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For known parental haplotypes, count $r_1$ recombinant transmissions among $m_1$ informative meioses. The conditional pedigree likelihood has the formwhere $C_1$ contains parameter-independent gamete and genotype factors. Differentiating its logarithm gives $r_1/\theta -(m_1-r_1)/(1-\theta )$, so the constrained maximum-likelihood estimate isWhen $r_1=0$ the maximum is at zero; if no meiosis is informative, the likelihood is constant and the parameter is not estimable from that family. This formula presumes the family phase has been resolved; otherwise use phase averaging in a linkage likelihood first.

The family-specific value cannot be calculated because its pedigree, phases and transmission counts are absent from the supplied PDF. The missing data cannot be reconstructed from the requested estimate alone.

Let $s$ index compatible parental phases in family 2, with fixed prior conditional weights ${\pi }_s$. For a given phase, let $r_s$ be the recombinant count and $m_s$ the informative-transmission count, absorbing any fixed genotype factors into $C_s$. The appropriate pedigree likelihood isThis is phase averaging in a linkage likelihood; maximizing over a phase after observing the offspring would not be the same likelihood. In the common special case of two equally likely phases that exchange recombinant and nonrecombinant labels across $m_2$ transmissions, it reduces, up to a constant, toThe constrained maximum-likelihood estimate compares all stationary points and endpoints on $[0,1/2]$.

The actual phase set, weights, counts and numerical maximum for family 2 cannot be determined without its missing pedigree. The special two-phase expression is a conditional example, not an assertion about this unidentified family.

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: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 likelihoodsAssuming 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 givesAny factors for the parental genotypes omitted by this conditioning are constant in $\theta$. The derivative is $(1-2\theta )/16$, so the maximum likelihood estimate is $\hat{\theta }=1/2$ 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).