OCR A Level Biology A: DNA sequencing, profiling, PCR and electrophoresis (H420) – Revision Notes

These notes are for OCR A Level Biology A (H420), written to the OCR specification Version 4.1 (April 2026), for first assessment in 2025. They cover module 6.1.3 learning outcomes (a) to (e): sequencing, bioinformatics, DNA profiling, PCR and electrophoresis. Full explanations and worked examples are in the study guide, and exam-style questions are in the practice set. Course pages: OCR A Level Biology hub, the printable checklist and the free 10-minute diagnostics.

Key terms

Term	Meaning
Sequencing	Finding the order of bases in DNA
Dideoxynucleotide	Modified nucleotide with no 3′ hydroxyl group; stops the growing strand
High throughput sequencing	Millions of reactions at once; software assembles the short reads
Bioinformatics	Software and databases to store, search and analyse biological data
Computational biology	Computer models and algorithms applied to biological systems
Synthetic biology	Designing and building new biological parts, systems or organisms, or redesigning existing ones
STR	Short tandem repeat: a short base sequence repeated head to tail, with a variable number of repeats
DNA profile	The combination of repeat lengths at several loci
PCR	In vitro amplification of a chosen region of DNA
Primer	Short single-stranded DNA that binds to the template and gives the polymerase a start point
Electrophoresis	Separation of charged molecules in an electric field through a gel

6.1.3(a) Sequencing

Sanger method in steps

1. Mix template DNA, primer, DNA polymerase, ordinary nucleotides and a small amount of four dideoxynucleotides, each with a different fluorescent label.
2. The polymerase builds new strands. When a dideoxynucleotide is added, no phosphodiester bond can form with the next nucleotide, so that strand stops.
3. Termination is random, so the mixture holds strands of every length, each ending in a labelled base.
4. Capillary electrophoresis separates the strands by length.
5. A laser and detector record the label of each strand, shortest first.
6. The labels in order, shortest to longest, give the new strand 5′ to 3′. The template is complementary.

Sanger to high throughput. Sanger sequencing reads one short stretch per reaction. High throughput sequencing runs millions of reactions together. It is faster and cheaper and makes whole-genome sequencing routine. You are not required to know the details of high throughput methods.

6.1.3(b) Bioinformatics

Use	What sequence data allow
Genome-wide comparison between species	Percentage identity measures relatedness. More similar suggests a more recent common ancestor
Comparison between individuals	Variants more common in people with a trait link genotype to phenotype
Epidemiology	Pathogen genomes show chains of transmission and spreading resistance alleles
Predicting polypeptides	Find the start codon, read triplets to a stop codon, translate with the genetic code
Synthetic biology	Known sequences are used to design DNA, which is then made chemically and checked by sequencing

Percentage identity = (bases that match ÷ bases compared) × 100.

Predicting a polypeptide in steps: write the mRNA complementary to the template strand (U replaces T), split it into codons, then look each one up.

Limits. Introns mean the coding sequence is found by comparing with mRNA or cDNA. A sequence does not show folding or later modification. A predicted function is a hypothesis until tested.

6.1.3(c) DNA profiling

Method in steps

1. Extract DNA.
2. Amplify STR loci by PCR with labelled primers.
3. Separate the products by electrophoresis.
4. Compare band or peak patterns between samples.

Uses. Forensics (match a scene sample to a suspect or database), paternity and other family relationships, and analysis of disease risk (carrier status, or risk linked to repeat length).

Match probability. Multiply the frequencies of the matching pattern at each independent locus. Example: 0.25 × 0.2 × 0.04 = 0.002, which is 1 in 500. More loci make a chance match less likely.

Repeat number from a product size. Repeats = (product size − flanking length) ÷ repeat unit length.

Caution. A match is not proof of guilt. Contamination, mixed samples and degraded DNA reduce reliability. Risk results are probabilities.

6.1.3(d) PCR

Component	Role
Template DNA	Contains the target
Two primers	Bind to opposite strands at the ends of the target
Free DNA nucleotides	Raw material for new strands
Thermostable DNA polymerase (Taq)	Builds new strands and survives about 95 °C

Step	Temperature	Event
Denaturation	about 95 °C	The two strands come apart
Annealing	about 50 to 65 °C	Primers bind by complementary base pairing
Extension	about 72 °C	Polymerase adds nucleotides to the 3′ end of each primer

Copies after n cycles (ideal) = starting copies × 2ⁿ. For example, 5 copies after 8 cycles gives 5 × 256 = 1280. Ten cycles give a 1024-fold increase.

Applications in DNA analysis: amplifying tiny samples for profiling, preparing DNA for sequencing, and testing for particular alleles. Run a no-template control to detect contamination.

6.1.3(e) Electrophoresis

Nucleic acids

• DNA is negatively charged (phosphate groups), so it moves towards the anode.
• Agarose gel is a mesh, so short fragments travel further than long ones in the same time.
• A ladder of known sizes is run alongside. Plot distance against log₁₀(length) to read off unknown sizes.
• A band is made up of many fragments of one size.

Proteins

• Polyacrylamide gel is used.
• With SDS (SDS-PAGE), proteins are unfolded and coated with negative charge, so they separate by mass.
• Without SDS, charge, size and shape all matter.

Reading a gel: the smallest fragments are furthest from the wells, and two bands at the same distance are fragments of the same size. If a sample shows no band, the DNA may be absent or degraded, or the sample may not have loaded.

Practical (PAG6): wear gloves, handle stains with care, keep mains equipment dry, and stop the run before small fragments leave the gel.

Must-know distinctions

• Dideoxynucleotide and ordinary nucleotide. The first lacks the 3′ hydroxyl group, so it ends the chain.
• PCR and DNA replication in a cell. PCR uses a thermal cycler, a heat-stable polymerase and primers made in advance. It copies a chosen region only.
• Sequencing and profiling. Sequencing gives the base order. Profiling compares repeat patterns to identify or relate individuals.
• Hydrogen bonds and phosphodiester bonds in PCR. Heating at denaturation breaks hydrogen bonds between the strands. The polymerase forms phosphodiester bonds when it extends the primers. Primers are held on the template by hydrogen bonds between complementary bases.
• SDS-PAGE and DNA gels. SDS gives proteins a uniform negative charge. DNA already has one.

Quick self-test

1. Why does a dideoxynucleotide stop strand synthesis?
2. Fragment terminal bases, shortest to longest, are C, A, G, G, T. State the new strand and the template (both 5′ to 3′).
3. Two aligned 20-base sequences differ at 3 positions. Give the percentage identity.
4. A template strand reads 3′ TACGGTCAT 5′. Give the mRNA and the first three amino acids (AUG Met, CCA Pro, GUA Val).
5. A sample has 5 copies of the target. How many after 8 ideal PCR cycles?
6. Why is Taq polymerase used?
7. A PCR product across a repeat is 142 bp. Flanking sequence is 100 bp and the repeat unit is 3 bp. How many repeats?
8. Which travels further in agarose, a 200 bp or an 800 bp fragment?
9. Why is SDS added for protein electrophoresis?
10. Three independent loci have matching-pattern frequencies 0.25, 0.2 and 0.04. Give the combined probability and its 1 in n form.

Answers

1. It lacks the 3′ hydroxyl group, so no phosphodiester bond can form with the next nucleotide.
2. New strand 5′ CAGGT 3′. The template is complementary and antiparallel: 5′ ACCTG 3′.
3. 17 ÷ 20 × 100 = 85%.
4. mRNA 5′ AUGCCAGUA 3′, giving Met, Pro, Val.
5. 5 × 2⁸ = 1280.
6. It withstands the heating to about 95 °C, so it does not need replacing each cycle.
7. (142 − 100) ÷ 3 = 14 repeats.
8. The 200 bp fragment, because it moves more easily through the gel mesh.
9. It unfolds the proteins and gives them a negative charge in proportion to their length, so they separate by mass.
10. 0.25 × 0.2 × 0.04 = 0.002, or 1 in 500.

Where marks are usually lost

• Reading the template from the detector instead of the new strand, or forgetting that the template is antiparallel.
• Saying dideoxynucleotides “stop the polymerase” and not saying why the next bond cannot form.
• Giving a match probability as a decimal when the question asks for “1 in n”, or stopping before the final step.
• Stating that PCR doubles copies every cycle without saying “ideally”, or using 2n instead of 2ⁿ.
• Saying DNA moves “because it is small” without charge, or forgetting that DNA moves to the anode.
• Writing that a DNA profile match “proves” guilt.
• Saying Taq polymerase is used because it works “fast”, not because it is heat stable.
• Plotting fragment length (not its logarithm) against distance and reading a curve badly.
• Describing synthetic biology as simply “genetic engineering” without mentioning design and synthesis.

Official syllabus

OCR Advanced GCE in Biology A (H420) specification, Version 4.1, April 2026, for first assessment in 2025, published by OCR (Cambridge University Press and Assessment): module 6.1.3, learning outcomes (a) to (e).