Estimating Transmission Probabilities From Offspring Genotypes

Why Estimate a Transmission Probability?

When a parent is heterozygous at a genetic locus, each offspring receives one of that parent's two alleles. Under Mendelian transmission, either allele has probability one-half of being transmitted. In family data, however, the observed genotypes may not always reveal which allele came from which parent. Estimating a transmission probability therefore starts by asking what the parental genotypes allow us to infer from each offspring.

This tutorial considers estimation of the probability that a specified allele is transmitted by a specified heterozygous parent. The focus is the likelihood and its uncertainty when parental genotypes are known. The estimate describes the transmission pattern under the stated sampling model; by itself, it does not establish why a pattern occurred.

Definition: Let a parent have known genotype \(A/a\), and designate \(A\) as the allele of interest. The transmission probability \(\theta\) is the probability that this parent transmits \(A\) in a meiosis to an offspring. For \(n\) independent, informative meioses, let \(X_i=1\) if \(A\) was transmitted in meiosis \(i\), and \(X_i=0\) otherwise. If all meioses share the same probability \(\theta\), then \(X_i\sim\operatorname{Bernoulli}(\theta)\), and the number of transmissions \(K=\sum_{i=1}^n X_i\) has a binomial distribution.

A meiosis is informative for this estimation when the observed offspring genotype, together with the known parental genotypes, identifies whether the specified parent transmitted \(A\) or \(a\). For example, if the other parent is \(a/a\), an \(A/a\) offspring must have received \(A\) from the heterozygous parent, while an \(a/a\) offspring must have received \(a\). If the other parent is also \(A/a\), an \(A/a\) offspring does not by itself reveal which parent supplied which allele.

Likelihood and Maximum-Likelihood Estimate

Suppose \(n\) informative offspring transmissions are observed and \(k\) involve transmission of \(A\). Under independent meioses with a common transmission probability \(\theta\), the likelihood for the count is the binomial likelihood. The factor \(\binom{n}{k}\) is included when writing the probability of the count; it does not depend on \(\theta\).

L(\theta\mid k,n)=\binom{n}{k}\theta^k(1-\theta)^{n-k},\qquad 0\leq\theta\leq1.

The maximum-likelihood estimate (MLE) is the value of \(\theta\) that makes the observed data most likely under this model. For \(0<k<n\), maximize the log-likelihood. The binomial coefficient is constant in \(\theta\), so it can be omitted when differentiating.

\ell(\theta)=k\log(\theta)+(n-k)\log(1-\theta)+\text{constant}.

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

The second derivative is \(-k/\theta^2-(n-k)/(1-\theta)^2\), which is negative for \(0<\theta<1\); thus the stationary point is a maximum. If \(k=0\), the likelihood is maximized at \(\hat{\theta}=0\); if \(k=n\), it is maximized at \(\hat{\theta}=1\). The estimate is the observed proportion of informative transmissions of \(A\), not a count of all offspring regardless of whether their genotypes reveal the transmission.

Worked Example: Estimating From a Heterozygous Parent and a Homozygous Parent

In a hypothetical family, one parent has genotype \(A/a\), and the other has genotype \(a/a\). Among 10 offspring, 6 have genotype \(A/a\) and 4 have genotype \(a/a\). Assume offspring genotypes are measured correctly. The parental genotypes make every offspring informative: \(A/a\) means the heterozygous parent transmitted \(A\), and \(a/a\) means that parent transmitted \(a\).

Let \(k=6\) and \(n=10\). The likelihood is \(L(\theta)=\binom{10}{6}\theta^6(1-\theta)^4\). The MLE is \(\hat{\theta}=k/n=6/10=0.6000\).

For example, the likelihood at \(\theta=0.6000\) is \(\binom{10}{6}(0.6)^6(0.4)^4 =210(0.046656)(0.0256)\approx0.2508\). This is the probability of observing exactly 6 transmissions of \(A\) out of 10 under that value of \(\theta\). The estimate indicates that 6 of the 10 observed informative meioses transmitted \(A\); it does not imply that the true transmission probability is known without uncertainty.

Uncertainty: A Wilson Score Interval

A point estimate alone does not show how much the estimate could vary across repeated samples. Under the binomial model, the estimated standard error of \(\hat{\theta}\) is \(\sqrt{\hat{\theta}(1-\hat{\theta})/n}\). This plug-in estimate can be unstable for small samples, especially when \(\hat{\theta}\) is near 0 or 1. A Wilson score interval is a useful binomial interval that avoids some poor behavior of the simple Wald interval.

For a two-sided confidence level of approximately 95%, use \(z=1.96\), the standard-normal critical value. With \(\hat{\theta}=k/n\), the Wilson interval has center \(C\) and margin \(M\) as follows.

C=\frac{\hat{\theta}+z^2/(2n)}{1+z^2/n}, \qquad M=\frac{z}{1+z^2/n} \sqrt{\frac{\hat{\theta}(1-\hat{\theta})}{n}+\frac{z^2}{4n^2}}, \qquad (C-M,\ C+M).

The interval is obtained by inverting the binomial score test for each candidate value of \(\theta\). For a candidate \(\theta\), the score statistic, expressed using \(\hat{\theta}\), is \((\hat{\theta}-\theta)/\sqrt{\theta(1-\theta)/n}\). Retain candidate values for which its absolute value is at most \(z\). Squaring and multiplying by the positive denominator gives the quadratic inequality \(n(\hat{\theta}-\theta)^2\leq z^2\theta(1-\theta)\). Solving this inequality for \(\theta\) gives the two endpoints \(C-M\) and \(C+M\) above. The endpoints are in the probability range from 0 to 1.

Worked Example: A Wilson Interval With Nine Informative Offspring

In a hypothetical sample, a heterozygous parent has known genotype \(A/a\), and the other parent is \(a/a\). Seven of nine offspring have genotype \(A/a\); the remaining two have genotype \(a/a\). Thus all nine offspring are informative, \(k=7\), and \(\hat{\theta}=7/9\approx0.7778\). Assume independent transmissions and accurate genotypes.

Using \(z=1.96\), \(z^2=3.8416\), and \(n=9\), the Wilson center is \(C=[0.7778+3.8416/(18)]/[1+3.8416/9]\approx0.6947\). Using the unrounded value \(7/9\) in the margin calculation gives \(M\approx0.2421\).

Therefore, the 95% Wilson interval is approximately \((0.694680-0.242096,\ 0.694680+0.242096)\approx(0.4526,\ 0.9368)\). In repeated sampling under the binomial model, the Wilson procedure would produce intervals containing the true transmission probability about 95% of the time. This particular interval is not a statement that there is a 95% probability that the fixed parameter lies between these endpoints.

Using Parental Genotypes to Decide What Counts

Known parental genotypes determine how offspring genotypes contribute to the likelihood. When the offspring genotype uniquely identifies the allele transmitted by the parent of interest, that offspring provides a Bernoulli observation. When it does not, it should not be labelled a success or failure merely to make a count. A more general genotype likelihood can sum over possible parental transmissions consistent with the observed offspring genotype.

For instance, suppose the parent of interest is \(A/a\), the other parent is also \(A/a\), and the other parent's transmission follows Mendelian segregation with probability one-half for each allele. If the target parent transmits \(A\) with probability \(\theta\), then the offspring genotype probabilities are \(P(AA)=\theta/2\), \(P(Aa)=\theta(1/2)+(1-\theta)(1/2)=1/2\), and \(P(aa)=(1-\theta)/2\). An \(Aa\) offspring is uninformative about \(\theta\) in this setup: its probability is one-half for every \(\theta\). The homozygous offspring genotypes do contain information about the target parent's transmission.

Worked Example: A Heterozygous Offspring Does Not Identify the Transmitting Parent

Consider a hypothetical set of 10 offspring of two known \(A/a\) parents. The parent being studied transmits \(A\) with probability \(\theta\); assume the other parent transmits each allele with probability one-half. The observed offspring genotypes are 3 \(AA\), 5 \(Aa\), and 2 \(aa\). Assume offspring transmissions are independent and genotypes are correctly recorded.

For each child, the genotype probabilities are \(P(AA)=\theta/2\), \(P(Aa)=1/2\), and \(P(aa)=(1-\theta)/2\). The likelihood for these genotype counts, including the multinomial coefficient, is \(\frac{10!}{3!5!2!}(\theta/2)^3(1/2)^5((1-\theta)/2)^2\). As a function of \(\theta\), this is proportional to \(\theta^3(1-\theta)^2\): the five \(Aa\) observations contribute a factor that does not depend on \(\theta\).

Maximizing this likelihood gives \(\hat{\theta}=3/(3+2)=0.6000\). Equivalently, among the five offspring whose homozygous genotypes identify which allele the target parent transmitted, three identify \(A\) and two identify \(a\). The five heterozygous offspring are not five extra failures or successes; their genotypes do not distinguish the target parent's contribution in this model.

Assumptions, Pooling, and Interpretation

The simple binomial likelihood is appropriate when the observations being counted are informative transmissions with the same probability \(\theta\), and the transmissions are independent under the sampling model. In practice, this requires attention to several features of the data:

Known and correctly specified parental genotypes: these are used to determine which offspring genotypes reveal the transmission. Genotype error can misclassify transmissions or alter their probabilities.
Independence: each meiosis contributes a transmission observation that is independent of the others under the model. Dependence or a design that selects families in a way related to the observed transmissions can require a different likelihood.
A common probability: pooling observations across parents or families assumes they share the same \(\theta\). If their probabilities differ, a single pooled proportion may not describe any one parent's transmission probability.
A defined informative set: the denominator \(n\) includes only observations whose genotype likelihood supplies the relevant transmission information, as defined by the model. It is not automatically the total number of offspring genotyped.

Under Mendelian segregation at an autosomal locus, the specified allele from a heterozygous parent has \(\theta=1/2\). The estimate and interval can be compared with that value descriptively, but a formal test of Mendelian transmission is a separate inferential procedure. Also, a departure of \(\hat{\theta}\) from one-half does not on its own show a biological cause: sampling variation, genotype error, model misspecification, or the way families entered the study may matter.

Common Mistakes

Counting all offspring as informative. The wrong reasoning is that every observed child contributes one success or failure. The correction is to use the parental genotypes and offspring genotype to determine whether transmission from the parent of interest is identifiable, or to use the appropriate genotype likelihood when it is not.
Calling the estimate the true probability. The wrong reasoning is that \(k/n\) is the population transmission probability. It is the MLE from the observed sample; uncertainty remains and should be summarized with an interval or other appropriate inferential method.
Interpreting a confidence interval as a probability statement about a fixed parameter. The wrong reasoning is that the computed 95% interval has a 95% chance of containing \(\theta\). The frequentist interpretation concerns the long-run coverage of the procedure under its assumptions.
Pooling without a common-parameter assumption. The wrong reasoning is that counts from different families can always be combined. Pooling estimates one common probability only if that is a reasonable model for the transmissions being combined.

Key takeaway: With \(k\) observed transmissions of a specified allele among \(n\) independent, informative meioses that share probability \(\theta\), the binomial MLE is \(\hat{\theta}=k/n\). Known parental genotypes determine which observations are informative; a Wilson interval describes sampling uncertainty under the same model.

Check Your Understanding

Answer each question using the transmission model and assumptions stated.

A parent has genotype \(B/b\), the other parent is \(b/b\), and 9 of 15 offspring have genotype \(B/b\). What is the MLE of the probability that the heterozygous parent transmits \(B\)?
Why is an offspring with genotype \(A/a\) not necessarily informative when both parents have genotype \(A/a\)?
Write the binomial likelihood for \(k\) transmissions of the specified allele among \(n\) informative meioses.
What does a 95% Wilson interval mean under repeated sampling, and what does it not mean?
Name two assumptions that should be considered before pooling transmission observations across families.