OxfordAQA International AS and A-level Biology: Recombinant DNA technology (9610) – Revision Notes

These notes condense section 3.4.10 (Recombinant DNA technology) of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, Unit 4. Read the study guide for the full explanations and worked examples, then use the practice questions. The course hub, the printable checklist and the free diagnostics help you decide what to revise first.

Key terms

Term	Meaning
Recombinant DNA	DNA containing fragments from more than one organism or species
cDNA	DNA made as a copy of mRNA by reverse transcriptase, with no introns
Reverse transcriptase	Enzyme that makes DNA using an RNA template
Restriction endonuclease	Enzyme that cuts DNA only where it meets its own recognition sequence
Sticky ends	Short single-stranded overhangs left by an offset cut
Blunt ends	Ends left by a straight cut, with no overhang
DNA ligase	Joins sugar-phosphate backbones by forming phosphodiester bonds
Vector	DNA, such as a plasmid, that carries a fragment into a host cell
Transformation	Taking a vector into a host cell
Marker gene	Gene whose product reveals which cells have taken in the vector
Promoter	Region where RNA polymerase attaches so transcription can begin
Terminator	Region that ends transcription
Primer	Short single-stranded DNA that starts copying in PCR

3.4.10.1 Principles

• Recombinant DNA technology transfers DNA fragments between different organisms or species.
• The transferred DNA can be translated in the recipient because the genetic code and the mechanisms of transcription and translation are universal.
• Write it as: a codon specifies the same amino acid in donor and recipient cells.

3.4.10.2 Three ways to make a fragment

Method	Starting material	Key points
cDNA	mRNA from cells that make the protein	Reverse transcriptase makes one strand, DNA polymerase the second. No introns, no promoter
Restriction endonucleases	Genomic DNA	Cut either side of the gene. May leave introns and other sequences
Artificial gene synthesis	Known amino acid or base sequence	Built in a machine. No source organism needed. Codons chosen for the host

Method: cut and compare.

1. Find the recognition sites on the DNA.
2. Subtract positions to get fragment lengths.
3. Check that the lengths add up to the total.
4. Number of fragments: linear n + 1, circular n, for n sites.
5. Larger fragments are nearer the wells of a gel.

Reminder. A circular plasmid of 7,300 bp has sites at 900, 4,100 and 5,600 bp. Fragments: 4,100 − 900 = 3,200; 5,600 − 4,100 = 1,500; 7,300 − 5,600 + 900 = 2,600. Total 7,300.

Reminder. A linear 6,300 bp molecule with sites at 1,500 and 2,800 bp gives 1,500, 1,300 and 3,500 bp.

Reminder. A synthetic gene for 120 amino acids and one stop codon has 120 × 3 + 3 = 363 bases.

3.4.10.3 PCR

Stage	Temperature	Event
Denature	about 95 °C	Heat breaks the hydrogen bonds between strands
Anneal	about 50 to 65 °C	Primers pair with their complementary sequences
Extend	about 72 °C	Heat-stable DNA polymerase extends the primers

• Ingredients: template DNA, two primers, nucleotides, heat-stable polymerase, buffer.
• Copies after n cycles from one template: 2ⁿ. With 3 starting molecules and 18 cycles: 3 × 2¹⁸ = 786,432.
• Real runs fall short of this. A cycle that is 90% efficient multiplies the amount by 1.9, not 2.
• PCR needs no cells, and it is quick, but it needs some sequence knowledge for the primers.
• Identifying a fragment: primers that match only the target sequence give a product only if that sequence is present. A no-template tube checks for contamination.

3.4.10.4 In vivo amplification

Method: from fragment to transformed culture.

1. Add a promoter before the gene and a terminator after it.
2. Cut the fragment and the vector with the same restriction endonuclease (or two matched enzymes).
3. Mix, let the sticky ends pair, and add DNA ligase.
4. Transform host cells, for bacteria by calcium ions with a brief heat shock, or by an electric pulse.
5. Culture the cells on a medium that shows which carry the marker.
6. Select, then grow the selected cells to make more of the DNA or the protein.

Markers.

• Antibiotic resistance: cells holding the vector survive on agar that contains the antibiotic, and the rest die.
• Green fluorescent protein: transformed colonies glow under UV light.
• Insertional inactivation: a gene inserted inside the marker stops it working. Colonies that lose the marker contain the insert.

Reminder: efficiency. 0.030 µg of plasmid, cells recovered in 1,000 µl, 50 µl plated, 90 colonies. Total 90 × 20 = 1,800 colonies. Efficiency = 1,800 ÷ 0.030 = 60,000 colonies per µg.

Interpreting data

Gel. Compare each band with the size markers. A band that is absent where you expect one means no cut, no product, or no target. Fragments whose lengths add up to the size of the starting DNA show a complete digest.

Plates. Count the colonies that grow where only transformed cells can grow. Divide to find the share that are recombinant. Say what the control plate shows, such as growth with no antibiotic, which confirms the cells were alive.

Tables of costs and yields. Work out the change as a percentage, then state whether the gain outweighs the cost.

Reminder. A GM crop yields 5.0 tonnes per hectare against 3.8 for the conventional crop. Increase = (5.0 − 3.8) ÷ 3.8 × 100 = 31.6%.

Evaluation: a short framework

Angle	Prompt
Ethical	Is it right to move genes between species? What happens to animals and wild populations?
Financial	Who pays for development, who owns the patent, and who can afford the product?
Social	Who gains health or food, who lacks access, and what choice does the public have?

State a benefit and a concern for each angle you use. Then give a judgement, with a condition, such as “worth it if the product stays affordable for small farms”. Use the data in the question, and say which group each point affects: patients, farmers, companies or the environment.

Must-know distinctions

• cDNA and genomic fragment: cDNA has no introns. A fragment cut from the genome may have them.
• PCR and in vivo: PCR is in a tube and needs no cells. In vivo uses living cells that divide and can also make the protein.
• Sticky and blunt ends: only sticky ends give complementary overhangs that pair by hydrogen bonds before ligase seals them.
• Restriction endonuclease and ligase: one cuts the backbone, the other joins it.
• Promoter and terminator: start and end of transcription.
• Annealing and extension: primers pair at annealing, polymerase works at extension.

Quick self-test

1. Give the reason a human gene can be translated in a bacterium.
2. What does reverse transcriptase make, and from what?
3. Why does cDNA lack introns?
4. A linear DNA of 5,000 bp has two restriction sites, at 800 and 3,100 bp. State the fragment lengths.
5. Name the bond ligase forms.
6. Why is Taq polymerase used in PCR?
7. How many copies after 10 cycles from one molecule, with perfect copying?
8. What is the role of the terminator?
9. A gene is inserted into a gene for GFP. How do recombinant colonies look under UV?
10. Give one financial issue linked to recombinant DNA technology.
11. State why a negative control is used in PCR for identification.

Answers

1. The genetic code and the mechanisms of transcription and translation are universal.
2. cDNA, from an mRNA template.
3. The mRNA used as the template has already had its introns removed.
4. 800, 2,300 and 1,900 bp (3,100 − 800 = 2,300; 5,000 − 3,100 = 1,900).
5. Phosphodiester bond, between the sugar-phosphate backbones.
6. It is heat-stable: it keeps its shape at about 95 °C and works at about 72 °C.
7. 2¹⁰ = 1,024.
8. It ends transcription.
9. They do not fluoresce, because the GFP gene is disrupted.
10. For example, development costs, patent-protected seed or drug prices.
11. A tube with no template shows whether contamination gave a false product.

Where marks are usually lost

• Writing “the code is the same” instead of saying the genetic code is universal.
• Saying restriction endonucleases cut “at specific places” without naming the recognition sequence.
• Leaving out DNA ligase or saying it joins the bases. It joins the sugar-phosphate backbones.
• Mixing up the PCR stages or their temperatures, or forgetting the primers.
• Using 2ⁿ for the number of cycles instead of the number of copies.
• Forgetting that a plasmid with no insert can also be taken up and shows the marker.
• Listing advantages and disadvantages of GM technology with no link to the data given.
• Evaluating with a one-sided answer and no conclusion.

Official syllabus

The OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, for International A-level exams May/June 2018 onwards, published by OxfordAQA.