Pearson Edexcel International A Level Biology: Recombinant DNA, Genetic Engineering and Bioinformatics (YBI11) – Revision Notes

These notes condense statements 8.17 to 8.22 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2. Topic 8 is part of International A2 (IA2); Topics 1 to 4 are International AS. For full explanations use the study guide, and for exam-style work use the practice questions. See also the hub, the checklist and the free diagnostics. PCR is covered in the PCR and electrophoresis notes.

Key terms

Term	Meaning
Recombinant DNA	DNA formed by joining DNA from different sources
GMO	Organism whose genome contains DNA from another source or altered by technology
Restriction endonuclease	Enzyme that cuts DNA at a specific recognition sequence
Sticky end	Short single-stranded overhang left by a staggered cut
DNA ligase	Joins sugar-phosphate backbones by phosphodiester bonds
Vector	Carrier used to move a gene into a host cell, such as a plasmid or virus
cDNA	DNA copied from mRNA by reverse transcriptase
Marker gene	Gene that shows which cells took up the vector
Microarray	Slide with thousands of DNA probe spots, used to show which genes are active
Bioinformatics	Computer-based storage and analysis of biological data such as sequences

Making recombinant DNA (8.18)

Method in steps:

1. Obtain the gene: cut from the genome, make cDNA from mRNA, or synthesise from a known sequence.
2. Cut gene and plasmid with the same restriction endonuclease.
3. Mix: complementary sticky ends anneal by hydrogen bonds.
4. Add DNA ligase: phosphodiester bonds seal the backbone.
5. Check the construct has a promoter that works in the host.

Points to remember:

• The enzyme is specific because its active site is complementary in shape to one recognition sequence.
• Recognition sequences are usually palindromic. EcoRI: GAATTC, cut between G and A.
• Why cDNA for bacteria: it contains no introns, and bacteria cannot splice them.
• Reverse transcriptase makes DNA from RNA; DNA polymerase then makes the second strand.

Fragment-size reminder:

• Circular plasmid, one site: one linear piece of the full length.
• Circular plasmid, two sites a apart in a plasmid of total length L: pieces a and L − a.
• Linear DNA with n cuts gives n + 1 pieces.
• Recombinant size = vector length + insert length.
• On a gel the smallest fragments travel furthest.

Worked reminder: a 5,000 bp circular plasmid with two sites 1,100 bp apart gives 1,100 bp and 3,900 bp.

Getting DNA into cells (8.19)

Host	Method
Bacteria	Transformation: calcium chloride, then heat shock; or electroporation
Plants	Agrobacterium tumefaciens carrying the gene in its Ti plasmid; or gene gun
Animals	Microinjection into a fertilised egg nucleus; viral vectors; liposomes

Selection:

• Antibiotic-resistance marker: only transformed cells grow on that antibiotic.
• Fluorescent marker (green fluorescent protein): transformed cells glow under UV light.
• Insertional inactivation: gene inserted into a resistance gene, so recombinants lose that resistance.

Method for replica plating, step by step:

1. Plate on agar with antibiotic 1 and count colonies, which contain plasmid.
2. Copy onto agar with both antibiotics.
3. Colonies on the first plate but absent from the second carry the insert.
4. Recombinant = first count minus second count.

Calculations:

• Colonies in the whole sample = colonies counted × (total volume ÷ volume plated).
• Transformation efficiency = total colonies ÷ mass of DNA in µg.
• Worked reminder: 85 colonies from 0.1 cm³ of 1.0 cm³ is 850 colonies; with 0.02 µg DNA that is 42,500 per µg.

Drugs from GMOs (8.17)

Organism type	Example idea	Why used
Microorganism	Bacteria or yeast making human insulin in a fermenter	Fast growth, cheap, human sequence, scalable
Animal	Goats making human antithrombin in milk	Large amounts of protein, easy to collect
Plant	Plants or plant cells making a human protein	Cheap to grow, no human pathogens

A gene for milk production needs a mammary-gland-specific promoter so that the protein appears in milk. Yield calculation: volume × concentration, so 10,000 dm³ × 0.8 g per dm³ = 8.0 kg.

Microarrays (8.20)

Method in steps:

1. Extract mRNA from the tissue.
2. Make fluorescent cDNA with reverse transcriptase.
3. Add to the array so cDNA hybridises to complementary probes.
4. Wash off unbound cDNA.
5. Scan: brighter spot, more mRNA, more active gene.

Two-sample arrays: red only means more active in sample 1, green only more active in sample 2, yellow means similar. A ratio above 1 favours red and below 1 favours green. Worked reminder: red 1,200 and green 600 gives a ratio of 2.0, so the gene is twice as active in the red sample.

Bioinformatics (8.21)

• Computer analysis of large biological datasets, mainly sequences.
• Uses: compare a new sequence with database entries, locate genes, build evolutionary trees, predict links between variants and disease, and interpret microarray results.
• Percentage identity = identical aligned bases ÷ bases compared × 100. Worked reminder: 18 of 20 gives 90%.

Answer frameworks

Describe how a human gene can be put into bacteria so they make a protein:

1. Isolate the gene, or make cDNA from mRNA using reverse transcriptase.
2. Cut gene and plasmid with the same restriction endonuclease to give complementary sticky ends.
3. Anneal the sticky ends, then seal with DNA ligase.
4. Include a promoter that works in bacteria, and a marker gene.
5. Chill bacteria with calcium chloride and plasmids, then heat shock.
6. Grow on agar with the antibiotic and select transformed colonies.
7. Culture them in a fermenter and purify the protein.

Interpret a microarray result:

• Name the colour or intensity for each spot and say what it means for the gene.
• Quote the ratio, or the intensities, from the table.
• Link to the biology, such as a gene more active in a tumour being a possible drug target.
• Add a limit: mRNA level does not prove how much protein is made.

Evaluate a GM crop:

• One benefit with data, for example a percentage rise in yield.
• One risk with a reason, for example resistance evolving.
• A clear judgement, such as “benefits outweigh risks if the crop is monitored”.

Risks and benefits of GMOs (8.22)

Benefits	Risks
Higher yield	Gene flow to wild relatives
Less pesticide	Harm to non-target species
Added nutrients	Resistance evolving in pests or weeds
Cheap medicines	Allergy concerns
Stress tolerance	Patents and cost to farmers
	Ethical concerns for animals

Always make a judgement in “evaluate” answers, using data if given.

Must-know distinctions

• Restriction endonuclease cuts; ligase joins.
• Hydrogen bonds hold sticky ends temporarily; phosphodiester bonds are permanent.
• cDNA is made from mRNA; genomic DNA contains introns.
• Marker gene shows uptake; insertional inactivation shows the gene was inserted.
• Microarray measures transcription (mRNA), not protein.
• Bioinformatics analyses data; it does not make recombinant DNA.

Quick self-test

1. What is the role of DNA ligase?
2. Why does cutting a gene and a plasmid with the same enzyme help?
3. A linear DNA molecule is cut at 3 sites. How many fragments?
4. Why is cDNA used to make a human protein in bacteria?
5. How are bacterial cells made to take up plasmids?
6. Why is Agrobacterium used for plants?
7. A 3,000 bp circular plasmid has two sites 800 bp apart. What fragments result?
8. 120 colonies grow on plate 1 and 45 on the replica plate with both antibiotics. How many are recombinant?
9. Red intensity 2,400 and green intensity 600 for one spot. What does it show?
10. Give one benefit and one risk of herbicide-resistant crops.

Answers

1. Forms phosphodiester bonds between the sugar-phosphate backbones, sealing the gene into the plasmid.
2. Both have complementary sticky ends, which anneal by base pairing.
3. Four fragments.
4. It has no introns and bacteria cannot splice them out; it is made from mRNA so carries the coding sequence only.
5. Chilling in calcium chloride then heat shock, or electroporation.
6. It naturally transfers DNA from its Ti plasmid into plant cells.
7. 800 bp and 3,000 − 800 = 2,200 bp.
8. 120 − 45 = 75.
9. Ratio 4.0: the gene is four times more active in the red sample.
10. Benefit: weeds can be controlled without harming the crop. Risk: resistance may spread to wild weeds through gene flow.

Where marks are usually lost

• Writing that ligase “joins the sticky ends” without naming the bond made.
• Saying the enzyme “cuts DNA” without stating it cuts at a specific base sequence.
• Forgetting the promoter, so the gene is not expressed.
• Describing antibiotic resistance genes as making bacteria “immune” or “kill the antibiotic”.
• Reading a microarray as showing protein level or DNA amount.
• Giving only risks or only benefits in an evaluation.
• Mixing up the first and second plate counts in a replica-plating subtraction.
• Dropping units, such as bp or µg, in calculations.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Biology (YBI11), Specification, Issue 2, February 2021, Pearson Education Limited.