Offspring Genotype Probabilities From Parental Genotypes

From Parental Genotypes to Offspring Genotypes

A specified pair of parental genotypes determines which alleles each parent can pass to an offspring and how likely each transmission is. The offspring receives one allele from each parent. Combining the two transmission probabilities gives the probability of each possible offspring genotype. This is a calculation for a cross at a single locus; it is not a statement about how common genotypes are in a population.

We will use \(A\) and \(a\) for the two alleles at one autosomal locus. The genotypes are \(AA\), \(Aa\), and \(aa\). In this tutorial, “carrier” refers to an individual with genotype \(Aa\); whether that genotype is associated with a particular phenotype depends on additional biological assumptions.

Transmission and the Combination Rule

At a single locus, a parent with genotype \(AA\) transmits \(A\) with probability 1. A parent with genotype \(aa\) transmits \(a\) with probability 1. A heterozygous parent \(Aa\) transmits \(A\) with probability \(1/2\) and \(a\) with probability \(1/2\), under Mendelian segregation. These are probabilities for the allele in one gamete, not probabilities for the offspring’s full genotype.

Conditional on the two parental genotypes, the allele transmitted by one parent and the allele transmitted by the other are treated as independent. Thus, the probability of one particular pair of transmitted alleles is the product of the two transmission probabilities. If either parental transmission can take more than one route to the same offspring genotype, the route probabilities are added.

For an offspring with genotype \(Aa\), there are generally two possible routes: parent 1 transmits \(A\) while parent 2 transmits \(a\), or parent 1 transmits \(a\) while parent 2 transmits \(A\). The two routes are mutually exclusive, so their probabilities are added:

Here \(P_1(A)\), for example, means the probability that parent 1 transmits allele \(A\). This calculation concerns an unordered offspring genotype: \(Aa\) and \(aA\) describe the same genotype.

A Short Derivation

The combination rule follows directly from independence and addition of probabilities for disjoint events. Let \(G_1\) and \(G_2\) be the alleles transmitted by parents 1 and 2. The event that the offspring is \(Aa\) is the union of the events \((G_1=A,G_2=a)\) and \((G_1=a,G_2=A)\). Those events cannot both occur in the same conception. Therefore:

The second equality uses independence of the parental transmissions. For a homozygous offspring genotype, there is only one allele-pair route: for example, \(AA\) requires both parents to transmit \(A\), so \(P(AA)=P_1(A)P_2(A)\).

Homozygous Parents and Carrier Crosses

When one parent is homozygous, that parent’s gamete allele is fixed. This makes many crosses especially direct: the offspring genotype depends on the allele transmitted by the other parent. The same product-and-addition rules still apply; no separate method is needed.

A Reliable Calculation Procedure

For any specified cross at this locus, the calculation can be organized in the following sequence.

A Punnett square is a compact way to carry out these steps. Its cells represent combinations of parental gametes; when gametes have unequal probabilities, the cells must be weighted accordingly. The square is an enumeration device, not a reason to assume that every cell has equal probability in every cross.

Assumptions and Interpretation

The calculations above assume that the parental genotypes at the locus are known, alleles segregate according to Mendelian transmission, and there is no mutation altering the transmitted allele. The two parental gamete transmissions are independent conditional on the parental genotypes, and fertilization combines one gamete from each parent. These assumptions define the simple single-locus cross being analyzed.

The results apply to each conception under those assumptions. For example, a 50% probability of genotype \(Aa\) does not mean that exactly half of a couple’s children must have genotype \(Aa\). Nor does one child’s genotype change the transmission probabilities for a later conception under the model. The probability describes the outcome distribution for an individual offspring, not a guaranteed family count.

Genotype probabilities also do not automatically equal phenotype probabilities. To translate genotypes into a trait or disease outcome requires a model relating genotype to phenotype, including any relevant dominance, penetrance, or environmental assumptions. A cross calculation alone supplies the genotype distribution and should not be presented as a disease-risk calculation unless that relationship is specified.

Common Mistake

• Wrong reasoning: “The genotype \(Aa\) has two arrangements, \(Aa\) and \(aA\), so it is a different genotype from each arrangement, or it should be counted without weighting the parental gamete probabilities.”
• Correction: \(Aa\) is one unordered genotype. Add the probabilities of the distinct routes that produce it, multiplying the two parental transmission probabilities within each route. For the \(Aa\)-by-\(Aa\) cross, those two routes each have probability \(1/4\), giving \(P(Aa)=1/4+1/4=1/2\).
• Another useful check: If the genotype probabilities do not sum to 1, a possible transmission route may have been omitted or a route may have been counted more than once.