= Solution
A case chromosome is transmitted and a pseudo-control chromosome is untransmitted. Summing the paired table by columns for cases and by rows for controls gives
$$
\begin{array}{c|rr}
\text{Allele}&\text{Case}&\text{Pseudo-control}\\\hline
1&27&33\\
2&73&67\\\hline
\text{Total}&100&100
\end{array}.
$$
For the ordinary unpaired <chi-squared test> of independence, the expected cells are 30,30,70,70. Its uncorrected Pearson statistic is
$$
\boxed{X^2=\frac{(27-30)^2}{30}+\frac{(33-30)^2}{30}
+\frac{(73-70)^2}{70}+\frac{(67-70)^2}{70}
=\frac67\approx0.8571.}
$$
For the paired <McNemar's test>, only the discordant transmissions matter, with counts 24 and 18. Thus
$$
\boxed{X^2_{\mathrm{McN}}=\frac{(24-18)^2}{24+18}=\frac67\approx0.8571.}
$$
Both uncorrected statistics happen to coincide and have the same approximate one-degree-of-freedom <chi-squared distribution> reference, giving $p\approx0.355$. The data do not show persuasive transmission imbalance at the conventional 5% level.
The equality is a numerical coincidence, not a justification for ignoring pairing. To see the <equality criterion for paired and unpaired allele tests>, write the original matched cells as $a,b,c,d$ and $n=a+b+c+d$. The unpaired marginal-table statistic is
$$
\frac{2n(b-c)^2}{(2a+b+c)(2d+b+c)},
$$
whereas McNemar's is $(b-c)^2/(b+c)$. Here $n=100$, $b+c=42$, and $(2a+b+c)(2d+b+c)=60\times140=2n(b+c)$, giving equality. Changing concordant counts can change the unpaired statistic without changing McNemar's. The <transmission disequilibrium test> should retain its conditional paired interpretation.
If a continuity correction is used, state it explicitly: McNemar's corrected statistic is $(|24-18|-1)^2/42=25/42\approx0.5952$; the corresponding Yates-corrected marginal statistic also coincides here. An exact conditional test uses $\operatorname{Binomial}(42,1/2)$ and gives the two-sided <probability> $2\mathbb P(K\leq18)\approx0.4408$.
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