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. All genes, cells and data sets are fictional. Lengths are given in nucleotides (nt). Write every nucleic acid sequence with its 5′ and 3′ ends labelled.
Question 1 (multiple choice · foundation)
A fictional toxin destroys the rRNA in a eukaryotic cell but leaves its DNA, mRNA and tRNA undamaged. Which process is most directly stopped?
- (A) Transcription of genes into pre-mRNA in the nucleus
- (B) Addition of the poly-A tail to new mRNA
- (C) Building of polypeptides, because working ribosomes can no longer form
- (D) Attachment of amino acids to their tRNAs
Answer and explanation
Answer: (C). rRNA is a main building block of ribosomes. Without it, mRNA cannot be translated.
- (A) Transcription uses RNA polymerase and DNA, not ribosomes.
- (B) The poly-A tail is added by nuclear enzymes during processing, not by ribosomes.
- (D) Each tRNA binds its amino acid before it reaches the ribosome; the tRNA itself is undamaged.
Question 2 (multiple choice · core)
The template strand of a short region of DNA is 3′-ATGCCTAGGTAC-5′. Which is the RNA transcribed from it?
- (A) 5′-UACGGAUCCAUG-3′
- (B) 5′-AUGCCUAGGUAC-3′
- (C) 5′-GUACCUAGGCAU-3′
- (D) 5′-TACGGATCCATG-3′
Answer and explanation
Answer: (A). The template is already written 3′ → 5′, so pair each base in turn (A→U, T→A, G→C, C→G). The RNA is antiparallel, so it starts with its 5′ end: 5′-UACGGAUCCAUG-3′.
- (B) copies the template and swaps T for U. That is the template’s own sequence, not its complement.
- (C) is the correct RNA written backwards, so its 5′ and 3′ labels are on the wrong ends.
- (D) has the right pairing order but uses T. RNA contains uracil, not thymine; this is the coding strand’s sequence.
Question 3 (multiple choice · core)
Researchers inject equal amounts of two versions of a fictional mature mRNA into egg cells. Version 1 has a normal 5′ cap; version 2 has no cap. Both have the same poly-A tail. After 2 hours, both mRNAs are still present in equal amounts, but eggs given version 2 have made less than a tenth as much of the encoded protein. Which statement best explains these results?
- (A) Without the cap, ribosomes do not recognise the mRNA well, so it is translated far less.
- (B) Without the cap, the mRNA is broken down much faster, so less is available to translate.
- (C) The cap contains the coding sequence for the first part of the protein.
- (D) Without the cap, the introns in the mRNA cannot be removed.
Answer and explanation
Answer: (A). The mRNA amounts are equal, so the uncapped mRNA is not being lost; it is being used less. The cap helps the ribosome recognise the mRNA, so without it much less protein is made.
- (B) is contradicted by the data: both mRNAs are present in equal amounts after 2 hours. (In this course, stability is linked mainly to the poly-A tail.)
- (C) The cap is a modified nucleotide added to the end of the mRNA. It is not part of the coding sequence.
- (D) Mature mRNAs contain no introns to remove.
Question 4 (visual representation · core)
The diagram shows transcription of part of a gene. The RNA is drawn as a dashed line; its two ends are labelled X and Y.
(a) Identify the template strand, and give one piece of evidence from the diagram. (b) State whether end X is the 5′ or the 3′ end of the RNA, and explain. (c) State the direction (left or right) in which the RNA polymerase is moving. (d) Inside the bubble, the template strand reads 5′-CGCATT-3′ from left to right. Write the RNA made from this section and the matching section of the coding strand, each written 5′ → 3′.
Worked solution and suggested Marlbridge rubric
(a) Strand 2 is the template. The RNA is lying alongside it and base-paired to it inside the bubble.
(b) X is the 5′ end. RNA is antiparallel to its template. Strand 2 runs 5′ (left) to 3′ (right), so the RNA runs 3′ (Y, left) to 5′ (X, right). Also, X has already left the enzyme, and the 5′ end is made first.
(c) Left. The template is read 3′ → 5′, which on strand 2 is from right to left. The RNA grows at Y, on the left.
(d) Rewrite the template 3′ → 5′: 3′-TTACGC-5′. Pair each base: RNA 5′-AAUGCG-3′. Coding strand 5′-AATGCG-3′ (same as the RNA with T for U).
| Point | What earns it |
|---|---|
| 1 | Strand 2, with evidence (RNA paired to it) |
| 1 | X = 5′, justified by antiparallel pairing or by the 5′ end being made first |
| 1 | Moving left, linked to reading the template 3′ → 5′ |
| 1 | RNA 5′-AAUGCG-3′ and coding strand 5′-AATGCG-3′, both with correct end labels |
Do not award (d) for 5′-GCGUAA-3′ (ends reversed).
Question 5 (data analysis · core)
A researcher makes two versions of a fictional reporter mRNA. Both have the same 5′ cap and the same coding sequence. Version T has a poly-A tail; version N has none. She adds equal amounts of each to separate batches of cultured cells, blocks all new transcription, and measures how much of each mRNA remains.
| Time / h | 0 | 1 | 2 | 4 | 6 |
|---|---|---|---|---|---|
| Version T remaining / % | 100 | 92 | 84 | 71 | 59 |
| Version N remaining / % | 100 | 63 | 40 | 16 | 6 |
(a) Calculate the percentage of each mRNA lost during the first 4 hours. (b) Estimate the time taken for half of version N to be lost. Show how you used the data. (c) Make a claim about the role of the poly-A tail, and support it with evidence from the table. (d) Explain why the researcher blocked new transcription and why both versions had the same cap. (e) Predict which version would lead to more of the reporter protein being made over 6 hours, and explain.
Worked solution and suggested Marlbridge rubric
(a) Version T: 100 − 71 = 29% lost. Version N: 100 − 16 = 84% lost.
(b) 50% lies between 1 h (63%) and 2 h (40%). The fall over that hour is 63 − 40 = 23 percentage points, and 63 − 50 = 13 of them are needed. So the time is about 1 + 13 ÷ 23 ≈ 1.6 h (anything from 1.5 to 1.6 h is reasonable).
(c) Claim: the poly-A tail makes the mRNA more stable. Evidence: after 4 h, 71% of version T remains against 16% of version N, about 4.4 times as much; version T is still above half (59%) after 6 h.
(d) Blocking transcription stops new mRNA being made, so any fall measures breakdown only. With the same cap and coding sequence, the poly-A tail is the only difference between the versions.
(e) Version T. An mRNA can be translated many times while it lasts; version T lasts longer, so it is translated more.
| Point | What earns it |
|---|---|
| 1 | 29% and 84% |
| 1 | About 1.5–1.6 h, with the interpolation between 1 h and 2 h shown |
| 1 | Claim (tail increases stability) supported by at least one pair of compared values |
| 1 | Blocking transcription isolates breakdown and same cap isolates the tail as the variable |
| 1 | Version T, linked to the mRNA lasting longer and being translated more times |
Question 6 (constructed response · stretch)
A fictional gene has six exons. Exons 1 and 6 are always kept. Exons 2 and 3 can each be kept or skipped, independently. Exons 4 and 5 are mutually exclusive: every mature mRNA contains exactly one of them. Exon lengths:
| Exon | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Length / nt | 150 | 60 | 90 | 108 | 114 | 240 |
(a) Calculate how many different mature mRNAs this gene can produce. (b) Calculate the lengths of the shortest and the longest possible mature mRNA (ignore the cap and tail). (c) Muscle cells always use exon 4 and nerve cells always use exon 5. Both cell types contain identical DNA. Explain how they end up with different versions of the protein. (d) A student says: “A cell can make more kinds of protein than it has genes.” Use this gene to support the statement.
Model answer and suggested Marlbridge rubric
(a) Exon 2: 2 choices (in or out). Exon 3: 2 choices. Exons 4/5: 2 choices (one or the other). Total = 2 × 2 × 2 = 8 mature mRNAs.
(b) Shortest: skip exons 2 and 3, choose exon 4 (shorter than 5): 150 + 108 + 240 = 498 nt. Longest: keep exons 2 and 3, choose exon 5: 150 + 60 + 90 + 114 + 240 = 654 nt.
(c) Both cell types transcribe the same pre-mRNA. The difference arises in splicing: muscle cells keep exon 4 and remove exon 5; nerve cells do the reverse. The mature mRNAs differ in base sequence in that region, so the proteins differ in amino acid sequence there.
(d) This one gene can give up to 8 mature mRNAs, and so up to 8 proteins. If many genes are spliced like this, the cell’s proteins can far outnumber its genes.
| Point | What earns it |
|---|---|
| 1 | 8, with the 2 × 2 × 2 reasoning |
| 1 | 498 nt and 654 nt |
| 1 | Same DNA and same pre-mRNA; difference arises in splicing (alternative splicing) |
| 1 | Different exon gives a different mRNA sequence and so a different amino acid sequence |
| 1 | Uses “one gene, several mRNAs/proteins” to support the claim |
Common error in (a): 2⁴ = 16 treats exons 4 and 5 as independent.
Question 7 (constructed response · stretch)
(a) Copy and complete the table to compare DNA replication (Topic 6.2) with transcription.
| Feature | DNA replication | Transcription |
|---|---|---|
| Strands used as template | ||
| Main enzyme that builds the new strand | ||
| Primer needed? | ||
| Base paired with adenine on the template | ||
| How much of the DNA is copied |
(b) A fictional drug, compound R, blocks RNA polymerase in human cells. In treated cells, mRNA X has a half-life of 2 hours and mRNA Y a half-life of 6 hours. Calculate the percentage of each mRNA left 6 hours after the drug is added. (c) Predict how the amounts of protein X and protein Y in the cells will change over the next few hours, and justify your prediction.
Model answer and suggested Marlbridge rubric
(a)
| Feature | DNA replication | Transcription |
|---|---|---|
| Strands used as template | both strands | one strand (the template strand) |
| Main enzyme that builds the new strand | DNA polymerase | RNA polymerase |
| Primer needed? | yes, an RNA primer | no |
| Base paired with adenine on the template | thymine (T) | uracil (U) |
| How much of the DNA is copied | the whole genome, once per cell cycle | one gene (or a few) at a time, as needed |
Both build 5′ → 3′ and read the template 3′ → 5′.
(b) For X, 6 h is 3 half-lives: 100% → 50% → 25% → 12.5%. For Y, 6 h is exactly 1 half-life, so 50% is left: four times as much as X.
(c) No new mRNA is made, and existing mRNA is broken down. mRNA X disappears quickly, so little protein X is made and its amount falls as existing molecules are broken down. Protein Y falls more slowly, because mRNA Y is still translated for hours. The drug does not destroy proteins already made, so protein levels lag behind mRNA levels.
| Point | What earns it |
|---|---|
| 1 | At least four table rows correct |
| 1 | 12.5% for X and 50% for Y |
| 1 | No new mRNA is made and existing mRNA is broken down |
| 1 | Protein X falls sooner or faster than protein Y, linked to the mRNA half-lives |
| 1 | Recognises that proteins already made persist for a while, so protein lags behind mRNA |
How did you do?
- Q1 wrong: re-read “Three kinds of RNA, three jobs” in the study guide.
- Q2 or Q4 wrong: rework Worked example 1 and look again at Figure 1; check the RNA is antiparallel to the template and contains U.
- Q3 or Q5 incomplete: re-read “RNA processing in eukaryotes”: cap for recognition, tail for stability.
- Q6 wrong: rework Worked example 2 and Figure 2 on alternative splicing.
- Q7 table incomplete: revisit Topic 6.2, DNA Replication and the section “Transcription: copying one strand”.
Then tick off the topic checklist.