 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/5k/1/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The displayed analysis of variance compares Model 2 with the unrestricted three-mean Model 1. It tests

H_{0} : μ_{A a} = μ_{AA},

which is precisely full dominance: once an

A

allele is present, a second copy has no further effect. Under

H_{0}

F = \frac{( 54.148 - 48.487 ) /1}{48.487/57} = 6.6551 \sim F_{1, 57} .

0.01249

rejects full dominance at the

5%

level, providing evidence that the heterozygote and

AA

means differ.

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