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 for every question: an A–T pair is held by 2 hydrogen bonds and a C–G pair by 3. Sequences are written 5′ to 3′ unless both ends are labelled. All data sets are fictional.
Question 1 (multiple choice · foundation)
Which comparison of DNA and RNA is correct?
- (A) DNA contains ribose and uracil; RNA contains deoxyribose and thymine.
- (B) DNA contains deoxyribose and thymine and is typically double-stranded; RNA contains ribose and uracil and is typically single-stranded.
- (C) DNA and RNA contain the same sugar, but DNA has thymine where RNA has uracil.
- (D) DNA nucleotides have a phosphate group, but RNA nucleotides do not.
Answer and explanation
Answer: (B). The two nucleic acids differ in three ways: the sugar (deoxyribose against ribose), one base (thymine against uracil) and the usual number of strands (two against one).
- (A) swaps every feature between the two molecules.
- (C) is half right about the bases, but the sugars differ: deoxyribose lacks the –OH on carbon 2′ that ribose has.
- (D) is wrong: every nucleotide, in DNA or RNA, has a phosphate. The phosphates form part of the backbone of both.
Question 2 (multiple choice · core)
One strand of a DNA molecule has the sequence 5′-GATTCA-3′. Which is the sequence of the complementary strand, written 5′ to 3′?
- (A) 5′-CTAAGT-3′
- (B) 5′-ACTTAG-3′
- (C) 5′-TGAATC-3′
- (D) 5′-UGAAUC-3′
Answer and explanation
Answer: (C). Pair each base (G–C, A–T): under 5′-GATTCA-3′ you get 3′-CTAAGT-5′, because the strands are antiparallel. Read from its 5′ end (right to left), this is 5′-TGAATC-3′.
- (A) has the right partner bases but the wrong labels: CTAAGT runs 3′ to 5′ from left to right, so labelling it 5′-CTAAGT-3′ describes a different molecule.
- (B) is the original strand reversed, with no base pairing at all.
- (D) contains uracil, which is found in RNA, not DNA.
Question 3 (multiple choice · core)
A sample of double-stranded DNA from a fictional bacterium contains 18% guanine. What percentage of its bases are adenine?
- (A) 18%
- (B) 32%
- (C) 36%
- (D) 64%
Answer and explanation
Answer: (B). G pairs with C, so C = 18% and G + C = 36%. That leaves 100 − 36 = 64% for A and T. A pairs with T, so A = 64 ÷ 2 = 32%.
- (A) assumes A = G. The equal pairs are A = T and G = C.
- (C) is G + C, not adenine.
- (D) is A + T together; it has not been split between the two bases.
Question 4 (visual representation · core)
The diagram shows a short nucleic acid strand of two nucleotides. Some parts are labelled X, Y and Z. The ends are labelled End 1 and End 2.
(a) Identify the parts labelled X, Y and Z. (b) State which end is the 5′ end and which is the 3′ end. Give a reason. (c) Is this strand DNA or RNA? Give two pieces of evidence from the diagram. (d) A third nucleotide is to be added to this strand. State at which end it is added and what it bonds to.
Model answer and suggested Marlbridge rubric
(a) X = phosphate group; Y = five-carbon sugar (ribose here); Z = nitrogenous base (guanine).
(b) End 1 is the 5′ end: it has a free phosphate attached to carbon 5′ of the sugar. End 2 is the 3′ end: it has a free –OH on carbon 3′ of the last sugar.
(c) RNA. Evidence: each sugar has an –OH on carbon 2′, so the sugar is ribose, not deoxyribose; and the strand contains the base uracil (U), which is found in RNA, not DNA.
(d) At End 2, the 3′ end. The phosphate of the new nucleotide forms a covalent bond with the free 3′ –OH of the last sugar.
| Point | What earns it |
|---|---|
| 1 | X, Y and Z all correctly identified |
| 1 | Both ends correct with a reason based on the free phosphate or free 3′ –OH |
| 1 | RNA, with two valid pieces of evidence (2′ –OH / ribose and uracil) |
| 1 | New nucleotide added at the 3′ end, bonding to the 3′ –OH (covalent bond) |
Do not award point 3 for “it is single-stranded”: DNA can also be single-stranded, and two nucleotides cannot show whether a molecule is paired.
Question 5 (data analysis · core)
A researcher measures the base composition of four fictional nucleic acid samples.
| Sample | A / % | T / % | G / % | C / % | U / % |
|---|---|---|---|---|---|
| W | 30 | 30 | 20 | 20 | 0 |
| X | 26 | 0 | 23 | 20 | 31 |
| Y | 25 | 33 | 18 | 24 | 0 |
| Z | 28 | 0 | 22 | 22 | 28 |
(a) Identify which samples are DNA and which are RNA. Justify your answer. (b) Identify which sample is most likely single-stranded DNA. Use the data to justify your answer. (c) A student says sample Z cannot be double-stranded because “RNA is single-stranded”. Evaluate this claim using the data.
Model answer and suggested Marlbridge rubric
(a) W and Y contain thymine and no uracil, so they are DNA. X and Z contain uracil and no thymine, so they are RNA.
(b) Sample Y. In double-stranded DNA every A pairs with a T and every G with a C, so A = T and G = C. In Y, A (25%) ≠ T (33%) and G (18%) ≠ C (24%), so the bases are not all paired. Sample W does follow the rule (A = T = 30%, G = C = 20%), consistent with double-stranded DNA.
(c) The claim is not supported. RNA is typically single-stranded, but sample Z has A = U (28% each) and G = C (22% each). That is exactly the pattern expected if every base is paired, so Z is consistent with double-stranded RNA (some viruses carry it). Sample X, where A ≠ U and G ≠ C, fits single-stranded RNA. Equal ratios are consistent with pairing but do not prove it; a direct test of structure would be stronger evidence.
| Point | What earns it |
|---|---|
| 1 | DNA (W, Y) and RNA (X, Z) identified, using T versus U |
| 1 | Y chosen, with A ≠ T and G ≠ C quoted from the data |
| 1 | Recognises that Z’s A = U and G = C fits pairing, so the claim is not supported |
| 1 | Notes that “single-stranded” is typical, not universal, or that equal ratios support but do not prove pairing |
Question 6 (constructed response · stretch)
Two fictional double-stranded DNA fragments, each 200 base pairs long, are heated slowly. The table shows their G + C content and the temperature at which half of the molecules have separated into single strands.
| Fragment | G + C / % | Temperature for half separation / °C |
|---|---|---|
| P | 38 | 83 |
| Q | 58 | 91 |
(a) Calculate the number of C–G pairs and A–T pairs in each fragment. (b) Calculate the total number of hydrogen bonds between the strands of each fragment. (c) Explain the difference in separation temperature. (d) A student predicts that heating breaks each fragment into single nucleotides. Evaluate this prediction.
Worked solution and suggested Marlbridge rubric
(a) P: C–G pairs = 0.38 × 200 = 76; A–T pairs = 200 − 76 = 124. Q: C–G pairs = 0.58 × 200 = 116; A–T pairs = 200 − 116 = 84.
(b) P: (76 × 3) + (124 × 2) = 228 + 248 = 476 hydrogen bonds. Q: (116 × 3) + (84 × 2) = 348 + 168 = 516 hydrogen bonds.
(c) Q has a higher proportion of C–G pairs, and each C–G pair is held by three hydrogen bonds rather than two. Q therefore has 40 more hydrogen bonds holding its strands together, so more energy (a higher temperature) is needed to separate them: 91 °C against 83 °C.
(d) The prediction is wrong. Heating breaks the weak hydrogen bonds between the strands. The nucleotides within each strand are linked by strong covalent bonds in the sugar–phosphate backbone, which stay intact at these temperatures. So heating gives two separate single strands, each still 200 nucleotides long.
| Point | What earns it |
|---|---|
| 1 | Correct pair counts for both fragments |
| 1 | 476 and 516 hydrogen bonds |
| 1 | Higher separation temperature linked to more C–G pairs and three hydrogen bonds per C–G pair |
| 1 | Hydrogen bonds between strands break; covalent backbone bonds within strands do not |
Accept answers to (b) carried forward from an error in (a). More advanced courses add that stacking between neighbouring base pairs also stabilises C–G-rich DNA; the hydrogen-bond explanation is enough here.
Question 7 (constructed response · stretch)
A new DNA strand is being built along a template strand that reads 3′-TACGGATC-5′.
(a) Write the sequence of the completed new strand, labelling both ends. (b) State in which direction the new strand grows, and explain why. (c) Calculate how many covalent links join the nucleotides of the completed new strand, and how many hydrogen bonds join it to the template. (d) A fictional modified nucleotide has –H instead of –OH on its 3′ carbon. It can still pair with its partner base. Predict what happens to synthesis of the new strand if this nucleotide is added, and explain.
Worked solution and suggested Marlbridge rubric
(a) Pair each base and keep the strands antiparallel: under 3′-TACGGATC-5′ write 5′-ATGCCTAG-3′.
(b) It grows 5′ → 3′. Each new nucleotide is joined by a covalent bond between its phosphate and the free 3′ –OH at the end of the growing strand. Only the 3′ end has that free –OH, so nucleotides are added only there.
(c) Covalent links: 8 nucleotides − 1 = 7. Hydrogen bonds: the template has 4 C–G pairs (C, G, G, C) and 4 A–T pairs, so (4 × 3) + (4 × 2) = 12 + 8 = 20.
(d) Synthesis stops once the modified nucleotide is added. It can join the strand, because its own phosphate can bond to the previous 3′ –OH. But it has no 3′ –OH of its own, so the next nucleotide has nothing to bond to. The new strand ends at that point and is shorter than the template.
| Point | What earns it |
|---|---|
| 1 | 5′-ATGCCTAG-3′ with both ends labelled correctly |
| 1 | Growth 5′ → 3′, explained by addition to the free 3′ –OH |
| 1 | 7 covalent links and 20 hydrogen bonds |
| 1 | Synthesis stops, explained by the missing 3′ –OH |
Common error in (a): the right bases with the wrong labels, 3′-ATGCCTAG-5′. That describes a different molecule, so it does not earn the point.
How did you do?
- Q1 or Q4 wrong: re-read “The nucleotide: three parts” and “DNA and RNA compared” in the study guide, and look again at Figure 1.
- Q2 or Q7(a) wrong: rework Worked example 1; reverse the complement so it reads 5′ to 3′.
- Q3 or Q5 wrong: rework Worked example 2; the pairing rules apply only to paired bases.
- Q6 or Q7(c) wrong: rework Worked example 3 and re-read “Reading Figure 2”.
- Q7(b) or (d) wrong: re-read “Building a strand: the 5′ and 3′ ends”.
Then tick off the topic checklist.