Biology

1. 1.

   Distinguish genotype and phenotype, explaining why a dominant phenotype need not reveal a unique genotype.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A visible trait can map to more than one allele combination. A test cross or molecular assay can distinguish those combinations where observation alone cannot.

   Marking points

   • Genotype describes the alleles present.
   • Phenotype is the observable characteristic influenced by genotype and environment.
   • Under complete dominance, both AA and Aa show the dominant phenotype.

   Examiner tip: Dominant does not mean common in a population.

2. 2.

   Explain how independent assortment during meiosis increases variation, specifying the condition under which two genes can be treated as assorting independently.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Each pair's orientation is independent of another pair's. Linked genes on the same chromosome need separate consideration of recombination rather than automatically using independent assortment.

   Marking points

   • Homologous chromosome pairs orient independently at metaphase I.
   • Gametes receive different combinations of maternal and paternal chromosomes.
   • Genes on different chromosome pairs are treated as independently assorting in the simple model.

   Examiner tip: Independent assortment concerns chromosome pairs; crossing over is a distinct source of variation.

3. 3.

   For AaBb x AaBb with complete dominance and independent assortment, calculate the probability of aabb and the expected number among 320 offspring.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Find recessive homozygosity at each locus separately and multiply the independent probabilities. Multiply the joint probability by 320; an expectation need not equal the observed count exactly.

   Marking points

   • P(aa) = 1/4 and P(bb) = 1/4.
   • P(aabb) = 1/16.
   • Expected number = 20.

   Examiner tip: Independence justifies multiplication; it is a stated assumption, not guaranteed for linked genes.

4. 4.

   In a Hardy-Weinberg population, the frequency of recessive homozygotes is 0.09. Calculate recessive allele frequency and heterozygote frequency.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The recessive phenotype represents q^2 under the model, not q itself. Take its square root, subtract from one to obtain p, then calculate 2pq for heterozygotes.

   Marking points

   • q = sqrt(0.09) = 0.30.
   • p = 0.70.
   • 2pq = 0.42.

   Examiner tip: State that the calculation relies on the equilibrium assumptions.

5. 5.

   A small island population loses an allele after a storm, although survivors show no trait-dependent survival difference. Compare genetic drift with natural selection as explanations.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Chance removal can eliminate an allele without making it disadvantageous. The absence of measured trait-dependent survival weakens selection as the immediate explanation, while later reproduction would still need study.

   Marking points

   • Random survival creates a bottleneck and sampling change in allele frequency.
   • Drift is stronger in small populations.
   • Selection requires heritable trait differences linked to reproductive success.
   • The evidence favours drift but does not exclude unmeasured selective differences.

   Examiner tip: Loss of an allele is not automatically evidence that it was harmful.

6. 6.

   A carrier mother and unaffected father have children for an X-linked recessive trait. Calculate the affected probability among sons and among all children, assuming equal sex probability; explain why the father's normal X does not protect sons.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Conditioning on being a son leaves only the maternal X choice. For the probability over all births, multiply the son probability by the probability of receiving the affected maternal X.

   Marking points

   • Mother transmits her affected X with probability 1/2.
   • Affected probability among sons = 1/2.
   • Affected probability among all children = 1/4.
   • Sons inherit the father's Y, while daughters receive his normal X.

   Examiner tip: Keep conditional risk among sons separate from risk per birth.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.