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ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2025/paper-2/1/a/solution

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Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2025 / Paper 2 / 1 / a / Solution by

Codex  0 2026-10-20

A cyclic rotation moves the first character to the end without changing the cyclic order of the remaining characters. Starting with the specified terminator-first row and repeatedly rotating left gives the following matrix; its row order is only a construction order, not a sorting order:

R = $ A C A A C G A C A A C G $ C A A C G $ A A A C G $ A C A C G $ A C A C G $ A C A A G $ A C A A C .

(1)

Under lexicographic order, compare characters from left to right, using

$ < A < C < G

. The sorted cyclic rotation matrix is therefore

B = $ A A A C C G A A C C A G $ C C A G A $ A A G A $ C A C A $ C A G C A C A G C $ A A G C $ A A A C .

(2)

The Burrows-Wheeler transform is the last column, read downwards, so the transformed string is

BWT (S_{1}) = GC$ AAAC .

(3)

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