Practice questions · Topic 5.3

Unit 5 · Practice for the Topic 5.3 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Mendelian genetics, summed up

Video coming soon

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.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

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.

Question 1

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

In goats, the coat-color gene has two alleles, D and d. One goat's homologous pair is drawn below.

What is this goat's genotype, and is it homozygous or heterozygous?

A. D, homozygous B. DD, homozygous C. Dd, heterozygous D. DD, heterozygous

Question 2

In goats, a Dd goat has a dark coat and a dd goat has a light coat.

Which allele is dominant, and why?

A. d, because a light coat needs two copies to show B. D, because dark is the common coat color in goats C. D, because a dark coat shows in the heterozygote Dd D. Neither, because Dd goats show a mix of both colors

Question 3

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?

A. The eight-year line only B. The bought plant only C. Both the line and the bought plant D. Neither the line nor the bought plant

Question 4

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?

A. The P generation B. The F1 generation C. The F2 generation D. The F3 generation

Question 5

A breeder crosses two Dd goats (dark coat, D, dominant to light, d).

What is the probability that an offspring has a light coat?

A. 0.25 B. 0.5 C. 0.75 D. 1 : 3

Question 6

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?

A. It is certainly DD, because a Dd parent would have given at least one light offspring among eight B. Nothing at all, because crossing with a light dd goat reveals nothing about a dark goat's genotype C. It is certainly Dd, because a dark goat crossed with a light one is always heterozygous D. Probably DD, but Dd is still possible: eight dark from Dd × dd has a probability of about 1 in 256

Question 7

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?

A. A pollen grain and an ovule B. A pollen grain and an egg C. A sperm and an ovule D. A sperm and an egg

Question 8

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?

A. Autosomal dominant B. Autosomal recessive C. Dominant, not autosomal D. Recessive, not autosomal

Question 9

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 ?.

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.

Which genotype belongs in the marked cell?

A. JjQQ B. JjQq C. jjQq D. jQ

Question 10

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?

A. No sex chromosomes and 46 autosomes B. One sex chromosome, the Y, and 45 autosomes C. XY and 44 autosomes D. XX and 44 autosomes

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

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.

(a) State the null hypothesis for her chi-square test. (1 point)

Hint: Which statement says there is no real difference between the counts and the model?

A full-credit 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.

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 being due to chance.

Common 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 point)

Hint: What share of a 3 : 1 ratio is the striped class, and what is that share of 280?

Write down the values in the question:

total offspring = 280

predicted ratio = 3 striped : 1 plain

Write down the equation:

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

Substitute the values into the equation:

e_{striped} = 280 \times \frac{3}{4} = 210

e_{plain} = 280 \times \frac{1}{4} = 70

A full-credit answer: Expected striped = total × the striped share of the ratio = 280 × $\frac{3}{4}$ = 210 (and 70 plain are expected).

(c) Calculate the chi-square value for her counts. (1 point)

Hint: Use the equation on the formula sheet, one class per line, each class divided by its own expected count.

Write down the values in the question:

o = 200 striped, 80 plain

e = 210 striped, 70 plain

Write down the equation:

χ^{2} = \sum \frac{{(o - e)}^{2}}{e}

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

\frac{{(200 - 210)}^{2}}{210} = \frac{100}{210} = 0.476

\frac{{(80 - 70)}^{2}}{70} = \frac{100}{70} = 1.429

χ^{2} = 0.476 + 1.429 = 1.90

A full-credit answer: χ² = 0.476 + 1.429 = 1.90.

(d) Identify the degrees of freedom and the critical value at p = 0.05. (1 point)

Hint: How many classes of fish are there, and which row and column of the table do you read?

A full-credit 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.

Check the box for each point your answer earns

1 degree of freedom and a critical value of 3.84.

Common 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 point)

Hint: Compare the calculated value with the critical value: which is larger, and what does that comparison decide?

A full-credit 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.

Check the box for each point your answer earns

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’).

Common slip: Writing ‘accept the null hypothesis’. Failing to reject is not accepting.

Free-response score: 0 of 5

Free response 2 · Scientific Investigation · 4 points

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.

(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 point)

A full-credit answer: DE, De, dE and de, each with probability $\frac{1}{4}$ .
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.

Check the box for each point your answer earns

DE, De, dE, de, each 25%, because the two homologous pairs line up independently at metaphase I.

Common 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 point)

Write down the values in the question:

total offspring = 160

predicted ratio = 9 : 3 : 3 : 1

Write down the equation:

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

Substitute the values into the equation:

e_{light, short} = 160 \times \frac{1}{16} = 10

e_{dark, long} = 160 \times \frac{9}{16} = 90

e_{dark, short} = e_{light, long} = 160 \times \frac{3}{16} = 30

A full-credit answer: 10 light-coated, short-eared goats are expected (the 1-in-16 class).

(c) Calculate the chi-square value for the breeder's counts against the 9 : 3 : 3 : 1 prediction. (1 point)

Write down the values in the question:

o = 88, 26, 30, 16

e = 90, 30, 30, 10

Write down the equation:

χ^{2} = \sum \frac{{(o - e)}^{2}}{e}

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

\frac{{(88 - 90)}^{2}}{90} = 0.044

\frac{{(26 - 30)}^{2}}{30} = 0.533

\frac{{(30 - 30)}^{2}}{30} = 0

\frac{{(16 - 10)}^{2}}{10} = 3.6

χ^{2} = 0.044 + 0.533 + 0 + 3.6 = 4.18

A full-credit answer: χ² = 4.18.

(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 point)

A full-credit 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.

Check the box for each point your answer earns

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).

Common slip: Saying the test proves the genes are on different chromosomes. Failing to reject leaves the belief standing; it does not prove it.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

Take the end-of-topic test