Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 37 / 6

Let $G$ count copies of allele $2$. Under Hardy-Weinberg equilibrium, $G$ has a binomial distribution with parameters $2,\varphi$, and the penetrance is $q_G=p{\theta }^G$. ConsequentlyThus the overall penetrance isThe multiplicative diallelic penetrance model requires $0\le \varphi \le 1$, $p\ge 0$, $\theta \ge 0$, and each occupied genotype's value $p{\theta }^G$ to be at most one. These are probabilities, not unrestricted relative-risk scores.

Since the binary trait $Y$ has marginal Bernoulli distribution with probability $K=p{\alpha }^2$, its total variance isWithin 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:Another Hardy-Weinberg equilibrium average givesHenceBy the law of total variance, the remaining between-genotype component is the genetic variance of penetrance, namelyIt is nonnegative, as a variance must be. It also follows from $\beta -{\alpha }^2=\varphi (1-\varphi )(\theta -1{)}^2\ge 0$. The single-genotype variance and its population average are different quantities; both have been specified.

Use the usual genetic model in which disease outcomes are conditionally independent given the two individuals' genotypes, and unshared founder alleles are independently drawn from the same outbred population. Let $J$ denote their number of allele copies shared by identity by descent. Conditional independence then givesWhen $J=2$, both individuals have the same genotype $G$. Its distribution remains the population genotype distribution under Hardy-Weinberg equilibrium, so their individual trait means are $K=p{\alpha }^2$ andThereforeWhen $J=0$, all four unshared copies are independent, making $G_1,G_2$ independent population genotypes. Thus $\mathbb{E}(q_{G_1}{q}_{G_2})=(p{\alpha }^2{)}^2$ andThe residual-independence assumption is necessary. Marginal penetrances alone do not determine joint disease probabilities. For example, with $\theta =1$, setting both traits equal to one shared Bernoulli variable of probability $p$ preserves every marginal penetrance, but gives covariance $p(1-p)$ even with no shared alleles. Shared environmental disease determinants or population structure therefore require an expanded model. The displayed answers are the intended locus-only covariance results, not a consequence of the penetrance table without that usual assumption.

For binary traits, $Y_1{Y}_2$ is the indicator that both individuals are affected. Hence the pair probability is its expectation, or equivalently the covariance plus the product of individual means. Using the preceding results,These statements retain the same conditional-independence and common-population assumptions as the multiplicative diallelic penetrance covariance calculation.

For full siblings in an outbred pedigree, Mendelian IBD sharing of full siblings gives prior sharing probabilities $(1/4,1/2,1/4)$ for $J=0,1,2$. Let $A$ be the event that both siblings are affected. Bayes' rule weights those probabilities by the pair penetrances, giving immediatelyTo obtain the one-shared-copy probability, assign each allele a risk multiplier $R$, equal to $1$ with probability $1-\varphi$ and $\theta$ with probability $\varphi$. Then $\mathbb{E}R=\alpha$ and $\mathbb{E}{R}^2=\beta$. With $J=1$, write the siblings' penetrances as $p{R}_s{R}_1$ and $p{R}_s{R}_2$, where the shared multiplier $R_s$ and the two unshared multipliers are independent. Conditional independence of the disease outcomes givesThe total pair probability is therefore $p^2(\beta +{\alpha }^2{)}^2/4$, and the three ascertained sharing probabilities areIn particular,The one-IBD proportion is below one half for a polymorphic locus with a genuine penetrance effect. Equality holds when $\beta ={\alpha }^2$, namely when $\varphi \in \{0,1\}$ or $\theta =1$; there is then no variation in the allele risk multiplier. The inequality follows from the different ascertainment weights for zero, one and two shared copies, rather than from applying the unascertained sibling probabilities to affected pairs. This is sibling IBD ascertainment under multiplicative penetrance. All affected-pair conditional probabilities presume $\mathrm{Pr}(A)>0$; if disease never occurs, conditioning on affected pairs is undefined.