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S5 Biology: DNA and Protein Synthesis (Paper 4)
S5 Biology: DNA and Protein Synthesis (Paper 4)
The fourth paper concentrates on base ratios, strand counting and coding length, and repairs the common replication muddle.
Direct answer A equals T and G equals C, so work out one pair then split the remainder; n rounds of replication give two to the power n molecules; three bases code one amino acid and the stop codon adds none.
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)
- FoundationA DNA segment has 600 bases, 150 of them guanine. How many adenine bases are there?
- FoundationA human body cell has 46 chromosomes, each holding one DNA molecule. After one round of replication, how many DNA molecules are there?
- FoundationWhich tRNA anticodon pairs with the start codon AUG on mRNA?
- FoundationWhich bonds join the base pairs, and which bonds join neighbouring nucleotides along the backbone?
- IntermediateA double-stranded DNA has 1 200 base pairs with adenine making up 20 per cent of all bases. How many guanine bases are there?
- IntermediateOne DNA molecule replicates three times. How many molecules result, and how many strands are newly made?
- IntermediateA protein has 250 amino acids. What is the minimum number of mRNA bases in its coding region, including the stop codon?
- ChallengeIn a DNA sample guanine and cytosine together make up 60 per cent of all bases. What percentage is adenine?
- ChallengeA coding region has 3 000 bases and no introns. At most how many amino acids can it encode?
- Error redoA student calls DNA replication conservative because both daughters keep an old strand. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 150 adenine bases | Guanine and cytosine take 150 each, so the remaining 300 bases are shared equally by adenine and thymine. |
| 2 | 92 molecules | Each molecule is copied into two, so forty-six becomes ninety-two. |
| 3 | UAC | The anticodon is complementary, so A pairs with U and G with C, giving UAC for AUG. |
| 4 | Hydrogen bonds between bases, phosphodiester covalent bonds along the backbone | Hydrogen bonds between bases are weak enough to be opened during replication, while covalent phosphodiester bonds keep each strand intact. |
| 5 | 720 guanine bases | Twelve hundred pairs give 2 400 bases; A and T take 480 each, leaving 1 440 shared by G and C at 720 each. |
| 6 | Eight molecules and fourteen new strands | Three rounds give eight molecules and sixteen strands; only the two original strands are parental, so fourteen are new. |
| 7 | 753 bases | Two hundred and fifty amino acids need 750 bases, plus three for the stop codon, giving 753. |
| 8 | 20 per cent | Guanine and cytosine take sixty per cent, leaving forty per cent shared equally by adenine and thymine at twenty per cent. |
| 9 | 999 amino acids | Three thousand bases give 1 000 codons; the final stop codon codes for no amino acid, leaving 999. |
| 10 | It is semi-conservative: each daughter keeps one old and one new strand | Conservative replication would keep both old strands in one daughter, but each daughter actually keeps one old and one new strand, which is semi-conservative. |
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.
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