= Solution
Let $I,J$ denote the two ordered <alleles> at the <genetic locus>. Under <Hardy-Weinberg equilibrium> they are independent with <probabilities> $\pi_i,\pi_j$. Write the multiplicative <penetrance> as $\mathbb P(D\mid I=i,J=j)=k\psi_i\psi_j$, with parameters chosen so these are valid <probabilities>, and set $Z=\sum_u\pi_u\psi_u$. Then
$$
\mathbb P(D)=k\sum_{i,j}\pi_i\pi_j\psi_i\psi_j=kZ^2.
$$
The <Bayes' theorem> gives
$$
\boxed{\mathbb P(I=i,J=j\mid D)
=\frac{k\pi_i\pi_j\psi_i\psi_j}{kZ^2}
=\pi_i^*\pi_j^*,\qquad \pi_i^*=\frac{\pi_i\psi_i}{Z}.}
$$
Thus <multiplicative penetrance preserves Hardy-Weinberg equilibrium>: the affected subjects still have two independent <allele> draws, with tilted <allele> frequencies. The displayed product uses ordered <allele> slots. For the usual unordered <genotypes>, the corresponding <probabilities> are
$$
\boxed{\mathbb P(i/i\mid D)=(\pi_i^*)^2,\qquad
\mathbb P(i/j\mid D)=2\pi_i^*\pi_j^*\quad(i\ne j).}
$$
The factor of two is required for heterozygotes and must not be dropped when translating the ordered notation into <genotype> counts.
For <genetic association> analysis, the case <genotype> <probabilities> factor into <allele> frequencies, so under this model an allelic comparison with representative population controls is appropriate; separate dominance departures are not required by the <penetrance> model. The ratio of case <allele> frequency to population <allele> frequency is proportional to $\psi_i$, and relative risk multipliers satisfy $\psi_i/\psi_j=(\pi_i^*/\pi_i)/(\pi_j^*/\pi_j)$. Population controls are important to this exact statement. Among specifically unaffected controls,
$$
\mathbb P(I=i,J=j\mid D^c)=\frac{\pi_i\pi_j(1-k\psi_i\psi_j)}{1-kZ^2},
$$
which generally does not factor. For a rare disease it is close to the population <genotype> law, but no rare-disease assumption was needed for the case factorization itself. <Population stratification> and sampling dependence must also be addressed before using an ordinary allelic <chi-squared test>.
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