A Level Biology H3 Evolution Diversity Quiz

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A-Level Biology H3 Quiz - Evolution Diversity

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Duration: 60 minutes
Total Marks: 40
Topic: Evolution Diversity (Biological Evolution – Core Idea 4)

Instructions:

• This quiz contains 20 questions mapped to the H3 Biology syllabus (Core Idea 4: Biological Evolution).
• Section A: Knowledge and understanding (1 mark each).
• Section B: Application and explanation (2 marks each).
• Section C: Data handling and synthesis (3 marks each).
• Answer all questions. Write clearly in the spaces provided.
• This is syllabus-first practice content generated from inferred patterns; it is not derived from past-year exam papers.

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Section A: Knowledge and Understanding (Questions 1–5, 1 mark each)

1. State one defining feature of sexual selection.

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2. Give one example of adaptive radiation in island ecosystems.

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3. Define a ring species in one sentence.

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4. Name one mechanism by which polyploidy arises in plants.

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5. State the term for the transfer of genes from one species to another via hybridisation and backcrossing.

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Section B: Application and Explanation (Questions 6–15, 2 marks each)

6. Explain how sexual selection can lead to traits that reduce survival but increase mating success.

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7. Describe how adaptive radiation is supported by the finches of the Galápagos Islands.

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8. Explain why a ring species challenges the biological species concept.

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9. Describe how hybridisation can contribute to evolution in plants.

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10. Explain the role of introgression in reconstructing phylogenies.

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11. State how atmospheric oxygen concentration changes during early evolution facilitated aerobic metabolism.

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12. Explain how biomolecules such as ribosomal RNA are used to infer evolutionary relationships.

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13. Describe one significance of polyploidy in crop plant evolution.

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14. Explain how biochemical processes (e.g. photosynthesis pathways) provide evidence for evolution.

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15. Outline how sexual selection differs from natural selection in mechanism.

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Section C: Data Handling and Synthesis (Questions 16–20, 3 marks each)

16. A population of snails has two colour morphs controlled by a single gene with alleles $C$ (dominant, striped) and $c$ (recessive, plain). In a sample of 200 snails, 128 are striped and 72 are plain. Using the Hardy–Weinberg equation $p^2 + 2pq + q^2 = 1$, calculate the frequency of allele $c$ and the frequency of heterozygous individuals. Show your working.

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17. The diagram below shows a hypothetical ring species distribution around a mountain range.

Image pending generation: map for Q17.

Using the map, explain how this pattern demonstrates a ring species and identify the evolutionary implication. (3 marks)

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18. A researcher compares cytochrome c amino acid sequences among five vertebrate species. The number of differences from Species X are: Y=2, Z=5, W=9, V=12. Explain how this data can be used to construct a phylogenetic tree and state one caution. (3 marks)

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19. Polyploidy events are common in ferns. Describe a scenario where autopolyploidy arises and explain its evolutionary consequence for speciation. (3 marks)

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20. Evaluate how integrating biomolecular evidence with fossil records improves our understanding of adaptive radiation compared with using fossils alone. (3 marks)

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A-Level Biology H3 Quiz - Evolution Diversity: Answer Key

Topic: Evolution Diversity (Biological Evolution – Core Idea 4)
Total Marks: 40
Syllabus-first content; not past-year derived.

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Section A (1 mark each)

1. Sexual selection is selection based on traits that increase mating success rather than survival.
Mark: 1 for correct definition or feature (e.g. mate choice, intrasexual competition).

2. Example: Darwin’s finches on Galápagos Islands diversified into multiple feeding niches.
Mark: 1 for any valid example (e.g. Hawaiian honeycreepers).

3. A ring species is a series of neighbouring populations that can interbreed with adjacent ones but where the two ends are reproductively isolated.
Mark: 1 for single-sentence definition.

4. Mechanism: non-disjunction during meiosis producing unreduced gametes.
Mark: 1 for any correct mechanism (e.g. somatic chromosome doubling).

5. Introgression.
Mark: 1.

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Section B (2 marks each)

6. Traits favoured by mates (e.g. bright plumage) may attract predators; however, increased mating offsets survival cost.
Marks: 1 for trait example, 1 for trade-off explanation.

7. Finches colonised islands, exploited different foods, leading to beak diversification from common ancestor.
Marks: 1 for colonisation/ancestor, 1 for niche divergence.

8. Biological species concept defines species by interbreeding; ring species show continuous interbreeding yet terminal forms do not mate.
Marks: 1 for concept, 1 for contradiction.

9. Hybridisation merges genomes, creating novel combinations and sometimes fertile polyploids.
Marks: 1 for genome mixing, 1 for outcome.

10. Introgression leaves genetic traces, confusing strict tree-like phylogeny; shows networks.
Marks: 1 for gene transfer trace, 1 for phylogenetic implication.

11. Rising O₂ allowed aerobic respiration, higher ATP yield, enabling complex life.
Marks: 1 for O₂ rise, 1 for metabolic benefit.

12. rRNA sequences mutate slowly; similarities indicate common ancestry, differences estimate divergence time.
Marks: 1 for conserved sequence use, 1 for inference.

13. Polyploidy gives larger cells/ organs (gigas effect), aiding domestication (e.g. wheat).
Marks: 1 for example, 1 for significance.

14. Shared biochemical pathways (e.g. Calvin cycle) imply descent from common ancestor.
Marks: 1 for pathway, 1 for evolutionary inference.

15. Sexual selection acts on reproductive success via mates; natural selection on survival/environment.
Marks: 1 each mechanism.

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Section C (3 marks each)

16.
$q^2 = 72/200 = 0.36 \Rightarrow q = \sqrt{0.36} = 0.6$ (frequency of $c$).
$p = 1 - 0.6 = 0.4$.
$2pq = 2 \times 0.4 \times 0.6 = 0.48$ (heterozygotes).
Marks: 1 for q² calc, 1 for q and p, 1 for 2pq.

17. Map shows adjacent interbreeding, A–F isolated despite contact. This is ring species; implies speciation can be gradual around barrier.
Marks: 1 for map reading, 1 for ring species demo, 1 for implication.

18. Fewer differences = closer relative; tree: X–Y–Z–W–V. Caution: convergent amino acid change may mislead.
Marks: 1 for tree logic, 1 for ordering, 1 for caution.

19. Autopolyploidy: unreduced gamete self-fertilises → AAA (triploid) or AAAA. Consequence: instant reproductive isolation from diploid.
Marks: 1 for scenario, 2 for consequence/speciation.

20. Biomolecules give genetic dates/resolutions; fossils give morphology/time; combined reduces missing-record bias.
Marks: 1 biomolecular benefit, 1 fossil benefit, 1 integration value.