Practice questions · Topic 7.5

Unit 7 · Practice for the Topic 7.5 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: Hardy–Weinberg equilibrium, summed up

Video coming soon

Genotype and allele frequencies are shares of different things; p and q; two draws from the pool give p², 2pq and q²; backwards from the recessive share; the five conditions and the model as a null hypothesis; reading and constructing the frequency graph; percent change; expected counts against observed; the census-to-verdict chain.

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. The first free-response question walks you through one data set 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 shows the chi-square table, it is the formula sheet’s table, as the exam prints it.

Question 1

A student reads a census of 200 tortoises at a shell gene and says: “The genotype frequency of Tt is 0.35, so 35% of the gene copies are Tt.”

Which statement about the student’s claim is correct?

A. The student is right: every frequency in a census is a share of the gene copies B. The student is wrong: 0.35 is the frequency of allele T, a share of the copies C. The student is wrong: 35% of the tortoises carry at least one copy of T D. The student is wrong: 0.35 is a share of the tortoises, so 35% of the tortoises are Tt

Question 2

A census of hamsters gives these genotype frequencies at a coat gene with alleles G and g: GG 0.15, Gg 0.50 and gg 0.35.

Which of the following is the frequency of allele g?

A. 0.25 B. 0.35 C. 0.60 D. 0.85

Question 3

In a population of orchids, a lip-color gene has two alleles. The spotted-lip allele is recessive to the plain-lip allele. A biologist calls the spotted-lip allele allele 2 and writes its frequency as q. In the gene pool, q = 0.85.

Which of the following is p?

A. 0.0225 B. 0.15 C. 0.7225 D. 0.85

Question 4

The gene pool of the moth population as a Hardy–Weinberg square, p = 0.7 for B.

The square shows a population of moths that mates at random at a wing gene, with p = 0.7 for allele B and q = 0.3 for allele b. The two shaded cells are both Bb.

Which of the following is the expected share of Bb offspring, and why do the two shaded cells add?

A. 0.42, because the two cells are two ways of forming Bb, one for each order of draw B. 0.21, because the two cells show one genotype and are counted once C. 0.42, because a Bb offspring carries two alleles, so one cell is counted twice D. 0.91, because the two shaded cells and the BB cell all carry allele B

Question 5

Of the 800 bluebells in a wood, 49% have white flowers. White (w) is recessive to blue (W), the plants mate at random, and no force acts on this gene.

Which of the following is the expected number of carriers, the Ww plants?

A. 72 plants B. 336 plants C. 392 plants D. 560 plants

Question 6

The dark-wing allele’s frequency after twelve generations in eight enclosures of one beetle stock.

Researchers set up eight enclosures of beetles from one stock, four with 40 adults and four with 4,000 adults. Every enclosure started with the dark-wing allele at a frequency of 0.50. For twelve generations no beetle entered or left, no new mutation arose, the beetles mated at random, and wing color made no difference to survival or offspring number. The table shows the frequency after twelve generations.

Which Hardy–Weinberg condition did the 40-adult enclosures break, and why?

A. The population is large: with only 40 adults, the beetles had too few mates and most of them stopped breeding B. No condition failed: an allele frequency in any enclosure wanders this far by chance in twelve generations C. The population is large: with 40 adults, chance decides which copies reach the next generation D. No natural selection: chance differences in survival among 40 adults are a form of natural selection

Question 7

One census of a chipmunk population beside the model’s expectation from its own p.

Biologists census a population of chipmunks at a stripe gene with alleles A and a. In the gene pool p = 0.60 and q = 0.40. The table shows the census’s genotype frequencies beside the frequencies the Hardy–Weinberg model expects.

Which conclusion does the census support?

A. Natural selection is acting on this gene B. The population is in Hardy–Weinberg equilibrium, because the three frequencies add to 1 C. No force acts on this gene, because p and q add to 1 D. A force is acting on this gene, though the census does not say which

Question 8

The frequency of allele F in one red squirrel population, generations 0 to 8.

A biologist graphs the frequency of allele F in a population of red squirrels over eight generations.

Which conclusion does the graph support?

A. Consistent with Hardy–Weinberg equilibrium at this gene: the frequency of F held at 0.35 for eight generations B. In Hardy–Weinberg equilibrium at every one of the population’s genes, not only this one C. No force acts on this gene, because the frequency did not change D. Evolving slowly at this gene, because the frequency stayed at 0.35

Question 9

The frequency of an allele in a population of kestrels was 0.50 ten generations ago and is 0.38 today.

Which of the following is the percent change in the allele’s frequency?

A. −31.6% B. −24.0% C. −12.0% D. +24.0%

Question 10

The chi-square table from the AP Biology formula sheet.

For a census of 300 skinks, chi-square for the three genotype classes against the Hardy–Weinberg expectation is 1.42. Use the chi-square table above.

Which of the following is the verdict at p = 0.05, and what does it say?

A. Fail to reject the null hypothesis: the counts fit the model, which does not prove that no force acts B. Reject the null hypothesis: 1.42 is less than 5.99, so the counts depart from the model C. Accept the null hypothesis: the counts prove that no force acts on the gene D. Fail to reject the null hypothesis: the population is proved to be in Hardy–Weinberg equilibrium

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 · Analyze Data · 6 points

Biologists genotype 600 barnacles on one shore at a shell-thickness gene with alleles T and t. A storm then strikes the shore. Afterwards the biologists genotype the 250 barnacles that survived. The table shows both censuses.

Two censuses of one shore’s barnacles: before the storm, and the survivors after it.

(a) Calculate p, the frequency of allele T, before the storm. (1 point)

Hint: Count the copies of T: two per TT barnacle and one per Tt barnacle, out of the 1,200 copies the 600 barnacles carry.

Write down the values in the question:

TT barnacles = 216

Tt barnacles = 288

N = 600

Write down the equation:

p = \frac{2 \times TT barnacles + Tt barnacles}{2 N}

Substitute the values into the equation:

p = \frac{2 \times 216 + 288}{1200} = 0.60

A full-credit answer: Before the storm, p, the frequency of T, was 0.60.

(b) Calculate the number of TT barnacles expected among the 250 survivors if the storm had killed barnacles regardless of their genotype. (1 point)

Hint: Use the pre-storm p: the expected TT share is p², and the expected count is that share times 250.

Write down the values in the question:

p = 0.60

survivors = 250

Write down the equation:

expected TT count = p^{2} \times N

Substitute the values into the equation:

expected TT count = {0.60}^{2} \times 250 = 90

A full-credit answer: If the storm had spared and killed barnacles by chance alone, the survivors would be a random draw from the same gene pool, so 90 TT barnacles are expected.

(c) Describe how the survivors’ counts differ from the expected counts. (1 point)

Hint: Work out the expected Tt and tt counts the same way as in part (b), then set the three observed counts beside the three expected counts.

A full-credit answer: Compared with the expected counts, the survivors include many more TT barnacles, 135 against 90, slightly fewer Tt barnacles, 105 against 120, and far fewer tt barnacles, 10 against 40.

Check the box for each point your answer earns

more TT than expected (135 against 90) AND fewer tt than expected (10 against 40). Accept with or without the Tt comparison (105 against 120).

Common slip: Comparing the survivors with the pre-storm counts (216, 288, 96). The survivors are 250 barnacles, so the comparison is with the counts expected among 250.

(d) Calculate chi-square for the three genotype classes of the survivors, to three significant figures. (1 point)

Hint: One class per line: the observed count minus the expected count, squared, divided by the expected count; then add the three terms.

Write down the values in the question:

observed TT, Tt, tt = 135, 105, 10

expected TT, Tt, tt = 90, 120, 40

Write down the equation:

χ^{2} = \sum \frac{{(o - e)}^{2}}{e}

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

TT: \frac{{(135 - 90)}^{2}}{90} = 22.5

Tt: \frac{{(105 - 120)}^{2}}{120} = 1.88

tt: \frac{{(10 - 40)}^{2}}{40} = 22.5

χ^{2} = 22.5 + 1.88 + 22.5 = 46.9

A full-credit answer: chi-square = 22.5 + 1.88 + 22.5 = 46.9.

(e) Determine, using the table, whether the survivors are consistent with a storm that killed regardless of genotype, at p = 0.05. (1 point)

Hint: How many degrees of freedom do three classes give? Read that column of the p = 0.05 row and compare it with your chi-square.

The chi-square table from the AP Biology formula sheet.

A full-credit answer: The survivors are not consistent with a storm that killed regardless of genotype, because chi-square, 46.9, is greater than 5.99, the critical value for two degrees of freedom.
So the null hypothesis is rejected.

Check the box for each point your answer earns

not consistent (reject the null hypothesis), because 46.9 is greater than 5.99 at two degrees of freedom.

Common slip: Reading the column for three degrees of freedom, 7.81. Three classes give 3 − 1 = 2 degrees of freedom.

(f) Identify the Hardy–Weinberg condition that the pattern in the survivors suggests was broken. (1 point)

Hint: Which genotype survived far more often than chance would give, and which far less?

A full-credit answer: The pattern suggests that the condition broken was no natural selection: TT barnacles survived far more often than expected and tt barnacles far less, so survival depended on genotype.

Check the box for each point your answer earns

no natural selection (survival depended on genotype). Accept: natural selection acted on the gene, with or without the survivors’ counts as evidence.

Common slip: Naming a large population, because only 200 survived. The pattern is not a random wander: one genotype survived far more often than the others.

Free-response score: 0 of 6

Free response 2 · Conceptual Analysis · 4 points

Suppose a large island population of tortoises has mated at random for many generations at a shell-shape gene with alleles D and d. Domed shell (D) is dominant to flat shell (d). The frequency of D is 0.90, and every census has matched the genotype frequencies predicted from that p. Then the tortoises begin to choose mates with the same shell shape as their own. No tortoise arrives or leaves, no new mutation arises, and shell shape makes no difference to survival or to the number of offspring.

(a) Describe the genotype make-up expected while the tortoises mate at random, giving the three genotype frequencies. (1 point)

A full-credit answer: With p = 0.90 and q = 0.10, the model expects DD at p² = 0.81, Dd at 2pq = 0.18 and dd at q² = 0.01.

Check the box for each point your answer earns

DD 0.81, Dd 0.18 and dd 0.01.

Common slip: Giving 0.90 and 0.10 as the genotype make-up. Those are the allele frequencies; the genotype shares are p², 2pq and q².

(b) Explain why the change in mate choice moves the population away from Hardy–Weinberg equilibrium. (1 point)

A full-credit answer: Choosing a mate by shell shape breaks the random-mating condition.
A flat tortoise now mates only with a flat tortoise, and every offspring of two dd parents is dd.
Domed tortoises mate among themselves, so fewer Dd offspring form than random pairing would give.
So the heterozygote share falls and both homozygote shares rise, away from p², 2pq and q².

Check the box for each point your answer earns

random mating is broken, and like-with-like mating makes heterozygotes rarer and homozygotes commoner, so the genotype frequencies leave p², 2pq and q².

Common slip: Saying the mate choice is natural selection. Every tortoise survives and breeds as before; only who mates with whom has changed.

(c) Predict how the frequency of allele D changes over the following generations. (1 point)

A full-credit answer: The frequency of D stays at 0.90.

Check the box for each point your answer earns

the frequency of D does not change (stays at 0.90).

Common slip: Predicting that D rises because DD tortoises become commoner. More DD tortoises and fewer Dd tortoises hold the same number of copies of D.

(d) Justify your prediction in part (c). (1 point)

A full-credit answer: Mate choice only re-arranges the copies of D and d into genotypes.
No tortoise arrives or leaves, no mutation makes a new copy, and no genotype leaves more offspring, so no copy of D or d is added or removed.
The number of copies of D and the total number of copies are unchanged, so p stays at 0.90.
The population has not evolved at this gene, although its genotype frequencies changed.

Check the box for each point your answer earns

no copy of D or d is added or removed (no migration, mutation or selection), so the allele frequencies are unchanged although the genotype frequencies moved. Accept with or without the statement that the population has not evolved at this gene.

Common slip: Justifying with the falling Dd count alone. The Dd count shows the genotype change; the allele frequency needs the copies counted.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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