End-of-topic test: Meiosis and Genetic Diversity

Unit 5 · Topic 5.2 end-of-topic test

Suggested time: about 44 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question. For each multiple-choice question, pick one option. When you have answered every question, press Submit the test; the feedback then gives the reasoning for each. 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.

Question 1

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?

A. It is impossible: a gamete carries all of one parent’s chromosomes B. It is a possible gamete: one member of each pair, each from either parent C. It is a possible gamete: crossing over swapped the paternal long chromosome into a maternal set D. It is impossible: a gamete carries the maternal member of every pair

Question 2

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 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.

Which set of chromosomes does the cell that forms at the left pole receive?

A. All three maternal B. Maternal long, maternal medium, paternal short C. Maternal long, paternal medium, maternal short D. Paternal long, maternal medium, paternal short

Question 3

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?

A. 10 B. 16 C. 25 D. 32

Question 4

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.

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.

Which of the following pairs of chromatids could also exchange pieces by crossing over?

A. Chromatids 1 and 2 B. Chromatids 1 and 4 C. Chromatids 3 and 4 D. Any two of the four

Question 5

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?

A. The student is wrong: crossing over moves alleles the parent already had into new combinations B. The student is right: crossing over rewrites the DNA sequence at the point where the two chromatids break C. The student is wrong: independent orientation, not crossing over, made the new allele at metaphase I D. The student is right, because a new allele can only arise in prophase I, while the homologs are paired

Question 6

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.

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.

Which chromatid is recombinant?

A. A chromatid reading E with G B. A chromatid reading e with g C. A chromatid reading E with g D. All four chromatids

Question 7

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?

A. 256 B. 512 C. 65,536 D. 131,072

Question 8

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?

A. Each egg received its chromosomes by chance inside the female, before it was shed B. The sperm and the egg exchange pieces of their chromosomes at the moment they fuse C. The sperm enters the egg before the egg’s meiosis has begun, and starts that meiosis D. Which sperm enters each egg is chance, so each zygote pairs one kind of egg with one kind of sperm

Question 9

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?

A. An egg carrying the maternal copy of chromosome 2 and the paternal copy of chromosome 5 B. A chromatid carrying the mother’s allele of one gene and the father’s allele of the next gene along the same chromosome C. A zygote made of one of the mother’s kinds of egg and one of the father’s kinds of sperm D. A sperm carrying an allele with a base that neither parent carried

Question 10

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?

A. The parents’ alleles changed by mutation between one clutch and the next B. The chicks of one clutch grew from one egg and one sperm, and crossing over in their own cells made them differ C. The female’s eggs were all alike, so the chicks differ only in which sperm reached each egg D. Each chick grew from a different egg and a different sperm, each carrying its own mix of alleles

Question 11

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?

A. Why does protein matter so much to meiosis in voles? B. Is a low-protein diet bad for the voles? C. What is the purpose of chiasmata in a vole’s meiosis in prophase I? D. Does a low-protein diet lower the number of chiasmata per cell?

Question 12

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?

A. Nondisjunction in meiosis I B. Nondisjunction in meiosis II C. Crossing over in prophase I D. Independent orientation at metaphase I

Question 13

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’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.

What are the chromosome counts of the four cells produced?

A. 13, 11, 12 and 12 B. 13, 13, 11 and 11 C. 13, 12, 12 and 12 D. 24, 24, 12 and 12

Question 14

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?

A. The student is wrong: the extra copy is an ordinary chromosome, and three copies of each gene make more product than two B. The student is right: a chromosome that fails to separate is damaged as it is pulled to the wrong pole 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 D. The student is right: the extra copy carries new alleles that neither parent’s chromosomes had

Question 15

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?

A. 54 chromosomes: one chromosome missing from both sets B. 55 chromosomes: one copy of one of them, a monosomy C. 56 chromosomes, because the sperm makes up the missing one D. 57 chromosomes: three copies of one of them, a trisomy

Question 16

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?

A. A gamete with 29 chromosomes, missing that chromosome B. A sperm with 30 chromosomes, one of them carrying a new allele C. An egg or sperm with 31 chromosomes, two copies of that chromosome D. A sperm carrying all 60 of the bull’s chromosomes

Question 17

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?

A. All 100 B. None C. About 50 D. About 25

Question 18

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?

A. Recombinant B. Parental C. Made by independent orientation D. Carried by a sister chromatid of the Q R chromosome

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Scientific Investigation · 4 points

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.

(a) Determine the division in which the failure happened, using the counts. (1 point)

A full-credit 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.

Check the box for each point your answer earns

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).

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

at anaphase II both sister chromatids of one chromosome went to the same pole (nondisjunction).

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

47, a monosomy (one copy of one chromosome).

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

25, 25, 23, 23 (any order).

Common 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.

Free-response score: 0 of 4

Free response 2 · Conceptual Analysis · 3 points

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 point)

A full-credit 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.

Check the box for each point your answer earns

independent orientation, each pair facing either way at metaphase I on its own, so A goes with B or with b.

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

crossing over between non-sister chromatids in prophase I, breaking at the same point between the genes and exchanging pieces.

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

any egg can fuse with any sperm, so the combinations multiply beyond the gametes’ own variety.

Common slip: Saying fertilization makes the gametes vary. The gametes varied already; fertilization multiplies the combinations by pairing them at random.

Free-response score: 0 of 3

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Multiple choice checked: 0 of 18 correct.

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