These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data: take 0.34 nm of DNA length per base pair. Purines are A and G; pyrimidines are C, T and U. A calculator is assumed. All organisms and data sets are fictional.
Question 1 (multiple choice · foundation)
Which statement correctly describes base pairing in a DNA double helix?
- (A) A purine always pairs with a pyrimidine, so every base pair contains three rings in total.
- (B) A purine always pairs with another purine, because two double rings fit together most tightly.
- (C) Adenine pairs with guanine and cytosine pairs with thymine, because each pair shares the same ring structure.
- (D) Any base can pair with any other base, provided that hydrogen bonds form between them.
Answer and explanation
Answer: (A). A (purine) pairs with T (pyrimidine), and G (purine) pairs with C (pyrimidine). Each pair has 2 + 1 = 3 rings, so every rung of the ladder is the same length and the helix has a constant width.
- (B) A purine–purine pair has four rings and would be too wide to fit between the backbones.
- (C) pairs like with like. A and G are both purines; C and T are both pyrimidines. These pairs do not form in DNA.
- (D) ignores the specific pairing rules. Only A–T and G–C line up their hydrogen-bonding groups correctly, which is why one strand fixes the sequence of the other.
Question 2 (multiple choice · core)
A biologist compares a bacterial cell with a cell from a fern leaf. Which comparison of their genetic material is most likely to be correct?
- (A) The bacterium has several linear chromosomes in a nucleus; the fern cell has one circular chromosome in its cytoplasm.
- (B) The bacterium has one circular chromosome in the cytoplasm and may have plasmids; the fern cell has several linear chromosomes in a nucleus, wound around histones.
- (C) Both cells have a single circular chromosome, but only the fern cell wraps it around histones.
- (D) The bacterium has no chromosome, only plasmids; the fern cell has several linear chromosomes.
Answer and explanation
Answer: (B). Prokaryotes typically have one circular chromosome in the nucleoid region and often carry plasmids. Eukaryotes such as ferns typically have several linear chromosomes in a nucleus, condensed by histones.
- (A) swaps the two cell types.
- (C) is wrong for the fern: eukaryotes typically have several linear chromosomes.
- (D) confuses “no nucleus” with “no chromosome”. Bacteria do have a chromosome; plasmids are extra to it.
Question 3 (multiple choice · core)
A double-stranded DNA molecule from a fictional alga is 5 000 base pairs long. How many of its nucleotides contain a purine base?
- (A) 2 500
- (B) 5 000
- (C) 10 000
- (D) It cannot be found without knowing the percentage of adenine.
Answer and explanation
Answer: (B). 5 000 base pairs contain 2 × 5 000 = 10 000 nucleotides. Every pair has exactly one purine, so there are 5 000 purines (half of 10 000).
- (A) halves the number of base pairs instead of the number of nucleotides.
- (C) counts every nucleotide as a purine.
- (D) The percentage of adenine would be needed to find the number of A alone, but the total of purines (A + G) is fixed by the pairing rule.
Question 4 (model evaluation · core)
A student builds a cardboard model of DNA. Each purine is a strip 2.0 cm long and each pyrimidine is a strip 1.0 cm long. The two “backbones” are rulers fixed 3.0 cm apart.
(a) Calculate the total length of an A–G rung and of a C–T rung, and explain what would happen to the rulers if each were added to the model. (b) Explain why the real double helix has the same width all along its length. (c) Give one way in which the model does not fully represent real DNA.
Worked solution and suggested Marlbridge rubric
(a) A–G: 2.0 + 2.0 = 4.0 cm, which is 1.0 cm more than the 3.0 cm gap, so the rulers would be pushed apart (a bulge). C–T: 1.0 + 1.0 = 2.0 cm, 1.0 cm short, so the rulers would be pulled together (a pinch) or the strips would not meet.
(b) Every real base pair is one double-ringed purine plus one single-ringed pyrimidine (A–T or G–C), so every rung is the same length. That keeps the distance between the two sugar–phosphate backbones constant (about 2 nm).
(c) Any one: the model shows nothing that makes A choose T rather than C (real specificity comes from matching hydrogen-bonding groups); it is flat, not a twisted helix; the strips are not shown with direction, so the antiparallel 5′–3′ arrangement is missing; the backbones are rigid rulers rather than chains of covalently linked sugars and phosphates.
| Point | What earns it |
|---|---|
| 1 | Both rung lengths (4.0 cm and 2.0 cm) with the effect on the rulers |
| 1 | Purine + pyrimidine in every pair gives equal rung length |
| 1 | Links equal rung length to constant helix width |
| 1 | A valid limitation of the model, explained |
Question 5 (data analysis · core)
A fictional bacterium grows well on plates containing the antibiotic tetracycline. A researcher thinks the resistance gene is on a plasmid, not on the chromosome. She grows the bacteria with a treatment that makes cells lose their plasmids but leaves the chromosome unchanged. She spreads equal samples on plates with and without tetracycline, before and after the treatment, and counts colonies.
| Sample | Colonies, no antibiotic | Colonies, with tetracycline |
|---|---|---|
| Before treatment | 212 | 205 |
| After treatment | 198 | 3 |
(a) Calculate the percentage of colonies that survive tetracycline before and after treatment. (b) Do the results support her claim? Use the data in your answer. (c) Explain the purpose of the plates without antibiotic. (d) Suggest one explanation for the 3 colonies after treatment.
Model answer and suggested Marlbridge rubric
(a) Before: 205 ÷ 212 × 100 = 96.7%. After: 3 ÷ 198 × 100 = 1.5%.
(b) Yes. Removing plasmids, while leaving the chromosome unchanged, cut survival on tetracycline from 96.7% to 1.5%. If the gene were on the chromosome, losing the plasmids should not have changed resistance. So the gene is most likely on the plasmid.
(c) They show how many living cells were spread. The similar counts (212 and 198) show the treatment did not simply kill the bacteria, so the fall on tetracycline is due to loss of resistance.
(d) Any one: a few cells kept their plasmid (the treatment is not 100% effective); a few cells gained a new mutation giving resistance; contamination.
| Point | What earns it |
|---|---|
| 1 | Both percentages correct |
| 1 | Claim supported, with both values quoted |
| 1 | Reasoning: chromosome unchanged, so the change points to the plasmid |
| 1 | Correct role of the no-antibiotic plates |
| 1 | A plausible explanation for the 3 colonies |
Question 6 (calculation · core)
A fictional eukaryotic chromosome contains 9.0 × 10⁷ base pairs. During cell division it condenses to a structure 6 µm long.
(a) Calculate the length of its DNA, in mm. (b) By what factor has the DNA been shortened? (c) If one nucleosome plus linker uses about 200 base pairs, how many nucleosomes does the chromosome contain? (d) Explain why DNA binds tightly to histones.
Worked solution and suggested Marlbridge rubric
(a) 9.0 × 10⁷ × 0.34 nm = 3.06 × 10⁷ nm = 3.06 × 10⁴ µm = 30.6 mm (about 31 mm).
(b) 30.6 mm = 30 600 µm. Factor = 30 600 ÷ 6 = 5 100 (about 5 × 10³).
(c) 9.0 × 10⁷ ÷ 200 = 4.5 × 10⁵ nucleosomes.
(d) The phosphate groups in DNA’s backbone are negatively charged. Histones have many positively charged amino acid side chains. Opposite charges attract, so DNA wraps tightly around the histone core.
| Point | What earns it |
|---|---|
| 1 | Length 30.6 mm (or 3.06 × 10⁷ nm) |
| 1 | Factor about 5 100, with consistent units |
| 1 | 4.5 × 10⁵ nucleosomes |
| 1 | Negative phosphate backbone attracted to positive histones |
Common error: dividing nm by µm without converting, which gives an answer 1 000 times too large.
Question 7 (constructed response · stretch)
One strand of a short double-stranded DNA region is 5′-AAGGAGCTGA-3′.
(a) Write the complementary strand from 5′ to 3′. (b) Calculate the percentage of purines in each strand separately, and in the whole double-stranded region. (c) A student says: “Purines are always 50% of DNA.” Evaluate this claim using your answers.
Worked solution and suggested Marlbridge rubric
(a) Pair each base (A↔T, G↔C), then reverse so it reads 5′ to 3′: 5′-TCAGCTCCTT-3′.
(b) Given strand: A, A, G, G, A, G, G, A are purines: 8 of 10 = 80%. Complementary strand: only A and G at positions 3 and 4 (5′-TCAGCTCCTT-3′): 2 of 10 = 20%. Whole region: 8 + 2 = 10 purines out of 20 nucleotides = 50%.
(c) The claim is only true for the double-stranded region. Each purine on one strand pairs with a pyrimidine on the other, so the totals balance at 50%. A single strand can have any proportion of purines (here 80% or 20%). The claim is false for single strands, single-stranded viral genomes or most RNA.
| Point | What earns it |
|---|---|
| 1 | Complementary strand correct and written 5′ to 3′ |
| 1 | 80%, 20% and 50% |
| 1 | Explains that the 50% rule comes from pairing, so it holds for the duplex |
| 1 | States that it does not hold for a single strand, using the data |
How did you do?
- Q1 or Q4 wrong: re-read “Purines and pyrimidines” in the study guide and look again at Figure 2.
- Q2 or Q5 wrong: re-read “Organising the DNA” and “Plasmids”, with Figure 1.
- Q3, Q6 or Q7 calculations wrong: rework Worked examples 1 and 2; check whether you are counting base pairs or nucleotides, and convert units before dividing.
Then tick off the topic checklist.