OCR A Level Biology A: Genetic engineering and gene therapy (H420) – Revision Notes

These notes condense 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. The full explanations are in the study guide, and exam-style questions are in the practice set. The OCR A Level Biology hub and printable checklist show the whole course, and the free diagnostics help you decide what to revise first.

Key terms

Term	Meaning
Genetic engineering	Transferring a gene from one organism to another so the recipient makes the product
Recombinant DNA	DNA made by joining sequences from different sources
Transgenic	An organism that contains a gene from another species
Vector	A carrier (plasmid or virus) that moves DNA into a host cell
Plasmid	Small circular DNA in bacteria, separate from the chromosome and able to copy itself
Restriction enzyme	Enzyme that cuts DNA only at its recognition sequence
Sticky end	A few unpaired bases left at the end of each fragment after a staggered cut
DNA ligase	Enzyme that seals the gene into the plasmid by making phosphodiester bonds in the backbone
cDNA	DNA made from mRNA by reverse transcriptase; it has no introns
Marker gene	A gene in the vector (antibiotic resistance or GFP) that shows which cells took it up
Pharming	Genetically modified animals making pharmaceuticals, often in their milk
Somatic gene therapy	Treating body cells; the change is not inherited
Germ line gene therapy	Changing gametes, a zygote or an early embryo; the change is inherited

Method in steps: making and using recombinant DNA (6.1.3(f))

1. Isolate the gene. Cut it out with restriction enzymes, make cDNA from mRNA with reverse transcriptase, or build it in a gene machine.
2. Cut the vector. Use the same restriction enzyme so that the sticky ends are complementary.
3. Anneal. Gene and plasmid sticky ends pair by hydrogen bonds between complementary bases.
4. Seal. DNA ligase makes phosphodiester bonds in the backbone.
5. Add a promoter that works in the host, so the gene is transcribed.
6. Transfer into host cells. Electroporation, heat shock with calcium ions, Agrobacterium (plants) or a viral vector.
7. Select. Grow on antibiotic or look for GFP fluorescence; use insertional inactivation to separate recombinant from non-recombinant plasmids.

Why cDNA for a bacterial host

A human gene in the genome contains introns, and bacteria cannot splice them out. cDNA made from mature mRNA is the coding sequence only, so the bacterium can make the correct protein.

Why the same enzyme

An enzyme cuts only at its own recognition sequence because its active site is complementary to it. Using one enzyme on both the gene and the plasmid gives matching sticky ends.

Electroporation

A short electrical pulse makes temporary pores in the cell surface membrane, and DNA moves in through them. The pores then close. It works on bacteria, yeast and many other cell types.

Selection logic

Result	What it tells you
No growth on antibiotic A	The cell has no plasmid
Growth on A	The cell took up a plasmid (resistance gene A)
Growth on A, no growth on B, where the gene was inserted into gene B	Recombinant plasmid: insertional inactivation of B
Growth on A and B	Plasmid that closed up without the insert
Fluorescent colonies	Cells expressing the GFP marker gene

Worked reminder: fragment sizes

A linear piece of DNA is 5,000 bp. A restriction enzyme cuts at 1,200 bp and 3,700 bp. Three fragments result: 1,200 bp, 3,700 − 1,200 = 2,500 bp and 5,000 − 3,700 = 1,300 bp. In a circular plasmid, n cuts give n fragments; in a linear molecule, n cuts give n + 1 fragments.

Worked reminder: transformation efficiency

Efficiency = colonies in the whole tube ÷ mass of DNA used. If 120 colonies grow from 0.1 cm³ of a 1.0 cm³ recovery culture, the whole tube would give 120 × 10 = 1,200 colonies. With 0.03 µg of DNA, 1,200 ÷ 0.03 = 40,000 colonies per µg.

Ethical issues (6.1.3(g))

Case	Positive	Negative
Insect-resistant soya	Less crop loss, higher yield, less insecticide	Non-target insects; pest resistance; gene flow to relatives
Modified pathogens for research	Understanding disease, developing treatments	Escape or misuse (biosecurity)
Pharming	Human proteins made in large amounts at lower cost	Animal welfare; low success rate; objections to using animals as producers
Patenting	Rewards and funds development	High seed prices; control by a few companies; farmers’ dependence
Technology transfer	Can reduce hunger when seed reaches poor farmers	Variety may not suit local conditions; dependence on suppliers
Humans	Cure for inherited disease	Consent of future people; “designer” children; safety

Answer structure for an ethics question

1. State the benefit and say who gains.
2. State the harm or risk and say who bears it.
3. Add what is uncertain (long-term effects, resistance, spread).
4. Finish with a reasoned judgement that follows from the points you made.

Gene therapy (6.1.3(h))

Aim: supply a functional allele to cells that lack it, to treat the cause of a genetic disorder. It works best for single-gene, often recessive, disorders.

Vectors: modified viruses (disease genes removed) or liposomes. Cells may be treated outside the body and returned (ex vivo) or treated in the body (in vivo).

	Somatic	Germ line
Target	Body cells	Gametes, zygote, early embryo
Inherited	No	Yes
Consent	Patient can consent	Future individuals cannot
Duration	May fade as treated cells are replaced	Passed on in every cell and to later generations
Position	Used in medicine	Heavily restricted; ethically controversial

Risks: insertion in the wrong place (which can switch on a cancer-causing gene or disrupt a healthy one); an immune response to the vector; effect wearing off; high cost.

Worked reminder: interpreting a decline

A patient has 40% of T cells carrying a working gene at 1 month and 20% at 12 months. The fall is 40 − 20 = 20 percentage points, which is (20 ÷ 40) × 100 = 50% of the earlier value. Explain with reference to treated cells being lost or replaced. Say whether the result suggests repeat treatment is needed.

Quick self-test

1. State the role of DNA ligase in making recombinant DNA.
2. Why must a gene and its plasmid be cut with the same restriction enzyme?
3. A linear DNA molecule of 8,000 bp is cut at 2,100 bp and 5,600 bp. Give the fragment sizes.
4. A vector of 3,900 bp receives an insert of 850 bp. What is the size of the recombinant plasmid?
5. Of 160 colonies on an ampicillin plate, 40 also grow on streptomycin, with the gene of interest inserted in the streptomycin resistance gene. What percentage are recombinant?
6. Give two uses of marker genes in genetic engineering.
7. Suggest why a promoter must be included next to the inserted gene.
8. A pharming goat gives 500 dm³ of milk a year containing 2.0 g of the human protein per dm³. How much protein is that in kg?
9. State two differences between somatic and germ line gene therapy.
10. Give one negative ethical argument relating to patenting of modified seed.

Answers

1. It makes phosphodiester bonds in the backbone where gene and plasmid meet, joining them permanently.
2. The same enzyme leaves complementary sticky ends, which anneal by hydrogen bonds between bases.
3. Three fragments: 2,100 bp, 5,600 − 2,100 = 3,500 bp and 8,000 − 5,600 = 2,400 bp.
4. 3,900 + 850 = 4,750 bp.
5. Recombinants are those that grow on ampicillin but not streptomycin: 160 − 40 = 120. 120 ÷ 160 × 100 = 75%.
6. Any two: show which cells took up the vector, allow selection on antibiotic, allow identification by fluorescence (GFP), show whether a gene was inserted.
7. The host’s RNA polymerase binds to the promoter and starts transcription, so the gene is expressed.
8. 500 × 2.0 = 1,000 g = 1.0 kg.
9. Somatic treats body cells and is not inherited; germ line alters gametes or embryos and is inherited. Somatic effects may fade, while germ line effects are passed on. The patient can consent to somatic therapy, but future people cannot consent to germ line therapy.
10. Patent holders can set high prices, so poor farmers may not afford the seed, or they become dependent on one company.

Where marks are usually lost

• Saying ligase “sticks the bases together” instead of forming phosphodiester bonds in the backbone.
• Forgetting that a restriction enzyme is specific to one sequence because of its active site shape.
• Not stating that cDNA has no introns when explaining why it is used with bacteria.
• Describing electroporation without saying that the pores are temporary and let DNA pass through the membrane.
• Mixing up the two antibiotic results in insertional inactivation.
• Giving only advantages or only disadvantages in an ethics answer.
• Writing that gene therapy “cures” a person without saying which cells are treated and whether the change is inherited.
• Calculation answers without units or without the working that scales plated volumes up to the whole tube.

Official syllabus

These notes 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).