In a population of 100 individuals, we have 25 people that are homozygous AA, 50 people that are heterozygous Aa, and 25 people that are homozygous aa. Assuming that the population is in Hardy-Weinberg equilibrium, match the frequency of the heterozygous Aa genotype to the appropriate part of the Hardy-Weinberg equilibrium equation: Select one: a. q² b. 2pq c. p²

b.  2pq

Explanation:

If there exist a population of 100 individuals and we have a cross between two heterozygous individuals

i.e    Aa    ×    Aa

             A              a

A          AA             Aa

a           Aa              aa

we can see that the proportion of the offspring are:

AA = 1/4       which is equivalent to 25 %

Aa = 2/4      which is equivalent to 50 %

aa = 1/4        which is equivalent to 25 %

Now, in solving Hardy Weinberg Equilibrium

There are two equation involved.

p  +  q = 1

p²  + 2pq + q² = 1

where:

p² exist as the frequency of individual with homozygous dominant genotype(AA) = 25%

2pq as the frequency of individual with the heterozygous genotype (Aa) = 50%

q² exist as the frequency of individual with homozygous recessive genotype(aa) = 25%

As such, the frequency of the heterozygous Aa genotype to the appropriate part of the Hardy-Weinberg equilibrium equation is  2pq.

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