End-of-topic test: Mendelian Genetics

Unit 5 · Topic 5.3 end-of-topic test

Suggested time: about 51 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

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?

A. The student is right: the yellow allele was lost in the F1, and a new mutation brought yellow back in the F2 plants B. The student is wrong: every F1 plant carried the yellow allele without showing it, and yellow came back in the F2 C. The student is wrong: the F1 plants kept yellow by being yellow inside their fruit and red on the outside D. The student is right: the yellow allele was lost in the F1, and the F2 yellow plants came from red pollen only

Question 2

One horse's homologous pair carrying the coat-color gene; the allele each chromosome carries is written on it.

In horses, the coat-color gene has two alleles, K and k. One horse's homologous pair is drawn below.

What is this horse's genotype for coat color, and is the horse homozygous or heterozygous?

A. KK, homozygous B. kk, homozygous C. Kk, heterozygous D. Kk, homozygous

Question 3

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?

A. Rr and Rr B. RR and rr C. Rr and rr D. rr and rr

Question 4

A Punnett square for two Rr pepper plants, with one cell still to fill, marked ?.

Two Rr pepper plants are crossed. The Punnett square below has one cell still to fill, marked ?.

Which genotype belongs in the marked cell?

A. Rr B. rr C. RR D. R

Question 5

Two Rr pepper plants (red fruit dominant to yellow) are crossed.

Which of these is the genotypic ratio of their offspring?

A. 3 red : 1 yellow B. 1 RR : 2 Rr : 1 rr C. 1 RR : 1 rr D. 2 red : 1 yellow

Question 6

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?

A. No; two Kk horses give three black foals for every chestnut one, so six foals cannot be five and one B. No; five to one shows that one parent must be KK C. Yes; each foal has the same one-in-four chance of chestnut, and six foals is a small batch D. No; six foals from two Kk horses always include at least two chestnut ones

Question 7

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?

\frac{3}{4}

\frac{1}{4}

\frac{1}{16}

\frac{1}{2}

Question 8

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?

\frac{1}{4}

\frac{1}{2}

\frac{3}{4}

1

Question 9

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?

\frac{1}{12}

\frac{3}{4}

\frac{1}{64}

\frac{1}{4}

Question 10

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?

A. RR, because more than half of the offspring are red B. Rr, because yellow offspring received an r allele from it C. RR, because an Rr parent would give only yellow offspring with rr D. It stays undecided, because a cross with a yellow-fruited plant shows nothing

Question 11

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?

A. The striped allele: all the F1 and about 75% of the F2 are striped B. The plain allele, because plain is by far the more common pattern in her tanks C. The plain allele, because the plain fish vanished and came back in the F2 D. Neither allele, because striped and plain fish are both present in the F2

Question 12

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?

A. A black foal with genotype Kk B. A chestnut foal with genotype kk C. A chestnut foal whose black parent gave it k D. A black foal with genotype KK

Question 13

A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.

A rare condition is recorded in the family below.

Is the condition dominant or recessive, and which family decides it?

A. Dominant: II-2 has the condition although neither of her parents has it B. Dominant: the condition appears in generation II and again in generation III C. Recessive: an affected child, II-2, of two unaffected parents, I-1 and I-2 D. Recessive: the condition skips generation I

Question 14

A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.

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.

Which genotype must II-4 have?

A. AA B. Aa C. aa D. AA or Aa; the pedigree cannot decide

Question 15

A family pedigree for a rare condition across three generations. Squares are males, circles females; a filled shape shows the condition.

In the pedigree below, II-1 has a child with a man who has the condition.

What is the probability that the child shows the condition?

\frac{1}{4}

\frac{1}{3}

\frac{1}{2}

\frac{2}{3}

Question 16

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?

A. MR only B. MM and Rr, half each C. MR and Mr, half each D. MR, Mr, mR and mr, each 25%

Question 17

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?

A. 30 B. 90 C. 120 D. 270

Question 18

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?

\frac{1}{16}

\frac{1}{8}

\frac{1}{2}

\frac{3}{4}

Question 19

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?

A. The law applies to neither case, because it describes only the seven traits of Mendel's peas B. The law applies to both cases, because every gene assorts independently of every other gene C. The law holds for the pepper genes on different chromosomes; two genes on one chromosome need not follow it D. The law applies to the two genes on one chromosome, because alleles inherited together assort independently

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 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.

(a) State the null hypothesis for the breeder's chi-square test. (1 point)

A full-credit 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.

Check the box for each point your answer earns

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.

Common 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 point)

Write down the values in the question:

total offspring = 300

predicted ratio = 3 red : 1 yellow

Write down the equation:

expected count = total \times that class's share of the ratio

Substitute the values into the equation:

e_{red} = 300 \times \frac{3}{4} = 225

e_{yellow} = 300 \times \frac{1}{4} = 75

A full-credit answer: 225 red-fruited plants are expected (and 75 yellow-fruited).

(c) Calculate the chi-square value for the breeder's counts. (1 point)

Write down the values in the question:

o = 214 red, 86 yellow

e = 225 red, 75 yellow

Write down the equation:

χ^{2} = \sum \frac{{(o - e)}^{2}}{e}

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

\frac{{(214 - 225)}^{2}}{225} = \frac{121}{225} = 0.538

\frac{{(86 - 75)}^{2}}{75} = \frac{121}{75} = 1.613

χ^{2} = 0.538 + 1.613 = 2.15

A full-credit answer: χ² = 2.15.

(d) Identify the degrees of freedom and the critical value at p = 0.05, and determine whether the null hypothesis is rejected. (1 point)

A full-credit 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.

Check the box for each point your answer earns

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’.

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

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.

Common 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.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

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) Support the claim that the cystic fibrosis allele is recessive with evidence from one set of parents and their children in the pedigree. (1 point)

A full-credit 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.

Check the box for each point your answer earns

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).

Common 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 point)

A full-credit 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.

Check the box for each point your answer earns

I-1 Ff, I-2 Ff and II-1 ff (all three).

Common 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 point)

Write down the values:

II-3 and II-4 are both Ff (unaffected parents of an ff child)

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

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

A full-credit 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: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ , which is 0.25.

(d) Explain why II-2's genotype stays undecided from the pedigree, and give the two genotypes it could be. (1 point)

A full-credit 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.

Check the box for each point your answer earns

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.

Common 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.

Free-response score: 0 of 4

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Multiple choice checked: 0 of 19 correct.

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