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Autozygosity is homozygosity by descent: the maternal and paternal copies in one individual derive from the same ancestral copy. Unlike ordinary homozygosity, it concerns the ancestry of chromosome copies rather than only their observed states. In a pedigree with one ancestral disease-allele copy and fully penetrant recessive inheritance, every affected individual is autozygous for that copy.

Homozygosity means that the two allele states at a diploid genetic locus are the same; an individual with that property is a homozygote. Observed homozygosity does not by itself establish identity by descent.

The inbreeding coefficient is the probability that the two allele copies at a locus within an individual are identical by descent. It equals the kinship coefficient of the individual's parents. It concerns ancestral identity, rather than merely homozygosity of observed allelic states.

// Target: biology.bigb

Take “cousins” to mean first cousins with unrelated, outbred founders and unrelated spouses in the intervening generation. Interpret a homozygous chromosome as two copies identical by descent, and first use an idealization of independent, nonrecombining chromosome blocks. Ordinary sequence homozygosity is a different quantity: unrelated copies can have the same allele, so its probability also requires allele frequencies. Similarly, identity of an entire recombining chromosome requires a genetic map, not just a pedigree.

At a single autosomal locus, the probability of autozygosity in their child is the parents' kinship coefficient. Each of the two common grandparents contributes a path of length two from each parent, with probability $(1/2{)}^{2+2+1}$. Thus the first-cousin inbreeding coefficient isIf there are $L$ independent autosomal blocks, the probability that none is autozygous is $(1-F{)}^L$. Hence the usual 23-block idealization givesThe exponent describes the stated independent-block approximation; it is not a claim that all 23 human chromosome pairs inherit like autosomes.

Humans have 22 autosomal pairs. A son has one X and one Y and cannot have two homologous X copies. A daughter can have an autozygous X, with probability $F_X$ determined by the sex-linked cousin kinship calculation in part (b). Thus a sex-aware version iswhere $p_D$ is the probability of a daughter, or is one or zero for a specified daughter's or son's sex. The independent assortment approximation applies between the autosomes and the X. Opposite-sex parentage alone does not determine $F_X$: the parents may be cousins through their mothers, their fathers or one of each. Without these pedigree and chromosome-block assumptions, the wording does not specify a unique probability of an entire homozygous human chromosome.

Let $E_q$ be the event that all eight selected subjects carry two copies of the same specified ancestral copy $q$, and that no other generation-4 subject is autozygous at this genetic locus. The selected subjects occur in every sibship, so this event forces both generation-2 transmissions and all six relevant generation-3 transmissions. Applying the preceding conditional probabilities,Once all six parents carry $q$, their other copies come from the two unrelated outside pedigree founders, one on each side of the pedigree. A child not receiving two $q$ copies therefore cannot become autozygous for a different founding copy. This verifies the “only” condition, rather than merely excluding homozygosity for $q$.

For the event $E$ without specifying which of the original four copies is shared, the four alternatives $E_q$ are disjoint: a selected child cannot simultaneously have both its copies descended from two different ancestral copies. HenceAll selected individuals are then autozygous for the same copy and mutually 2-IBD. The distinction from the $3/4$ probability in the alternative interpretation of part (a) is precisely this disjointness.

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.