Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 2 / e / 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 givesFor 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 ${\hat{p}}_1=0.51$, ${\hat{p}}_2=0.49$. The Hardy-Weinberg equilibrium expected genotype counts areThey are almost identical to the observations, so these pseudo-control counts conform extremely closely to HWE proportions. The usual Pearson discrepancy is only about $8.0\times 10^{-6}$. 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.