End-of-topic test: Non-Mendelian Genetics

Unit 5 · Topic 5.4 end-of-topic test

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback tells you what a wrong choice assumed. For the two free-response questions, write your answer in full sentences and 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.

Question 1

In snapdragons, 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 conclusion do the two generations support?

A. The red allele hides the white allele, and the soil made some F2 plants pink or white B. The red and white alleles stayed separate in the pink plants, and neither hid the other C. Red, pink and white are three alleles of the gene, and each F1 plant carried all three D. The two alleles merged into one pink allele in the F1, and new mutations gave red and white in the F2

Question 2

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?

A. 90 B. 180 C. 270 D. 360

Question 3

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?

A. All type A B. All type O C. About three quarters type A and one quarter type O D. About half type A and half type O

Question 4

In mice, a female whose cells hold a mixture of ordinary and faulty mitochondria has offspring that carry very different shares of the faulty ones, with no 3 : 1 and no 1 : 1 among them. The faulty mitochondria lack one protein of the electron transport chain.

Where does the gene for that protein sit, and why do the offspring carry such different shares?

A. On an autosome; the two alleles part at anaphase I, one per egg B. In the mitochondrion's own DNA; the mitochondria fall into each egg at random C. In the mitochondrion's own DNA; each egg copies the mother's mixture exactly D. On the X chromosome; a son has one X and a daughter has two

Question 5

In fruit flies, gray body is dominant to black and normal wings to vestigial. A true-breeding gray, normal-winged fly is crossed with a true-breeding black, vestigial-winged fly, and two F1 flies are then crossed with each other. Their 1,000 offspring are 610 gray normal, 140 gray vestigial, 140 black normal and 110 black vestigial.

Which explanation best accounts for the departure from the 9 : 3 : 3 : 1 that two genes on different chromosomes would give?

A. The two genes sit on one chromosome, so the parental combinations, gray with normal and black with vestigial, are in excess B. The two genes sit on different chromosomes, and chance alone pushed both parental combinations above their shares C. Complete dominance lets the dominant alleles enter gametes together more often than the recessive ones D. The two genes sit on different chromosomes, but crossing over made the parental combinations more common

Question 6

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.

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.

Which two classes are the recombinants?

A. Black spotted and brown plain B. Black spotted and black plain C. Brown plain and brown spotted D. Black plain and brown spotted

Question 7

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.

For the beetle cross in the table above, 236 black spotted, 228 brown plain, 21 black plain and 15 brown spotted offspring were counted, 500 in all.

What is the map distance between the shell-color gene and the spotting gene?

A. 0.072 map units B. 7.2 map units C. 7.8 map units D. 92.8 map units

Question 8

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?

A. 0 %, because two genes on one chromosome always travel together B. 15 %, because a crossover separates the two only half of the time it falls between them C. 30 %, because a crossover falls between two genes more often the farther apart they sit D. 60 %, because a crossover between the genes separates them in both chromatids

Question 9

In a plant, three genes sit on one chromosome: stem height (H), leaf shape (L) and seed color (S). Their map distances are H to L 9 map units, L to S 22 map units, and H to S 13 map units.

In what order do the three genes sit along the chromosome?

A. L, H, S B. H, L, S C. L, S, H D. H, S, L

Question 10

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 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.

What is the verdict, and what does it say about the two genes?

A. Fail to reject the null hypothesis, because 9.20 is smaller than 9.49; the genes assort independently B. Accept the null hypothesis: the counts prove that the genes sit on different chromosomes C. Reject the null hypothesis, because 9.20 is larger than 7.81; the genes do not assort independently D. Reject the null hypothesis, and the two genes therefore sit 9.20 map units apart

Question 11

A human cell in meiosis I has its chromosomes paired. In twenty-two of the pairs the two chromosomes match in size. In the last pair, one chromosome is much smaller than the other, and the two carry different genes.

Which chromosomes make up the last pair, and what does it show about the person the cell came from?

A. Two autosomes; one of them lost a piece, so the pair says nothing about the person's sex B. Two X chromosomes; the person is female, because the last pair is always two X chromosomes C. An autosome and the X; the person is male, because the X pairs with a leftover autosome D. The X and the Y; the person is male, because only a male carries a Y

Question 12

A woman carries an allele on one of her two X chromosomes.

Which of her children can receive that allele from her?

A. Any child, son or daughter, each with a probability of one half B. Only her daughters, because the allele sits on an X chromosome C. Only her sons, because a son's X always comes from his mother D. Every child, because a mother gives an X to every child

Question 13

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?

A. She is color-blind, because a daughter always receives her father's X and the allele on it B. She sees colors as most people do: her mother's X carries the ordinary allele, which hides her father's C. She sees colors as most people do, because a daughter receives no X from her father D. She is color-blind, because the allele shows in anyone who carries one copy

Question 14

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?

\frac{1}{4}

\frac{1}{2}

\frac{3}{4}

\frac{1}{2} + \frac{1}{2} = 1

Question 15

In birds, a female is ZW and a male is ZZ. A chick receives a W chromosome.

Which parent gave the chick the W, and what sex is the chick?

A. The father, and the chick is male B. The father, and the chick is female C. The mother, and the chick is male D. The mother, and the chick is female

Question 16

In people, one allele of one gene gives very long limbs, loose joints and a weakened wall of the main artery leaving the heart, and the three appear together in the people who inherit the allele.

Which pattern of inheritance is this?

A. Incomplete dominance B. Linked genes C. Pleiotropy D. Codominance

Question 17

A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition. II-2, II-4 and II-6 married into the family.

A rare condition is recorded in the family below. II-2, II-4 and II-6 married into the family.

Which mode of inheritance best fits this pedigree?

A. Autosomal dominant B. Maternal inheritance C. X-linked recessive D. Autosomal recessive

Question 18

In four o'clock plants, some plants have green leaves and some have white leaves. A green plant supplying the ovules is crossed with a white plant supplying the pollen: all 120 seedlings are green. In the reciprocal cross, a white plant supplying the ovules and a green plant supplying the pollen: all 120 seedlings are white. A grower now takes one white seedling from the second cross to supply the ovules and one green seedling from the first cross to supply the pollen.

Which explanation fits the two crosses, and what leaf color will the planned cross give?

A. Both gametes bring chloroplasts equally, so every seedling of the planned cross carries both kinds and is patchy green and white B. Chloroplasts come through the pollen, so the planned cross gives green seedlings C. A dominant nuclear allele sets green leaves, so the planned cross gives about half green and half white seedlings D. Chloroplasts come through the ovule, so the planned cross gives white seedlings

Question 19

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 trait is recorded in the family below. II-1 married into the family.

Which feature of this pedigree shows the route an X-linked recessive allele takes through a family?

A. The trait appears in generation I and again in generation III, so a dominant allele passed straight down the family B. Only males are affected, which shows that the allele sits on the X chromosome and can never reach a female C. The allele went from the affected man I-1 through his unaffected daughter II-2 to her affected son III-1 D. II-2 is unaffected although her father is affected, which shows that she carries no copy of the allele

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 · 5 points

A rare 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 rare condition across three generations. Squares are males, circles females; a filled shape shows the condition. II-4 married into the family.

(a) Identify the mode of inheritance of the condition. (1 point)

A full-credit answer: The condition is autosomal recessive.

Check the box for each point your answer earns

autosomal recessive.

Common slip: Writing X-linked recessive because two of the affected people are hard to tell from a carrier pattern. The affected daughter II-1 decides it (see part c).

(b) Justify the claim that the allele is recessive, using one family in the pedigree. (1 point)

A full-credit answer: I-1 and I-2 do not have the condition, yet their daughter II-1 does. A dominant allele shows in everyone who carries it, so unaffected parents could not pass it on; two unaffected parents with an affected child are possible only if both carry a recessive allele without showing it. The same is true of II-3 and II-4 with their son III-1.

Check the box for each point your answer earns

two unaffected parents (I-1 and I-2, or II-3 and II-4) have an affected child, which a dominant allele cannot produce, so the allele is recessive.

Common slip: Arguing from the condition skipping a generation. That is a hint, not a proof.

(c) Justify the claim that the gene sits on an autosome rather than on the X chromosome, naming the individual who decides it. (1 point)

A full-credit answer: II-1 is an affected daughter of an unaffected father, I-1. If the recessive allele were on the X, a daughter would need an X carrying it from each parent, and her father, with one X carrying the ordinary allele, would himself be affected if his X carried the allele. An unaffected father cannot give a daughter an X-linked recessive allele, so the gene is on an autosome.

Check the box for each point your answer earns

II-1 (an affected daughter of an unaffected father) rules X-linked recessive out, because she would need an X with the allele from her father, who would then show the condition himself.

Common slip: Pointing at III-1, an affected son. An affected son of unaffected parents fits X-linked recessive as well as autosomal recessive; only the affected daughter decides.

(d) Write A for the ordinary allele and a for the allele that causes the condition. Identify the genotypes of II-3 and II-4, and explain how the pedigree fixes them. (1 point)

A full-credit answer: II-3 and II-4 are both Aa. Their son III-1 has the condition, so he is aa and received an a from each parent; neither parent has the condition, so each also carries an A. Both are carriers, Aa.

Check the box for each point your answer earns

II-3 Aa and II-4 Aa, because their affected son III-1 is aa and received an a from each of them, while neither parent shows the condition.

Common slip: Writing II-4 as AA because he 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-4 has the condition. (1 point)

Write down the values in the question:

P (a from II-3) = \frac{1}{2}

P (a from II-4) = \frac{1}{2}

Write down the equation:

P (A and B) = P (A) \times P (B)

Substitute the values into the equation:

P (a a) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25

A full-credit answer: The probability is $\frac{1}{4}$ , which is 0.25.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 5 points

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, whose chromosomes carry D with L and d with l, 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 (D with L on one chromosome, d with l on the other) crossed with a ddll moth. Bottom: the chi-square table from the AP Biology formula sheet.

(a) Identify the deviation from Mendel's ratios that these counts show, and justify it from the four classes. (1 point)

A full-credit answer: The two genes are linked. If they assorted independently the four classes would be about 200 each; instead the two parental combinations, dark long (342) and pale short (338), are large and the two new combinations, dark short (62) and pale long (58), are small, because the two genes sit on one chromosome and pass into a gamete together unless a crossover falls between them.

Check the box for each point your answer earns

linked genes, because the two parental combinations (dark long, pale short) are far above the 200 each that 1 : 1 : 1 : 1 predicts and the two recombinant combinations are far below.

Common slip: Calling the two large classes the recombinants. The large classes carry the combinations the parent's chromosomes had.

(b) Calculate the recombination frequency between the two genes, and state the map distance between them with its unit. (1 point)

Write down the values in the question:

recombinant offspring = 62 + 58 = 120

total offspring = 800

Write down the equation:

recombination frequency = \frac{recombinant offspring}{total offspring} \times 100 %

Substitute the values into the equation:

recombination frequency = \frac{120}{800} \times 100 % = 15.0 %

map distance = 15.0 map units

A full-credit answer: The recombination frequency is 15.0 %, so the genes are 15.0 map units apart.

(c) Calculate the chi-square value for the four counts against the 1 : 1 : 1 : 1 ratio that independent assortment predicts. (1 point)

Write down the values in the question:

o = 342, 338, 62, 58

e = 800 \times \frac{1}{4} = 200 for each class

Write down the equation:

χ^{2} = \sum \frac{{(o - e)}^{2}}{e}

Substitute the values into the equation, one class per line:

\frac{{(342 - 200)}^{2}}{200} = \frac{20164}{200} = 100.82

\frac{{(338 - 200)}^{2}}{200} = \frac{19044}{200} = 95.22

\frac{{(62 - 200)}^{2}}{200} = \frac{19044}{200} = 95.22

\frac{{(58 - 200)}^{2}}{200} = \frac{20164}{200} = 100.82

χ^{2} = 100.82 + 95.22 + 95.22 + 100.82 = 392.08 \approx 392

A full-credit answer: χ² = 392.

(d) Identify the degrees of freedom and the critical value at p = 0.05, and state the verdict of the test and what it says about the two genes. (1 point)

A full-credit answer: Four classes give 3 degrees of freedom, and the critical value at p = 0.05 is 7.81. χ² = 392 is far larger than 7.81, so the breeder rejects the null hypothesis: the counts do not fit 1 : 1 : 1 : 1, and the two genes are not assorting independently. They are linked.

Check the box for each point your answer earns

3 degrees of freedom, critical value 7.81, reject the null hypothesis because 392 is larger than 7.81, so the genes do not assort independently.

Do not award: ‘accept’ or ‘prove’, or a verdict that names a distance.

Common 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 point)

A full-credit answer: In prophase I, while 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 the pieces, so those chromatids carried D with l and d with L, combinations neither homolog had. Gametes made from those recombinant chromatids met 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, which is why the two classes are small.

Check the box for each point your answer earns

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.

Common 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.

Free-response score: 0 of 5

Multiple choice checked: 0 of 19 correct.

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