Mendelian Transmission From Heterozygous Parents

Introduction

At an autosomal locus, a diploid parent carries two alleles but contributes only one copy of that locus to each gamete. A useful starting point for Mendelian probability is therefore to ask a narrowly defined question: what is the probability that this parent transmits a specified allele? The answer depends on whether the parent is homozygous or heterozygous.

This tutorial focuses on transmission from one parent at one autosomal locus. It establishes the allele-transmission probabilities and the conditions behind them. It does not combine the transmissions from two parents to obtain an offspring genotype distribution.

Definition: Let $G$ denote a parent’s genotype at a single autosomal locus, and let $T_A$ denote the event that a gamete from that parent carries allele $A$. The allele-transmission probability for $A$ is $P(T_A\mid G)$. For a diploid parent with alleles $A$ and $a$, each gamete carries one of the two alleles at this locus.

The Mendelian Segregation Rule

Suppose a parent is heterozygous, with genotype $Aa$. The two homologous copies of the autosome carry different alleles at the locus: one carries $A$, and the other carries $a$. In ordinary meiosis, the homologues separate, so a gamete receives one of them. If neither allele is preferentially transmitted, the two alternatives have equal probability.

Rule: Under Mendelian segregation at a single autosomal locus, with no transmission distortion, a heterozygous parent transmits each allele with probability one-half: $$ P(T_A\mid G=Aa)=\frac{1}{2}, \qquad P(T_a\mid G=Aa)=\frac{1}{2}. $$ A homozygous parent transmits the allele present on both homologues with probability one: $$ P(T_A\mid G=AA)=1, \qquad P(T_a\mid G=aa)=1. $$

These probabilities describe the allele in a gamete, conditional on the parent’s genotype. They do not describe how common an allele is in a population. They also do not imply that a heterozygote produces exactly equal numbers of the two gamete types in any particular finite set of meioses.

Why a Heterozygote Has a One-Half Probability

The probability follows from the two possible homologues that can carry the locus into a gamete. For an $Aa$ parent, one homologue carries $A$ and the other carries $a$. Under ordinary segregation, each homologue has the same chance of being represented in a gamete. The event $T_A$ occurs when the homologue carrying $A$ is represented; $T_a$ occurs when the one carrying $a$ is represented.

P(T_A\mid G=Aa) = \frac{\text{number of equally likely homologues carrying }A} {\text{number of equally likely homologues}} = \frac{1}{2}.

The same reasoning gives $P(T_a\mid G=Aa)=1/2$. The two events are mutually exclusive and exhaustive for the allele at this locus: a gamete from this parent carries either $A$ or $a$, but not both and not neither. Their probabilities therefore sum to one.

For $AA$, both homologues carry $A$, so either homologue contributes the same allele at the locus. Thus $P(T_A\mid G=AA)=2/2=1$. For $aa$, both carry $a$, giving $P(T_a\mid G=aa)=2/2=1$. These statements concern the allele at this locus; they do not claim that a gamete is identical across all loci.

Worked Examples

Worked Example: Transmission From Each Possible Genotype

Suppose, hypothetically, that a parent’s genotype at an autosomal locus is known to be $Aa$. What is the probability that a gamete from this parent carries $A$, and what is the probability that it carries $a$?

The parent is heterozygous, so each of the two homologues carries a different allele. Under Mendelian segregation, the homologues have equal chance of contributing the locus to a gamete: $$ P(T_A\mid G=Aa)=\frac{1}{2}=0.500, \qquad P(T_a\mid G=Aa)=\frac{1}{2}=0.500. $$

Thus, each specified allele has a 0.500 probability of transmission in a single gamete. The probabilities add to $0.500+0.500=1.000$, as required because the gamete receives one of the two alleles at the locus. These are probabilities, not a promise that any particular pair of gametes will contain one copy of each allele.

Worked Example: Transmission From a Homozygous Parent

In a hypothetical case, a parent has genotype $aa$ at one autosomal locus. Calculate the probability that a gamete carries $a$, and the probability that it carries $A$, assuming $A$ is the other allele considered at this locus.

Both homologues in this parent carry $a$. Whichever homologue contributes the locus, the gamete receives $a$. Therefore, $$ P(T_a\mid G=aa)=\frac{2}{2}=1.000. $$

There is no $A$ allele in this parent’s genotype to transmit, so $$ P(T_A\mid G=aa)=0. $$

The two probabilities sum to one. In this model, every gamete carries $a$ at the locus. This conclusion assumes the genotype is correctly specified and excludes mutation or other processes that could change the allele before or during transmission.

Worked Example: A Finite Set of Transmissions

Suppose, hypothetically, that an $Aa$ parent contributes gametes in eight independent meioses, and the question is whether exactly five of the eight gametes carry $A$. Assume each meiosis has the Mendelian transmission probability $P(T_A\mid G=Aa)=1/2$, with independent transmission outcomes across the eight meioses.

For any specified sequence with five $A$ transmissions and three $a$ transmissions, the probability is $$ \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^8 = \frac{1}{256}. $$

There are $\binom{8}{5}=56$ ways to choose which five of the eight transmissions carry $A$. These mutually exclusive sequences all have the same probability, so $$ P(\text{exactly five }A\text{ transmissions}) = \binom{8}{5}\left(\frac{1}{2}\right)^8 = \frac{56}{256} = 0.21875. $$

The probability is approximately 0.219. Even though the transmission probability is one-half, the probability of exactly four $A$ transmissions is not one, and exactly five is a possible outcome. A one-half transmission probability describes each meiosis; it does not force a balanced count in a finite sample.

Assumptions and Interpretation

The basic rule applies to a single autosomal locus and a specified parental genotype. It relies on ordinary Mendelian segregation: the alleles carried on the homologous chromosomes separate in meiosis, with no allele having a systematic transmission advantage. In interpreting the examples, assume the genotype is known, there is no mutation at the locus, and the gametes under consideration are not selectively removed in a way that changes the observed transmission probabilities.

A probability of one-half is a model statement about repeated transmissions under those assumptions. It does not say that the next gamete is half $A$ and half $a$; one gamete carries one allele at this locus. Nor does it say that every finite collection of gametes will show a 1:1 count. Random variation can produce unequal counts even when each transmission has probability one-half.

The independence assumption used in the eight-transmission example is appropriate when considering outcomes across independent meioses under the stated model. It should not be added automatically to every biological sampling setup. The one-half probability for an individual heterozygous transmission is the central segregation result; claims about a sequence or a count require additional assumptions about how those transmissions are sampled.

A Practical Calculation Strategy

Specify the locus and allele.
State which autosomal locus is under consideration and name the allele whose transmission event is being evaluated, such as $T_A$.

Condition on the parent’s genotype.
Use the stated genotype, rather than a population frequency or an offspring outcome, to determine which alleles the parent can transmit.

Apply segregation.
A homozygote transmits its allele with probability one. A heterozygote transmits either allele with probability one-half under Mendelian segregation.

Check the event and assumptions.
For the two alleles at this locus, the transmission probabilities should sum to one. For multiple transmissions, state any additional independence assumption before multiplying probabilities.

Common Mistake

A frequent error is to reason, “The parent is heterozygous, so exactly half of the gametes must carry each allele.” This confuses a probability for one transmission with a guaranteed count across a finite number of transmissions. For instance, in the eight-meiosis example, each meiosis has probability one-half of transmitting $A$, but exactly five $A$ transmissions still has probability $0.21875$, and other counts are possible.

The correction is to distinguish the levels of the claim. For one gamete from an $Aa$ parent, $P(T_A\mid G=Aa)=1/2$. For a set of gametes, the count varies; calculating its probability requires specifying the number of transmissions and the assumptions about their dependence. For a homozygote, by contrast, transmission of its allele at the locus is certain under the stated model.

Key takeaway: At a single autosomal locus, a homozygous parent transmits the allele it carries with probability one, while a heterozygous parent transmits either allele with probability one-half under Mendelian segregation. These are conditional transmission probabilities, not guarantees about finite counts.

Check Your Understanding

Answer each question using the parent’s genotype and the stated Mendelian transmission model.

A parent has genotype $BB$ at an autosomal locus. What is the probability that a gamete carries $B$? State the assumption behind your answer.
A parent has genotype $Bb$. What are the transmission probabilities for $B$ and $b$, and why do they sum to one?
Explain why a transmission probability of one-half does not guarantee exactly equal counts of the two alleles in a finite set of gametes.
For an $Aa$ parent, what is the probability of the particular sequence $A,a,A$ across three independent meioses?
What extra assumption is needed to calculate the probability of a sequence of transmissions by multiplying the probability of each transmission?