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

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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 tells you what a wrong choice assumed. For the free-response questions, write your answer in full sentences and 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, because only one kind of allele is present B. DD; homozygous, because the two alleles are the same C. Dd; heterozygous, because the two chromosomes differ D. DD; heterozygous, because there are two alleles

Question 2

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

Which allele is dominant, and how do you know?

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 student's Punnett square for a Dd goat crossed with a dd goat.

A student draws the Punnett square below for a Dd goat crossed with a dd goat.

Which statement about the square is correct?

A. The square is right: gametes on its edges, each cell its row and column alleles B. The edges are wrong: they should carry Dd and dd, the two parents' genotypes C. The two dd cells are wrong: a dark-coated parent cannot give light offspring D. The cells are wrong: a square needs four different genotypes, one in each cell

Question 4

Two Dd goats (dark coat, D, dominant to light, d) are crossed and 120 offspring are counted.

How many dark-coated offspring are expected?

A. 30 B. 60 C. 120 D. 90

Question 5

Two Dd goats are crossed.

What is the probability that an offspring is heterozygous, Dd?

\frac{1}{4}

\frac{1}{2}

\frac{3}{4}

\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

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?

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

In zebrafish, striped (T) is dominant to plain (t). A pet shop's card says two plain fish produced a striped offspring.

What does the single-gene model say the two plain parents can produce?

A. Only plain offspring, so the card's striped offspring cannot be theirs as written B. Mostly plain offspring, with about one striped in four C. Half striped and half plain offspring D. Striped offspring whenever a hidden T is present in a plain parent

Question 8

A pedigree for Huntington's disease across three generations. Squares are males, circles females; a filled shape shows the condition.

The pedigree below records Huntington's disease in one family.

Is the allele that causes Huntington's disease dominant or recessive, and which family shows it?

A. Recessive, because it appears in every generation B. Dominant, because I-1 and I-2 both have the condition and their daughter II-2 does not C. Recessive, because II-3 and II-4 have an unaffected daughter, III-1 D. Dominant, because most of the family has the condition

Question 9

In goats, a dark coat (D) is dominant to a light coat (d) and long ears (E) to short ears (e); the genes are on different chromosomes. A DdEe goat is crossed with a ddEe goat.

What is the probability that an offspring is light-coated with short ears?

\frac{1}{16}

\frac{1}{4}

\frac{1}{8}

\frac{3}{4}

Question 10

A student has two data sets from a goat breeder: the numbers of dark and light offspring from a Dd × Dd cross, and the mean body mass of the dark offspring against the mean body mass of the light offspring, each with its standard error.

Which data set can be tested with chi-square, and why?

A. The body masses, because chi-square is the test that compares two means with each other B. Both data sets, because chi-square works on any two numbers that can be compared C. Neither data set, because chi-square needs at least four categories to compare D. The offspring counts, because chi-square compares counts in categories with a predicted ratio

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: 210 striped fish are expected (and 70 plain).

(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 = 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) State the verdict of the test on the null hypothesis, and 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 she fails to reject the null hypothesis; the counts are consistent with the 3 : 1 model. That does not prove the model, it means the counts give no reason to doubt it.

Check the box for each point your answer earns

fail to reject the null hypothesis because 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 one quarter. 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 one quarter, 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, state the verdict, and justify what it says about the breeder's belief that the genes are on different chromosomes. (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 = 4.18 is smaller than 7.81, so she fails to reject the null hypothesis: the counts are consistent with 9 : 3 : 3 : 1, and so with the two genes assorting independently on different chromosomes. The test does not prove that they are; it finds no reason to doubt it.

Check the box for each point your answer earns

3 degrees of freedom, critical value 7.81, fail to reject because 4.18 is less than 7.81, so the counts are consistent with independent assortment (not proved).

Common slip: Using 1 degree of freedom. There are four classes, so three degrees of freedom.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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