Cambridge International AS & A Level Biology 9700: Genetic technology – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover topic 19, Genetic technology, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, sections 19.1 to 19.3. It is A Level content, examined on Paper 4. Calculators are allowed throughout.

Revise first with the study guide and the revision notes. The 9700 Biology hub and the printable checklist cover the whole course.

Questions

1.

(a) Define the term recombinant DNA. [1] (b) Explain what is meant by gene editing. [1]

2. A human gene is to be inserted into a bacterial plasmid. Explain the role of each of the following.

(a) restriction endonuclease [2] (b) DNA ligase [1] (c) the plasmid [1]

3. A company wants bacteria to make a human enzyme.

(a) Explain why the gene is made from the enzyme’s mRNA rather than cut from human chromosomal DNA. [2] (b) State one advantage of synthesising the gene chemically from nucleotides instead. [1] (c) The enzyme has 240 amino acids. Calculate the minimum number of nucleotides in the coding sequence, including one stop codon. [1]

4.

(a) Explain why a promoter may have to be transferred with the desired gene. [2] (b) Explain how a marker gene coding for a fluorescent product can confirm that the gene is being expressed. [2]

5. PCR is used to amplify DNA from a small tissue sample.

(a) Describe what happens in each of the three steps of one PCR cycle. [3] (b) Explain why Taq polymerase is used. [2] (c) The sample contains 8 copies of the target. Assuming perfect doubling, calculate the number of copies after 22 cycles. Give your answer in standard form to 3 significant figures. [2] (d) State the minimum number of cycles needed to produce more than one million copies from a single molecule. [1]

6. A linear DNA molecule 9.0 kb long has recognition sites for one restriction enzyme at 2.0 kb, 3.5 kb and 7.2 kb from one end.

(a) Calculate the length of each fragment produced. [2] (b) The fragments are separated by gel electrophoresis. List them in order from the band nearest the well to the band furthest away. [1] (c) Explain why DNA moves through the gel and why the fragments separate. [3] (d) A circular plasmid of 7 200 bp has one site for enzyme X at 1 000 bp and one site for enzyme Y at 3 400 bp. State the number and lengths of the fragments when both enzymes are used together. [2]

7. Outline how a microarray could be used to compare gene expression in tumour cells and normal cells from the same tissue. [4]

8. Outline two benefits of databases of nucleotide sequences, amino acid sequences and protein structures. [2]

9.

(a) Explain the advantages of treating diabetes with recombinant human insulin rather than insulin extracted from animals. [3] (b) State one advantage of recombinant factor VIII over factor VIII purified from donated blood. [1]

10. Genetic technology is used in diagnosis and treatment.

(a) Outline one advantage of screening a woman for BRCA1 and BRCA2 alleles. [2] (b) Huntington’s disease is caused by a dominant allele. A man is heterozygous and his partner is homozygous recessive. State the probability that a child inherits the allele, and discuss one ethical issue of testing the child. [3] (c) Outline how gene therapy can be used to treat SCID caused by ADA deficiency. [3] (d) Suggest two reasons why inherited eye diseases are suitable for gene therapy. [2]

11.

(a) Explain how Bt cotton increases crop productivity. [2] (b) Explain how GM salmon could help to meet global food demand. [2] (c) Discuss the ethical and social implications of growing herbicide-resistant soybean. [4]

Answers

1. (a) DNA made by joining DNA from two or more different sources. [1] (b) Genetic engineering that inserts, deletes or replaces DNA at a specific site in the genome. [1] Examiner insight: In (b), “at specific sites” is the marking point; “changing genes” alone scores nothing.

2. (a) Cuts DNA at a specific recognition sequence. [1] Using the same enzyme on the gene and plasmid gives complementary sticky ends. [1] (b) Forms phosphodiester bonds, sealing the gene into the plasmid backbone. [1] (c) A vector that carries the gene into the bacterium and replicates there. [1] Examiner insight: “Ligase joins the DNA” is too vague; name the phosphodiester bond.

3. (a) mRNA (via reverse transcriptase and DNA polymerase) gives cDNA with no introns. [1] Bacteria cannot remove introns, so the chromosomal gene would not give the correct protein. [1] (b) No donor cells needed; or the sequence can be designed or altered as required. [1] (c) 240 × 3 + 3 = 723 nucleotides [1] Examiner insight: In (a) both points are needed: “no introns” and why that matters to a bacterium.

4. (a) RNA polymerase must bind to a promoter to transcribe the gene. [1] The host may not recognise the donor’s promoter, so the gene would not be expressed. [1] (b) The fluorescent marker gene is inserted with the desired gene, under the same promoter. [1] Cells that fluoresce under UV or blue light have taken up the gene and are expressing it. [1] Examiner insight: Answers that describe antibiotic-resistance markers do not match this outcome, which is limited to fluorescent products.

5. (a) Denaturation at about 95 °C: hydrogen bonds break and strands separate. [1] Annealing at about 55–65 °C: primers bind to complementary sequences at the ends of the target. [1] Extension at about 72 °C: Taq polymerase adds nucleotides to the primers, forming new strands. [1] (b) Taq polymerase is from a thermophilic bacterium and is not denatured at 95 °C. [1] So it survives every cycle and does not need adding again; its optimum is near 72 °C. [1] (c) 8 × 2²² = 33 554 432 [1] = 3.36 × 10⁷ [1] (d) 2²⁰ = 1 048 576, so 20 cycles. [1] Examiner insight: In (c), a correct unrounded value gains the method mark even if the standard form is wrong; give both.

6. (a) Any two correct lengths from 2.0 kb, 1.5 kb, 3.7 kb and 1.8 kb [1]; all four correct, adding to 9.0 kb. [1] (b) 3.7 kb, 2.0 kb, 1.8 kb, 1.5 kb. [1] (c) Phosphate groups make DNA negatively charged. [1] It moves towards the anode (positive electrode). [1] The gel acts as a sieve, so shorter fragments move faster and further. [1] (d) Two fragments [1]: 2 400 bp and 4 800 bp. [1] Examiner insight: In (d), a circle cut twice gives two fragments, not three; allow ECF in (b) from wrong lengths in (a).

7. Extract mRNA from tumour and normal cells. [1] Reverse transcribe it to cDNA and label each sample with a different fluorescent dye. [1] Apply both to the microarray, where cDNA hybridises to complementary probes. [1] The colour and brightness of each spot show which genes are more or less active in tumour cells. [1] Examiner insight: Missing the mRNA-to-cDNA step loses a mark; microarray probes are DNA.

8. Any two: free and fast access worldwide, so work is not repeated [1]; new sequences can be compared with known ones to identify genes, organisms or evolutionary relationships; primers can be designed from known sequences; protein structures help drug design. [1] Examiner insight: Two distinct benefits are needed for both marks; two versions of “easy to access” score once.

9. (a) It is identical to human insulin, so fewer immune or allergic reactions. [1] It can be made in large quantities, giving a reliable supply. [1] It avoids ethical or religious objections to animal products. [1] (b) No risk of infection such as HIV or hepatitis from donated blood. [1] Examiner insight: “It is cheaper” alone rarely scores; the syllabus focus is on identical structure, supply and safety.

10. (a) Women with these alleles have a higher risk of breast cancer. [1] They can have more frequent checks for early diagnosis or choose preventive surgery. [1] (b) Probability 0.5 (50%). [1] A child cannot give informed consent. [1] A positive result for an incurable, late-onset disease could cause anxiety and affect choices about insurance or employment. [1] (c) White blood cells or bone-marrow stem cells are removed from the patient. [1] A normal ADA allele is inserted into them using a viral vector. [1] The cells are returned, where they make functional ADA. [1] (d) The eye is small and enclosed, so a small dose of vector can be injected directly. [1] The other eye can be kept as an untreated control. [1] Examiner insight: In (b), “discuss” needs the issue and its consequence; a single word such as “privacy” is not enough.

11. (a) The plant makes Bt toxin, which kills insect larvae that eat it. [1] Less crop damage, and less insecticide is needed, so yields rise. [1] (b) Extra growth hormone gene with an always-active promoter makes the fish grow faster. [1] They reach market size sooner, giving more food per unit time and feed. [1] (c) Benefit: weeds can be controlled with herbicide without harming the crop, raising yields. [1] Concern: resistance genes could spread to wild relatives, producing herbicide-resistant weeds. [1] Concern: seed is costly and may need buying each year, which is hard for poorer farmers. [1] Concern: some consumers distrust GM food and want labelling; or heavy herbicide use may harm other species. [1] Examiner insight: “Discuss” needs both sides; a list of only concerns usually caps below full marks.

Where marks are usually lost

• Restriction enzymes described as cutting at random.
• Ligase said to form hydrogen bonds.
• Promoter answers that do not mention RNA polymerase.
• PCR steps without temperatures, or with the wrong event at each.
• Taq polymerase: “works at high temperature” without “not denatured”.
• Longer fragments said to travel further.
• Circular DNA fragment counts given as cuts + 1.
• Microarray answers that skip reverse transcription of mRNA.
• One-sided “discuss” answers on GM crops and screening.
• Confusing genetic screening with gene therapy.

Next steps

• Revision notes for topic 19
• Study guide for topic 19
• 9700 Biology course hub
• Printable 9700 checklist
• 9700 A Level diagnostic
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus, for examination in 2025, 2026 and 2027 (version 1), published by Cambridge University Press & Assessment. Topic 19, sections 19.1 to 19.3.