OCR A Level Biology A: Nucleotides, ATP, DNA and replication (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 follow the OCR A Level Biology A specification (H420, Version 4.1, April 2026), section 2.1.3(a) to (e): nucleotides, phosphodiester bonds, ADP and ATP, DNA structure, the DNA purification practical and semi-conservative replication. Learn the content first from the study guide and the revision notes. Course pages: OCR A Level Biology hub and the printable checklist. Not sure of your level? Take a free diagnostic first.

Questions

1. (a) Give two ways in which the nucleotides of DNA differ from those of RNA. [2] (b) Name the two purines found in nucleic acids and say how a purine molecule differs in shape from a pyrimidine. [2]

2. A single-stranded polynucleotide contains 35 nucleotides. (a) State the number of phosphodiester bonds in the strand and the number of water molecules released when it was made. [2] (b) State the number of phosphodiester bonds in a double-stranded DNA fragment of 35 base pairs. [1]

3. A DNA fragment has 25 base pairs and 69 hydrogen bonds in total. Calculate the number of A-T pairs and the number of G-C pairs, and state the number of cytosine bases. Show your working. [4]

4. (a) Describe the structure of an ATP molecule. [3] (b) State the products of the hydrolysis of ATP. [1]

5. A cell process needs 183 kJ of energy per mole of product. Hydrolysis of one mole of ATP supplies 30.5 kJ (value supplied). (a) Calculate the minimum number of moles of ATP needed per mole of product. [2] (b) Explain why the real amount is likely to be greater. [1]

6. In a DNA purification from fruit, the tissue is mashed, then mixed with detergent, salt and a protease, and filtered. Ice-cold ethanol is then poured slowly on top of the filtrate. Explain the purpose of the detergent, the salt, the protease and the ice-cold ethanol. [5]

7. A student compared the DNA obtained from three plant tissues using the same method. The dried DNA was weighed.

Tissue	Mass of tissue (g)	Mass of dried DNA (mg)
Banana	12.0	3.6
Kiwi fruit	15.0	5.7
Spinach leaf	24.0	8.4

(a) Calculate the mass of DNA in mg per gram of tissue for kiwi fruit and spinach. [2] (b) Banana gave 0.30 mg per gram. Calculate the percentage more DNA per gram that kiwi fruit gave compared with banana, to the nearest whole number. [2] (c) Suggest two reasons why the student cannot be confident that kiwi fruit has the most DNA per gram. [2]

8. Describe how DNA is replicated semi-conservatively, including the roles of helicase and DNA polymerase. [6]

9. A bacterial culture was kept for many generations on a food source labelled with heavy nitrogen (¹⁵N), so all its DNA was heavy (HH). It was then moved to light nitrogen (¹⁴N) for two generations, then to ¹⁵N again for one generation. Hybrid DNA has one heavy and one light strand (HL). (a) Give the percentages of HH, HL and LL DNA after the two generations in ¹⁴N. [2] (b) Give the percentages of each type after the following generation in ¹⁵N, showing your working. [2] (c) A conservative model of replication predicts that one wholly original and one wholly new molecule are made each time. Use the first generation in ¹⁴N to explain why the result does not support that model. [2]

10. A yeast genome has 1.2 × 10⁷ base pairs. Assume that 3.0 × 10⁻¹⁰ errors occur per base pair per replication (value supplied). (a) Calculate the expected number of errors per replication of the genome. [2] (b) Calculate the expected number of errors across 5,000 replications. [1] (c) Explain what is meant by saying that spontaneous mutations are random. [2] (d) State one way a spontaneous mutation can arise during replication. [1]

11. Explain how the structure of DNA makes it both a stable store of genetic information and able to be copied accurately. [7]

Answers

1. (a) DNA nucleotide has deoxyribose and RNA nucleotide has ribose [1]; DNA has thymine where RNA has uracil [1]. [2] (b) Adenine and guanine [1]; purines have two rings whereas pyrimidines have one ring [1]. [2] Examiner insight: “Contains deoxyribose” earns credit only if the comparison is clear, so name the sugar in each nucleotide.

2. (a) 34 bonds, since 35 − 1 [1]; 34 water molecules, one per bond [1]. [2] (b) 2 × 34 = 68 [1]. [1] Examiner insight: Answers that say 35 or 70 have counted the units instead of the links between them.

3. Let a = A-T pairs and g = G-C pairs; a + g = 25 and 2a + 3g = 69 [1]. Substituting gives 50 − 2g + 3g = 69, so g = 19 [1]. a = 25 − 19 = 6 [1]. Cytosine pairs with guanine, so there are 19 cytosines [1]. [4] Examiner insight: Writing both equations first makes your method clear and lets you check the arithmetic against them.

4. (a) A ribose sugar [1]; the base adenine [1]; three inorganic phosphates [1]. [3] (b) ADP and inorganic phosphate [1]. [1] Examiner insight: “Three phosphates” is stronger than “a phosphate group”, because the number is what separates ATP from ADP.

5. (a) Moles = 183 ÷ 30.5 [1] = 6.0 mol [1]. [2] (b) Some energy is lost as heat, so more ATP is needed than the minimum [1]. [1] Examiner insight: Include the unit with the numerical answer; a bare number leaves it unclear what has been calculated.

6. Detergent dissolves the phospholipids in membranes (cell surface membrane and nuclear envelope), releasing DNA [1]. Salt (sodium ions) masks the negative charge of the phosphate groups so the DNA strands clump together [1]. Protease digests proteins attached to the DNA [1]. DNA is insoluble in ethanol, so it precipitates [1]. Layering slowly lets the DNA appear as white strands at the boundary [1]. [5] Examiner insight: For each reagent, state the effect on the cell or on the DNA, not just “it helps extract DNA”.

7. (a) Kiwi: 5.7 ÷ 15.0 = 0.38 mg per g [1]; spinach: 8.4 ÷ 24.0 = 0.35 mg per g [1]. [2] (b) (0.38 − 0.30) ÷ 0.30 × 100 [1] = 27% [1]. [2] (c) Only one result per tissue, so no repeats or mean [1]; the differences between 0.30, 0.35 and 0.38 are small [1]. (Equally creditable: the product is crude and contains RNA and protein, so the mass is not all DNA.) [2] Examiner insight: A percentage answer should show which value is the baseline; dividing by the wrong tissue gives a different figure.

8. Helicase breaks the hydrogen bonds between complementary base pairs [1]. The two strands separate and the helix unwinds [1]. Each strand acts as a template [1]. Free nucleotides line up by complementary base pairing, A with T and G with C [1]. DNA polymerase joins the nucleotides by forming phosphodiester bonds [1]. Both daughter molecules are made of an original strand paired with a new strand, which is what semi-conservative means [1]. [6] Examiner insight: Say what each enzyme does to which bond; “helicase unzips DNA” on its own is too vague.

9. (a) After the first generation in ¹⁴N all are HL; after the second, 2 HL and 2 LL out of 4, so 50% HL and 50% LL [1]; 0% HH [1]. [2] (b) Each of the 2 HL gives 1 HH and 1 HL; each of the 2 LL gives 2 HL, so 8 molecules: 2 HH, 6 HL [1]. 25% HH, 75% HL, 0% LL [1]. [2] (c) A conservative model predicts one heavy and one light molecule, with no hybrid [1]; the result was all hybrid (HL), which fits only semi-conservative replication [1]. [2] Examiner insight: Give the working molecule by molecule, so a slip in the final percentage is traceable.

10. (a) 1.2 × 10⁷ × 3.0 × 10⁻¹⁰ [1] = 3.6 × 10⁻³ [1]. [2] (b) 3.6 × 10⁻³ × 5,000 = 18 [1]. [1] (c) It is not possible to predict which base or which gene will change [1]; each is equally likely to occur at any replication (the change is not directed by need) [1]. [2] (d) A wrong nucleotide is paired with the template and joined into the new strand [1]. [1] Examiner insight: “Random” needs an explanation in terms of prediction or direction; repeating “it happens by chance” gains little.

11. The sugar-phosphate backbone is held by strong covalent phosphodiester bonds [1]. The bases are on the inside, protected by the backbone [1]. Many hydrogen bonds between the bases hold the strands together, which makes the molecule stable [1]. Each hydrogen bond is weak, so helicase can separate the strands for copying [1]. The sequence of bases carries the genetic information [1]. Complementary base pairing (A with T, G with C) means each strand can be used as a template to give the same sequence [1]. In semi-conservative replication each daughter molecule keeps one original strand, which conserves information [1]. [7] Examiner insight: A strong answer links each feature to a function; listing features without the consequence gains less.

Where marks are usually lost

• Writing hydrogen bonds for the link between nucleotides in the same strand.
• Giving the number of nucleotides when asked for bonds.
• Forgetting the unit or a stated baseline in a calculation.
• Describing helicase or DNA polymerase without naming the bond affected.
• Using “DNA” and “gene” or “base” interchangeably.
• Explaining DNA purification as a list of steps without saying why.
• Confusing the pyrimidine and purine families.
• Treating a single run as reliable when comparing yields.
• Mixing up conservative and semi-conservative predictions.

Next steps

• Re-read the revision notes and the study guide.
• Tick off outcomes on the printable checklist and visit the course hub.
• Take all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, Level 3 Advanced GCE in Biology A (H420), Version 4.1 (April 2026), first assessment 2025, section 2.1.3(a) to (e).