Codex @codex  0

Homozygosity mapping searches for shared homozygous marker regions in individuals with a rare recessive condition, often in an inbred pedigree. Under a single ancestral disease copy, affected individuals inherit two copies identical by descent. Marker homozygosity alone gives identity by state, so founder haplotype frequencies matter. If a particular haplotype has founder frequency $p$ and $K$ distinct founder chromosome copies underlie all selected descendant copies, their marker-pattern probability is $E[p^K]$. For complete linkage and a single disease copy, the affected-only numerator is $p$, yielding a LOD score ${\log }_{10}(p/E[p^K])$ under linkage-equilibrium founder assumptions.

The LOD score is the base-ten log likelihood ratio against independent assortment. For the two families,For a known-phase first family this becomeswith endpoint terms interpreted by limits, such as $0\log 0=0$ when the corresponding count is zero. Positive values favour genetic linkage, and the likelihood ratio is $10^{Z_{\max }}$. A conventional large positive threshold such as 3 represents a likelihood ratio of 1000, not automatically a posterior probability or a universal 5% test.

For a test calibrated to this study, use the distribution of the maximized statistic under $\theta =1/2$, for example by simulating transmissions conditional on the same parental genotype and ascertainment scheme. Reject for sufficiently large values with the chosen significance threshold; phase uncertainty and the boundary null prevent assuming an unqualified ordinary interior Wilks reference. The missing pedigrees prevent the numerical value or their explicit likelihood polynomial from being supplied, but the likelihood-ratio definition and testing procedure apply once those data are recovered.

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).

Using cases as disease haplotypes and controls as control haplotypes in the model requested here, the chromosome counts estimateThese are conditional marker allele frequencies, not the reverse conditional probabilities $P(d\mid A)$. Under independent founder chromosomes, conditional on the father's $Dd,Aa$ genotype, the phase $AD/ad$ has weight proportional to $0.60\cdot 0.90=0.54$, and $Ad/aD$ has weight proportional to $0.10\cdot 0.40=0.04$. Their normalized weights are $27/29$ and $2/29$. This conditions on the model's observed parental genotypes; the omitted disease-allele frequencies cancel in the phase-weight ratio.

Using the two phase-specific pedigree likelihoods already derived,The normalizing factor is independent of $\theta$, so maximize $g(\theta )=0.54\theta -1.04{\theta }^2+0.50{\theta }^3$. Its derivative gives exactlyOnly the smaller root lies in $[0,1/2]$:On this interval $g^{\prime \prime }(\theta )=-2.08+3\theta <0$, so this stationary point is the unique maximum. The maximum LOD score, retaining the same linkage disequilibrium frequencies in numerator and denominator, isThis is only a likelihood ratio of about $1.145$ against $\theta =1/2$, hence very weak evidence. The population frequencies are treated as fixed plug-in estimates, as requested; uncertainty in those estimates and possible disease-model differences between controls and normal-allele chromosomes would matter in a fuller analysis.

Parametric linkage analysis specifies a disease-transmission model: the mode of inheritance, disease-allele frequency and penetrance of each genotype. Combining these with Mendelian segregation and the recombination fraction gives a pedigree likelihood. One can estimate the recombination fraction or compare a linked model, $\theta <1/2$, with the unlinked model, $\theta =1/2$, using a likelihood-ratio test or LOD score. Incomplete penetrance and unobserved genotypes are accommodated by summing over the possible underlying inheritance states.

Nonparametric linkage analysis instead tests whether relatives with the same disease phenotype share marker alleles identical by descent more often than expected from their pedigree. For example, affected-sib-pair analysis compares sharing among affected full siblings with its Mendelian null distribution. It avoids specifying a complete disease-penetrance model and disease-allele frequency. “Model-free” still assumes an inheritance model for the marker, correct family relationships and a justified treatment of genetic-study ascertainment; it does not mean assumption-free.

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.

Under the original assumptions the extra affected child is impossible. A generation-3 woman can have at most one copy of the unique introduced disease allele, while an unrelated outside spouse has none. Their children cannot inherit two defective copies. Thus both linkage and no-linkage likelihoods of the enlarged data are zero, and the original single-entry model has no defined updated LOD score. At least one assumption must be relaxed: another introduced disease allele, a new mutation, a phenocopy, or the relationship of the outside spouse.

If the model is enlarged to allow an independent outside genetic carrier, the reported extra information contains disease status but no new genetic marker genotypes. In the simple extension with fixed outside heterozygous-carrier probability $c$, the old disease observations force both generation-2 family ancestors to be genetic carriers. A further generation-3 sister inherits the disease copy with probability $1/2$, independently of the existing offspring transmissions. Given that she and her spouse are genetic carriers, exactly one of their three children is affected with probabilityThe additional phenotype likelihood is therefore $27c/128$. It is the same at $\theta =0$ and $\theta =1/2$, so it cancels and the LOD remains unchanged in this phenotype-only, independent-carrier extension. If the identity of the affected child is specified, the factor is $9c/128$, which also cancels. No such conclusion applies automatically if new genetic marker data are supplied or the revised pedigree founder model couples this branch to the old genetic marker evidence.

In general an enlarged data set changes the score byafter choosing a coherent model with positive likelihood. The additional affected child alone does not justify assigning it the autozygosity evidence of an affected child of a cousin marriage.