OCR A Level Biology A: Populations, Hardy-Weinberg, speciation and artificial selection (H420) – 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 section 6.1.2(e) to (h) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: selection, genetic drift, the bottleneck and founder effect, Hardy-Weinberg, speciation and artificial selection. Revise first with the study guide and the revision notes. See also the OCR A Level Biology hub, the checklist and the free diagnostics. In the Hardy-Weinberg questions, assume p + q = 1 and p² + 2pq + q² = 1 are given.

Questions

1. Distinguish between stabilising selection and directional selection. [4]

2. A recessive allele causes a trait that is shown by 169 plants in a meadow population of 10 000. Assume the population is in Hardy-Weinberg equilibrium.

(a) Calculate the frequencies of the recessive allele, q, and the dominant allele, p. [2] (b) Calculate the number of heterozygous plants. [2] (c) Calculate the frequency of plants showing the dominant phenotype. [1]

3. In a sample of 200 rabbits, 98 are BB, 76 are Bb and 26 are bb.

(a) Calculate the frequency of allele B. [2] (b) Calculate the number of Bb rabbits expected if the sample were in Hardy-Weinberg equilibrium. [2] (c) Suggest what the difference between the expected and observed Bb numbers might show. [1]

4. In a population the frequency of recessive allele a is 0.6. Individuals of genotype aa die before they can breed.

(a) Calculate the frequency of a among the breeding adults. [2] (b) Calculate the frequency of aa individuals in the next generation. [1] (c) Explain why allele a is not removed from the population in one generation. [1]

5. A population of 6000 voles had a frequency of 0.40 for allele T. A flood left 24 survivors: 6 TT, 12 Tt and 6 tt.

(a) Calculate the frequency of T in the survivors. [2] (b) Name this event and explain why the frequency of T has changed. [2] (c) Suggest one consequence of this event for the population’s future. [1]

6. A river changed course and split a population of freshwater fish into two groups. The groups live in different lakes. Explain how this could lead to the formation of two species. [6]

7. Two populations of a plant grow in the same meadow. One flowers in May and the other in July.

(a) Name this isolating mechanism. [1] (b) Explain how it could lead to two separate species. [3] (c) Name one other reproductive isolating mechanism and say how it prevents gene flow. [1]

8. A grower wants a tomato variety with firm fruit and resistance to a fungal disease. Both features are found in existing plants, but not together. Describe how the grower could produce the new variety. [5]

9. Explain why seed banks and populations of wild relatives of crops are kept. [3]

10. Some breeds of dog have been selected for very short noses and have breathing difficulties. Discuss the ethical considerations of using artificial selection in this way. [6]

11. The table shows the mean number of eggs laid per hen per year in a flock selected for egg production. Only the top-laying hens and a few cockerels from their families were bred in each generation.

Generation	0	2	4	6	8	10
Mean eggs per hen per year	238	251	262	270	275	276

(a) Calculate the percentage increase in mean egg number between generation 0 and generation 10. [2] (b) Compare the mean increase per generation between generations 0 and 4 with that between generations 6 and 10. [2] (c) Suggest two reasons why the increase slows. [3]

Answers

1. Stabilising selection favours individuals near the mean of a feature, and selects against both extremes [1]. The mean stays the same and the range of values narrows [1]. Directional selection favours one extreme of the feature [1]. The mean shifts towards that extreme over generations [1]. [4] Examiner insight: Answers that name both the favoured phenotype and the effect on the mean are easy to credit. Saying only that “the graph changes” does not tell the examiner which type you mean.

2. (a) q² = 169 ÷ 10 000 = 0.0169, so q = 0.13 [1]. p = 1 − 0.13 = 0.87 [1]. [2] (b) 2pq = 2 × 0.87 × 0.13 = 0.2262 [1]. Number = 0.2262 × 10 000 = 2262 [1]. [2] (c) p² + 2pq = 1 − 0.0169 = 0.9831 [1]. [1] Examiner insight: The recessive phenotype gives q², so take the square root before doing anything else. Showing the line q² = 0.0169 makes your method clear even if you slip later.

3. (a) Total alleles = 2 × 200 = 400 [1]. B alleles = (2 × 98) + 76 = 272, so frequency = 272 ÷ 400 = 0.68 [1]. [2] (b) q = 0.32 [1]. Expected Bb = 2 × 0.68 × 0.32 × 200 = 87.04, about 87 [1]. [2] (c) There are fewer heterozygotes than expected (76 against 87). This could suggest non-random mating, such as breeding between relatives, or selection against heterozygotes [1]. [1] Examiner insight: Count alleles, not individuals. A fair comparison between observed and expected numbers states both values.

4. (a) Breeders are AA (0.16) and Aa (0.48) [1]. Allele a is half of the Aa breeders, so q = 0.24 ÷ (0.16 + 0.48) = 0.24 ÷ 0.64 = 0.375 [1]. [2] (b) 0.375² = 0.14 (0.141) [1]. [1] (c) Heterozygotes (Aa) carry the allele but do not show the phenotype, so selection cannot act against them [1]. [1] Examiner insight: In (c), the key idea is that the allele is hidden in carriers. An answer such as “mutation keeps producing it” does not explain why it persists in this generation.

5. (a) Total gene copies = 2 × 24 = 48 [1]. T copies = (2 × 6) + 12 = 24, so frequency = 24 ÷ 48 = 0.50 [1]. [2] (b) A genetic bottleneck [1]. The survivors were a small chance sample of the gene pool, not chosen by an advantage, so their allele frequency differs from the original 0.40 by chance (genetic drift) [1]. [2] (c) Fewer alleles are left in the gene pool, so there is less variation for natural selection to act on if conditions change [1]. [1] Examiner insight: Naming the event is one point and explaining the change is another. A frequency that changes “because T was better” would describe selection, not a bottleneck.

6. The river is a geographical barrier that stops the groups interbreeding, so there is no gene flow between them [1]. Mutations occur independently in each group [1]. The lakes differ in food, predators or temperature, so different alleles are selected in each [1]. In a small group, chance (genetic drift) may also change allele frequencies [1]. Over many generations the allele frequencies diverge [1]. If the groups meet again they cannot interbreed to produce fertile offspring, so they are separate species [1]. [6] Examiner insight: A six-mark answer needs a chain of linked steps. “They adapted to their lakes” with no mention of gene flow or selection earns little.

7. (a) Temporal isolation [1]. [1] (b) The two groups flower at different times, so pollen cannot pass between them and there is no gene flow [1]. Mutations and selection then act independently on each group, so allele frequencies diverge [1]. Eventually they cannot interbreed to produce fertile offspring, so they are different species (sympatric speciation) [1]. [3] (c) Behavioural isolation: courtship signals, such as a song or display, are not recognised by the other group, so mating does not occur [1]. (Other correct mechanisms are credited.) [1] Examiner insight: In (c), a name alone is incomplete. State how it stops gene flow.

8. Choose plants with firm fruit and plants with disease resistance, and cross them [1]. Grow the offspring and select those showing both features [1]. Breed from these selected plants [1]. Repeat the selection over several generations [1]. The frequency of the alleles for firmness and resistance rises until the variety breeds true [1]. [5] Examiner insight: Show that selection is repeated over generations. A single cross, followed by “the offspring are used”, does not describe selective breeding.

9. Wild relatives and old varieties carry alleles, for example for disease resistance or drought tolerance, that have been lost from modern varieties [1]. If a new disease or a changed climate appears, these alleles can be crossed into crops [1]. Once an allele is lost it cannot be recovered, so the stores keep genetic diversity for future selective breeding [1]. [3] Examiner insight: Say what is stored (alleles) and what it is for (future breeding). “In case they are needed” is too vague.

10. For: dogs give companionship, and some owners value the appearance [1]. Breeders can supply what buyers want [1]. Against: a short nose can cause breathing difficulty, so the dog suffers [1]. Dogs cannot consent to being bred for human preferences [1]. A narrow gene pool raises the frequency of harmful recessive alleles [1]. A balanced conclusion weighs the benefit to humans against the harm to the animal, and may suggest breed standards that put health first [1]. [6] Examiner insight: A discussion question rewards both sides and a final judgement. A list of only harms, with no weighing, is limited.

11. (a) Increase = 276 − 238 = 38 [1]. 38 ÷ 238 × 100 = 16.0% [1]. [2] (b) Generations 0 to 4: (262 − 238) ÷ 4 = 6.0 eggs per generation [1]. Generations 6 to 10: (276 − 270) ÷ 4 = 1.5 eggs per generation, four times slower [1]. [2] (c) Alleles for lower egg number are lost, so the variation available for selection falls [1]. Alleles for high egg number become fixed, so there is little difference between hens [1]. There may be a physiological limit to egg production, or inbreeding in the small breeding group may reduce fitness [1]. [3] Examiner insight: Quote values from the table when you compare rates. The slowing is best explained by the reduced variation, not by “the hens get tired”.

Where marks are usually lost

• Taking the recessive phenotype frequency as q rather than q².
• Counting individuals instead of gene copies when finding an allele frequency.
• Forgetting that the dominant phenotype is p² + 2pq.
• Describing a bottleneck or the founder effect as selection.
• Writing that “the population adapted” in a speciation answer.
• Leaving out gene flow or fertile offspring in a speciation answer.
• Naming an isolating mechanism without saying how it stops gene flow.
• Describing one cross as selective breeding, with no repeated selection.
• Giving only one side in a discussion of ethics.

Next steps

• Review the revision notes and the study guide.
• Use the OCR A Level Biology hub and the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class at /trial/.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: section 6.1.2(e) to (h), Patterns of inheritance (stabilising and directional selection, genetic drift, the genetic bottleneck and founder effect, Hardy-Weinberg, isolating mechanisms and artificial selection).