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A founder is an individual whose parents are outside the recorded pedigree. The standard pedigree likelihood assigns unrelated founders independent genotype distributions based on population allele or haplotype frequencies. If founders are related, their joint prior must instead account for that relationship.

Ascertainment in a genetic study specifies how individuals or families enter a genetic study. If $A$ is the event of inclusion, then $P(X\mid A)=P(X)P(A\mid X)/P(A)$. Selection on disease phenotype can change an unconditional pedigree likelihood. When comparing genetic marker probabilities conditional on a specified disease pattern, selection depending only on that pattern cancels from the linkage likelihood ratio.

Assign each factor exactly once to a containing clique and multiply the factors there into ${\psi }_i$. The displayed sum-product belief propagation message sums out variables of $C_i$ outside the separator $S_{ij}=C_i\cap C_j$. By the running intersection property, eliminating all variables in the subtree on the $i$ side creates precisely this function of the separator. An inward pass computes the normalizing constant at the root; an outward pass produces marginal distributions. For pedigree likelihoods, observed genotypes and penetrance factors simply multiply the relevant potentials.

Parametric linkage analysis forms a pedigree likelihood from a specified disease model, including penetrance, allele frequencies and recessive inheritance or another transmission model. It estimates or tests the recombination fraction between a disease locus and a genetic marker.

// Target: biology.bigb

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.

For independent families, multiply their pedigree likelihoods. If family 1 has the known-phase counts above, the combined log-likelihood derivative at an interior point isThus the score equation for a phase-averaged linkage likelihood isFor the general phase sum, let $w_s(\theta )={\pi }_s{C}_s{\theta }^{r_s}(1-\theta {)}^{m_s-r_s}/L_2(\theta )$. Since the weights sum to one, the equivalent interior equation isThe finite likelihood sum is continuous on the compact parameter interval $[0,1/2]$, so a constrained maximum exists. This does not by itself prove an interior score root: complete nonrecombinant data, for example, maximize at zero. To show the requested root lies in the appropriate range for the two particular families, one must insert their transmission counts and evaluate the score or the resulting polynomial at the relevant endpoints.

The source lacks those pedigree counts, so the data-specific equation and its asserted interior-root verification remain undetermined. No universal interior-root claim is substituted for the missing calculation.

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.

Label the full siblings at the bottom $X,Y$ and their cousin parents $F,M$. Let $F$ have parents $A,B$ and $M$ parents $C,D$, where $A,C$ are siblings with parents $U,V$. Include the other great-grandparents as $P,Q$, parents of $B$, and $R,S$, parents of $D$. This gives fourteen people, with six independent founders in a pedigree $U,V,P,Q,R,S$.

The pedigree graphical model has the following Directed acyclic graph, with both members of each parent pair pointing to every listed child:Let $f(g)$ be the founder genotype law under Hardy-Weinberg equilibrium, and let $t(g\mid h,k)$ be the Mendelian segregation kernel. The joint probability of all genotypes isFor a biallelic locus with allele-1 frequency $p$, $f(11)=p^2$, $f(12)=2p(1-p)$ and $f(22)=(1-p{)}^2$. The kernel $t$ assigns probability $1/4$ to each labelled pair of parental transmissions and adds these probabilities when they produce the same unordered genotype. Summing this Bayesian network factorization over unobserved genotypes gives their marginal distribution or a pedigree likelihood.

The nine maximal cliques of this triangulation areEvery other nonempty clique is a subset of one of these. One junction tree, listing its separators after the edges, isEach variable appears in a connected subtree, verifying the running intersection property. For example, the cliques containing $M$ are $K_1,K_2,K_4,K_5,K_6$, and their connecting paths contain $M$ throughout.

The junction tree algorithm assigns every founder factor and every child-with-parents factor to one containing clique, and multiplies any observation or penetrance factors there. The junction-tree sum-product message from clique $i$ to clique $j$ sums out $C_i\setminus (C_i\cap C_j)$ after multiplying the local potential and all incoming messages except $j$'s. An inward pass sums the joint probability or pedigree likelihood; a subsequent outward pass gives each person's posterior genotype probabilities. This is the same variable-elimination principle as the Elston-Stewart algorithm, organized to handle the pedigree loop. Its largest clique has four genotype variables, much smaller than summing naively over fourteen at once.

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.

Let $a$ denote the disease allele. Complete penetrance with recessive inheritance makes each affected child $aa$. The unaffected parents must therefore both be $Aa$ genetic carriers. Condition on these parental genotypes and the fully informative marker; the distinct paternal and maternal marker copies identify which parent's copies are shared. Average over the two equally likely marker-disease haplotype phases in each parent. This is the phase-symmetric model implicit in the requested pedigree likelihood identities.

For one parent, suppose first that both children receive the same marker copy. If that marker copy is on the parental chromosome carrying $a$, both children receive $a$ with probability $(1-\theta {)}^2$; in the opposite phase the probability is ${\theta }^2$. The phase-averaged probability isIf the children receive different marker copies, one disease-allele transmission is recombinant and the other is nonrecombinant in either phase. The probability that both receive $a$ is thereforePaternal and maternal transmissions are independent. Sharing zero marker copies means the different-copy case for both parents; sharing two means the same-copy case for both; sharing one means one of each. Thus the recessive affected-sib-pair likelihood isHere $L_j$ is the probability that both children are affected conditional on the marker sharing state $J=j$, not the probability of $J=j$ conditional on their being affected. Since $u,v\ge 0$,At $\theta =1/2$, all three conditional disease probabilities equal $1/16$, as expected from two independent children of an $Aa\times Aa$ mating. At $\theta =0$, only the two-copy-sharing state can produce two affected children. Phase weights other than $1/2$ would require a different likelihood function and need not obey these identities.

Treat $G_i$ as the entire phase-known genotype of individual $i$ at the genetic loci being studied, and let $\overline{F}$ denote the nonfounders. Assume unrelated pedigree founders have independent genotype priors, each nonfounder's genotype depends only on its parents' genotypes, and the observed phenotypes are conditionally independent given the genotypes. These are the standard assumptions behind the pedigree likelihood; shared environmental effects or related pedigree founders would require additional factors.

The directed ancestry graph is acyclic, even when there is inbreeding. Order individuals with parents preceding offspring and apply the chain rule for probabilities. The parental conditional independence assumption givesConditional phenotype independence gives $P(X_1,\dots ,X_n\mid G_1,\dots ,G_n)={\prod }_{i}P(X_i\mid G_i)$. Multiplying these two expressions and summing over every possible latent genotype configuration provesThis is the law of total probability, not a claim that relatives have independent unconditional genotypes.

An unphased multilocus genotype can have several pairs of haplotypes. Its pedigree likelihood must average over these possible phases with prior weights ${\pi }_s$, rather than maximize over phases after observing the offspring. Under linkage disequilibrium, the weights need not be equal. They can be obtained by conditioning population haplotype frequencies on the observed parental genotype; different choices of what is conditioned on change only parameter-independent factors when those frequencies are held fixed.

A phenotype is an observable characteristic of an organism, arising from its genotype, environment and other influences. Seed shape is a phenotype; $Aa$ is a genotype. The distinction prevents appearance from being mistaken for a complete inherited description. See NHGRI: phenotype.

For a specified binary phenotype $T$, let its penetrance in genotype $g$ be $r_g=P(T\mid g)$. The law of total probability gives $P(T)={\sum }_g{r}_{g}P(g)$. Thus the ideal $Aa\times Aa$ example has $P(T)=r_{AA}/4+r_{Aa}/2+r_{aa}/4$: the familiar $3/4$ follows only when these penetrances are $1,1,0$. A pedigree likelihood keeps inheritance and the conditional phenotype model as separate factors.