 Solution

The transmission disequilibrium test is the paired McNemar test statistic $(b-c{)}^2/(b+c)$. From the raw unambiguous-transmission table,The formal one-degree-of-freedom chi-squared distribution approximation gives $p\simeq 0.071$, so the calculation does not reject at 5%. Its ordinary null calibration would require a valid missing-parent inclusion rule; the raw arithmetic alone does not establish one.

Using the complete-trio table gives ${\chi }_{\mathrm{TDT}}^2=36/14=18/7\simeq 2.571$, with $p\simeq 0.109$. It also does not reject. Homozygous-parent transmissions in $a,d$ supply no discordant contrast for this test.