Mendelian Transmission Likelihoods in Parent-Offspring Trios

Why a Trio Likelihood Has Two Parts

A parent-offspring trio records three genotypes, but those observations arise through two related processes. The parents have genotypes drawn from a population distribution, and each parent transmits one allele to the child according to Mendelian inheritance. A trio likelihood represents both processes when the parents are treated as sampled individuals. If the analysis conditions on the parents' observed genotypes, only the transmission part remains.

The distinction matters because the same child genotype can be likely or impossible depending on the parental genotypes. Conversely, parental genotype probabilities depend on population allele frequencies, while the child's genotype conditional on the parents is determined by transmission. Separating these pieces makes the likelihood interpretable and helps reveal which observations actually inform transmission.

Definition: Consider a diploid autosomal locus with two alleles, \(A\) and \(a\). Let \(p\) be the population frequency of \(A\), and let \(q=1-p\) be the frequency of \(a\). Code a genotype by \(k\), the number of copies of \(A\): \(k=0\) denotes \(aa\), \(k=1\) denotes \(Aa\), and \(k=2\) denotes \(AA\). A trio likelihood is the probability of the observed father, mother, and child genotypes under a specified population-genotype and transmission model.

Assumptions and the Parental Genotype Distribution

For the basic joint likelihood below, assume the locus is autosomal, each person is diploid, genotypes are observed without error, and transmission follows Mendelian inheritance with no mutation. Assume also that the parents are unrelated founders sampled independently from a population in Hardy–Weinberg equilibrium with allele frequency \(p\). These assumptions give each parent's genotype probability:

f_k(p)=P(G=k)=\binom{2}{k}p^kq^{2-k},\qquad k\in\{0,1,2\}.

Thus \(P(aa)=q^2\), \(P(Aa)=2pq\), and \(P(AA)=p^2\). The independent-parent assumption means the probability of the two parental genotypes is the product of their separate probabilities. If the analysis instead conditions on the parents' observed genotypes, these parental probabilities are not part of the conditional transmission likelihood.

The conditional transmission calculation does not require Hardy–Weinberg equilibrium. Given a parent's genotype, each of the two alleles has probability one-half of being transmitted. The Hardy–Weinberg and independent-parent assumptions enter when assigning probabilities to the parental genotypes in the joint trio likelihood.

Calculating Mendelian Transmission Probabilities

Let \(k_f\), \(k_m\), and \(k_c\) denote the numbers of \(A\) alleles in the father, mother, and child. A parent with genotype count \(k\) transmits \(A\) with probability \(k/2\): probability zero for \(aa\), one-half for \(Aa\), and one for \(AA\). Conditional on the parental genotypes, the two transmissions are independent.

For a compact derivation, let \(Z_f\) and \(Z_m\) equal 1 if the father or mother transmits \(A\), respectively, and 0 otherwise. Then \(P(Z_f=1\mid k_f)=k_f/2\), and \(P(Z_m=1\mid k_m)=k_m/2\). The child's count is \(k_c=Z_f+Z_m\). Summing the probabilities of all transmission pairs that produce the observed child genotype gives:

T(k_c\mid k_f,k_m) = \sum_{z_f=0}^{1}\sum_{z_m=0}^{1} \mathbf{1}(z_f+z_m=k_c) \left(\frac{k_f}{2}\right)^{z_f} \left(1-\frac{k_f}{2}\right)^{1-z_f} \left(\frac{k_m}{2}\right)^{z_m} \left(1-\frac{k_m}{2}\right)^{1-z_m}.

Here \(\mathbf{1}(\cdot)\) equals 1 when its condition holds and 0 otherwise. The indicator keeps only transmission pairs that yield the observed child genotype; the remaining factors give the probability of each pair. For example, with two heterozygous parents, an \(Aa\) child can arise by either parent transmitting \(A\). Each of those two distinct transmission pairs has probability \(1/2\times1/2=1/4\), so their probabilities add to \(1/2\).

Father and mother genotypes	\(P(aa\text{ child})\)	\(P(Aa\text{ child})\)	\(P(AA\text{ child})\)
\(aa\) and \(aa\)	1	0	0
\(aa\) and \(Aa\)	1/2	1/2	0
\(aa\) and \(AA\)	0	1	0
\(Aa\) and \(Aa\)	1/4	1/2	1/4
\(Aa\) and \(AA\)	0	1/2	1/2
\(AA\) and \(AA\)	0	0	1

The table covers the distinct genotype pairings; reversing which parent has which genotype leaves the probabilities unchanged. In particular, when both parents are homozygous, the child's genotype is fixed. When at least one parent is heterozygous, one or more child genotypes can have nonzero probability, so the child's observed genotype can provide information about which allele was transmitted.

The Joint Likelihood for an Observed Trio

Under the stated population assumptions, multiply the two parental genotype probabilities by the conditional probability of the child's genotype. This factorization is the product rule for probabilities: first assign probabilities to the observed parents, then assign the probability of the child conditional on those parents.

Trio likelihood: For observed genotype counts \(k_f,k_m,k_c\), the joint likelihood as a function of \(p\) is

L(p;k_f,k_m,k_c) = f_{k_f}(p)f_{k_m}(p)T(k_c\mid k_f,k_m).

This is a likelihood because the observed genotypes are held fixed and the expression is viewed as a function of the model parameter \(p\). The transmission factor depends on the parental genotypes but not on \(p\) in this model. It can be zero if the observed child genotype cannot result from the observed parental genotypes.

If the question concerns transmission conditional on the observed parents, use \(T(k_c\mid k_f,k_m)\) alone. Including \(f_{k_f}(p)f_{k_m}(p)\) in that conditional likelihood would count the parental genotype distribution even though the parents' genotypes are being conditioned upon. The joint likelihood and the conditional transmission likelihood answer different questions; neither is automatically preferable without specifying the sampling and conditioning.

Worked Example: One Heterozygous Parent

Suppose a hypothetical trio has an \(AA\) father, an \(Aa\) mother, and an \(Aa\) child. Use \(p=0.30\), so \(q=0.70\), and calculate both the conditional transmission probability and the joint trio likelihood.

The father transmits \(A\) with probability \(2/2=1\). The mother transmits \(A\) with probability \(1/2\) and \(a\) with probability \(1/2\). An \(Aa\) child requires the father to transmit \(A\) and the mother to transmit \(a\). Therefore:

T(1\mid2,1)=1\times\frac{1}{2}=0.5.

For the parental genotype probabilities, \(P(AA)=p^2=(0.30)^2=0.09\), and \(P(Aa)=2pq=2(0.30)(0.70)=0.42\). Thus the joint likelihood is:

$$ L=(0.09)(0.42)(0.5)=0.0189. $$

The conditional likelihood of the child's genotype given these parents is 0.5. The joint likelihood of all three observed genotypes under the specified population model is 0.0189. The child genotype distinguishes between the mother's possible transmissions here because the father must transmit \(A\): an \(Aa\) child corresponds to the mother transmitting \(a\), whereas an \(AA\) child corresponds to her transmitting \(A\).

Worked Example: Two Heterozygous Parents

For a second hypothetical trio, suppose both parents are \(Aa\) and the child is \(Aa\), again with \(p=0.30\) and \(q=0.70\). Find the transmission probability and the joint trio likelihood.

An \(Aa\) child can result from the father transmitting \(A\) and the mother \(a\), or from the father transmitting \(a\) and the mother \(A\). Each path has probability \((1/2)(1/2)=0.25\). Adding the two mutually exclusive paths gives:

T(1\mid1,1)=0.25+0.25=0.50.

Each parent has genotype probability \(2pq=2(0.30)(0.70)=0.42\). The joint likelihood is therefore:

$$ L=(0.42)(0.42)(0.50)=0.0882. $$

The observed child genotype is compatible with two transmission paths. Because the child is heterozygous and the parents have the same genotype, the child's unphased genotype does not reveal which parent transmitted \(A\). It informs the combined outcome—one \(A\) and one \(a\) were transmitted—but does not identify the parent of origin.

Worked Example: Deterministic Transmission and an Impossible Genotype

Consider a third hypothetical trio with an \(aa\) father and an \(AA\) mother, with \(p=0.30\) and \(q=0.70\). First suppose the child is \(Aa\). Then consider what happens if the recorded child genotype is instead \(AA\).

The father can transmit only \(a\), and the mother can transmit only \(A\). Thus the child must be \(Aa\), with conditional probability one. The parental genotype probabilities are \(P(aa)=q^2=(0.70)^2=0.49\) and \(P(AA)=p^2=(0.30)^2=0.09\). For an \(Aa\) child:

T(1\mid0,2)=1,\qquad L=(0.49)(0.09)(1)=0.0441.

If the recorded child genotype were \(AA\), the transmission probability would instead be zero because the father cannot transmit \(A\):

T(2\mid0,2)=0,\qquad L=(0.49)(0.09)(0)=0.

The compatible trio has no random variation in transmission conditional on the parents: its child genotype was fixed before observing the child. The incompatible trio has zero likelihood under the stated no-mutation, error-free model. A zero likelihood flags a conflict between the observed genotypes and the assumptions; it does not by itself identify which assumption or observation is responsible.

Which Genotypes Inform Transmission?

The parental genotypes determine whether transmission is variable or deterministic. A heterozygous parent has two possible transmitted alleles, each with probability one-half. A homozygous parent has only one possible transmitted allele. Consequently, when both parents are homozygous, the child's genotype is predetermined: an observed compatible child contributes no variation-based information about which allele was transmitted. An incompatible child has probability zero under the model.

When at least one parent is heterozygous, the offspring genotype can distinguish among possible transmission outcomes. With \(Aa\) and \(AA\) parents, for example, the child is \(Aa\) if the heterozygous parent transmits \(a\), and \(AA\) if that parent transmits \(A\). Both outcomes have probability one-half. With two \(Aa\) parents, the child genotype has probabilities \(1/4\), \(1/2\), and \(1/4\) for \(aa\), \(Aa\), and \(AA\), respectively.

This is information about transmission conditional on the parents, not automatically information about the population allele frequency. In the joint likelihood, the parental genotype factors depend on \(p\), while the transmission factor in this basic model does not. Keeping those roles separate prevents a transmission observation from being misinterpreted as a direct estimate of allele frequency.

Common Mistakes

Using the child's genotype probability without conditioning on the parents. A probability such as \(2pq\) describes a randomly sampled genotype under Hardy–Weinberg equilibrium; it is not generally the child's probability given two observed parental genotypes. Use the Mendelian transmission probability for that conditional question.
Forgetting to add distinct transmission paths. With two heterozygous parents, an \(Aa\) child can result from either parent transmitting \(A\). Each path has probability \(1/4\), so the total is \(1/2\), not \(1/4\).
Multiplying in parental genotype probabilities after conditioning on the parents. The product \(f_{k_f}(p)f_{k_m}(p)\) belongs to the joint trio likelihood under the sampling assumptions. It is not part of the likelihood conditional on the observed parental genotypes.
Calling deterministic compatibility evidence of a random transmission outcome. If both parents are homozygous, the compatible child genotype is fixed. It is consistent with Mendelian inheritance but does not distinguish among variable transmission paths.

Key takeaway: For an observed trio, calculate the parental genotype probabilities under the chosen population model, calculate the child's genotype probability conditional on the parents by summing compatible Mendelian transmission paths, and multiply the two parts only when constructing the joint likelihood.

Check Your Understanding

Use the genotype coding and likelihood factorization from this tutorial to answer the following.

What is the probability that an \(Aa\) parent transmits \(A\)? What is the probability for an \(AA\) parent?
For \(aa\) and \(Aa\) parents, what are the probabilities of an \(aa\), \(Aa\), and \(AA\) child?
For two \(Aa\) parents and an \(Aa\) child, how many distinct transmission paths produce the child's genotype, and what is the probability of each path?
When does the parental genotype probability factor enter the trio likelihood, and when should it be omitted?
Why does a compatible child genotype from two homozygous parents not distinguish among variable transmission outcomes?