Transcription, Translation & Gene Expression

Question 1 (10 marks)

(a) (4 marks — 1 per correct labelled arrow, max 4) $\text{DNA} \xrightarrow[\text{DNA polymerase}]{\text{replication}} \text{DNA}$ $\text{DNA} \xrightarrow[\text{RNA polymerase}]{\text{transcription}} \text{RNA}$ $\text{RNA} \xrightarrow[\text{ribosome/rRNA + tRNA}]{\text{translation}} \text{Protein}$

• Replication (DNA→DNA), DNA polymerase (1)
• Transcription (DNA→RNA), RNA polymerase (1)
• Translation (RNA→Protein), ribosome (1)
• Correct direction of overall flow DNA→RNA→Protein (1)

(b) (3 marks)

• Reverse transcription, RNA→DNA (1)
• Enzyme: reverse transcriptase (1)
• Context: retroviruses (e.g. HIV) / telomerase / lab cDNA synthesis (1)

(c) (3 marks)

• There is no known enzyme that reads an amino acid sequence and synthesises a corresponding nucleic acid (1).
• The genetic code is effectively "one-way": codon→amino acid, but no cellular machinery reverses it (1).
• Because degeneracy means several codons map to one amino acid, the original codon could not be uniquely recovered anyway (1).

Question 2 (12 marks)

(a) (2 marks) Coding strand = complement of template, same sequence as mRNA but with T. Template 3'-TAC GGA TTA CGC ATT ACT-5' Coding strand 5'→3': ATG CCT AAT GCG TAA TGA (1 for correct bases, 1 for correct 5'→3' direction).

(b) (4 marks)

• mRNA is complementary/antiparallel to the template strand, U replaces T (rule stated) (1).
• RNA uses ribose sugar and uracil instead of deoxyribose/thymine (1).
• Transcript 5'→3': AUG CCU AAU GCG UAA UGA — wait, check pairing.

Template read 3'→5': T A C | G G A | T T A | C G C | A T T | A C T Complementary RNA (5'→3'), pairing A-U, T-A, G-C, C-G:

• T→A, A→U, C→G → AUG
• G→C, G→C, A→U → CCU
• T→A, T→A, A→U → AAU
• C→G, G→C, C→G → GCG
• A→U, T→A, T→A → UAA
• A→U, C→G, T→A → UGA

mRNA 5'→3': AUG CCU AAU GCG UAA UGA (2 for correct sequence).

(c) (4 marks)

• 5' cap (7-methylguanosine): protects from exonucleases / aids ribosome binding & export (1 name + 1 function).
• Poly-A tail (string of adenines at 3'): protects from degradation / increases stability & aids export (1 name + 1 function).

(d) (2 marks) RNA polymerase binds the promoter (a DNA sequence upstream of the gene) which positions it at the start site and defines the direction, then it synthesises RNA 5'→3' by reading the template 3'→5' (1 for promoter recruitment/positioning, 1 for RNA pol synthesising the transcript).

Question 3 (12 marks)

(a) (6 marks) mRNA: AUG CCU AAU GCG UAA UGA

• Start codon: AUG = Met (1)
• CCU = Pro (1) (note: CCU→Pro; question's reference gave GGA=Gly etc. Use standard table for CCU=Pro)
• AAU = Asn (1)
• GCG = Ala (1)
• UAA = STOP → translation ends (1)

Peptide: Met – Pro – Asn – Ala (STOP) (1)

(Accept correct translation from the standard codon table; markers award per correct codon→amino acid mapping. The peptide is Met-Pro-Asn-Ala.)

(b) (3 marks)

• Codon CGC (from earlier) → anticodon must be complementary & antiparallel: 5'-GCG-3' (1).
• The anticodon pairs antiparallel to the codon: codon 5'-CGC-3' aligns with anticodon 3'-GCG-5' (1).
• Base pairing C-G, G-C, C-G confirms the match; antiparallel means the strands run in opposite 5'→3' directions (1).

(c) (3 marks)

• Peptide length = 4 amino acids (Met-Pro-Asn-Ala) (1).
• The reading frame contains a stop codon (UAA) which codes for no amino acid — it only signals release (1).
• So although there are ~5–6 triplets, the stop codon(s) add no residue, giving one fewer amino acid than the last coding position (1).

Question 4 (10 marks)

(a) (4 marks)

• A site (aminoacyl): accepts the incoming aminoacyl-tRNA whose anticodon matches the codon (1–2).
• P site (peptidyl): holds the tRNA carrying the growing polypeptide chain (1).
• E site (exit): holds the deacylated (empty) tRNA about to leave the ribosome (1).

(b) (4 marks)

• Aminoacyl-tRNA delivered to A site, anticodon pairs with codon (1).
• Peptide bond formation: peptidyl transferase (rRNA) transfers the growing chain from the P-site tRNA onto the A-site amino acid (1).
• Translocation: ribosome moves one codon along; A-site tRNA shifts to P site, P-site tRNA shifts to E site (1).
• E-site tRNA released; A site now free for the next aminoacyl-tRNA (1).

(c) (2 marks)

• Termination is triggered when a stop codon (UAA/UAG/UGA) reaches the A site (1).
• A release factor (protein), not a tRNA, enters the A site, causing hydrolysis and release of the polypeptide (1).

Question 5 (10 marks)

(a) (4 marks)

• Degeneracy: most amino acids are encoded by more than one codon (1).
• Average = 64/20 = 3.2 codons per amino acid (1).
• Distribution uneven: e.g. Met and Trp have only 1 codon each, while Leu/Ser/Arg have 6 (1).
• Reason: 3 codons are stop codons (61 code for amino acids), and the "wobble" position allows redundancy but not uniformly (1).

(b) (3 marks)

• A base change in the third (wobble) position often still codes for the same amino acid (1).
• Therefore the protein is unchanged and function preserved (1).
• Such a mutation is a silent (synonymous) mutation (1).

(c) (3 marks)

• Exons = coding sequences retained in mature mRNA; introns = non-coding sequences removed during splicing (1).
• Alternative splicing: different combinations of exons are joined (some exons skipped) (1).
• One gene → multiple mRNA variants → multiple protein isoforms, increasing proteome diversity (1).

Question 6 (6 marks)

(a) (2 marks) PTM = chemical modification of a protein AFTER translation (1), altering its structure, activity, location, or stability (1).

(b) (4 marks) — any two, 2 marks each (example + consequence):

• Phosphorylation (add phosphate) → switches enzyme activity on/off; signalling (2).
• Glycosylation (add sugar) → cell surface recognition / protein folding / stability (2).
• Proteolytic cleavage (e.g. proinsulin→insulin) → activates the protein (2).
• Ubiquitination → targets protein for degradation (2).
• Disulfide bond formation / acetylation / methylation → folding/stability/regulation (2).

[
  {"claim":"Average codons per amino acid = 64/20 = 3.2","code":"result = (64/20 == 3.2)"},
  {"claim":"61 codons code for amino acids (64 total minus 3 stop)","code":"result = (64 - 3 == 61)"},
  {"claim":"mRNA from template 3'-TAC-5' start is AUG (transcribe & complement)","code":"comp = {'A':'U','T':'A','G':'C','C':'G'}; template_3to5='TAC'; mrna=''.join(comp[b] for b in template_3to5); result = (mrna == 'AUG')"},
  {"claim":"Anticodon (5'->3') for codon 5'-CGC-3' is GCG","code":"pair={'A':'U','U':'A','G':'C','C':'G'}; codon='CGC'; anti_antiparallel=''.join(pair[b] for b in codon)[::-1]; result = (anti_antiparallel=='GCG')"},
  {"claim":"Peptide length is 4 (Met-Pro-Asn-Ala), one less than 5 coding triplets before stop","code":"codons=['AUG','CCU','AAU','GCG','UAA']; peptide=[c for c in codons if c not in ('UAA','UAG','UGA')]; result = (len(peptide)==4)"}
]