Bioinformatics & Computational Biology

Question 1 (14 marks)

(a) DP Matrix (6 marks) — Rows indexed by A = H,E,A,G; columns by B = P,A,W,G. Gap = −2.

Initialisation: first row/column = multiples of −2.

Each cell = max(diag + match/mismatch, up − 2, left − 2).

• (6 marks: full correct matrix; award ~1 per row correctly filled; −1 per arithmetic slip.)

(b) Optimal score = −1 (bottom-right cell). (2)

(c) Traceback (3 marks) — One optimal alignment:

H E A G
| . | . 
P A W G     →  H-E-A-G vs P-A-W-G

Aligned (no gaps needed on this path):

H E A G
P A W G

Score check: H/P (−1) + E/A (−1) + A/W (−1) + G/G (+2) = −1. ✓ (3 marks)

(d) BLOSUM advantage (3 marks) — A fixed −1 penalises all mismatches equally. BLOSUM62 assigns higher (less negative or positive) scores to conservative substitutions (chemically similar residues that are frequently interchanged in real evolution) and heavier penalties to unlikely ones. This makes the alignment biologically realistic, improving detection of distant homologs whose residues have diverged but retain function. (3 marks: mention differential scoring + evolutionary basis + improved remote homology detection.)

Question 2 (12 marks)

(a) Ranking X > Y > Z (4 marks)

• X most significant: lowest E-value (3e-140), highest bit score, high identity, near-full-length alignment.
• Y intermediate: E-value 2e-08 still highly significant (≪ typical 1e-5 threshold), good coverage (210 res).
• Z insignificant: E-value 4.0 (> 1, expected by chance), short alignment (45 res), low identity.
• Two criteria: E-value (statistical significance) and alignment length/coverage + % identity (biological substance). (4 marks)

(b) Twilight zone (3 marks) — Identity ~20–35% where sequence similarity alone cannot reliably distinguish homology from chance. Y at 31% falls here. However, the E-value (2e-08) accounts for alignment length and score against database size, so a low E-value over a substantial alignment provides statistical confidence of homology beyond raw % identity. (3 marks)

(c) Database ×10 (3 marks) — E-value scales approximately linearly with database size (E ∝ N). So Z's E-value ≈ 4.0 × 10 = ~40. Hit Z becomes less trustworthy (even more expected by chance). (3 marks)

(d) Heuristic (2 marks) — BLAST seeds on short exact/near-exact words then extends, avoiding a full DP scan → much faster. It sacrifices the guarantee of finding the mathematically optimal alignment that Needleman–Wunsch provides; some weak true hits may be missed. (2 marks)

Question 3 (13 marks)

(a) UPGMA (7 marks)

Smallest distance = A–B (4). Cluster (AB) at height 4/2 = 2.

Recompute distances (average):

• (AB)–C = (10+10)/2 = 10
• (AB)–D = (12+12)/2 = 12
• C–D = 6

Smallest = C–D (6). Cluster (CD) at height 6/2 = 3.

Recompute:

• (AB)–(CD) = (10+12)/2 = 11

Final join ((AB),(CD)) at height 11/2 = 5.5. (7 marks: 3 for correct join order, 2 for correct averaging, 2 for correct heights.)

(b) Tree (3 marks)

             ___________ 5.5
            |           |
        __2_|        __3|__
       |    |       |     |
       A    B       C     D

Heights: (A,B) join at 2; (C,D) join at 3; root at 5.5. (3 marks)

(c) Molecular clock failure (3 marks) — UPGMA assumes equal evolutionary rates so root-to-tip distances are equal. If one lineage evolves faster (long-branch, e.g. D accumulating changes rapidly), UPGMA can group taxa by rate rather than true ancestry, misplacing the fast lineage. Alternative: Neighbour-Joining (NJ), which does not assume a clock (produces unrooted, additive trees). (3 marks)

Question 4 (11 marks)

(a) Correct order (5 marks)

1. Quality control (trimming) — remove low-quality bases/adapters.
2. Read alignment/quantification — map reads to reference/transcriptome.
3. Raw read counts — count reads per gene.
4. Normalisation — correct for library size/composition to allow cross-sample comparison.
5. Statistical testing for differential expression — identify significantly changed genes. (5 marks: 1 per correctly ordered step with purpose.)

(b) Interpretation (3 marks) — log₂FC +3.2 → gene expressed 2^3.2 ≈ 9.2-fold higher in tumour than normal. Adjusted p = 0.002 → statistically significant after correction. Adjusted p-value (multiple-testing correction, e.g. Benjamini–Hochberg) is required because thousands of genes are tested simultaneously; raw p-values would produce many false positives by chance. (3 marks)

(c) 8% variant (3 marks) — Biological: somatic mosaicism / tumour subclone / low-frequency variant present in a fraction of cells. Artefactual: sequencing/PCR error or misalignment creating spurious calls. (3 marks)

Question 5 (10 marks)

(a) Accuracy misleading (4 marks) — With 95% non-enzymes, a classifier predicting "non-enzyme" for everything scores 95% accuracy while detecting zero enzymes. Accuracy ignores class imbalance. Better metrics: precision & recall (sensitivity), F1-score, or AUC-ROC / Matthews correlation coefficient (any two). (4 marks)

(b) Overfitting (3 marks) — Model learns noise/idiosyncrasies of training data, giving high training but poor unseen performance. Detect: gap between training and validation/test accuracy (cross-validation). Reduce: regularisation, more data, dropout, simpler model, or early stopping. (3 marks)

(c) AlphaFold rationale (3 marks) — Structure is determined by sequence (Anfinsen), so a mapping exists to learn. MSAs reveal co-evolution: residues in physical contact tend to mutate in a correlated way across homologous sequences; these correlations are strong features predicting spatial proximity, which the network exploits to infer 3D contacts and fold. (3 marks)

[
  {"claim":"NW optimal alignment score for HEAG/PAWG equals -1","code":"match=2\nmis=-1\ngap=-2\nA='HEAG'\nB='PAWG'\nimport functools\n@functools.lru_cache(None)\ndef nw(i,j):\n    if i==0 and j==0: return 0\n    opts=[]\n    if i>0 and j>0:\n        s=match if A[i-1]==B[j-1] else mis\n        opts.append(nw(i-1,j-1)+s)\n    if i>0: opts.append(nw(i-1,j)+gap)\n    if j>0: opts.append(nw(i,j-1)+gap)\n    return max(opts)\nresult = (nw(4,4)==-1)"},
  {"claim":"UPGMA root height for given matrix is 5.5","code":"ab_cd=(10+12)/2\nresult = (ab_cd/2==5.5)"},
  {"claim":"log2FC of 3.2 corresponds to ~9.2 fold change","code":"fc=2**3.2\nresult = (abs(fc-9.19)<0.1)"},
  {"claim":"Database expansion x10 raises E-value of Z from 4.0 to ~40","code":"E=4.0*10\nresult = (E==40.0)"}
]