Question

Consider a population in Hardy-Weinburg equilibrium, with the following genotypic frequencies:

0.15 hh
0.48 Hh
0.37 HH
The allele frequency of H is ___.

Answer 1

The allele frequency of H in a Hardy-Weinburg equilibrium population with the given genotypic frequencies is calculated to be 0.61.

Step-by-step explanation:

To calculate the allele frequency of H in a population in Hardy-Weinburg equilibrium with genotypic frequencies of hh = 0.15, Hh = 0.48, and HH = 0.37, we can use the Hardy-Weinberg equation.

We know that the frequency of the homozygous dominant (HH) plus half of the frequency of the heterozygous (Hh) will give us the frequency of the dominant allele H (p).

p (frequency of H) = frequency of HH + 1/2(frequency of Hh)

= 0.37 + 1/2(0.48)

= 0.37 + 0.24

= 0.61.

Thus, the allele frequency of the dominant allele H is 0.61.