Question

in a gene pool, the Z allele frequency is 89% and the z allele frequency is 11%. Determine the frequency of heterozygous (Zz) individuals in the population

Answer 1

Step-by-step explanation:

The frequency of heterozygous individuals can be calculated using the Hardy-Weinberg equation:

p^2 + 2pq + q^2 = 1

where p is the frequency of one allele (Z in this case) and q is the frequency of the other allele (z).

Given that the Z allele frequency is 89%, we can calculate q as:

q = 1 - p

q = 1 - 0.89

q = 0.11

Using this value for q and the given value for p, we can calculate the frequency of heterozygous individuals as:

2pq = 2(0.89)(0.11) = 0.196

So, the frequency of heterozygous (Zz) individuals in the population is 0.196 or 19.6%.