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 Incoming links: Genetic variance of penetrance

Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 37 / 6 / ii / Solution Created 2026-10-03 Updated 2026-10-14

Since the binary trait

Y

has marginal Bernoulli distribution with probability

K = p α^{2}

, its total variance is

Var (Y) = p α^{2} (1 - p α^{2}) .

Within a specified genotype

G = g

, the conditional variance is

q_{g} (1 - q_{g})

. The within-genotype contribution to total variance is its average over the population:

V_{W} = E [q_{G} (1 - q_{G})] = E (q_{G}) - E (q_{G}^{2}) .

Another Hardy-Weinberg equilibrium average gives

E (q_{G}^{2}) = p^{2} [(1 - ϕ) + ϕ θ^{2}]^{2} = p^{2} β^{2}, β = 1 + ϕ (θ^{2} - 1) .

Hence

V_{W} = p α^{2} - p^{2} β^{2} .

By the law of total variance, the remaining between-genotype component is the genetic variance of penetrance, namely

V_{G} = Var (q_{G}) = Var (Y) - V_{W} = p^{2} (β^{2} - α^{4}) .

It is nonnegative, as a variance must be. It also follows from

β - α^{2} = ϕ (1 - ϕ) (θ - 1)^{2} \geq 0

. The single-genotype variance and its population average are different quantities; both have been specified.

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