Transcription, Translation & Gene Expression

Question 1

(a) [3] Central dogma: DNA → (transcription) → RNA → (translation) → Protein. (1) Transcription = DNA→RNA (1); RNA processing acts on the RNA before translation; translation = RNA→protein (1).

(b) [6] Pre-mRNA contains introns (non-coding) interspersed with exons. During splicing by the spliceosome, introns are excised and exons ligated (2), removing large stretches (here 4×500 = 2000 nt) → much shorter mRNA (1). The other two processing events: 5′ capping (7-methylguanosine cap — protects from exonucleases, aids ribosome binding/export) (1.5) and 3′ polyadenylation (poly-A tail — stability, export, translation) (1.5).

(c) [6] (i) Retained exons = 90+120+60+150+90 = 510 nt (2). (ii) Coding nt = 510 − 30 (5′UTR) − 42 (3′UTR) = 438 nt (2). (iii) 438/3 = 146 codons; subtract 1 stop codon → 145 amino acids in the polypeptide (2). (The start Met is included in the 146 codons; polypeptide length = 146 − 1 stop = 145 residues.)

(d) [9] (i) Each of the $k$ internal exons has 2 independent states (include/skip). By product rule, number of isoforms $= 2^k$. (2) Proof by induction: Base $k=0$: no internal exons ⇒ exactly 1 isoform $=2^0=1$. ✓ (1) Inductive step: assume $k$ internal exons give $2^k$ isoforms. Add one more internal exon; each existing isoform splits into two (new exon included or skipped), giving $2\cdot 2^k = 2^{k+1}$. ✓ (3) Hence by induction $N(k)=2^k$ for all $k\ge 0$. (1) (ii) $k=3 \Rightarrow N = 2^3 = \mathbf{8}$ isoforms. (2)

Question 2

(a) [3] Degeneracy: multiple codons can specify the same amino acid (more than one codon per amino acid) (2). Advantage: buffers point mutations — many single-base changes (esp. 3rd position "wobble") are synonymous, reducing deleterious effects (1).

(b) [6]

• AUG → Met (Start); anticodon 5′-CAU-3′
• GCC → Ala; anticodon 5′-GGC-3′
• UUU → Phe; anticodon 5′-AAA-3′
• GGA → Gly; anticodon 5′-UCC-3′
• UAA → STOP (no tRNA) Peptide: Met-Ala-Phe-Gly (stop). (1 mark each correct codon translation ×4 = 4; anticodons ×2 pooled = 2). Anticodon = reverse complement of codon read 5′→3′.

(c) [6] (i) Codon info $=\log_2 64 = 6$ bits (1.5). Amino-acid-level (21 symbols) $=\log_2 21 \approx 4.392$ bits (1.5). (ii) 61 sense codons for 20 amino acids ⇒ average $61/20 = 3.05$ codons per amino acid (2). This >1 ratio confirms redundancy/degeneracy — on average ~3 codons encode each amino acid (1).

(d) [5]

def translate(mrna, table):
    i = mrna.find("AUG")          # locate start
    if i == -1: return ""
    prot = []
    while i + 3 <= len(mrna):
        codon = mrna[i:i+3]
        aa = table[codon]
        if aa == "*":             # stop codon
            break
        prot.append(aa)
        i += 3
    return "".join(prot)

(logic 3, correct start/stop handling 1) Complexity: single pass, O(n) in input length $n$ (dict lookup O(1)) (1).

Question 3

(a) [4]

• A (aminoacyl) site: receives incoming aminoacyl-tRNA whose anticodon matches the codon (1.5).
• P (peptidyl) site: holds the tRNA bearing the growing polypeptide; peptide bond forms transferring chain to A-site tRNA (1.5).
• E (exit) site: holds the now-deacylated tRNA before it leaves the ribosome (1).

(b) [6] (i) Time = 420 / 15 = 28 s (2). (ii) At steady state each ribosome finishes and, since loading interval = 2 s, one protein completes every 2 s once the pipeline is full. Rate = 60/2 = 30 proteins per minute (3). (The 8 ribosomes fill the pipeline; steady-state throughput is set by the loading interval, 1 per 2 s.) (1)

(c) [6] Independent codons ⇒ probability all 420 correct: $P = p^{420} = 0.9997^{420}.$ (2) $\ln P = 420\ln(0.9997) = 420\times(-0.000300045) = -0.12602$; $P = e^{-0.12602} \approx \mathbf{0.8816}$ (~88.2%). (2) Comment: error probability per protein $\approx 1-0.882 = 11.8\%$; for longer proteins $p^n$ falls, so proofreading (kinetic + aminoacyl-tRNA synthetase editing) is essential to keep functional protein yield high. (2)

[
  {"claim":"Retained exon nt = 510, coding = 438, amino acids = 145",
   "code":"exons=90+120+60+150+90; coding=exons-30-42; aa=coding//3-1; result=(exons==510 and coding==438 and aa==145)"},
  {"claim":"Alternative splicing isoforms for k=3 is 8",
   "code":"k=3; result=(2**k==8)"},
  {"claim":"Average codons per amino acid = 61/20 = 3.05",
   "code":"result=(Rational(61,20)==Rational(305,100))"},
  {"claim":"Zero-error probability 0.9997^420 approx 0.8816",
   "code":"P=Float('0.9997')**420; result=(abs(P-Float('0.8816'))<Float('0.001'))"},
  {"claim":"Polysome throughput 30 proteins/min at 2s loading interval",
   "code":"result=(60/2==30)"}
]