= Solution
Distinguish the disease <alleles> $D$ (normal) and $d$ (recessive disease) from the marker <alleles> $A,a$. The affected mother and two affected children are $dd$. The unaffected father must be $Dd$, since he has affected children; the unaffected child is $Dd$, since the mother always transmits $d$. The mother is $AA$ at the marker, so the marker <genotypes> identify the father's transmissions to the three children as $a,A,A$. Their disease transmissions from the father are $d,d,D$.
The father's unphased double <heterozygosity> has two possible pairs of <haplotypes>:
$$
s_1:AD/ad,\qquad s_2:Ad/aD.
$$
Under $s_1$, the three paternal transmissions $(ad,Ad,AD)$ are respectively nonrecombinant, recombinant and nonrecombinant. Under $s_2$, they are recombinant, nonrecombinant and recombinant. Each particular gamete has <probability> $(1-\theta)/2$ or $\theta/2$, giving the conditional <pedigree likelihoods>
$$
L_1(\theta)=\frac{\theta(1-\theta)^2}{8},\qquad L_2(\theta)=\frac{\theta^2(1-\theta)}8.
$$
Assuming linkage equilibrium among founder <haplotypes>, the two phases have equal prior weights conditional on the observed paternal <genotype>. <Phase averaging in a linkage likelihood> therefore gives
$$
\boxed{L(\theta)=\tfrac12L_1(\theta)+\tfrac12L_2(\theta)=\frac{\theta(1-\theta)}{16},\qquad0\le\theta\le\tfrac12.}
$$
Any factors for the parental <genotypes> omitted by this conditioning are constant in $\theta$. The derivative is $(1-2\theta)/16$, so \b[the maximum likelihood estimate is $\widehat\theta=1/2$ and there is no evidence for linkage]: the maximum <LOD score> is zero. Choosing the best phase after observing the offspring would give a different and inappropriate calculation; the unknown phase must be averaged. The linkage-equilibrium assumption is modified explicitly in part (v).
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