S5 Biology: DNA and Protein Synthesis (Paper 3)
- 2026-11-01
S5 Biology: DNA and Protein Synthesis (Paper 3)
The third paper works on mutation, laboratory techniques and base-ratio reasoning, judging consequences from given data.
Direct answer A substitution may leave the amino acid unchanged, while insertions shift the reading frame; higher GC content raises the melting temperature, and 20 per cent adenine leaves 30 per cent guanine.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S5 Biology: DNA and Protein Synthesis |
| Level | Four basic, three advanced, two challenge and one error-repair item |
02 Questions (10 items)
- FoundationWhat is meant by a gene mutation?
- FoundationDoes replacing one base always change the protein?
- FoundationWhat is DNA fingerprinting used for?
- FoundationA DNA segment has 1 200 base pairs of 0.34 nm each. What is its total length?
- IntermediateWhy does GC-rich DNA need a higher temperature to separate its strands?
- IntermediateWhy is a one-base insertion usually worse than a substitution?
- IntermediateWhat are the three steps of the polymerase chain reaction?
- ChallengeAdenine makes up 20 per cent of the bases in a DNA segment. What percentage is guanine?
- ChallengeA gene has 900 base pairs, 300 of them introns. If the mRNA keeps only exons, how many amino acids result?
- Error redoA student says each daughter molecule is built from two brand-new strands. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | A change in the base sequence of DNA | The change may be a substitution, an insertion or a deletion, and the message changes with it. |
| 2 | Not always, because the code is degenerate and the amino acid may stay the same | The code is degenerate, so several codons can specify the same amino acid. |
| 3 | Identifying individuals and establishing family relationships | Repeat lengths differ between individuals, so the band pattern after electrophoresis is unique. |
| 4 | 408 nm | Twelve hundred times 0.34 nm gives 408 nanometres. |
| 5 | G and C share three hydrogen bonds, one more than A with T | More hydrogen bonds mean more energy is needed to separate the strands, so the melting point rises. |
| 6 | Every codon after the insertion shifts, so the amino acid sequence changes drastically | The reading frame shifts by one, so nearly every later amino acid is altered. |
| 7 | Denaturation, annealing and extension | Heat separates the strands, cooling lets primers bind, then polymerase extends the new strand. |
| 8 | Thirty per cent | A and T together take 40 per cent, so the remaining 60 per cent is split equally at 30 per cent each. |
| 9 | Four hundred amino acids | The exons hold 600 pairs, that is 1 200 bases, which divide into 400 codons. |
| 10 | Each daughter keeps one parental strand plus one new strand | That is what semi-conservative means: each daughter keeps one parental strand. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Book an S5 Biology tutorTutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.


