Pedigree Transmission Probabilities

From Parental Genotypes to Offspring Probabilities

At a genetic locus, an offspring receives one allele from each parent. If the parental genotypes are known, this gives a direct way to calculate the probabilities of the offspring's possible genotypes. The calculation is useful in family-based analysis because it describes the transmission mechanism itself, before considering disease status or any other observed trait.

This tutorial focuses on a single autosomal locus with two alleles, \(A\) and \(a\). The same probability logic applies to other allele labels. The calculations below assume that each parent contributes one allele, that each allele copy in a heterozygous parent has an equal chance of transmission, and that there is no mutation or segregation distortion. Parental genotypes are treated as known.

Definition: A transmission probability is the probability that a parent passes a particular allele to one offspring, conditional on the parent's genotype. An offspring-genotype probability is the probability of the genotype formed by combining the allele transmitted by each parent, conditional on both parental genotypes.

For example, an \(Aa\) parent has one copy of \(A\) and one copy of \(a\), so either allele is transmitted with probability \(1/2\). An \(AA\) parent can transmit only \(A\), and an \(aa\) parent can transmit only \(a\). These are probabilities of the allele received from one parent, not probabilities of a phenotype.

Formal Calculation

Let \(T_g(A)\) and \(T_g(a)\) denote the probabilities that a parent with genotype \(g\) transmits \(A\) or \(a\), respectively. Under ordinary Mendelian segregation at this locus, the transmission probabilities are:

Parental genotype \(g\)	\(T_g(A)\)	\(T_g(a)\)
\(AA\)	1	0
\(Aa\)	1/2	1/2
\(aa\)	0	1

Suppose parent 1 has genotype \(g_1\) and parent 2 has genotype \(g_2\). Assume their allele transmissions to an offspring are independent conditional on those genotypes. The offspring is \(AA\) only if both parents transmit \(A\); it is \(aa\) only if both transmit \(a\). An \(Aa\) offspring can arise in either of two mutually exclusive ways: parent 1 transmits \(A\) while parent 2 transmits \(a\), or parent 1 transmits \(a\) while parent 2 transmits \(A\).

\begin{aligned} P(AA \mid g_1,g_2) &= T_{g_1}(A)T_{g_2}(A),\\ P(Aa \mid g_1,g_2) &= T_{g_1}(A)T_{g_2}(a)+T_{g_1}(a)T_{g_2}(A),\\ P(aa \mid g_1,g_2) &= T_{g_1}(a)T_{g_2}(a). \end{aligned}

The two terms for \(Aa\) are added because they describe distinct ways to obtain that genotype. The probabilities across the three possible offspring genotypes sum to 1: every offspring receives one allele from each parent, and the resulting pair is \(AA\), \(Aa\), or \(aa\).

Key result: Given specified parental genotypes, calculate each parent's allele-transmission probabilities first. Multiply probabilities for the required transmissions from both parents, and add the mutually exclusive transmission routes that produce the same offspring genotype.

Worked Matings

Worked Example: Two Homozygous Parents

Suppose parent 1 is \(AA\) and parent 2 is \(aa\). Under the stated model, the first parent transmits \(A\) with probability 1, while the second transmits \(a\) with probability 1. Assume the genotypes are known, the locus is autosomal, and there is no mutation or segregation distortion.

For an \(AA\) offspring, both parents would need to transmit \(A\), but parent 2 cannot do so. For an \(aa\) offspring, both would need to transmit \(a\), but parent 1 cannot do so. The probability of \(Aa\) is:

\begin{aligned} P(Aa \mid AA,aa) &= T_{AA}(A)T_{aa}(a)+T_{AA}(a)T_{aa}(A)\\ &= (1)(1)+(0)(0)\\ &=1. \end{aligned}

Thus the offspring genotype probabilities are \(P(AA \mid AA,aa)=0\), \(P(Aa \mid AA,aa)=1\), and \(P(aa \mid AA,aa)=0\). Every offspring is heterozygous under these assumptions.

Worked Example: A Heterozygous Parent and a Homozygous Parent

Suppose parent 1 is \(Aa\) and parent 2 is \(aa\). The \(Aa\) parent transmits either allele with probability \(1/2\); the \(aa\) parent always transmits \(a\). Assume the specified genotypes are correct and Mendelian transmission applies.

An \(AA\) offspring is impossible because the \(aa\) parent cannot transmit \(A\). For an \(Aa\) offspring, parent 1 must transmit \(A\) and parent 2 must transmit \(a\):

P(Aa \mid Aa,aa) = T_{Aa}(A)T_{aa}(a)+T_{Aa}(a)T_{aa}(A) = (1/2)(1)+(1/2)(0) =1/2.

For an \(aa\) offspring, both parents must transmit \(a\):

P(aa \mid Aa,aa)=T_{Aa}(a)T_{aa}(a)=(1/2)(1)=1/2.

The probabilities are therefore \(0\) for \(AA\), \(1/2\) for \(Aa\), and \(1/2\) for \(aa\). They sum to 1, as required. Each offspring has a one-half probability of receiving \(A\) from the heterozygous parent, and a one-half probability of receiving \(a\) from that parent.

Worked Example: Carrier-by-Carrier Mating

Let \(a\) be a specified recessive variant, and suppose both parents are heterozygous carriers, \(Aa\). Here, “carrier” means heterozygous at this locus; it does not by itself specify disease status. Each parent transmits \(A\) with probability \(1/2\) and \(a\) with probability \(1/2\). Assume no mutation or segregation distortion and independent transmissions from the two parents.

For \(AA\), both parents must transmit \(A\). For \(aa\), both must transmit \(a\). The two distinct ways to obtain \(Aa\) are included in its probability:

\begin{aligned} P(AA \mid Aa,Aa) &= (1/2)(1/2)=1/4,\\ P(Aa \mid Aa,Aa) &= (1/2)(1/2)+(1/2)(1/2)=1/2,\\ P(aa \mid Aa,Aa) &= (1/2)(1/2)=1/4. \end{aligned}

The possible genotypes therefore have probabilities \(1/4\), \(1/2\), and \(1/4\), respectively. If the question is whether an offspring has at least one copy of \(a\), the relevant genotypes are \(Aa\) and \(aa\), so the probability is \(1/2+1/4=3/4\). If the question is whether the offspring is homozygous \(aa\), the probability is \(1/4\). These are different events and must not be confused.

Why the Formula Works

The calculation follows from the structure of the inheritance event. At this locus, each parent contributes exactly one of their two allele copies. Conditional on the parental genotypes, the probability of a particular pair of parental transmissions is the product of the two individual transmission probabilities, by the assumed independence of the parents' transmissions. An offspring genotype is the union of the transmission routes that produce that genotype.

For instance, in an \(Aa \times Aa\) mating, the \(Aa\) genotype has two routes: parent 1 transmits \(A\) and parent 2 transmits \(a\), or parent 1 transmits \(a\) and parent 2 transmits \(A\). Each route has probability \((1/2)(1/2)=1/4\). Since the routes cannot occur simultaneously for one offspring, their probabilities add to \(1/2\). This justifies the heterozygote formula rather than treating it as a single transmission route.

Record each parental genotype.
Use the specified genotypes at the same autosomal locus, such as \(Aa\) and \(aa\).

List each parent's possible transmitted alleles.
A homozygote transmits its only allele with probability 1; a heterozygote transmits each allele with probability \(1/2\).

Combine transmissions into genotypes.
Multiply the probabilities for each pair of parental transmissions, then add the probabilities of distinct pairs that give the same offspring genotype.

Check the result.
Confirm that genotype probabilities are between 0 and 1 and sum to 1.

Interpretation and Scope

These probabilities are conditional on the parental genotypes and the stated transmission model. They do not account for uncertainty about a parent's genotype, the probability that a genotype produces a trait, or the process by which a family was selected for study. They also do not assign a disease probability unless a separate model connects genotype to disease status.

For example, if \(a\) is associated with a recessive condition, the \(1/4\) probability of \(aa\) in a carrier-by-carrier mating is a probability of the offspring genotype under the assumptions above. It is not automatically a \(1/4\) probability of observing disease. Translating genotype into disease status requires additional assumptions about penetrance and other causes; genotype transmission alone does not supply them.

The calculations also rely on the model being appropriate. If transmission is not Mendelian, mutation is relevant, or the locus is not autosomal, the simple table of transmission probabilities may need modification. A probability of 0 or 1 here is conditional on the model and specified genotypes; it should not be presented as a claim that biological exceptions are impossible in every real setting.

Common Mistakes

Confusing transmission with offspring genotype. Saying “the \(Aa\) parent transmits \(a\) with probability \(1/2\)” describes one parent's contribution. The offspring genotype also depends on the allele transmitted by the other parent.
Forgetting the second route to a heterozygote. In an \(Aa \times Aa\) mating, either parent can provide \(A\), so the two routes to \(Aa\) must both be counted.
Calling every offspring with \(a\) a carrier. Under the usual single-locus terminology, a heterozygote \(Aa\) is a carrier of \(a\); an \(aa\) offspring has two copies. The event “has at least one \(a\)” includes both genotypes, while “is heterozygous” includes only \(Aa\).
Interpreting genotype probabilities as disease probabilities. Transmission probabilities describe inherited alleles conditional on parental genotypes. A phenotype requires an additional genotype-to-phenotype model.
Leaving out the conditioning assumptions. State the parental genotypes and the Mendelian assumptions. Otherwise, a calculated probability can be mistaken for an unconditional probability about a family.

Key takeaway: At a single autosomal locus, specify each parent's allele-transmission probabilities, multiply across parents for each transmission pair, and add distinct pairs that produce the same offspring genotype. Interpret the result as conditional on the parental genotypes and the transmission assumptions.

Check Your Understanding

For each question, treat the parental genotypes as known and assume Mendelian transmission at a single autosomal locus with no mutation or segregation distortion.

What allele-transmission probabilities apply to a parent with genotype \(AA\)? What about a parent with genotype \(Aa\)?
For an \(AA \times Aa\) mating, calculate the probabilities of \(AA\), \(Aa\), and \(aa\) offspring.
For an \(Aa \times aa\) mating, explain why an \(AA\) offspring has probability 0.
In an \(Aa \times Aa\) mating, why are there two transmission routes to genotype \(Aa\), and what is the total probability of that genotype?
In a carrier-by-carrier mating, distinguish the probability of an \(aa\) offspring from the probability of an offspring with at least one \(a\) allele. Why are these not the same event?