OxfordAQA International AS and A-level Biology: Allele frequencies, drift and Hardy-Weinberg (9610) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.3.7.1 and 3.3.7.2 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, for International A-level exams May/June 2018 onwards. Unit 3 is listed under the International A-level assessments. Figures in questions are data supplied. Learn the ideas in the study guide first.

Questions

1. Define each of the following terms: population, gene pool and allele frequency. [3]

2. A gene controlling throat colour in a lizard has two codominant alleles, S and T. A sample of 160 lizards contains 56 SS, 72 ST and 32 TT.

(a) Calculate the frequency of allele S. [2] (b) If the population were in Hardy-Weinberg equilibrium, calculate the number of ST lizards expected in the sample. [2]

3. An allele has a frequency of 0.10 in population A, which has 10 individuals, and in population B, which has 500 individuals.

(a) Calculate the number of copies of the allele in each population. [2] (b) Explain why genetic drift is more likely to remove this allele from population A than from population B. [2]

4. A population of 6000 salamanders contains a rare allele at a frequency of 0.01. A landslide destroys most of the habitat and leaves 12 survivors.

(a) Calculate the copies of the allele in the original population. [1] (b) Calculate the number of copies of this gene in the survivors, and the number of copies of the rare allele you would expect among them if they were a random sample. [2] (c) The probability that the rare allele is absent from the survivors is 0.99 raised to the power of the number of gene copies in the survivors. Calculate this probability. [1] (d) Suggest why the allele frequency in the survivors may differ from 0.01. [1]

5. Dairy farmers have bred cattle for high milk yield for many generations.

(a) Explain how repeated selection of the highest-yielding cows and bulls as parents changes the gene pool of the breed. [3] (b) Suggest two disadvantages of having a gene pool with little variety. [2]

6. (a) Give three conditions required for a population to be in Hardy-Weinberg equilibrium. [3] (b) A population of plants is grazed by animals that eat only plants with one particular phenotype. Explain why the Hardy-Weinberg equation would not correctly predict the genotype frequencies in the next generation. [2]

7. A recessive disorder affects 1 in 625 individuals in a large, randomly mating population with no selection, mutation or migration.

(a) Work out the frequency of the recessive allele. [2] (b) Work out the frequency of the dominant allele. [1] (c) Work out the frequency of carriers. [1] (d) Work out how many carriers are expected in a population of 50 000. [1]

8. In a meadow of 2500 plants, 94.24% have hairy leaves. Hairy (H) is dominant to smooth (h). Hardy-Weinberg conditions hold.

(a) Find the frequencies of alleles H and h. [3] (b) Find how many plants are heterozygous. [2] (c) Find how many plants have smooth leaves. [1]

9. A population of 300 beetles contains 135 AA, 120 Aa and 45 aa individuals.

(a) Calculate the frequency of allele a. [2] (b) Calculate the numbers of AA, Aa and aa beetles expected if the population were in Hardy-Weinberg equilibrium. [3] (c) Compare the observed and expected numbers and suggest one reason for the difference. [2]

10. A population of 4000 tree frogs has a recessive allele d at a frequency of 0.30. A disease outbreak leaves 20 survivors: 6 DD, 10 Dd and 4 dd. The disease affects all genotypes equally.

(a) Calculate the frequency of allele d among the survivors. [2] (b) Name the type of event the outbreak represents for this population. [1] (c) Explain why the change in the frequency of d is an example of genetic drift and not natural selection. [2] (d) The survivors breed and produce a generation of 400 frogs. Assuming Hardy-Weinberg conditions from now on, calculate the number of dd frogs expected. [2] (e) Suggest one reason why Hardy-Weinberg conditions are unlikely to hold in a population this small. [1]

11. The table shows illustrative data for wheat.

	Wild relatives	Modern high-yield variety
Mean number of alleles per gene studied	7.4	2.9
Mean grain yield (tonnes per hectare)	3.1	8.4

(a) Calculate the percentage decrease in the mean number of alleles per gene. [2] (b) Calculate the percentage increase in mean grain yield. [2] (c) Explain how selection for high yield could produce these results, and describe one risk of growing the modern variety over a large area. [3]

Answers

1. Population: organisms of one species that could interbreed, living in the same place at the same time [1]. Gene pool: every allele of a gene present in one population [1]. Allele frequency: the share of the gene copies in the population that are a chosen allele [1]. [3] Examiner insight: Include the qualifiers: one species, same place and time, every allele, a proportion of all copies.

2. (a) Copies of S = (2 × 56) + 72 = 184 [1]; total copies = 2 × 160 = 320, so frequency = 184 ÷ 320 = 0.575 [1]. (b) q = 1 − 0.575 = 0.425, so 2pq = 2 × 0.575 × 0.425 = 0.48875 [1]; 0.48875 × 160 = 78.2, so about 78 lizards [1]. [4] Examiner insight: Show 2N as the denominator. Dividing 184 by 160 gives a value above 1: a sign of mixing up alleles and individuals.

3. (a) Population A: 2 × 10 × 0.10 = 2 copies [1]. Population B: 2 × 500 × 0.10 = 100 copies [1]. (b) In A there are few copies, so a chance event removes a large fraction of all copies [1]. In B there are many copies, so chance events affect a small fraction and the effects average out, so the allele is much less likely to be lost [1]. [4] Examiner insight: Tie your explanation to the number of copies: each chance event is a larger proportion of them.

4. (a) 2 × 6000 × 0.01 = 120 copies [1]. (b) Copies of the gene in the survivors = 2 × 12 = 24 [1]; expected copies of the rare allele = 24 × 0.01 = 0.24 [1]. (c) 0.99²⁴ = 0.79 (to 2 significant figures) [1]. (d) The survivors form a small random sample, so the allele may by chance be over- or under-represented [1]. [5] Examiner insight: In (d), say the survivors are a chance sample, not that the allele was “weaker”.

5. (a) Cows and bulls with a high milk yield are chosen to breed, and the rest are not [1]. Alleles for high yield are passed on more often, so their frequency rises [1]. Other alleles become less frequent or are lost, so the gene pool becomes smaller and less varied [1]. (b) Any two from: little variation for selection to act on if conditions change [1]; more individuals susceptible to a new disease at the same time [1]; breeding between relatives makes harmful homozygous recessive genotypes more likely. [5] Examiner insight: Link the farmer’s choice to a change in allele frequency, not just to “better cows”.

6. (a) Any three from: a large population [1]; mating at random [1]; mutation absent [1]; no gene flow into or out of the population; no selection. (b) The condition of no selection is broken [1]. The plants with the eaten phenotype are less likely to survive and reproduce, so the allele frequencies change and the genotype frequencies do not match p² + 2pq + q² for the original frequencies [1]. [5] Examiner insight: Name the condition broken, then say what it does to allele frequency.

7. (a) q² = 1 ÷ 625 = 0.0016 [1]; q = √0.0016 = 0.04 [1]. (b) p = 1 − 0.04 = 0.96 [1]. (c) 2pq = 2 × 0.96 × 0.04 = 0.0768 [1]. (d) 0.0768 × 50 000 = 3840 carriers [1]. [5] Examiner insight: “1 in 625” is q², not q. Write q² = 1/625 before taking the square root.

8. (a) The smooth phenotype frequency is 1 − 0.9424 = 0.0576, so q² = 0.0576 [1]; q = 0.24 [1]; p = 0.76 [1]. (b) 2pq = 2 × 0.76 × 0.24 = 0.3648 [1]; 0.3648 × 2500 = 912 plants [1]. (c) 0.0576 × 2500 = 144 plants [1]. [6] Examiner insight: The 94.24% is the dominant phenotype, which is p² + 2pq, not p². Subtract it from 1 to get q² first.

9. (a) Copies of a = (2 × 45) + 120 = 210 [1]; frequency = 210 ÷ 600 = 0.35 [1]. (b) p = 0.65 [1]. AA = 0.65² × 300 = 126.75 [1]; Aa = 2 × 0.65 × 0.35 × 300 = 136.5 and aa = 0.35² × 300 = 36.75 [1]. (c) There are fewer heterozygotes than expected (120 against 136.5) and more of both homozygotes (135 against 126.75; 45 against 36.75), so the population is not in equilibrium [1]. A possible reason is non-random mating, such as mating between relatives, or selection against the heterozygotes [1]. [7] Examiner insight: Quote observed and expected numbers and say how they differ.

10. (a) Copies of d = (2 × 4) + 10 = 18 [1]; total copies = 2 × 20 = 40, so frequency = 18 ÷ 40 = 0.45 [1]. (b) A genetic bottleneck [1]. (c) The frequency rose from 0.30 to 0.45, but the disease affected all genotypes equally, so d gave no survival advantage [1]. The change arose because the 40 surviving gene copies were a small random draw from the alleles present, which is genetic drift [1]. (d) q² = 0.45² = 0.2025 [1]; 0.2025 × 400 = 81 frogs [1]. (e) In a very small population drift is large (or mating between relatives is likely), so the large-population and random-mating conditions are not met [1]. [8] Examiner insight: In (c), “the disease did not select” is not enough alone. Say what did cause it: a small random sample. In (d), use the new q of 0.45.

11. (a) Decrease = 7.4 − 2.9 = 4.5 [1]; (4.5 ÷ 7.4) × 100 = 60.8% [1]. (b) Increase = 8.4 − 3.1 = 5.3 [1]; (5.3 ÷ 3.1) × 100 = 171% [1]. (c) Breeders bred from the highest-yielding plants, so alleles for high yield increased in frequency [1] while other alleles were lost, so the gene pool became less varied [1]. Risk: with little variation, a new disease or pest could affect most of the crop at once [1]. [7] Examiner insight: For percentage change, divide by the starting value (the wild relatives for (a), the lower yield for (b)).

Where marks are usually lost

• Using a recessive phenotype frequency (such as 1 in 625) as q rather than q².
• Dividing by the number of individuals instead of 2N when finding allele frequency.
• Missing the 2 in 2pq.
• Describing drift as “survival of the fittest”.
• Giving a vague Hardy-Weinberg condition instead of a named one.
• Using the wrong base value for a percentage change.

Next steps

• Revisit the study guide and the revision notes.
• Tick off the printable checklist and see the course hub.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, for International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, published by OxfordAQA. Sections 3.3.7.1 and 3.3.7.2.