Question: Calculate the frequency of the AA, Aa, and aa genotypes after one generation if the initial population consists of 0.2 AA, 0.6 Aa, and 0.2 aa.

What I have done is:

P = (0.2)+(0.5)(0.6)= 0.50

q = 1-p = 1 - 0.50 = 0.50

Can someone check my answer plz.

Looks correct. There are good google articles under genotype probabilities

To calculate the frequency of the AA, Aa, and aa genotypes after one generation, you can use the Hardy-Weinberg Equilibrium equation. This equation allows you to determine genotype frequencies based on the allele frequencies in the population.

The Hardy-Weinberg equation states that p^2 + 2pq + q^2 = 1, where p represents the frequency of the dominant allele (in this case, A) and q represents the frequency of the recessive allele (in this case, a).

In your case, you have already determined the initial allele frequencies using the given population information:
p = 0.2 + 0.5(0.6) = 0.5 (frequency of the dominant allele A)
q = 1 - p = 1 - 0.5 = 0.5 (frequency of the recessive allele a)

Now, let's calculate the genotype frequencies after one generation using the values of p and q obtained:

Frequency of AA genotype (p^2):
AA genotype frequency = p^2 = 0.5^2 = 0.25 (or 25%)

Frequency of Aa genotype (2pq):
Aa genotype frequency = 2pq = 2(0.5)(0.5) = 0.5 (or 50%)

Frequency of aa genotype (q^2):
aa genotype frequency = q^2 = 0.5^2 = 0.25 (or 25%)

So, after one generation, the frequencies of the AA, Aa, and aa genotypes in the population will be 25%, 50%, and 25%, respectively.

It seems your calculation and answer are correct! Well done!