Genotype Configurations in Nuclear Families

Why Enumerate Family Genotypes?

A nuclear family may have recorded genotypes for only some relatives. Even so, the family’s possible genetic states can be described by assigning a genotype to each parent and each child. Each such assignment is a genotype configuration. Its probability follows from two ingredients: the probabilities of the parental genotypes and the Mendelian transmission probabilities from those parents to their children.

This representation is useful because it makes the family structure explicit. Rather than treating a child’s genotype as an isolated observation, it records which parental genotypes could have produced it and how likely each complete assignment is. The focus here is on probabilities of specified configurations, not on disease status or on combining all configurations compatible with incomplete data.

Genotypes and Transmission Probabilities

Consider one autosomal locus with two alleles, $A$ and $a$, and assume no mutation. The possible diploid genotypes are $AA$, $Aa$, and $aa$. Let $G_f$ and $G_m$ denote the father’s and mother’s genotypes, and let $G_j$ denote the genotype of child $j$. A parent transmits one of their two alleles to each child. A heterozygous parent transmits either allele with probability $1/2$; a homozygous parent transmits their only allele with probability 1.

Definition: A genotype configuration for a nuclear family is a specified assignment of genotypes to both parents and to each child under consideration. Its joint probability is the probability of that complete assignment under a stated model for parental genotypes and allele transmission.

The table gives the possible child genotype probabilities for each ordered pair of parental genotypes. The order is father first and mother second; because this locus is autosomal, reversing the parents gives the same probabilities.

Father	Mother	Child genotype probabilities
AA	AA	$P(AA)=1$
AA	Aa	$P(AA)=1/2,\;P(Aa)=1/2$
AA	aa	$P(Aa)=1$
Aa	Aa	$P(AA)=1/4,\;P(Aa)=1/2,\;P(aa)=1/4$
Aa	aa	$P(Aa)=1/2,\;P(aa)=1/2$
aa	aa	$P(aa)=1$

For example, an $Aa$ parent produces an $A$-bearing gamete with probability $1/2$ and an $a$-bearing gamete with probability $1/2$. When both parents are $Aa$, the child receives $A$ from each with probability $(1/2)(1/2)=1/4$, receives one of each allele with probability $1/4+1/4=1/2$, and receives $a$ from each with probability $1/4$.

Joint Probability of a Family Configuration

First consider parental genotypes treated as known and fixed. Conditional on those genotypes, the probability of a specified configuration of $k$ children is the product of the children’s genotype probabilities, provided the children’s meioses are independent given the parental genotypes. Each child receives one allele from each parent, and the transmissions to different children are independent under the standard Mendelian model.

P(G_1=g_1,\ldots,G_k=g_k\mid G_f=g_f,G_m=g_m) =\prod_{j=1}^{k}P(G_j=g_j\mid G_f=g_f,G_m=g_m).

If parental genotypes are themselves unobserved, a probability for a complete configuration also needs a model for their genotypes. For illustration, suppose the parents are unrelated founders sampled independently from a population in Hardy–Weinberg equilibrium. Let $r$ be the frequency of allele $A$, so the frequency of $a$ is $1-r$. Under this assumption, the genotype probabilities for each founder are $r^2$, $2r(1-r)$, and $(1-r)^2$ for $AA$, $Aa$, and $aa$, respectively. These frequencies are used for the parents; offspring genotype probabilities are then determined by the parental genotypes and transmissions.

Result: Under independent founder sampling and independent Mendelian transmissions to children, the probability of a complete configuration is $$ P(G_f=g_f,G_m=g_m,G_1=g_1,\ldots,G_k=g_k) =P(G_f=g_f)P(G_m=g_m)\prod_{j=1}^{k}P(G_j=g_j\mid G_f=g_f,G_m=g_m). $$ If parental genotypes are conditioned on as known rather than assigned founder probabilities, use only the product of the child transmission probabilities.

Worked Example: Two Children of a Heterozygous and a Homozygous Parent

Suppose a hypothetical family has a father with genotype $Aa$ and a mother with genotype $aa$. Calculate the probability, conditional on these parental genotypes, that their first child is $Aa$ and their second child is $aa$.

The father transmits $A$ or $a$ with probability $1/2$ each, while the mother always transmits $a$. Thus, for each child, $P(Aa\mid Aa,aa)=1/2$ and $P(aa\mid Aa,aa)=1/2$. Assuming independent transmissions across the two children, the specified configuration has probability

P(G_1=Aa,G_2=aa\mid G_f=Aa,G_m=aa) =\frac{1}{2}\times\frac{1}{2}=0.25.

So, conditional on these parental genotypes, the probability of this ordered configuration—first child $Aa$, second child $aa$—is 0.25. Reversing which child has which genotype also has probability 0.25. These are distinct configurations because the children are identified by their order in the question.

Configurations with Unobserved Parents

When a parent’s genotype is unobserved, it is still possible to calculate the probability of a specified complete configuration by assigning that parent a candidate genotype and including its founder probability. An unobserved genotype is not a fourth genotype category: it is a genotype whose value is unknown in the record. Different candidate assignments define different complete configurations.

Worked Example: A Candidate Configuration for Two Unobserved Parents

In a hypothetical population, the frequency of allele $A$ is $r=0.7$, so the frequency of $a$ is $0.3$. Assume unrelated founder parents are independently sampled from a population in Hardy–Weinberg equilibrium. Calculate the probability of the complete configuration in which the father is $Aa$, the mother is $AA$, the first child is $Aa$, and the second child is $AA$.

The founder probabilities are

P(G_f=Aa)=2(0.7)(0.3)=0.42, \qquad P(G_m=AA)=(0.7)^2=0.49.

For an $Aa$ father and an $AA$ mother, each child has probability $1/2$ of $Aa$ and $1/2$ of $AA$. The probability of the specified ordered pair of child genotypes, conditional on the parents, is $(1/2)(1/2)=0.25$. Therefore, the complete configuration has probability

$$ P(Aa,AA,Aa,AA) =(0.42)(0.49)(0.25) =0.05145. $$

The result, about 0.05145, is the probability of this one fully specified configuration under the stated population and transmission assumptions. It is not a probability that the observed family record has those genotypes, because the parents’ and children’s genotypes in this example were specified as a complete assignment for calculation.

The assumptions matter. If the parents are related, drawn from different populations, or sampled under a mating pattern that changes genotype frequencies, the independent Hardy–Weinberg founder probabilities may not apply. In that case, a different parental genotype model is needed. The child transmission calculation can still be made conditional on parental genotypes under the Mendelian assumptions.

Several Complete Configurations Can Fit Partial Information

Partial records often leave more than one possible complete configuration. A useful first step is to list the candidate genotypes and calculate each configuration’s probability using the same rules. The calculations should keep clear which genotypes are fixed as observed and which are candidate assignments for unobserved relatives.

Worked Example: An Unobserved Father and Two Recorded Children

Suppose a hypothetical family’s mother is known to be $Aa$, and the two children are recorded as $aa$ and $Aa$, in that order. The father’s genotype is unobserved. Let the frequency of $A$ among unrelated founders be $0.7$, with Hardy–Weinberg equilibrium. Calculate the probabilities of two complete configurations: one with an $aa$ father and one with an $Aa$ father, conditional on the mother’s known genotype.

For the candidate $aa$ father, the founder probability is $P(G_f=aa)=(0.3)^2=0.09$. An $aa$ father and $Aa$ mother produce either $aa$ or $Aa$ with probability $1/2$ each. Thus,

P(G_f=aa,G_1=aa,G_2=Aa\mid G_m=Aa) =(0.09)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) =0.0225.

For the candidate $Aa$ father, the founder probability is $2(0.7)(0.3)=0.42$. With both parents $Aa$, the first child’s $aa$ probability is $1/4$, and the second child’s $Aa$ probability is $1/2$. Therefore,

P(G_f=Aa,G_1=aa,G_2=Aa\mid G_m=Aa) =(0.42)\left(\frac{1}{4}\right)\left(\frac{1}{2}\right) =0.0525.

These are probabilities for two separate complete assignments, conditional on the known maternal genotype and the stated founder model. They show how the same recorded child genotypes can be paired with different candidate paternal genotypes, each with its own probability. The calculations do not by themselves turn the incomplete record into one probability; doing that requires combining probabilities across all compatible assignments under an explicitly defined observation model.

A Short Justification of the Product Formula

The family probability formula follows from the chain rule and the conditional independence assumptions. For a fixed pair of parental genotypes, the chain rule writes the joint child probability as successive conditional probabilities. If the transmissions to different children are independent once the parents’ genotypes are given, each conditional probability depends on the parental genotypes and the corresponding child genotype, but not on the other children’s genotypes. The product therefore follows.

\begin{aligned} &P(G_1=g_1,\ldots,G_k=g_k\mid G_f=g_f,G_m=g_m)\\ &=\prod_{j=1}^{k}P(G_j=g_j\mid G_f=g_f,G_m=g_m). \end{aligned}

If the parental genotypes are also random founders, independent sampling gives $P(G_f=g_f,G_m=g_m)=P(G_f=g_f)P(G_m=g_m)$. Applying the chain rule to the parents and children then multiplies this founder probability by the conditional child product. This derivation depends on the stated assumptions: it is not a claim that family members’ genotypes are unconditionally independent.

Common Pitfall and Practical Checks

A frequent error is to multiply child genotype probabilities without conditioning on the parents, or to treat siblings as independent draws from the population genotype frequencies. Siblings share their parents, so their genotype probabilities are linked through those parental genotypes. The correct calculation first fixes or assigns the parental genotypes, obtains each child’s transmission probability conditional on them, and only then multiplies across children under independent meioses.

A second error is to confuse the probability of one complete assignment with the probability of incomplete observations. If a parent’s genotype is unknown, a calculation for a particular candidate genotype is only that candidate configuration’s probability. Other candidate assignments remain possible. Keep the distinction explicit, and state whether a result is conditional on known parental genotypes or includes founder genotype probabilities.

Before accepting a configuration calculation, check that each offspring genotype can be formed from one allele transmitted by each parent, that the transmission probabilities for a given parental pair sum to 1, and that every founder probability follows from the stated genotype-frequency model. These checks catch impossible assignments, omitted alternatives, and mismatched assumptions.

Key takeaway: A family configuration probability is built from parental genotype probabilities, when needed, and Mendelian transmission probabilities conditional on those parents. Unobserved genotypes should be represented as alternative complete assignments, with the conditioning and probability model stated clearly.

Check Your Understanding

For each question, state the conditioning assumptions needed for your calculation.

If the parents are $Aa$ and $aa$, what is the probability that two ordered children are $aa$ and $Aa$, respectively, conditional on the parents?
Under Hardy–Weinberg equilibrium with $A$-allele frequency $r=0.6$, what is the founder probability of genotype $Aa$?
Why is the probability of a specified two-child configuration calculated as a product conditional on the parental genotypes?
With an unobserved parent, what does the probability for one candidate parental genotype and specified child genotypes describe?
Why should siblings not be treated as independent draws from the population genotype frequencies?

Father	Mother	Child genotype probabilities
AA	AA	\(P(AA)=1\)
AA	Aa	\(P(AA)=1/2,\;P(Aa)=1/2\)
AA	aa	\(P(Aa)=1\)
Aa	Aa	\(P(AA)=1/4,\;P(Aa)=1/2,\;P(aa)=1/4\)
Aa	aa	\(P(Aa)=1/2,\;P(aa)=1/2\)
aa	aa	\(P(aa)=1\)