= Solution
Under the null put $p=q=r$. The <likelihood> becomes the same two binomial constants times $r^{m+n}(1-r)^{2M+2N-m-n}$. Maximization gives the pooled estimate
$$
\boxed{\hat r=\frac{m+n}{2(M+N)}.}
$$
Under an interior null, (ii) has zero mean and <variance> $r(1-r)(1/(2M)+1/(2N))$. <Statistical consistency> of $\hat r$ and <Slutsky's theorem> permit replacement of $r$ by $\hat r$ in the denominator, giving the stated $Z\Rightarrow N(0,1)$. For a two-sided level-$\alpha$ test, reject when $\boxed{|Z|>\Phi^{-1}(1-\alpha/2)}$. The large-sample approximation requires both expected counts of each type to grow; if all observed alleles are identical the stated denominator degenerates and this approximation is not applicable.
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