Before you start
This review joins the eight topics of Unit 6. Do the unit diagnostic first and close its gaps. Then try the questions without looking at the answers.
Big ideas of the unit
- Base pairing is the master rule. A purine pairs with a pyrimidine: A with T (U in RNA), G with C, in all living things. Prokaryotes usually have one circular chromosome and may carry plasmids; eukaryotes have linear chromosomes wound around histones (Topic 6.1).
- Copying keeps one old strand. Replication is semiconservative. Helicase opens the helix, topoisomerase relieves twisting, DNA polymerase builds 5′ → 3′ from an RNA primer, and ligase joins lagging-strand fragments (Topic 6.2).
- Every polymerase works 5′ → 3′. RNA polymerase reads one template strand 3′ → 5′ and builds mRNA 5′ → 3′. In eukaryotes, a cap, a poly-A tail and splicing turn pre-mRNA into mature mRNA (Topic 6.3).
- The code is read in triplets. Ribosomes start at AUG and stop at a stop codon; tRNA anticodons match codons. The code is nearly universal, which points to common ancestry. Retroviruses run information backwards, RNA → DNA (Topic 6.4).
- Expression is controlled, not fixed. Operons switch groups of bacterial genes on or off; epigenetic marks and shared transcription factors control eukaryotic genes (Topic 6.5).
- Same genome, different cells. Activators, repressors and small RNAs decide which products a cell makes, so cells specialise (Topic 6.6).
- Changes feed evolution. Mutations and chromosome errors change the type or amount of a product. Horizontal transfer and viral recombination add variation that selection can act on (Topic 6.7).
- Tools copy the cell. PCR uses primers and DNA polymerase; gels separate fragments by size; transformation works because the code is shared (Topic 6.8).
Key relationships and methods
| Idea | Rule or method | Where it appears |
|---|---|---|
| Pairing | A–T (A–U in RNA), G–C; purines = pyrimidines in any double strand | DNA, double-stranded RNA, primers, anticodons |
| Direction | new strand built 5′ → 3′; template read 3′ → 5′; write answers 5′ → 3′ | replication, transcription, PCR |
| Reading frame | start at AUG; triplets; length change not a multiple of 3 → frameshift | translation, mutations, retained introns |
| Codons to amino acids | (coding nucleotides ÷ 3) − 1 stop codon | counting polypeptide length |
| Control step | no pre-mRNA → transcription; mRNA normal but little protein → translation; different mRNA forms → splicing | eukaryotic regulation data |
| Phenotype | change in type or amount of product → change in phenotype; silent changes do not | mutations, enhancers |
| Copies | errors per copy = genome length × error rate | replication, reverse transcription |
| Gel | shorter fragments travel further; cut site lost → fewer, longer bands | fingerprints, allele tests |
These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not official scoring guidelines. All genes, organisms and data sets are fictional.
Question 1 (multiple choice · mixed)
A mutation at a splice site of a fictional gene stops one intron from being removed, so it stays in the mature mRNA. The intron is 100 nucleotides long and lies inside the coding sequence. What is the most likely effect on the protein?
- (A) Extra amino acids are added, and the reading frame after the intron is shifted, so later amino acids change and an early stop codon is likely.
- (B) There is no effect, because introns never code for amino acids.
- (C) Exactly 33 extra amino acids are added, and the rest of the protein is normal.
- (D) There is no effect, because the ribosome skips the intron.
Answer and explanation
Answer: (A). The ribosome reads the mature mRNA in triplets. 100 = 3 × 33 + 1, so the frame shifts by one nucleotide. Every later codon is misread, and a stop codon usually appears soon.
- (B) holds only when introns are removed.
- (C) divides 100 by 3 but ignores the leftover nucleotide.
- (D) gives the ribosome a splicing job. Splicing happens in the nucleus, before translation.
Question 2 (multiple choice · mixed)
During replication of one DNA molecule, DNA polymerase places a T opposite a G in the template, and the error is not repaired. Both daughter molecules then replicate once more, correctly. Of the four DNA molecules now present, how many have an A–T pair where the original molecule had a G–C pair?
- (A) 0
- (B) 1
- (C) 2
- (D) 4
Answer and explanation
Answer: (B). Replication is semiconservative. After round 1, one daughter has a mismatched G–T pair; the other is normal. In round 2, the G strand gets a C and the T strand gets an A. So one of four molecules has the new A–T pair, now fixed in both strands.
- (A) assumes the mismatch is lost. Unrepaired, the T acts as a template.
- (C) assumes both daughters of the mismatched molecule are mutant. Only the one built on the T strand is.
- (D) forgets that the other original strand was copied correctly.
Question 3 (multiple choice · mixed)
A test for a fictional virus with a single-stranded RNA genome uses PCR. Before PCR, each sample is treated with reverse transcriptase. Why?
- (A) Reverse transcriptase cuts the RNA into fragments small enough to move through the gel.
- (B) Reverse transcriptase makes the RNA double-stranded so that primers can bind.
- (C) The DNA polymerase used in PCR builds DNA on a DNA template, so the viral RNA must first be copied into DNA.
- (D) Reverse transcriptase translates the RNA into protein, which PCR then detects.
Answer and explanation
Answer: (C). Retroviruses use reverse transcriptase to copy RNA into DNA. The test borrows it: once a DNA copy exists, PCR can amplify it.
- (A) describes a restriction enzyme’s kind of job, not reverse transcriptase.
- (B) is tempting because primers need a partner strand, but the product is DNA, not double-stranded RNA.
- (D) confuses reverse transcription with translation. PCR copies nucleic acids, not proteins.
Question 4 (constructed response · mixed)
A student designs PCR primers to copy a 480 bp region of a fictional gene. The top strand of the region starts 5′-ACGTTGCAGGCA… and ends …TTCAGGATCCGA-3′. The forward primer is 5′-ACGTTGCAGGCA-3′.
(a) Write the reverse primer, 5′ → 3′. Show your working. (b) Explain why each primer must bind so that its 3′ end points into the target region. (c) In cells, replication starts from RNA primers. State why both PCR and the cell need a primer, and one difference between the two. (d) By mistake the reverse primer is made as 5′-TTCAGGATCCGA-3′. Predict the effect on the PCR, and explain.
Worked solution and suggested Marlbridge rubric
(a) The reverse primer pairs with the end of the top strand, so it is complementary and antiparallel to it. The partner of top-strand …TTCAGGATCCGA-3′, read below it, is 3′-AAGTCCTAGGCT-5′. Written 5′ → 3′: 5′-TCGGATCCTGAA-3′.
(b) DNA polymerase adds nucleotides only to the 3′ end of a primer and builds 5′ → 3′. If each 3′ end points inward, the two new strands each cover the whole target and contain the other primer’s site. Every copy can then act as a template in the next cycle.
(c) DNA polymerase cannot start a strand; it can only extend one. The cell makes short RNA primers with an enzyme as replication goes on. PCR uses short DNA primers made in advance and chosen to flank one region, so only that region is copied. (Heat separating the strands, in place of helicase, is also an acceptable difference.)
(d) No 480 bp product builds up. The faulty primer has the same sequence as the top strand, so it cannot pair with it. It can pair with the bottom strand, but there its 3′ end points away from the target. Only the forward primer copies the region, so each cycle adds just one new copy per original bottom strand instead of doubling the copies.
| Point | What earns it |
|---|---|
| 1 | Reverse primer 5′-TCGGATCCTGAA-3′ |
| 1 | Extension only from the 3′ end, 5′ → 3′, so copies include both primer sites |
| 1 | Primer needed because polymerase cannot start a new strand |
| 1 | One correct difference (RNA vs DNA primers, or heat vs helicase) |
| 1 | Faulty primer cannot pair usefully or points the wrong way, so no amplification |
Total: 5 points.
Question 5 (constructed response · mixed)
Retrovirus V9 is fictional. Its RNA genome is 9,600 nucleotides long. Its reverse transcriptase makes about 1 error per 32,000 nucleotides copied. In one infected person, about 10⁹ new cells are infected each day, and each infection needs one reverse transcription of the genome.
(a) Calculate the mean number of errors per reverse-transcribed genome, and per day. (b) Explain why an error made by reverse transcriptase ends up in every viral protein made from that DNA copy. (c) The person is treated with drug Q, which blocks reverse transcriptase. The table shows the percentage of viruses carrying mutation R.
| Week of treatment | 0 | 2 | 4 | 8 |
|---|---|---|---|---|
| Viruses with mutation R / % | 0.01 | 2 | 35 | 92 |
A student claims: “Drug Q caused mutation R.” Evaluate the claim using the data and your answer to (a). (d) A strain resistant only to drug Q and a strain resistant only to drug S infect the same cell. Explain how a strain resistant to both could arise.
Worked solution and suggested Marlbridge rubric
(a) Per genome: 9,600 ÷ 32,000 = 0.30 errors. Per day: 0.30 × 10⁹ = 3.0 × 10⁸ errors, about 31,000 at each of the 9,600 positions. Any single-base mutation probably arises many times a day.
(b) The DNA copy integrates into a host chromosome. Host RNA polymerase transcribes it and host ribosomes translate the mRNA, so every mRNA, and every protein, made from it carries the change.
(c) Not supported. R was already present at week 0 (0.01%), before the drug acted, and with 3.0 × 10⁸ errors a day it arises by chance anyway. The drug selects for R: viruses without R cannot copy their genome, while R viruses keep reproducing, so R rises from 0.01% to 92%.
(d) Related viruses infecting the same cell can recombine their genetic information. A new virus can then carry the Q-resistance region from one strain and the S-resistance region from the other.
| Point | What earns it |
|---|---|
| 1 | 0.30 errors per genome and 3.0 × 10⁸ per day |
| 1 | Integration, then host transcription and translation, so the change is in every product |
| 1 | R present before treatment (week 0) used as evidence |
| 1 | Mutation random; drug selects, with the rise to 92% explained |
| 1 | Recombination between related viruses in the same host cell |
Total: 5 points.
Question 6 (constructed response · mixed)
Gene Kel in a fictional fish is studied in three tissues. A microRNA, miR-k, is found only in fin cells. It is complementary to a sequence near the 3′ end of Kel mRNA. Promoter methylation is the percentage of sites carrying methyl groups.
| Tissue | Promoter methylation / % | Mature mRNA form | Kel mRNA (relative) | Kel protein (relative) |
|---|---|---|---|---|
| Gill | 4 | exons 1-2-3-5 | 100 | 100 |
| Fin | 6 | exons 1-2-4-5 | 95 | 18 |
| Eye | 88 | too little to identify | 1 | 0 |
(a) Calculate protein per unit of mRNA in gill and fin cells. (b) For eye and for fin, identify the main step at which Kel expression is reduced, with evidence. (c) Explain how gill and fin cells make different versions of Kel protein from identical DNA. (d) Predict the effect of deleting the miR-k target sequence from the Kel gene in fin cells. Explain.
Worked solution and suggested Marlbridge rubric
(a) Gill: 100 ÷ 100 = 1.00. Fin: 18 ÷ 95 = 0.19.
(b) Eye: transcription. Almost no mRNA is made, and the promoter is 88% methylated. This epigenetic mark keeps the gene from being transcribed. Fin: translation, after transcription. Fin cells have almost as much mRNA as gill cells (95 against 100) but much less protein per mRNA (0.19 against 1.00). miR-k base-pairs with Kel mRNA and blocks its translation or leads to its breakdown.
(c) Alternative splicing. Both cells transcribe the same pre-mRNA. Gill cells keep exon 3 and fin cells keep exon 4, so the mature mRNAs differ and are translated into proteins with different amino acid sequences.
(d) Kel protein in fin cells should rise towards the gill level, with mRNA still about 95, because miR-k can no longer pair with the mRNA. This would show that miR-k causes the low protein, not just that it is present.
| Point | What earns it |
|---|---|
| 1 | 1.00 and 0.19 |
| 1 | Eye: transcription blocked, with methylation and lack of mRNA as evidence |
| 1 | Fin: control after transcription, with mRNA similar but protein low, linked to miR-k pairing |
| 1 | Alternative splicing: different exons kept from the same pre-mRNA |
| 1 | Protein rises with mRNA unchanged, as a test of cause |
Total: 5 points.
Question 7 (constructed response · mixed)
The coding strand at the start of a fictional plant gene, Bx, reads 5′-ATG AAG CTT GGT CAG-3′. A restriction enzyme cuts DNA only at the sequence 5′-AAGCTT-3′. A mutant allele has 5′-ATG AAG CTC GGT CAG-3′.
Codons: AUG Met · AAG Lys · CUU Leu · CUC Leu · GGU Gly · CAG Gln
(a) Write the mRNA for the wild-type sequence and the amino acids it codes for. (b) Classify the mutation, and explain its effect on the protein. (c) A 900 bp PCR product containing this region is cut with the enzyme. In the wild type, the site lies 240 bp from one end. Predict the bands for a homozygous wild-type plant, a homozygous mutant and a heterozygote. (d) A student claims: “The mutation shows up on the gel, so it must change the plant’s phenotype.” Evaluate the claim.
Worked solution and suggested Marlbridge rubric
(a) The mRNA has the coding-strand sequence with U for T: 5′-AUG AAG CUU GGU CAG-3′, giving Met–Lys–Leu–Gly–Gln.
(b) A point mutation (substitution of T by C at the third base of codon 3). It is silent: CUU and CUC both code for leucine, so the amino acid sequence is unchanged.
(c) Wild type: the site is cut, giving 240 and 660 bp. Mutant: AAGCTC is not the site, so the product stays 900 bp. Heterozygote: 900, 660 and 240 bp.
(d) The claim is not supported. The gel detects a change in the DNA sequence. The protein is identical, so this change alone should not alter the phenotype. It is likely neutral, but useful to scientists as a marker for telling the two alleles apart.
| Point | What earns it |
|---|---|
| 1 | Correct mRNA and Met–Lys–Leu–Gly–Gln |
| 1 | Substitution identified as silent, with the redundancy of the code |
| 1 | Wild type 240 + 660; mutant 900 |
| 1 | Heterozygote shows all three bands |
| 1 | Claim rejected: DNA changed, protein unchanged |
Total: 5 points.
How did you do?
Add up your points: 3 for multiple choice and 20 for the constructed responses. Compare the total only with your own next attempt; it does not predict an exam score. Where did you lose points?
- Strands and direction (Questions 2, 4, 7(a)): rework the worked examples in Topic 6.2 and Topic 6.3.
- Reading frames and codons (Questions 1, 7(b)): go back to Topic 6.4 and Topic 6.7.
- Control steps (Question 6): reread Topic 6.5 and Topic 6.6.
- Selection or tools (Questions 3, 5, 7(c)): reread Topic 6.7 and Topic 6.8.
- Want a quick recheck? Retake the unit diagnostic.
Then tick off each topic checklist: