OCR GCSE Biology A: DNA structure, protein synthesis and enzyme action (J247) – 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 cover module B1.2 of OCR GCSE (9-1) Gateway Science Biology A (J247), specification Version 4.0 (August 2026), for first assessment in 2018. Questions 4, 5, 6 and 12(a) are Higher Tier only (statements B1.2d and B1.2e) and are labelled. The rest suit both tiers. Learn the content first from the study guide or the revision notes. Course pages: OCR GCSE Biology hub and checklist.

Questions

1. Describe the structure of DNA. [3]

2. (a) Describe the parts of one DNA nucleotide. [2] (b) One strand of DNA reads GTCCAGT. Write the sequence of the strand paired with it. [2]

3. A sample of double-stranded DNA contains 31% cytosine. Calculate the percentage of guanine, adenine and thymine. [3]

4. (Higher Tier only) Describe how a gene in the nucleus is used to make a protein. [5]

5. (Higher Tier only) The template strand of a gene reads TAC GGT CTT AAA. Data supplied: AUG = Met, CCA = Pro, GAA = Glu, UUU = Phe, GAU = Asp, UUC = Phe. (a) Write the mRNA sequence. [2] (b) Give the order of amino acids. [2]

6. (Higher Tier only) Use the gene and data in question 5. (a) The third triplet changes from CTT to CTA. State the new amino acid order and explain how this could stop an enzyme working. [3] (b) Instead the last triplet changes from AAA to AAG. Explain why the protein is unchanged. [2]

7. Explain why one enzyme will catalyse only one reaction. [4]

8. Catalase is heated in boiling water, then cooled. It no longer breaks down hydrogen peroxide. Explain why. [4]

9. An amylase and starch mixture is tested at six pH values. The table shows the time for the starch to disappear.

pH	4	5	6	7	8	9
Time / s	240	120	60	40	80	200

(a) Calculate 1000 ÷ time for pH 8 and pH 9 as a measure of rate. [2] (b) The value for pH 7 is 25. Calculate the percentage decrease in rate from pH 7 to pH 9. [2] (c) State the best pH tested and suggest how to find the optimum more precisely. [2]

10. An enzyme is tested with different substrate concentrations.

Substrate / mmol per dm³	0	10	20	40	80	160
Rate / cm³ per min	0	4.5	8.8	15.6	18.9	19.2

(a) Describe the pattern. [2] (b) Calculate the percentage increase in rate when substrate concentration doubles from 20 to 40. [2] (c) Explain the pattern between 80 and 160. [2]

11. Plan an experiment to find how temperature changes the speed of oxygen production when catalase acts on hydrogen peroxide. [8]

12. A mutation changes one amino acid in an enzyme made by a bacterium. The normal enzyme forms 24 units of product per minute. The mutant forms 6 units per minute. (a) (Higher Tier only) Explain how a change in the gene can change the enzyme’s activity. [3] (b) Calculate the percentage decrease in activity. [2] (c) Suggest why a different base change in the gene might have no effect on the enzyme. [2]

Answers

1. DNA is a polymer [1] made of nucleotides, in two strands [1] forming a double helix [1]. [3] Examiner insight: Answers that name the monomer and say how many strands there are tend to gain credit; just “a double helix” is a shape, not a description of what it is made from.

2. (a) Each nucleotide has a sugar and a phosphate group [1], with one of four bases attached to the sugar [1]. [2] (b) Pair G with C, T with A, then C, C, A, G, T with G, G, T, C, A [1]: CAGGTCA [1]. [2] Examiner insight: Write the answer as a single clear string. A line of letters that is hard to read can be misread, and a pairing slip such as C with A is a common way to lose both marks.

3. Guanine pairs with cytosine, so G = 31% [1]. A + T = 100 - 62 = 38% [1]. They are equal, so A and T are each 19% [1]. [3] Examiner insight: Show the subtraction. Writing 38% for A plus T makes your use of the pairing rule visible to the reader.

4. The DNA double helix unzips around the gene [1]. A copy of the gene is made as mRNA [1] in the nucleus (transcription), matching the base sequence of the DNA [1]. The mRNA leaves the nucleus for the cytoplasm [1]. The mRNA is translated: tRNA brings amino acids and they are joined in the order the mRNA specifies [1]. [5] Examiner insight: Name the stage and where it happens. Answers that blur transcription and translation, or leave the nucleus out, tend to lose marks on this kind of sequence question.

5. (a) T→A, A→U, C→G gives AUG [1]; the rest give CCA GAA UUU, so AUG CCA GAA UUU [1]. [2] (b) Looking up each triplet [1]: Met - Pro - Glu - Phe [1]. [2] Examiner insight: Work in triplets and write each step. A swap of A for T instead of A for U in the mRNA is a frequent slip.

6. (a) CTA gives mRNA GAU, so the order is Met - Pro - Asp - Phe [1]. A different amino acid changes how the chain folds [1], so the active site shape changes and the substrate no longer fits [1]. [3] (b) AAG gives UUC [1], which codes for Phe, the same amino acid as UUU, so the protein is unchanged [1]. [2] Examiner insight: Link the change step by step: base, then amino acid, then shape, then function. Stating only “the enzyme changes” skips the reasoning.

7. The enzyme has an active site of a specific shape [1]. Only a substrate with a complementary shape fits [1], like a key in a lock [1], so only that reaction is catalysed [1]. [4] Examiner insight: Use the word “complementary” or “matching” about the shape of the active site and the substrate. Saying the enzyme “changes to fit” is not the model asked for here.

8. The high temperature changes the shape of the enzyme [1], including the active site [1]. The enzyme is denatured, so the substrate no longer fits [1]. Cooling does not restore the shape, because denaturation is permanent [1]. [4] Examiner insight: Use “denatured” and explain what it means. “Killed” is wrong, because enzymes are not living.

9. (a) pH 8: 1000 ÷ 80 = 12.5 [1]. pH 9: 1000 ÷ 200 = 5.0 [1]. [2] (b) (25 - 5) ÷ 25 × 100 [1] = 80% decrease [1]. [2] (c) pH 7 gives the shortest time [1]. Test more values close to it, such as pH 6.5 and 7.5 [1]. [2] Examiner insight: Give “the best value tested” rather than “the optimum” unless the data pin it down. Saying which values to test next shows you understand the limits of the data.

10. (a) The rate rises as substrate concentration increases [1], at a slowing rate, then levels off at about 19 cm³ per min [1]. [2] (b) (15.6 - 8.8) ÷ 8.8 × 100 [1] = 77% [1]. [2] (c) All the active sites are occupied [1], so extra substrate cannot be converted faster; the enzyme concentration is limiting [1]. [2] Examiner insight: Pair each description with a figure from the table and each explanation with active sites. Answers that say the rate “stops” are weaker than ones that say it levels off at a stated value.

11. Any eight from:

• Set up water baths at different temperatures, for example 15, 25, 35, 45 and 55 °C [1].
• Measure the volume of oxygen collected in a fixed time, using a gas syringe or measuring cylinder [1].
• Put the enzyme and the hydrogen peroxide into the bath separately until each reaches the temperature, then mix [1].
• Keep the hydrogen peroxide volume and concentration identical in every run [1].
• Use the same amount of catalase source, such as liver, and the same pH [1].
• Repeat each temperature and calculate a mean [1].
• Calculate rate = volume ÷ time, and plot rate against temperature [1].
• Wear eye protection because hydrogen peroxide can irritate [1]. [8] Examiner insight: Plans gain credit for being specific: a named instrument, a named control and a stated measurement. “Keep everything else the same” with nothing named is too vague.

12. (a) A change in a base alters a triplet, so a different amino acid may be added [1]. This changes how the protein folds, including the active site [1]. The substrate no longer fits as well, so the activity falls [1]. [3] (b) (24 - 6) ÷ 24 × 100 [1] = 75% decrease [1]. [2] (c) The new triplet might code for the same amino acid [1], or the changed amino acid might be far from the active site, leaving its shape unchanged [1]. [2] Examiner insight: When a question asks for a percentage change, divide by the original value. Dividing by the new value is a frequent slip and gives a different figure.

Where marks are usually lost

• Describing DNA as made of protein or sugar.
• Pairing bases incorrectly, or using T in mRNA.
• Mixing up transcription and translation, or forgetting where each happens.
• Skipping a link in the chain from gene to base sequence to amino acids to shape to function.
• Saying enzymes are “killed” rather than denatured.
• Saying cold temperatures denature enzymes.
• Forgetting that the rate levels off because active sites are occupied.
• Dividing by the wrong value when working out a percentage change.
• Leaving variables uncontrolled, or naming none, in a practical plan.

Next steps

• Review the revision notes and the study guide.
• See the OCR GCSE Biology hub and tick off the checklist.
• Try all the free 10-minute diagnostics.
• Revisit the wider topic in Cell level systems practice.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Gateway Science Biology A (J247), Version 4.0 (August 2026), for first assessment in 2018, published by OCR: module B1.2.