 Solution

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

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

Under Hardy-Weinberg equilibrium, the two parental allele copies are sampled independently from the same pool. Multiplying their frequencies, and adding the two orderings for a heterozygote, gives

Pr (a / a) = p^{2}, Pr (a / b) = 2 p (1 - p), Pr (b / b) = (1 - p)^{2} .

Their sum is

(p + (1 - p))^{2} = 1

. The genotypes are unordered even though the paternal and maternal transmissions may be distinguished in the calculation.

 New to topics? Read the docs here!