Calculate the frequencies of the genotypes in future generations if the initial

Question

Calculate the frequencies of the genotypes in future generations if the initial population consists of the following genotype frequencies: 0.6AA, 0.2Aa, and 0.2aa and meets the requirements of the Hardy–Weinberg relationship.

A. What will be the frequency of AA in the first generation?

B. What will be the frequency of Aa in the first generation?

C. What is the frequency of aa in the first generation?

D. What will be the frequency of AA in the second generation?

E. What will be the frequency of Aa in the second generation?

F. What will be the frequency of aa in the second generation?

Round your answers to three decimal places.

Explanation / Answer

To calculate genotype frequencies, we use HW equation:

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

Now, we have to calculate allelic frequencies.

Imagine that there are 10 individuals in this population. 60% (0.6) have the AA genotype, 20%(0.2) have the Aa genotype, and 20%(0.2) have the aa genotype.

This works out to around 6 individuals possessing two AA alleles, 2 individuals possessing both an A and "a" allele, and 2 individuals possessing 2 aa alleles.

So, we can calculate that the frequency of A= 0.5, and the frequency of a = 0.5. We substitute these values into our equation.

Therefore, in the second generation:
p^2 + 2pq + q^2 = 1
(0.5)^2 + 2(0.5)(0.5) + (0.5)^2 = 1.

p^2 = frequency of homozygous A genotype= 0.25
2pq = frequency of heterozygous genotype = 0.5
q^2 = frequency of homozygous a genotype = 0.25

Keep in mind that as the population abides to the HW relationship, mating is completely random. You would expect to end with the frequencies listed above.

Frequency of A = p = 0.6 + 0.2 /2 = 0.6 + 0.1 = 0.7

q = 1- p = 0.3

In the second generation,

Frequency of AA =p^2 = (0.5)^2 = 0.25 0r 25%
Frequency of Aa = 2pq = 2 (0.5) (0.5) = 0.5 or 50%
Frequency of aa = q^2 = (0.5)2 = 0.25 = 25%

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