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Natural selection: exam questions

8 questions, 24 marks. Write your answers on paper, then open each mark scheme.

Spec D4.1Questions and mark scheme (PDF)5 pages

Question 1

Paper 1A style

Which process generates new alleles?

  1. Mutation
  2. Crossing over
  3. Random orientation of bivalents
  4. Random fertilisation
[1 mark]
Show mark scheme for question 1

Answer: A [1]

Question 2

Paper 1A style

Which is a density-independent selection pressure?

  1. Competition for nesting sites
  2. A very cold winter
  3. Predation by owls
  4. Spread of disease in a crowded population
[1 mark]
Show mark scheme for question 2

Answer: B [1]

Question 3

Paper 2A style

A gardener grows a plant in rich, fertile soil and it grows very tall. Explain why its offspring will not inherit this extra height.

[2 marks]
Show mark scheme for question 3
  • height was acquired due to an environmental factor / soil, not genes [1]
  • acquired characteristics are not encoded in the base sequence of genes / DNA of gametes [1]
  • so they cannot be passed on / only heritable traits can evolve [1]
  • max 2

Question 4

Paper 1B style

In an experiment modelled on Endler's work, guppies were kept in replicate ponds with no predator, with the weak predator Rivulus, or with the strong predator Crenicichla. The table shows the mean number of coloured spots per male (practice data). (a) Calculate the percentage change in mean spots per male in the no-predator ponds from 0 to 14 months. [2] (b) Compare the results for Rivulus and Crenicichla ponds. [2] (c) Explain the trend in the Crenicichla ponds. [2]

Mean number of spots per male guppy
Treatment0 months5 months14 months
No predator10.011.212.5
Rivulus10.011.012.2
Crenicichla10.09.18.4
[6 marks]
Show mark scheme for question 4
  • (a) (12.5 − 10.0) ÷ 10.0 × 100 [1]
  • (a) +25% / increase of 25% [1]
  • (b) Rivulus: spots increased (10.0 to 12.2) whereas Crenicichla: spots decreased (10.0 to 8.4) [1]
  • (b) both start at the same value / both change in the same time / difference grows with time [1]
  • (c) brightly coloured males are more conspicuous / more likely to be eaten by Crenicichla [1]
  • (c) males with fewer spots survive and reproduce more, passing on alleles for fewer spots [1]
  • (c) predation outweighs sexual selection / female preference for colourful males OWTTE [1]
  • max 6

Question 5

Paper 2B style

Explain how natural selection can lead to evolutionary change in a population.

[6 marks]
Show mark scheme for question 5
  • variation exists between individuals in a population [1]
  • mutation generates new alleles [1]
  • sexual reproduction / meiosis and fertilisation generate new combinations of alleles [1]
  • more offspring are produced than the environment can support / overproduction [1]
  • population exceeds carrying capacity so there is competition for limited resources, e.g. food [1]
  • selection pressures, e.g. predation / extreme temperature, affect individuals differently [1]
  • better-adapted individuals / those with higher fitness survive and reproduce more [1]
  • they pass on alleles for the favourable traits to offspring [1]
  • only heritable traits / not acquired characteristics are passed on [1]
  • over many generations the frequency of the favourable trait / allele increases [1]
  • max 6

Question 6

Paper 1A style

HL only (what this means)HL only: additional Higher Level content, only for HL students. SL students can skip it. What the labels mean

After repeated use of an antibiotic, most bacteria in a hospital population are resistant to it. What type of selection is this?

  1. Artificial selection
  2. Sexual selection
  3. Natural selection
  4. Stabilizing selection
[1 mark]
Show mark scheme for question 6

Answer: C [1]

Question 7

Paper 2A style

HL only (what this means)HL only: additional Higher Level content, only for HL students. SL students can skip it. What the labels mean

In a population in Hardy–Weinberg equilibrium, 16% of individuals show a recessive phenotype. Calculate the percentage of individuals that are heterozygous.

[3 marks]
Show mark scheme for question 7
  • q² = 0.16 so q = 0.4 [1]
  • p = 1 − 0.4 = 0.6 [1]
  • 2pq = 2 × 0.6 × 0.4 = 0.48, so 48% [1]
  • Allow ECF from an incorrect value of q

Question 8

Paper 1B style

HL only (what this means)HL only: additional Higher Level content, only for HL students. SL students can skip it. What the labels mean

A gene in a population of 1000 plants has two alleles, A and a. The table shows the observed genotypes (practice data). (a) Calculate the frequency of allele A. [1] (b) Calculate the expected number of heterozygotes if the population were in Hardy–Weinberg equilibrium. [1] (c) Compare the observed and expected numbers of heterozygotes and suggest which Hardy–Weinberg condition is not met. [2]

Observed genotypes
GenotypeAAAaaa
Number of plants300400300
[4 marks]
Show mark scheme for question 8
  • (a) p = (2 × 300 + 400) ÷ 2000 = 0.5 [1]
  • (b) 2pq = 2 × 0.5 × 0.5 = 0.5, so 500 heterozygotes expected [1] Allow ECF
  • (c) fewer heterozygotes observed (400) than expected (500) [1]
  • (c) mating is non-random, e.g. self-pollination / similar plants mating / survival of heterozygotes is lower [1]
  • (c) Accept population divided / not a single interbreeding population OWTTE
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