Question

In a population under Hardy-Weinberg equilibrium with alleles A and a, where the frequency of homozygous recessive individuals (aa) is 0.04, calculate the value of q (frequency of allele a), then find the value of p (frequency of allele A). Finally, determine the frequency of the heterozygote genotype (Aa).

Answer 1

The frequency of allele a (q) is 0.2, and the frequency of allele A(p) is 0.8. The frequency of the heterozygote genotype (Aa) is 0.32.

Step-by-step explanation:

In a population under Hardy-Weinberg equilibrium, the frequencies of alleles A and a are represented by p and q, respectively. The sum of these frequencies is always 1. Given that the frequency of homozygous recessive individuals (aa) is 0.04, we can directly equate this to the square of the frequency of a (q^2). Therefore, q^2 = 0.04. Taking the square root of both sides, we find that the frequency of a (q) is 0.2.

Now, to find the frequency of A (p), we subtract q from 1 (p = 1 - q). Therefore, p = 1 - 0.2, giving us a value of 0.8 for the frequency of A.

The frequency of heterozygotes (Aa) can be determined using the formula 2 * p * q. Substituting the values we found earlier, we get 2 * 0.8 * 0.2 = 0.32. Thus, the frequency of the heterozygote genotype is 0.32.

In summary, the Hardy-Weinberg equilibrium allows us to calculate allele frequencies and genotype frequencies in a population. In this case, the frequency of allele a (q) is 0.2, the frequency of allele A (p) is 0.8, and the frequency of the heterozygote genotype (Aa) is 0.32. These calculations provide insights into the genetic composition of a population and can be crucial in understanding evolutionary processes.

Answer 2

The frequency of allele a (q) is 0.2, the frequency of allele A (p) is 0.8, and the frequency of the heterozygote genotype (Aa) is 0.32 in a population under Hardy-Weinberg equilibrium where the frequency of homozygous recessive individuals is 0.04.

Step-by-step explanation:

Hardy-Weinberg Equilibrium Calculation

When a population is under Hardy-Weinberg equilibrium with alleles A and a, and the frequency of homozygous recessive individuals (aa) is given as 0.04, we can deduce the following:

The frequency of allele a (q) is the square root of the frequency of aa, which is √0.04 = 0.2.

Using the equation p + q = 1, we can find the frequency of allele A (p) by subtracting q from 1, giving us 1 - 0.2 = 0.8.

To determine the frequency of the heterozygote genotype (Aa), we use the equation 2pq. Plugging in the values for p and q, we get 2(0.8)(0.2) = 0.32.

In conclusion, the frequency of allele a (q) is 0.2, the frequency of allele A (p) is 0.8, and the frequency of the heterozygote genotype (Aa) is 0.32.