Mutations & Gene Regulation
Level 3 Paper: Production (Derivation & Explain-Out-Loud)
Time limit: 45 minutes Total marks: 60
Instructions: Answer all questions. Where a process is asked for, reason step-by-step from first principles. Use the genetic code convention that stop codons are UAA, UAG, UGA.
Question 1 — Reconstruct mutation effects from a sequence (12 marks)
A gene's template (non-coding) strand reads, in the region of interest:
3'- TAC GGA CCT ATT ... -5'
(a) Write the corresponding mRNA sequence (5'→3') and translate the first four codons using: AUG=Met, CCU=Pro, GGA=Gly, UAA=Stop. (4)
(b) A substitution changes the template DNA so the second codon of the mRNA becomes CCC. Given CCC also = Pro, classify this mutation and explain why it has no effect on the protein. (3)
(c) A single nucleotide is now inserted at the start of the third mRNA codon. Explain, from scratch, what type of mutation this is and why its downstream consequences are typically more severe than a substitution. (5)
Question 2 — Build the sickle-cell explanation (10 marks)
Starting from the DNA level and ending at the phenotype, construct a complete causal chain explaining how sickle-cell anaemia arises. Your answer must explicitly link: the type of point mutation, the specific amino acid change (glutamic acid → valine), the molecular consequence for haemoglobin, and the cellular/clinical phenotype. (10)
Question 3 — Derive lac operon behaviour (12 marks)
(a) From memory, describe the state of the lac operon (repressor, operator, RNA polymerase, structural genes) in each of the following, explaining the logic in each case: (8) (i) lactose absent (ii) lactose present, glucose absent
(b) Explain why the lac operon is described as an inducible system, and contrast this one-sentence definition with a repressible system. (4)
Question 4 — Contrast two operons and predict a mutant (10 marks)
(a) Compare the trp operon and the lac operon by stating, for each, the normal ligand and whether that ligand switches transcription on or off. (4)
(b) Predict and justify the phenotype of a trp operon mutant in which the repressor protein cannot bind tryptophan (the corepressor). Will tryptophan-synthesis genes be transcribed when tryptophan is abundant? Explain. (6)
Question 5 — Explain epigenetic regulation out loud (10 marks)
(a) Explain, mechanistically, how DNA methylation and histone acetylation each affect transcription, stating the direction (increase/decrease) of gene expression for each. (6)
(b) Two genetically identical twins differ in a trait not caused by any DNA sequence change. Explain how epigenetics accounts for this, and state one reason such a change can be described as heritable yet reversible. (4)
Question 6 — Reconstruct the RNAi pathway (6 marks)
Describe, step by step, how a microRNA (miRNA) reduces the expression of a target gene. Identify the protein complex involved and the two possible fates of the target mRNA. (6)
Answer keyMark scheme & solutions
Question 1 (12 marks)
(a) Template 3'-TAC GGA CCT ATT-5'. mRNA is complementary and antiparallel, read 5'→3':
5'- AUG CCU GGA UAA -3' (2 marks: correct base pairing + antiparallel orientation)
Translation: Met – Pro – Gly – Stop (2 marks)
Why: mRNA is synthesised complementary to the template read 3'→5', so mRNA reads 5'→3' matching the coding strand; UAA terminates translation. (0.5 rounded into above)
(b) Codon changes CCU→CCC, both code Pro. This is a silent (synonymous) substitution (1). No effect because the genetic code is degenerate — multiple codons specify the same amino acid, so the encoded protein sequence is unchanged (2).
(c) Inserting one nucleotide is a frameshift mutation (1). Because the ribosome reads mRNA in non-overlapping triplets, adding one base shifts the reading frame for every codon downstream (2). This alters all subsequent amino acids and frequently introduces a premature stop codon, usually destroying protein function (1). By contrast, a substitution changes at most one codon, so its effect is local (1).
Question 2 (10 marks)
Marking: 2 marks per correctly-ordered link (5 links).
- DNA/point mutation: A single base substitution in the β-globin gene: template change causing mRNA codon GAG→GUG (2).
- Amino acid change: codon 6 now codes valine instead of glutamic acid (missense mutation) (2).
- Protein consequence: valine is hydrophobic; it creates a sticky patch so deoxygenated haemoglobin (HbS) molecules polymerise/aggregate into fibres (2).
- Cellular phenotype: red blood cells become rigid and sickle-shaped (2).
- Clinical phenotype: sickled cells block capillaries (vaso-occlusion) and are destroyed early → pain crises and anaemia (2).
Question 3 (12 marks)
(a)(i) Lactose absent (4): Repressor protein (active) binds the operator (1); RNA polymerase is blocked from transcribing (1); structural genes (lacZ, Y, A) are off (1). Logic: no need to make lactose-metabolising enzymes when no lactose is present — saves energy (1).
(a)(ii) Lactose present, glucose absent (4): Lactose (allolactose) is the inducer; it binds the repressor, changing its shape so it cannot bind the operator (1); operator is free, RNA polymerase transcribes the structural genes (1); enzymes for lactose metabolism are made (1). (With glucose absent, cAMP-CAP further boosts transcription) (1).
(b) (4): Inducible = normally off, switched on by the presence of a substrate (inducer) (2). Contrast: a repressible system is normally on, switched off when its product (corepressor) accumulates (2).
Question 4 (10 marks)
(a) (4, 1 each):
- trp operon: ligand = tryptophan; switches transcription OFF (repressible).
- lac operon: ligand = lactose/allolactose; switches transcription ON (inducible).
(b) (6): Normally, when tryptophan is abundant it acts as a corepressor: it binds the trp repressor, activating it so it binds the operator and switches the genes off (2). In this mutant the repressor cannot bind tryptophan, so the repressor remains inactive and cannot bind the operator (2). Therefore the trp-synthesis genes are transcribed even when tryptophan is abundant — the operon is constitutively (wastefully) expressed (2).
Question 5 (10 marks)
(a) (6):
- DNA methylation: addition of methyl groups (typically to cytosine in CpG islands of promoters) compacts chromatin / blocks transcription factors, so gene expression decreases (3).
- Histone acetylation: acetyl groups neutralise the positive charge of histone tails, loosening DNA–histone binding → open chromatin (euchromatin), so gene expression increases (3).
(b) (4): The twins have identical DNA but different epigenetic marks (methylation/histone patterns) established by environment/chance, altering which genes are expressed → different phenotype (2). It is heritable because the marks are copied to daughter cells during mitosis (and can persist), yet reversible because they do not change the underlying DNA sequence and can be enzymatically added/removed (2).
Question 6 (6 marks)
Step by step (6):
- miRNA is transcribed and processed into a short single-stranded RNA (1).
- It is loaded into the RISC (RNA-induced silencing complex) (2).
- The miRNA guides RISC to a complementary sequence on the target mRNA (1).
- Depending on degree of complementarity, the mRNA is either cleaved/degraded or its translation is blocked (2). Net effect: reduced protein output = gene silencing.
[
{"claim":"Template 3'-TAC-5' pairs to give mRNA codon 5'-AUG-3' (start codon Met)", "code":"comp={'A':'U','T':'A','G':'C','C':'G'}; templ='TAC'; mrna=''.join(comp[b] for b in templ); result = (mrna=='AUG')"},
{"claim":"Template 3'-ATT-5' gives mRNA codon UAA which is a stop codon", "code":"comp={'A':'U','T':'A','G':'C','C':'G'}; codon=''.join(comp[b] for b in 'ATT'); result = (codon=='UAA' and codon in ['UAA','UAG','UGA'])"},
{"claim":"CCU and CCC both encode proline (degenerate code => silent mutation)", "code":"code={'CCU':'Pro','CCC':'Pro','CCA':'Pro','CCG':'Pro'}; result = (code['CCU']==code['CCC'])"},
{"claim":"Single insertion shifts reading frame: frame index changes by 1 (frameshift)", "code":"orig=0; ins=(orig+1)%3; result = (ins!=orig)"}
]