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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-37/5/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 37 / 5 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Treat the two parents as unordered, so three possible genotypes give six mating types. A homozygous parent transmits its allele with probability one; a heterozygous parent transmits either allele with probability one half by Mendelian segregation. Independent parental transmissions give the following complete offspring distributions:

parental genotypes a / a \times a / a a / a \times a / b a / a \times b / b a / b \times a / b a / b \times b / b b / b \times b / b Pr (a / a) 1 1/2 0 1/4 00 Pr (a / b) 0 1/2 1 1/2 1/2 0 Pr (b / b) 000 1/4 1/2 1

(1)

Each row sums to one. In particular, heterozygote-by-heterozygote mating can produce all three offspring genotypes, whereas the two opposite homozygotes produce only heterozygotes. These conditional offspring probabilities do not require Hardy-Weinberg equilibrium of the parents.

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