 Solution

Write $D$ for the disease observations and $M$ for the genetic marker observations. The LOD score for complete linkage isThe disease marginal cancels because the single-locus disease segregation law does not depend on the genetic marker–disease recombination fraction. The ten genetic markers form one nonrecombining haplotype; they do not supply ten independent copies of the pedigree's segregation evidence.

There is a distinction between identity by state and identity by descent. The reported observation that affected subjects all have $h/h$ establishes the former. An exact genetic marker likelihood requires pedigree founder haplotype frequencies, or the actual pedigree founder genotypes if conditioning on them. Here is an explicit expression using independent pedigree founder haplotypes, with frequency $p$ for the observed haplotype $h$, and linkage equilibrium between pedigree founder genetic marker and disease states. There are eight genetic marker copies in the four pedigree founders: the original couple and the two generation-2 outside spouses. For genetic marker inheritance configuration $v$, let $K(v)$ be the number of distinct pedigree founder copies ancestral to the sixteen chromosome copies in the eight selected subjects. Thenis the unlinked probability that all those copies have state $h$. At complete linkage the affected subjects all inherit the disease pedigree founder copy twice, and the probability that this copy has genetic marker state $h$ is $p$. Therefore

The denominator can be evaluated without an unspecified inheritance sum. Let $B_a=\left(\genfrac{}{}{0ex}{}{2}{a}\right)p^a(1-p{)}^{2-a}$ for $a=0,1,2$. If parental genetic marker genotypes contain $a,b$ copies of $h$, the child count distribution isFor a generation-2 couple defineFor a particular generation-3 parent with $j$ genetic marker copies, the chance it transmits $h$ to all its $r$ selected offspring is $(j/2{)}^r$, which explains each factor. Conditional on the original founding couple, its two generation-2 siblings and their independent outside spouses give independent left and right contributions. ThusExpanding givesIt satisfies $R(1)=1$, as it must for a monomorphic haplotype. This expression supplies a numerical LOD once $p$ is specified; the problem does not specify that frequency.

In the idealized limit where a matching rare haplotype identifies one ancestral copy, the leading coefficient is particularly simple. Dropping restrictions on the eight unaffected subjects, the probability that all selected subjects are autozygous for one common copy isAccordingly $R(p)/p\to 2^{-22}$ as $p\to 0$, and the affected-only IBD calculation gives $Z(0)=22{\log }_{10}2\simeq 6.623$.

If the genetic marker observations additionally establish the exact pattern $E$ of part (d), including that the eight unaffected subjects are not autozygous, its idealized IBD likelihood ratio is insteadThis latter value requires the additional negative genetic marker observations. It cannot be inferred merely from homozygosity in the affected subjects, and neither simplified IBD score is an exact identity-by-state likelihood for an unspecified common haplotype.