OxfordAQA International AS and A-level Biology: Natural selection and speciation (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 3.3.8.1 and 3.3.8.2 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1 (Unit 3, International A-level only). Read the study guide and revision notes first. Course links: the OxfordAQA A-level Biology hub and the checklist.

Questions

1. Define evolution, and name two selection pressures that cause differential survival. [2]

2. A plant grows on soil contaminated with metal. Tolerance is controlled by allele T, with genotypes TT, Tt and tt. Seedlings were genotyped on sowing and again after one season on contaminated soil.

Genotype	TT	Tt	tt
Seedlings sown	24	56	20
Survivors	22	45	3

(a) Calculate the frequency of T among the seedlings sown. [2] (b) Calculate the frequency of T among the survivors, to 2 decimal places. [2] (c) Calculate the percentage of tt seedlings that survived. [1]

3. Use your answers to question 2 to explain why the frequency of T is expected to be higher in the next generation. [4]

4. State which type of selection each situation shows.

(a) After a drought, the mean body mass of a lizard population is unchanged, but the range is narrower. [1] (b) The mean beak depth of a seed-eating bird rises as large, hard seeds become the main food. [1] (c) In one meadow, plants flower very early or very late, and plants flowering in the middle are eaten before setting seed. [1]

5. Leaf hairs deter insects. The number of hairs per leaf section was counted on 100 plants before and after several generations of grazing.

Hairs per section	0	2	4	6	8
Plants before	10	30	35	20	5
Plants after	2	12	31	35	20

(a) Calculate the mean number of hairs before grazing. [2] (b) The mean after grazing is 5.2. Show how this is obtained. [2] (c) Name the type of selection shown. [1]

6. A bird feeds on seeds. Small seeds and very large seeds are plentiful, but medium seeds are scarce. (a) Describe how the distribution of beak sizes is likely to change over many generations. [2] (b) Suggest how this could lead to two species in one area. [2]

7. Explain why two populations of one species, separated for many generations, develop different gene pools. [3]

8. A river changes course and splits a population of ground-dwelling rodents into two groups. Explain how this could lead to allopatric speciation. [5]

9. A grass has 2n = 18. A tetraploid grass arises in the same field. (a) State the chromosome number of the tetraploid. [1] (b) State the chromosome number of a hybrid between the tetraploid and a diploid. [1] (c) Explain why the hybrid is sterile. [2] (d) Explain why this is sympatric speciation. [1]

10. Populations A and B of a small fish were kept separate for many generations. Forty pairs were set up for each cross.

Cross	Pairs producing offspring	Pairs whose offspring were fertile
A × A	39	38
B × B	37	36
A × B	11	0

(a) Calculate the percentage of pairs producing fertile offspring in each cross. [3] (b) Decide whether A and B are separate species and justify your answer. [2] (c) Suggest two reasons why only 11 A × B pairs produced offspring. [2]

11. A resistance allele to a pesticide was tracked in two insect populations. One field was sprayed each generation; the other was not.

Generation	0	10	20	30
Frequency of resistance allele, sprayed field	0.05	0.14	0.36	0.71
Frequency of resistance allele, unsprayed field	0.05	0.05	0.06	0.05

(a) Calculate the change in frequency in the sprayed field from generation 0 to 30, and how many times larger the final frequency is than the starting one. [2] (b) Calculate the mean change in frequency per generation from 0 to 10 and from 20 to 30. [2] (c) Explain the pattern in the sprayed field, and the purpose of the unsprayed field. [4]

Answers

1. Evolution is a change in the allele frequencies in a population [1]; any two of predation, disease, competition for resources [1]. [2] Examiner insight: answers tend to gain credit for “allele frequencies in a population”; “organisms change over time” does not show the genetic meaning.

2. (a) T = (2 × 24 + 56) ÷ (2 × 100) [1] = 104 ÷ 200 = 0.52 [1] (b) Survivors = 70; T = (2 × 22 + 45) ÷ (2 × 70) [1] = 89 ÷ 140 = 0.64 [1] (0.636) (c) 3 ÷ 20 × 100 = 15% [1] Examiner insight: show the count of gene copies (twice the individuals) as well as the final value, as the working explains the denominator.

3. Variation existed and tt plants were poorly adapted to the metal [1]; only 15% of tt survived compared with 92% of TT [1]; survivors with T reproduce and pass on T [1]; so the frequency of T in the gene pool rises (0.52 to 0.64) in the next generation [1]. [4] Examiner insight: a full explanation links survival, reproduction and inheritance, and quotes figures from the table.

4. (a) Stabilising [1] (b) Directional [1] (c) Disruptive [1] Examiner insight: each answer is a single word, so the name must match the pattern: narrower range with an unchanged mean is stabilising.

5. (a) Sum of (hairs × plants) = 0 + 60 + 140 + 120 + 40 [1] = 360, so mean = 360 ÷ 100 = 3.6 [1] (b) 0 + 24 + 124 + 210 + 160 [1] = 518, and 518 ÷ 100 = 5.2 (5.18) [1] (c) Directional selection [1] Examiner insight: “show that” answers need the substituted working, not just the final figure.

6. (a) Beaks of medium size become less frequent [1]; small and large beak sizes become more frequent, giving two peaks [1]. [2] (b) If small-beaked and large-beaked birds feed in different places or breed mainly with their own type [1], gene flow between them is reduced and the gene pools diverge until they cannot interbreed to produce fertile offspring [1]. [2] Examiner insight: name the process (disruptive selection) and describe the result; “they split” without a mechanism is not enough.

7. No gene flow between the populations [1]; different mutations occur and different alleles are selected in each environment [1]; chance changes in allele frequency also differ, so allele frequencies diverge [1]. [3] Examiner insight: each mark is a separate cause, so list mutation, selection and chance rather than repeating “different environment”.

8. The river acts as a physical barrier, which stops gene flow [1]; mutations arise independently in each group [1]; selection pressures differ, so different alleles are favoured [1]; allele frequencies and gene pools diverge over generations [1]; if the groups later meet and fail to produce fertile offspring together, they are different species [1]. [5] Examiner insight: the barrier is not the cause of the new species, so state that it stops gene flow and then give the divergence.

9. (a) 36 [1] (b) 27 [1] (18 + 9) (c) An odd number of chromosome sets means chromosomes cannot pair in homologous sets at meiosis [1]; so viable gametes are not made, giving no fertile offspring [1]. [2] (d) The tetraploid arises in the same place as the diploids, with no geographical barrier [1]. Examiner insight: for chromosome numbers, state which parent gives which gametes (the tetraploid gives 18, the diploid 9) so the arithmetic can be followed.

10. (a) A × A = 38 ÷ 40 × 100 = 95% [1]; B × B = 36 ÷ 40 × 100 = 90% [1]; A × B = 0% [1] (b) They are separate species [1], because crosses between them produce no fertile offspring although within-population crosses do [1]. [2] (c) Any two: different courtship behaviour; different breeding times; incompatible gametes or reproductive structures; embryos fail to develop [1] [1]. [2] Examiner insight: the conclusion must use both parts of the species idea, fertile offspring and inability to interbreed, with a figure.

11. (a) 0.71 − 0.05 = 0.66 [1]; 0.71 ÷ 0.05 = 14.2 times [1] (b) 0 to 10: (0.14 − 0.05) ÷ 10 = 0.009 per generation [1]; 20 to 30: (0.71 − 0.36) ÷ 10 = 0.035 per generation [1] (c) Spraying kills susceptible insects [1]; resistant insects survive and reproduce more [1]; the resistance allele is passed on and its frequency rises, faster as it becomes more common [1]; the unsprayed field shows no change without the pressure, so the rise is due to spraying [1]. [4] Examiner insight: compare the sprayed field with the control using figures from both, and use “allele frequency” not “the insects became resistant”.

Where marks are usually lost

• Writing “adapt to” or “need to” instead of “variation existed and was selected”.
• Omitting that survivors reproduce and pass on alleles.
• Calculating a phenotype frequency when the question asks for an allele frequency.
• Forgetting to double the number of individuals when counting gene copies.
• Giving a rise in frequency without units, or confusing absolute change and percentage change.
• Naming the selection type without describing the change in the distribution.
• Saying “cannot interbreed” without “produce fertile offspring”.
• Treating the geographical barrier as the cause of speciation, rather than the reason for no gene flow.

Next steps

• Review with the revision notes and the study guide.
• Return to the OxfordAQA A-level Biology hub and tick off the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class at /trial/.

Official syllabus

OxfordAQA, International AS and A-level Biology (9610), specification Version 5.1, sections 3.3.8.1 and 3.3.8.2 (Unit 3, Populations and genes).