OCR A Level Biology A: Genetic engineering and gene therapy (H420) – 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 OCR A Level Biology A (H420), Version 4.1 (April 2026), module 6.1.3 outcomes (f), (g) and (h): genetic engineering, ethical issues and gene therapy. Read the study guide first and use the revision notes to check recall.

Questions

1. Give the role of each of the following in making recombinant DNA.

(a) A restriction enzyme [1] (b) DNA ligase [1] (c) Reverse transcriptase [1]

2. A circular plasmid of 6,300 base pairs (bp) has two recognition sites for the restriction enzyme HindIII. The sites are 2,300 bp apart. A separate vector of 3,600 bp, cut once with HindIII, receives a 2,300 bp insert.

(a) Give the sizes of the fragments produced when the 6,300 bp plasmid is digested. [2] (b) State which fragment would travel further in an electrophoresis gel and give a reason. [1] (c) Calculate the size of the recombinant vector. [1] (d) Explain why the insert can join to the cut vector. [1]

3. A laboratory engineers bacteria to produce a human hormone from DNA copied from mRNA. Explain why this cDNA is used rather than a gene cut from a human chromosome. [4]

4. A vector carries a gene for ampicillin resistance and a gene for chloramphenicol resistance. A gene of interest was placed inside the chloramphenicol resistance gene. Treated bacteria were spread on ampicillin agar, where 412 colonies grew. Each was transferred to chloramphenicol agar, and 97 colonies grew.

(a) Explain why all 412 colonies contain a plasmid. [1] (b) Calculate the percentage of the 412 colonies that contain the gene of interest. Give your answer to 3 significant figures. [2] (c) Explain why colonies containing the gene of interest do not grow on chloramphenicol. [2] (d) Suggest one advantage of using a gene for green fluorescent protein as the marker instead. [1]

5. Describe how electroporation can be used to introduce recombinant DNA into bacterial cells. [3]

6. A student mixes 0.2 µg of plasmid with bacteria and suspends them in 5.0 cm³ of broth. Spreading 0.5 cm³ gives 64 colonies. Work out how many transformed colonies the tube would give per µg of plasmid. [4]

7. A trial compared conventional soya with a variety carrying an insect-resistance gene. Conventional soya gave a mean yield of 2.8 tonnes per hectare and needed insecticide costing 54 dollars per hectare. The modified variety gave 3.2 tonnes per hectare and needed insecticide costing 21 dollars per hectare.

(a) By what percentage is the yield of the modified variety higher? [2] (b) Work out the insecticide saving per hectare, and express it as a percentage of the conventional cost. [2] (c) Give two negative issues that might be raised about growing the modified variety. [2]

8. A genetically modified sheep makes a human protein in its milk. It produces 400 dm³ of milk a year, containing 2.5 g of the protein per dm³.

(a) Calculate the mass of protein from a flock of 15 such sheep in one year, in kg. [2] (b) The gene was attached to a promoter that works only in milk-producing cells. Give the reason for this choice. [1] (c) Give one ethical objection to this use of animals. [1]

9. A company patents a crop variety it has genetically modified. Discuss one benefit and one concern of patenting, and define technology transfer here. [3]

10. Describe two differences between somatic cell gene therapy and germ line cell gene therapy, and explain why germ line therapy is considered more ethically controversial. [4]

11. A patient with an inherited immune deficiency receives somatic gene therapy. Blood samples measure the percentage of T cells that carry a working copy of the gene.

Time after treatment (months)	0	1	6	12	24
T cells with working gene (%)	0	40	30	20	20

(a) Calculate the percentage decrease in the proportion of corrected T cells between 1 month and 12 months. [2] (b) Describe the trend shown in the table. [2] (c) Suggest an explanation for the stable value between 12 and 24 months. [2] (d) Explain why the patient’s children are unlikely to inherit the working gene. [1]

12. Two carriers of a recessive allele for a serious single-gene disorder ask about two treatments: somatic cell gene therapy for a child born with the disorder, or germ line gene therapy on an early embryo. Evaluate the two approaches. [9]

Answers

1. (a) It cuts DNA only where it finds its recognition sequence [1]. (b) It makes phosphodiester bonds in the backbone, fixing the gene into the vector [1]. (c) It makes DNA (cDNA) from an mRNA template [1]. Examiner insight: Name the bond or substrate, such as “phosphodiester bonds”, rather than a vague word such as “joins”.

2. (a) Two cuts in a circular molecule give two fragments [1]: 2,300 bp and 6,300 − 2,300 = 4,000 bp [1]. (b) The 2,300 bp fragment, because smaller fragments move through the gel pores faster [1]. (c) 3,600 + 2,300 = 5,900 bp [1]. (d) The cut vector and the insert have complementary sticky ends, which pair by hydrogen bonds between bases (ligase then seals them) [1]. Examiner insight: Count the cuts first: a circular plasmid cut twice gives two pieces, and a reason in part (b) should link size to movement through the gel.

3. The gene in the chromosome contains introns [1]. A bacterial cell cannot remove introns from the transcript [1]. cDNA is made from mature mRNA, which has had introns removed [1]. So the bacterium can translate the cDNA into the correct protein [1]. Examiner insight: “Easier to get” gains little; the point that bacteria cannot remove introns is what answers “explain why”.

4. (a) All colonies grew on ampicillin, so each must have the ampicillin resistance gene on a plasmid [1]. (b) Colonies with the insert do not grow on chloramphenicol: 412 − 97 = 315 [1]. 315 ÷ 412 × 100 = 76.5% [1]. (c) The gene of interest sits inside the chloramphenicol resistance gene, which disrupts it [1], so no functional resistance protein is made and the bacteria cannot grow on chloramphenicol [1]. (d) Fluorescent colonies can be identified without being transferred to antibiotic plates or killed (or without replica plating) [1]. Examiner insight: In part (b) the recombinants are the colonies that fail to grow on chloramphenicol; using the 97 would give the wrong percentage.

5. Bacteria and recombinant DNA are mixed in a solution [1]. A short pulse of high voltage is applied, which makes temporary pores in the cell surface membrane [1]. DNA passes through the pores into the cells, and the pores close again afterwards [1]. Examiner insight: “Uses electricity” alone does not describe the mechanism; say that the pulse makes temporary pores.

6. Colonies in the whole 5.0 cm³: 64 × (5.0 ÷ 0.5) [1] = 640 colonies [1]. Efficiency = 640 ÷ 0.2 [1] = 3,200 colonies per µg [1]. Examiner insight: Write out the scaling step from sample to whole tube, and give the unit “colonies per µg”.

7. (a) 3.2 − 2.8 = 0.4 [1]. 0.4 ÷ 2.8 × 100 = 14.3% [1]. (b) 54 − 21 = 33 dollars per hectare [1]. 33 ÷ 54 × 100 = 61.1% [1]. (c) Any two: harm to non-target insects; pests evolving resistance; gene flow to wild relatives; dependence on seed suppliers [1] [1]. Examiner insight: For percentage change, divide by the original value (the conventional crop); dividing by the new value is a common slip.

8. (a) One sheep: 400 × 2.5 = 1,000 g [1]. Flock: 1,000 × 15 = 15,000 g = 15 kg [1]. (b) The protein is then made only in the mammary glands and secreted in the milk, where it can be collected and purified, and not made in other tissues [1]. (c) Any one: animal welfare; use of animals as production systems; risk to the animal’s health [1]. Examiner insight: An ethical objection earns credit when specific; “it is wrong” on its own is too vague.

9. Benefit: a patent gives the company exclusive rights, which rewards and funds the cost of developing the variety [1]. Concern: seed may be expensive and farmers may become dependent on the company [1]. Technology transfer: making the modified seed or knowledge available to poor farmers or developing countries, for example at low cost [1]. Examiner insight: Make one clear point on each part requested; a second benefit does not replace the technology transfer definition.

10. Two differences, each with a contrast: somatic therapy treats body cells whereas germ line therapy alters gametes, a zygote or an early embryo [1]; somatic changes are not inherited whereas germ line changes are passed on to offspring [1]. Reason: the people affected by germ line changes are future generations who cannot consent [1], and any mistake or unforeseen harm would be inherited, so it cannot easily be reversed [1]. Examiner insight: “Describe two differences” needs both sides of each contrast.

11. (a) 40 − 20 = 20 percentage points [1]. 20 ÷ 40 × 100 = 50% [1]. (b) The percentage rises from 0 to 40% by 1 month [1], then falls to 20% by 12 months and stays at 20% to 24 months [1]. (c) Corrected stem cells (or long-lived cells) keep producing corrected T cells [1], so the proportion stays steady once other corrected cells have been replaced by uncorrected ones [1]. (d) The treated cells are body (somatic) cells, not gametes, so the working gene is not passed to gametes and children [1]. Examiner insight: Quote values with units at stated times; a trend described without numbers is usually too general.

12. Credit any nine points, covering both approaches and a reasoned judgement.

• Somatic therapy treats the affected child’s own cells [1].
• The patient, or their parents, can consent [1].
• The effect may fade as treated cells are replaced, so repeat treatment may be needed [1].
• Risks include the vector inserting the gene in the wrong place, causing cancer, or an immune response [1].
• The change is not inherited, so the child’s own children are at the same risk as before [1].
• Germ line therapy would correct the allele in every cell and could stop the disorder being passed on [1].
• Future people cannot consent to the change [1].
• Errors or unforeseen harm would be inherited [1].
• A judgement supported by the points above, for example that somatic therapy is preferable at present because the risk is limited to one monitored person [1]. Examiner insight: Finish with a conclusion that follows from your points; a list of arguments with no decision loses the last mark.

Where marks are usually lost

• Writing “joins” or “sticks” for ligase without naming phosphodiester bonds or the backbone.
• Giving three fragments for a circular plasmid cut twice, or two for a linear molecule cut twice.
• Calculating a percentage of the wrong total, or using the colonies that grow on both antibiotics as the recombinants.
• Leaving out the scaling step from the plated volume to the whole culture, or the unit.
• Treating somatic and germ line therapy as the same, or forgetting that only germ line changes are inherited.
• Listing arguments with no judgement when asked to evaluate.

Next steps

• Review the facts with the revision notes and the study guide.
• Go to the OCR A Level Biology hub and the printable checklist.
• Take all free 10-minute diagnostics to find your weakest topics.
• Book a free trial class: /trial/.

Official syllabus

These questions follow OCR Level 3 Advanced GCE in Biology A (H420), Version 4.1 (April 2026), for first assessment in 2025, published by OCR (Cambridge University Press & Assessment), module 6.1.3(f) to (h).