Two-Point Pedigree Likelihoods

Why a Pedigree Likelihood Sums Over Hidden Inheritance

A pedigree records people and their observed marker genotypes and disease phenotypes, but it usually does not reveal exactly which chromosome each person inherited from each parent. A two-point likelihood evaluates how probable those observations are under a specified recombination fraction, \(\theta\), for a disease locus and a nearby marker locus. The calculation must allow for every unobserved inheritance path and phase configuration that could have produced the data.

The key operation is a sum, not a choice of the single most plausible path. If two different inheritance paths both agree with a child’s observed marker genotype, both contribute probability. Likewise, when a parent’s phase is uncertain, the likelihood averages the data probability over the possible phases using their prior weights under the stated model.

Definition: A two-point pedigree likelihood at \(\theta\) is the probability of the observed pedigree data under a specified disease-and-marker model and recombination fraction \(\theta\). It is computed by summing over unobserved inheritance paths and phase configurations, weighting each configuration by its probability and by the probability of the observations given that configuration. If parental genotypes are treated as fixed, this is a likelihood conditional on those genotypes.

Transmission Probabilities and the Likelihood Sum

At the disease locus, write the alleles as \(D\) and \(d\); at the marker locus, write them as \(M\) and \(m\). A haplotype is the pair of alleles on one chromosome, such as \(D\)-\(M\). The phase specifies which disease and marker alleles are paired on each of a person’s two chromosomes. For a parent with haplotypes \(D\)-\(M\) and \(d\)-\(m\), the phase is called coupling; the same genotypes could instead have haplotypes \(D\)-\(m\) and \(d\)-\(M\).

For a heterozygous parent, a gamete is nonrecombinant if it carries one of the parental haplotypes and recombinant if it carries a new pairing of the disease and marker alleles. Under the standard two-locus model, \(0\leq\theta\leq 1/2\). Each of the two nonrecombinant gametes has probability \((1-\theta)/2\), and each of the two recombinant gametes has probability \(\theta/2\). Thus, for a parent with coupling phase \(D\)-\(M/d\)-\(m\), the four gamete probabilities are:

P(D\text{-}M)=P(d\text{-}m)=\frac{1-\theta}{2}, \qquad P(D\text{-}m)=P(d\text{-}M)=\frac{\theta}{2}.

The likelihood combines these transmission probabilities with the observation model. For example, the transmitted marker alleles determine the child’s marker genotype, while the transmitted disease alleles determine the child’s disease genotype. The specified disease model then assigns a probability to the observed phenotype given that genotype. In examples below, a fully penetrant dominant model with no phenocopies is used: a child is affected if and only if the child has at least one \(D\) allele.

Let \(\phi\) denote a phase configuration, \(z\) a complete inheritance path (the gametes transmitted in all relevant meioses), and \(O\) the observed marker genotypes and phenotypes. If phase is uncertain, let \(w(\phi)\) be its specified probability, conditional on the genotypes on which the analysis is conditioned. Then the likelihood has the form:

L(\theta) = \sum_{\phi} w(\phi) \sum_{z} P(z\mid \phi,\theta)\,P(O\mid z,\phi).

In this expression, paths inconsistent with an observed marker genotype have \(P(O\mid z,\phi)=0\) when marker genotypes are assumed error-free. The phenotype contribution is the model probability of the observed phenotype given the disease genotype implied by the path. If the phenotype observations are conditionally independent given disease genotypes, their joint contribution is the product of their individual probabilities. If the founder genotypes are not conditioned on, their probabilities under the specified population model must also be included in the sum.

Why the Sum Is the Correct Calculation

The sum follows directly from the law of total probability. The possible complete paths and phases form mutually exclusive alternatives: in the realized pedigree, one particular configuration occurred, even if it is not observed. For any fixed \(\theta\), partition the event “the observed data are \(O\)” according to those alternatives. The probability of \(O\) is the sum of the probability of each alternative multiplied by the probability of \(O\) conditional on it.

\begin{aligned} P(O\mid\theta) &=\sum_{\phi,z}P(O,\phi,z\mid\theta)\\ &=\sum_{\phi,z}P(\phi\mid\theta)\, P(z\mid\phi,\theta)\, P(O\mid z,\phi,\theta). \end{aligned}

When phase weights are specified conditional on observed parental genotypes and do not otherwise depend on \(\theta\), \(P(\phi\mid\theta)\) is represented by \(w(\phi)\). The last conditional probability is the observation model, including marker-genotype compatibility and disease-phenotype probabilities. This derivation also explains why an inheritance path that is not observed cannot simply be discarded: its probability contributes to the total probability of the data.

Worked Example: One Child With a Known Coupling Phase

Suppose an affected parent has haplotypes \(D\)-\(M/d\)-\(m\), and the other parent is \(d\)-\(m/d\)-\(m\). The parents’ genotypes and the first parent’s phase are treated as known. Their child is observed to be \(M/m\) at the marker and affected. Assume complete dominant penetrance with no phenocopies. Find the conditional likelihood contribution for this child at \(\theta=0.10\).

The \(d\)-\(m/d\)-\(m\) parent must transmit \(d\)-\(m\). For the child to be \(M/m\), the heterozygous parent must transmit a gamete carrying \(M\). There are two possible \(M\)-bearing gametes: \(D\)-\(M\) and \(d\)-\(M\). But the child is affected under the stated model only if the first parent transmits \(D\). Therefore only the \(D\)-\(M\) path is compatible with both observations.

Given the known coupling phase, \(D\)-\(M\) is nonrecombinant and has probability \((1-\theta)/2\). The other parent’s transmission is fixed, and the phenotype is certain given the resulting \(D/d\) disease genotype. Thus:

L(\theta)=\frac{1-\theta}{2}, \qquad L(0.10)=\frac{1-0.10}{2} =\frac{0.90}{2} =0.4500.

The value \(0.4500\) is the probability of this child’s marker-genotype and phenotype observations conditional on the stated parental genotypes and phase, under the model at \(\theta=0.10\). It is not the probability that the phase or the model is true.

Worked Example: Two Siblings With an Uncertain Shared Phase

A disease-heterozygous parent and a noncarrier \(d\)-\(m/d\)-\(m\) parent have two children. Both children are \(M/m\) at the marker; the first is affected and the second is unaffected. The disease-heterozygous parent’s phase is unknown, with coupling and repulsion each assigned probability \(1/2\). Assume complete dominant penetrance with no phenocopies, and condition on the parental genotypes. Calculate the likelihood at \(\theta=0.10\).

Under coupling phase \(D\)-\(M/d\)-\(m\), the affected child must receive \(D\)-\(M\), a nonrecombinant gamete with probability \((1-\theta)/2\). The unaffected child must receive \(d\)-\(M\), a recombinant gamete with probability \(\theta/2\). Because the two transmissions are independent conditional on this shared phase, their joint probability is the product.

P(O\mid\text{coupling},\theta) = \frac{1-\theta}{2}\frac{\theta}{2} = \frac{\theta(1-\theta)}{4}.

Under repulsion phase \(D\)-\(m/d\)-\(M\), the affected child’s \(D\)-\(M\) gamete is recombinant, with probability \(\theta/2\). The unaffected child’s \(d\)-\(M\) gamete is nonrecombinant, with probability \((1-\theta)/2\). The joint probability is again \(\theta(1-\theta)/4\). Average the two phase-specific joint probabilities using their weights:

\begin{aligned} L(\theta) &=\frac{1}{2}\left(\frac{\theta}{2}\frac{1-\theta}{2}\right) +\frac{1}{2}\left(\frac{1-\theta}{2}\frac{\theta}{2}\right)\\ &=\frac{\theta(1-\theta)}{4},\\ L(0.10) &=\frac{0.10(1-0.10)}{4} =\frac{0.10(0.90)}{4} =\frac{0.0900}{4} =0.02250. \end{aligned}

The likelihood is \(0.02250\), conditional on the parental genotypes and the stipulated phase weights. The same unknown parental phase applies to both children, so the phase-specific probabilities for the two transmissions are multiplied first and then averaged over phase.

Worked Example: Summing Over Two Marker-Allele Origins

A mother has known coupling phase \(D\)-\(M/d\)-\(m\). The father is \(d/d\) at the disease locus and \(M/m\) at the marker. Their child is affected and has marker genotype \(M/m\). Assume the same complete dominant model with no phenocopies. Condition on the parental genotypes and phase. Calculate the child’s likelihood contribution at any \(\theta\).

The child must receive \(D\) from the mother to be affected, and must receive \(d\) from the father. There are two inheritance paths that also produce marker genotype \(M/m\). In the first, the mother transmits \(D\)-\(M\) and the father transmits \(m\). In the second, the mother transmits \(D\)-\(m\) and the father transmits \(M\). The marker genotype alone does not reveal which parent supplied the \(M\) allele, so both paths contribute.

The father’s marker transmission is \(M\) or \(m\) with probability \(1/2\) each. The maternal \(D\)-\(M\) gamete has probability \((1-\theta)/2\); the maternal \(D\)-\(m\) gamete has probability \(\theta/2\). Summing the two compatible paths gives:

\begin{aligned} L(\theta) &=\left(\frac{1-\theta}{2}\right)\left(\frac{1}{2}\right) +\left(\frac{\theta}{2}\right)\left(\frac{1}{2}\right)\\ &=\frac{1-\theta+\theta}{4} =\frac{1}{4}. \end{aligned}

For example, at \(\theta=0.10\), the two path probabilities are \(0.90/2\times 1/2=0.2250\) and \(0.10/2\times 1/2=0.02500\); their sum is \(0.2500\). This calculation illustrates why the observed marker genotype must be matched against the alleles transmitted by both parents, rather than assigned to one parent without evidence.

Practical Construction and a Common Pitfall

For a small pedigree, a useful calculation proceeds in a consistent order:

State what is conditioned on.
Specify which parental genotypes and phases are fixed, which phases are uncertain, and the assumed disease-phenotype model.

List compatible latent configurations.
For each phase, enumerate the inheritance paths that can produce the observed marker genotypes. Include every compatible path, not just the most obvious one.

Assign each path a probability.
Use \((1-\theta)/2\) for each nonrecombinant gamete and \(\theta/2\) for each recombinant gamete in a heterozygous parent, and multiply transmission probabilities across meioses conditional on the phase.

Apply the observation model and sum.
Retain paths compatible with the marker data, multiply by the corresponding phenotype probabilities, sum paths within phase, and then weight and sum over uncertain phases.

A common error is to average the possible phase-specific transmission probability for each child separately and then multiply those averages. In the sibling example at \(\theta=0.10\), the probability of the affected child’s \(M/m\) observation averaged over phase is \((0.4500+0.05000)/2=0.2500\). The unaffected child’s corresponding average is also \(0.2500\). Multiplying gives \(0.06250\), not the correct joint likelihood \(0.02250\). The two children share one parental phase; their observations are therefore not independent after phase has been averaged out. The correct calculation multiplies within each shared phase and only then averages across phases.

Another mistake is to treat a likelihood at \(\theta\) as a probability that \(\theta\) is correct. The likelihood is the probability of the observed data under a specified value of \(\theta\) and the stated model. Comparing likelihoods across values is useful for inference, but a likelihood value by itself is not a posterior probability or a measure of the probability that the model is true. Similarly, if the analysis is conditional on parental genotypes, those genotypes should not be multiplied into the likelihood a second time; if they are not conditioned on, their probabilities must be incorporated explicitly.

Key takeaway: A two-point pedigree likelihood at a specified \(\theta\) is obtained by summing the probabilities of all inheritance paths and phase configurations compatible with the observed marker genotypes and phenotypes. Keep shared phase configurations intact across relatives, and state clearly which genotypes are conditioned on.

Check Your Understanding

Use the stated model assumptions in each question; where a numerical value is needed, assume error-free marker genotypes.

For a parent with haplotypes \(D\)-\(M/d\)-\(m\), what is the probability of transmitting \(d\)-\(M\) at recombination fraction \(\theta\), and is that gamete recombinant?
A child’s observed marker genotype is compatible with two different parental transmission paths. Should the likelihood use only the more probable path, or sum both? Explain.
In the two-sibling example, why must the likelihood be averaged over the shared phase after multiplying the two transmission probabilities within each phase?
At \(\theta=0.10\), verify that \(\theta(1-\theta)/4=0.02250\).
What additional probability terms may be needed if parental genotypes are not conditioned on in the pedigree likelihood?

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