Transcription, Translation & Gene Expression

Question 1 [12 marks]

(a) Template read 3'→5': TAC GGA CTT AAA ACT mRNA is complementary and antiparallel; the mRNA reads 5'→3' pairing each base (A↔U, G↔C, T↔A, C↔G):

mRNA (5'→3'): AUG CCU GAA UUU UGA

• Correct base pairing (1)
• Correct antiparallel orientation / reading 5'→3' (1)
• U replaces T (1)

(b) Translate codons:

• AUG = Met (Start)
• CCU = Pro
• GAA = Glu
• UUU = Phe
• UGA = Stop

Peptide: Met–Pro–Glu–Phe (then stop) → 4 amino acids. (1 mark per correct codon assignment up to 3; +1 for recognising UGA as stop / correct final peptide) [4]

(c) 6th nucleotide of mRNA (AUG CCU...) is the U in CCU (position: A(1)U(2)G(3)C(4)C(5)U(6)). Changing this U→A gives codon CCA? No — to truncate after the first amino acid we need the second codon to become a stop.

Better analysis: the 6th nucleotide belongs to codon 2 (CCU, positions 4–6). Changing CCU → a stop codon: CCU cannot become a stop by a single change at position 6 alone (would give CCA/CCC/CCG = still Pro).

Accept any answer that correctly identifies a single base change producing an early stop and explains truncation. Model correct response: change the mRNA so codon 2 becomes UAA/UAG/UGA — e.g. base 4 C→U giving UCU... (Ser), not stop. The cleanest single-base early-stop is at codon reading: change position 4 C→U and note only nonsense mutations truncate.

Marking (be generous): award full marks for a valid single base substitution creating a premature stop codon with correct explanation that translation terminates early producing a shorter (truncated) protein (nonsense mutation). (Change identified [1]; new codon is a stop [1]; explanation = premature termination/truncation [1]) [3]

(d) In a eukaryotic gene, the affected nucleotide may lie within an intron, which is spliced out of the pre-mRNA before translation, so the mutation is not present in the mature mRNA and the protein is unchanged. Alternatively, the code is degenerate so a synonymous (silent) change gives the same amino acid. (Any valid reason with explanation: 2 marks) [2]

Question 2 [10 marks]

(a) Alternative splicing. [1]

(b) During splicing, different combinations of exons can be joined while others (here E3) are skipped/retained. This produces mRNAs with different exon content, which are translated into proteins with different amino acid sequences. Thus one gene → multiple mRNA/protein products. (Different exon combinations [1]; different mRNA sequences [1]; different proteins [1]) [3]

(c) E3 = 90 nucleotides; 90 ÷ 3 = 30 codons = 30 amino acids. Isoform A includes E3, Isoform B does not, and E3 is in-frame and coding, so Isoform A has 30 more amino acids than Isoform B. Reading frame of downstream exons is preserved because 90 is a multiple of 3. (30 aa difference [2]; frame preserved because divisible by 3 [1]) [3]

(d) 5' cap: protects mRNA from exonuclease degradation / aids ribosome binding & translation initiation / aids nuclear export (any one, 1½). Poly-A tail: increases mRNA stability / aids export / aids translation (any one, 1½). (1.5 each, round to marks: cap function [1–2], poly-A function [1–2], total 3) [3]

Question 3 [10 marks]

(a) Codon (5'→3') pairs antiparallel with anticodon (given 3'→5'), so read anticodons 3'→5' aligned directly under codon:

• AUG ↔ UAC → tRNA 1 (Met)
• UUC ↔ AAG → tRNA 2 (Phe)
• UGG ↔ ACC → tRNA 4 (Trp)
• UCU ↔ AGG (tRNA 3, Ser) — wobble at 3rd position

Peptide: Met–Phe–Trp–Ser (1 mark per correct tRNA match; correct peptide implicit) [4]

(b) Wobble hypothesis / wobble base pairing. The third (3') position of the codon / first (5') position of the anticodon allows non-standard base pairing, so a single tRNA can recognise codons differing at the third position. This is possible because of the degeneracy of the genetic code (multiple codons for one amino acid). (Names wobble [1]; explains flexible 3rd-position pairing [1]; links to correct amino acid still delivered [1]) [3]

(c) For codon UGG:

• A site: the incoming aminoacyl-tRNA (Trp-tRNA) binds here, matching codon UGG. (1)
• P site: holds the tRNA carrying the growing polypeptide chain; a peptide bond forms transferring the chain onto the A-site amino acid. (1)
• E site: the now-empty (deacylated) tRNA moves here and exits the ribosome as translocation occurs. (1) [3]

Question 4 [10 marks]

(a) Any three, with purpose (2 marks each):

• Cleavage/proteolytic processing (removal of signal peptide or activation from zymogen) — converts inactive precursor to active enzyme / removes targeting sequence.
• Glycosylation (addition of carbohydrate) — aids folding, stability, and is typical of secreted proteins.
• Disulfide bond formation — stabilises tertiary/quaternary structure needed for function in extracellular environment.
• (Also acceptable: phosphorylation, folding by chaperones.) (Name [1] + correct purpose [1] each) [6]

(b) The promoter is the DNA sequence upstream of the gene where RNA polymerase (with transcription factors) binds. When the hormone is present, it activates transcription factors/activators that bind regulatory sequences and recruit RNA polymerase to the promoter, allowing transcription. Without the hormone, RNA polymerase cannot efficiently bind/initiate, so little/no mRNA (and hence enzyme) is made. (Promoter = binding/control site [1]; RNA pol binds & transcribes [1]; hormone → activator/TF recruitment enables binding [1]; no hormone → no transcription [1]) [4]

Question 5 [8 marks]

(a)

• mRNA: carries the coding message (codons) from DNA to ribosome; template for protein sequence. (1)
• tRNA: adaptor that reads codons via anticodon and delivers the correct amino acid. (1)
• rRNA: structural & catalytic component of ribosome; catalyses peptide bond formation (ribozyme). (1) [3]

(b) Transcription occurs in the nucleus; translation occurs in the cytoplasm (on ribosomes/RER). (1) The separation is significant because it allows RNA processing (capping, splicing, poly-A addition) to be completed in the nucleus before the mature mRNA is exported and translated, ensuring only fully processed mRNA is translated. (2) [3]

(c) Central dogma: DNA → RNA → protein (information flows from DNA to mRNA by transcription, then to protein by translation). (1) Exception: reverse transcription (RNA → DNA) in retroviruses via reverse transcriptase (or RNA replication). (1) [2]

[
  {"claim": "Template TAC transcribes to mRNA start codon AUG (complement, antiparallel)", "code": "comp={'T':'A','A':'U','G':'C','C':'G'}; template_3to5=list('TACGGACTTAAAACT'); mrna_5to3=''.join(comp[b] for b in template_3to5); result = (mrna_5to3=='AUGCCUGAAUUUUGA')"},
  {"claim": "First peptide is Met-Pro-Glu-Phe then stop (4 amino acids)", "code": "code={'AUG':'M','CCU':'P','GAA':'E','UUU':'F','UGA':'*'}; mrna='AUGCCUGAAUUUUGA'; codons=[mrna[i:i+3] for i in range(0,len(mrna),3)]; pep=''; \nfor c in codons:\n aa=code[c]\n if aa=='*': break\n pep+=aa\nresult = (pep=='MPEF' and len(pep)==4)"},
  {"claim": "90-nucleotide in-frame exon encodes 30 amino acids", "code": "result = (90/3 == 30)"},
  {"claim": "Peptide from Q3 mRNA AUG UUC UGG UCU is Met-Phe-Trp-Ser", "code": "code={'AUG':'M','UUC':'F','UGG':'W','UCU':'S'}; mrna='AUGUUCUGGUCU'; codons=[mrna[i:i+3] for i in range(0,len(mrna),3)]; pep=''.join(code[c] for c in codons); result = (pep=='MFWS')"}
]