End-of-topic test: Non-Mendelian Genetics

Unit 5 · Topic 5.4 end-of-topic test

Suggested time: about 54 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. Where a question needs a critical value, the chi-square table is drawn with it; read the p = 0.05 row.

Question 1

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?

A. The red allele hid the white allele B. The two alleles stayed separate, and neither hid the other C. Pink is a third allele, dominant to red and white D. The red and white alleles blended into one pink allele

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 75% type A and 25% type O D. About half type A and half type O

Question 4

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?

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 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?

A. The two genes sit on one chromosome, so the grandparents’ two combinations are in excess B. The two genes sit on different chromosomes, and chance alone pushed the grandparents’ two 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 a BbDd snail (B with d on one chromosome, b with D on the other) crossed with a bbdd snail.

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.

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

A. 0.12 map units B. 12.0 map units C. 13.6 map units D. 88.0 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 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?

A. 22 map units B. 13 map units C. 9 map units D. 4 map units

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

Bristle shape of the daughters and the sons of two crosses between true-breeding fruit-fly lines.

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.

Which conclusion do the two crosses support?

A. The gene is on the X chromosome: the trait is X-linked B. The gene is on the Y chromosome: the trait is Y-linked C. The gene is on an autosome, and forked bristles are recessive D. The gene is in mitochondrial DNA: the trait shows non-nuclear inheritance

Question 12

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?

A. Only the two daughters, each with probability one B. Only the two sons, each with probability one C. Any of the four, each with probability one half D. All four, each with probability one

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}

D. 1

Question 15

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?

A. Z B. Z and W C. Either Z or W D. W

Question 16

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?

A. The student is right: each of the three problems comes from a different gene, inherited separately B. The student is right: the anemia and the pain are separate traits, so a child can inherit one alone C. The student is wrong: the three problems come from three different alleles of the one hemoglobin gene D. The student is wrong: one changed protein makes the red cells rigid, and rigid cells cause all three

Question 17

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 rare condition is recorded in the family below. I-1, II-2, II-4 and III-10 married into the family.

Which mode of inheritance best fits this pedigree?

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

Question 18

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?

A. Green, because a nuclear allele for green leaves is dominant, so the direction of the cross does not matter B. Green, because the chloroplasts come through the pollen, and the pollen came from a green branch C. Patchy green and white, because the ovule and the pollen each bring their own chloroplasts D. White, because the chloroplasts come through the ovule, and the ovule came from a white branch

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.

The trait is X-linked recessive. Which unaffected individual in this pedigree must carry the allele?

A. I-2 B. II-1 C. II-2 D. III-2

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 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) Make a claim about the mode of inheritance of the condition. (1 point)

A full-credit answer: Claim: the condition is autosomal recessive.

Check the box for each point your answer earns

the claim, autosomal recessive; no reasoning is required for this point.

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

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).

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

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.

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

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).

Common 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 point)

Write down the values in the question:

P (a from II-3) = 1

P (a from II-5) = \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) = 1 \times \frac{1}{2} = \frac{1}{2} = 0.5

A full-credit answer: The probability is $\frac{1}{2}$ , which is 0.5.

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, 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.

(a) Evaluate the claim that these two genes assort independently, using the four classes. (1 point)

A full-credit 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.

Check the box for each point your answer earns

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.

Common 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 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 determine whether the null hypothesis of independent assortment is rejected. (1 point)

A full-credit 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.

Check the box for each point your answer earns

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.

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 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.

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

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