Why Compare Disease Models?
A pedigree with several affected relatives may be compatible with more than one explanation. A disease allele could act dominantly, require two copies to act recessively, or be unrelated to the observed pattern, with cases arising sporadically. The useful comparison is not based on a pedigree’s appearance alone: each model assigns probabilities to genotypes and phenotypes, and those probabilities determine the likelihood of the observations.
This tutorial compares specified models. Each model’s allele frequency, penetrances, and phenocopy risk are treated as fixed inputs. The likelihood measures how well that particular specification predicts the observed data; it does not by itself estimate the model’s parameters or establish that a model is true.
Specify the Models Before Calculating
Let \(D\) denote a disease-associated allele and \(d\) the alternative allele. An individual’s genotype \(g\) is one of \(DD\), \(Dd\), or \(dd\). Let \(Y=1\) indicate affected status and \(Y=0\) indicate unaffected status. The penetrance for genotype \(g\) is \(f_g=P(Y=1\mid g)\); the probability of being unaffected is therefore \(1-f_g\).
For example, let \(q=P(D)\) be the frequency of \(D\) in the population. If founders are randomly sampled from a population in Hardy–Weinberg equilibrium, their genotype probabilities are \(P(DD)=q^2\), \(P(Dd)=2q(1-q)\), and \(P(dd)=(1-q)^2\). These are founder probabilities; genotypes of descendants are generated by Mendelian transmission from their parents, not by drawing each descendant independently from those frequencies.
The comparison also requires explicit assumptions. In the examples below, the models use the same \(q\) where a disease allele is specified, founders are unrelated and in Hardy–Weinberg equilibrium, and transmissions follow Mendelian inheritance without mutation. Conditional on genotype and the model, relatives’ phenotypes are independent. The data are assumed not to have been selected by observing an affected proband, or else the likelihood is understood to include the appropriate conditioning on ascertainment.
Pedigree Likelihoods and Unobserved Genotypes
A pedigree likelihood is the probability of the observed phenotypes and any observed genotypes under a specified model. If some genotypes are unknown, calculate the likelihood for every possible genotype configuration and add those probabilities. For each configuration, include founder genotype probabilities, transmission probabilities, and phenotype probabilities.
Here \(M\) denotes a particular model, \(\mathbf g\) is a possible assignment of genotypes to the pedigree members, and the product runs over observed phenotypes. The term \(P(\mathbf g\mid M)\) is built from founder genotype probabilities and Mendelian transmissions. For an affected individual the phenotype factor is \(f_{g_i}\); for an unaffected individual it is \(1-f_{g_i}\). If genotypes are observed, the sum is restricted to configurations consistent with those observations. If the analysis conditions on observed genotypes, their probability is not included in the conditional phenotype likelihood.
The summation is essential: an affected person with unknown genotype is not automatically a carrier, and an unaffected person is not automatically free of a disease allele. The likelihood accounts for uncertainty by weighting each possible genotype configuration by its probability.
Worked Example: One Affected Founder
Suppose \(q=0.01\), so the founder genotype probabilities are \(0.0001\) for \(DD\), \(0.0198\) for \(Dd\), and \(0.9801\) for \(dd\). Compare these fixed penetrances: dominant \(f_{DD}=f_{Dd}=0.80,\ f_{dd}=0.01\); recessive \(f_{DD}=0.80,\ f_{Dd}=f_{dd}=0.01\); sporadic \(f_g=0.02\) for every genotype. What is the likelihood of observing one affected founder?
Under the dominant model, sum the genotype-specific probabilities of being affected:
Under the recessive model, only \(DD\) has penetrance \(0.80\); the other genotypes have affected risk \(0.01\):
Under the sporadic model the risk is constant, so \(L_{\mathrm{spor}}=0.02\), regardless of the genotype distribution. For this one affected founder, the dominant specification has the largest likelihood, \(0.025721\), followed by the sporadic specification, \(0.02000\), and then the recessive specification, \(0.010079\). This ordering is evidence about these fixed specifications for these data, not a statement that the dominant model has a 2.5721% probability of being true.
Why Marginalization Matters: A Derivation
The likelihood formula follows from the law of total probability. For a pedigree with unobserved genotypes, the observed phenotype can arise through any genotype configuration. These mutually exclusive configurations exhaust the possibilities, so their joint probabilities with the phenotypes must be summed.
The last equality uses conditional independence of phenotypes given genotypes and the model. The genotype probability factors into founder probabilities and the conditional transmission probabilities of descendants. Thus a pedigree likelihood combines population allele frequencies with inheritance and phenotype evidence, while avoiding the unsupported assumption that any unobserved genotype is known.
Worked Example: An Assayed Parent–Child Trio
Consider a hypothetical trio in which genotyping shows that the father is \(Dd\) and the mother is \(dd\). The father is affected, the mother is unaffected, and their child is affected; the child’s genotype is not observed. Use the penetrances from the preceding example. Calculate phenotype likelihoods conditional on the known parental genotypes.
The father passes \(D\) or \(d\) with probability \(1/2\); the mother always passes \(d\). Thus the child has probability \(1/2\) of being \(Dd\) and \(1/2\) of being \(dd\). Under the dominant model, the probability the child is affected is
The father’s affected probability is \(0.80\), and the mother’s unaffected probability is \(1-0.01=0.99\). Therefore the trio’s conditional phenotype likelihood is
Under the recessive model, the father’s \(Dd\) genotype has affected risk \(0.01\), the \(dd\) mother is unaffected with probability \(0.99\), and neither possible child genotype (\(Dd\) or \(dd\)) has affected risk above \(0.01\). Hence
Under the sporadic model, the father’s affected probability is \(0.02\), the mother’s unaffected probability is \(0.98\), and the child’s affected probability is \(0.02\):
Conditional on the parental genotypes, the dominant model has the largest phenotype likelihood for this trio. The genotype probabilities of the assayed parents are not multiplied into these values because the calculation is explicitly conditional on their observed genotypes. If instead the analysis used the unconditional probability of the complete data, those genotype probabilities would also be included.
Combining Independent Family Observations
When observations come from independent founders sampled under the same model, their likelihood contributions multiply. The resulting comparison may differ from a comparison based on one person, because both affected and unaffected observations contribute evidence. Within a pedigree, however, relatives are not independent genotype draws: their shared inheritance must be represented through transmission probabilities.
Worked Example: One Affected and One Unaffected Founder
Use the allele frequency and penetrances from the first example. Suppose two unrelated founders are observed: one is affected and the other is unaffected. Their genotypes are unknown, and the founders are independent. The affected person’s likelihoods are \(0.025721\) (dominant), \(0.010079\) (recessive), and \(0.02\) (sporadic).
For the unaffected founder, sum the probabilities of each genotype multiplied by its unaffected probability. Under the dominant model:
Under the recessive model:
The sporadic model gives \(L_{\mathrm{spor,unaff}}=1-0.02=0.98\). Thus the recessive model has the largest unaffected-founder likelihood, \(0.989921\), followed by the sporadic model, \(0.9800\), and the dominant model, \(0.974279\). (In numerical order, the recessive value is actually greater than the sporadic value: \(0.989921>0.9800\).) This comparison illustrates why the ordering should be checked against the calculated values rather than inferred from a model label.
Because the two founders are independent, multiply their contributions:
For this pair, the dominant model has the greatest likelihood, followed by the sporadic model and then the recessive model. The unaffected founder’s contribution alone favored the recessive model, but combining both observations changes the ordering. A complete comparison uses the whole observed data set.
Interpreting a Model Comparison
For a fixed data set, the model with the larger likelihood assigns greater probability to the observations under its stated assumptions and parameter values. A likelihood ratio, such as \(L(M_1)/L(M_2)\), expresses how many times as likely the data are under one specified model as under another. It is not the probability that either model is true. Posterior model probabilities require additional ingredients, including prior probabilities for models.
A raw likelihood comparison also does not automatically account for model complexity or justify a formal significance test. Here the purpose is to calculate and interpret likelihoods under explicitly stated, fixed specifications. Parameter estimation and formal tests of segregation patterns require additional procedures.
Ascertainment deserves particular attention. If families entered a study because a proband was affected, the observed sample is not an unselected sample of pedigrees. The probability of the observed data must then be conditioned on the selection event, or corrected using the appropriate ascertainment model. Otherwise, the affected proband’s phenotype can be counted as though it were an independent, randomly observed outcome.
Common Mistakes
- Assigning a genotype from the phenotype. An affected person need not carry the modeled disease genotype, because phenocopies are allowed; an unaffected carrier is also possible when penetrance is incomplete. Sum over possible unobserved genotypes.
- Using founder frequencies for every relative. Founder genotype frequencies describe population sampling. Descendant genotypes must follow from parental genotypes and Mendelian transmission.
- Forgetting unaffected probabilities. An unaffected person contributes \(1-f_g\), not \(f_g\), conditional on genotype \(g\).
- Comparing numbers without checking their order. For example, \(0.989921\) exceeds \(0.9800\). State which likelihood is largest only after comparing the computed values.
- Calling the largest likelihood the probability a model is true. It is the probability of the data under that model and its fixed parameters; it is not a posterior probability.
- Ignoring selection of families. If ascertainment depends on affection status, use a likelihood that accounts for that selection rather than treating the sample as unselected.
Check Your Understanding
For each question, identify the probability factors needed before comparing the models.
- Under a recessive model, an unaffected \(Dd\) individual has penetrance \(0.70\) for genotype \(DD\) and phenocopy risk \(0.02\) for \(Dd\) and \(dd\). What phenotype probability contributes for this unaffected individual?
- A \(Dd\) parent and a \(dd\) parent have a child whose genotype is unknown. What are the child’s possible genotypes and their transmission probabilities?
- Why should an affected person with an unobserved genotype not automatically be treated as \(DD\) under a recessive model?
- Two unrelated founders have independent phenotype observations. How are their likelihood contributions combined, and why does the same multiplication rule not treat pedigree relatives as independent genotype draws?
- A model assigns the observed data likelihood \(0.06\), while another assigns likelihood \(0.03\). State the correct likelihood-ratio interpretation and one conclusion that cannot be drawn from these values alone.