit an X-linked recessive allele carried by the mother just as well; unaffected parents rule a dominant trait out, not the X.
  3. C. X-linked recessive
    One affected son of unaffected parents fits an autosomal recessive trait as well, with both parents carriers; a son alone does not place the allele on the X.
  4. D. ✓ Recessive, autosomal or X-linked

Why: II-1 is an affected son of two unaffected parents.
Under autosomal recessive inheritance both parents are carriers; under X-linked recessive inheritance his mother is a carrier.
Nothing here separates the two: an affected daughter of I-1 would rule X-linked out.

76
Practice writing an answer

A trait is recorded in the family below. The pedigree fits autosomal recessive or X-linked recessive.

A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded square, and II-2, an unshaded square
A two-generation pedigree: I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded square, and II-2, an unshaded square

(a) Explain why the pedigree fits both modes so far, and identify the kind of child that would decide between them. (1 pt)

Model answer II-1 is an affected son of two unaffected parents.
Under autosomal recessive inheritance both parents are carriers; under X-linked recessive inheritance his mother is a carrier.
Nothing here separates the two: an affected daughter of I-1 would rule X-linked out.
Rubric
  • Award 1 point for: the affected child is a son of unaffected parents, which fits both modes; an affected daughter of an unaffected father would rule X-linked recessive out.
77
Check q28

In one family, an affected woman has six children with an unaffected man, and all six, three sons and three daughters, are affected. Her affected brother has five children with an unaffected woman, and all five are unaffected.

Which mode of inheritance does this family show?

  1. A. ✓ Maternal
  2. B. Autosomal dominant
    Under a dominant trait the affected brother would pass the allele to about half of his children, and he passed it to none.
  3. C. Autosomal recessive
    Her unaffected partner would have to be a carrier and pass the allele to all six children, about a 1-in-64 chance; her brother's five unaffected children fit a recessive trait.
  4. D. X-linked recessive
    An X-linked recessive allele from an affected father reaches every daughter as a carrier, and an affected mother’s allele would show in her sons, not in all six children.

Why: All six children of the affected mother are affected, sons and daughters alike, and none of the affected father’s five children is.
That is the maternal pattern: the egg supplies the mitochondria, so a mother passes such a trait to every child and a father to none.

78
Practice writing an answer

A trait is recorded in the family below. II-4 married into the family.

A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded square, II-2, a shaded circle, and II-3, an unshaded circle. II-3 is joined to II-4, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded circle
A three-generation pedigree. I-1, an unshaded square, and I-2, an unshaded circle, have II-1, a shaded square, II-2, a shaded circle, and II-3, an unshaded circle. II-3 is joined to II-4, an unshaded square; their children are III-1, a shaded square, and III-2, an unshaded circle

(a) Make a claim about the mode of inheritance of the trait. (1 pt)

Model answer Claim: the trait is autosomal recessive.
Rubric
  • Award 1 point for: the claim, autosomal recessive; no reasoning is required for this point.

Slip Answering X-linked recessive because the affected people are mostly sons. II-2 is an affected daughter, and her father is unaffected.

(b) Support the autosome-or-X part of your claim with evidence from one individual in the pedigree, and name that individual. (1 pt)

Model answer Evidence: II-2 is an affected daughter of an unaffected father, I-1.
Under X-linked recessive inheritance she would carry the allele on both of her X chromosomes, one from each parent.
Her father I-1 is unaffected, so his only X carries the ordinary allele, and he could not have given her the allele.
So the trait cannot be X-linked recessive, and the gene sits on an autosome.
Rubric
  • Award 1 point for: the evidence (II-2, an affected daughter of an unaffected father) AND the reasoning (she would need the allele on the X from her father, and an unaffected father’s X does not carry it).

Slip Naming II-1 or III-1, the affected sons. Affected sons of unaffected mothers fit X-linked recessive inheritance; only the affected daughter breaks it.

(c) Using A for the dominant allele and a for the recessive allele, identify the genotypes of I-1 and I-2, and explain how the pedigree fixes them. (1 pt)

Model answer I-1 and I-2 are both Aa.
Their daughter II-2 has the trait, so she is aa, and each of her two a alleles came from one parent.
Each parent is unaffected, so each carries an A as well: both are carriers, Aa.
Rubric
  • Award 1 point for: I-1 and I-2 are both Aa, because their affected child is aa and received an a from each of them while they show the ordinary trait.

Slip Writing AA for a parent. An AA parent has no a to pass on, and their child II-2 is aa.

(d) Calculate the probability, as a decimal, that the next child of II-3 and II-4 has the trait. (1 pt)

Answer: 0.25  (tolerance ±0.005)

Model answer II-3 and II-4 are both Aa, because their son III-1 has the trait and received an a from each of them.
The probability that their next child has the trait is 0.25.
Working
Write down the values in the question:
II-3 and II-4 are both Aa, because their son III-1 is aa
P(a from II-3)=12
P(a from II-4)=12
Write down the equation:
P(A and B)=P(A)×P(B)
Substitute the values into the equation:
P(aa)=12×12=14=0.25
Rubric
  • Award 1 point for: 1/4 (0.25), from II-3 and II-4 both Aa.

APBIO-U05-L28B Which deviation is this? Decide from the counts

Topic 5.4 · Non-Mendelian Genetics · 44 steps

Four small tables of offspring counts in a row, none labeled. First: 421, 407, 46, 42. Second: 29, 63, 28. Third: daughters 118 and 0, sons 0 and 112. Fourth: two reciprocal crosses, green ovule parent by pale pollen parent giving 96 green seedlings, and pale by green giving 104 pale
Four small tables of offspring counts in a row, none labeled. First: 421, 407, 46, 42. Second: 29, 63, 28. Third: daughters 118 and 0, sons 0 and 112. Fourth: two reciprocal crosses, green ovule parent by pale pollen parent giving 96 green seedlings, and pale by green giving 104 pale

Here are four sets of offspring counts from four different crosses, and no labels. In one, two classes are large and two are small. In one, three colors come out 1 : 2 : 1. In one, the sons and daughters show different results. In one, a pair of reciprocal crosses gives offspring that match the mother every time.

Which deviation from Mendel’s ratios is each, and what settles it?

Unit 5 · Heredity

1The signatures

2

Video: Watch: The signatures

Each of the four tables in turn, its signature named and the deviation written beside it: two large classes and two small; 1 : 2 : 1 by color; sons unlike daughters; offspring that match the mother. Then the five signatures gathered into one table.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L28Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L28Ba.mp4

3

How do you name a deviation from Mendel’s ratios when all you have is counts? Each deviation leaves its own signature in the counts.

4

Take the four tables in turn and ask of each: which of Mendel’s ratios did the breeder expect, and in what way do the counts depart from it?

5
Check q1

A fly heterozygous for two genes is test-crossed, and the two genes assort independently.

Which of the following appears among the offspring?

  1. A. One class of offspring only
    A heterozygote makes more than one kind of gamete, so its offspring fall into more than one class.
  2. B. Two large classes and two small ones
    Two large classes and two small ones appear when the two genes travel together on one chromosome.
  3. C. ✓ Four classes of offspring, all equal

Why: Under independent assortment the heterozygote makes four kinds of gamete in equal numbers.
Each kind of gamete fuses with the partner’s one kind of gamete.
So the offspring fall into four equal classes.

6

The first table. Suppose a fly heterozygous for body color and wing shape was test-crossed, and 916 offspring were counted.

A table of offspring counts from a fruit-fly test cross: gray body and normal wings 421, black body and vestigial wings 407, gray body and vestigial wings 46, black body and normal wings 42
A table of offspring counts from a fruit-fly test cross: gray body and normal wings 421, black body and vestigial wings 407, gray body and vestigial wings 46, black body and normal wings 42
7

Independent assortment predicted four equal classes. Two classes are large and two are small.

8

So the two genes are linked. The small classes are the recombinants.

9
Check q2

Two heterozygotes are crossed for a gene with complete dominance.

Which ratio of phenotypes does Mendel’s model predict for their offspring?

  1. A. 1 : 1
    A 1 : 1 ratio comes from a heterozygote crossed with a homozygous recessive partner, the test cross.
  2. B. ✓ 3 : 1
  3. C. 1 : 2 : 1
    1 : 2 : 1 is the ratio of genotypes; with complete dominance the heterozygote looks like the dominant homozygote, so the phenotypes come out 3 : 1.

Why: The square for two heterozygotes holds one dominant homozygote, two heterozygotes and one recessive homozygote.
Under complete dominance the heterozygote shows the dominant trait.
So three cells show the dominant trait and one shows the recessive: 3 : 1.

10

The second table shows two pink snapdragons crossed. Mendel’s ratio for two heterozygotes is 3 : 1.

A table of offspring counts from two pink snapdragons crossed: red 29, pink 63, white 28
A table of offspring counts from two pink snapdragons crossed: red 29, pink 63, white 28
11

The plants came out red, pink and white, close to 1 : 2 : 1. The pink heterozygote is in between red and white, so flower color shows incomplete dominance.

12

A heterozygote can also show both traits side by side, as a roan calf shows red hairs and white hairs. Two such heterozygotes give the same 1 : 2 : 1, and the deviation is codominance.

13
Check q3

A mother and a father have a son.

Which parent gave the son his X chromosome?

  1. A. ✓ His mother
  2. B. His father
    A father gives his X to each daughter and his Y to each son.

Why: A mother gives an X to every child.
A father gives his X to each daughter and his Y to each son.
So a son’s X always came from his mother.

14

The third table: a white-eyed female fly was crossed with a red-eyed male, and every daughter has red eyes and every son has white eyes.

A table of offspring counts from a fruit-fly cross, split by sex: daughters with red eyes 118, daughters with white eyes 0, sons with red eyes 0, sons with white eyes 112
A table of offspring counts from a fruit-fly cross, split by sex: daughters with red eyes 118, daughters with white eyes 0, sons with red eyes 0, sons with white eyes 112
15

Each son’s only X came from his white-eyed mother, and each daughter also received her father’s X. So the eye-color gene sits on the X: a sex-linked trait.

16
Check q4

A plant grows from a seed.

Which parent supplied the seedling’s chloroplasts?

  1. A. ✓ The ovule parent
  2. B. The pollen parent
    The pollen grain carries the sperm and no chloroplasts that last.
  3. C. Both parents equally
    The egg cell in the ovule brings the chloroplasts; the pollen brings none.

Why: The egg cell sits in the ovule and carries the chloroplasts.
The sperm in the pollen grain brings none.
So the seedling’s chloroplasts came from the ovule parent.

17

The fourth table. Suppose a green plant and a pale plant are crossed both ways, once with each plant as the ovule parent.

A table of two reciprocal crosses in a garden plant: a green ovule parent with a pale pollen parent gave 96 seedlings, all green; a pale ovule parent with a green pollen parent gave 104 seedlings, all pale
A table of two reciprocal crosses in a garden plant: a green ovule parent with a pale pollen parent gave 96 seedlings, all green; a pale ovule parent with a green pollen parent gave 104 seedlings, all pale
18

Every seedling matched the ovule parent, whichever plant supplied the pollen. So the trait is written in chloroplast DNA: non-nuclear inheritance, which shows in a family as maternal inheritance.

19

Here is a table of the five signatures: the ratio Mendel’s model predicts, what the counts show instead, and the deviation.

A table of the five signatures. A dihybrid test cross: Mendel predicts 1 : 1 : 1 : 1, the counts show two large classes and two small, the deviation is linked genes. Two heterozygotes: Mendel predicts 3 : 1, the counts show 1 : 2 : 1 with a blended middle form, incomplete dominance. Two heterozygotes: Mendel predicts 3 : 1, the counts show 1 : 2 : 1 with both forms side by side, codominance. One cross read by sex: Mendel predicts the same in both sexes, the counts show sons unlike daughters, a sex-linked trait. Reciprocal crosses: Mendel predicts the same result both ways, the counts show every offspring like the mother, non-nuclear inheritance
A table of the five signatures. A dihybrid test cross: Mendel predicts 1 : 1 : 1 : 1, the counts show two large classes and two small, the deviation is linked genes. Two heterozygotes: Mendel predicts 3 : 1, the counts show 1 : 2 : 1 with a blended middle form, incomplete dominance. Two heterozygotes: Mendel predicts 3 : 1, the counts show 1 : 2 : 1 with both forms side by side, codominance. One cross read by sex: Mendel predicts the same in both sexes, the counts show sons unlike daughters, a sex-linked trait. Reciprocal crosses: Mendel predicts the same result both ways, the counts show every offspring like the mother, non-nuclear inheritance
20

What you are expected to know Name the deviation from Mendel’s ratios that a set of offspring counts shows, linked genes, incomplete dominance, codominance, a sex-linked trait or non-nuclear inheritance, from its signature in the counts.

21
Check q5

In Andalusian chickens, a black bird crossed with a splashed-white bird gives blue chicks, a shade between black and white. Two blue chickens are crossed and their chicks are counted in the table below.

A table of offspring counts from two blue Andalusian chickens crossed: black 24, blue 51, splashed white 25
A table of offspring counts from two blue Andalusian chickens crossed: black 24, blue 51, splashed white 25

Which deviation from Mendel’s ratios does this cross show?

  1. A. Codominance
    A codominant heterozygote shows both parents’ traits side by side, black feathers and white feathers, and a blue chick is a shade in between.
  2. B. ✓ Incomplete dominance
  3. C. Linked genes
    Linked genes leave two large classes and two small ones in a two-gene test cross; here one gene gives three color classes from two heterozygotes.
  4. D. Sex-linked trait
    A sex-linked trait shows up as sons and daughters with different results, and the table does not split by sex.

Why: Two blue heterozygotes give black, blue and white chicks close to 1 : 2 : 1. The heterozygote’s blue sits between black and white, so neither allele masks the other.
That is incomplete dominance.

22Quick quiz: the signatures mixed practice

23
Check q6

A garden plant comes in a green-leaved form and a pale-leaved form. A gardener crossed the two forms both ways and counted the seedlings in the table below.

A table of two reciprocal crosses in a garden plant with green or pale leaves: a green ovule parent with a pale pollen parent gave 140 seedlings, all green; a pale ovule parent with a green pollen parent gave 136 seedlings, all pale
A table of two reciprocal crosses in a garden plant with green or pale leaves: a green ovule parent with a pale pollen parent gave 140 seedlings, all green; a pale ovule parent with a green pollen parent gave 136 seedlings, all pale

Which deviation from Mendel’s ratios do these crosses show?

  1. A. Sex-linked trait
    A sex-linked trait shows as sons and daughters with different results.
    Here every seedling from one pair of parents is alike.
  2. B. ✓ Non-nuclear inheritance
  3. C. Incomplete dominance
    Incomplete dominance would give one in-between color from a green × pale cross, whichever plant supplied the pollen.
    Here the result depends on which parent supplied the ovule.
  4. D. Linked genes
    Linked genes show as two large and two small classes from a two-gene test cross.
    Here one trait is followed and each cross gives a single class.

Why: Every seedling has the leaf color of the ovule parent: green in one cross and pale in the reciprocal cross.
The ovule supplies the chloroplasts, so a trait written in chloroplast DNA follows the ovule parent.
That is non-nuclear inheritance.

24
Check q7

Suppose a corn plant heterozygous for kernel color (purple or yellow) and kernel texture (smooth or shrunken) was test-crossed with a yellow, shrunken plant. The kernels are counted in the table below.

A table of offspring counts from a corn test cross: purple smooth 386, yellow shrunken 379, purple shrunken 118, yellow smooth 117
A table of offspring counts from a corn test cross: purple smooth 386, yellow shrunken 379, purple shrunken 118, yellow smooth 117

Which deviation from Mendel’s ratios do these counts show?

  1. A. Incomplete dominance
    Incomplete dominance is a one-gene pattern with a heterozygote in between, and these are two genes in a test cross with four classes.
  2. B. Non-nuclear inheritance
    An organelle trait gives offspring that all match the ovule parent, and here four classes appear.
  3. C. ✓ Linked genes
  4. D. Sex-linked trait
    A sex-linked trait shows as different results in the two sexes, and kernels are not sorted by sex; independent assortment predicted four equal classes and two came out small.

Why: Independent assortment predicts four equal classes from this test cross.
Purple smooth and yellow shrunken are large, and the other two classes are small.
So the two genes sit on one chromosome, and the small classes are the recombinants.
That is linked genes.

25
Check q8

In radishes, a long-rooted plant crossed with a round-rooted plant gives oval-rooted plants. A grower crosses two oval-rooted radishes and counts the offspring in the table below.

A table of offspring counts from two oval-rooted radishes crossed: long 26, oval 49, round 25
A table of offspring counts from two oval-rooted radishes crossed: long 26, oval 49, round 25

Which deviation from Mendel’s ratios does this cross show?

  1. A. ✓ Incomplete dominance
  2. B. Codominance
    A codominant heterozygote shows both parents’ traits side by side; an oval root is a shape in between, not a root part long and part round.
  3. C. Linked genes
    Linked genes show as two large and two small classes from a two-gene test cross.
    Here one gene gives three classes from two heterozygotes.
  4. D. Non-nuclear inheritance
    An organelle trait gives offspring that all match the ovule parent.
    Here two oval parents give three classes.

Why: Two oval heterozygotes give long, oval and round roots close to 1 : 2 : 1. The heterozygote’s oval root sits between long and round, so neither allele masks the other.
That is incomplete dominance.

26
Check q9

In a bird, a black bird crossed with a white bird gives speckled chicks, whose feathers are each either black or white. Two speckled birds are crossed and their chicks are counted in the table below.

A table of offspring counts from two speckled birds crossed: black 22, speckled black-and-white 45, white 23
A table of offspring counts from two speckled birds crossed: black 22, speckled black-and-white 45, white 23

Which deviation from Mendel’s ratios does this cross show?

  1. A. ✓ Codominance
  2. B. Incomplete dominance
    An incompletely dominant heterozygote is a blend, gray feathers between black and white.
    A speckled chick shows black feathers and white feathers side by side.
  3. C. Linked genes
    Linked genes show as two large and two small classes of a two-gene test cross.
    Here one gene gives three classes from two heterozygotes.
  4. D. Non-nuclear inheritance
    An organelle trait gives offspring that match the mother.
    Here two speckled parents give three classes.

Why: Two speckled heterozygotes give black, speckled and white chicks close to 1 : 2 : 1. The heterozygote shows both traits side by side, black feathers and white feathers, rather than a blend.
That is codominance.

27
Check q10

Suppose a female fruit fly with normal wings was crossed with a male with normal wings. Their offspring are counted by sex in the table below.

A table of offspring counts from a fruit-fly cross, split by sex: daughters with normal wings 94, daughters with miniature wings 0, sons with normal wings 47, sons with miniature wings 49
A table of offspring counts from a fruit-fly cross, split by sex: daughters with normal wings 94, daughters with miniature wings 0, sons with normal wings 47, sons with miniature wings 49

Which deviation from Mendel’s ratios do these counts show?

  1. A. Linked genes
    Linked genes show as two large and two small classes of a two-gene cross, and here one gene is followed and the split is by sex.
  2. B. Non-nuclear inheritance
    An organelle trait passes from the mother to every child, sons and daughters alike, and here the sons and daughters differ.
  3. C. Incomplete dominance
    Incomplete dominance gives a heterozygote in between two parental forms, and no in-between wing appears; the tell is the difference between sons and daughters.
  4. D. ✓ Sex-linked trait

Why: Two normal-winged parents gave miniature wings in about half the sons and in no daughter.
A recessive allele on the mother’s X shows in the sons who receive it.
Every daughter also receives the father’s X, which masks it.
That is a sex-linked trait.

28
Check q11

In fruit flies, red eyes and white eyes were crossed both ways, and the offspring are recorded in the table below.

A table of two reciprocal fruit-fly crosses: red-eyed female with white-eyed male gave 210 offspring, all red-eyed; white-eyed female with red-eyed male gave 104 red-eyed daughters and 98 white-eyed sons
A table of two reciprocal fruit-fly crosses: red-eyed female with white-eyed male gave 210 offspring, all red-eyed; white-eyed female with red-eyed male gave 104 red-eyed daughters and 98 white-eyed sons

Which deviation from Mendel’s ratios do these crosses show?

  1. A. Non-nuclear inheritance
    An organelle trait passes to sons and daughters alike.
    In the second cross the daughters and the sons differ.
  2. B. Incomplete dominance
    Incomplete dominance gives an in-between eye color in the heterozygote.
    Every fly here is red-eyed or white-eyed.
  3. C. Linked genes
    Linked genes show as two large and two small classes from a two-gene test cross.
    One gene is followed here.
  4. D. ✓ Sex-linked trait

Why: The reciprocal crosses differ.
In the second cross the daughters are all red-eyed and the sons all white-eyed.
A son’s only X came from his white-eyed mother, so he is white-eyed.
A daughter also got her father’s red-eye X, so she is red-eyed.
That is a sex-linked trait.

29Close counts: chi-square against 1 : 1 : 1 : 1

30

When the counts are stark, as in the four opening tables, the signature decides on its own.

31

When two classes are only a little larger than the other two, you need a number for the gap: chi-square against 1 : 1 : 1 : 1.

32
Check q12

A dihybrid test cross gives 1,000 offspring in four classes. The null hypothesis is a 1 : 1 : 1 : 1 ratio.

How is the expected count of one class found?

  1. A. Take the observed count of that class
    The expected count comes from the model, not from what was counted.
  2. B. ✓ Multiply the total by that class’s share of the ratio
  3. C. Divide the total by the observed count of that class
    The observed count plays no part in the expected count; the ratio’s share of the total does.

Why: The expected count, e, is the count the ratio predicts for the total.
So e is the total times that class’s share of the ratio.

33
Check q13

A breeder’s chi-square value comes out larger than the critical value at p = 0.05.

What does that say about the gap between her counts and the 1 : 1 : 1 : 1 ratio?

  1. A. Chance alone could make a gap this size
    A gap that chance alone could make gives a chi-square below the critical value.
  2. B. ✓ The gap is too large to be chance

Why: A chi-square larger than the critical value means the gap between the counts and the ratio is too large to be chance.
So the null hypothesis is rejected.

34

Four classes give three degrees of freedom. Here is the chi-square table from your formula sheet: the critical value at three degrees of freedom and p = 0.05 is 7.81.

The chi-square table from the formula sheet: degrees of freedom 1 to 8 across the top; the p = 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the p = 0.01 row reads 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09; the cell for 3 degrees of freedom at p = 0.05, 7.81, is ringed
The chi-square table from the formula sheet: degrees of freedom 1 to 8 across the top; the p = 0.05 row reads 3.84, 5.99, 7.81, 9.49, 11.07, 12.59, 14.07, 15.51 and the p = 0.01 row reads 6.63, 9.21, 11.34, 13.28, 15.09, 16.81, 18.48, 20.09; the cell for 3 degrees of freedom at p = 0.05, 7.81, is ringed
35

What you are expected to know Test close dihybrid test-cross counts against 1 : 1 : 1 : 1 with chi-square, and state whether the null hypothesis is rejected.

36
Check q14 numeric entry

Suppose a tomato plant heterozygous for fruit color (red or yellow) and skin texture (smooth or fuzzy) was test-crossed and 1,000 plants were counted, as in the table below. The null hypothesis is a 1 : 1 : 1 : 1 ratio.

A table of offspring counts from a tomato test cross: red smooth 262, yellow fuzzy 258, red fuzzy 241, yellow smooth 239; total 1,000
A table of offspring counts from a tomato test cross: red smooth 262, yellow fuzzy 258, red fuzzy 241, yellow smooth 239; total 1,000

Calculate the chi-square value for these counts.

Answer: 1.64  (tolerance ±0.01)

Working
Write down the values in the question:
o = 262, 258, 241, 239
total = 1,000
ratio under the null hypothesis = 1 : 1 : 1 : 1, so each class is 1 share of 4
Write down the equations:
e=total×that class's share of the ratio
χ2=∑(o−e)2e
Substitute the values into the first equation:
e=1,000×14=250 for every class
Substitute the values into the second equation, one class per line:
(262−250)2250=144250=0.576
(258−250)2250=64250=0.256
(241−250)2250=81250=0.324
(239−250)2250=121250=0.484
χ2=0.576+0.256+0.324+0.484=1.64
Compare chi-square with the critical value: it is smaller, so the null hypothesis is not rejected, and the counts are consistent with independent assortment.
degrees of freedom = 4 − 1 = 3
critical value at p = 0.05 = 7.81
1.64<7.81
37

Back to the four tables of the opening, four crosses and no labels: each gave itself away by its signature, and chi-square decided the close case. Here is a table of the four: the counts, the signature and the deviation.

A table of the four opening sets of counts. First: 421, 407, 46 and 42 flies from a test cross; two large classes and two small; linked genes. Second: 29 red, 63 pink and 28 white from pink by pink; 1 : 2 : 1 with pink in between; incomplete dominance. Third: 118 red-eyed daughters and 112 white-eyed sons; sons unlike daughters; a sex-linked trait. Fourth: 96 green and 104 pale seedlings from reciprocal crosses; every seedling like the ovule parent; non-nuclear inheritance
A table of the four opening sets of counts. First: 421, 407, 46 and 42 flies from a test cross; two large classes and two small; linked genes. Second: 29 red, 63 pink and 28 white from pink by pink; 1 : 2 : 1 with pink in between; incomplete dominance. Third: 118 red-eyed daughters and 112 white-eyed sons; sons unlike daughters; a sex-linked trait. Fourth: 96 green and 104 pale seedlings from reciprocal crosses; every seedling like the ovule parent; non-nuclear inheritance

38Mixed practice mixed practice

39
Check q15 numeric entry

Suppose a sweet-pea plant heterozygous for flower color (purple or red) and pollen shape (long or round) was test-crossed and 1,000 plants were counted, as in the table below. The null hypothesis is a 1 : 1 : 1 : 1 ratio.

A table of offspring counts from a sweet-pea test cross: purple flowers, long pollen 305; red flowers, round pollen 295; purple flowers, round pollen 205; red flowers, long pollen 195; total 1,000
A table of offspring counts from a sweet-pea test cross: purple flowers, long pollen 305; red flowers, round pollen 295; purple flowers, round pollen 205; red flowers, long pollen 195; total 1,000

Calculate the chi-square value for these counts.

Answer: 40.4  (tolerance ±0.05)

Working
Write down the values in the question:
o = 305, 295, 205, 195
e=1,000×14=250 for every class
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
(305−250)2250=12.1
(295−250)2250=8.10
(205−250)2250=8.10
(195−250)2250=12.1
χ2=12.1+8.10+8.10+12.1=40.4
Compare chi-square with the critical value: it is larger, so the null hypothesis is rejected; the two genes are linked, and the two smaller classes are the recombinants.
degrees of freedom = 3
critical value = 7.81
40.4>7.81
40
Check q16

For a dihybrid test cross, a student calculates χ² = 15.2 against a 1 : 1 : 1 : 1 ratio, with three degrees of freedom. The critical value at p = 0.05 for three degrees of freedom is 7.81.

What does this result say about the two genes?

  1. A. The counts prove the two genes sit on different chromosomes
    Rejecting the null hypothesis says the counts do not fit independent assortment; genes on different chromosomes would fit it.
  2. B. The two genes show incomplete dominance, one partly masking the other
    Incomplete dominance is a one-gene pattern read from a heterozygote’s appearance, and chi-square on a two-gene test cross does not test it.
  3. C. ✓ The counts do not fit 1 : 1 : 1 : 1, so the genes do not assort independently
  4. D. The breeder miscounted one of the four classes of offspring
    A large chi-square value measures a real gap between the counts and the model, and is not a sign of counting errors.

Why: 15.2 is larger than the critical value 7.81 for three degrees of freedom.
So the null hypothesis of a 1 : 1 : 1 : 1 ratio is rejected: the counts do not fit independent assortment.
Two large and two small classes are the signature of linked genes.

41
Check q17

Suppose a rabbit heterozygous for coat color (black or brown) and hair length (short or long) was test-crossed and 1,000 kits were counted, as in the table below. Against a 1 : 1 : 1 : 1 ratio, χ² = 0.232 with three degrees of freedom. The critical value at p = 0.05 for three degrees of freedom is 7.81.

A table of offspring counts from a rabbit test cross: black short 248, brown long 252, black long 255, brown short 245; total 1,000
A table of offspring counts from a rabbit test cross: black short 248, brown long 252, black long 255, brown short 245; total 1,000

What is the verdict of the test?

  1. A. Reject
    0.232 is smaller than the critical value 7.81, so the null hypothesis stands; and the four classes are nearly equal, with no two-large-two-small signature.
  2. B. ✓ Fail to reject

Why: 0.232 is smaller than 7.81, the critical value for three degrees of freedom, so the breeder fails to reject the null hypothesis: the counts are consistent with 1 : 1 : 1 : 1 and with independent assortment.
That does not prove the genes are on different chromosomes.

42
Practice writing an answer

Suppose a rabbit heterozygous for coat color (black or brown) and hair length (short or long) was test-crossed and 1,000 kits were counted, as in the table below. Against a 1 : 1 : 1 : 1 ratio, χ² = 0.232 with three degrees of freedom. The verdict is to fail to reject the null hypothesis.

A table of offspring counts from a rabbit test cross: black short 248, brown long 252, black long 255, brown short 245; total 1,000
A table of offspring counts from a rabbit test cross: black short 248, brown long 252, black long 255, brown short 245; total 1,000

(a) Explain the verdict, and describe what it says about the two genes. (1 pt)

Model answer 0.232 is smaller than 7.81, the critical value for three degrees of freedom, so the breeder fails to reject the null hypothesis: the counts are consistent with 1 : 1 : 1 : 1 and with independent assortment.
That does not prove the genes are on different chromosomes.
Rubric
  • Award 1 point for: χ² = 0.232 is far below the critical value, so the counts fit 1 : 1 : 1 : 1, independent assortment; the data give no evidence of linkage, though they do not prove the genes are on different chromosomes.
43
Practice writing an answer

A tomato breeder sets up two experiments. First, she crosses a green-leaved plant with a pale-yellow-leaved plant both ways, keeping a record of which plant supplied the ovules and which the pollen. Second, she test-crosses a plant heterozygous for fruit color (red or yellow) and leaf shape (potato leaf or cut leaf) with a yellow-fruited, cut-leaf plant and counts 1,000 plants. Her records are in the two tables below, with the chi-square table from the formula sheet beneath them.

Two tables and the chi-square table. First table, reciprocal crosses in tomato: a green ovule parent with a pale-yellow pollen parent gave 88 seedlings, all green; a pale-yellow ovule parent with a green pollen parent gave 92 seedlings, all pale yellow. Second table, a test cross for fruit color and leaf shape: red potato-leaf 275, yellow cut-leaf 268, red cut-leaf 232, yellow potato-leaf 225; total 1,000. Beneath them the formula sheet's chi-square table, degrees of freedom 1 to 8, rows p = 0.05 and p = 0.01
Two tables and the chi-square table. First table, reciprocal crosses in tomato: a green ovule parent with a pale-yellow pollen parent gave 88 seedlings, all green; a pale-yellow ovule parent with a green pollen parent gave 92 seedlings, all pale yellow. Second table, a test cross for fruit color and leaf shape: red potato-leaf 275, yellow cut-leaf 268, red cut-leaf 232, yellow potato-leaf 225; total 1,000. Beneath them the formula sheet's chi-square table, degrees of freedom 1 to 8, rows p = 0.05 and p = 0.01

(a) Identify the deviation from Mendel’s ratios that the reciprocal crosses show, and justify your answer from the counts. (1 pt)

Model answer Non-nuclear inheritance, seen here as maternal inheritance.
Every seedling matched the ovule parent: the green ovule parent gave 88 green seedlings and the pale-yellow ovule parent gave 92 pale-yellow seedlings, so swapping the parents swapped the result.
The ovule supplies the chloroplasts and the pollen does not, so a trait written in chloroplast DNA follows the ovule parent.
Rubric
  • Award 1 point for: non-nuclear inheritance or maternal inheritance, justified by every seedling matching the ovule parent in both reciprocal crosses.

Slip Calling it dominance of green. A dominant allele would give the same result whichever parent supplied the ovule, and here the two crosses differ.

(b) Predict the leaf color of the seedlings from a cross of a green ovule parent with a green pollen parent that itself grew from a pale-yellow pollen parent. (1 pt)

Model answer All green.
The seedlings’ chloroplasts come from the ovule parent, which is green, so every seedling is green; the pollen parent contributes no chloroplasts, so the pale-yellow color of its own pollen parent has no route into the seedlings.
Rubric
  • Award 1 point for: all green, because the chloroplasts come from the ovule parent and the pollen parent contributes none.

Slip Predicting some pale-yellow seedlings because the pollen parent’s own pollen parent was pale yellow. The pollen parent is green, and even its own chloroplasts do not reach the seedlings.

(c) Calculate the chi-square value for the test-cross counts against a 1 : 1 : 1 : 1 ratio. (1 pt)

Answer: 7.59  (tolerance ±0.01)

Model answer χ² = 7.59.
Working
Write down the values in the question:
o = 275, 268, 232, 225
e=1,000×14=250 for every class
Write down the equation:
χ2=∑(o−e)2e
Substitute the values into the equation, one class per line:
(275−250)2250=2.50
(268−250)2250=1.296
(232−250)2250=1.296
(225−250)2250=2.50
χ2=2.50+1.296+1.296+2.50=7.59
Rubric
  • Award 1 point for: χ² = 7.59 (accept 7.58 to 7.60).

(d) State the verdict of the chi-square test at p = 0.05 on the null hypothesis, with the reason, and state what the verdict means for the two genes. (1 pt)

Model answer There are four classes, so three degrees of freedom, and the critical value at p = 0.05 is 7.81.
The calculated χ² of 7.59 is smaller than 7.81, so she fails to reject the null hypothesis: the counts are consistent with a 1 : 1 : 1 : 1 ratio and with independent assortment.
The critical value is a line: a χ² below 7.81 fails to reject, however close, and a χ² above 7.81 rejects.
That does not prove the two genes sit on different chromosomes; chance alone can account for the gap.
Rubric
  • Award 1 point for: fail to reject the null hypothesis because 7.59 is smaller than 7.81 (three degrees of freedom), so the counts are consistent with independent assortment.
  • Do not award: ‘accept the null hypothesis’ or ‘the genes are proved to be on different chromosomes’.
  • Do not award a verdict from a two-class test of parental against recombinant plants (543 : 457 against 500 : 500): that test has a different null hypothesis. Part (c) set the null hypothesis as 1 : 1 : 1 : 1 across four classes, so the verdict scored is the four-class one.

Slip Rejecting because two classes are larger than the other two. The gap is measured against the critical value, and 7.59 falls short of 7.81.

APBIO-U05-P54 Practice questions: Topic 5.4

Topic 5.4 · Non-Mendelian Genetics · 10 MCQ · 2 FRQ · for APBIO-U05-T54

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation. The first free-response question walks you through a linked test cross one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a question needs a critical value, the chi-square table is drawn with it; read the p = 0.05 row.

Video: Watch first: Non-Mendelian genetics, summed up

Heterozygotes that show, 1 : 2 : 1 in phenotypes; linked genes, recombinants and map units; the X and the Y, and why recessive X-linked traits are mostly male; one gene with many effects; organelle genes from the mother; deciding the mode from a pedigree or from counts.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T54-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T54-summary.mp4

Q1 P54-q01

Suppose that in one breed of dog, true-breeding curly-coated dogs mated with true-breeding straight-coated dogs give only wavy-coated puppies, and two wavy-coated dogs mated give curly, wavy and straight puppies.

Which kind of dominance does coat texture show?

  1. A. Complete dominance
    Under complete dominance the heterozygote looks like one of the homozygotes, curly or straight; here it is wavy.
  2. B. Codominance
    Codominance shows both traits separately in the heterozygote, curly patches beside straight patches; a wavy coat is a blend.
  3. C. ✓ Incomplete dominance
  4. D. A merging of the two alleles into one wavy-coat allele
    Curly and straight come back among the wavy dogs’ puppies, so the alleles were carried unchanged; nothing merged.

Why: The heterozygote’s coat is between the two homozygotes’ coats, and curly and straight return when two wavy dogs are mated: incomplete dominance.

Q2 P54-q02

Suppose that in a species of lizard one gene sets scale color. The allele Sᴮ gives black scales and the allele Sʸ gives yellow scales, and the two alleles are codominant.

What do the scales of an SᴮSʸ lizard look like?

  1. A. Every scale is black
    Under codominance both alleles’ traits show.
    Every scale black would mean Sᴮ hid Sʸ: complete dominance.
  2. B. Every scale is yellow
    Under codominance both alleles’ traits show.
    Every scale yellow would mean Sʸ hid Sᴮ: complete dominance.
  3. C. Every scale is one shade in between black and yellow
    A shade in between is a blend: incomplete dominance.
    Under codominance each trait shows whole and separate, so no scale is in between.
  4. D. ✓ Black scales and yellow scales, each scale one color

Why: Codominance means both alleles are dominant together.
So both alleles’ traits show in the heterozygote, each whole and separate.
Sᴮ gives black scales and Sʸ gives yellow scales.
So an SᴮSʸ lizard has black scales and yellow scales, and no scale is in between.

Q3 P54-q03

A type B child has a type A mother and a type B father.

Which two alleles of the ABO gene does the mother carry?

  1. A. Iᴬ and Iᴬ
    An IᴬIᴬ mother gives every child an Iᴬ, and a type B child carries no Iᴬ.
    So the mother gave an i.
  2. B. ✓ Iᴬ and i
  3. C. Iᴮ and i
    An Iᴮi person is type B, and the mother is type A, so she carries an Iᴬ.
    She gave the child an i, so she is Iᴬi.
  4. D. Iᴬ and Iᴮ
    An IᴬIᴮ person is type AB, and the mother is type A.
    She gave the child an i, so her alleles are Iᴬ and i.

Why: The child is type B: Iᴮ and Iᴮ, or Iᴮ and i, so no Iᴬ.
The mother is type A, so she carries an Iᴬ, yet she gave the child no Iᴬ.
So she gave the child her other allele, an i.
So the mother is Iᴬi.

Q4 P54-q04

In a species of gecko, a dark body (D) is dominant to a pale body (d) and a long tail (L) to a short tail (l). A DdLl gecko is crossed with a ddll gecko. One offspring has a pale body and a long tail.

Which gamete did the DdLl parent give that offspring?

  1. A. DL
    The offspring has a pale body, so it carries no D.
    The DdLl parent gave it a d, not a D.
  2. B. Dl
    The offspring has a pale body and a long tail, so it carries d and L.
    A Dl gamete would give a dark body and a short tail.
  3. C. ✓ dL
  4. D. dl
    The offspring has a long tail, so it carries an L.
    The ddll parent gave only l, so the L came from the DdLl parent.

Why: The ddll parent gave the offspring d and l.
The offspring has a pale body, so its other allele for that gene is d.
It has a long tail, so its other allele for that gene is L.
So the DdLl parent gave it dL.

Q5 P54-q05

In humans, a recessive allele on the X chromosome, written Xᵃ, causes a trait, and Xᴬ is the ordinary allele. A woman who carries the allele, XᴬXᵃ, and a man with the trait, XᵃY, have children. Four squares drawn for this cross are numbered 1 to 4 below, with the woman’s gametes meant to sit along the top edge and the man’s down the left edge.

Four Punnett squares drawn for the cross, numbered 1 to 4.
Four Punnett squares drawn for the cross, numbered 1 to 4.

Which square is drawn correctly?

  1. A. Square 1
    Square 1 gives the man two X sperm and no Y sperm.
    A man is XY, so his edge reads Xᵃ and Y.
  2. B. Square 2
    Square 2 writes an allele on the Y.
    The Y carries no allele of this gene, so it is written bare, and a Y cell reads XᴬY or XᵃY.
  3. C. Square 3
    Square 3 puts Xᴬ and Xᴬ along the top.
    The woman is XᴬXᵃ, so half of her eggs carry Xᴬ and half carry Xᵃ.
  4. D. ✓ Square 4

Why: The woman’s eggs carry Xᴬ or Xᵃ: the top edge.
The man’s sperm carry his Xᵃ or a bare Y: the left edge.
Each cell holds the chromosome above it with the one beside it: XᴬXᵃ, XᵃXᵃ, XᴬY, XᵃY.
Only square 4 has those edges and cells.

Q6 P54-q06

In a species of mammal, a gene sits on the Y chromosome, and one of its alleles gives a patterned coat. A patterned male has two sons and two daughters. One of his daughters later has a son.

Which of these five descendants carry the patterned-coat allele?

  1. A. ✓ The two sons only
  2. B. The two daughters only
    A daughter receives her father’s X, never his Y.
    So no daughter carries an allele on the Y, not even hidden.
  3. C. The two sons and the daughter’s son
    The daughter’s son received his Y from his own father, not from his grandfather.
    His mother has no Y to pass on.
  4. D. All four children and the daughter’s son
    A father gives his Y to every son and to no daughter.
    The daughters have no Y, so they cannot pass one to the grandson.

Why: A father gives his Y to every son and to no daughter.
So both sons carry the allele, and neither daughter does, not even hidden.
The daughter’s son received his Y from his own father.
So only the two sons carry the allele.

Q7 P54-q07

Red-green color blindness comes from a recessive allele on the X chromosome. In a large town the allele is uncommon.

Which of the following describes the people in the town who are color-blind?

  1. A. Men only
    A woman whose two X chromosomes both carry the allele is color-blind.
    Rare is not never: about 1 in 200 women are.
  2. B. ✓ Many men and a few women
  3. C. About as many women as men
    A man shows the allele with one copy, on his only X.
    A woman shows it only with a copy on each of her two X’s, which is far rarer.
  4. D. Many women and a few men
    A woman has a second X, and an ordinary allele on it hides the color-blindness allele.
    A man has no second X, so one copy shows.

Why: A man has one X, so one copy of the allele shows: about 1 in 12 men.
A woman’s second X usually carries an ordinary allele, which hides the first.
She is color-blind only when both X’s carry the allele: about 1 in 200 women.
So: many men, few women.

Q8 P54-q08

A family's records follow a rare disorder across three generations. Every child of an affected woman is affected, sons and daughters alike. Affected men also have sons and daughters, and all of their children are unaffected.

Which explanation fits the records?

  1. A. Both parents pass the disorder on equally through the mitochondria in their gametes
    If both parents passed mitochondria on equally, affected fathers would pass the disorder on too, and none of theirs is affected.
  2. B. The disorder is autosomal dominant, because every affected parent passes it on
    An autosomal dominant allele passes from affected fathers as well as affected mothers, to about half of the children of each.
  3. C. ✓ The disorder is written in mitochondrial DNA, which the egg supplies and the sperm does not
  4. D. The disorder is X-linked recessive, because affected mothers pass an X to every child
    An X-linked recessive allele reaches every daughter of an affected father, yet here affected fathers pass nothing on; and a carrier mother's sons would show it more than her daughters.

Why: Every child of an affected mother is affected and no child of an affected father is.
The egg supplies the zygote's mitochondria; the sperm's do not persist.
So a trait in mitochondrial DNA passes from every affected mother to every child and from no father: exactly this pattern.

Q9 P54-q09

In humans, a recessive allele on the X chromosome, written Xᵈ, causes a trait, and Xᴰ is the ordinary allele. The Punnett square below is for a woman who carries the allele, XᴰXᵈ, and a man whose X carries the ordinary allele, XᴰY.

A Punnett square for XᴰXᵈ × XᴰY: the woman’s gametes along the top edge, the man’s down the left edge.
A Punnett square for XᴰXᵈ × XᴰY: the woman’s gametes along the top edge, the man’s down the left edge.

Which of the couple’s children show the trait?

  1. A. ✓ Half of the sons and none of the daughters
  2. B. Half of the daughters and none of the sons
    The cells with two X’s are the daughters: XᴰXᴰ and XᴰXᵈ.
    Each daughter has an Xᴰ, which hides an Xᵈ, so no daughter shows the trait.
  3. C. Half of the sons and half of the daughters
    Every daughter received her father’s Xᴰ, which hides an Xᵈ from her mother.
    Only a son, with one X, shows one copy.
  4. D. All of the sons and none of the daughters
    The cells with a Y are the sons: XᴰY and XᵈY.
    Only the XᵈY son shows the trait, so half of the sons.

Why: The cells with two X’s are the daughters, XᴰXᴰ and XᴰXᵈ; each Xᴰ hides an Xᵈ.
So no daughter shows the trait.
The cells with a Y are the sons, XᴰY and XᵈY; the XᵈY son’s one X carries the allele.
So half of the sons show it.

Q10 P54-q10

In honeybees, a worker develops from an egg that a sperm fertilized, and a drone, a male bee, develops from an egg that no sperm fertilized. A worker’s body cells each hold 32 chromosomes.

How many chromosomes does each of a drone’s body cells hold?

  1. A. 8
    An egg carries one set, half of the worker’s 32: 16 chromosomes.
    8 would be half of a set.
  2. B. ✓ 16
  3. C. 32
    32 is two sets, a worker’s count.
    No sperm added a second set to the drone’s egg, so the drone has one set.
  4. D. 64
    64 would be four sets.
    An egg carries one set of 16, and no sperm added another.

Why: A worker’s 32 chromosomes are two sets of 16.
An egg is haploid: it carries one set, 16 chromosomes.
A drone develops from an egg that no sperm fertilized, so no second set was added.
So each of a drone’s body cells holds 16 chromosomes.

FRQ 1 P54-frq1 · Conceptual Analysis scaffolded

Suppose a breeder works with a plant in which tall (T) is dominant to dwarf (t) and a hairy stem (H) to a smooth stem (h). The breeder crosses a TtHh plant, bred from a true-breeding tall hairy line and a true-breeding dwarf smooth line, with a tthh plant and counts 600 offspring. The counts are in the table below, with the chi-square table from the formula sheet beneath it. Against a 1 : 1 : 1 : 1 ratio the breeder has already calculated χ² = 311.

Top: the four classes among 600 offspring of a TtHh plant, bred from true-breeding tall hairy and dwarf smooth lines, crossed with a tthh plant. Bottom: the chi-square table from the AP Biology formula sheet.
Top: the four classes among 600 offspring of a TtHh plant, bred from true-breeding tall hairy and dwarf smooth lines, crossed with a tthh plant. Bottom: the chi-square table from the AP Biology formula sheet.

(a) Identify the two parental classes and the two recombinant classes, and justify the choice. (1 pt)

Frame The parental classes are … and …, because …; the recombinant classes are … and …, because …

Hint Which two combinations of alleles did the TtHh plant inherit, one from each of its true-breeding parents?

Model answer The parental classes are tall hairy (261) and dwarf smooth (255), because the TtHh plant inherited T with H from one true-breeding parent and t with h from the other, so those are the combinations its chromosomes carried; the recombinant classes are tall smooth (44) and dwarf hairy (40), because those combinations were on neither of its chromosomes and are the two small classes.
Rubric
  • Award 1 point for: tall hairy and dwarf smooth as parental (the combinations the parent's chromosomes carried, the two large classes); tall smooth and dwarf hairy as recombinant (new combinations, the two small classes).

Slip Calling the two large classes the recombinants.

(b) Calculate the number of recombinant offspring. (1 pt)

Frame Recombinant offspring = … + … = …

Hint Add the counts of the two recombinant classes you named in part (a).

Answer: 84  (tolerance ±0)

Model answer Recombinant offspring = 44 + 40 = 84.
Working
Add the two recombinant classes:
recombinant offspring = 44 + 40 = 84
Rubric
  • Award 1 point for: 84 recombinants.

(c) Calculate the recombination frequency between the two genes. (1 pt)

Frame Recombination frequency = recombinant offspring ÷ total offspring × 100 % = …

Hint Which count goes on top of the fraction, and which total goes underneath, before you multiply by 100?

Answer: 14 %  (tolerance ±0.05)

Model answer Recombination frequency = 84 ÷ 600 × 100 % = 14.0 %.
Working
Write down the values in the question:
recombinant offspring = 84
total offspring = 600
Write down the equation:
tex:\text{recombination frequency} = \frac{\text{recombinant offspring}}{\text{total offspring}} \times 100\,\%
Substitute the values into the equation:
tex:\text{recombination frequency} = \frac{84}{600} \times 100\,\% = 14.0\,\%
Rubric
  • Award 1 point for: 14.0 % (accept 14 %).

(d) State the map distance between the two genes, with its unit. (1 pt)

Frame The two genes are … apart.

Hint How many map units is one percent of recombination?

Answer: 14 map units  (tolerance ±0.05)

Model answer The two genes are 14.0 map units apart, because one percent of recombination is one map unit.
Working
Convert the recombination frequency to a map distance:
1 % recombination = 1 map unit
map distance = 14.0 map units
Rubric
  • Award 1 point for: 14.0 map units (accept 14 map units), with the unit.

(e) Identify the degrees of freedom and the critical value at p = 0.05, and determine whether the null hypothesis of independent assortment is rejected and what that says about the two genes. (1 pt)

Frame There are … classes, so … degrees of freedom; the critical value is …; χ² = 311 is … than it, so the null hypothesis is …, and the two genes …

Hint How many classes are there, which row and column of the table do you read, and which is larger, the calculated value or the critical value?

Model answer There are four classes, so 3 degrees of freedom.
The critical value at p = 0.05 is 7.81.
χ² = 311 is far larger than 7.81, so the null hypothesis is rejected.
So the counts do not fit 1 : 1 : 1 : 1, and the two genes are not assorting independently.
Therefore the two genes are linked.
Rubric
  • Award 1 point for: 3 degrees of freedom and a critical value of 7.81, AND the decision (reject the null hypothesis) with its ground (311 is larger than 7.81), so the genes do not assort independently (not ‘accept’ or ‘prove’).

Slip Reading the column for four degrees of freedom. Degrees of freedom are the number of classes minus one.

FRQ 2 P54-frq2 · Scientific Investigation

A condition caused by one gene is recorded in one family in the pedigree below; a filled shape shows the condition. II-4 married into the family.

A family pedigree for a condition across three generations. Squares are males, circles females; a filled shape shows the condition. II-4 married into the family.
A family pedigree for a condition across three generations. Squares are males, circles females; a filled shape shows the condition. II-4 married into the family.

(a) Make a claim about the mode of inheritance of the condition. (1 pt)

Model answer Claim: the condition is autosomal dominant.
Rubric
  • Award 1 point for: the claim, autosomal dominant; no reasoning is required for this point.
  • Accept: dominant, on an ordinary pair of chromosomes.
  • Accept as extra reasoning, if given: II-2, an unaffected daughter of two affected parents, rules out X-linked recessive and non-nuclear inheritance; as the daughter of the affected I-1, who gave her his one X, she rules out X-linked dominant too.

Slip Writing recessive because the condition appears in every generation. Appearing in every generation is a hint, not a proof; the decisive family decides. Writing X-linked or maternal: II-2 is an unaffected daughter of two affected parents, which neither an X-linked recessive allele nor mitochondrial DNA can produce, and her affected father I-1 gave her his one X, so an X-linked dominant allele would have made her affected as well; the allele is on an ordinary pair.

(b) Support the dominant-or-recessive part of your claim with evidence from one family in the pedigree. (1 pt)

Model answer Evidence: I-1 and I-2 both have the condition, yet their daughter II-2 does not.
If the allele were recessive, both affected parents would be homozygous for it.
Then every child would receive the allele from both parents and be affected.
An unaffected child of two affected parents is possible only if the allele is dominant and both parents are heterozygous.
So this family supports the claim that the allele is dominant.
Rubric
  • Award 1 point for: the evidence (two affected parents, I-1 and I-2, have an unaffected child, II-2) AND the reasoning that links it to the claim (a recessive allele in two affected parents would reach every child, so the allele must be dominant).

Slip Giving the evidence without the link, or the reasoning without naming the family. Support a claim needs both.

(c) Using A for the dominant allele and a for the recessive allele, identify the genotypes of I-1, I-2 and II-2, and explain how the pedigree fixes them. (1 pt)

Model answer II-2 does not have the condition, so she carries no A: she is aa, and she received an a from each parent.
I-1 and I-2 have the condition, so each carries an A, and each also passed an a to II-2, so both are Aa.
Rubric
  • Award 1 point for: II-2 aa; I-1 Aa and I-2 Aa, because each shows the condition (carries A) and each passed an a to their unaffected daughter.

Slip Writing I-1 and I-2 as AA. An AA parent has no a to pass on, so an aa child rules AA out.

(d) Calculate the probability that the next child of II-3 and II-4 has the condition. (1 pt)

Answer: 0.5  (tolerance ±0.005)

Model answer The probability is 12, which is 0.5.
II-3 has the condition and has an unaffected daughter, III-2, so she is Aa; II-4 is unaffected, so he is aa.
Half of II-3's gametes carry A, and every child who receives it has the condition.
Working
Write down the values in the question:
II-3 is Aa, II-4 is aa
tex:P(A \text{ from II-3}) = \frac{1}{2}
tex:P(a \text{ from II-4}) = 1
Write down the equation:
tex:P(A \text{ and } B) = P(A) \times P(B)
Substitute the values into the equation:
tex:P(Aa) = \frac{1}{2} \times 1 = \frac{1}{2} = 0.5
Rubric
  • Award 1 point for: 12 (0.5), from II-3 Aa and II-4 aa.

APBIO-U05-T54 End-of-topic test: Non-Mendelian Genetics

Topic 5.4 · Non-Mendelian Genetics · 19 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Show any calculation, then open the scoring guide and mark your own work against it. Where a question needs a critical value, the chi-square table is drawn with it; read the p = 0.05 row.

Q1 T54-q01

Suppose that in one kind of plant a true-breeding red-flowered line is crossed with a true-breeding white-flowered line, and every F1 plant flowers pink. Two F1 plants are crossed, and their 240 F2 offspring are 61 red, 123 pink and 56 white.

Which statement about the alleles in the pink F1 plants is correct?

  1. A. The red allele hid the white allele
    If the red allele hid the white allele, every F1 plant would be red, not pink.
    The F1 plants are pink.
    So neither allele hid the other.
  2. B. ✓ The two alleles stayed separate, and neither hid the other
  3. C. Pink is a third allele, dominant to red and white
    A dominant pink allele would give pink in 75% of the F2; here red and white each came back in about 25%, so pink is the heterozygote's look.
  4. D. The red and white alleles blended into one pink allele
    Two alleles blended into one pink allele could give nothing but pink; red and white came back in the F2, about 25% each, so the two alleles stayed separate.

Why: The pink F1 shows that neither allele hides the other, and the red and white F2 plants show that the alleles were carried unchanged inside the pink plants: incomplete dominance.

Q2 T54-q02

In carnations, red (Cᴿ) and white (Cᵂ) show incomplete dominance, and a CᴿCᵂ plant is pink. Two pink carnations are crossed and 360 offspring are grown to flowering.

How many of the 360 offspring are expected to be pink?

  1. A. 90
    90 is one share of the four, the count expected for red or for white; pink is the heterozygote, which takes two shares of the four.
  2. B. ✓ 180
  3. C. 270
    270 is 75% of 360, the dominant share of a 3 : 1 ratio; here the heterozygote shows its own color, pink, two shares of four.
  4. D. 360
    360 is the total; only the CᴿCᵂ offspring are pink, two cells of the four.

Why: The CᴿCᵂ × CᴿCᵂ square gives 1 CᴿCᴿ : 2 CᴿCᵂ : 1 CᵂCᵂ, and only the heterozygotes are pink, two shares of four: 360×24=180.

Q3 T54-q03

A man is type A, and his mother was type O. He has children with a woman who is type O.

Which blood types are expected among their children, and in what shares?

  1. A. All type A
    The man's type O mother gave him an i, so he is Iᴬi; a child that receives his i and its mother's i is ii, type O.
  2. B. All type O
    The man carries an Iᴬ too, and half of his gametes carry it; a child that receives it is Iᴬi, type A, because i is recessive to Iᴬ.
  3. C. About 75% type A and 25% type O
    3 : 1 needs two heterozygotes; here only the man is heterozygous, Iᴬi, and the woman is ii, so the square's cells are Iᴬi and ii in equal shares.
  4. D. ✓ About half type A and half type O

Why: The man's type O mother gave him an i, so he is Iᴬi; the woman is ii.
Half of his gametes carry Iᴬ and half carry i, and every one of hers carries i, so the children are Iᴬi (type A) and ii (type O) in equal shares.

Q4 T54-q04

A female mouse's cells hold a mixture of ordinary mitochondria and faulty ones that lack one electron-transport protein. Her offspring carry very different shares of the faulty mitochondria, with no 3 : 1 and no 1 : 1 among them.

Where does the gene for that protein sit, and why do the offspring carry such different shares?

  1. A. On an autosome; the two alleles part at anaphase I, one per egg
    A gene on an autosome follows the nuclear rules and gives Mendelian ratios, and this trait gives none.
  2. B. ✓ In the mitochondrion's own DNA; the mitochondria fall into each egg at random
  3. C. In the mitochondrion's own DNA; each egg copies the mother's mixture exactly
    An egg does not copy the mother's mixture; when a cell divides, its mitochondria fall into the two halves however they lie, so each egg gets a different share.
  4. D. On the X chromosome; a son has one X and a daughter has two
    A gene on the X is on a chromosome in the nucleus and gives sons and daughters fixed proportions; this trait gives no fixed ratio at all.

Why: Mitochondria carry their own small DNA, so a gene there is inherited outside the nuclear rules.
Chromosomes are counted into gametes one per pair; mitochondria are not counted at all.
They fall into each egg however they lie, so the offspring carry a range of shares, not a Mendelian ratio.

Q5 T54-q05

In a species of cricket, a black body is dominant to a brown body, and long wings are dominant to short wings. A breeder crosses two F1 crickets from a true-breeding black long-winged × true-breeding brown short-winged cross. Their 1,000 offspring are 610 black long-winged, 140 black short-winged, 140 brown long-winged and 110 brown short-winged.

Which explanation best accounts for the departure from the 9 : 3 : 3 : 1 that two genes on different chromosomes would give?

  1. A. ✓ The two genes sit on one chromosome, so the grandparents’ two combinations are in excess
  2. B. The two genes sit on different chromosomes, and chance alone pushed the grandparents’ two combinations above their shares
    Independent assortment predicts about 563, 188, 188 and 63; both parental classes sit far above and both recombinant classes far below, a pattern chance does not give.
  3. C. Complete dominance lets the dominant alleles enter gametes together more often than the recessive ones
    Dominance decides which trait shows in a heterozygote, not which alleles a gamete receives; every gamete gets one allele of each gene whatever its dominance.
  4. D. The two genes sit on different chromosomes, but crossing over made the parental combinations more common
    Crossing over happens between genes on one chromosome, and it makes the recombinant combinations, not the parental ones; genes on different chromosomes give four gamete types in equal shares.

Why: Black with long wings and brown with short wings are the combinations the true-breeding grandparents carried.
They are in excess, and the two new combinations are scarce, because the two genes sit on one chromosome and pass into a gamete together unless a crossover falls between them.

Q6 T54-q06

In a species of beetle, black shell (S) is dominant to brown (s) and spotted (P) to plain (p). An SsPp beetle whose chromosomes carried S with P and s with p is crossed with an sspp beetle. The offspring are counted in the table below.

The four classes of offspring of an SsPp beetle (S with P on one chromosome, s with p on the other) crossed with an sspp beetle.
The four classes of offspring of an SsPp beetle (S with P on one chromosome, s with p on the other) crossed with an sspp beetle.

Which two classes are the recombinants?

  1. A. Black spotted and brown plain
    Black spotted and brown plain are the two large classes, the combinations the heterozygous parent's chromosomes carried: the parental combinations.
  2. B. Black spotted and black plain
    Black spotted is a parental combination, S with P as the parent's chromosome carried them; only black plain is a new combination.
  3. C. Brown plain and brown spotted
    Brown plain is a parental combination, s with p as the parent's chromosome carried them; only brown spotted is a new combination.
  4. D. ✓ Black plain and brown spotted

Why: The heterozygous parent's chromosomes carried S with P and s with p, so black spotted and brown plain are the parental combinations, the two large classes.
Black plain (S with p) and brown spotted (s with P) are combinations neither chromosome had.
So they are the recombinants, from a crossover.

Q7 T54-q07

In a species of snail, a banded shell (B) is dominant to plain (b) and a dark shell (D) to pale (d). A BbDd snail whose chromosomes carried B with d and b with D is crossed with a bbdd snail. The offspring are counted in the table below.

The four classes of offspring of a BbDd snail (B with d on one chromosome, b with D on the other) crossed with a bbdd snail.
The four classes of offspring of a BbDd snail (B with d on one chromosome, b with D on the other) crossed with a bbdd snail.

What is the map distance between the banding gene and the shell-color gene?

  1. A. 0.12 map units
    0.12 is the recombinant share of the offspring before it is multiplied by 100; a recombination frequency is a percentage.
  2. B. ✓ 12.0 map units
  3. C. 13.6 map units
    13.6 comes from dividing the 48 recombinants by the 352 parental offspring; the frequency is recombinants over all 400 offspring.
  4. D. 88.0 map units
    88.0 % is the share of parental offspring; the map distance is the recombinant share, 48 of 400.

Why: The parent's chromosomes carried B with d and b with D, so banded pale and plain dark are the parental classes.
Banded dark and plain pale are the recombinants: 48 of 400.
48400×100=12.0%, and 1% recombination is one map unit: 12.0 map units.

Q8 T54-q08

On one chromosome of a plant, genes A and B sit 30 map units apart. A plant carries A with B on one homolog and a with b on the other.

In what share of its gametes is the A allele separated from the B allele, and why?

  1. A. 0 %, because two genes on one chromosome always travel together
    Linked genes travel together only when no crossover falls between them, and a crossover can fall anywhere along the chromosome.
  2. B. 15 %, because a crossover separates the two only half of the time it falls between them
    The map distance already is the share of gametes in which the two are separated; nothing is halved.
  3. C. ✓ 30 %, because a crossover falls between two genes more often the farther apart they sit
  4. D. 60 %, because a crossover between the genes separates them in both chromatids
    30 map units means 30 % recombinant gametes; the distance is not doubled.

Why: A map unit is one percent of recombination, so 30 map units means the A allele is separated from the B allele in 30 % of gametes; a crossover can fall anywhere, and the longer the stretch between two genes, the more often one falls between them.

Q9 T54-q09

In a plant, three genes sit in the order L, H, S along one chromosome. L to H is 9 map units and H to S is 13 map units.

What is the map distance between L and S?

  1. A. ✓ 22 map units
  2. B. 13 map units
    13 map units is the distance from H to S alone; L lies beyond H, so L to S includes the stretch from L to H as well.
  3. C. 9 map units
    9 map units is the distance from L to H alone; S lies beyond H, so L to S includes the stretch from H to S as well.
  4. D. 4 map units
    4 is 13 minus 9. Subtracting fits a gene that sits between the other two.
    H sits between L and S, so the two distances add.

Why: H sits between L and S.
So the distance from L to S is the distance from L to H plus the distance from H to S.
The working below gives 22 map units.

Q10 T54-q10

A breeder test-crosses a plant heterozygous for two genes and sorts the 1,000 offspring into four classes. Against the 1 : 1 : 1 : 1 that independent assortment predicts, chi-square comes out at 9.20. The table below is the formula sheet's.

The chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.
The chi-square table from the AP Biology formula sheet: critical values for 1 to 8 degrees of freedom at p = 0.05 and p = 0.01.

What is the verdict, and what does it say about the two genes?

  1. A. Fail to reject the null hypothesis, because 9.20 is smaller than 9.49; the genes assort independently
    Four classes give three degrees of freedom, and the 0.05 row for three degrees of freedom reads 7.81, not 9.49.
  2. B. Accept the null hypothesis: the counts prove that the genes sit on different chromosomes
    A chi-square verdict is reject or fail to reject, never accept or prove; and 9.20 is above 7.81, so the counts do not fit.
  3. C. ✓ Reject the null hypothesis, because 9.20 is larger than 7.81; the genes do not assort independently
  4. D. Reject the null hypothesis, and the two genes therefore sit 9.20 map units apart
    A chi-square value is not a distance; the map distance comes from the recombinant share of the offspring, a separate calculation.

Why: Four classes give three degrees of freedom; the working is below.
The 0.05 row reads 7.81.
Chi-square, 9.20, is larger than 7.81, so the null hypothesis is rejected.
So the counts do not fit 1 : 1 : 1 : 1: the genes are not assorting independently.

Q11 T54-q11

In fruit flies, forked bristles are recessive to the ordinary straight bristles. A breeder crosses true-breeding straight-bristled females with forked-bristled males (cross 1), and true-breeding forked-bristled females with straight-bristled males (cross 2). The offspring of each cross are recorded by sex in the table below.

Bristle shape of the daughters and the sons of two crosses between true-breeding fruit-fly lines.
Bristle shape of the daughters and the sons of two crosses between true-breeding fruit-fly lines.

Which conclusion do the two crosses support?

  1. A. ✓ The gene is on the X chromosome: the trait is X-linked
  2. B. The gene is on the Y chromosome: the trait is Y-linked
    A Y-linked allele passes from a father to every son.
    In cross 2 the fathers have straight bristles, yet every son is forked, so the allele came from the mother.
  3. C. The gene is on an autosome, and forked bristles are recessive
    For a gene on an autosome the reciprocal cross gives the same offspring as the first cross.
    Here cross 2 gives forked sons and cross 1 gives none.
  4. D. The gene is in mitochondrial DNA: the trait shows non-nuclear inheritance
    A trait in mitochondrial DNA passes from the mother to every child.
    In cross 2 the daughters have straight bristles, so the trait did not follow the mother.

Why: In cross 2 every son is forked like his mother, and every daughter is straight like her father.
A son's only X comes from his mother, and his Y carries no allele of the gene.
So the allele the sons show sits on the X: the trait is X-linked.

Q12 T54-q12

A woman carries an allele on one of her two X chromosomes. She has two sons and two daughters.

Which of the four children could have received that allele from her?

  1. A. Only the two daughters, each with probability one
    A mother gives one of her two Xs to every child, sons included; the Y comes from the father, so a son can receive the X with the allele too.
  2. B. Only the two sons, each with probability one
    A daughter also receives one X from her mother.
    It is the father's X that goes only to daughters.
  3. C. ✓ Any of the four, each with probability one half
  4. D. All four, each with probability one
    Every child receives an X from her, but her two Xs part at anaphase I, so only half of her eggs carry the X with the allele.

Why: A mother gives an X to every child, son or daughter, and her two X chromosomes part at anaphase I, so half of her eggs carry the X with the allele: any child can receive it, each with probability one half.

Q13 T54-q13

Red-green color blindness comes from a recessive allele on the X chromosome. A man is color-blind. His wife's two X chromosomes both carry the ordinary allele. They have a daughter.

What does the daughter see, and why?

  1. A. She is color-blind, because a daughter always receives her father's X and the allele on it
    She does receive her father's X, but she also receives an X from her mother carrying the ordinary allele, which is dominant and hides the color-blindness allele.
  2. B. ✓ She sees colors as most people do: her mother's X carries the ordinary allele, which hides her father's
  3. C. She sees colors as most people do, because a daughter receives no X from her father
    A father gives his X to every daughter, so she does carry his color-blindness allele; it is hidden by the ordinary allele on her other X.
  4. D. She is color-blind, because the allele shows in anyone who carries one copy
    One copy shows only when there is no second X: in an XY individual.
    She has two X chromosomes and the ordinary allele on one of them hides the other.

Why: The daughter receives her father's X, carrying the color-blindness allele, and her mother's X, carrying the ordinary allele; the ordinary allele is dominant and hides it, so she sees colors as most people do and carries the allele.

Q14 T54-q14

In humans, a recessive allele h on the X chromosome causes a bleeding disorder called hemophilia. A woman who carries the allele, XᴴXʰ, and a man with the disorder, XʰY, have a child.

What is the probability that the child is a son with the disorder?

  1. A. ✓ 14
  2. B. 12
    12 is the probability that the child is a son, or the probability that a son has the disorder; the question asks for both at once.
  3. C. 34
    34 is the complement of 14, the probability of every other outcome together.
  4. D. 1
    1 comes from adding 12 and 12.
    Being a son and receiving the Xʰ from the mother must both happen in one child, so the two probabilities are multiplied.

Why: A son receives the Y from his father and one X from his mother; half of the children are sons, and half of the mother's eggs carry Xʰ, so a son with the disorder is 12×12=14.

Q15 T54-q15

In birds, a female is ZW and a male is ZZ. A hen and a rooster have a female chick.

Which sex chromosome did the chick receive from the hen?

  1. A. Z
    A female chick is ZW; the rooster is ZZ, so every sperm carries a Z and the chick's Z came from him, so the hen gave the W.
  2. B. Z and W
    An egg carries one sex chromosome, not two.
    The hen is ZW, and her two sex chromosomes part at anaphase I, so each egg carries a Z or a W.
  3. C. Either Z or W
    Either is possible before the chick's sex is known.
    This chick is female, ZW.
    The rooster can give only a Z. So the hen must have given the W.
  4. D. ✓ W

Why: A female chick is ZW.
The ZZ rooster can give only a Z, so the chick's Z came in the sperm.
The W must have come in the egg: the hen gave the W.
In birds the egg decides the chick's sex.

Q16 T54-q16

A child inherits the sickle-cell allele from both parents. The child has anemia, episodes of pain and organ damage. A student says: “Three problems means three faulty genes.”

Which statement about the student’s claim is correct?

  1. A. The student is right: each of the three problems comes from a different gene, inherited separately
    Every child with two sickle-cell alleles has all three problems together.
    Three separate genes would let the three problems be inherited apart.
  2. B. The student is right: the anemia and the pain are separate traits, so a child can inherit one alone
    The anemia and the pain both come from rigid red cells.
    A child who inherits the allele inherits the rigid cells, and with them every problem they cause.
  3. C. The student is wrong: the three problems come from three different alleles of the one hemoglobin gene
    The child carries two copies of one allele, not three alleles.
    One allele changes one hemoglobin, and that one change causes all three problems.
  4. D. ✓ The student is wrong: one changed protein makes the red cells rigid, and rigid cells cause all three

Why: The sickle-cell allele changes one amino acid in hemoglobin.
The changed hemoglobin forms stiff fibers, and the fibers make the red blood cells rigid.
Rigid cells break down (anemia), block small vessels (pain) and damage organs.
So one gene causes all three problems, and the student is wrong.

Q17 T54-q17

A rare condition is recorded in the family below. I-1, II-2, II-4 and III-10 married into the family.

A family pedigree for a rare condition across four generations. Squares are males, circles females; a filled shape shows the condition. I-1, II-2, II-4 and III-10 married into the family.
A family pedigree for a rare condition across four generations. Squares are males, circles females; a filled shape shows the condition. I-1, II-2, II-4 and III-10 married into the family.

Which mode of inheritance best fits this pedigree?

  1. A. Autosomal dominant
    A dominant allele passes to half of an affected parent's children; here affected mothers passed it to all 9 of theirs and affected father II-3 to none of his 5.
  2. B. ✓ Non-nuclear inheritance
  3. C. X-linked recessive
    An affected X-linked daughter needs the allele from her father, yet II-1, III-2 and III-3 are affected daughters of unaffected fathers, and II-3's daughters III-6 and III-8 are unaffected.
  4. D. Autosomal recessive
    The married-in spouses I-1, II-2 and III-10 would all have to carry the rare recessive allele, and even then only about half of the children would show it.

Why: I-2, II-1 and III-2, affected women, have 9 children, all affected.
II-3, an affected man, has 5 children, none affected.
The egg supplies the zygote's mitochondria and the sperm does not.
So a mitochondrial-DNA trait passes from every affected mother to every child and from no father, exactly this pattern.

Q18 T54-q18

In four o’clock plants, one variegated plant carries green branches and white branches. A grower crosses a flower on a green branch, supplying the ovules, with pollen from a flower on a white branch. All 120 seedlings are green. The grower then makes the reciprocal cross: a flower on a white branch supplies the ovules, and the pollen comes from a flower on a green branch.

What leaf color will the seedlings of the reciprocal cross have, and why?

  1. A. Green, because a nuclear allele for green leaves is dominant, so the direction of the cross does not matter
    Both branches grow on one plant, so every cell of the plant carries the same nuclear alleles.
    A nuclear allele cannot make one branch green and another white.
  2. B. Green, because the chloroplasts come through the pollen, and the pollen came from a green branch
    In the first cross the pollen came from a white branch, yet every seedling was green.
    So the pollen did not supply the seedlings’ chloroplasts.
  3. C. Patchy green and white, because the ovule and the pollen each bring their own chloroplasts
    If the ovule and the pollen both brought chloroplasts, the first cross would have given patchy seedlings.
    Every one of its 120 seedlings was green.
  4. D. ✓ White, because the chloroplasts come through the ovule, and the ovule came from a white branch

Why: In the first cross every seedling matched the branch that supplied the ovules, not the pollen.
So the seedlings’ chloroplasts, and the chloroplast DNA, come through the ovule.
The reciprocal cross takes its ovules from a white branch.
So its seedlings are white, whatever the pollen brings.

Q19 T54-q19

A trait is recorded in the family below. II-1 married into the family.

A family pedigree for a trait across three generations. Squares are males, circles females; a filled shape shows the trait. II-1 married into the family.
A family pedigree for a trait across three generations. Squares are males, circles females; a filled shape shows the trait. II-1 married into the family.

The trait is X-linked recessive. Which unaffected individual in this pedigree must carry the allele?

  1. A. I-2
    II-2 also received an X from the affected I-1, whose X carries the allele; nothing in the pedigree requires I-2's X to carry it.
  2. B. II-1
    II-1 married into the family and is unaffected.
    A man has one X, so an unaffected man's X carries the ordinary allele; he cannot carry the recessive allele unseen.
  3. C. ✓ II-2
  4. D. III-2
    III-2 is a daughter of II-2; she received one of her mother's two Xs, so she has a one-half chance of carrying the allele; the pedigree does not fix it.

Why: A father gives his one X to every daughter, so II-2 received the affected I-1’s X with the allele.
Her other X carries the ordinary allele, so she is unaffected.
Her son III-1 received that X and shows the trait.
So II-2 carries the allele unseen.

FRQ 1 T54-frq1 · Scientific Investigation

A condition caused by one gene is recorded in one family in the pedigree below; a filled shape shows the condition. II-5 married into the family.

A family pedigree for a condition across three generations. Squares are males, circles females; a filled shape shows the condition. II-5 married into the family.
A family pedigree for a condition across three generations. Squares are males, circles females; a filled shape shows the condition. II-5 married into the family.

(a) Make a claim about the mode of inheritance of the condition. (1 pt)

Model answer Claim: the condition is autosomal recessive.
Rubric
  • Award 1 point for: the claim, autosomal recessive; no reasoning is required for this point.

Slip Writing X-linked recessive because II-3 is an affected son of unaffected parents. That family fits X-linked recessive, but II-2, an affected daughter of an unaffected father, rules it out (see part c).

(b) Support the dominant-or-recessive part of your claim with evidence from one family in the pedigree. (1 pt)

Model answer Evidence: I-1 and I-2 do not have the condition, yet their daughter II-2 and their son II-3 do.
A dominant allele shows in everyone who carries it.
So two unaffected parents cannot carry a dominant allele, and cannot pass one on.
Two unaffected parents with an affected child are therefore possible only if both carry a recessive allele without showing it.
So this family supports the claim that the allele is recessive.
Rubric
  • Award 1 point for: the evidence (two unaffected parents, I-1 and I-2, have an affected child, II-2 or II-3) AND the reasoning that links it to the claim (a dominant allele would show in its carriers, so the allele must be recessive).

Slip Arguing from the condition skipping a generation. That is a hint, not a proof.

(c) Support the autosome-or-X part of your claim with evidence from one individual in the pedigree, and name that individual. (1 pt)

Model answer Evidence: II-2 is an affected daughter of an unaffected father, I-1.
If the recessive allele were on the X, an affected daughter would need an X carrying it from each parent.
Her father has one X.
If his X carried the allele, he would be affected himself.
He is unaffected, so his X carries the ordinary allele.
So I-1 cannot have given II-2 an X-linked recessive allele, and the gene is on an autosome.
Rubric
  • Award 1 point for: the evidence (II-2, an affected daughter of an unaffected father) AND the reasoning (she would need an X with the allele from her father, who would then show the condition himself), so the gene is on an autosome. II-2 alone earns the point; no second individual is required.
  • Do not award: III-1. Her father II-3 is affected, so an affected daughter there fits X-linked recessive as well; she decides nothing.

Slip Pointing at II-3, an affected son. An affected son of unaffected parents fits X-linked recessive as well as autosomal recessive; only an affected daughter of an unaffected father decides.

(d) Using A for the dominant allele and a for the recessive allele, identify the genotypes of II-3 and II-5, and explain how the pedigree fixes them. (1 pt)

Model answer II-3 has the condition, so he is aa.
II-5 does not have the condition, so she carries at least one A.
Their daughter III-1 has the condition, so she is aa and received an a from each parent.
So II-5 gave her an a, and II-5 carries an a as well as an A: II-5 is Aa, a carrier.
Rubric
  • Award 1 point for: II-3 aa (he shows the condition) and II-5 Aa (she is unaffected, so she has an A, and her affected daughter III-1 received an a from her).

Slip Writing II-5 as AA because she married in. A married-in parent of an affected child must carry the allele too.

(e) Calculate the probability that the next child of II-3 and II-5 has the condition. (1 pt)

Answer: 0.5  (tolerance ±0.005)

Model answer The probability is 12, which is 0.5.
Working
Write down the values in the question:
tex:P(a \text{ from II-3}) = 1
tex:P(a \text{ from II-5}) = \frac{1}{2}
Write down the equation:
tex:P(A \text{ and } B) = P(A) \times P(B)
Substitute the values into the equation:
tex:P(aa) = 1 \times \frac{1}{2} = \frac{1}{2} = 0.5
Rubric
  • Award 1 point for: 12 (0.5), from II-3 aa and II-5 Aa.
FRQ 2 T54-frq2 · Conceptual Analysis

In a species of moth, dark wings (D) are dominant to pale (d) and long antennae (L) to short (l). A breeder crosses a DdLl moth, bred from a true-breeding dark long line and a true-breeding pale short line, with a ddll moth and counts 800 offspring. The counts are in the table below, with the chi-square table from the formula sheet beneath it.

Top: the four classes among 800 offspring of a DdLl moth, bred from true-breeding dark long and pale short lines, crossed with a ddll moth. Bottom: the chi-square table from the AP Biology formula sheet.
Top: the four classes among 800 offspring of a DdLl moth, bred from true-breeding dark long and pale short lines, crossed with a ddll moth. Bottom: the chi-square table from the AP Biology formula sheet.

(a) Evaluate the claim that these two genes assort independently, using the four classes. (1 pt)

Model answer If the genes assorted independently, the four classes would be about 200 each.
Instead the two parental combinations, dark long (342) and pale short (338), are far above 200.
The two new combinations, dark short (62) and pale long (58), are far below 200.
So the claim is not supported: the two genes are linked.
Rubric
  • Award 1 point for: the judgement (the claim is not supported: the two genes are linked) AND the ground (the parental classes, dark long and pale short, are far above the 200 each that independent assortment predicts, and the recombinant classes far below).
  • Accept: ‘the two genes sit on one chromosome’ or ‘the genes do not assort independently’ for the judgement.

Slip Calling the two large classes the recombinants. The large classes carry the combinations the parent's chromosomes had.

(b) If the genes are linked, calculate the recombination frequency between the two genes, and state the map distance between them with its unit. (1 pt)

Answer: 15 %  (tolerance ±0.05)

Model answer The recombination frequency is 15.0 %, so the genes are 15.0 map units apart.
Working
Write down the values in the question:
recombinant offspring = 62 + 58 = 120
total offspring = 800
Write down the equation:
tex:\text{recombination frequency} = \frac{\text{recombinant offspring}}{\text{total offspring}} \times 100\,\%
Substitute the values into the equation:
tex:\text{recombination frequency} = \frac{120}{800} \times 100\,\% = 15.0\,\%
map distance = 15.0 map units
Rubric
  • Award 1 point for: 15.0 % recombination and a map distance of 15.0 map units (accept 15 map units).

(c) Calculate the chi-square value for the four counts against the 1 : 1 : 1 : 1 ratio that independent assortment predicts. (1 pt)

Answer: 392  (tolerance ±1)

Model answer χ² = 392.
Working
Write down the values in the question:
o = 342, 338, 62, 58
tex:e = 800 \times \frac{1}{4} = 200 \text{ for each class}
Write down the equation:
tex:\chi^2 = \sum \frac{(o - e)^2}{e}
Substitute the values into the equation, one class per line:
tex:\frac{(342 - 200)^2}{200} = \frac{20164}{200} = 100.82
tex:\frac{(338 - 200)^2}{200} = \frac{19044}{200} = 95.22
tex:\frac{(62 - 200)^2}{200} = \frac{19044}{200} = 95.22
tex:\frac{(58 - 200)^2}{200} = \frac{20164}{200} = 100.82
tex:\chi^2 = 100.82 + 95.22 + 95.22 + 100.82 = 392.08 \approx 392
Rubric
  • Award 1 point for: χ² = 392 (accept 391 to 393), with each class divided by its own expected count of 200.

(d) Identify the degrees of freedom and the critical value at p = 0.05, and determine whether the null hypothesis of independent assortment is rejected. (1 pt)

Model answer Four classes give 3 degrees of freedom.
The critical value at p = 0.05 is 7.81.
χ² = 392 is far larger than 7.81, so the null hypothesis is rejected.
So the counts do not fit the 1 : 1 : 1 : 1 ratio that independent assortment predicts.
Rubric
  • Award 1 point for: 3 degrees of freedom and a critical value of 7.81, AND the decision (reject the null hypothesis) with its ground (392 is larger than 7.81).
  • Do not award: ‘accept’ or ‘prove’, or a verdict that names a distance. The linkage conclusion is scored in part (a), not here.

Slip Taking the degrees of freedom as 800 or as 4. Degrees of freedom count the classes minus one.

(e) Explain how the 62 dark short and 58 pale long moths arose in the heterozygous parent's meiosis. (1 pt)

Model answer In prophase I the homologs were paired.
A chromatid of one homolog and a chromatid of the other broke at the same point between the two genes and exchanged pieces.
So those two chromatids carried D with l and d with L, combinations neither homolog had.
Gametes carrying those recombinant chromatids fused with the ddll parent's gametes and gave the dark short and pale long moths.
A crossover falls between the two genes in a minority of meioses, so the two classes are small.
Rubric
  • Award 1 point for: a crossover between the two genes in prophase I exchanged pieces between a chromatid of each homolog, giving chromatids with D and l, or d and L, that neither homolog had; gametes carrying them gave the recombinant classes.

Slip Saying the recombinants come from independent orientation at metaphase I. Independent orientation shuffles whole chromosomes; two genes on one chromosome are separated only by a crossover.

APBIO-U05-L29 Same plant, two pots

Topic 5.5 · Environmental Effects on Phenotype · 110 steps

Two potted hydrangea plants side by side: the left one with a head of blue flowers, labeled blue flowers, acid soil; the right one with a head of pink flowers, labeled pink flowers, alkaline soil; a dashed line joins the two pots with the words same DNA
Two potted hydrangea plants side by side: the left one with a head of blue flowers, labeled blue flowers, acid soil; the right one with a head of pink flowers, labeled pink flowers, alkaline soil; a dashed line joins the two pots with the words same DNA

Here are two hydrangea plants grown from cuttings of one parent plant, so their DNA is the same, cell for cell.

One pot has blue flowers. The other has pink. Why do the two plants differ?

Unit 5 · Heredity

1Same genes, different looks

2

Video: Watch: Same genes, different looks

The two pots with a dashed same-DNA line between them; the rabbit with its cool patch and its warm patch; the twins; the one condition that differs named under each.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L29a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L29a.mp4

3

How can two plants with the same DNA look different?

4

The two pots differ in one condition: the soil. Acid soil gives blue hydrangea flowers, and alkaline soil gives pink.

5

The genes did not change. What changed is what reached the petal cells from the soil.

6

Move a plant, and the color follows the soil. The plant's offspring inherit the genes, not the color.

7
Check q1

A hydrangea plant has blue flowers.

Which of the following is part of the plant's phenotype?

  1. A. The plant's two alleles for flower color
    The two alleles are the plant's genotype.
    The phenotype is what the plant shows: its blue flowers.
  2. B. ✓ The plant's blue flowers

Why: A plant's phenotype is its set of features that can be seen.
The blue flowers are a feature of the plant that can be seen.
So the blue flowers are part of its phenotype, and the two alleles are its genotype.

8

Here are the two plants again, drawn between a photograph of a blue hydrangea and a photograph of a pink one. The two drawn plants grew from cuttings of one hydrangea.

Left, a photograph of a head of blue hydrangea flowers. Middle, two drawn potted hydrangeas: the left has blue flowers and its pot is labeled acid soil; the right has pink flowers and its pot is labeled alkaline soil; a dashed line between the pots reads same DNA. Right, a photograph of a head of pink hydrangea flowers. A credit line names the two photographers
Left, a photograph of a head of blue hydrangea flowers. Middle, two drawn potted hydrangeas: the left has blue flowers and its pot is labeled acid soil; the right has pink flowers and its pot is labeled alkaline soil; a dashed line between the pots reads same DNA. Right, a photograph of a head of pink hydrangea flowers. A credit line names the two photographers
9

A cutting is a piece cut from the parent plant and grown into a new plant. So every cell of both plants carries the parent's DNA, unchanged.

10

Plants grown from pieces of one parent, cuttings or runners, all carry the same DNA, cell for cell.

11

A set of organisms copied from one parent like that is called a , because each organism is a clone, a copy, of the parent.

12

The blue plant of the clonal line stands in acid soil. The pink plant of the clonal line stands in alkaline soil.

13

The two plants have the same DNA. So the difference between them is not in their genes.

14

The one thing that differs between the two pots is the soil.

15

Now consider a Himalayan rabbit. It shows the same thing on one animal.

16

A patch of its skin kept cool grows dark fur. A patch of the same rabbit's skin kept warm grows white fur.

Left, a photograph of a Himalayan rabbit: white body, dark ears, nose and tail. Right, a rabbit drawn from the side as a pale body with a head and long ears; two square patches are marked on its back: the left patch is dark and labeled kept cool, dark fur; the right patch is pale and labeled kept warm, white fur. A credit line names the photographer
Left, a photograph of a Himalayan rabbit: white body, dark ears, nose and tail. Right, a rabbit drawn from the side as a pale body with a head and long ears; two square patches are marked on its back: the left patch is dark and labeled kept cool, dark fur; the right patch is pale and labeled kept warm, white fur. A credit line names the photographer
17

Now consider identical twins. They began as one zygote, so their DNA is the same.

18

Twins fed differently as children can reach different heights.

19

In all three cases the genes are the same. One condition differs: the soil, the skin's temperature, the food.

20

The same genes gave different looks.

21

What you are expected to know Describe a case in which organisms with the same DNA show different phenotypes in different conditions.

22
Check q2

A gardener grows two plants from cuttings of one rosemary bush. The plant on a sunny ledge grows dense and bushy; the plant in a shaded corner grows thin and sparse.

Which of the following differs between the two plants?

  1. A. Their DNA
    Cuttings of one bush carry the bush's DNA, unchanged.
    So the two plants' DNA is the same.
  2. B. ✓ The light they get
  3. C. Their DNA and the light they get
    The light differs and the DNA does not.
    Cuttings of one bush carry the same DNA.

Why: The two plants are cuttings of one rosemary bush.
So they carry the same DNA.
One plant stands in sun and one in shade.
So the light is the one thing that differs.

23
Check q3

A gardener grows two plants from cuttings of one lavender plant. The plant in a sunny bed grows tall and bushy; the plant in deep shade grows short and thin.

What do the two plants show?

  1. A. The two plants differ in genotype
    Cuttings of one plant carry the same DNA, so the two plants have one genotype.
    A difference in phenotype is not evidence of a difference in genotype.
  2. B. The shade changed the DNA of the second plant
    A change in how a plant grows is not evidence that its DNA changed.
    The shade changed the conditions, not the genes.
  3. C. ✓ One genotype gave two phenotypes under two conditions
  4. D. Growth in lavender is set by the conditions alone
    The two plants share one genotype, and that one genotype gave both the tall plant and the short plant.
    So the genes took part as well as the light.

Why: Cuttings of one plant have the same DNA, so the two plants share one genotype.
They grew in different light, and they came out different.
So one genotype produced two phenotypes, because one condition differed.

24
Check q4

A gardener grows genetically identical maize seedlings for three weeks with the same soil, water and temperature. The seedlings in bright light reach a mean height of 16 cm; the seedlings in shade reach 25 cm.

Which conclusion do these results support?

  1. A. The shaded seedlings carried alleles for tallness that the others lacked
    The seedlings are genetically identical, so no group carried alleles the other lacked.
  2. B. ✓ The light level changed the seedlings' height
  3. C. The shade changed the DNA of the seedlings
    A difference in height is not evidence of a change in DNA; the seedlings differed only in the light they grew in.
  4. D. Height in maize is set by genes alone
    The genes were the same in both groups.
    The heights still differed by 9 cm.
    Only the light differed, so the light made the shaded seedlings taller.

Why: The seedlings had the same DNA, so the two groups shared one genotype.
Only the light differed between them.
So the light level, not the genes, made the shaded seedlings taller.

25Quick quiz: clonal line mixed practice

26
Check q5

A grower has a set of plants.

Which of the following makes the set a clonal line?

  1. A. The plants grew from seeds of one parent, each with its own mix of alleles
    Each seed carries its own mix of the parent's alleles.
    Plants with different DNA are not a clonal line.
  2. B. ✓ The plants grew from cuttings of one parent, all with the same DNA
  3. C. The plants are one species growing side by side in one field
    One species in one field can hold many genotypes.
    A clonal line is plants with the same DNA, cell for cell.

Why: A clonal line is a set of organisms copied from one parent without fertilization.
Each organism is a clone, a copy, of the parent.
So every one in the set carries the same DNA, cell for cell.

27
Check q6

A grower roots 20 cuttings taken from one basil plant.

Are the 20 plants a clonal line?

  1. A. ✓ Yes
  2. B. No
    Each cutting is a piece of the one basil plant, so all 20 carry the same DNA.

Why: Each cutting is a piece of the one basil plant.
So all 20 plants carry the parent's DNA, unchanged.
Plants with the same DNA, from one parent, are a clonal line.

28
Check q7

A grower sows 20 seeds taken from one apple tree.

Are the 20 plants a clonal line?

  1. A. Yes
    Each seed formed from its own egg cell and pollen grain, so each seed carries its own mix of alleles.
  2. B. ✓ No

Why: Each seed formed from one egg cell and one pollen grain.
Meiosis gave each egg cell and each pollen grain its own mix of alleles.
So the 20 plants differ in their DNA, and they are not a clonal line.

29
Check q8

A strawberry plant sends out runners, thin stems that root where they touch the soil, and 20 new plants grow from them.

Are the 20 plants a clonal line?

  1. A. ✓ Yes
  2. B. No
    Each runner is a stem of the one strawberry plant, so every new plant carries the parent's DNA.

Why: Each runner is a stem of the one strawberry plant.
A new plant grown from a runner carries the parent's DNA, unchanged.
So the 20 plants are a clonal line.

30
Check q9

A gardener sows a packet of 20 bean seeds collected from many bean plants.

Are the 20 plants a clonal line?

  1. A. Yes
    Seeds from many plants carry many different mixes of alleles.
  2. B. ✓ No

Why: The 20 seeds came from many bean plants.
Each seed carries its own mix of alleles.
Plants that differ in their DNA are not a clonal line.

31
Practice writing an answer

A gardener roots 12 cuttings from one rose bush and grows them into 12 plants.

(a) Explain why the 12 plants are a clonal line. (1 pt)

Model answer Each cutting is a piece of the one rose bush.
So each of the 12 plants carries the bush's DNA, unchanged.
Plants grown from one parent, all with the same DNA, are a clonal line.
Rubric
  • Award 1 point for: each plant grew from a piece of one parent, so all 12 carry the same DNA.

32The genes did not change

33

Video: Watch: The genes did not change

A petal cell with its pigment: aluminum arriving from acid soil and binding the pigment blue, none arriving from alkaline soil and the same pigment pink; the cutting moved from one pot to the other and its next flowers changing color; a skin cell using its pigment genes more under UV light.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L29b.mp4

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34

How can the same genes give blue petals in one pot and pink in the other?

35

Both plants make the same pigment in their petals. On its own, that pigment is pink.

36

Acid soil frees aluminum, a metal held in the soil. The roots take the aluminum up.

Two petal cells side by side, each a rounded box holding the same pigment, drawn as a circle. Left, labeled petal cell, acid soil, with an arrow marked aluminum from acid soil entering from above: a box marked aluminum is attached to the pigment circle, which is shaded dark; below the cell the words pigment with aluminum, blue. Right, labeled petal cell, alkaline soil, marked above no aluminum from alkaline soil, with no arrow: the pigment circle stands alone, shaded light; below the cell the words pigment alone, pink
Two petal cells side by side, each a rounded box holding the same pigment, drawn as a circle. Left, labeled petal cell, acid soil, with an arrow marked aluminum from acid soil entering from above: a box marked aluminum is attached to the pigment circle, which is shaded dark; below the cell the words pigment with aluminum, blue. Right, labeled petal cell, alkaline soil, marked above no aluminum from alkaline soil, with no arrow: the pigment circle stands alone, shaded light; below the cell the words pigment alone, pink
37

The aluminum reaches the petal cells and binds to the pigment. Pigment bound to aluminum is blue.

38

Alkaline soil keeps the aluminum locked in the soil. So no aluminum reaches the petal cells, and the same pigment stays pink.

39

The genes did not change. What reached the petal cells did.

40

Now suppose the blue plant is moved into alkaline soil. Its next flowers open pink.

A potted hydrangea with blue flowers in acid soil on the left; an arrow labeled moved points right to the same plant in alkaline soil, now with pink flowers
A potted hydrangea with blue flowers in acid soil on the left; an arrow labeled moved points right to the same plant in alkaline soil, now with pink flowers
41

The color changed and the DNA did not. So a change in phenotype is not evidence of a change in DNA.

42

The soil changed what reached the petal cells. Most conditions work another way: they change which genes the cells use.

43
Check q10

A stomach cell carries the gene for the protein pepsin and makes pepsin. A muscle cell carries the same gene and makes no pepsin.

Which cell is expressing the pepsin gene?

  1. A. ✓ The stomach cell
  2. B. The muscle cell
    The muscle cell carries the gene and makes no pepsin.
    A cell expresses a gene when it uses the gene's instructions to make the protein.

Why: A cell is expressing a gene when it uses that gene's instructions to make the protein.
The stomach cell makes pepsin from the pepsin gene.
So the stomach cell is expressing the gene.

44
Check q11

A hormone reaches a liver cell, and the cell starts making more of one protein.

Which of the following did the hormone change?

  1. A. ✓ Which genes the cell expresses and how much
  2. B. The order of bases in the cell's DNA
    A signal does not rewrite the DNA.
    The cell made more of a protein it already had the gene for.

Why: The cell already carried the gene for the protein.
The hormone is a signal.
A signal changes which genes a cell expresses and how much.
So the cell sped up making that protein, and its DNA stayed the same.

45

Now consider a summer tan. Ultraviolet light, UV, makes skin cells use their pigment genes more.

46

So the skin cells make more of the dark pigment, melanin, with the same genes. The tan is not a change in any allele.

47

In Chinese primroses, one plant opens red flowers at 20 °C and white flowers at 30 °C. The warmth changes which pigment genes the petal cells use and how much.

48

The rule in one sentence: a condition outside the organism changes which genes its cells use and how much, so the phenotype changes while the DNA does not.

49

The rabbit's patches show the same thing another way. Every skin cell of the rabbit makes the pigment enzyme.

50

Cool skin lets the enzyme work. So the cool patch makes pigment, and the fur grows dark.

51

Warm skin stops the enzyme working. So the warm patch makes no pigment, and the fur grows white.

52

The genes are the same in both patches.

53

What you are expected to know Explain why one genotype gives different phenotypes in different conditions.

54
Check q12

In Chinese primroses, plants grown at 20 °C open red flowers and the same plants grown at 30 °C open white flowers. A red-flowered primrose is moved into a greenhouse kept at 30 °C, and its next flowers open white.

What changed in the plant's petal cells?

  1. A. The order of bases in their DNA
    A change in flower color is not evidence of a change in DNA.
    The warmth changed what the petal cells did with the same genes.
  2. B. ✓ Which genes they used and how much

Why: The plant carries the same DNA at 30 °C as it did at 20 °C.
The warmth is a condition outside the plant.
A condition changes which genes the cells use and how much.
So the petal cells used their pigment genes differently, and the flowers opened white.

55
Practice writing an answer

In Chinese primroses, plants grown at 20 °C open red flowers and the same plants grown at 30 °C open white flowers. A red-flowered primrose is moved into a greenhouse kept at 30 °C. It opens white flowers.

(a) Explain how the primrose shows that a condition changes gene expression while the DNA stays the same. (1 pt)

Model answer The moved plant carries the same alleles at 30 °C as it did at 20 °C.
The warmth is a condition outside the plant.
A condition outside the organism changes which genes its cells use and how much.
So the petal cells at 30 °C use their pigment genes differently and open white flowers.
The DNA did not change; the gene expression did.
Rubric
  • Award 1 point for: the plant's DNA is the same at 20 °C and at 30 °C, and the warmth changed which genes the petal cells used, not the genes themselves.
56
Check q13

A gardener grows one cutting of a coleus plant in full sun and another cutting of the same plant in shade. The sun cutting has red leaves; the shade cutting has green leaves. A student says the sun changed the plant's leaf-color alleles.

Which of the following is right about the student's claim?

  1. A. The sun did change the leaf-color alleles
    A change in leaf color is not evidence of a change in DNA.
    The sun changed what the leaf cells did with the same alleles.
  2. B. ✓ The sun changed which genes the leaf cells used, not the alleles
  3. C. The two cuttings carried different alleles from the start
    Cuttings of one plant carry the same DNA, so their alleles were the same.

Why: The two cuttings have the same alleles.
The sun is a condition outside the plant.
So the sun changed which pigment genes the leaf cells used and how much, and the leaves came out red.
The DNA did not change.

57Not inherited

58

Video: Watch: Not inherited

A tanned parent beside a newborn with untanned skin; the parent's skin cells changed, the parent's gametes unchanged; the arrow from gametes to child carrying genes, not the tan.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L29c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L29c.mp4

59

Now consider a parent who spends the summer outdoors and tans. UV light made the parent's skin cells make more melanin.

60
Check q14

A parent has a child.

Which DNA does the child inherit from that parent?

  1. A. ✓ The DNA in the parent's gametes
  2. B. The DNA in the parent's skin cells
    A skin cell takes no part in fertilization.
    The child forms from one gamete of each parent.

Why: The child began as a zygote.
The zygote formed from one gamete of each parent.
So the child's DNA came from the parent's gametes, not from any other cell.

61

The parent then has a child. The child is not born tanned.

62

UV light changed the parent's skin cells. It did not change the DNA in the parent's gametes.

63

The child inherits the DNA in the gametes. So the child inherits the parent's genes, not the parent's tan.

64

A change a condition makes in the body's cells is not passed to offspring. Only the DNA in the gametes passes on, and that DNA did not change.

65

The moved hydrangea follows the same rule.

66

The alkaline soil changed the petals, not the DNA in the plant's gametes. So the plant's seeds carry the genes, not the pink color.

A potted hydrangea with blue flowers in acid soil on the left; an arrow labeled moved points right to the same plant in alkaline soil, now with pink flowers
A potted hydrangea with blue flowers in acid soil on the left; an arrow labeled moved points right to the same plant in alkaline soil, now with pink flowers
67

What you are expected to know Explain why a change a condition makes in a parent's body cells is not inherited.

68
Check q15

A man's skin darkens over a summer of outdoor work. He then has a child.

Is the child born with the darker skin?

  1. A. Yes
    The sun changed the man's skin cells, not the DNA in his gametes.
  2. B. ✓ No

Why: The sun changed what the man's skin cells did with their genes.
The DNA in his gametes did not change.
The child inherits the DNA in the gametes.
So the child inherits his genes, not his tan.

69
Practice writing an answer

A woman's skin darkens over years of outdoor work. She then has a baby. The baby's skin color at birth is the same as that of a baby of hers born before the years outdoors.

(a) Explain why the baby born after the years outdoors is no darker at birth than the baby born before them. (1 pt)

Model answer The sun changed what the woman's skin cells did with their genes.
The DNA in her gametes did not change.
Each baby inherits the DNA in her gametes.
So both babies inherit the same genes, and neither inherits the tan.
Rubric
  • Award 1 point for: the tan changed the woman's skin cells, not the DNA in her gametes, so both babies inherit the same genes.
70
Check q16

A gardener takes a cutting from a blue-flowered hydrangea growing in acid soil and plants the cutting in alkaline soil.

Predict the color of the cutting's first flowers.

  1. A. ✓ Pink
  2. B. Blue
    The cutting carries the parent's genes, not its color.
    The color follows the soil, and the soil is alkaline.

Why: The cutting carries the parent's DNA.
The blue color was made by the acid soil, not by the DNA.
The cutting stands in alkaline soil, and alkaline soil gives pink flowers.
So its first flowers open pink.

71Quick quiz: inherited or not inherited mixed practice

72
Check q17

A cyclist's leg muscles grow large over years of training.

Is this change passed to the cyclist's offspring?

  1. A. Yes
    Training changed the muscle cells, not the DNA in the gametes.
  2. B. ✓ No

Why: Training changed what the muscle cells did with their genes.
The DNA in the cyclist's gametes did not change.
Only the DNA in the gametes passes on, so the large muscles are not passed on.

73
Check q18

A change in the DNA of a mouse's egg cell gives one gene a new version.

Is this change passed to the offspring that grows from that egg cell's fertilization?

  1. A. ✓ Yes
  2. B. No
    The change sits in the DNA of a gamete, and the DNA in the gametes is what passes on.

Why: The egg cell is a gamete.
The DNA in the gametes is what passes to the offspring.
So a change in that egg cell's DNA is passed to the offspring that grows from it.

74
Check q19

A hydrangea's flowers open pink because of the soil in its garden.

Is the pink color passed to the plant's offspring?

  1. A. Yes
    The soil changed the petal cells, not the DNA in the plant's gametes.
  2. B. ✓ No

Why: The soil changed what reached the petal cells.
The DNA in the plant's gametes did not change.
Only the DNA in the gametes passes on, so the color is not passed on.

75
Check q20

A change in the DNA of one pollen grain of a pea plant gives one gene a new version.

Is this change passed to the offspring that grows from that pollen grain's fertilization?

  1. A. ✓ Yes
  2. B. No
    The pollen grain carries a gamete, and the DNA in the gametes is what passes on.

Why: The pollen grain carries the plant's male gamete.
The DNA in the gametes is what passes to the offspring.
So a change in that DNA is passed on.

76
Check q21

A carpenter's hands grow thick calluses over years of work.

Is this change passed to the carpenter's children?

  1. A. Yes
    The work changed the skin cells of the hands, not the DNA in the gametes.
  2. B. ✓ No

Why: The work changed what the skin cells of the hands did with their genes.
The DNA in the carpenter's gametes did not change.
Only the DNA in the gametes passes on, so the calluses are not passed on.

77
Check q22

A change in the DNA of one sperm cell of a bull gives one gene a new version.

Is this change passed to the calf that grows from that sperm cell's fertilization?

  1. A. ✓ Yes
  2. B. No
    The sperm cell is a gamete, and the DNA in the gametes is what passes on.

Why: The sperm cell is a gamete.
The DNA in the gametes is what passes to the offspring.
So a change in the sperm cell's DNA is passed on to the calf.

78The name

79

Video: Watch: The name

The hydrangea line, the Daphnia clone, the two Daphnia strains and the two dog breeds in turn, each gaining the label phenotypic plasticity or genetic difference; the four-row table filling in.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L29d.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L29d.mp4

80

One genotype can produce different phenotypes under different conditions. That ability is called , because something plastic can be shaped, and here the conditions shape the phenotype.

81

For example, the hydrangea clonal line flowered blue in acid soil and pink in alkaline soil. This is phenotypic plasticity, because one genotype produced two phenotypes under two conditions.

82

And one clone of Daphnia, a small water animal, grows a defensive crest in water that has held predators and no crest in water that has not. This is phenotypic plasticity, because one genotype produced two phenotypes under two conditions.

83

But two Daphnia strains kept in one water differ in crest size. This is not phenotypic plasticity, because two genotypes differ under one condition: a genetic difference.

84

And two breeds of dog kept in one house differ in size. This is not phenotypic plasticity, because two genotypes differ under one condition: a genetic difference.

85

Here is a table of the four cases: whether the genotype is one, whether the conditions differ, and the verdict.

A table of four cases with three columns: one genotype, conditions differ, verdict. Hydrangea clonal line in two soils: yes, yes, phenotypic plasticity. One Daphnia clone in two waters: yes, yes, phenotypic plasticity. Two Daphnia strains in one water: no, no, genetic difference. Two dog breeds in one house: no, no, genetic difference
A table of four cases with three columns: one genotype, conditions differ, verdict. Hydrangea clonal line in two soils: yes, yes, phenotypic plasticity. One Daphnia clone in two waters: yes, yes, phenotypic plasticity. Two Daphnia strains in one water: no, no, genetic difference. Two dog breeds in one house: no, no, genetic difference
86

So phenotypic plasticity is the range of phenotypes one genotype can produce.

87

A population full of different-looking individuals may vary because its genotypes differ. That variety is not phenotypic plasticity.

88

What you are expected to know Classify a described case as phenotypic plasticity or as a genetic difference between individuals.

89

Back to the two pots: two hydrangea plants from cuttings of one parent, the same DNA cell for cell, one blue in acid soil and one pink in alkaline soil.

90

The acid soil let aluminum reach the petal cells, and the alkaline soil did not. So the same genes, and the same pigment, gave blue in one pot and pink in the other.

91

Move a plant and the color follows the soil. The offspring inherit the genes, not the color.

92

One genotype, two phenotypes under two conditions: the two pots show phenotypic plasticity.

93Quick quiz: phenotypic plasticity mixed practice

94
Check q23

Which of the following is phenotypic plasticity?

  1. A. A condition changing the order of bases in an organism's DNA
    Phenotypic plasticity never involves a change in DNA.
    The genes stay the same and the cells use them differently.
  2. B. ✓ One genotype producing different phenotypes in different conditions
  3. C. Two genotypes showing different phenotypes under one condition
    Two genotypes differing under one condition is a genetic difference.

Why: Phenotypic plasticity is the ability of one genotype to produce different phenotypes under different conditions.
The genotype is one; the conditions differ; the phenotypes differ.

95
Check q24

One clonal line of water lettuce grows long roots in water poor in nutrients and short roots in water rich in nutrients.

Which of the following does this case show?

  1. A. ✓ Phenotypic plasticity
  2. B. A genetic difference
    The plants are one clonal line, so they share one genotype; two different waters gave two phenotypes, and that is phenotypic plasticity.

Why: The plants are one clonal line, so they share one genotype.
The two waters are two conditions.
One genotype produced two root lengths in two conditions.
That is phenotypic plasticity.

96
Check q25

Two varieties of a grass grown side by side in one field differ in leaf width.

Which of the following does this case show?

  1. A. Phenotypic plasticity
    The two varieties are two genotypes, and they grew in the same field, the same conditions.
    Phenotypic plasticity is one genotype producing two phenotypes in two conditions.
  2. B. ✓ A genetic difference

Why: The two varieties are two different genotypes.
They grew side by side in one field, so the conditions were the same.
Two genotypes differing under the same conditions is a genetic difference, not phenotypic plasticity.

97
Check q26

Two breeds of sheep kept on one farm grow wool of different thickness.

Which of the following does this case show?

  1. A. Phenotypic plasticity
    Two breeds are two genotypes, and the sheep are kept on one farm, in the same conditions.
    Phenotypic plasticity is one genotype producing two phenotypes in two conditions.
  2. B. ✓ A genetic difference

Why: Two breeds are two different genotypes.
They are kept on one farm, so the conditions are the same.
Two genotypes differing under the same conditions is a genetic difference.

98
Check q27

Genetically identical mice fed different diets reach different body masses.

Which of the following does this case show?

  1. A. ✓ Phenotypic plasticity
  2. B. A genetic difference
    The mice are genetically identical, so they share one genotype.
    Their diets differed and their body masses differed.
    One genotype producing different phenotypes in different conditions is phenotypic plasticity.

Why: The mice share one genotype.
The diets are two conditions.
One genotype produced different body masses in different conditions.
That is phenotypic plasticity.

99
Check q28

Cuttings of one ivy plant grow large leaves in shade and small leaves in full sun.

Which of the following does this case show?

  1. A. ✓ Phenotypic plasticity
  2. B. A genetic difference
    Cuttings of one plant share one genotype; two light levels gave two leaf sizes, and one genotype producing two phenotypes is phenotypic plasticity.

Why: Cuttings of one plant share one genotype.
Shade and full sun are two conditions.
One genotype produced two leaf sizes in two conditions.
That is phenotypic plasticity.

100
Check q29

Two varieties of tomato grown in one greenhouse ripen fruit of different colors.

Which of the following does this case show?

  1. A. Phenotypic plasticity
    Two varieties are two genotypes, and they grew in one greenhouse, the same conditions.
    Phenotypic plasticity is one genotype producing two phenotypes in two conditions.
  2. B. ✓ A genetic difference

Why: Two varieties are two different genotypes.
They grew in one greenhouse, so the conditions were the same.
Two genotypes differing under the same conditions is a genetic difference.

101
Practice writing an answer

One clonal line of water lettuce grows long roots in water poor in nutrients and short roots in water rich in nutrients. Two breeds of chicken kept in one coop lay eggs of different shell colors, one breed brown and the other white.

(a) Identify which case shows phenotypic plasticity and which shows a genetic difference, and justify each choice. (1 pt)

Model answer The water lettuce plants are one clonal line, so they share one genotype.
They grew in two conditions, poor water and rich water, and came out with two root lengths.
One genotype producing two phenotypes in two conditions is phenotypic plasticity.
The two chicken breeds are two genotypes.
They are kept in one coop, so the conditions are the same.
So their difference in shell color comes from their genes: a genetic difference.
Rubric
  • Award 1 point for: the water lettuce is one genotype (a clonal line) giving two phenotypes in two conditions; the chicken breeds are two genotypes under the same conditions.

102Mixed practice mixed practice

103
Check q30

A Siamese cat has dark fur on its ears, paws and tail, the coolest parts of its body, and pale fur on its warm body.

Which explanation fits the pattern?

  1. A. The ears, paws and tail carry different pigment alleles from the rest of the body's cells
    Every cell of the cat carries the same alleles.
    The parts differ in temperature, not in genes.
  2. B. The cold changed the DNA sequence in the cells of the ears, paws and tail
    A difference in fur color is not evidence of a change in DNA.
    The cold changed what the skin cells did with the same genes.
  3. C. ✓ The enzyme that makes the pigment works only in cool skin
  4. D. The dark parts and the pale body were inherited from different parents of the cat
    Every cell of the cat received the same two sets of chromosomes, one from each parent.
    No part of the cat came from one parent alone.

Why: Every skin cell of the cat carries the same pigment genes and makes the same pigment enzyme.
That enzyme works only in cool skin.
So the cool ears, paws and tail make pigment and grow dark fur.
The warm body makes no pigment and stays pale.

104
Check q31

Flamingos fed a diet containing little pigment grow new feathers that are pale pink instead of vivid pink.

Which conclusion does this observation support?

  1. A. ✓ The diet changed the birds' phenotype
  2. B. The pale birds inherited different feather-color alleles
    The same birds paled when their diet changed; no new alleles arrived.
  3. C. Feather color is set by genes alone
    The same birds with the same genes changed color when the diet changed, so the diet was the cause.
  4. D. The diet changed the birds' genotype
    A change in a trait is not evidence of a change in DNA; the same birds, with the same genes, grew paler feathers when their food changed.

Why: The pink comes from pigment in the birds' food, which the feather cells lay into new feathers.
With less pigment in the food, the new feathers grow paler.
The diet changed the phenotype, and the genes stayed the same.

105
Check q32

A person trains hard for a year, and the muscles of their legs grow much larger.

Which statement about the change is right?

  1. A. Training changed the DNA of the muscle cells
    A change in a trait is not evidence of a change in DNA.
    The muscle cells used their genes differently under a new condition.
  2. B. The person's children will be born with larger leg muscles than other children
    A change a condition makes in body cells is not passed to offspring.
    The DNA in the gametes did not change.
  3. C. The larger muscles show that the person inherited new alleles during the year
    A person's alleles are the ones inherited at fertilization, and training does not add new ones.
  4. D. ✓ Training changed which genes the muscle cells used, not their DNA

Why: Training is a condition outside the cells that changes which genes the muscle cells use and how much, so the muscles grow while the DNA stays the same.

106
Check q33

In Chinese primroses, plants grown at 20 °C open red flowers and the same plants grown at 30 °C open white flowers. A white-flowered primrose grown at 30 °C is moved into a greenhouse kept at 20 °C.

Predict the color of the flowers it opens next.

  1. A. ✓ Red
  2. B. Pink
    The plant is moved, not crossed with a red plant.
    At 20 °C the same plants open red flowers, not pink.
  3. C. White
    The white color was made by the warmth, and the plant has left the warmth.
    At 20 °C the same plants open red flowers.

Why: The same plants open red flowers at 20 °C, so the cooler greenhouse changes which pigment genes the petal cells use and how much.
The plant's DNA stays as it was.
So its next flowers open red.

107
Check q34

Two breeds of horse kept in one field differ in height.

Which of the following does this case show?

  1. A. Phenotypic plasticity
    Two breeds are two genotypes, and they are kept in one field, the same conditions.
    Phenotypic plasticity is one genotype producing two phenotypes in two conditions.
  2. B. ✓ A genetic difference

Why: Two breeds are two different genotypes.
They are kept in one field, so the conditions are the same.
Two genotypes differing under the same conditions is a genetic difference.

108
Check q35

A gardener moves a pink-flowered hydrangea from alkaline soil into acid soil.

Predict the color of its next flowers.

  1. A. ✓ Blue
  2. B. Pink
    The pink color was made by the alkaline soil, and the plant has left that soil.
    Acid soil gives blue flowers.

Why: The plant's DNA is unchanged by the move.
The soil's acidity is the condition that sets the color.
Acid soil gives blue hydrangea flowers.
So the next flowers open blue.

109
Practice writing an answer

Flamingos fed a diet containing little pigment grow new feathers that are pale pink instead of vivid pink. When the birds return to their usual diet, their new feathers grow vivid pink again.

(a) Explain what the two changes in feather color show about the flamingos' phenotype and their genotype. (1 pt)

Model answer The same birds grew pale feathers on the low-pigment diet and vivid feathers on the usual diet.
The birds' DNA was the same on both diets.
So the diet, a condition, changed the phenotype.
The genotype did not change, because the vivid color returned with the usual diet.
Rubric
  • Award 1 point for: the same birds with the same DNA changed feather color as the diet changed, so the diet changed the phenotype and the genotype stayed the same.

Glossary

clonal line
A set of organisms all copied from one parent without fertilization, such as plants from cuttings or runners, or aphids that breed without mating, so every one carries the same DNA, cell for cell. Each organism is a clone, a copy, of the parent. Two cuttings of one hydrangea in two soils are a clonal line in two conditions.
phenotypic plasticity
The ability of one genotype to produce different phenotypes under different conditions: one clone of a hydrangea flowers blue in acid soil and pink in alkaline soil. The DNA does not change; the conditions change the phenotype.

APBIO-U05-L30 Six cases

Topic 5.5 · Environmental Effects on Phenotype · 31 steps

Six panels in a row: two stick figures of different heights labeled twins; a blue and a pink flower head labeled hydrangeas; a white fox and a brown fox labeled arctic fox; a turtle on sand labeled turtle egg; a dark arm and a pale arm labeled skin; two round cells with an arrow between them labeled yeast
Six panels in a row: two stick figures of different heights labeled twins; a blue and a pink flower head labeled hydrangeas; a white fox and a brown fox labeled arctic fox; a turtle on sand labeled turtle egg; a dark arm and a pale arm labeled skin; two round cells with an arrow between them labeled yeast

Here are six pictures. A tall twin and a short one. A blue hydrangea and a pink one. An arctic fox in white and in brown. A turtle hatching from warm sand. A tanned arm beside a pale one. A yeast cell putting out a signal toward another.

In all six, a trait answers to a condition. In each picture, what changed, and what answered?

Unit 5 · Heredity

1What changed, and what answered

2

Video: Watch: Six cases

The six panels, each gaining a label for the condition and a label for the trait, one case at a time, and the same rule spoken under each.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L30a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L30a.mp4

3

Here are six cases of one genotype giving different phenotypes: height, flower color, fur color, a turtle's sex, skin color, and a yeast cell's signal.

4

In each case one condition changes and one trait answers. In none of the six does the DNA change.

5

Here are two arctic foxes photographed in two seasons: one in its white winter coat, one in a dark summer coat.

Top left, a photograph of an arctic fox in its white winter coat sitting on snow; top right, a photograph of a different arctic fox with a dark summer coat on a grassy slope. Below, the same fox drawn twice: on the left in winter, its coat white, labeled short days, cold, white fur; on the right in summer, its coat brown, labeled long days, warm, brown fur. A credit line names the two photographers
Top left, a photograph of an arctic fox in its white winter coat sitting on snow; top right, a photograph of a different arctic fox with a dark summer coat on a grassy slope. Below, the same fox drawn twice: on the left in winter, its coat white, labeled short days, cold, white fur; on the right in summer, its coat brown, labeled long days, warm, brown fur. A credit line names the two photographers
6

Below the photographs one fox is drawn in both seasons. Its DNA is the same in winter and in summer; its fur is not.

7

As the days shorten and the air cools, the cells in the fox's skin change which genes they use. The skin cells use their pigment genes less, so the new fur grows in white.

8

The rule: a condition outside the organism changes which genes its cells use and how much, so the phenotype changes while the DNA does not.

9

Each of the six pictures follows that rule. Here is the condition and the trait in each:

  1. Height and weight in humans: nutrition changes them, and many genes set the range a person can reach, so food moves a person within that range.
  2. Flower color in hydrangeas: the acidity of the soil changes it, blue in acid soil and pink in alkaline.
  3. Seasonal fur color in arctic foxes and hares: day length and temperature change it, white in winter and brown in summer.
  4. Sex in many turtles and alligators: the temperature at which the egg develops sets it.
  5. Melanin, the dark pigment in skin: exposure to UV light raises it, and the skin darkens.
  6. Pheromone production in yeast: a yeast cell puts out far more of its signal molecule, a pheromone, when a cell of the opposite mating type, the yeast's version of the other sex, is nearby.

10

Here are the six in one table, the trait beside the condition that changes it.

A table of the six cases: human height and weight, nutrition; flower color in hydrangeas, soil acidity; seasonal fur color in arctic foxes and hares, day length and temperature; sex in many turtles and alligators, the temperature the egg develops at; melanin in skin, UV light; pheromone production in yeast, a cell of the opposite mating type nearby
A table of the six cases: human height and weight, nutrition; flower color in hydrangeas, soil acidity; seasonal fur color in arctic foxes and hares, day length and temperature; sex in many turtles and alligators, the temperature the egg develops at; melanin in skin, UV light; pheromone production in yeast, a cell of the opposite mating type nearby
11

Human height has a twist. Many genes set the range a person can reach, and nutrition moves the person within that range.

12

Identical twins share one genotype, so they share one range. Different food can put the two twins at different heights within it.

A photograph of two young girls standing side by side in matching school uniforms; they are identical twins. A credit line names the photographer
A photograph of two young girls standing side by side in matching school uniforms; they are identical twins. A credit line names the photographer
13

In none of the six does the DNA change.

14

Warm sand does not change the DNA in a turtle egg. The warmth changes which genes the embryo's cells use, and the hatchling's sex follows.

A photograph of a dark sea turtle hatchling, about the size of a hand, crawling over shell-strewn sand. A credit line names the source
A photograph of a dark sea turtle hatchling, about the size of a hand, crawling over shell-strewn sand. A credit line names the source
15

That is what the six pictures have in common: in every case the genes stayed the same, and the condition changed what the cells did with them.

16

What you are expected to know Name, for each of the six cases, the condition that changes and the trait that answers.

17
Check q1

Winter comes: the days shorten and the air cools around an arctic fox.

Which way does the color of its new fur move?

  1. A. ✓ Toward white
  2. B. Toward brown
    Brown fur is the summer coat.
    As the days shorten and cool, the fox's skin cells use their pigment genes less, so the new fur grows in white.

Why: Short days and cold are the winter condition.
The fox's skin cells answer by using their pigment genes less.
So the new fur grows in white.

18
Check q2

A gardener moves a blue-flowered hydrangea from acid soil into alkaline soil.

Which way does the color of its next flowers move?

  1. A. Toward blue
    Acid soil gives blue flowers, and the plant has left the acid soil.
    In alkaline soil no aluminum reaches the petals, so the pigment shows pink.
  2. B. ✓ Toward pink

Why: The soil's acidity is the condition.
Acid soil gives blue flowers and alkaline soil gives pink.
The plant is now in alkaline soil, so its next flowers move toward pink.

19
Check q3

A person spends two weeks in strong sunshine.

Which way does the amount of melanin in their skin move?

  1. A. ✓ Up
  2. B. Down
    UV light makes skin cells use their pigment genes more, not less.
    More melanin is made, and the skin darkens.
  3. C. No change
    Melanin does answer to UV light.
    The skin cells use their pigment genes more and make more melanin.

Why: UV light is the condition.
The skin cells answer by using their pigment genes more.
So the amount of melanin goes up, and the skin darkens.

20
Check q4

A child eats far too little food for several years while growing.

Which way does the child's adult height move, within the range the child's genes allow?

  1. A. Up
    Nutrition moves a person within the range their genes allow, and too little food moves them toward the bottom of that range.
  2. B. ✓ Down
  3. C. No change
    Human height does answer to nutrition.
    Many genes set the range, and food moves the person within it.

Why: Nutrition is the condition.
Many genes set the range of heights the child can reach, and food moves the child within that range.
Too little food moves the child toward the bottom of the range.
So the adult height moves down.

21
Check q5

Spring comes: the days lengthen and the air warms around a snowshoe hare.

Which way does the color of its new fur move?

  1. A. Toward white
    White fur is the winter coat.
    As the days lengthen and warm, the hare's skin cells use their pigment genes more, so the new fur grows in brown.
  2. B. ✓ Toward brown

Why: Long days and warmth are the summer condition.
The hare's skin cells answer by using their pigment genes more.
So the new fur grows in brown.

22
Check q6

A researcher adds a yeast cell of the opposite mating type to a dish holding one yeast cell.

Which way does the first cell's output of pheromone move?

  1. A. ✓ Up
  2. B. Down
    A cell of the opposite mating type nearby is the condition that raises pheromone output.
    The first cell uses its pheromone genes more.
  3. C. No change
    Pheromone output does answer to a nearby cell of the opposite mating type.
    The first cell uses its pheromone genes much more.

Why: A cell of the opposite mating type nearby is the condition.
The first cell answers by using its pheromone genes much more.
So its output of pheromone goes up.

23

Back to the six pictures: a tall twin beside a short one, a blue hydrangea beside a pink one, an arctic fox in white fur and in brown, a turtle hatching from warm sand, a tanned arm beside a pale one, a yeast cell putting out a signal toward another.

24

In each picture one condition changed and one trait answered: food and height, soil and flower color, season and fur, warmth and sex, UV light and skin, a nearby cell and a signal.

25

The DNA never changed.

26Quick quiz: the condition behind each case mixed practice

27
Check q7

A sea turtle lays a clutch of eggs in sand. Some of the hatchlings are male and some are female.

What set the sex of each hatchling?

  1. A. The alleles the egg received from the father
    In many turtles the temperature of the developing egg sets its sex, not an allele from the father.
  2. B. ✓ The temperature at which the egg developed
  3. C. The food the mother turtle ate before laying the eggs
    Food is the condition behind human height and weight; a turtle's sex answers to the temperature of the developing egg.
  4. D. The day length during the weeks the egg lay in the sand
    Day length is the condition that changes an arctic fox's fur, not a turtle's sex.

Why: In many turtles and alligators the temperature at which the egg develops sets the hatchling's sex.
The warmth changes which genes the embryo's cells use.

28
Check q8

A snowshoe hare's coat turns white as winter comes and brown again in spring.

Which conditions change the hare's fur color?

  1. A. The acidity of the soil it lives on
    Soil acidity is the condition that changes a hydrangea's flower color, not a hare's coat.
  2. B. The food it finds in each season
    Food is the condition behind human height and weight; the hare's coat answers to day length and temperature.
  3. C. ✓ Day length and temperature
  4. D. The number of hares living nearby
    Other hares are not the condition; a yeast cell answers to a nearby cell, a hare's coat does not.

Why: Seasonal fur color in arctic foxes and hares answers to day length and temperature.
As the days shorten and cool, the skin cells use their pigment genes less, and the new fur grows white.

29
Check q9

A yeast cell of one mating type sits in a dish. It starts to put out far more pheromone than before.

Which condition changed?

  1. A. ✓ A cell of the opposite mating type came near
  2. B. The liquid in the dish became more acid
    Acidity is the condition behind a hydrangea's flower color; a yeast cell's pheromone answers to a nearby cell of the opposite mating type.
  3. C. The dish moved into stronger light
    UV light is the condition behind melanin in skin; a yeast cell's pheromone answers to a nearby cell of the opposite mating type.
  4. D. The food in the dish was nearly gone
    Food is the condition behind human height and weight; a yeast cell's pheromone answers to a nearby cell of the opposite mating type.

Why: Pheromone production in yeast answers to one condition: a cell of the opposite mating type nearby.
The cell's genes stay the same.
The cell uses its pheromone genes much more when that cell is near, so the cell puts out far more pheromone.

30
Check q10

An arctic fox's coat turns white for the winter. A hydrangea's flower color follows its soil. A person's skin darkens in the sun. Food changes how tall a person grows. A yeast cell makes far more pheromone when a cell of the opposite mating type is near. The temperature of the sand sets a turtle hatchling's sex.

What do these six cases have in common?

  1. A. In each, the condition changes one allele into another
    In none of the six does the DNA change.
    The alleles stay as they were, and the cells use them differently.
  2. B. In each, the trait is passed to the offspring in its changed form
    A change a condition makes in body cells is not inherited.
    A tanned parent's child is not born tanned.
  3. C. In each, the trait is set by genes alone and the condition is chance
    In each case a named condition, not chance, changes the trait.
    The genes stay the same.
  4. D. ✓ In each, one condition changes what the cells do with the same genes

Why: In every one of the six, the genes stayed the same.
A condition outside the organism changed which genes the cells used and how much.
So the trait changed.

APBIO-U05-L30B How to test one

Topic 5.5 · Environmental Effects on Phenotype · 65 steps

Two potted hydrangeas on the left, one with blue flowers and one with pink; above them the two claims, the soil did it, and, the cuttings must differ; on the right a tray of 24 small cuttings split into two groups of 12, labeled 12 at pH 5 and 12 at pH 7, under the heading 24 cuttings of one plant
Two potted hydrangeas on the left, one with blue flowers and one with pink; above them the two claims, the soil did it, and, the cuttings must differ; on the right a tray of 24 small cuttings split into two groups of 12, labeled 12 at pH 5 and 12 at pH 7, under the heading 24 cuttings of one plant

Here are the two hydrangea pots again, blue and pink, and beside them a tray of 24 cuttings taken from one plant.

A gardener says the soil did it. A neighbor says the two cuttings must have differed all along. Design the test that settles it.

Unit 5 · Heredity

1The design

2

Video: Watch: The design

The tray of 24 cuttings split 12 and 12 between two soils, everything else labeled the same, the control named; the neighbor's seeds shown failing the test.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L30Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L30Ba.mp4

3

How do you prove that a condition, not a gene, caused a difference?

4

Split genetically identical material, cuttings or clones of one parent, between two conditions. Hold everything else the same.

5

One of the two conditions is the control. Then measure the trait.

6

If the cuttings in one condition come out different from the cuttings in the other, the condition caused the difference, because the genes were the same.

7

When the trait is a count, blue against pink, you compare the numbers. When the trait is a measurement, such as height in centimeters, you compare the mean of each group with ±2SE error bars and the overlap rule.

8
Check q1

A grower roots 24 cuttings taken from one hydrangea.

Which of the following is true of the 24 plants?

  1. A. ✓ The 24 plants carry the same DNA, cell for cell
  2. B. Each plant carries its own mix of the parent's alleles
    Each cutting is a piece of the one parent, not a seed from it.
    A piece of the parent carries the parent's DNA, unchanged.

Why: Each cutting is a piece of the one hydrangea.
So every plant carries the parent's DNA, unchanged.
The 24 plants are a clonal line.

9
Check q2

In a test, a grower changes one condition and measures one trait.

What is the one condition the grower changes called?

  1. A. The dependent variable
    The dependent variable is the trait measured, the one that depends on the change.
  2. B. ✓ The independent variable

Why: The condition the grower changes is the independent variable.
The trait the grower measures is the dependent variable, because it depends on the change.

10
Check q3

A grower gives one group of plants a change and leaves a second group as it was. Light, water and temperature are the same for every plant.

Which group is the control?

  1. A. ✓ The group left as it was
  2. B. The group given the change
    The group given the change is the test group.
    The control is the group treated alike but left as it was.

Why: The control is the group given the same treatment as the others but lacking the factor under test.
So the group left as it was is the control.

11

The gardener names the soil, and the neighbor names the cuttings' genes. A test settles their argument only if it lets one of those two causes vary and holds the other fixed.

12

Here is the test, in five steps:

  1. Start from genetically identical material: 24 cuttings taken from one hydrangea, so the genes cannot differ between any two of them.
  2. Split the cuttings between two conditions: 12 in acid soil at pH 5 and 12 in soil at pH 7. The soil's acidity is the one condition you change, so it is the independent variable.
  3. Hold everything else the same: light, water and temperature are the control variables, kept the same for every pot.
  4. Keep one condition as the control: the pots at pH 7, the soil the parent plant grew in, show the color the cuttings give without the change.
  5. Measure the trait: record the flower color of every plant at first flowering. Flower color is the dependent variable.

Two trays of twelve cuttings each, drawn as small stems: the left tray is labeled 12 cuttings, soil at pH 5; the right tray is labeled 12 cuttings, soil at pH 7, the control. A band above both trays reads: same light, same water, same temperature. A caption reads: only the soil's acidity differs between the two trays
Two trays of twelve cuttings each, drawn as small stems: the left tray is labeled 12 cuttings, soil at pH 5; the right tray is labeled 12 cuttings, soil at pH 7, the control. A band above both trays reads: same light, same water, same temperature. A caption reads: only the soil's acidity differs between the two trays
13

Starting from cuttings of one plant matters more than any other step. With seeds, or with plants bought at random, a difference in color could be genes or soil, and the test could not tell which.

14

If the 12 pots at pH 5 flower blue and the 12 at pH 7 flower pink, the soil changed the color, because nothing else was allowed to differ.

15

The neighbor's claim needed the genes to differ, and they could not.

16

Now consider a human case, where nobody can be split into two pots. People who move from sea level to a high mountain town have more red blood cells after a few weeks.

17

Moving changed more than the air's oxygen. Food, exercise and temperature changed too: four conditions at once.

18

So the extra red blood cells cannot be credited to any one of the four.

19

To credit the oxygen, change only the oxygen: some people breathe air with less oxygen in a room, others breathe ordinary air in the same room, and everything else is held the same.

20

What you are expected to know Design a test that separates a genetic cause from an environmental one.

21
Check q4

In the hydrangea test the parent plant grew in soil at pH 7, and 12 cuttings are potted at pH 5 and 12 at pH 7.

Which pots are the control?

  1. A. The pH 5 pots
    The pH 5 pots are the group given the change; the control is the group given the same treatment without it.
  2. B. ✓ The pH 7 pots
  3. C. The parent plant
    The parent is one plant in one soil, not a group treated like the test group without the change.
  4. D. All 24 pots together
    The control is one group, the one without the change, against which the other is compared.

Why: The control is the group given the same treatment as the others but lacking the factor under test.
The pH 7 pots, in the parent's own soil, show the color the cuttings give without the acid.
So any difference in the pH 5 pots is credited to the acid.

22
Check q5

A grower wants to know whether the leaf size of a basil plant depends on the light it gets.

Which design separates the light from the genes?

  1. A. Seeds from one packet, half in bright light and half in shade, same water and temperature
    Seeds from one packet are not genetically identical, so a difference in leaf size could be genes or light.
  2. B. Two different basil plants, one in bright light and one in shade, same water and temperature
    Two different plants have two genotypes, so a difference in leaf size could be genes or light.
  3. C. Cuttings of one basil plant, half in bright light with extra water and half in shade
    Two conditions change at once, light and water, so neither can be credited with a difference.
  4. D. ✓ Cuttings of one basil plant, half in bright light and half in shade, same water and temperature

Why: Cuttings of one plant share one genotype, so genes cannot differ.
With water and temperature held the same, the light is the only condition that differs.
So any difference in leaf size is credited to the light.

23
Check q6

A student plans the hydrangea test and says: 'Use seeds from two different hydrangea plants, one lot in each soil. Two plants make it a fairer test.'

Is the student's plan right?

  1. A. Yes
    Seeds from two plants carry two genotypes.
    So the genes differ as well as the soil.
  2. B. ✓ No

Why: Seeds from two plants carry two different genotypes.
A color difference between the two soils could then be genes or soil.
So the test could not tell which cause did it.
Cuttings of one plant fix the genes, so only the soil can differ.

24
Practice writing an answer

A grower takes 24 cuttings of one hydrangea and pots 12 in soil at pH 5 and 12 in soil at pH 7, with water, light and temperature the same for every pot.

(a) Explain why the grower starts from cuttings of one plant rather than from a packet of hydrangea seeds. (1 pt)

Model answer Every cutting of one plant carries the parent's DNA.
So the 24 plants share one genotype, and a color difference between the two soils cannot come from their genes.
Seeds from a packet carry different genotypes.
With seeds, a color difference could be genes or soil.
Rubric
  • Award 1 point for: cuttings of one plant share one genotype, so the genes cannot differ between the groups; seeds would let the genes differ.
25
Practice writing an answer

A grower has one tomato plant of a variety whose fruit is sometimes sweet and sometimes bland. She thinks the hours of light the plants get each day decide the sweetness. Her plant has always had 12 hours of light a day, and she can root as many cuttings from it as she needs.

(a) Describe a test that would show whether the hours of light, rather than the plants' genes, set the sweetness of the fruit. (4 pt)

Model answer Root 30 cuttings from the one plant, so every plant has the same DNA.
Give 15 plants 12 hours of light a day and 15 plants 16 hours.
Give every plant the same water, temperature and soil, so the hours of light are the only condition that differs.
The 12-hour plants, at the plant's usual light, are the control.
Measure the sugar in the fruit of every plant and compare the 12-hour plants with the 16-hour plants.
Rubric
  • Award 1 point for: every plant grown as a cutting of the one plant (a clonal line). Accept with or without: the reason stated, that the genes cannot then differ between the groups.
  • Award 1 point for: two groups given two amounts of light a day, with water, temperature and soil held the same for every plant.
  • Award 1 point for: one group kept at the plant's usual 12 hours of light as the control.
  • Award 1 point for: the sweetness of the fruit measured in every plant, by its sugar content or by a blind taste score, and the two groups compared. Accept with or without: sugar in grams per 100 grams as the measure.

Slip Using seeds from the plant, or plants bought from a shop. Seeds and shop plants carry different genotypes, so a difference in sweetness could be genes or light.

26When the trait is a measurement

27

Video: Watch: When the trait is a measurement

Two bars of means with their ±2SE error bars drawn from the working, values to formula to substitution; the two ends read off the axis; the overlap rule applied.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L30Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-L30Bb.mp4

28

Flower color is a trait you count: so many blue, so many pink. Height, mass or sugar content is a trait you measure.

29

A measured trait is compared through its mean. A number you count for each plant, such as berries per branch, is compared through its mean too.

30

The tools are the mean of each group, as its one representative value, and an error bar of ±2SE around each mean.

31
Check q7

A grower compares two means, each with a ±2SE error bar, and the two error bars overlap.

What do the data show?

  1. A. ✓ The data do not show a difference between the means
  2. B. The difference between the means is very unlikely to be chance
    Bars that do not overlap make chance very unlikely.
    These bars overlap, so the two true means could be the same.

Why: Each ±2SE bar is the range its true mean is likely to lie in.
The two bars overlap, so the two true means could be the same.
So the data do not show a difference.

32

Now consider maize seedlings from one genetically identical line, grown for 21 days: 20 in bright light and 20 in shade. The grower measures every stem and finds the mean stem length of each group.

33
Worked example

The bright-light group has a mean stem length of 16.4 cm with a standard deviation s of 2.7 cm; the shade group has a mean of 25.1 cm with s of 3.6 cm; n is 20 in each group. Calculate the ±2SE error bar of each group.

Write down the values in the question:
mean (bright light)=16.4cm
s(bright light)=2.7cm
mean (shade)=25.1cm
s(shade)=3.6cm
n=20
Write down the equations:
SE=sn
error bar=mean±2SE
Substitute the values into the equations:
SE(bright light)=2.720=0.60cm
2SE=1.2cm
bright-light bar=16.4±1.2=15.2cm to 17.6cm
SE(shade)=3.620=0.80cm
2SE=1.6cm
shade bar=25.1±1.6=23.5cm to 26.7cm
34

Here are the two means drawn as bars, each with its ±2SE error bar.

Two bars of mean stem length in centimeters for genetically identical maize seedlings: 16.4 cm in bright light and 25.1 cm in shade, each with an error bar, from 15.2 to 17.6 cm and from 23.5 to 26.7 cm; gridlines every 5 cm; the legend reads: error bars represent ±2SE
Two bars of mean stem length in centimeters for genetically identical maize seedlings: 16.4 cm in bright light and 25.1 cm in shade, each with an error bar, from 15.2 to 17.6 cm and from 23.5 to 26.7 cm; gridlines every 5 cm; the legend reads: error bars represent ±2SE
35

Bars overlap when the bottom of the higher bar sits below the top of the lower bar.

36

When two ±2SE bars do not overlap, the difference between the two means is very unlikely to be chance. When two ±2SE bars overlap, the data do not show a difference.

37

Read at the top of the bright-light bar's error bar: 17.6 cm. Read at the bottom of the shade bar's error bar: 23.5 cm.

38

17.6 cm sits below 23.5 cm, so the two bars do not overlap. So the difference between the means is very unlikely to be chance.

39

The seedlings' genes were the same, and only the light differed. So the light level moved the stem length.

40

The bars make chance very unlikely, and the cuttings rule out the genes.

41

Back to the two pots and the tray of 24 cuttings of one plant.

42

Split 12 and 12 between acid soil and the parent's soil, everything else the same, and record the color of every plant.

43

If the acid-soil cuttings differ in color from the parent-soil cuttings, the soil changed the color, because the genes could not.

44

Had the trait been a height, the two means with their ±2SE bars would settle it the same way. That test settles the gardener's argument with the neighbor, whichever of them is right.

45

What you are expected to know Compare a measured trait between two conditions with the mean of each group and its ±2SE error bar.

46
Check q8 numeric entry

A grower grows cuttings of one geranium at two temperatures, 10 at 14 °C and 10 at 24 °C, and measures every stem. The 14 °C group has a mean stem height of 11.0 cm with a standard deviation s of 1.3 cm; the 24 °C group has a mean of 14.5 cm with s of 1.4 cm.

Calculate the bottom of the 24 °C group's ±2SE error bar.

Part 1. Calculate the standard error of the 14 °C group's mean.

Answer: 0.41 cm  (tolerance ±0.01)

Working
Divide the standard deviation by the square root of the group size:
SE=sn=1.310=0.41cm

Part 2. Calculate the top of the 14 °C group's ±2SE error bar.

Answer: 11.8 cm  (tolerance ±0.1)

Working
Add two standard errors to the mean:
top=mean+2SE=11.0+2(0.41)=11.8cm

Answer: 13.6 cm  (tolerance ±0.1)

Working
Write down the values in the question:
mean=14.5cm
s=1.4cm
n=10
Write down the equations:
SE=sn
bottom=mean−2SE
Substitute the values into the equations:
SE=1.410=0.44cm
bottom=14.5−2(0.44)=13.6cm
47
Check q9

The two geranium means, 11.0 cm at 14 °C and 14.5 cm at 24 °C, are drawn below with their ±2SE error bars. Bars overlap when the bottom of the higher bar sits below the top of the lower bar.

Two bars of mean stem height in centimeters for cuttings of one geranium grown at two temperatures: 11.0 cm at 14 degrees Celsius and 14.5 cm at 24 degrees Celsius, each with an error bar, from 10.2 to 11.8 cm and from 13.6 to 15.4 cm; gridlines every 2 cm; the legend reads: error bars represent ±2SE
Two bars of mean stem height in centimeters for cuttings of one geranium grown at two temperatures: 11.0 cm at 14 degrees Celsius and 14.5 cm at 24 degrees Celsius, each with an error bar, from 10.2 to 11.8 cm and from 13.6 to 15.4 cm; gridlines every 2 cm; the legend reads: error bars represent ±2SE

Do the two bars overlap?

  1. A. Yes
    Read at the bottom of the 24 °C bar's error bar: 13.6 cm.
    13.6 cm sits above the 14 °C bar's top, 11.8 cm.
  2. B. ✓ No

Why: Read at the top of the 14 °C bar's error bar: 11.8 cm.
Read at the bottom of the 24 °C bar's error bar: 13.6 cm.
11.8 cm sits below 13.6 cm.
So the two bars do not overlap.

48
Check q10

A grower grows cuttings of one geranium at two temperatures and records the mean stem height of each group. The bars below carry ±2SE error bars. Bars overlap when the bottom of the higher bar sits below the top of the lower bar.

Two bars of mean stem height in centimeters for cuttings of one geranium grown at two temperatures: 11.0 cm at 14 degrees Celsius and 14.5 cm at 24 degrees Celsius, each with an error bar, from 10.2 to 11.8 cm and from 13.6 to 15.4 cm; gridlines every 2 cm; the legend reads: error bars represent ±2SE
Two bars of mean stem height in centimeters for cuttings of one geranium grown at two temperatures: 11.0 cm at 14 degrees Celsius and 14.5 cm at 24 degrees Celsius, each with an error bar, from 10.2 to 11.8 cm and from 13.6 to 15.4 cm; gridlines every 2 cm; the legend reads: error bars represent ±2SE

What do the bars show about the difference between the two means?

  1. A. ✓ The difference is very unlikely to be chance
  2. B. The difference could be chance
    The bottom of the 24 °C bar's error bar, 13.6 cm, sits above the 14 °C bar's top, 11.8 cm.
    Bars that do not overlap make chance very unlikely.

Why: Read at the top of the 14 °C bar's error bar: 11.8 cm.
Read at the bottom of the 24 °C bar's error bar: 13.6 cm.
11.8 cm sits below 13.6 cm, so the bars do not overlap.
So the difference between the means is very unlikely to be chance.

49Quick quiz: read a whole result mixed practice

50
Check q11

A grower grows cuttings of one sage plant in two soils and records the mean leaf length of each group. The bars below carry ±2SE error bars. Bars overlap when the bottom of the higher bar sits below the top of the lower bar.

Two bars of mean leaf length in centimeters for cuttings of one sage plant grown in two soils: 3.2 cm in sandy soil and 3.5 cm in clay soil, each with an error bar, from 2.6 to 3.8 cm and from 2.9 to 4.1 cm; gridlines every 1 cm; the legend reads: error bars represent ±2SE
Two bars of mean leaf length in centimeters for cuttings of one sage plant grown in two soils: 3.2 cm in sandy soil and 3.5 cm in clay soil, each with an error bar, from 2.6 to 3.8 cm and from 2.9 to 4.1 cm; gridlines every 1 cm; the legend reads: error bars represent ±2SE

What do the bars show about the difference between the two means?

  1. A. The difference is very unlikely to be chance
    Read at the bottom of the clay-soil bar's error bar: it sits below the top of the sandy-soil bar's error bar.
    Overlapping ±2SE bars mean the gap could be chance.
  2. B. ✓ The difference could be chance

Why: Read at the bottom of the higher bar's error bar, the clay-soil bar: it sits below the top of the sandy-soil bar's error bar.
So the two bars overlap.
Overlapping ±2SE bars mean the data do not show a difference, so the gap could be chance.

51
Check q12

A grower grows plantlets of one banana plant in two light levels and records the mean height of each group. The bars below carry ±2SE error bars. Bars overlap when the bottom of the higher bar sits below the top of the lower bar.

Two bars of mean plant height in centimeters for plantlets of one banana plant grown in two light levels: 52.0 cm in dim light and 62.0 cm in bright light, each with an error bar, from 49.0 to 55.0 cm and from 59.0 to 65.0 cm; gridlines every 5 cm; the legend reads: error bars represent ±2SE
Two bars of mean plant height in centimeters for plantlets of one banana plant grown in two light levels: 52.0 cm in dim light and 62.0 cm in bright light, each with an error bar, from 49.0 to 55.0 cm and from 59.0 to 65.0 cm; gridlines every 5 cm; the legend reads: error bars represent ±2SE

What do the bars show about the difference between the two means?

  1. A. ✓ The difference is very unlikely to be chance
  2. B. The difference could be chance
    Read at the bottom of the bright-light bar's error bar: it sits above the top of the dim-light bar's error bar.
    Bars that do not overlap make chance very unlikely.

Why: Read at the bottom of the higher bar's error bar, the bright-light bar: it sits above the top of the dim-light bar's error bar.
So the two bars do not overlap.
So the difference between the means is very unlikely to be chance.

52
Check q13

A grower gives cuttings of one fig tree two amounts of water and records the mean fruit mass of each group. The bars below carry ±2SE error bars. Bars overlap when the bottom of the higher bar sits below the top of the lower bar.

Two bars of mean fruit mass in grams for cuttings of one fig tree grown with two amounts of water: 38.0 g with the usual watering and 41.0 g with double watering, each with an error bar, from 34.0 to 42.0 g and from 36.0 to 46.0 g; gridlines every 5 g; the legend reads: error bars represent ±2SE
Two bars of mean fruit mass in grams for cuttings of one fig tree grown with two amounts of water: 38.0 g with the usual watering and 41.0 g with double watering, each with an error bar, from 34.0 to 42.0 g and from 36.0 to 46.0 g; gridlines every 5 g; the legend reads: error bars represent ±2SE

What do the bars show about the difference between the two means?

  1. A. The difference is very unlikely to be chance
    Read at the bottom of the double-watering bar's error bar: it sits below the top of the usual-watering bar's error bar.
    Overlapping ±2SE bars mean the gap could be chance.
  2. B. ✓ The difference could be chance

Why: Read at the bottom of the higher bar's error bar, the double-watering bar: it sits below the top of the usual-watering bar's error bar.
So the two bars overlap.
Overlapping ±2SE bars mean the data do not show a difference, so the gap could be chance.

53
Check q14

A grower grows plantlets of one orchid at two humidities and records the mean flower diameter of each group. The bars below carry ±2SE error bars. Bars overlap when the bottom of the higher bar sits below the top of the lower bar.

Two bars of mean flower diameter in centimeters for plantlets of one orchid grown at two humidities: 6.0 cm at low humidity and 8.0 cm at high humidity, each with an error bar, from 5.5 to 6.5 cm and from 7.5 to 8.5 cm; gridlines every 1 cm; the legend reads: error bars represent ±2SE
Two bars of mean flower diameter in centimeters for plantlets of one orchid grown at two humidities: 6.0 cm at low humidity and 8.0 cm at high humidity, each with an error bar, from 5.5 to 6.5 cm and from 7.5 to 8.5 cm; gridlines every 1 cm; the legend reads: error bars represent ±2SE

What do the bars show about the difference between the two means?

  1. A. ✓ The difference is very unlikely to be chance
  2. B. The difference could be chance
    Read at the bottom of the high-humidity bar's error bar: it sits above the top of the low-humidity bar's error bar.
    Bars that do not overlap make chance very unlikely.

Why: Read at the bottom of the higher bar's error bar, the high-humidity bar: it sits above the top of the low-humidity bar's error bar.
So the two bars do not overlap.
So the difference between the means is very unlikely to be chance.

54
Check q15

A grower grows cuttings of one blueberry bush in two soils and records the mean number of berries per branch in each group. The bars below carry ±2SE error bars. Bars overlap when the bottom of the higher bar sits below the top of the lower bar.

Two bars of mean number of berries per branch for cuttings of one blueberry bush grown in two soils: 21.0 in the usual soil and 23.0 in soil with added compost, each with an error bar, from 18.0 to 24.0 berries and from 20.0 to 26.0 berries; gridlines every 3; the legend reads: error bars represent ±2SE
Two bars of mean number of berries per branch for cuttings of one blueberry bush grown in two soils: 21.0 in the usual soil and 23.0 in soil with added compost, each with an error bar, from 18.0 to 24.0 berries and from 20.0 to 26.0 berries; gridlines every 3; the legend reads: error bars represent ±2SE

What do the bars show about the difference between the two means?

  1. A. The difference is very unlikely to be chance
    Read at the bottom of the compost bar's error bar: it sits below the top of the usual-soil bar's error bar.
    Overlapping ±2SE bars mean the gap could be chance.
  2. B. ✓ The difference could be chance

Why: Read at the bottom of the higher bar's error bar, the compost bar: it sits below the top of the usual-soil bar's error bar.
So the two bars overlap.
Overlapping ±2SE bars mean the data do not show a difference, so the gap could be chance.

55
Check q16

A grower grew cuttings of one mint plant in two soils, 15 in each, and recorded the mean root length of each group. The bars below carry ±2SE error bars.

Two bars of mean root length in centimeters for cuttings of one mint plant: 6.1 cm in sandy soil and 6.6 cm in loam, each with an error bar, from 5.2 to 7.0 cm and from 5.8 to 7.4 cm; gridlines every 1 cm; the legend reads: error bars represent ±2SE
Two bars of mean root length in centimeters for cuttings of one mint plant: 6.1 cm in sandy soil and 6.6 cm in loam, each with an error bar, from 5.2 to 7.0 cm and from 5.8 to 7.4 cm; gridlines every 1 cm; the legend reads: error bars represent ±2SE

What do the bars show about the two soils?

  1. A. The two soils differ
    Two means almost always differ a little by chance.
    The error bars overlap, so the gap could be chance.
  2. B. The loam changed the plants' genes
    A taller bar is not evidence of a change in genes.
    The bars overlap in any case.
  3. C. No comparison can be made
    A comparison is exactly what the two bars allow.
    The bars are compared, and they overlap.
  4. D. ✓ The data do not show a difference

Why: Read at the bottom of the higher bar's error bar, the loam bar: it sits below the top of the sandy-soil bar's error bar.
So the two bars overlap.
Overlapping ±2SE bars mean the data do not show a difference between the two soils.

56
Practice writing an answer

A grower grew cuttings of one mint plant in two soils, 15 in each, and recorded the mean root length of each group. The bars below carry ±2SE error bars. The data do not show a difference between the two soils.

Two bars of mean root length in centimeters for cuttings of one mint plant: 6.1 cm in sandy soil and 6.6 cm in loam, each with an error bar, from 5.2 to 7.0 cm and from 5.8 to 7.4 cm; gridlines every 1 cm; the legend reads: error bars represent ±2SE
Two bars of mean root length in centimeters for cuttings of one mint plant: 6.1 cm in sandy soil and 6.6 cm in loam, each with an error bar, from 5.2 to 7.0 cm and from 5.8 to 7.4 cm; gridlines every 1 cm; the legend reads: error bars represent ±2SE

(a) Explain how the error bars lead to that verdict. (1 pt)

Model answer Each error bar represents ±2SE around its mean, the range the true mean is likely to lie in.
Read at the bottom of the loam bar's error bar: it sits below the top of the sandy-soil bar's error bar, so the two bars overlap.
So the two true means could be the same value, and the gap between the two sample means could be chance.
So the data do not show a difference between the two soils.
Rubric
  • Award 1 point for: the ±2SE bars overlap, so the two true means could be the same and the gap between the sample means could be chance.

57Mixed practice mixed practice

58
Check q17

A student sows marigold seeds from one packet: 20 pots in soil at pH 5 and 20 pots in soil at pH 7, with the same light, water and temperature. More of the pH 5 plants have orange flowers.

What is wrong with the design?

  1. A. Nothing is wrong with the design
    One thing was left free to differ besides the soil: the seeds' genotypes.
  2. B. ✓ The seeds from one packet carry different genotypes
  3. C. Twenty pots in each group are too few for any fair comparison of the soils
    Twenty pots in each group is a fair number; the fault is in the material, which carries different genotypes.
  4. D. The two soils should have differed in more than their acidity alone
    A test changes one condition at a time; changing more would make the result harder to read.

Why: Seeds are different individuals with different genotypes.
A difference in flower color between the groups could then be genes or soil, and the design cannot tell which.
Cuttings of one plant would fix the genes.

59
Check q18

A grower splits cuttings of one tomato plant: 12 in a warm, bright greenhouse and 12 in a cool, dim shed. The greenhouse plants grow taller.

What can be concluded about the cause of the difference in height?

  1. A. The warmth of the greenhouse made the greenhouse plants grow taller
    The light changed as well as the warmth, so the warmth cannot be singled out.
  2. B. The brighter light of the greenhouse made the greenhouse plants grow taller
    The warmth changed as well as the light, so the light cannot be singled out.
  3. C. ✓ Neither the warmth nor the light can be credited with the difference
  4. D. The greenhouse changed the genes of the plants that grew in it
    Cuttings of one plant share one genotype and no condition rewrites DNA; the plants differ in what their cells did.

Why: A test credits a condition only when it is the one thing that differs.
Here warmth and light both differed between the groups.
So the height difference cannot be assigned to either.

60
Check q19

A student plants all 24 cuttings of a hydrangea in soil at pH 5, and all of them flower blue. The student concludes that acid soil makes hydrangeas blue.

What is missing from the test?

  1. A. ✓ A second group of cuttings in a different soil
  2. B. A larger number of cuttings in the acid soil
    More cuttings in the same soil still give nothing to compare with; every plant had the same treatment.
  3. C. Seeds in place of the cuttings
    Seeds would let the genes differ and spoil the test; the material is right, the comparison is missing.
  4. D. A measurement of each plant's height at flowering
    Height is not the trait in question; the color needs a comparison group.

Why: With every cutting in acid soil there is no control: nothing shows what color the same cuttings give without the acid.
A second group in another soil, everything else the same, is what the comparison needs.

61
Check q20

A grower grows cuttings of one plant at two temperatures, 15 at 15 °C and 15 at 25 °C, and records the mean height of each group. The bars below carry ±2SE error bars.

Two bars of mean plant height in centimeters for cuttings of one plant grown at two temperatures: 21.0 cm at 15 degrees Celsius and 27.5 cm at 25 degrees Celsius, each with an error bar, from 19.5 to 22.5 cm and from 25.7 to 29.3 cm; gridlines every 5 cm; the legend reads: error bars represent ±2SE
Two bars of mean plant height in centimeters for cuttings of one plant grown at two temperatures: 21.0 cm at 15 degrees Celsius and 27.5 cm at 25 degrees Celsius, each with an error bar, from 19.5 to 22.5 cm and from 25.7 to 29.3 cm; gridlines every 5 cm; the legend reads: error bars represent ±2SE

What do the bars show?

  1. A. The data do not show a difference
    Read at the bottom of the 25 °C bar's error bar: it sits above the top of the 15 °C bar.
    Bars that do not overlap show a difference.
  2. B. The difference is chance
    Read at the bottom of the 25 °C bar's error bar: it sits above the top of the 15 °C bar.
    Bars that do not overlap make chance very unlikely.
  3. C. The warmer plants changed their DNA
    A taller bar is not evidence of a change in DNA.
    The warmth changed what the cells did with the same genes.
  4. D. ✓ The difference is very unlikely to be chance

Why: Read at the bottom of the 25 °C bar's error bar: it sits above the top of the 15 °C bar.
So the bars do not overlap, and the difference is very unlikely to be chance.
Only the temperature differed between the cuttings, so the temperature caused it.

62
Check q21

Four traits are measured or counted in tests on genetically identical material.

Which trait is compared through a mean with ±2SE error bars?

  1. A. ✓ The mass of fruit from tomato plants in two light levels
  2. B. The sex of turtle hatchlings from warm sand and from cool sand
    Male and female are counted, not measured.
  3. C. The presence or absence of a crest on Daphnia in two waters
    Crest or no crest is a count of individuals in two classes.
  4. D. The flower color, blue or pink, of plants in two soils
    Blue or pink is a class each plant falls into.
    The plants are counted into the two classes, not measured.

Why: A mass is a measurement.
So each group's fruit masses are averaged into a mean, and the mean carries a ±2SE error bar.
Colors, sexes and crests are counted into classes, not measured.

63
Check q22

A grower compares the mean height of cuttings of one lavender plant in two soils. The two ±2SE error bars overlap. A student says: 'The bars overlap, so the two soils give the same height.'

Is the student right?

  1. A. Yes
    Overlapping bars mean the data do not show a difference.
    That is not the same as showing the two heights are equal.
  2. B. ✓ No

Why: Each ±2SE bar is the range its true mean is likely to lie in.
Overlapping bars mean the two true means could be the same, or could differ a little.
So the data do not show a difference; they do not show that the heights are the same.

64
Practice writing an answer

A grower has 30 tomato plants from one clonal line, so every plant has the same DNA. She wants to know whether the sweetness of the fruit depends on the hours of light the plants get each day. Her plants have always had 12 hours of light a day. She splits them into two groups of 15, keeps one group at 12 hours of light and gives the other 16 hours from lamps on timers, holds water, temperature and soil the same for every plant, and measures the sugar in the fruit of every plant in grams per 100 grams. Her results are drawn below; each error bar represents ±2SE.

Two bars of mean sugar content of tomato fruit in grams per 100 grams for one clonal line of plants: 3.8 under 12 hours of light a day and 5.0 under 16 hours, each with an error bar, from 3.5 to 4.1 grams per 100 grams and from 4.6 to 5.4 grams per 100 grams; gridlines every 0.5 grams per 100 grams; the legend reads: error bars represent ±2SE
Two bars of mean sugar content of tomato fruit in grams per 100 grams for one clonal line of plants: 3.8 under 12 hours of light a day and 5.0 under 16 hours, each with an error bar, from 3.5 to 4.1 grams per 100 grams and from 4.6 to 5.4 grams per 100 grams; gridlines every 0.5 grams per 100 grams; the legend reads: error bars represent ±2SE

(a) Justify the grower's decision to hold water, temperature and soil the same for both groups. (1 pt)

Model answer With water, temperature and soil the same for every plant, the hours of light are the only condition that differs between the 12-hour plants and the 16-hour plants.
So a difference in sweetness can be credited to the light.
If the water had differed as well, neither the light nor the water could be credited with the difference.
Rubric
  • Award 1 point for: everything but the light held the same, so the light is the only condition that differs between the groups and a difference in sweetness is credited to it.

Slip Saying the conditions are held the same so that the plants grow to the same size. The point is that only one condition may differ, so that a difference can be credited to it.

(b) Predict what the two means and their ±2SE error bars would look like if the hours of light had no effect on sweetness. (1 pt)

Model answer The two mean values would be about the same.
The ±2SE error bars of the 12-hour group and the 16-hour group would overlap.
So the data would show no difference between 12 and 16 hours of light.
Rubric
  • Award 1 point for: the two ±2SE error bars would overlap, so the data would show no difference. Accept with or without: the two means about the same.

Slip Predicting that the two means would be exactly equal. Two means almost always differ a little by chance; the test is whether the bars overlap.

(c) Evaluate the claim that the hours of light changed the sweetness of the fruit, using the results drawn above. (1 pt)

Model answer The mean under 16 hours of light, about 5.0 g per 100 g, is higher than the mean under 12 hours, about 3.8 g per 100 g.
Read at the bottom of the 16-hour bar's error bar: it sits above the top of the 12-hour bar, so the ±2SE bars do not overlap.
So the difference is very unlikely to be chance.
The plants share one genotype and only the light differed.
So the claim is supported.
Rubric
  • Award 1 point for: the judgement (the claim is supported) AND the ground (the ±2SE bars do not overlap, so the difference is very unlikely to be chance). Accept with or without: the plants share one genotype and only the light differed, so the light is credited.

Slip Arguing only that 5.0 is bigger than 3.8. Two means can differ by chance; the error bars are what make chance very unlikely.

APBIO-U05-P55 Practice questions: Topic 5.5

Topic 5.5 · Environmental Effects on Phenotype · 9 MCQ · 2 FRQ · for APBIO-U05-T55

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. The first free-response question walks you through one investigation one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a question shows means with error bars, read the legend first: every error bar here represents ±2SE.

Video: Watch first: Environmental effects on phenotype, summed up

The same genes, different looks; conditions change which genes the cells use, not the genes; phenotypic plasticity; the six named cases; how clones in two conditions separate genes from environment.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T55-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U05-T55-summary.mp4

Q1 P55-q01

A nine-banded armadillo always gives birth to four young with the same DNA, because all four grow from one fertilized egg. A wildlife center rears one set of four. It feeds two of them twice as much food as the other two, in one enclosure at the same temperature. The two fed more grow much heavier.

What do the four young show?

  1. A. The two young fed less inherited alleles for a small body
    The four young grew from one fertilized egg, so all four carry the same alleles.
    No pair inherited alleles the other pair lacks.
  2. B. The smaller ration changed the DNA of the two young that ate it
    A difference in body mass is not evidence of a change in DNA.
    The ration changed the condition the young grew in, not their genes.
  3. C. ✓ The same genes gave two body masses, one on each ration
  4. D. Body mass in armadillos is set by the food alone
    The genes set the range of body mass an armadillo can reach, and the food decides where in that range each animal falls: both play a part.

Why: The four young grew from one fertilized egg, so they share one genotype.
Two ate twice as much as the other two, in the same enclosure.
So one condition, the food, differed, and the same genes gave two body masses.

Q2 P55-q02

A sea snail moved from water without crabs into water where crabs that eat snails live grows its new shell thicker. Its DNA is the same in both waters.

What does the presence of the crabs change inside the snail's cells?

  1. A. The sequence of the DNA in the snail's shell-building genes
    The snail's DNA is the same in both waters; the water with crabs is a condition, and a condition does not rewrite DNA.
  2. B. Which alleles for shell thickness the snail's cells carry
    The alleles a snail inherited were fixed at fertilization and the crabs cannot swap them.
  3. C. The number of chromosomes in each cell of the snail
    Every body cell keeps the same chromosome number whatever lives in the water.
  4. D. ✓ How strongly the cells turn on their shell-building genes

Why: The crabs' presence is a condition outside the snail that changes which genes its cells use and how much; the cells build more shell, and the DNA stays the same.

Q3 P55-q03

Which of the following changes to a ram can be passed on to a lamb the ram fathers?

  1. A. The ram’s fleece grows thick over a cold winter
    The cold changed what the skin cells did with their genes.
    The DNA in the ram’s gametes did not change, so the thick fleece is not passed on.
  2. B. ✓ A change in the DNA of one of the ram’s sperm cells gives one gene a new version
  3. C. A change in the DNA of one of the ram’s skin cells gives one gene a new version
    A skin cell takes no part in fertilization.
    Only the DNA in a gamete passes to the lamb, and this change is in a skin cell.
  4. D. The ram’s leg muscles grow large from climbing steep pasture
    Climbing changed what the muscle cells did with their genes.
    The DNA in the ram’s gametes did not change, so the large muscles are not passed on.

Why: A lamb inherits only the DNA in the gametes that formed it.
A change in one sperm cell’s DNA is in a gamete, so that lamb inherits it.
The fleece, the muscles and the skin cell’s DNA changed in body cells, which take no part in fertilization.

Q4 P55-q04

Marbled crayfish are all one clonal line: every marbled crayfish has the same DNA.

Which of the following observations shows phenotypic plasticity in marbled crayfish?

  1. A. Marbled crayfish kept in one tank at one temperature all grow to about the same size
    One genotype in one condition gave one phenotype.
    Plasticity is one genotype producing different phenotypes under different conditions, and here the conditions did not differ.
  2. B. Marbled crayfish and a second crayfish species kept in one tank grow to different sizes
    Two species are two genotypes, so a difference between them under one condition is a genetic difference.
  3. C. A marbled crayfish grows larger as it gets older
    Growing with age is not a different phenotype under a different condition; every animal grows.
    Plasticity needs two conditions giving two phenotypes from one genotype.
  4. D. ✓ Marbled crayfish kept in warm water grow larger than those kept in cool water

Why: Every marbled crayfish has the same DNA, so the line is one genotype.
Warm water and cool water are two conditions.
One genotype gave two sizes in two conditions: phenotypic plasticity.

Q5 P55-q05

A nursery planted cuttings of one hydrangea in two beds. The soil in the first bed is acid, and its plants flower blue; the soil in the second bed is alkaline, and its plants flower pink. A gardener takes cuttings from the pink-flowered plants in the alkaline bed and plants them in her own garden, where the soil is acid.

Predict the flower color of the gardener's plants, and give the reason.

  1. A. Pink, because a cutting keeps the alleles of the plant it was cut from
    Every cutting of the one hydrangea carries the same alleles; the alkaline bed's plants flowered pink because of their soil, not because of a pink allele.
  2. B. ✓ Blue, because the soil decides the color and the gardener's soil is acid
  3. C. Blue, because the acid soil changes the plants' color alleles
    Soil does not rewrite a plant's alleles; the same alleles gave blue in the acid bed and pink in the alkaline bed.
  4. D. Pink, because a plant keeps the color of the bed it came from, whatever soil it is moved to
    Nothing of the old bed travels with the cutting; the color follows the soil the plant grows in now, and the gardener's soil is acid.

Why: All the cuttings share one genotype, so the pink of the alkaline bed came from its soil, not from the plants.
In the gardener's acid soil the same genes give blue flowers, as they did in the acid bed.

Q6 P55-q06

A grower tests whether the hours of light a day change the leaf length of spider plants. The parent plant has always had 8 hours of light a day. Her design is in the table below.

The grower's design: two groups of twelve spider-plant plantlets from one plant, with the hours of light each group gets; water, soil and temperature are the same for every plant.
The grower's design: two groups of twelve spider-plant plantlets from one plant, with the hours of light each group gets; water, soil and temperature are the same for every plant.

Which group is the control, and why?

  1. A. ✓ Group A, because it grows under the parent plant's usual light
  2. B. Group B, because it gets the most light
    The control is not the group with the most of the condition; it is the group kept at the usual condition to compare against.
  3. C. Both groups, because both are plantlets of one plant
    The control is one group, the one that lacks the change under test; plantlets of one plant make the genes the same but do not make both groups the control.
  4. D. Neither group, because a control must get no light at all
    A control does not remove the condition entirely; it keeps the usual condition so that the changed group has something to be compared against.

Why: The control is the group kept at the usual condition, here the parent plant's 8 hours of light a day, so any difference in group B can be credited to the longer light.

Q7 P55-q07

A student grows 16 cuttings of one blackcurrant bush in shade and 16 in sun and measures every stem. The sun group has a mean stem length of 30.0 cm with a standard deviation s of 4.0 cm.

Which of the following is the ±2SE error bar of the sun group’s mean?

  1. A. 29.0 cm to 31.0 cm
    29.0 cm to 31.0 cm is the mean ± 1SE.
    SE=4.016=1.0cm, and the bar is the mean ± 2SE: 2.0 cm each side.
  2. B. ✓ 28.0 cm to 32.0 cm
  3. C. 26.0 cm to 34.0 cm
    26.0 cm to 34.0 cm is the mean ± s.
    The bar uses the standard error, SE=sn=1.0cm, and is the mean ± 2SE.
  4. D. 22.0 cm to 38.0 cm
    22.0 cm to 38.0 cm is the mean ± 2s.
    The bar is the mean ± 2SE, and SE=sn=4.016=1.0cm.

Why: The error bar is the mean ± 2SE, and SE is s divided by the square root of n.
With s = 4.0 cm and n = 16, 2SE = 2.0 cm, so the bar reaches from 28.0 cm to 32.0 cm.

Q8 P55-q08

A grower grows cuttings of one plant in two conditions with everything else the same, and measures a trait in each group. The two means carry ±2SE error bars, and the bars stay clear of each other.

What can be concluded?

  1. A. The two groups must have carried different alleles
    Cuttings of one plant carry the same alleles; that is why a difference between the groups can be credited to the condition and not the genes.
  2. B. The difference between the two means is exactly its true size
    An error bar shows the range within which each mean is likely to sit, not the exact size of the difference.
  3. C. ✓ The difference is unlikely to be chance, so the condition is credited
  4. D. Nothing about the condition
    Two ±2SE bars that stay clear of each other do say something about the two groups together: the difference between their means is unlikely to be chance.

Why: Bars of ±2SE that do not overlap make the difference between the means unlikely to be chance; with one genotype and one condition changed, the condition is the cause left.

Q9 P55-q09

A student says that because the environment can change a phenotype, an organism's genotype must change over its lifetime too.

What is wrong with the student's statement?

  1. A. ✓ The set of genes in use changes, not the genes themselves; the genotype stays fixed
  2. B. Nothing; when the cells change which genes they use, the genotype must change with them
    A phenotype can change while the DNA does not; the cells use the same genes differently.
  3. C. The environment changes the genotype only in the body cells, and the gametes keep the original
    A condition does not rewrite the DNA of any cell, body cell or gamete.
    It changes which genes the body cells use; their genotype stays as it was.
  4. D. The genotype does change over a lifetime, and the changed genotype is passed to the offspring
    A condition changes what the cells do with their genes, not the genes.
    Nothing in the DNA changed, so no changed genotype exists to pass on.

Why: A condition outside the organism changes which genes its cells use and how much, so the phenotype changes while the DNA does not; the genotype stays as it was fixed at fertilization.

FRQ 1 P55-frq1 · Scientific Investigation scaffolded

Does compost in the soil change how many flower heads a chrysanthemum makes? A student tests this with 20 plants she rooted from cuttings of one chrysanthemum. She pots 10 in the ordinary soil the parent plant grows in and 10 in the same soil with compost mixed in, keeps every pot in one greenhouse with the same light, water and temperature, and after twelve weeks counts the flower heads on each plant and finds each group’s mean. The means are drawn below; each error bar represents ±2SE.

Mean number of flower heads per plant for chrysanthemum cuttings from one plant grown in ordinary soil and in soil with compost, 10 plants per group, with the ends of each error bar printed beside it; error bars represent ±2SE.
Mean number of flower heads per plant for chrysanthemum cuttings from one plant grown in ordinary soil and in soil with compost, 10 plants per group, with the ends of each error bar printed beside it; error bars represent ±2SE.

(a) Identify the independent variable in the student’s test. (1 pt)

Frame The independent variable is …, because it is the condition the student …

Hint Which one condition did the student deliberately change between the two groups?

Model answer The independent variable is the soil, ordinary or with compost mixed in, because it is the one condition the student deliberately changed.
Rubric
  • Award 1 point for: the soil (ordinary or with compost) as the independent variable.

Slip Naming the number of flower heads. That is what was counted, the dependent variable.

(b) Identify the dependent variable and one control variable. (1 pt)

Frame The dependent variable is …, the quantity counted; one control variable is …, kept the same for every plant.

Hint What did the student count, and which conditions did she hold the same in both groups?

Model answer The dependent variable is the number of flower heads per plant, the quantity counted; one control variable is the light, kept the same for every plant.
The water, the temperature and the twelve weeks of growth are control variables too, and any one of them counts.
Rubric
  • Award 1 point for: flower heads per plant as the dependent variable and any one condition held the same for every plant (light, water, temperature, the growing time) as a control variable.

Slip Giving the compost as the dependent variable. The compost is what the student changed; the flower heads are what she counted.

(c) Identify the control group and describe what it shows. (1 pt)

Frame The control group is the … plants, and it shows …

Hint Which group lacks the change under test, and what does its result let the student compare against?

Model answer The control group is the 10 plants in ordinary soil.
It shows how many flower heads the cuttings make without compost, so the compost group’s result can be compared against it and any difference credited to the compost.
Rubric
  • Award 1 point for: the ordinary-soil group as the control, giving the count without compost to compare against.

Slip Calling the compost group the control. The control lacks the change under test.

(d) Justify the student’s use of cuttings from one chrysanthemum rather than plants grown from a packet of seeds. (1 pt)

Frame Cuttings from one plant all have the same …, so a difference between the groups cannot come from …

Hint What would plants from many seeds differ in that cuttings of one plant do not?

Model answer Cuttings from one plant all have the same DNA, so both groups share one genotype and a difference in flower heads between them cannot come from their genes.
Plants from a packet of seeds would carry different genotypes, and a difference could be genes or compost.
Rubric
  • Award 1 point for: cuttings of one plant share one genotype, so a difference between the groups cannot be genetic and is credited to the soil.

Slip Saying cuttings are used because they flower sooner. The point is that their genes are the same.

(e) Evaluate the claim that the compost raised the flower-head count, using the error bars drawn above. (1 pt)

Frame The mean was … in ordinary soil and … in soil with compost; the bottom of the compost bar’s error bar is at … and the top of the ordinary-soil bar’s error bar is at …, so the bars …; so the claim is …

Hint Read the bottom of the higher bar’s error bar and the top of the lower bar’s error bar, and ask whether the two bars overlap.

Model answer The mean was 12.0 flower heads per plant in ordinary soil and 17.5 with compost.
Read at the bottom of the compost bar’s error bar: 15.9. Read at the top of the ordinary-soil bar’s: 13.4.
13.4 sits below 15.9, so the ±2SE bars do not overlap, and the difference is very unlikely to be chance.
The plants share one genotype, and only the soil differed.
So the claim is supported: the compost raised the flower-head count.
Rubric
  • Award 1 point for: the judgement (the claim is supported) AND the ground (the ±2SE bars do not overlap, so the difference is very unlikely to be chance). Accept with or without: one genotype and one condition changed, so the compost is credited.

Slip Arguing only that 17.5 is larger than 12.0. Two means can differ by chance; the non-overlapping bars are what make chance very unlikely.

FRQ 2 P55-frq2 · Scientific Investigation

A grower has 30 rose plants grown from cuttings of one rose. He wants to know whether the brightness of the light changes the size of the flowers. He grows 15 plants in bright light and 15 in dim light, with the same water, soil and temperature for every plant, and at flowering he measures the diameter of the first flower on each plant and finds each group's mean. The means are drawn below; each error bar represents ±2SE.

Mean flower diameter in centimeters of rose plants from one clonal line grown in bright light and in dim light, 15 plants per group; error bars represent ±2SE.
Mean flower diameter in centimeters of rose plants from one clonal line grown in bright light and in dim light, 15 plants per group; error bars represent ±2SE.

(a) Identify the independent variable and the dependent variable. (1 pt)

Model answer The independent variable is the brightness of the light, bright or dim, the one condition changed.
The dependent variable is the flower diameter, the quantity measured.
Rubric
  • Award 1 point for: light brightness as the independent variable and flower diameter as the dependent variable.

Slip Swapping the two. The grower changed the light and measured the flowers.

(b) Describe the feature of the grower's design that lets a difference between the groups be credited to the light rather than to the plants' genes. (1 pt)

Model answer Every plant is a cutting of one rose, so all 30 plants share one genotype.
A difference between the bright-light group and the dim-light group therefore cannot come from their genes, and with water, soil and temperature the same the light is the only condition that differs.
Rubric
  • Award 1 point for: the plants are cuttings of one rose, so both groups share one genotype and a difference cannot be genetic; with everything else the same the light is the only difference.

Slip Naming the equal group sizes as the feature. Fifteen in each group helps the comparison, but it is the shared genotype that rules out the genes.

(c) Predict what the two means and their ±2SE error bars would look like if dim light gave larger flowers. (1 pt)

Model answer The dim-light mean would sit above the bright-light mean.
The two ±2SE error bars would not overlap: the bottom of the dim-light bar would sit above the top of the bright-light bar.
Then the difference would be unlikely to be chance, and with one genotype and one condition changed the light would be credited.
Rubric
  • Award 1 point for: the dim-light mean higher and the two ±2SE bars not overlapping.

Slip Describing the means farther apart with the bars still overlapping. Overlapping bars leave the difference open however far apart the means look; the bars must clear each other.

(d) Evaluate the claim that dim light gives larger flowers, using the error bars drawn above. (1 pt)

Model answer The mean flower diameter was 7.2 cm in bright light and 7.6 cm in dim light.
The error bars represent ±2SE.
The bright-light bar runs from 6.6 to 7.8 cm and the dim-light bar from 7.0 to 8.2 cm.
The two bars overlap, so the difference between the mean values could be chance.
So the claim is not supported by these data: they do not show a difference between the two light levels.
Rubric
  • Award 1 point for: the judgement (the claim is not supported) AND the ground (the ±2SE bars overlap, so the difference between the means could be chance).

Slip Arguing that 7.6 is larger than 7.2, so dim light gives bigger flowers. Two means can differ by chance; the overlapping bars mean chance has not been ruled out.

APBIO-U05-T55 End-of-topic test: Environmental Effects on Phenotype

Topic 5.5 · Environmental Effects on Phenotype · 16 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. Where a question shows means with error bars, read the legend first: every error bar here represents ±2SE.

Q1 T55-q01

A water buttercup grows with part of its stem under water and part in the air. On the one plant, the leaves under water are cut into fine threads and the leaves in the air are broad and flat.

Which of the following differs between the leaves under water and the leaves in the air?

  1. A. The alleles for leaf shape that each leaf inherited
    Every leaf on one plant grew by mitosis from one zygote.
    So every leaf carries the same alleles, and the shapes differ for another reason.
  2. B. ✓ The condition each leaf grew in
  3. C. Both the alleles each leaf inherited and the condition each leaf grew in
    The leaves grew in different conditions, water and air.
    Their alleles are the same: every cell of one plant carries the same DNA.
  4. D. The DNA in each leaf’s cells, rewritten by the water
    A difference in leaf shape is not evidence of a change in DNA.
    The water changed the condition the leaf grew in, not its genes.

Why: All the leaves belong to one plant, so every leaf cell carries the same DNA.
The leaves under water grew in one condition and the leaves in the air in another.
So the one thing that differs between the two kinds of leaf is the condition each grew in.

Q2 T55-q02

Two potato plants grown from pieces of one tuber are drawn below. One grew at 18 °C and formed large tubers; the other grew at 28 °C and formed small tubers. Light, water and soil were the same for both. A student says the plant at 28 °C must carry different alleles from the plant at 18 °C.

Two potato plants grown from pieces of one tuber: the plant at 18 °C with large tubers and the plant at 28 °C with small tubers.
Two potato plants grown from pieces of one tuber: the plant at 18 °C with large tubers and the plant at 28 °C with small tubers.

Which statement about the student's claim is correct?

  1. A. The student is right: only plants with different alleles can give tubers of different sizes
    Both plants grew from pieces of one tuber, and a piece of a tuber carries the parent plant's DNA unchanged, so the same alleles gave two tuber sizes.
  2. B. The student is right: the warmth changed the alleles of the plant at 28 °C
    A difference in tuber size is not evidence that any allele changed.
    Warmth is a condition outside the plant.
    A condition changes which genes the cells use, not the DNA.
  3. C. The student is wrong: tuber size is set by chance alone, not by alleles or by temperature
    Only the temperature differed between the two plants, and 18 °C gave large tubers while 28 °C gave small ones, so the temperature, not chance, made the difference.
  4. D. ✓ The student is wrong: both plants carry the same alleles, and the temperature set the size

Why: Pieces of one tuber share one genotype, so the two plants carry the same alleles.
Only the temperature differed between them.
So the temperature, not a difference in alleles, set the tuber size, and the student is wrong.

Q3 T55-q05

A racehorse trained hard for years has much larger leg muscles than its untrained brother. The horse then sires a foal.

Predict the leg muscles of the foal at birth compared with a foal of the untrained brother, and give the reason.

  1. A. Larger, because a parent's acquired change is passed to the foal
    A change a condition makes in the body's cells is not passed to offspring.
    The foal inherits the horse's genes, not his muscles.
  2. B. Larger, because the training changed the alleles in his gametes
    The training changed which genes the muscle cells used and how much.
    It did not rewrite the DNA in the horse's gametes.
  3. C. Ordinary, because training cannot change which genes a muscle cell uses, so nothing in the horse changed
    Training did change something in the horse: his muscle cells used some genes more, and the muscles grew.
    Only the DNA in his gametes stayed as it was.
  4. D. ✓ Ordinary, because the change in the horse's muscle cells left the DNA in his gametes as it was

Why: The large muscles are a change a condition made in the horse's muscle cells.
The DNA in his gametes is unchanged, so the foal inherits the genes and not the trained muscles.

Q4 T55-q06

A gardener grows cuttings of one grass plant in salty soil and in ordinary soil. The plants in salty soil grow short, with thick leaves. A student writes: the salty soil switched the leaf genes off for good, so cuttings taken from the short plants will stay short in ordinary soil.

What is wrong with the student's statement?

  1. A. Nothing; a condition that changes gene use changes it permanently, for the life of the plant
    The salt changed which genes the leaf cells used for as long as the leaves grew in salty soil.
    Nothing was switched off for good.
  2. B. ✓ The salt changed the leaf cells’ gene use only while they grew in salty soil, not for good
  3. C. The salt rewrote the leaf alleles, so the change is inherited
    An allele is a version of a gene written in the DNA, and salt does not rewrite DNA; the salt changed which genes the leaf cells used, nothing heritable.
  4. D. Leaf form is set by alleles alone, so the salt had no effect
    Cuttings with the same alleles came out different in the two soils.
    So the salt did have an effect: it changed which genes the leaf cells used and how much.

Why: The cuttings share one set of alleles.
The salt changed which genes the leaf cells used, not the DNA.
The salt acts on the leaves growing in it.
So new leaves grown in ordinary soil take the ordinary form, and the cuttings do not stay short.

Q5 T55-q09

One clonal line of duckweed grows large fronds in a warm pond and small fronds in a cold pond. Two duckweed strains grown side by side in one warm tank differ in frond size.

Which of the following shows phenotypic plasticity?

  1. A. Neither
    Plasticity never involves a change in DNA.
    It is one genotype producing different phenotypes in different conditions, which the clonal line in the two ponds shows.
  2. B. ✓ The clonal line in the two ponds only
  3. C. The two strains in the tank only
    The two strains are two genotypes grown in the same conditions.
    So their difference is genetic, not plasticity.
  4. D. Both the clonal line and the two strains
    Plasticity is one genotype producing two phenotypes.
    In the tank two genotypes produced two phenotypes under the same conditions.

Why: The clonal line is one genotype, and it produced two frond sizes in two conditions, the warm pond and the cold pond: plasticity.
In the tank the conditions were the same and the genotypes differed: a genetic difference.

Q6 T55-q10

A pond holds snails of one species with shells of many colors and patterns. A student writes: this variety shows phenotypic plasticity.

What is wrong with the student's statement?

  1. A. ✓ Plasticity is the range one genotype can produce; the snails’ variety may be genetic
  2. B. Nothing; variety of phenotype across a population is what phenotypic plasticity means
    Phenotypic plasticity is a property of one genotype: the ability of individual genotypes to produce different phenotypes.
    Variety across a population is not that.
  3. C. The variety must be a genetic difference, because shell color in snails is set by alleles alone
    Shell color may well respond to conditions, and the case does not say.
    Variety among many snails shows neither plasticity nor a genetic difference on its own.
  4. D. Plasticity needs a change in the snails’ DNA, which the pond water cannot make
    Plasticity never involves a change in DNA.
    It is one genotype producing different phenotypes under different conditions, with the DNA unchanged.

Why: Phenotypic plasticity is the ability of individual genotypes to produce different phenotypes under different conditions.
Many differently colored snails may simply be many genotypes; only one genotype giving different shells in different conditions would show plasticity.

Q7 T55-q07

On one oak tree, the leaves on a shaded branch grow larger and thinner than the leaves on a branch in full sun.

What differs inside the leaf cells of the two branches?

  1. A. ✓ The set of genes in use, and how much each one is used
  2. B. The sequence of the DNA, and so the set of genes the cells carry
    A difference in leaf size is not evidence of a change in DNA; both branches grew from the same tree and carry the same sequence.
  3. C. Which alleles the cells inherited at fertilization
    Every cell of one tree grew by mitosis from one zygote and carries the same alleles.
  4. D. The number of chromosomes each cell carries
    The chromosome number is the same in every cell of the tree; a branch's light does not change it.

Why: Both branches belong to one tree and share one genotype.
The light is a condition outside the cells that changes which genes they use and how much, so the leaves differ while the DNA does not.

Q8 T55-q11

Which of the following cases is phenotypic plasticity?

  1. A. A mutation in a pigment gene gives a mouse a white coat
    A mutation is a change in the DNA, which plasticity never is.
  2. B. A red-flowered plant and a white-flowered plant give pink offspring
    Pink offspring from a red and a white parent are heterozygotes of an incompletely dominant pair, a matter of the alleles inherited, not of conditions acting on one genotype.
  3. C. ✓ Cuttings of one shrub grow tall in a sheltered garden and short in a windy one
  4. D. Two breeds of cattle kept on one farm give different amounts of milk
    Two breeds are two genotypes, so their difference in milk is genetic.

Why: Only the shrub’s cuttings show one genotype producing two phenotypes under two conditions, sheltered and windy.
The mouse and the pink flowers are set by the alleles present, and the two breeds are two genotypes.

Q9 T55-q12

A researcher splits alligator eggs from one nest between two incubators, at 30 °C and 34 °C, with the same moisture and light. All the 30 °C hatchlings are female and all the 34 °C hatchlings are male.

What set the sex of the hatchlings, and how do the results show it?

  1. A. ✓ The temperature at which each egg developed, the one condition that differed between the two groups
  2. B. The alleles each egg received from its father, because sex is set by the sex chromosomes in every animal
    No allele fixes an alligator egg's sex; eggs from the same parents came out all female at one temperature and all male at the other, so the temperature decided it.
  3. C. The moisture around each egg, because eggs in a nest of rotting plants develop in damp conditions
    The moisture and the light were the same for every egg, so they cannot account for a difference between the two groups.
  4. D. Chance alone, because about half of the eggs in any clutch develop as males whatever the conditions
    Chance would give a mixture of males and females in each incubator; here each incubator gave one sex only, following its temperature.

Why: The eggs came from one nest and had the same moisture and light; only the temperature differed, and it sorted the hatchlings by sex, so the temperature at which the egg develops sets the sex.
The warmth changes which genes the embryo's cells use, and no allele decides it.

Q10 T55-q13

After two weeks of strong sunshine a person's forearms are noticeably darker than they were in winter.

Which condition changed, and which trait responded?

  1. A. The temperature of the air changed, and the skin's thickness responded
    The condition that changed was the UV light in strong sunshine, not the air temperature, and the trait that responded is the skin’s color, not its thickness.
  2. B. The person's diet changed, and the skin's melanin responded
    Nothing in the case says the person's diet changed; skin darkens in sunshine because UV light makes the skin cells make more melanin.
  3. C. ✓ UV light from the sun changed, and the skin's melanin responded
  4. D. UV light from the sun changed, and the skin's alleles responded
    The skin cells made more melanin with the same genes; the tan is not a change in any allele.

Why: Increased UV light is the condition, and melanin in the skin is the trait that responds: the skin cells use their pigment genes more, and no gene changes.

Q11 T55-q14

Two children of the same parents grow up with very different amounts of food and reach different adult heights.

Which statement about human height fits the case?

  1. A. Height is set by one gene alone, whatever a child eats
    Many genes, not one, set the range of heights a person can reach.
    Nutrition then moves a person within that range, so food does have an effect.
  2. B. Nutrition sets height entirely, whatever alleles a child carries
    The genes set the range that food can move a person within.
    Two children fed alike still differ in height if their genes differ.
  3. C. Food changes the height alleles a child carries, so the taller child’s alleles changed
    Food changes what the body's cells do with the same genes.
    It does not change any allele.
  4. D. ✓ Many genes set the range, and nutrition moves the person within it

Why: Human height and weight respond to nutrition, with many genes setting the range a person can reach; food moves a person within that range, and no gene changes.

Q12 T55-q08

A cutting of one plant grown in shade has large, thin leaves. The cutting is moved into full sun and grows new leaves.

Predict the new leaves, and give the reason.

  1. A. Large and thin, because the fixed genotype makes the cells use the same genes in any light
    A fixed genotype can still give different phenotypes in different conditions; the cutting's cells respond to the new light.
  2. B. ✓ Smaller and thicker, because in sun the leaf cells use some genes more and others less
  3. C. Smaller and thicker, because the sun changed the cutting's leaf alleles
    The new leaves do come out smaller and thicker, but no allele changed; sunlight does not rewrite DNA.
  4. D. Large and thin, because the leaf form set in shade is kept by every leaf grown afterward
    A leaf's form is not fixed by the conditions an earlier leaf grew in; each new leaf takes the form of the light it grows in, here sun.

Why: Moved into sun, the cutting's leaf cells change which genes they use and how much, so the new leaves take the sun form; the DNA stays as it was.

Q13 T55-q15

A grower wants to know whether temperature changes the height of sunflowers. The design is in the table below. After four weeks the plants in group A are taller.

The grower's design: two groups of sunflower plantlets from one clonal line, with the temperature and the watering of each group.
The grower's design: two groups of sunflower plantlets from one clonal line, with the temperature and the watering of each group.

What is wrong with the design?

  1. A. The two groups came from different plants, so their genes may differ
    Both groups are plantlets of one clonal line, so their genes are the same.
  2. B. ✓ Two conditions differ between the groups, so neither can be credited with the difference
  3. C. There is no control, because neither group was left completely unwatered
    A control is the group grown under the condition already in use, to compare against; it need not receive none of the water, the fault in this design is elsewhere.
  4. D. Nothing; the taller plants in group A show that 20 °C makes sunflowers taller
    Group B differs from group A in watering as well as temperature, so the height difference could be the water's doing.

Why: A test credits a condition only if it is the one condition that differs.
Here the temperature and the watering both differ between the groups, so the height difference cannot be credited to either.

Q14 T55-q18

A grower gives 12 cuttings of one grapevine the vineyard's usual watering and 12 twice as much, with soil, light and temperature the same. She measures the sugar in each plant's grapes.

Which group is the control, and what does it show?

  1. A. ✓ The usual-watering group; it shows the sugar reached under the condition already in use
  2. B. The double-watering group; it shows the sugar reached under the changed condition, which is the result under test
    The control is not the group that receives the change; it is the group kept at the usual condition, so the changed group's result has something to compare against.
  3. C. Both groups; each shows what one vine's genes can do, because every cutting came from the same plant
    Cuttings of one plant hold the genes the same in both groups; that does not make both groups the control, which is the one group lacking the change under test.
  4. D. Neither group; a control must receive none of the condition, and both groups were given water
    A control keeps the usual condition rather than removing it; a group given no water at all would test something else.

Why: The control is the group kept at the usual condition, here the vineyard's usual watering, so any difference in the double-watered group's sugar can be credited to the extra water; with cuttings of one vine and the same soil, light and temperature, the water is the only condition that differs.

Q15 T55-q16

A student grew cuttings of one rosemary plant at two humidities, 15 in each, and recorded the mean leaf length of each group. The bars below carry ±2SE error bars.

Mean leaf length in millimeters of rosemary cuttings from one plant grown at 60 % and at 90 % humidity, 15 cuttings per group; error bars represent ±2SE.
Mean leaf length in millimeters of rosemary cuttings from one plant grown at 60 % and at 90 % humidity, 15 cuttings per group; error bars represent ±2SE.

What do the bars show about the two humidities?

  1. A. The two humidities give leaves of exactly the same length
    The two means differ, 21.3 mm and 22.4 mm.
    Overlapping bars mean the data do not show a difference; they do not show that the leaves are equal.
  2. B. The higher humidity makes leaves longer
    Two means almost always differ a little by chance.
    The ±2SE bars overlap, so this gap could be chance.
  3. C. ✓ The data do not show a difference between the two humidities
  4. D. The cuttings must have carried different alleles
    Cuttings of one plant carry the same alleles.
    A small gap between two means with overlapping bars needs no cause beyond chance.

Why: The bar is ±2SE around each mean.
The bars run from 19.9 to 22.7 mm and from 20.8 to 24.0 mm, so they overlap, and the data do not show a difference between the two humidities; that is not the same as showing the two are equal.

Q16 T55-q17

A gardener roots 20 cuttings from a rose bush of one variety and 20 cuttings from a bush of a different variety. She grows the first 20 in a cold frame and the second 20 in a warm greenhouse, with the same soil, water and light for every plant. The greenhouse roses grow taller.

What can the gardener conclude about the cause of the height difference?

  1. A. The warmth caused it, because the soil, water and light were the same for both groups of cuttings
    The two groups came from bushes of different varieties, so their genes differ as well as their warmth.
    The warmth cannot be credited on its own.
  2. B. The genes caused it, because the two groups came from bushes of different varieties
    The genes differ between the groups, but so does the temperature.
    The test cannot say which of the two did it.
  3. C. ✓ Neither the warmth nor the genes can be credited, because both differed between the groups
  4. D. Both the warmth and the genes caused it, each adding some of the extra height
    The test shows only that the greenhouse roses grew taller.
    With two causes differing at once, it cannot show that either one acted, let alone both.

Why: To credit a condition, the genes must be held the same, with cuttings of one bush in both groups.
Here the variety and the temperature both differ between the groups.
So the height difference could be genes, warmth or both, and the test cannot separate them.

FRQ 1 T55-frq1 · Scientific Investigation

A grower has 40 strawberry plants grown from runners of one plant. She wants to know whether the temperature the plants grow at changes the mass of fruit they produce. She grows 20 plants at 18 °C and 20 at 26 °C in one glasshouse divided by a partition, with the same light, water and soil for every plant. At the end of the season she weighs the fruit of each plant and finds the mean fruit mass per plant for each group. The means are drawn below; each error bar represents ±2SE.

Mean fruit mass per plant in grams for strawberry plants of one clonal line grown at 18 °C and at 26 °C, 20 plants per group; error bars represent ±2SE.
Mean fruit mass per plant in grams for strawberry plants of one clonal line grown at 18 °C and at 26 °C, 20 plants per group; error bars represent ±2SE.

(a) Identify the independent variable and the dependent variable in the grower's test. (1 pt)

Model answer The independent variable is the temperature the plants grow at, 18 °C or 26 °C, the one condition the grower changed.
The dependent variable is the mass of fruit per plant, the quantity she measured to see the effect.
Rubric
  • Award 1 point for: temperature as the independent variable and fruit mass per plant as the dependent variable.

Slip Naming the light, water or soil as the independent variable. Those were held the same; the one condition changed was the temperature.

(b) Predict what the two means and their ±2SE error bars would look like if temperature had no effect on fruit mass. (1 pt)

Model answer The two mean values would be about the same.
The ±2SE error bars of the 18 °C group and the 26 °C group would overlap.
So the data would show no difference between the two temperatures.
Rubric
  • Award 1 point for: means about the same with overlapping ±2SE error bars (the data show no difference).

Slip Predicting that the two means would be exactly equal. Two sample means almost always differ a little by chance; the prediction is that the bars overlap.

(c) Evaluate the claim that the temperature changed the fruit mass, using the error bars drawn above. (1 pt)

Model answer The mean fruit mass was 240 g per plant at 18 °C and 175 g at 26 °C.
The error bars represent ±2SE: 222 to 258 g at 18 °C and 160 to 190 g at 26 °C.
The two bars do not overlap, so the difference between the mean values is unlikely to be chance.
The plants share one genotype and only the temperature differed between the groups.
So the claim is supported: the temperature changed the fruit mass.
Rubric
  • Award 1 point for: the judgement (the claim is supported) AND the ground (the ±2SE bars do not overlap, so the difference is unlikely to be chance; with the same genes and one condition changed, the temperature is credited).

Slip Arguing only that 240 is larger than 175. Two means can differ by chance; the non-overlapping error bars are what make chance unlikely.

(d) Explain why any difference in fruit mass between the two groups can be credited to the temperature rather than to a genetic difference between the groups. (1 pt)

Model answer Every plant grew from runners of one plant, so all 40 plants carry the same DNA.
So the two groups cannot differ in their genes, and a difference in fruit mass cannot be genetic.
Light, water and soil were the same for every plant, so the temperature is the only condition that differed.
So any difference in fruit mass between the groups can be credited to the temperature.
Rubric
  • Award 1 point for: the plants grew from runners of one plant, so both groups share one genotype and a difference cannot be genetic; with everything else the same, the temperature is the only condition that differed.

Slip Pointing to the equal group sizes. Twenty in each group helps the comparison; it is the shared genotype that rules the genes out.

(e) Describe what the result shows about how the plants' genes and the temperature together set the mass of fruit. (1 pt)

Model answer The same genotype gave heavier fruit at 18 °C and lighter fruit at 26 °C.
The DNA is the same in both groups.
So the temperature changed which genes the plants' cells used and how much.
The genes set the range of fruit mass the plants can reach, and the temperature decided where in that range each group fell.
Rubric
  • Award 1 point for EITHER: the same genes gave two phenotypes because the condition changed which genes the cells used and how much, with the DNA unchanged; OR: the genes set the range of fruit mass and the temperature decided where in that range each group fell.

Slip Saying the warmth changed the plants' genes. The DNA is the same in both groups; the temperature changed which genes the cells used, not the genes themselves.

FRQ 2 T55-frq2 · Conceptual Analysis

Aphids on a bean plant reproduce without mating, so all the aphids on one plant are a clonal line with the same DNA. A researcher keeps 10 bean plants crowded with aphids of one line and 10 bean plants carrying only a few aphids of the same line, all in one greenhouse with the same light and temperature. On the crowded plants, 84 % of the newborn aphids develop wings; on the uncrowded plants, 2 % do. A winged aphid that flies to an empty bean plant produces wingless offspring there.

(a) Identify the condition that changed and the trait that responded to it in the aphids. (1 pt)

Model answer The condition is how crowded the plant is with aphids; the trait that responded is whether a newborn aphid develops wings.
Rubric
  • Award 1 point for: crowding as the condition and wing development as the trait.

Slip Naming the bean plant or the food as the condition. The aphids on the crowded and the uncrowded plants eat the same kind of plant; what differs is how many aphids share it.

(b) Explain how crowding can change whether an aphid develops wings while its DNA stays the same. (1 pt)

Model answer Crowding is a condition outside the aphid that acts as a signal to its cells.
A condition outside the organism changes which genes its cells use and how much, so in a crowded plant the developing aphid's cells use the genes that build wings, and in an uncrowded plant they do not; the DNA is the same in both.
Rubric
  • Award 1 point for: the condition changes which genes the cells use and how much (gene expression), so the phenotype changes while the DNA does not.

Slip Saying crowding causes a mutation that gives wings. The wingless offspring of a winged aphid show that its DNA still carries the same genes.

(c) Justify the claim that the winged aphid's wings were a response to conditions rather than a change in its genes, using the winged aphid that flew to the empty plant. (1 pt)

Model answer The winged aphid's offspring on the empty plant have no wings, although they are copies of her DNA.
If the wings had come from a change in her genes, her offspring would carry that change and be winged too.
They are wingless because the condition, crowding, is gone, so the wings were the same genes used differently under crowding.
Rubric
  • Award 1 point for: the offspring share the winged aphid's DNA yet are wingless, so the wings were not written in the genes but produced by the condition.
  • Accept also: aphids of one line carry the same DNA, yet 84 % of newborns grew wings on the crowded plants and 2 % on the uncrowded plants, so the condition and not the genes set the wings.

Slip Saying the offspring lost the wing allele. Offspring of a clonal line carry the same DNA as their mother; nothing was lost.

(d) Two different clonal lines of aphids feed on identical, uncrowded bean plants in one greenhouse, and the aphids of one line are green while those of the other are red. Make a claim about whether this color difference is phenotypic plasticity or a genetic difference, and support your claim with evidence from the greenhouse. (1 pt)

Model answer Claim: the color difference is a genetic difference.
Evidence: the two lines feed on identical, uncrowded plants in one greenhouse, so the conditions are the same for both.
Phenotypic plasticity is one genotype producing different phenotypes under different conditions.
Here the conditions do not differ, and the two lines are two genotypes.
So the difference in color must come from their genes.
Rubric
  • Award 1 point for: the claim (a genetic difference) AND the support (the conditions are the same for both lines, so the difference cannot be plasticity; the two lines are two genotypes).

Slip Calling any difference in phenotype plasticity. Plasticity is a property of one genotype; two lines that differ under the same conditions differ in their genes.

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