Testing for a Major-Gene Segregation Pattern

Why Test for a Major-Gene Segregation Pattern?

A family pattern of affected and unaffected relatives may be more compatible with inheritance through a major gene than with a sporadic model in which disease occurs without familial transmission. A likelihood-based test makes this comparison explicit: it evaluates how well each model predicts the observed family data, while accounting for the genotype uncertainty and penetrance assumptions included in the models.

The test does not establish that a causal gene has been identified. Its conclusion is narrower: under the specified sampling scheme and model assumptions, the observed data are more or less compatible with one model than another. The comparison is only as meaningful as the likelihoods being compared.

Definition: Let $D$ denote the observed family data, $M_0$ a specified sporadic or nonfamilial model, and $M_1$ a specified major-gene segregation model. The likelihood ratio in favor of $M_1$ over $M_0$ is $L(M_1;D)/L(M_0;D)$, where $L(M;D)$ is the probability of the observed data under model $M$. The likelihood-ratio statistic is $$ T=2\log\left(\frac{L(M_1;D)}{L(M_0;D)}\right). $$ A positive $T$ means the observed data have greater likelihood under $M_1$; it is not, by itself, a p-value or a probability that $M_1$ is true.

Specify the Models and the Sampling Scheme

A major-gene model describes how genotypes are transmitted and how genotype and other specified factors relate to phenotype through penetrance. A sporadic model describes disease occurrence without the modeled familial genetic transmission. Each model must specify, or estimate from the data, the parameters needed to calculate the probability of the observations. Earlier material on pedigree likelihoods provides the tools for summing over unobserved genotypes and combining contributions from relatives.

The data and sampling mechanism also determine the likelihood. If families are sampled independently and the likelihood contribution for family $i$ is $L_i(M)$, then the joint likelihood is the product of the contributions:

L(M;D)=\prod_{i=1}^{F}L_i(M).

If families were selected because they contained an affected proband, the probability of the observed data must reflect that selection. One common form is a conditional likelihood, $P(D\mid A,M)=P(D,A\mid M)/P(A\mid M)$, where $A$ denotes the ascertainment event. Comparing unconditioned likelihoods for ascertained families can produce misleading evidence because the selection process itself makes affected relatives more likely to appear in the sample.

Before interpreting a test, check that the same observed data and sampling scheme are represented under both models; the penetrance and transmission assumptions are explicit; and independent family contributions are justified by the design. If the data are clustered or the likelihood already accounts for pedigree dependence, relatives must not also be treated as independent observations.

Worked Example: Comparing Two Fully Specified Models

Consider 20 hypothetical, independently sampled families, each contributing one binary outcome: whether a specified at-risk offspring is affected. Assume the outcome is observed without ascertainment based on disease status. A specified sporadic model assigns probability $0.10$ to an affected outcome; a specified major-gene model assigns probability $0.30$. Seven offspring are affected and 13 are unaffected. The two models and the sampling assumptions are treated as fixed.

Under the independence assumption, the binomial coefficient in the likelihood is the same under both models and cancels in their ratio. Thus:

T=2\left[7\log\left(\frac{0.30}{0.10}\right) +13\log\left(\frac{0.70}{0.90}\right)\right] =2\left[7\log(3)+13\log\left(\frac{7}{9}\right)\right] \approx 8.846.

The likelihood ratio is $\exp(T/2)\approx 83.3$: these data are about 83 times as likely under the specified major-gene model as under the specified sporadic model. This ratio does not mean the major-gene model is 83 times as probable, because likelihoods are not posterior probabilities.

For this particular pair of fully specified models, a chi-square reference distribution is not justified merely by the formula for $T$. Under the sporadic model, $X$, the number affected, follows a binomial distribution with $n=20$ and $p=0.10$. Since the likelihood ratio increases with the number affected, the exact one-sided probability of an outcome at least as supportive of the major-gene model as the observed result is:

P(X\geq 7\mid M_0) =\sum_{k=7}^{20}\binom{20}{k}(0.10)^k(0.90)^{20-k} \approx 0.002386.

Under these assumptions, the observed count would be unusual if the sporadic model were correct. The p-value is a probability of data at least this extreme under $M_0$, not the probability that $M_0$ is true. It also does not address whether the specified major-gene model is a good description of all other features of the data.

Fitted Parameters and the Chi-Square Approximation

Often a model contains unknown parameters estimated by likelihood. Write $\widehat{\theta}_0$ and $\widehat{\theta}_1$ for the maximum-likelihood estimates under the null and alternative models. The likelihood-ratio statistic is then

T=2\left[\ell(\widehat{\theta}_1;D)-\ell(\widehat{\theta}_0;D)\right], \qquad \ell(\theta;D)=\log L(\theta;D).

For nested models, the alternative allows every parameter setting allowed by the null and may allow additional settings. Under regularity conditions—including a null value in the interior of the parameter space, identifiable parameters, and a sufficiently large sample—the null distribution of $T$ is approximately chi-square with degrees of freedom equal to the difference in the numbers of independently estimated parameters. This is an asymptotic approximation, not an automatic property of every likelihood ratio.

Worked Example: A Nested Bernoulli Comparison

For a simplified illustration of a fitted-parameter test, suppose 40 independent, unascertained families each contribute one binary outcome, and 10 outcomes are affected. Under $H_0$, the affected probability is fixed at $p=0.10$. Under $H_a$, $p$ is free to take any value between 0 and 1. This is a one-parameter nested comparison; it illustrates the likelihood calculation, not a complete genotype-based segregation model.

The maximum-likelihood estimate under $H_a$ is $\hat{p}=10/40=0.25$. Using the binomial likelihood, the common binomial coefficient cancels, giving:

T=2\left[10\log\left(\frac{0.25}{0.10}\right) +30\log\left(\frac{0.75}{0.90}\right)\right] =2\left[10\log(2.5)+30\log\left(\frac{5}{6}\right)\right] \approx 7.386.

If the usual large-sample regularity conditions are accepted, the null reference distribution is approximately chi-square with $df=1$, because the alternative adds one free parameter. This gives an approximate p-value of $0.0066$. In context, this is the approximate probability under $H_0$ of a likelihood-ratio statistic at least as large as 7.386. It is evidence against the fixed $p=0.10$ model under the stated assumptions, not proof of a major-gene mechanism.

A real major-gene model may have multiple parameters, latent genotypes, constraints, or a null value on the boundary of the parameter space. Its degrees of freedom and reference distribution must be justified for that particular comparison. One should not use $df=1$ simply because a test concerns one named genetic mechanism.

How Genotype Uncertainty Enters the Comparison

A family’s genotypes may be partly or wholly unobserved. For a specified phenotype pattern $D$, its probability under a major-gene model is obtained by summing across compatible genotype configurations $g$:

P(D\mid M_1)=\sum_g P(g\mid M_1)P(D\mid g,M_1).

The first factor accounts for genotype probabilities under the inheritance model; the second accounts for phenotype probabilities given those genotypes. The resulting marginal probability—not the probability for a single guessed genotype—is used in the likelihood. The sporadic model must likewise assign probabilities to the same observed data.

Worked Example: Marginalizing a Latent Genotype Configuration

Suppose a hypothetical major-gene model assigns probability $0.25$ to genotype configuration $g_1$ and $0.75$ to $g_2$. For a particular observed family phenotype pattern $D$, assume $P(D\mid g_1,M_1)=0.60$ and $P(D\mid g_2,M_1)=0.12$. A specified sporadic model assigns probability $0.12$ to that same pattern. For simplicity, suppose eight independent families each contribute either this pattern or its complement, with three showing $D$.

The probability of $D$ under the major-gene model is the genotype-weighted sum:

P(D\mid M_1)=0.25(0.60)+0.75(0.12)=0.15+0.09=0.24.

The complement therefore has probability $0.76$ under $M_1$, compared with $0.88$ under the sporadic model. The likelihood-ratio statistic is:

T=2\left[3\log\left(\frac{0.24}{0.12}\right) +5\log\left(\frac{0.76}{0.88}\right)\right] =2\left[3\log(2)+5\log\left(\frac{19}{22}\right)\right] \approx 2.693.

The likelihood ratio is $\exp(T/2)\approx 3.84$, so these data are about 3.84 times as likely under the specified major-gene model as under the specified sporadic model. This is a likelihood comparison, not a significance conclusion: a p-value requires an appropriate null distribution or a suitable simulation procedure. The calculation also depends on the assumed genotype weights, conditional phenotype probabilities, independence of families, and definition of the pattern.

Justifying the Likelihood-Ratio Calculation

The statistic $T$ follows directly from comparing log likelihoods. Taking the logarithm converts a product of independent family contributions into a sum; multiplying by two is conventional and is useful for the chi-square approximation in regular nested problems.

Write the likelihood under each model.
For independent families, each likelihood is the product of the probabilities assigned to their observed data.

Form the ratio and take logs.
$\log(L_1/L_0)=\log L_1-\log L_0$; each observed family contributes the log of its probability ratio.

Multiply by two and calibrate.
The statistic is $T=2\log(L_1/L_0)$. Its reference distribution comes from the design and model comparison—not from the formula alone.

For example, with $k$ affected outcomes among $n$ independent observations and fixed probabilities $p_1$ and $p_0$, the binomial likelihood ratio simplifies to $2[k\log(p_1/p_0)+(n-k)\log\{(1-p_1)/(1-p_0)\}]$. This simplification works because the combinatorial factor is identical under both models; it does not justify treating correlated relatives as independent.

Common Pitfalls and Interpretation

Using a chi-square p-value for every likelihood ratio. Fully specified alternatives, boundary parameters, or non-nested models need not have the usual chi-square null distribution. Use an exact reference distribution where available or simulate the statistic under the null while reproducing the study design and fitting procedure.
Ignoring ascertainment. If families entered the study because of an affected proband, the likelihood should account for that selection. A raw likelihood can overstate familial evidence.
Confusing likelihood with probability of a model. A likelihood ratio compares how well the models predict the observed data. It does not give the posterior probability of either model.
Overstating the conclusion. Evidence favoring a major-gene model over a specified sporadic model does not prove that a particular gene causes disease. Unmodeled factors, penetrance misspecification, and sampling features can affect the comparison.

A careful report names the models, states whether parameters were fixed or estimated, describes the sampling and ascertainment treatment, gives the statistic and its calibration, and interprets the result in the family-study context. Where a standard reference distribution is doubtful, that uncertainty is part of the result rather than a technical detail to omit.

Key takeaway: A likelihood-ratio test measures how much more compatible the observed family data are with one specified model than another. Sound inference requires correctly constructed likelihoods and a reference distribution appropriate to the models, parameter constraints, and family-sampling design.

Check Your Understanding

Try these questions before reviewing the explanations above.

What does a positive likelihood-ratio statistic indicate, and what does it not establish?
Why can a chi-square reference distribution be inappropriate when both competing models are fully specified?
In a pedigree likelihood, why must probabilities be summed across unobserved genotype configurations?
A study selects families because they have an affected proband. What feature of the likelihood must be addressed?
For nested regular models, how are the degrees of freedom for the asymptotic likelihood-ratio test determined?