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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions 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). Learn the content first in the study guide or the revision notes. Course pages: OCR A Level Biology hub, the printable checklist and the diagnostics.

Questions

1. Describe how dideoxynucleotides are used to find the base sequence of a DNA template in Sanger sequencing. [4]

2. In a Sanger reaction the terminal bases of the fragments, from shortest to longest, are T, C, A, G, G, C, A, T, T, C, A, G.

(a) State the sequence of the new strand, 5′ to 3′. [1] (b) Give the template sequence, written 5′ to 3′. [2] (c) Explain why the shortest fragment reaches the detector first. [1]

3. Aligned 24-base stretches of one gene from three species, A, B and C, are:

A  ATGAAGCTGGACTTCGGTCCAATG
B  ATGAAGCTGGATTTCGGTCCAATA
C  ATGAGGCTAGATTTCGATCCGATA

(a) Calculate the percentage identity for A and B, A and C, and B and C. [3] (b) State which two species are likely to share the most recent common ancestor. [1] (c) Suggest why a comparison of one gene gives limited evidence of evolutionary relationships. [1]

4. The template strand of a gene reads 3′ TACCTTGGCAAAATT 5′. Codons: AUG Met; GAA Glu; GAC Asp; CCG Pro; CCA Pro; UUU Phe; UAA stop.

(a) Give the mRNA sequence and the amino acids it codes for. [2] (b) A mutation changes the second codon to GAC. State the effect on the polypeptide. [1] (c) A different mutation changes the third codon to CCA. State and explain the effect on the polypeptide. [2] (d) Explain why a predicted amino acid sequence may not show the structure of the working protein. [1]

5. Sequences of a bacterium were obtained from four patients. The table shows the number of base differences between isolates.

	P1	P2	P3	P4
P1	0	1	2	38
P2		0	2	39
P3			0	38

(a) Identify the patients most likely to be part of one chain of infection and justify your answer. [2] (b) Suggest how patient P4 was infected. [1] (c) Explain how the sequences could also help a health authority respond to the outbreak. [1]

6. (a) Explain two advantages of high throughput sequencing over Sanger sequencing for sequencing a whole genome. [2] (b) Explain how sequencing has made the development of synthetic biology possible. [3]

7. DNA from a crime scene and from two suspects is profiled. The table shows repeat numbers (a single entry means one allele detected).

Locus	Scene	Suspect A	Suspect B
L1	11, 14	11, 14	11, 14
L2	8	8	8, 9
L3	15, 17	15, 17	15, 17
L4	9, 12	9, 12	9, 12
L5	6, 7	6, 7	6, 7

(a) State which suspect can be excluded and explain why. [2] (b) The frequencies of the matching pattern at the five loci are 0.07, 0.11, 0.09, 0.16 and 0.12. Calculate the probability that a random person matches at all five loci, and give it as 1 in n to the nearest thousand. [3] (c) Give two reasons why a match is not proof that suspect A left the sample. [2]

8. A forensic sample contains 6 copies of a target sequence. Assume copies double every cycle.

(a) Calculate the copies after 22 cycles. [2] (b) Find the smallest number of cycles giving more than 1 × 10⁸ copies. Show that one fewer cycle is not enough. [3] (c) Each cycle takes 140 seconds. Calculate the time for your answer to (b), in minutes. [1] (d) Describe the effect on the reaction of setting the annealing temperature too high. [1]

9. A ladder of DNA fragments ran as follows: 1200 bp 16 mm; 800 bp 23 mm; 600 bp 28 mm; 400 bp 35 mm; 200 bp 47 mm. An unknown fragment ran 31 mm.

(a) Describe how to use the ladder to estimate the size of the unknown fragment. [3] (b) The best-fit line gives log₁₀(size) = 3.48 − 0.0251 × distance (distance in mm). Calculate the size of the unknown fragment to the nearest 10 bp. [1] (c) Explain why shorter fragments travel further. [2]

10. Protein standards ran in SDS-PAGE: 120 kDa at 12 mm, 80 kDa at 19 mm, 50 kDa at 27 mm, 30 kDa at 38 mm, 20 kDa at 48 mm. A protein ran 33 mm.

(a) Explain the role of SDS. [2] (b) Estimate the mass of the protein, to the nearest 5 kDa. [2] (c) State what would determine separation if SDS were left out. [1]

11. A late-onset disorder is linked to the number of 3-base repeats at one locus. PCR with primers either side of the repeat gives a product of 96 bp plus 3 bp per repeat. A repeat count of 40 or more on one allele is classed as high risk (data supplied). A patient’s products are 150 bp and 237 bp.

(a) Calculate the number of repeats on each allele. [2] (b) State the patient’s risk category. [1] (c) Describe how PCR and electrophoresis give these sizes. [4] (d) Discuss one ethical issue raised by predictive testing for the disorder. [3]

Answers

1. The mixture has template, primer, DNA polymerase, ordinary nucleotides and four dideoxynucleotides with different fluorescent labels [1]. A dideoxynucleotide has no 3′ hydroxyl group, so no phosphodiester bond can form with the next nucleotide and the strand stops [1]. Termination is random, so strands of every length are made, each ending in a labelled base [1]. Fragments are separated by length and the labels read from shortest to longest give the sequence [1]. [4] Examiner insight: Say why the chain stops: the missing 3′ hydroxyl group means the next bond cannot form. “Modified” alone is vague.

2. (a) TCAGGCATTCAG [1]. (b) The template is complementary and antiparallel: 3′ AGTCCGTAAGTC 5′ [1], so 5′ to 3′ it is CTGAATGCCTGA [1]. (c) Shorter fragments move through the capillary gel faster [1]. [4] Examiner insight: State which end you are writing from. A complement left unreversed is an easy loss.

3. (a) A and B differ at 2 bases: 22/24 = 91.7% [1]. A and C differ at 6: 18/24 = 75.0% [1]. B and C differ at 4: 20/24 = 83.3% [1]. (b) A and B, because they are most similar [1]. (c) Genes change at different rates, so one gene may not represent the genome [1]. [5] Examiner insight: Show the count of differences before each percentage, so that your method is visible.

4. (a) mRNA 5′ AUGGAACCGUUUUAA 3′ [1]; Met, Glu, Pro, Phe, then stop [1]. (b) Glu becomes Asp, so the polypeptide has a different amino acid [1]. (c) CCA still codes for Pro, so there is no change [1], because the genetic code is degenerate (several codons can specify one amino acid) [1]. (d) The sequence does not show folding, or changes made after translation [1]. [6] Examiner insight: Link “no change” to the degenerate code, not only state it, because the explanation is what answers “explain”.

5. (a) P1, P2 and P3 [1], because they differ by only 1 or 2 bases, so they probably share a recent source [1]. (b) P4 differs by 38 or 39 bases, so it probably came from a different source [1]. (c) Sequences show whether resistance alleles are spreading, so treatment can be chosen [1]. [4] Examiner insight: Quote values from the table. “Closely related” with no figures is weaker than “1 to 2 differences”.

6. (a) Millions of reactions run together, so a genome is done much faster [1]. The cost per base is lower [1]. (b) Known sequences let designers plan new DNA [1], which can be made chemically [1] and checked by sequencing [1]. [5] Examiner insight: Give two distinct advantages. Repeating “faster” in two ways makes one point.

7. (a) Suspect B [1], because at L2 B has a second allele (9) that is not in the scene sample [1]. (b) 0.07 × 0.11 × 0.09 × 0.16 × 0.12 [1] = 1.33 × 10⁻⁵ [1], about 1 in 75 000 [1]. (c) Any two: contamination or a mixed sample; a degraded, incomplete profile; a relative with a similar pattern [1] [1]. [7] Examiner insight: One mismatch is enough to exclude. For the probability, show the product before the “1 in n” statement.

8. (a) 6 × 2²² [1] = 2.52 × 10⁷ (25 165 824) [1]. (b) 6 × 2²⁴ = 1.01 × 10⁸, more than 1 × 10⁸, so 24 cycles [1]. For 23 cycles, 6 × 2²³ = 5.03 × 10⁷ [1], below 1 × 10⁸ [1]. (c) 24 × 140 = 3360 s = 56 minutes [1]. (d) Primers cannot bind, so little or no product forms [1]. [7] Examiner insight: Test the cycle number below your answer as well, and compare it with 1 × 10⁸.

9. (a) Plot distance against log₁₀(fragment size) for the ladder [1]. Draw a best-fit straight line [1]. Read log₁₀(size) at 31 mm and take the antilog to find the size [1]. (b) log₁₀(size) = 3.48 − 0.0251 × 31 = 2.70, so the size is about 500 bp [1]. (c) DNA is negatively charged and moves through a gel mesh [1]. Shorter fragments pass through the pores more easily, so they move further in the same time [1]. [6] Examiner insight: Show the line and the reading you used, as a value read from a graph is only approximate.

10. (a) SDS unfolds (denatures) the protein [1] and coats it with negative charge in proportion to its length, so separation depends on mass [1]. (b) Plot distance against log₁₀(mass), draw a best-fit line [1] and read 33 mm: about 40 kDa [1]. (c) Charge, size and shape [1]. [5] Examiner insight: “Gives the proteins a charge” is incomplete. Say the charge is negative and proportional to length.

11. (a) (150 − 96) ÷ 3 = 18 [1]; (237 − 96) ÷ 3 = 47 [1]. (b) High risk, because one allele has 47 repeats, which is 40 or more [1]. (c) PCR amplifies the region across the repeat using primers on either side, heat-stable polymerase and repeated denaturation, annealing and extension [1]. The products are loaded into gel wells beside a ladder [1]. Under current, negatively charged DNA moves towards the anode, shorter fragments travelling further [1]. Band positions are compared with the ladder to find sizes [1]. (d) Any one issue, developed: state the issue, such as insurers or employers using the result, anxiety about a disorder with no certain onset date, or relatives who share alleles learning of their own risk without asking [1]; give a consequence for the person [1]; say who is affected or how the harm or benefit could be reduced [1]. [10] Examiner insight: For a discuss command, state the issue, then its consequence, then who is affected.

Where marks are usually lost

• Reading the template from the detector, or not reversing the complement to write it 5′ to 3′.
• Saying a dideoxynucleotide “stops DNA polymerase”, without the missing 3′ hydroxyl group.
• Giving a match probability as a decimal when “1 in n” is asked for.
• Forgetting that negative charge moves DNA, and that SDS gives proteins that charge.

Next steps

• Revision notes and study guide.
• OCR A Level Biology hub and printable checklist.
• All free 10-minute diagnostics.
• Book a free trial class.

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).