= Solution
Label each parent's two <allele> copies by its realized transmission: let $(A,B)$ be the transmitted and untransmitted <alleles> of the first parent and $(C,D')$ those of the second. Under <random mating> and <Hardy-Weinberg equilibrium>, the four parental <allele> draws are independent with population frequencies $\pi$. Random <Mendelian segregation> merely swaps the two independent copies within each parent, so the transmitted/untransmitted labelled pair still has <probability> $\pi_a\pi_b$; the second pair has <probability> $\pi_c\pi_d$.
The child receives $A,C$. If $\mathcal D$ is its disease event, its <penetrance> is $k\psi_A\psi_C$ and its overall disease <probability> is $kZ^2$. Therefore
$$
\boxed{\begin{aligned}
&\mathbb P(A=a,B=b,C=c,D'=d\mid\mathcal D)\\
&\qquad=\frac{k\psi_a\psi_c\pi_a\pi_b\pi_c\pi_d}{kZ^2}
=\pi_a^*\pi_c^*\pi_b\pi_d.
\end{aligned}}
$$
This proves the <transmitted and untransmitted alleles under multiplicative penetrance> factorization. Under these assumptions, the transmitted pair $a/c$ has the affected-case distribution, while the complementary pair $b/d$ has the population distribution and is independent of the transmitted pair. It is the <family pseudo-control genotype>.
As in part (a), the formula records labelled transmissions. When observed <genotypes> are unordered, sum over every compatible parental-origin transmission. This includes multiplicities for overlapping parental <alleles> or two heterozygous parents producing a heterozygous child; it avoids interpreting one product as the total <probability> of all observationally identical configurations. The homogeneous random-mating model is also essential to the unconditional independence: mixing ancestry strata can correlate the case and pseudo-control through their shared family background.
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