Marginalizing Unobserved Genotypes in Pedigree Likelihoods

Why Unobserved Genotypes Must Be Included

In a pedigree, a relative’s phenotype may be recorded even when their genotype is not. For example, we may know whether a child is affected but have no genetic test result for that child. The child still contributes information: each possible genotype implies a different probability of the observed phenotype, and each genotype has a probability determined by inheritance. A likelihood must account for all compatible possibilities, rather than treating the missing genotype as though it were known.

The key operation is marginalization: calculate the probability of the observed data for each possible genotype configuration, then add those probabilities. Each configuration is weighted by its probability under the pedigree’s inheritance model and by the conditional probability of the observed phenotypes.

Definition: A genotype configuration is an assignment of a genotype to every member of a pedigree. A genotype is observed if it is available from the data and latent if it is not observed. The pedigree likelihood for observed genotype and phenotype data is the probability of those data under a specified genetic and penetrance model, with probabilities for latent genotypes summed over all compatible configurations.

Building the Likelihood

For a simple autosomal locus with alleles \(A\) and \(a\), write \(G_i\in\{AA,Aa,aa\}\) for individual \(i\)’s genotype. Let \(Y_i=1\) indicate that the individual is affected and \(Y_i=0\) indicate that the individual is unaffected. The penetrance for genotype \(g\) is \(f_g=P(Y_i=1\mid G_i=g)\). Thus, for either phenotype status, the conditional probability of the observation is

P(Y_i=y_i\mid G_i=g_i) =f_{g_i}^{\,y_i}(1-f_{g_i})^{1-y_i}.

This expression equals \(f_{g_i}\) for an affected individual and \(1-f_{g_i}\) for an unaffected individual. It allows for incomplete penetrance and phenocopies: neither an associated genotype nor a non-associated genotype has to determine disease status with certainty.

Let \(q_i(g_i\mid g_{m(i)},g_{f(i)})\) be the probability that nonfounder \(i\) has genotype \(g_i\), given the genotypes of their mother and father. These are the Mendelian transmission probabilities. A founder has no parents represented in the pedigree, so the model must specify a founder genotype probability \(\pi_i(g_i)\), unless that founder’s genotype is fixed as known. Assume genotypes are measured without error, and that phenotypes are conditionally independent given the genotypes.

Likelihood: Let \(\mathbf{g}\) denote a complete genotype configuration, and let \(z_i\) denote an observed genotype when one is available. With error-free genotype observations, the likelihood is

L = \sum_{\mathbf{g}} \left[ \prod_{i\in F}\pi_i(g_i) \prod_{i\in C}q_i(g_i\mid g_{m(i)},g_{f(i)}) \prod_{i}P(Y_i=y_i\mid G_i=g_i) \prod_{i:\,z_i\ \mathrm{observed}}\mathbf{1}(g_i=z_i) \right],

where \(F\) is the set of founders, \(C\) is the set of nonfounders, and \(\mathbf{1}(g_i=z_i)\) equals 1 if the configuration agrees with the observed genotype and 0 otherwise. The sum is over all genotype configurations. The indicator makes configurations inconsistent with observed genotypes contribute zero. If a founder genotype is known and fixed, use that genotype rather than summing over founder genotypes with a prior. In a concrete calculation, this often leaves a short sum over only the unobserved genotypes.

Worked Examples

Worked Example: One Untyped Child With an Affected Phenotype

Suppose a mother has genotype \(Aa\) and a father has genotype \(aa\), both known. Their child has no genotype result but is affected. Under Mendelian inheritance, what is the likelihood contribution of the child’s phenotype, conditional on the parents’ genotypes?

Assume \(f_{Aa}=0.80\) and \(f_{aa}=0.01\). The child can be \(Aa\) or \(aa\), each with probability \(1/2\): the mother transmits either allele with probability \(1/2\), while the father always transmits \(a\). The \(AA\) genotype is not possible in this mating.

For the latent configuration in which the child is \(Aa\), the joint probability of that genotype and an affected phenotype is \(\frac{1}{2}\times 0.80=0.40\). For the configuration in which the child is \(aa\), it is \(\frac{1}{2}\times 0.01=0.005\). Summing gives

P(Y=1\mid Aa,aa) = \frac{1}{2}f_{Aa}+\frac{1}{2}f_{aa} = \frac{1}{2}(0.80)+\frac{1}{2}(0.01) = 0.405.

The likelihood contribution is 0.405. It is not the penetrance for either genotype alone: the child’s genotype is unobserved, so both possible genotypes contribute, weighted by their transmission probabilities.

Worked Example: A Shared Unobserved Parent Genotype

A father is known to have genotype \(aa\). The mother’s genotype is unobserved, with specified probabilities \(P(G_m=aa)=0.60\) and \(P(G_m=Aa)=0.40\). Both children are affected, and neither child has a genotype result. Assume \(f_{aa}=0.01\) and \(f_{Aa}=0.80\). Find the likelihood of the two observed phenotypes under this model.

First condition on the mother being \(aa\). Then both parents are \(aa\), so both children must be \(aa\). Given the genotypes, the probability that both are affected is \(0.01\times0.01=0.0001\).

If the mother is \(Aa\), each child has probability \(1/2\) of being \(Aa\) and \(1/2\) of being \(aa\). For either child, the affected probability after summing over that child’s genotype is \(\frac{1}{2}(0.80)+\frac{1}{2}(0.01)=0.405\). Given the parental genotypes, the children’s transmissions and phenotypes are independent, so the probability that both are affected is \(0.405^2=0.164025\).

Now sum over the two possible maternal genotypes, weighting each result by its probability:

\begin{aligned} L &=P(G_m=aa)(0.0001)+P(G_m=Aa)(0.164025)\\ &=(0.60)(0.0001)+(0.40)(0.164025)\\ &=0.00006+0.06561\\ &=0.06567. \end{aligned}

The likelihood of the two affected phenotypes is 0.06567 under the specified genotype and penetrance probabilities. The same maternal genotype applies to both children. Therefore, it must be summed over at the level of the joint two-child data, not replaced by separate independent draws of the mother’s genotype for each child.

Worked Example: Combining an Observed Genotype With a Latent One

Two parents are both known to be \(Aa\). Their first child is observed to be \(Aa\) and affected; their second child is unaffected but untyped. Assume \(f_{Aa}=0.80\), \(f_{aa}=0.01\), and \(f_{AA}=0.90\). Find the likelihood of these observations, conditional on the parents’ genotypes.

For the first child, the probability of genotype \(Aa\) from an \(Aa\times Aa\) mating is \(1/2\). Given that genotype, the probability of being affected is \(0.80\). Thus the first child’s joint contribution is \(\frac{1}{2}\times0.80=0.40\).

The second child can be \(AA\), \(Aa\), or \(aa\), with probabilities \(1/4\), \(1/2\), and \(1/4\). Because this child is unaffected, use \(1-f_g\) for each genotype’s conditional phenotype probability:

\begin{aligned} P(Y_2=0\mid Aa,Aa) &=\frac{1}{4}(1-f_{AA}) +\frac{1}{2}(1-f_{Aa}) +\frac{1}{4}(1-f_{aa})\\ &=\frac{1}{4}(0.10)+\frac{1}{2}(0.20)+\frac{1}{4}(0.99)\\ &=0.025+0.100+0.2475\\ &=0.3725. \end{aligned}

Conditional on the known parental genotypes, the two children’s transmissions and phenotypes are independent. The likelihood of all the stated observations is therefore \(0.40\times0.3725=0.1490\). The first child’s observed genotype selects a particular genotype probability; the second child’s latent genotype requires a weighted sum.

Why the Sum Is the Correct Operation

The calculation follows directly from the law of total probability. Let \(D\) denote all observed phenotype and genotype data, and let \(\mathbf{G}\) be the complete genotype configuration. The configurations form mutually exclusive and exhaustive possibilities under the model. Therefore,

P(D)=\sum_{\mathbf{g}}P(D,\mathbf{G}=\mathbf{g}) =\sum_{\mathbf{g}}P(\mathbf{G}=\mathbf{g})P(D\mid\mathbf{G}=\mathbf{g}).

The configuration probability contains the founder genotype probabilities and the transmission probabilities for descendants. The conditional data probability contains the phenotype probabilities given genotypes and the requirement that any observed genotypes match the configuration. This is exactly the weighting-and-summing procedure in the likelihood expression.

For an observed genotype, incompatible configurations have probability zero in the likelihood because of the indicator. For an unobserved genotype, no such constraint removes its possibilities: each compatible genotype remains in the sum. The likelihood is a probability for the observed data under the model, not a probability distribution over the latent genotypes. If the aim is instead to calculate the probability of a latent genotype after observing data, that is a different quantity and requires conditioning on the observed data.

Common Pitfalls and Interpretation

A frequent error is to give an untyped relative a single genotype—often the most likely genotype—and proceed as if it had been observed. That discards uncertainty and can change the likelihood. The correction is to include every genotype compatible with the pedigree and observed genotypes, weighted by its probability under the inheritance model and by the phenotype probability for that genotype.

A second error is to treat relatives as independent after averaging over an unknown shared parental genotype. In the second example, the children’s outcomes are independent conditional on the mother’s genotype, but their marginal outcomes are generally dependent because they share that uncertain genotype. The joint likelihood must retain the shared genotype in the joint calculation until the data for both children have been accounted for.

Finally, an observed unaffected phenotype does not mean that a disease-associated genotype is impossible. With incomplete penetrance, an affected-associated genotype can be paired with an unaffected probability \(1-f_g\). Likewise, an affected phenotype can occur for a genotype with low penetrance. The conditional phenotype probabilities—not a deterministic affected/unaffected label assigned to each genotype—determine those contributions.

Key takeaway: To obtain a pedigree likelihood with unobserved genotypes, enumerate the compatible complete genotype configurations, weight each by its founder and transmission probabilities, multiply by the conditional probabilities of the observed phenotypes, and sum. Keep shared latent genotypes shared across all relatives who inherit from them.

Check Your Understanding

Use the definitions and likelihood logic above to answer these questions.

For an \(Aa\times aa\) mating, an untyped child is unaffected. If \(f_{Aa}=0.80\) and \(f_{aa}=0.01\), what is the phenotype likelihood contribution?
Why does an observed genotype \(z_i\) contribute an indicator in the likelihood when genotypes are assumed to be measured without error?
In the shared-parent example, why is it incorrect to draw a new maternal genotype independently for each child?
What part of the likelihood changes when the observed phenotype is unaffected rather than affected?
What additional model quantity is needed to sum over an unobserved founder genotype?