IB DP Biology: Natural selection and allele frequencies: Hardy-Weinberg and artificial selection (DP Biology) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions follow the IB Biology guide, first assessment 2025, section D4.1 (Natural selection). Questions 1 to 6 cover SL and HL content. Questions 7 to 12 are HL only. Revise first with the study guide and revision notes. See also the IB Biology hub, the printable checklist and the free diagnostics.

Questions

1. (a) State what is meant by a paradigm shift. [1] (b) Lamarck suggested that a stretched neck gained in life is passed on to offspring. Explain why this is not a valid mechanism for evolution. [2]

2. Outline how mutation and sexual reproduction each contribute to the variation on which natural selection acts. [4]

3. A female fish releases 5000 eggs during her life. The population size stays constant. (a) Calculate the percentage of her eggs that must, on average, survive to reproduce, if she and her mate are replaced by two offspring. [2] (b) Explain why the survivors are unlikely to be a random sample of the eggs. [2]

4. In a lizard population, 80 fast-running and 80 slow-running adults were marked before a severe frost. Afterwards 60 fast and 20 slow lizards were found alive. (a) Calculate the percentage survival of each group. [2] (b) Explain why a frost is described as a density-independent selection pressure. [1] (c) Suggest what further evidence would show that the change in running speed is evolutionary change. [2]

5. Males of one bird species have very long tail feathers. The tails reduce flying speed. Females prefer males with the longest tails. (a) Explain how this could lead to longer tails in the population. [3] (b) Explain why tails do not become indefinitely longer. [2]

6. Guppies were moved into three artificial pools. The mean number of colour spots per male was 6.0 at the start in each pool. After several generations the means were: pool with no predators 8.1, pool with a weak predator 6.3, pool with a strong predator 4.2. (These figures are illustrative, not data from a published study.) (a) Calculate the percentage change in each pool. [3] (b) Explain the results. [3]

7. (HL only) A population of 120 plants has 54 RR, 48 Rr and 18 rr. (a) Define gene pool. [1] (b) Calculate the frequency of each allele. [3]

8. (HL only) Allele t has a frequency of 0.05 on one island and 0.40 on another island 200 km away. (a) Suggest two reasons why the frequencies differ. [2] (b) State one precaution when comparing allele frequencies from a database and name a human example you could search. [2]

9. (HL only) Three populations of a bird were measured for beak depth over many generations. In population A the mean rose from 9 mm to 11 mm. In population B the range narrowed around the mean. In population C two groups of birds with deep and shallow beaks became common while medium beaks became rare. (a) Identify the type of selection in each population. [3] (b) State the effect common to all three types on the gene pool. [1] (c) Suggest one environmental change that could cause the pattern in population C. [1]

10. (HL only) In a population of 2000 mice, 91% show a dominant coat colour. Coat colour is controlled by one gene with two alleles, and the population is in Hardy-Weinberg equilibrium. Calculate the frequency of each allele and the expected number of each genotype. [7]

11. (HL only) A sample of 250 flowers has 90 AA, 100 Aa and 60 aa. (a) Calculate the allele frequencies. [2] (b) Calculate the expected number of each genotype under Hardy-Weinberg. [3] (c) Compare observed and expected numbers, and suggest two Hardy-Weinberg conditions that may not be met. [3]

12. (HL only) In a population, q = 0.5 for a recessive allele. (a) Calculate the new value of q after one generation if homozygous recessive individuals have a relative survival of 0.6 and all others survive equally. [4] (b) Explain why q falls only slowly under this selection. [1] (c) Distinguish between the evolution of antibiotic resistance in bacteria and the improvement of milk yield in dairy cattle. [4]

Answers

1. (a) A fundamental change in the accepted framework of a field of science. [1] (b) A stretched neck is an acquired characteristic caused by the environment, not a change in the base sequence of genes. [1] Gametes carry the unchanged alleles, so offspring are not born with longer necks. [1] [3] Examiner insight: Stating that acquired traits are not inherited is stronger when you give the reason: the DNA of the gametes is unchanged.

2. Mutation changes the base sequence of DNA. [1] It creates new alleles. [1] Sexual reproduction creates new combinations of alleles, through crossing over and independent assortment in meiosis, and random fertilisation. [1] Offspring therefore differ from each other and from their parents, giving variation for selection. [1] [4] Examiner insight: Keep the roles separate: new alleles belong to mutation, new combinations to sexual reproduction.

3. (a) 2 ÷ 5000 [1] × 100 = 0.04% [1] (b) Eggs differ in heritable traits that affect survival. [1] Those with advantageous traits are more likely to survive in competition for limited resources. [1] [4] Examiner insight: State both the proportion that die and that survival depends on traits. A bare statement that “most die” does not explain selection.

4. (a) Fast: 60 ÷ 80 × 100 = 75%. [1] Slow: 20 ÷ 80 × 100 = 25%. [1] (b) It affects survival regardless of population density. [1] (c) Evidence that running speed is heritable, such as faster offspring from fast parents raised in the same conditions. [1] Evidence that the next generation has a higher proportion of fast lizards or a higher frequency of the relevant allele. [1] [5] Examiner insight: Survival figures alone do not show evolution. Credit comes from linking to heritability and the next generation.

5. (a) Females choose long-tailed males, so these males mate more. [1] They leave more offspring carrying alleles for long tails. [1] Allele frequency for long tails increases over generations. [1] (b) Long tails reduce flying speed, so males are more easily caught by predators or have lower survival. [1] The benefit in mating is balanced by the cost in survival. [1] [5] Examiner insight: A full answer shows both pressures, mate choice and survival, acting in opposite directions.

6. (a) No predators: (8.1 − 6.0) ÷ 6.0 × 100 = +35%. [1] Weak predator: (6.3 − 6.0) ÷ 6.0 × 100 = +5%. [1] Strong predator: (4.2 − 6.0) ÷ 6.0 × 100 = −30%. [1] (b) Females prefer brighter males, so sexual selection favours more spots. [1] Predators catch conspicuous males, so natural selection favours fewer spots. [1] Stronger predation outweighs female preference, so spots increase when predators are absent and fall when they are strong. [1] [6] Examiner insight: Give the sign with a percentage change and always divide by the starting value.

7. (a) All the genes and their different alleles in a population. [1] (b) Total alleles = 120 × 2 = 240. [1] R = (2 × 54 + 48) ÷ 240 = 156 ÷ 240 = 0.65. [1] r = (2 × 18 + 48) ÷ 240 = 84 ÷ 240 = 0.35. [1] [4] Examiner insight: Show the allele count method. Check that the two frequencies sum to 1 before moving on.

8. (a) Little or no gene flow between the islands. [1] Different selection pressures on each island, or independent mutations. [1] (b) Check the sample size or that the sample represents the population. [1] Example: lactase persistence allele (or sickle cell allele, or ABO blood group alleles). [1] [4] Examiner insight: Give two distinct reasons. Repeating “isolation” in different words gives only one reason.

9. (a) A: directional selection. [1] B: stabilizing selection. [1] C: disruptive selection. [1] (b) Allele frequencies change. [1] (c) For example, a habitat where seeds are either small and soft or large and hard, so medium beaks cannot feed efficiently. [1] [5] Examiner insight: Name the type, then justify it from the change in distribution, not from the example’s wording alone.

10. Recessive phenotype = 100 − 91 = 9%, so q² = 0.09. [1] q = √0.09 = 0.3. [1] p = 1 − 0.3 = 0.7. [1] Heterozygous 2pq = 2 × 0.7 × 0.3 = 0.42, so 0.42 × 2000 = 840. [1] Homozygous dominant p² = 0.49, so 980. [1] Homozygous recessive q² = 0.09, so 180. [1] Check: 980 + 840 + 180 = 2000. [1] [7] Examiner insight: Start from the recessive phenotype. Using 0.91 as p² is the common error.

11. (a) Alleles = 500. A = (2 × 90 + 100) ÷ 500 = 280 ÷ 500 = 0.56. [1] a = 0.44. [1] (b) AA = 0.56² × 250 = 78.4. [1] Aa = 2 × 0.56 × 0.44 × 250 = 123.2. [1] aa = 0.44² × 250 = 48.4. [1] (c) Observed heterozygotes (100) are fewer than expected (123.2), while both homozygotes are more common than expected. [1] Mating may not be random, as with inbreeding or mating between similar types. [1] Survival may differ between genotypes, or flowers from two separate populations may have mixed. [1] [8] Examiner insight: Compare numbers directly and name the conditions, rather than saying “the population is not in equilibrium”.

12. (a) p = 0.5, so AA = 0.25, Aa = 0.50, aa = 0.25. [1] aa after selection = 0.25 × 0.6 = 0.15. [1] Total = 0.25 + 0.50 + 0.15 = 0.90. [1] New q = (0.50 ÷ 2 + 0.15) ÷ 0.90 = 0.40 ÷ 0.90 = 0.444. [1] (b) Most copies of the recessive allele are in heterozygotes, which are not selected against. [1] (c) Antibiotic resistance: the antibiotic is a selection pressure, resistant bacteria survive and reproduce more, so natural selection. [1] Nobody chooses which bacteria breed. [1] Milk yield: humans select cows with high yield to breed, so artificial selection. [1] Humans choose the parents deliberately. [1] [9] Examiner insight: Define who or what chooses the breeders. Show that the denominator is the sum of the surviving relative numbers.

Where marks are usually lost

• Writing that individuals adapt, rather than populations changing across generations.
• Describing differences in survival without saying the trait is heritable.
• Not showing the starting value in a percentage change calculation.
• Treating 91% dominant phenotype as p² instead of working from q² = 0.09.
• Forgetting to square-root before finding p.
• Not checking that genotype frequencies sum to 1.
• Naming a selection type without describing the change in the distribution.
• Calling antibiotic resistance artificial selection.
• Giving one reason twice when asked for two.

Next steps

• Review the revision notes and the study guide.
• Open the IB Biology hub and tick off the checklist.
• Try all free 10-minute diagnostics at diagnostics.
• Book a free trial class at trial.

Official syllabus

IB Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Section D4.1 Natural selection.