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Evolution, biodiversity and genetic technology

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AI-assisted English content review; independent teacher approval pending.

Practice scope and review

Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.

72 original questions across 15 selected topic areas. No account or payment required to practise.

Cambridge International AS & A Level Biology (9700), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.

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Original English concept practice addressing selected identified gaps. Not complete syllabus coverage or practical competence.

Selected subtopics: Speciation and conservation · Diversity indices and test choice · PCR and DNA separation · Medical and agricultural ethics

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Question 1Two sexually reproducing animal populations are separated by a new geographic barrier. After many generations they no longer produce fertile offspring when reunited under suitable conditions. Explain a possible allopatric-speciation pathway and the biological-species interpretation of this evidence. Give two distinct conservation approaches for a threatened population, explaining one limitation of each.5 indicative marks

Two sexually reproducing animal populations are separated by a new geographic barrier. After many generations they no longer produce fertile offspring when reunited under suitable conditions. Explain a possible allopatric-speciation pathway and the biological-species interpretation of this evidence. Give two distinct conservation approaches for a threatened population, explaining one limitation of each.

Cambridge 2027: Direct conceptual match

Original targeted concept exercise: Checked allopatric mechanism, conditional biological-species evidence and distinct in-/ex-situ limitations without claiming automatic species formation or conservation success. This is conceptual placement, not completion of the entire referenced subsection.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. Use the supplied post-reunion reproductive evidence rather than assuming a barrier automatically creates species. Isolation allows allele-frequency differences to accumulate; reproductive incompatibility may then prevent renewed gene flow.
  2. The conservation proposals address different scales: maintain a viable habitat in situ and preserve threatened stocks ex situ. Each requires monitoring and management rather than constituting a guaranteed rescue.

Self-marking points

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  1. The barrier reduces gene flow; mutation, drift and different selection can cause genetic divergence.
  2. Evolution of reproductive isolation can produce separate species; geographic separation alone does not guarantee speciation.
  3. Failure to produce fertile offspring supports separation under the biological-species concept for these sexual animals, not a universal test for asexual organisms.
  4. Habitat protection can preserve ecological interactions but may be difficult to enforce or may not remove all threats.
  5. Ex-situ breeding/genetic-resource storage can safeguard material or individuals but cannot alone recreate a functioning wild habitat or ensure reintroduction success.

Examiner tip

These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.

Question 2Two equally sampled communities each contain 10 individuals. A has species counts 5 and 5; B has counts 9 and 1. Using D=1-sum[n(n-1)]/[N(N-1)], calculate Simpson's diversity index for each and interpret the comparison. For a separate comparison of mean leaf lengths in two independent samples, state a suitable named test from this syllabus and one assumption to check; do not use the diversity counts as leaf-length data.4 indicative marks

Two equally sampled communities each contain 10 individuals. A has species counts 5 and 5; B has counts 9 and 1. Using D=1-sum[n(n-1)]/[N(N-1)], calculate Simpson's diversity index for each and interpret the comparison. For a separate comparison of mean leaf lengths in two independent samples, state a suitable named test from this syllabus and one assumption to check; do not use the diversity counts as leaf-length data.

Cambridge 2027: Direct conceptual match

Original targeted concept exercise: Recomputed the syllabus's 1-minus Simpson convention and separate t-test context; guards against confusing richness, evenness and statistical significance. This is conceptual placement, not completion of the entire referenced subsection.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. N(N-1)=10 x 9=90 for both. Calculate n(n-1) separately for each species; do not insert the square of the total abundance.
  2. Species richness alone is identical, so the difference reflects relative abundance. Test selection for continuous mean measurements requires its own data/assumption checks, not the claim that any two numbers justify a t-test.

Self-marking points

View marking points
  1. For A, D=1-(20+20)/90=0.556.
  2. For B, D=1-(72+0)/90=0.200.
  3. Both have two species, but A has more even abundance and a higher value of this specified diversity index.
  4. A suitable leaf-length comparison may use a t-test if independent samples and appropriate distribution/variance assumptions are justified; it is not a test of the two diversity counts.

Examiner tip

These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.

Question 3In an idealised PCR, 5 double-stranded target molecules are amplified for 8 perfect doubling cycles. Calculate the final target count. Explain the roles of primers and thermostable DNA polymerase, then predict which of DNA fragments 200, 500 and 1000 base pairs moves furthest in a conventional size-separating gel. Explain why expression of an inserted gene may require a compatible promoter.5 indicative marks

In an idealised PCR, 5 double-stranded target molecules are amplified for 8 perfect doubling cycles. Calculate the final target count. Explain the roles of primers and thermostable DNA polymerase, then predict which of DNA fragments 200, 500 and 1000 base pairs moves furthest in a conventional size-separating gel. Explain why expression of an inserted gene may require a compatible promoter.

Cambridge 2027: Direct conceptual match

Original targeted concept exercise: Verified 1280 ideal amplicons, primer/polymerase distinction, smaller-fragment migration and promoter recognition; no genetic-engineering protocol or exact thermal cycling recipe. This is conceptual placement, not completion of the entire referenced subsection.

  • 9700: 19.1: DNA technologies and sequence evidence

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. There are eight doublings of the starting target count, not eight copies added and not 252^{5}. Real amplification can be less efficient than the supplied ideal model.
  2. Compare fragment lengths under a shared gel/field condition, not fragments run in different gels. A gel band indicates DNA, while gene expression requires transcriptional control and would need separate evidence.

Self-marking points

View marking points
  1. Ideal target count = 5 x 282^{8} = 1280 double-stranded molecules.
  2. Primers provide complementary starting sites and delimit the amplified region; polymerase extends new strands.
  3. A thermostable polymerase tolerates repeated high-temperature strand separation.
  4. The 200-base-pair fragment typically travels furthest through the gel under the same run conditions because smaller fragments pass the matrix more readily.
  5. A promoter must be recognised by the host transcription machinery for transcription; inserting a coding sequence alone need not ensure expression.

Examiner tip

These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.

Question 4Compare the purpose of producing recombinant human insulin with screening for a BRCA variant and engineering insect-resistant cotton. For each, distinguish the biological intervention or information obtained from a guaranteed outcome. Discuss one relevant consent/privacy issue for screening and one ecological issue for the crop.5 indicative marks

Compare the purpose of producing recombinant human insulin with screening for a BRCA variant and engineering insect-resistant cotton. For each, distinguish the biological intervention or information obtained from a guaranteed outcome. Discuss one relevant consent/privacy issue for screening and one ecological issue for the crop.

Cambridge 2027: Direct conceptual match

Original targeted concept exercise: Checked the exact syllabus examples insulin/BRCA/cotton and their distinct purposes; uncertainty is explicit and there are no clinical recommendations or invented risk percentages. This is conceptual placement, not completion of the entire referenced subsection.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. The three cases act at different levels: protein production, risk information and an altered crop trait. Conflating screening with treatment or with deterministic prediction would misstate the biology.
  2. Ethical and ecological conclusions require case-specific evidence. Access, consent and privacy remain important even for technically accurate tests, and agricultural benefits do not remove the need for ecological assessment.

Self-marking points

View marking points
  1. Recombinant insulin production expresses a human protein for use as a treatment product; it is not by itself correction of a patient's genome.
  2. BRCA screening identifies a variant associated with altered risk, not certainty that a person will develop cancer.
  3. Insect-resistance genes can reduce damage from susceptible insects but do not guarantee resistance to every pest or permanent effectiveness.
  4. Screening requires informed consent and protection of sensitive genetic information, including implications for relatives.
  5. Crop assessment should consider effects on non-target organisms or selection for resistant pests, with evidence and monitoring rather than blanket claims of harmlessness or inevitable harm.

Examiner tip

These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.

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