Likelihood-Ratio Tests of Mendelian Transmission

Why Test a Mendelian Transmission Probability?

When a parent is heterozygous at a locus, the basic Mendelian model assigns probability one half to transmission of either allele in each meiosis. A sample of offspring will rarely show exactly half of each allele: random variation alone can produce an imbalance. A likelihood-ratio test asks whether the observed imbalance is sufficiently incompatible with the one-half model to count as evidence against it.

The test is about the probability of a particular allele being transmitted, not directly about disease risk or a causal effect. Its validity depends on how transmissions were identified and sampled. The simple model below applies when each transmission is informative, the transmission outcome is correctly classified, and the meioses can be treated as independent draws under the model. It also assumes that selection into the analysis has not distorted the transmission counts.

Definition: Let \(X_i=1\) if the allele of interest is transmitted in informative meiosis \(i\), and \(X_i=0\) otherwise. Under the simple transmission model, \(X_1,\ldots,X_n\) are independent Bernoulli random variables with a common transmission probability \(q\). Thus \(K=\sum_{i=1}^n X_i\), the number of transmissions of the allele, has a binomial distribution: \(K\sim\operatorname{Binomial}(n,q)\). The Mendelian null hypothesis is \(H_0:q=1/2\); the two-sided alternative is \(H_a:q\ne 1/2\).

The binomial model treats each informative meiosis as one observation. For example, two offspring of the same parent contribute two transmissions if the outcomes are informative and the model’s independence assumption is appropriate. A genotype that does not reveal which allele was transmitted is not automatically a zero or a one; it is uninformative for this count-based calculation. If transmission outcomes are dependent, misclassified, or selected in a way related to the outcome, the binomial likelihood and its calibration may not apply as written.

Likelihood and Test Statistic

For \(k\) observed transmissions of the allele among \(n\) informative meioses, the likelihood as a function of \(q\) is

L(q)=\binom{n}{k}q^k(1-q)^{n-k}.

The likelihood-ratio test compares the likelihood maximized under the null with the likelihood maximized over the unrestricted parameter space \(0\le q\le1\). Under \(H_0\), the likelihood is evaluated at \(q=1/2\). Under the alternative model, the maximum-likelihood estimate is \(\hat{q}=k/n\), the observed transmission proportion.

Definition: The likelihood-ratio statistic is \(\Lambda=L(1/2)/L(\hat{q})\), the null likelihood divided by the unrestricted maximum likelihood. The deviance form of the test statistic is \(T=-2\log(\Lambda)\). Larger values of \(T\) indicate that the observed data fit the unrestricted estimate better than they fit the Mendelian null.

The binomial coefficient occurs in both likelihoods and cancels in the ratio. Substituting the null and the maximum-likelihood estimate gives a convenient formula:

T=2\left[k\log\left(\frac{k}{n/2}\right)+(n-k)\log\left(\frac{n-k}{n/2}\right)\right].

If \(k=0\) or \(k=n\), the corresponding term \(0\log(0)\) is defined by its limit as zero. The statistic is nonnegative: the unrestricted model includes \(q=1/2\), so its maximized likelihood cannot be smaller than the null likelihood. It is zero when \(k=n/2\), and it grows as the observed count departs from the null expectation \(n/2\).

Worked Example: A Moderate Imbalance in 20 Transmissions

Suppose a hypothetical data set contains \(n=20\) informative meioses, with the allele of interest transmitted \(k=14\) times. Test \(H_0:q=1/2\) against \(H_a:q\ne1/2\), using the large-sample calibration and then an exact calibration.

The unrestricted estimate is \(\hat{q}=14/20=0.7000\). The likelihood-ratio statistic is

T=2\left[14\log\left(\frac{14}{10}\right)+6\log\left(\frac{6}{10}\right)\right] =2\left[14\log(1.4)+6\log(0.6)\right] \approx 3.291.

Under the usual large-sample calibration, \(T\) is compared with a chi-square distribution with one degree of freedom. The resulting upper-tail p-value is approximately \(0.0696\). At significance level \(\alpha=0.05\), this approximation does not provide sufficient evidence to reject the one-half transmission model.

For an exact likelihood-ratio test, calculate the probability under \(H_0\) of counts whose likelihood-ratio statistic is at least \(3.291\). For \(n=20\), those counts are \(K\le6\) or \(K\ge14\). Therefore,

P_{H_0}(K\le6\ \text{or}\ K\ge14) =\frac{2(1+20+190+1140+4845+15504+38760)}{2^{20}} \approx 0.1153.

The exact p-value is about \(0.1153\), also above \(0.05\). Both calibrations lead to the same decision here, though their p-values differ. Neither p-value is the probability that the Mendelian model is true.

Large-Sample Calibration and Its Justification

For a regular one-parameter likelihood model, Wilks’ theorem states that, under the null hypothesis and as the sample size increases, the likelihood-ratio statistic has an approximate chi-square distribution with degrees of freedom equal to the difference in the number of free parameters between the alternative and null models. Here the alternative estimates one parameter, \(q\), while the null fixes it at one half. Thus the difference is one, and the reference distribution is \(\chi^2_1\).

This result is an approximation, not an identity for every finite sample. Its standard conditions include independent observations from the stated model, an identifiable parameter, and a null value in the interior of the parameter space. In this binomial test, \(q=1/2\) is interior to \([0,1]\). The approximation is generally more dependable when the sample contains enough transmissions that the count is not concentrated in a few possible outcomes. For small samples, the exact binomial distribution can be used instead.

Theorem: If the independent Bernoulli transmission model is correctly specified and \(H_0:q=1/2\) holds, then as \(n\) grows, \(T\) converges in distribution to \(\chi^2_1\). The approximate p-value is \(P(\chi^2_1\ge T_{\text{obs}})\). This p-value describes how often a statistic at least this large would occur under the null model, using the chi-square approximation.

A short derivation shows why this statistic has the stated form. Write the log likelihood without its constant binomial coefficient as \(\ell(q)=k\log(q)+(n-k)\log(1-q)\). Differentiating and setting the derivative to zero gives the unrestricted maximum:

\frac{d\ell(q)}{dq}=\frac{k}{q}-\frac{n-k}{1-q}=0 \quad\Longrightarrow\quad k(1-q)=(n-k)q \quad\Longrightarrow\quad \hat{q}=\frac{k}{n}.

The second derivative is \(-k/q^2-(n-k)/(1-q)^2\), which is nonpositive for \(0<q<1\), so this stationary point is a maximum when it is interior. At \(k=0\) or \(k=n\), the likelihood is maximized at the corresponding boundary \(q=0\) or \(q=1\). Evaluating the log likelihood at \(1/2\) and at \(\hat q\), then multiplying their difference by two, yields

2\{\ell(\hat q)-\ell(1/2)\} =2\left[k\log\left(\frac{\hat q}{1/2}\right) +(n-k)\log\left(\frac{1-\hat q}{1/2}\right)\right],

which becomes the displayed formula for \(T\) after substituting \(\hat q=k/n\). This also shows that the statistic measures the improvement in log likelihood when the transmission probability is estimated rather than fixed at one half.

Worked Example: Small-Sample Calibration Can Change the Conclusion

In a second hypothetical data set, the allele is transmitted \(k=8\) times among \(n=10\) informative meioses. The estimate is \(\hat{q}=8/10=0.8000\), and

T=2\left[8\log\left(\frac{8}{5}\right)+2\log\left(\frac{2}{5}\right)\right] =2[8\log(1.6)+2\log(0.4)] \approx 3.855.

The chi-square approximation gives a p-value of about \(0.0496\), just below \(0.05\). On that approximate calibration, the test rejects \(H_0\) at the five-percent level.

The exact likelihood-ratio p-value instead sums the null probabilities of counts with \(T\ge3.855\). These are \(K\le2\) and \(K\ge8\), so

P_{H_0}(K\le2\ \text{or}\ K\ge8) =\frac{2(1+10+45)}{2^{10}} =\frac{112}{1024} =0.1094.

The exact p-value is above \(0.05\), so the exact test does not reject at that level. The difference is a consequence of the discrete binomial distribution and the roughness of the large-sample approximation with only 10 transmissions. It is not a contradiction in the data: the procedures use different calibrations.

Exact Calibration and Interpretation

An exact test uses the distribution of \(K\) under \(H_0\), namely \(K\sim\operatorname{Binomial}(n,1/2)\), rather than a limiting chi-square distribution. For a likelihood-ratio test, the exact p-value is the sum of null probabilities of all possible counts whose likelihood-ratio statistic is at least the observed statistic. This rule gives a two-sided test because unusually low and unusually high counts both conflict with a one-half transmission probability.

Because \(K\) takes only finitely many values, attainable exact p-values are discrete. An exact test at level \(\alpha\) therefore has rejection probability no greater than \(\alpha\), and can be more conservative than a test with a continuous reference distribution. In this symmetric one-half setting, counts equally far from \(n/2\) have the same likelihood-ratio statistic and are both included in the tail. The exact likelihood-ratio p-value need not equal every other possible two-sided binomial p-value, since two-sided tests can order outcomes by different rules.

Worked Example: A Modest Deviation With 40 Transmissions

Suppose a third hypothetical sample has \(n=40\) informative transmissions, of which \(k=23\) carry the allele. The estimated probability is \(\hat q=23/40=0.5750\). The statistic is

T=2\left[23\log\left(\frac{23}{20}\right)+17\log\left(\frac{17}{20}\right)\right] =2[23\log(1.15)+17\log(0.85)] \approx 0.9034.

Using the \(\chi^2_1\) reference distribution gives \(p\approx0.3418\). At \(\alpha=0.05\), this is not evidence against \(q=1/2\). The estimate is 0.5750, but that observed proportion alone does not establish that the underlying transmission probability differs from one half. The test accounts for the variability expected in 40 transmissions under the null.

Assumptions, Scope, and a Common Pitfall

The simple likelihood-ratio test is useful when the data genuinely represent independent, informative transmission outcomes sampled under the model. It does not automatically handle family ascertainment, correlated observations, uncertain genotype calls, or a more complex pedigree likelihood. If families were included because of a phenotype pattern related to the transmissions, the observed transmission counts may not follow the unconditional binomial model. The sampling mechanism must be considered before interpreting either the statistic or its p-value.

The test also does not show that a deviation is caused by a particular biological mechanism. A small p-value indicates that the observed imbalance would be unusual under the specified one-half model and its sampling assumptions. It does not identify why the imbalance occurred, nor does it quantify the practical importance of a departure from one half. The estimate \(\hat q\) gives the observed proportion; the likelihood-ratio test evaluates its compatibility with the null.

Mistake: treating the chi-square result as exact for any sample size. A p-value near a decision threshold may change when the finite-sample binomial distribution is used. Check the exact calibration when the sample is small or sparse.
Mistake: interpreting a p-value as the probability that Mendelian transmission is true. The p-value is calculated assuming \(H_0\) and describes the tail probability of the test statistic under that model; it is not a probability assigned to \(H_0\).
Mistake: counting every offspring as an informative transmission. Include an observation in this binomial count only if the allele’s transmission outcome is identifiable under the data and model. An ambiguous genotype is not evidence for either allele.
Mistake: treating a non-significant result as proof of one-half transmission. Failure to reject means the data do not provide sufficiently strong evidence against the null at the chosen level; it does not prove that the probability equals one half.

Key takeaway: For independent informative meioses, compare the binomial likelihood at \(q=1/2\) with its unrestricted maximum at \(\hat q=k/n\). Use \(T=-2\log\Lambda\), calibrate it with \(\chi^2_1\) when the large-sample approximation is reasonable, and use the exact binomial likelihood-ratio tail when finite-sample calibration matters.

Check Your Understanding

Use the model assumptions as well as the calculations when answering.

For \(n=30\) informative meioses with \(k=18\) transmissions of the allele, what is \(\hat q\), and which hypotheses are tested by the two-sided likelihood-ratio test?
Why does the unrestricted maximum-likelihood estimate equal \(k/n\) in the binomial transmission model?
What reference distribution is used for the large-sample likelihood-ratio test here, and why does it have one degree of freedom?
Why can the exact and chi-square-calibrated p-values differ substantially when \(n\) is small?
Give one example of a sampling or data feature that would make the simple independent-binomial calibration questionable.