= Solution
The <transmission disequilibrium test> is the paired <McNemar test> statistic $(b-c)^2/(b+c)$. From the raw unambiguous-transmission table,
$$
\boxed{\chi^2_{\rm TDT}=\frac{(11-4)^2}{11+4}=\frac{49}{15}\simeq3.267<3.84.}
$$
The formal one-degree-of-freedom <chi-squared distribution> approximation gives $p\simeq0.071$, so \b[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 $\chi^2_{\rm TDT}=36/14=18/7\simeq2.571$, with $p\simeq0.109$. It also does not reject. Homozygous-parent transmissions in $a,d$ supply no discordant contrast for this test.
Back to article page
