= Transmission disequilibrium test
{title2=$\chi^2_{\mathrm{TDT}}=\frac{(b-c)^2}{b+c}$}
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= TDT
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For affected offspring and heterozygous parents at a diallelic locus, let $b$ count transmissions of allele 1 with allele 2 untransmitted and $c$ count the reverse. Under the fair-transmission null, conditional on $b+c$, $b$ is binomial with probability $1/2$. The asymptotic statistic is $(b-c)^2/(b+c)$, with a one-degree-of-freedom chi-squared null distribution. Homozygous-parent transmissions give no contrast. Conditioning within families protects this test from ordinary <population stratification>.
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