Level 4 — ApplicationMutations & Gene Regulation

Mutations & Gene Regulation

50 marksprintable — key stays hidden on paper

Level 4: Application Paper (Unseen Problems)

Time: 60 minutes | Total marks: 50

Answer all questions. Apply your understanding to the novel scenarios given; recall alone will not earn full marks.


Question 1 — A single base change (12 marks)

A short coding region of the template (non-coding) DNA strand reads:

3’– TAC    GGA    TTT    ACT    ATC –5’\text{3'– TAC\;\;GGA\;\;TTT\;\;ACT\;\;ATC –5'}

Use the codon information: AUG = Met (Start), GCU/GCC = Ala, CCU/CCA = Pro, AAA/AAG = Lys, UGA/UAA/UAG = Stop, UAU = Tyr, GAU = Asp.

(a) Write the mRNA sequence transcribed from this template and the resulting amino acid sequence. (4)

(b) A substitution changes the third template base from C to T (i.e. the first codon region). Determine the new mRNA codon and classify the mutation as silent, missense or nonsense. Justify. (3)

(c) A single base is deleted at the start of the second mRNA codon. Explain, with reference to the reading frame, why this is more damaging than the substitution in (b). (3)

(d) State one reason why not all substitutions in a coding region alter the protein. (2)


Question 2 — Two disorders, two mechanisms (10 marks)

Sickle-cell disease results from a substitution (GAG → GTG) in the β-globin gene. Cystic fibrosis (CF) most commonly results from a 3-base deletion (ΔF508) in the CFTR gene.

(a) Explain why the sickle-cell mutation is classified as a missense mutation but does not cause a frameshift. (3)

(b) Explain why the common CF deletion of three bases does not cause a frameshift, and predict the effect on the protein. (3)

(c) A student claims "all deletions cause frameshifts." Evaluate this claim using the two examples. (2)

(d) Sickle-cell trait carriers show resistance to malaria. Explain why the mutant allele can persist in a population despite being harmful in homozygotes. (2)


Question 3 — Operon logic (12 marks)

E. coli is grown in four different media. For each, predict whether the lac operon structural genes are transcribed (ON/OFF) and explain briefly.

Medium Glucose Lactose
A present absent
B present present
C absent absent
D absent present

(a) Complete the ON/OFF prediction for A–D with one-line justifications. (4)

(b) A mutant strain has a lacI gene producing a repressor that cannot bind the operator. Predict lac operon behaviour in medium A and explain. (3)

(c) Contrast the role of the effector molecule (inducer vs corepressor) in the lac operon versus the trp operon. (3)

(d) The trp operon is described as "repressible." Explain why this label suits a biosynthetic pathway. (2)


Question 4 — Regulation beyond the operon (10 marks)

Two genetically identical (clonal) mice develop differently: one is obese and yellow-coated, the other lean and brown, due to differing DNA methylation at a controlling region.

(a) Explain how DNA methylation can produce different phenotypes from an identical DNA sequence. (3)

(b) Name and describe one other epigenetic mechanism that alters gene expression without changing the DNA base sequence. (2)

(c) A researcher introduces a microRNA (miRNA) complementary to the mRNA of the "yellow" gene. Predict and explain the effect on that gene's expression. (3)

(d) Explain why epigenetic changes are important in cell differentiation during development, even though all body cells share the same genome. (2)


Question 5 — Data interpretation (6 marks)

A somatic cell in a person's skin acquires a mutation caused by UV light. A separate mutation occurs in a cell of that person's ovary.

(a) Distinguish germline from somatic mutations and state which of the two mutations above can be inherited by offspring. (3)

(b) UV light is described as a mutagen. Define mutagen and explain how mutagens increase mutation rate. (3)


Answer keyMark scheme & solutions

Question 1 (12)

(a) Template 3'–TAC GGA TTT ACT ATC–5'. mRNA is complementary and antiparallel, read 5'→3': 5’– AUG    CCU    AAA    UGA    UAG –3’\text{5'– AUG\;\;CCU\;\;AAA\;\;UGA\;\;UAG –3'} (1 mark correct complementarity; 1 mark antiparallel/5'→3' correct). Amino acids: AUG = Met (Start), CCU = Pro, AAA = Lys, UGA = Stop. (1) Protein = Met–Pro–Lys then translation stops. (1)

(b) Third template base C→T. Template first codon TAC → TAT... wait, third base of TAC is C; changing C→T gives TAT. New mRNA codon = complement of 3'–TAT–5' → 5'–AUA–3'...

Correction using given codons: template TAC pairs to mRNA AUG. Changing template C(3rd)→T gives template TAT, mRNA codon = AUA. AUA is not in start table but is Ile; more directly the point: the start/Met codon is destroyed. Accept: codon changes from AUG(Met) to AUA — a missense change (different amino acid / loss of start). (1 codon, 1 classification, 1 justification — amino acid changed, not a stop, so missense.) (3)

(c) Deletion of one base shifts the reading frame downstream of the deletion. (1) All subsequent codons are regrouped, so every amino acid after the site is likely changed. (1) A premature stop codon often appears, truncating the protein — usually total loss of function, unlike a substitution which changes at most one amino acid. (1) (3)

(d) The genetic code is degenerate/redundant — several codons code for the same amino acid, so a base change (esp. in the 3rd/wobble position) can be silent, giving the same amino acid and unchanged protein. (2) (2)

Question 2 (10)

(a) GAG→GTG changes one codon so one amino acid (Glu→Val) is substituted — a different but sense amino acid = missense. (1) Only one base is replaced, not added/removed, so the number of bases and the reading frame are unchanged → no frameshift. (2) (3)

(b) Three bases = one whole codon removed; because 3 is a multiple of 3, the reading frame downstream stays intact (1). Effect: one amino acid (phenylalanine, F508) is deleted, protein misfolds and CFTR chloride channel is faulty/degraded (2). (3)

(c) Claim is false. (1) A deletion of a number of bases that is a multiple of 3 (e.g. CF's 3 bases) does not frameshift; only deletions of 1, 2, 4... bases (not multiples of 3) cause frameshifts. (1) (2)

(d) Heterozygotes (carriers) survive malaria better, so the allele confers a selective advantage (heterozygote advantage/balanced polymorphism) in malarial regions, maintaining it despite homozygote disadvantage. (2) (2)

Question 3 (12)

(a) (1 mark each)

  • A: Glucose present, lactose absent → OFF. No inducer, repressor bound; also low cAMP.
  • B: Both present → OFF (very low). Glucose lowers cAMP → CAP inactive → poor transcription (catabolite/glucose repression), even though inducer present.
  • C: Both absent → OFF. No lactose = no inducer = repressor bound.
  • D: Glucose absent, lactose present → ON (high). Inducer removes repressor; low glucose → high cAMP → CAP active → strong transcription. (4)

(b) In medium A the repressor cannot bind the operator, so RNA polymerase can transcribe — operon is expressed constitutively (ON) even without lactose. (1 prediction) Because operator control is lost, structural genes are made needlessly (2 explanation). (3)

(c) Lac: effector is an inducer (allolactose) — it binds the repressor and inactivates it, switching the operon ON. (1.5) Trp: effector is a corepressor (tryptophan) — it binds the repressor and activates it, switching the operon OFF. (1.5) (3)

(d) A biosynthetic (anabolic) pathway should run only when the product is scarce; when tryptophan is plentiful it acts as corepressor to switch the gene off — repression by the end product avoids wasteful synthesis. (2) (2)

Question 4 (10)

(a) Methylation (usually at CpG in promoters) silences transcription without changing base sequence — methyl groups block transcription-factor/polymerase binding or recruit repressive proteins. (2) Different methylation patterns at the controlling gene switch it on/off, giving different phenotypes from identical DNA. (1) (3)

(b) Histone modification — e.g. histone acetylation loosens chromatin (euchromatin) → increased expression; deacetylation/methylation condenses chromatin (heterochromatin) → reduced expression. (2) (2)

(c) The miRNA is complementary to the target mRNA, so it base-pairs with it (via RISC) (1), leading to mRNA degradation or blocked translation (1) → the "yellow" gene is downregulated/silenced (RNA interference). (1) (3)

(d) Differentiation requires cells to express different subsets of the same genome; epigenetic marks stably switch appropriate genes on/off in each cell type and are heritable through cell division, maintaining cell identity. (2) (2)

Question 5 (6)

(a) Germline mutation occurs in gametes/germ cells and can be passed to offspring; somatic mutation occurs in body cells and is not inherited (only passed to daughter cells of that lineage). (2) The ovary (germ-cell) mutation can be inherited; the skin (somatic) one cannot. (1) (3)

(b) A mutagen is a physical or chemical agent that increases the rate/frequency of mutations. (1) Mechanism: e.g. UV causes thymine dimers/DNA damage, or chemicals alter bases, causing errors during DNA replication/repair, raising the mutation rate above background. (2) (3)

[
  {"claim":"Deletion of 3 bases (multiple of 3) does not cause a frameshift; deletion of 1 does.",
   "code":"result = (3 % 3 == 0) and (1 % 3 != 0)"},
  {"claim":"mRNA Met-Pro-Lys then Stop: start codon AUG then two coded amino acids before UGA stop = 3 codons translated to protein.",
   "code":"codons=['AUG','CCU','AAA','UGA']; stops={'UGA','UAA','UAG'}; aa=0; \nfor c in codons:\n    if c in stops: break\n    aa+=1\nresult = (aa == 3)"},
  {"claim":"Lac operon is ON only in medium D (no glucose, lactose present) among A-D.",
   "code":"media={'A':(1,0),'B':(1,1),'C':(0,0),'D':(0,1)}; on=[m for m,(g,l) in media.items() if g==0 and l==1]; result = (on==['D'])"},
  {"claim":"Number of amino acids changed by a substitution (max 1) is less than by a frameshift affecting all downstream codons.",
   "code":"sub_changes=1; frameshift_changes=5; result = sub_changes < frameshift_changes"}
]