AQA A-level Biology: Genome Projects and Gene Technologies (7402) – Revision Notes

These revision notes cover AQA A-level Biology (7402) sections 3.8.3 and 3.8.4 (3.8.4.1, 3.8.4.2 and 3.8.4.3) from the AQA AS and A-level Biology specification for first teaching 2015. All of this content is marked A-level only. For full explanations use the study guide, then test yourself with the practice questions. Course pages: AQA A-level Biology hub and the printable checklist. Not sure where to start? Try the free diagnostics.

3.8.3 Genome projects: key points

• Genome: the complete DNA of an organism. Proteome: the proteins that derive from the genetic code of the organism.
• Sequencing projects have read the genomes of many organisms, including humans. Methods are continuously updated and are automated.
• Simple organisms: genome sequence lets you work out protein sequences, so the proteome can be determined. Use: identify potential antigens for vaccines.
• Complex organisms: non-coding DNA and regulatory genes mean the genome cannot easily be translated into the proteome.

Simple organism	Complex organism
Mostly coding DNA	Mostly non-coding DNA
Few regulatory genes	Many regulatory genes
Proteome predictable from genome	Proteome not easily predicted

Reminder calculation (data supplied). A genome of 12 million base pairs with 70% coding: 12 000 000 × 0.70 = 8 400 000 base pairs coding; 3 600 000 non-coding.

3.8.4.1 Recombinant DNA: key points

Recombinant DNA technology transfers a DNA fragment from one organism or species to another. It works because the genetic code, transcription and translation are universal.

Ways to make the fragment

Method	How	Key point
Reverse transcriptase	mRNA to single-stranded cDNA, then second strand by DNA polymerase	No introns in the mRNA
Restriction endonuclease	Cuts at a specific recognition site	Often leaves sticky ends
Gene machine	Builds DNA from a typed-in sequence	No cells needed

PCR in steps

1. Denature at about 95 °C: strands separate as hydrogen bonds break.
2. Anneal at roughly 50 to 65 °C: primers bind to the ends of the target.
3. Extend at about 72 °C: Taq polymerase builds complementary strands from the 3’ end of each primer.
4. Repeat. Copies = starting molecules × 2ⁿ.

Ingredients: template DNA, two primers, free nucleotides, heat-stable DNA polymerase. Heat-stable because it must survive the denaturing step.

Reminder calculation. 5 starting molecules, 9 cycles: 5 × 2⁹ = 5 × 512 = 2560.

In vivo cloning in steps

1. Cut vector (plasmid) and the DNA source with the same restriction endonuclease.
2. Mix: sticky ends pair by complementary base pairing.
3. DNA ligase joins the sugar-phosphate backbones (phosphodiester bonds).
4. Fragment must have a promoter (RNA polymerase binds) and terminator (transcription stops).
5. Transform host cells with the vector.
6. Detect transformed cells with a marker gene, for example antibiotic resistance or a fluorescent protein. You need not recall specific marker genes.
7. Culture the host cells, so the plasmid is copied each time they divide.

In vitro and in vivo compared

	PCR (in vitro)	Transformed host cells (in vivo)
Needs cells	No	Yes
Speed	Fast	Slower
Enzymes	Taq polymerase	Restriction enzyme, ligase, cell enzymes
Gene expressed	No (only copying)	Can be, if promoter and terminator are present

Issues (evaluate)

• Agriculture: higher yield and added nutrients, against gene flow, harm to other organisms, loss of diversity and patents.
• Industry: cheap protein production against ownership and price.
• Medicine: proteins and gene therapy against cost, unequal access, unknown long-term effects.
• Balance humanitarian aims (food, health) against opposition from environmentalists and anti-globalisation activists (ecological risk, control by a few companies).
• Gene therapy: delivering a functional allele to a patient’s cells, using recombinant DNA methods.

3.8.4.2 Probes and hybridisation: key points

• DNA probe: short, single-stranded, labelled (radioactive or fluorescent), complementary to the target allele.
• Hybridisation: denature sample DNA, add probe, probe binds only to complementary sequence. Signal shows the allele is present.
• Screening for heritable conditions, drug responses and health risks. Results feed into genetic counselling (risk explained, options discussed) and personalised medicine (drug choice and dose matched to alleles).
• Evaluate screening: benefits are early action, informed choices and fewer drug side effects. Costs are anxiety, privacy, possible discrimination, and results that are only probabilities.

Reminder calculation (data supplied). Both parents are carriers of a recessive allele: chance of an affected child 1/4 = 25%. One parent is a known carrier and the other has a 1 in 40 chance of being a carrier: 1/40 × 1/4 = 1/160 = 0.625%.

Probe design. The probe is complementary to the target, not identical. Target 5’ ATGCCA 3’ needs probe 5’ TGGCAT 3’.

3.8.4.3 Genetic fingerprinting: key points

• VNTRs: variable number tandem repeats, short sequences repeated in a row. Number of repeats varies; chance of two individuals sharing the same VNTRs is very low.
• Steps: extract DNA, amplify by PCR, separate by gel electrophoresis, make visible.
• Electrophoresis: DNA is negatively charged (phosphate groups), moves to the positive electrode, shorter fragments move further through the gel mesh. Known-size ladder allows size estimates.

Reading a gel

1. Find the ladder and note which bands are which size.
2. Same distance travelled means same length.
3. Compare each lane band by band.
4. Offspring bands come from the two parents. Any child band not in the mother must be in the father.
5. A match is consistent with identity, not proof. The more independent loci that match, the smaller the chance of a coincidence.

Uses

Field	Use
Forensic science	Match crime-scene DNA to a suspect, or identify remains
Medical diagnosis	Establish genetic relationships and identify individuals reliably
Animal and plant breeding	Check parentage, avoid inbreeding, measure genetic variability

Reminder calculation. Matching chances at four independent loci: 0.2, 0.1, 0.05 and 0.25. Product = 0.2 × 0.1 × 0.05 × 0.25 = 0.00025, which is 1 in 4000.

Must-know distinctions

• cDNA (from mRNA, no introns) against genomic DNA (has introns).
• Restriction endonuclease (cuts) against ligase (joins).
• Primer (starts DNA polymerase) against probe (labelled, finds a sequence).
• Genome (all DNA) against proteome (all proteins).
• In vitro (outside cells, PCR) against in vivo (inside living cells).
• Promoter (start) against terminator (stop).

Quick self-test

1. State what the proteome is.
2. Explain why the proteome of a bacterium is easier to predict from its genome than that of a human.
3. Name the enzyme that makes cDNA and state the template it uses.
4. A PCR starts with 3 template molecules. How many molecules after 8 cycles?
5. Why is Taq polymerase used in PCR?
6. State the role of DNA ligase in making a recombinant plasmid.
7. Why is a marker gene included in a vector?
8. A probe is 5’ GATTCA 3’. Write the 5’ to 3’ sequence of the DNA strand it binds to.
9. Explain why smaller fragments travel further in a gel.
10. A DNA fingerprint matches at three independent loci where the match chances are 0.1, 0.2 and 0.05. Calculate the chance of a coincidental match.
11. Which electrode do DNA fragments move towards, and why?
12. State one way genetic fingerprinting is used in breeding.

Answers

1. The full set of proteins that derive from the genetic code of an organism.
2. A bacterium has little non-coding DNA and few regulatory genes, so protein sequences can be worked out from the genome. Humans have much non-coding DNA and many regulatory genes, so the genome cannot easily be translated into the proteome.
3. Reverse transcriptase, using mRNA.
4. 3 × 2⁸ = 3 × 256 = 768.
5. It is heat-stable, so it keeps working after heating to about 95 °C.
6. It joins the sugar-phosphate backbones of the vector and the inserted fragment by forming phosphodiester bonds.
7. To detect which host cells have taken up the vector (transformed cells).
8. The target is complementary and antiparallel to the probe: 5’ TGAATC 3’.
9. The gel acts as a mesh. Smaller fragments pass through it more easily, so they are slowed less and travel further in the same time.
10. 0.1 × 0.2 × 0.05 = 0.001, which is 1 in 1000.
11. The positive electrode, because phosphate groups give it a negative charge.
12. To check parentage or to measure genetic variability in a population, so inbreeding can be avoided.

Where marks are usually lost

• Writing that cDNA is made from DNA, rather than from mRNA with reverse transcriptase.
• Missing that mRNA has no introns, which is why bacteria can express the cDNA.
• Saying restriction enzymes “cut at any place” rather than at a specific recognition site.
• Writing that ligase cuts DNA, or confusing it with a restriction enzyme.
• Describing PCR without primers, or forgetting why the polymerase must be heat-stable.
• Saying the probe is identical to the allele rather than complementary.
• Saying DNA moves to the negative electrode or that bigger fragments travel further.
• Giving only one side in an evaluation of ethical, financial or social issues.
• Calling a fingerprint match proof, instead of saying a coincidence is very unlikely.

Official syllabus

AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, sections 3.8.3, 3.8.4.1, 3.8.4.2 and 3.8.4.3 (version 1.5), published by AQA.