= Solution
Ignoring the multinomial coefficient, the likelihood is
$$
\begin{aligned}
L(\theta)
&=(1-\theta)^{2n_{AA}}\theta^{2n_{BB}}
[2\theta(1-\theta)]^{n_{AB}}\\
&=2^{n_{AB}}
\theta^{2n_{BB}+n_{AB}}
(1-\theta)^{2n_{AA}+n_{AB}}.
\end{aligned}
$$
The score equation is
$$
\frac{2n_{BB}+n_{AB}}{\theta}
-\frac{2n_{AA}+n_{AB}}{1-\theta}=0,
$$
and strict concavity of the log-likelihood gives
$$
\boxed{\ \widehat\theta_{\rm MLE}
=\frac{2n_{BB}+n_{AB}}{2n}
=\widehat\theta_{w^*}.\ }
$$
Using the variables $Z_i$ from part (b), the <central limit theorem> gives
$$
\sqrt n(\widehat\theta_{\rm MLE}-\theta)
=\frac1{\sqrt n}\sum_{i=1}^n(Z_i-\theta)
\xrightarrow{d}
N\left(0,\operatorname{Var}(Z_1)\right).
$$
Therefore the <allele-frequency estimator under Hardy-Weinberg equilibrium> has limiting law
$$
\boxed{\ \sqrt n(\widehat\theta_{\rm MLE}-\theta)
\xrightarrow{d}N\left(0,\frac{\theta(1-\theta)}2\right).\ }
$$
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