= Solution
Using cases as disease <haplotypes> and controls as control <haplotypes> in the model requested here, the chromosome counts estimate
$$
P(A\mid d)=20/200=0.10,\quad P(a\mid d)=0.90,\quad P(A\mid D)=120/200=0.60,\quad P(a\mid D)=0.40.
$$
These are conditional marker <allele> frequencies, not the reverse <conditional probabilities> $P(d\mid A)$. Under independent founder <chromosomes>, conditional on the father's $Dd,Aa$ <genotype>, the phase $AD/ad$ has weight proportional to $0.60\cdot0.90=0.54$, and $Ad/aD$ has weight proportional to $0.10\cdot0.40=0.04$. Their normalized weights are $27/29$ and $2/29$. This conditions on the model's observed parental genotypes; the omitted disease-allele frequencies cancel in the phase-weight ratio.
Using the two phase-specific <pedigree likelihoods> already derived,
$$
\begin{aligned}
L_{\rm LD}(\theta)&=\frac{0.54L_1(\theta)+0.04L_2(\theta)}{0.58}\\
&=\frac{\theta(1-\theta)}{8(0.58)}\{0.54(1-\theta)+0.04\theta\}\\
&=\frac{\theta(1-\theta)(0.54-0.50\theta)}{4.64}.
\end{aligned}
$$
The normalizing factor is independent of $\theta$, so maximize $g(\theta)=0.54\theta-1.04\theta^2+0.50\theta^3$. Its derivative gives exactly
$$
\boxed{1.5\widehat\theta^2-2.08\widehat\theta+0.54=0.}
$$
Only the smaller root lies in $[0,1/2]$:
$$
\boxed{\widehat\theta=\frac{2.08-\sqrt{2.08^2-4(1.5)(0.54)}}3\simeq0.34590.}
$$
On this interval $g''(\theta)=-2.08+3\theta<0$, so this stationary point is the unique maximum. The maximum <LOD score>, retaining the same <linkage disequilibrium> frequencies in numerator and denominator, is
$$
\boxed{Z_{\max}=\log_{10}\frac{\widehat\theta(1-\widehat\theta)(0.54-0.50\widehat\theta)}{(1/2)(1/2)(0.54-0.25)}\simeq0.05898.}
$$
This is only a <likelihood ratio> of about $1.145$ against $\theta=1/2$, hence very weak evidence. The population frequencies are treated as fixed plug-in estimates, as requested; uncertainty in those estimates and possible disease-model differences between controls and normal-allele chromosomes would matter in a fuller analysis.
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