In a fictional diploid organism, cross Aa with Aa at one autosomal locus. Assume complete dominance of A, equal gamete probabilities, random fertilisation and equal offspring viability. What are the genotype and phenotype probabilities?
Genotypes: 25% AA, 50% Aa, 25% aa. Phenotypes: 75% dominant, 25% recessive.
- Find each parent’s gametes
Each Aa parent contributes A with probability 1/2 or a with probability 1/2. A capital and lowercase letter represent different alleles of the same gene.
- Combine the gametes
The four equally likely combinations are A with A → AA, A with a → Aa, a with A → Aa, and a with a → aa. Each combination has probability 1/4.
- Group the genotypes
One box is AA, two are Aa and one is aa. Therefore AA = 1/4, Aa = 2/4 and aa = 1/4.
\(P(AA)+P(Aa)+P(aa)=\frac14+\frac24+\frac14=1\)
- Translate to phenotypes
Under complete dominance, AA and Aa have the dominant phenotype. Add their probabilities: 1/4 + 2/4 = 3/4. Only aa has the recessive phenotype.
The genotype probabilities total 100%, and the phenotype probabilities total 100%. Each independent offspring has these probabilities; four offspring are not guaranteed to show a 3:1 ratio.