Question

The frequencies for autosomal alleles A and a are $p = 0.5$ and $q = 0.5$, respectively, where A is dominant over a. Under the assumption of random mating, the mating frequency among dominant parents is ................ (Rounded off to two decimal places).

Hint:

Whenever asked about dominant phenotypes, include both homozygous dominant ($AA$) and heterozygous ($Aa$) genotypes. Square the frequency when considering random mating among that group.

Answer 1

Step 1: Genotype frequencies under Hardy–Weinberg equilibrium.
- $AA = p^2 = 0.25$
- $Aa = 2pq = 0.50$
- $aa = q^2 = 0.25$
Step 2: Dominant phenotype.
Dominant phenotype = $AA + Aa = 0.25 + 0.50 = 0.75$.
So, frequency of dominant parents = $0.75$. Step 3: Mating frequency among dominant parents.
Mating frequency = $(0.75)^2 = 0.5625 \approx 0.56$. Therefore, the mating frequency among dominant parents is about 0.56.