= Solution
Write $D$ for the disease observations and $M$ for the <genetic marker> observations. The <LOD score> for <complete linkage> is
$$
\boxed{Z(0)=\log_{10}\frac{P(M\mid D,\theta=0)}{P(M\mid D,\theta=1/2)}.}
$$
The 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. Then
$$
R(p)=\sum_v P(v)p^{K(v)}=E[p^K]
$$
is 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
$$
\boxed{Z(0)=\log_{10}\frac{p}{R(p)}.}
$$
The denominator can be evaluated without an unspecified inheritance sum. Let $B_a=\binom2a p^a(1-p)^{2-a}$ for $a=0,1,2$. If parental <genetic marker> <genotypes> contain $a,b$ copies of $h$, the child count distribution is
$$
Q_0(a,b)=(1-a/2)(1-b/2),\quad
Q_1(a,b)=(a/2)(1-b/2)+(1-a/2)(b/2),\quad
Q_2(a,b)=ab/4.
$$
For a generation-2 couple define
$$
F(a,b)=\prod_{r\in\{1,4,3\}}\left[\sum_{j=0}^2Q_j(a,b)(j/2)^r\right],\qquad
m_a=\sum_{b=0}^2B_bF(a,b).
$$
For 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. Thus
$$
R(p)=\sum_{u,v=0}^2B_uB_v\left[\sum_{a=0}^2Q_a(u,v)m_a\right]^2.
$$
Expanding gives
$$
R(p)=\frac{p}{2^{22}}\left(1+388p+38942p^2+480421p^3+1398448p^4+1540264p^5+650576p^6+85264p^7\right).
$$
It 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 is
$$
P(H)=4\cdot\frac14\cdot\frac1{64}\cdot\left(\frac14\right)^8=2^{-22}.
$$
Accordingly $R(p)/p\to2^{-22}$ as $p\to0$, and the affected-only <IBD> calculation gives $Z(0)=22\log_{10}2\simeq6.623$.
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 instead
$$
Z_E(0)=-\log_{10}P(E)=38\log_{10}2-8\log_{10}3\simeq7.622.
$$
This 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>.
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