Likelihood Contributions From Affected and Unaffected Relatives

From Genotypes to Family Disease-Status Contributions

A pedigree can contain both affected and unaffected relatives, and both kinds of observations matter. Under a penetrance model, an affected relative with a genotype that commonly leads to disease contributes a relatively large probability term. An unaffected relative with that same genotype contributes the complementary probability, which may be small. The likelihood uses these terms together with the probability of the family’s genotype configuration under the inheritance model.

The central distinction is between the disease-status probability given a genotype and the probability of that genotype configuration in the pedigree. The former comes from the penetrance and phenocopy model. The latter comes from the parents’ genotypes and the specified transmission model. Combining them gives a contribution for a particular genotype configuration; it does not, by itself, sum over alternative configurations.

Definition: Let \(D_i=1\) indicate that relative \(i\) is affected and \(D_i=0\) indicate that the relative is unaffected. Let \(G_i\) be that relative’s genotype, and let \(f_g=P(D_i=1\mid G_i=g)\) be the penetrance for genotype \(g\). The disease-status likelihood contribution is \(f_g\) for an affected relative and \(1-f_g\) for an unaffected relative.

For a binary disease status, the two cases can be written together as

P(D_i=d_i\mid G_i=g,\theta) =f_g^{d_i}(1-f_g)^{1-d_i}, \qquad d_i\in\{0,1\},

where \(\theta\) denotes the disease-model parameters, including the penetrances. If the person is affected, \(d_i=1\), so the expression reduces to \(f_g\). If the person is unaffected, \(d_i=0\), so it reduces to \(1-f_g\). This is a probability of the observed status conditional on genotype and model parameters; it is not the probability that the genotype or the model is true.

Combining Relatives’ Statuses

Suppose the pedigree contains \(m\) relatives with disease statuses recorded. The usual penetrance-model calculation assumes that, conditional on the relatives’ genotypes and the model parameters, their disease statuses are independent. This assumption means that any dependence in their modeled disease risks is accounted for through their genotypes and shared parameters; there is no additional residual dependence between statuses in this likelihood.

Under that assumption, the probability of all observed statuses for a specified genotype configuration is the product of the individual contributions:

P(\mathbf{D}=\mathbf{d}\mid \mathbf{G}=\mathbf{g},\theta) =\prod_{i=1}^{m} f_{g_i}^{d_i}(1-f_{g_i})^{1-d_i}.

The product includes every relative whose disease status is part of the data, not only affected relatives. An unaffected relative contributes \(1-f_{g_i}\); omitting unaffected relatives would discard observed information and produce a different likelihood.

Worked Example: One Affected and One Unaffected Person

Consider a hypothetical single-locus model with genotypes \(aa\), \(Aa\), and \(AA\), where \(A\) is the risk allele. Suppose the penetrances are \(f_{aa}=0.02\), \(f_{Aa}=0.60\), and \(f_{AA}=0.80\). A relative with genotype \(Aa\) is affected, and a second relative with genotype \(Aa\) is unaffected. Calculate the disease-status likelihood conditional on these genotypes.

For the affected relative, \(d=1\), so the contribution is \(f_{Aa}=0.60\). For the unaffected relative, \(d=0\), so the contribution is \(1-f_{Aa}=1-0.60=0.40\).

Given the stated conditional-independence assumption, the joint status contribution is

0.60\times 0.40=0.24.

Thus the probability of this particular affected/unaffected status pattern, conditional on both genotypes being \(Aa\), is \(0.24\) under the specified model. This calculation conditions on the genotypes; it does not include a probability for how likely those genotypes are in a pedigree.

Adding the Pedigree Genotype Probability

For a particular configuration of genotypes, a pedigree model supplies a probability for that configuration. For example, with no mutation at the locus, a parent with genotype \(Aa\) transmits either allele with probability \(1/2\). The probability of a child’s genotype follows from the alleles transmitted by both parents. Conditional on the parental genotypes, transmissions to different children are modeled as independent Mendelian transmissions.

Write \(Q(\mathbf{g})\) for the probability of a specified pedigree genotype configuration \(\mathbf{g}\) under the transmission model. The joint contribution of that configuration and the observed disease statuses is

Q(\mathbf{g})\, P(\mathbf{D}=\mathbf{d}\mid\mathbf{G}=\mathbf{g},\theta) = Q(\mathbf{g}) \prod_{i=1}^{m} f_{g_i}^{d_i}(1-f_{g_i})^{1-d_i}.

This expression separates inheritance from disease expression: \(Q(\mathbf{g})\) describes how the genotype configuration is generated, while the product describes how likely the observed statuses are if that configuration is present. If all genotypes are observed and the analysis is explicitly conditional on them, use the phenotype likelihood alone. If the contribution is for a specified configuration in a pedigree model, include its genotype probability.

Worked Example: A Child of a Heterozygous and a Noncarrier Parent

Suppose, hypothetically, that one parent is \(Aa\) and the other is \(aa\). Their child is affected, and the penetrances are \(f_{aa}=0.02\) and \(f_{Aa}=0.60\). Calculate the joint contribution for each of two specified child-genotype configurations. Assume Mendelian transmission with no mutation.

The \(Aa\) parent transmits \(A\) with probability \(1/2\) and \(a\) with probability \(1/2\); the \(aa\) parent always transmits \(a\). Therefore, the child has genotype \(Aa\) with probability \(1/2\) and \(aa\) with probability \(1/2\).

Configuration 1: child is \(Aa\) and affected. Its genotype probability is \(1/2\), and the affected-status contribution is \(f_{Aa}=0.60\). The joint contribution is

\frac{1}{2}\times 0.60=0.30.

Configuration 2: child is \(aa\) and affected. Its genotype probability is also \(1/2\), but the affected-status contribution is \(f_{aa}=0.02\). The joint contribution is

\frac{1}{2}\times 0.02=0.010.

The affected child’s status is more compatible with the specified \(Aa\) configuration under these penetrances. These are separate contributions for two specified configurations. The calculation has not combined them into a likelihood for the observed status when the child’s genotype is unknown.

Worked Pedigree Calculation

A useful way to check a family calculation is to list each person’s genotype probability and status contribution separately before multiplying. This makes it easier to detect a missing unaffected relative or a mistaken complement. In the next example, the parental genotypes are treated as specified, and the offspring genotypes are generated by Mendelian transmission.

Worked Example: A Couple and Three Children

Consider a hypothetical family with two \(Aa\) parents and three children. The children’s specified genotypes are \(Aa\), \(aa\), and \(AA\), respectively. The first and third children are affected; the second is unaffected. Both parents are unaffected. Use \(f_{aa}=0.02\), \(f_{Aa}=0.60\), and \(f_{AA}=0.80\) to calculate the joint contribution for this genotype configuration and these statuses. Assume no mutation, Mendelian transmission, independent transmissions to different children conditional on parental genotypes, and conditional independence of disease statuses given genotypes.

For two \(Aa\) parents, each child has genotype probabilities \(P(aa)=1/4\), \(P(Aa)=1/2\), and \(P(AA)=1/4\). Thus the probability of the specified ordered configuration for the three children is

Q(\mathbf{g})= \frac{1}{2}\times\frac{1}{4}\times\frac{1}{4} =\frac{1}{32}=0.03125.

Now calculate the disease-status contribution for each family member. Each unaffected \(Aa\) parent contributes \(1-f_{Aa}=1-0.60=0.40\), so the parents together contribute \(0.40\times0.40=0.16\). The affected \(Aa\) child contributes \(0.60\). The unaffected \(aa\) child contributes \(1-0.02=0.98\). The affected \(AA\) child contributes \(0.80\). The full status contribution is

0.16\times0.60\times0.98\times0.80=0.075264.

Multiplying by the probability of the specified child-genotype configuration gives

\frac{1}{32}\times0.075264=0.002352.

The value \(0.002352\) is the joint probability contribution for this specified configuration and observed status pattern under the stated assumptions. It includes both parents’ status information and the three children’s status information. It is not a parameter estimate or a p-value.

Why Unaffected Relatives Matter

An unaffected relative can make a genotype configuration more or less compatible with the observed family data. If a genotype has high penetrance, an unaffected person with that genotype contributes a relatively small term. For example, an unaffected \(AA\) relative under the values above contributes \(1-0.80=0.20\), whereas an unaffected \(aa\) relative contributes \(1-0.02=0.98\). Neither term is zero, because penetrance is not certainty and phenocopies are allowed by the model.

Likewise, an affected relative with a low-penetrance genotype still contributes a nonzero probability when the model permits disease in that genotype. This is why the status terms should be calculated from the specified penetrances rather than replaced by deterministic rules such as “risk genotype means affected.” The model describes probabilities, not guaranteed outcomes.

A Short Justification of the Product

The multiplication rule follows directly from the assumed conditional independence of the disease statuses. For fixed genotypes \(\mathbf{g}\), the probability of all statuses can be factored into successive conditional probabilities:

P(D_1=d_1,\ldots,D_m=d_m\mid\mathbf{G}=\mathbf{g},\theta) = \prod_{i=1}^{m} P(D_i=d_i\mid D_1=d_1,\ldots,D_{i-1}=d_{i-1}, \mathbf{G}=\mathbf{g},\theta).

Conditional independence says that each factor on the right is unchanged by the other relatives’ statuses once the genotypes and parameters are given. Therefore,

P(D_i=d_i\mid D_1=d_1,\ldots,D_{i-1}=d_{i-1}, \mathbf{G}=\mathbf{g},\theta) = P(D_i=d_i\mid G_i=g_i,\theta) = f_{g_i}^{d_i}(1-f_{g_i})^{1-d_i}.

Substituting these factors gives the product of the individual disease-status contributions. The joint contribution for a specified genotype configuration then follows from the multiplication rule: the probability of the configuration times the conditional probability of the observed statuses given that configuration.

Common Mistakes and a Practical Check

Using penetrance for an unaffected person. The wrong reasoning is that every relative contributes \(f_g\). An unaffected relative contributes \(1-f_g\), because the likelihood term must match the status actually observed.
Ignoring unaffected relatives. The wrong reasoning is that only affected relatives provide evidence about disease risk. Under the model, each recorded unaffected status contributes a probability and can change the likelihood.
Confusing a conditional phenotype likelihood with a joint contribution. If genotypes are conditioned on as known, use the status product. For a specified pedigree configuration whose probability is part of the calculation, multiply that product by the configuration’s genotype probability.
Treating a specified configuration’s contribution as the total likelihood when genotypes are unknown. A single configuration gives one contribution, not the combined likelihood over alternative configurations. Keep the calculation at the specified-configuration level here.
Interpreting a likelihood contribution as a probability that the model is true. It is the probability of specified data under specified model assumptions and parameter values; it does not assign a probability to the model itself.

Key takeaway: For each relative, use \(f_g\) if affected and \(1-f_g\) if unaffected. Multiply these status terms across relatives under conditional independence. When evaluating a specified pedigree genotype configuration, also multiply by its Mendelian genotype probability.

Check Your Understanding

For each question, identify which probabilities are conditional on genotype and which refer to the pedigree genotype configuration.

A relative with genotype \(Aa\) is unaffected and \(f_{Aa}=0.60\). What is the disease-status contribution?
Why does an affected relative with genotype \(aa\) have a nonzero contribution when \(f_{aa}=0.02\)?
Two \(Aa\) parents have a child specified to be \(AA\). What is the child’s genotype probability under Mendelian transmission with no mutation?
State the conditional-independence assumption that allows disease-status contributions to be multiplied across relatives.
In a family calculation, when should the genotype probability be included in addition to the disease-status terms?