Cambridge International AS & A Level Biology 9700: Selection and evolution – 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 topic 17, Selection and evolution, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027: sections 17.1 Variation, 17.2 Natural and artificial selection and 17.3 Evolution. This is A Level content, examined on Paper 4, with the t-test also possible on Paper 5. Calculators are allowed; formulae are given where needed.

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Questions

1. Explain the differences between continuous variation and discontinuous variation, including their genetic basis. [3]

2. Fruit mass in a plant is controlled by three unlinked genes, D/d, E/e and F/f. Each dominant allele adds the same mass to the fruit; recessive alleles add nothing. Two plants with genotype DdEeFf are crossed.

(a) State how many different fruit-mass classes are expected among the offspring. [1] (b) Calculate the fraction of offspring expected to have exactly three dominant alleles. [2] (c) Suggest why fruit mass in a field of these plants shows continuous variation rather than a set of distinct classes. [2]

3. Shell thickness of a species of sea snail was measured on an exposed shore and on a sheltered shore.

	Mean thickness / mm	Standard deviation / mm	n
Exposed shore	2.64	0.31	15
Sheltered shore	2.31	0.28	15

Use t = (x̄₁ − x̄₂) / √(s₁²/n₁ + s₂²/n₂) to decide whether the difference in mean shell thickness is significant. Critical values of t at p = 0.05: 26 df, 2.06; 28 df, 2.05; 30 df, 2.04. [5]

4. A recessive allele causes an inherited condition that affects 1 in 2500 people in a population. Use p + q = 1 and p² + 2pq + q² = 1.

(a) Calculate the expected number of heterozygous carriers in a population of 100 000. [4] (b) State two conditions that must apply for this calculation to be valid. [2]

5. A population of songbirds lives in a forest where the climate has been stable for a long time. Clutch size (number of eggs laid) ranges from 2 to 8, with a mean of 5. Nests with very small clutches produce few young, and nests with very large clutches often lose most of their young through starvation.

(a) Name the type of selection acting on clutch size. [1] (b) Describe and explain the effect of this selection on the distribution of clutch size over many generations. [3]

6. Outline how a population of bacteria becomes resistant to an antibiotic. [5]

7. On a mainland, the frequency of allele b is 0.01. Twenty mainland birds, three of which are heterozygous Bb and the rest BB, are blown to a remote island and found a new population.

(a) Calculate the frequency of allele b in the founding population. [2] (b) Name this effect and explain why the island population’s allele frequencies may continue to differ from those on the mainland. [2]

8. Explain how a farmer could use selective breeding to increase the milk yield of a herd of dairy cattle. [4]

9. This question is about selective breeding in crop plants.

(a) Explain how inbreeding and hybridisation are used to produce vigorous, uniform varieties of maize. [5] (b) Outline how disease resistance can be introduced into a high-yielding variety of rice. [2]

10. A 600-base section of the same gene was sequenced in three species, A, B and C. The number of base differences between each pair was: A–B, 12; A–C, 45; B–C, 43.

(a) Calculate the percentage difference between A and C. [1] (b) Explain what these data suggest about the evolutionary relationships of A, B and C. [3]

11. In a large lake, one species of fish feeds in the shallow margins. Some individuals begin to feed on snails on the deep lake floor. Over many generations the deep-feeding fish breed earlier in the year and in deeper water than the shallow-feeding fish.

(a) Name the type of speciation that could result. [1] (b) Explain how the two forms could become separate species. [5] (c) Suggest how DNA sequence data could be used to support the idea that the two forms are diverging. [2]

12. In a beetle population, green body colour (G) is dominant to brown (g). In year 1, 36% of beetles were brown. The vegetation then became greener, and by year 20 only 16% of beetles were brown. Use p + q = 1 and p² + 2pq + q² = 1.

(a) Calculate the frequencies of alleles G and g in year 1. [2] (b) Calculate the proportion of green beetles that were heterozygous in year 1 and in year 20. [3] (c) Name the type of selection shown and explain the change in allele frequency. [3]

Answers

1. Discontinuous: a few distinct classes; continuous: a range between two extremes [1]. Discontinuous: one or a few genes, alleles with large effects [1]. Continuous: many genes (polygenes) with small additive effects, plus a large environmental effect [1]. [3] Examiner insight: Each comparison mark needs both sides; describing only continuous variation cannot score the genetic-basis mark.

2. (a) Offspring can carry 0 to 6 dominant alleles, so 7 classes [1]. (b) Proportions 1 : 6 : 15 : 20 : 15 : 6 : 1 out of 64, so exactly three dominant alleles = 20/64 [1] = 5/16 (0.3125) [1]. (c) The environment (e.g. light, water, mineral ions) also affects fruit mass [1], blurring the genetic classes into a continuous range [1]. [5] Examiner insight: In (b) 20/64 unsimplified earns full credit; the method mark is for the 64 equally likely combinations.

3. s₁²/n₁ = 0.31²/15 = 0.00641 and s₂²/n₂ = 0.28²/15 = 0.00523 [1]. t = 0.33 / √0.01163 = 0.33 / 0.1079 = 3.06 [1]. Degrees of freedom = 15 + 15 − 2 = 28 [1]. 3.06 is greater than the critical value of 2.05 [1], so the difference is significant: reject the null hypothesis that there is no difference between the means [1]. [5] Examiner insight: The degrees-of-freedom mark is separate; using 30 df (n₁ + n₂) loses it even though the conclusion is unchanged.

4. (a) q² = 1/2500 = 0.0004 [1]; q = 0.02 and p = 0.98 [1]; 2pq = 2 × 0.98 × 0.02 = 0.0392 [1]; carriers = 0.0392 × 100 000 = 3920 [1]. (b) Any two, one mark each, from: large population, random mating, no mutation, no migration, no selection. For example: the population is large [1] and mating is random [1]. [6] Examiner insight: Error carried forward is allowed: a wrong q used correctly later can still earn the later method marks.

5. (a) Stabilising selection [1]. (b) Birds with clutches near 5 raise the most young, so their alleles are passed on more often [1]. Alleles for extreme clutch sizes become less frequent [1]. The mean stays about 5 but the range narrows [1]. [4] Examiner insight: “The mean stays the same” alone is not enough; the narrowing range is the key point.

6. A random mutation gives an allele for resistance, e.g. coding for an enzyme that breaks down the antibiotic [1]. The antibiotic kills susceptible bacteria [1]; resistant bacteria survive [1]. They reproduce by binary fission, passing on the allele, which may also spread on plasmids [1]. Over generations the resistance allele’s frequency increases [1]. [5] Examiner insight: No credit for saying the antibiotic causes the mutation or that bacteria become “immune”.

7. (a) 40 alleles in the founders, of which 3 are b [1]; frequency = 3/40 = 0.075 [1]. (b) The founder effect [1]. The island population is small and isolated, so genetic drift can change allele frequencies by chance, and no gene flow from the mainland restores them [1]. [4] Examiner insight: Dividing by 20 birds instead of 40 alleles loses both marks in (a).

8. Measure each cow’s milk yield and select the highest-yielding cows [1]. Bulls produce no milk, so choose bulls whose daughters have high yields (progeny testing) [1]. Inseminate the selected cows artificially with these bulls’ semen [1]. Select the best offspring and repeat for many generations, increasing the frequency of high-yield alleles [1]. [4] Examiner insight: “Choose the best bull” scores nothing without the reason or the method (daughters’ yield).

9. (a) Self-pollination over several generations produces inbred lines [1], homozygous at most loci, so each line is uniform [1]. Inbred lines are weak because harmful recessive alleles are homozygous (inbreeding depression) [1]. Crossing two inbred lines gives F1 hybrids heterozygous at many loci: hybrid vigour [1]. All F1 plants share one genotype, so the crop is uniform; new F1 seed is bought each year [1]. (b) Cross the high-yielding rice with a variety that carries an allele for disease resistance [1]. Select resistant offspring and cross them repeatedly with the high-yielding variety over several generations, keeping only resistant plants each time [1]. [7] Examiner insight: Credit needs “heterozygous” linked to vigour and “same genotype” linked to uniformity; vigour from homozygosity scores zero.

10. (a) 45/600 × 100 = 7.5% [1]. (b) A and B have the fewest differences, so are most closely related [1] and share the most recent common ancestor [1]. C differs from both similarly, so it diverged from the A–B line earlier [1]. [4] Examiner insight: Say “share a more recent common ancestor”, not “A evolved from B”; living species are not ancestors of one another.

11. (a) Sympatric speciation [1]. (b) The two forms are separated ecologically (food, depth) and in breeding time and place [1], so there is no gene flow between them [1]. Different selection pressures act on each, e.g. for feeding on snails [1]. Mutation and genetic drift also act independently, so the two gene pools diverge [1]. Eventually the forms can no longer interbreed to produce fertile offspring [1]. (c) Sequence the same genes in many fish of each form [1]; consistent base differences between the forms, increasing over generations, would show their gene pools are diverging [1]. [8] Examiner insight: Naming “reproductive isolation” without its consequence (no gene flow, diverging gene pools) earns at most one mark.

12. (a) q² = 0.36, so q (g) = 0.6 [1]; p (G) = 1 − 0.6 = 0.4 [1]. (b) Year 1: 2pq = 0.48 and p² = 0.16, so heterozygous green = 0.48/0.64 = 0.75 [1]. Year 20: q = 0.4, p = 0.6, so 2pq = 0.48 and p² = 0.36 [1]; heterozygous green = 0.48/0.84 = 0.571 [1]. (c) Directional selection [1]. Against greener vegetation, green beetles were better camouflaged, so fewer were taken by predators [1]; green beetles survived and reproduced more, passing on G, so the frequency of g fell from 0.6 to 0.4 [1]. [8] Examiner insight: In (b) divide by all green beetles (p² + 2pq); giving 0.48 loses the accuracy marks.

Where marks are usually lost

• Writing “genes” instead of “alleles” when describing changes in a population’s gene pool.
• Leaving out “survive and reproduce, passing on alleles”: survival alone does not complete a natural selection answer.
• Using n₁ + n₂ instead of n₁ + n₂ − 2 for t-test degrees of freedom.
• Starting a Hardy–Weinberg calculation from the dominant phenotype instead of q².
• Naming a type of speciation without explaining the lack of gene flow.
• Dividing by the whole population when asked about one phenotype.

Next steps

• Selection and evolution revision notes
• Selection and evolution study guide
• Cambridge A Level Biology course hub
• Printable 9700 checklist
• 9700 A Level diagnostic and all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027 (Version 1), Cambridge University Press & Assessment – topic 17, Selection and evolution.