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ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2025/paper-2/2/d/solution

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Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2025 / Paper 2 / 2 / d / Solution by

Codex  0 2026-10-20

For the same uniform background and base-two log-likelihood ratio, replace every emission probability

b_{i} (x)

e_{i} (x) = lo g_{2} [b_{i} (x) / (1/4)]

. The background specifies independent letters, not a competing hidden-state process. Its single emitting state has continuation factor one when scoring a fixed-length word, so each model transition contributes

s_{ij} = lo g_{2} a_{ij}

; dividing every transition by

1/4

would incorrectly count the background twice.

The complete transition-score diagram is

Start 0 M_{1} 0 M_{2} - 2 M_{3} - 1 M_{4} 0 End, M_{2} l o g_{2} (3/4) I_{2} l o g_{2} (3/4) M_{3}, I_{2} - 2 I_{2}, M_{3} - 1 I_{3} 0 M_{4} .

(1)

Every omitted transition has score

- \infty

. Silent Start and End supply no emission score. The complete table of emission scores is

state M_{1} M_{2} I_{2} M_{3} I_{3} M_{4} A 1 - \infty 0 - \infty - \infty 0 T - \infty - \infty 12 - \infty - \infty G 0 lo g_{2} 3 0 - \infty - \infty lo g_{2} 3 C 00 - \infty - \infty 2 - \infty .

(2)

For the unique nonzero path identified in (b), add the five emission scores and all transition scores:

score = emissions 0 + lo g_{2} 3 + 1 + 2 + 0 + transitions 2 lo g_{2} (3/4) - 1 = lo g_{2} 27 - 2 = lo g_{2} (27/4) .

(3)

This equals (c). For a sequence with several nonzero hidden paths, the full log-likelihood ratio is the log-sum-exp of their path scores; selecting only the largest score would instead compute a best-path approximation.

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