Cambridge International AS & A Level Biology 9700: Nucleic acids and protein synthesis – 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 topic 6 of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027: sections 6.1 (Structure of nucleic acids and replication of DNA) and 6.2 (Protein synthesis). This is AS Level content, examined on Paper 1 and Paper 2; the syllabus says AS knowledge is also required on Paper 4. The questions are written in the structured style of Paper 2.

Before you start, you may want the study guide or the revision notes. Course links: Cambridge A Level Biology hub, printable 9700 checklist and the AS Level diagnostic.

Codon table for Questions 8 and 9 (mRNA codons): AUG = Met (start), GCU = Ala, AAA = Lys, AAG = Lys, CCG = Pro, CUA = Leu, AAC = Asn, CGU = Arg, UAA / UAG / UGA = stop.

Questions

1. Name the two purine bases and state one structural difference between a purine and a pyrimidine. [2]

2. Describe the structure of a molecule of ATP. [3]

3. A sample of double-stranded DNA contains 31% guanine.

(a) Calculate the percentage of thymine in the sample. Show your working. [2] (b) A 40-base-pair section of this DNA contains 14 A–T pairs. Calculate the number of hydrogen bonds between the two strands in this section. [2]

4. State what is meant by a gene, and explain what is meant by the statement that the genetic code is universal. [3]

5. Explain what is meant by describing the two strands of DNA as antiparallel, and why the strands must be antiparallel for complementary base pairing. [3]

6. Bacteria were grown for many generations in a medium containing only heavy nitrogen (¹⁵N). They were then transferred to a medium containing only light nitrogen (¹⁴N). DNA was extracted after each generation and centrifuged, so that heavy, hybrid and light DNA formed separate bands.

(a) Describe the band or bands you would expect after one generation in ¹⁴N. [2] (b) Predict the proportion of DNA molecules in each band after three generations in ¹⁴N. [2] (c) Explain why the result after one generation rules out conservative replication. [2]

7. Explain why, during DNA replication, one new strand is made continuously and the other is made in fragments. Include the role of DNA ligase. [5]

8. The template strand of a short gene reads 3′-TAC CGA TTT GGC ATC-5′.

(a) Write the mRNA sequence transcribed from this strand. [1] (b) Use the codon table to give the sequence of amino acids. [2] (c) State the anticodon of the tRNA that carries the second amino acid. [1] (d) Write the sequence of the non-transcribed strand, 5′ to 3′. [1]

9. Three different mutations are found in the gene in Question 8.

(a) Mutation X changes the third triplet of the template strand from TTT to TTC. Explain the effect on the polypeptide. [2] (b) Mutation Y deletes the first base (C) of the second template triplet. Give the new amino acid sequence for the first four codons and explain why so many amino acids change. [3] (c) Mutation Z changes the third template triplet from TTT to ATT. Explain the effect on the polypeptide and suggest why the protein may not function. [3]

10. In a eukaryotic cell, the primary transcript of a gene is 2715 nucleotides long. Its introns total 1164 nucleotides.

(a) Calculate the length of the mRNA. [1] (b) In the mRNA, 132 nucleotides lie outside the section from the start codon to the stop codon and are not translated. Calculate the number of amino acids in the polypeptide. Show your working. [3] (c) Explain how the primary transcript is converted to mRNA and where this happens. [2]

11. Describe how the information in a gene is used to make a polypeptide in a eukaryotic cell. Refer to RNA polymerase, mRNA, codons, tRNA, anticodons and ribosomes in your answer. [9]

12. Compare DNA replication with transcription. [7]

Answers

1. Adenine and guanine [1]; purines have a double ring and pyrimidines have a single ring [1]. [2] Examiner insight: Both purines must be named for the first mark; “purines are bigger” alone does not earn the structural mark – ring number is the credited point.

2. A nucleotide / phosphorylated nucleotide [1]; containing the base adenine and the sugar ribose [1]; with three phosphate groups [1]. [3] Examiner insight: “Deoxyribose” loses the sugar mark; a structural formula is not required and earns nothing extra.

3. (a) C = 31%, so G + C = 62% and A + T = 38% [1]; thymine = 19% [1]. (b) 26 C–G pairs [1]; 14 × 2 + 26 × 3 = 106 hydrogen bonds [1]. Examiner insight: In (a) a bare 19% can gain both marks, but a wrong answer with no working gains none – showing A + T = 38% secures the method mark.

4. A gene is a sequence of nucleotides that forms part of a DNA molecule [1] and codes for a polypeptide [1]; universal means the same triplets code for the same amino acids (and start/stop signals) in almost all organisms [1]. [3] Examiner insight: “A gene codes for a characteristic” is not accepted; the syllabus definition ties a gene to a polypeptide.

5. One strand runs 5′ to 3′ and the other runs 3′ to 5′ [1]; in this orientation the bases on the two strands line up so that hydrogen bonds can form between A and T and between C and G [1]; each pair is a purine opposite a pyrimidine, keeping the width of the helix constant [1]. [3] Examiner insight: “The strands run in opposite directions” gains the first mark only if linked to the 5′ and 3′ ends somewhere in the answer.

6. (a) A single band [1]; in the hybrid (intermediate) position, because every molecule has one ¹⁵N strand and one ¹⁴N strand [1]. (b) Hybrid: 25% (1/4) [1]; light: 75% (3/4); no heavy band [1]. (c) Conservative replication would keep the original molecule intact, giving one heavy band and one light band [1]; only a hybrid band appeared, so each molecule contains one old strand [1]. Examiner insight: In (b), give both bands and state that no heavy DNA remains; a proportion for one band alone scores one mark at most.

7. DNA polymerase can only add nucleotides to the 3′ end of a new strand, so it builds in the 5′ to 3′ direction [1]; the strands are antiparallel [1]; on the leading strand polymerase moves towards the replication fork, so synthesis is continuous [1]; on the lagging strand it must move away from the fork, so short fragments are made as more template is exposed [1]; DNA ligase joins the fragments by forming phosphodiester bonds [1]. [5] Examiner insight: Without the 5′ to 3′ reason the answer describes rather than explains, and loses the first mark even if everything else is correct.

8. (a) 5′-AUG GCU AAA CCG UAG-3′ [1] (b) Met–Ala–Lys–Pro [1]; UAG is a stop codon, so there is no fifth amino acid [1]. (c) CGA [1] (d) 5′-ATG GCT AAA CCG TAG-3′ [1] Examiner insight: In (a) any T in the sequence scores zero, as it is no longer RNA; in (b) naming a “fifth amino acid” for UAG loses the second mark.

9. (a) The codon changes from AAA to AAG [1]; both code for lysine, so the polypeptide is unchanged [1]. (b) Met–Leu–Asn–Arg [1]; the deletion causes a frame shift [1]; every triplet after the deletion is read in a new grouping of three bases, so most codons change [1]. (c) The codon becomes UAA, a stop codon, so translation stops after Met–Ala [1]; the polypeptide is shorter [1]; missing amino acids alter the tertiary structure, so a binding or active site may not form [1]. Examiner insight: “Frame shift” must be explained in terms of triplets being read differently; the term alone is usually credited only once.

10. (a) 2715 − 1164 = 1551 nucleotides [1] (b) Translated section = 1551 − 132 = 1419 nucleotides [1]; 1419 ÷ 3 = 473 codons [1]; the last codon is a stop codon, so 472 amino acids [1]. (c) Introns are removed and exons are joined together [1]; in the nucleus, before the mRNA leaves [1]. Examiner insight: Allow error carried forward from (a) into (b), but only if the stop codon is subtracted – 473 loses the final mark.

11. In the nucleus, the DNA strands separate at the gene as hydrogen bonds break [1]; RNA polymerase moves along the template strand [1]; free RNA nucleotides pair by complementary base pairing (U opposite A) and are joined by phosphodiester bonds [1]; introns are removed from the primary transcript and exons joined to form mRNA, which leaves through a nuclear pore [1]; mRNA binds to a ribosome in the cytoplasm [1]; each codon of three bases codes for one amino acid, starting at AUG [1]; tRNA carries a specific amino acid and its anticodon pairs with a complementary codon [1]; a peptide bond forms between adjacent amino acids as the ribosome moves along one codon at a time [1]; translation ends at a stop codon and the polypeptide is released [1]. [9] Examiner insight: Each named molecule needs its role, not just its name; a list of terms without actions earns very little on a long “describe” question.

12. Similarity: both use a DNA strand as a template with complementary base pairing, forming phosphodiester bonds in the nucleus [1]. Differences: replication copies both strands, transcription copies only the template strand [1]; replication copies the whole DNA molecule, transcription copies one gene [1]; replication uses DNA polymerase (and DNA ligase), transcription uses RNA polymerase [1]; replication uses nucleotides containing deoxyribose, transcription uses nucleotides containing ribose [1]; adenine pairs with thymine in replication but with uracil in transcription [1]; replication produces double-stranded DNA, transcription produces single-stranded RNA [1]. [7] Examiner insight: A “compare” answer needs paired statements (“replication uses…, whereas transcription uses…”); a separate account of each process rarely earns full credit.

Where marks are usually lost

• Writing T instead of U in an mRNA sequence.
• Giving the codon when the question asks for an anticodon.
• Counting the stop codon as an amino acid.
• Leaving out “5′ to 3′” when explaining leading and lagging strands.
• Saying DNA ligase joins bases or breaks hydrogen bonds.
• Describing the Meselson–Stahl-type result without saying why each band appears.
• Explaining a deletion only as “a frame shift” with no mention of triplets.
• Missing “introns removed” and “exons joined” as two separate points.
• Writing about “protein” with no link from amino acid sequence to shape.

Next steps

• Recap the definitions: nucleic acids and protein synthesis revision notes
• Relearn a weak area: nucleic acids and protein synthesis study guide
• Course overview: Cambridge A Level Biology hub
• Track every outcome: printable 9700 checklist
• Test the whole AS course: 9700 AS Level diagnostic, or see all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International, Cambridge International AS & A Level Biology 9700 syllabus for 2025, 2026 and 2027 (Version 1, published September 2022), Cambridge University Press & Assessment. Topic 6, Nucleic acids and protein synthesis: sections 6.1 and 6.2.