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The unmatched haplotype-relative-risk counts satisfy $w=a+b$, $x=c+d$, $y=a+c$ and $z=b+d$. For the raw table this givesThe Pearson chi-squared test of independence in this two-by-two table has statisticThus the unmatched calculation gives nominal evidence at 5%, with $p\simeq 0.030$. It differs from the transmission disequilibrium test because it uses an unmatched table, and it also inherits the missing-parent selection issue from the raw counts.

For the ten complete trios, $(w,x,y,z)=(12,8,6,14)$, giving odds ratio $7/2$ and the unmatched Pearson chi-squared test of independence statistic $40/11\simeq 3.636$, with $p\simeq 0.057$. That calculation does not reject at 5%. The nominal significance from the raw table therefore should not be presented as a robust association conclusion without a justified model for incomplete families and the unmatched-table assumptions.