= Solution
For each family combine the two untransmitted parental <alleles> into one <family pseudo-control genotype>. A <heterozygous> child of two <heterozygous> parents has a <heterozygous> pseudo-control, whatever its parent-specific transmission phase. Combining the rows gives
$$
\boxed{(N_{11},N_{12},N_{22})=(13,25,12).}
$$
For example, pseudo-control <genotype> $1/1$ receives contributions $1,2,5,5$ from the relevant rows; <genotype> $2/2$ receives $3,1,2,6$, and the remaining 25 are <heterozygous>. These counts also give the untransmitted <allele> totals 51 and 49.
The estimated <allele> frequencies are $\widehat p_1=0.51$, $\widehat p_2=0.49$. The <Hardy-Weinberg equilibrium> expected <genotype> counts are
$$
50(\widehat p_1^2,2\widehat p_1\widehat p_2,\widehat p_2^2)=(13.005,24.990,12.005).
$$
They are almost identical to the observations, so \b[these pseudo-control counts conform extremely closely to <HWE> proportions]. The usual Pearson discrepancy is only about $8.0\times10^{-6}$. This numerical agreement is not a general guarantee of <HWE> for family pseudo-controls, whose construction involves mating structure and case <ascertainment in a genetic study>.
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