Question

In a hardy-weinberg population with two alleles, a and a, that are in equilibrium, the frequency of allele a is 0.2. what is the percentage of the population that is heterozygous for this allele

Answer 1

The Hardy-Weinberg equation is as follows:
$( {p + q})^{2} = 1$
${p}^{2} +2pq + {q}^{2} = 1$
Where:
(convert all % to decimals)

p= homozygous dominant
q= homozygous recessive
pq= heterozygous

While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.

Using the 1st equation, we can solve for the other dominant/recessive frequency:

1-0.2=0.8

Meaning that:

p= 0.8 & q=0.2

If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.

2(0.8)(0.2)= 2(0.16)= 0.32

So, the heterozygous frequency would be:
0.32

Hope this helps!

Answer 2

The percentage of the population that is heterozygous for the allele ; 32%

Applying Hardy-weinberg equation

P² + 2Pq + q² = 1   ------ ( 1 )

where ; P = h0m0zygous dominant  , q = h0m0zygous recessive,

            2Pq = heterozygous frequency

Frequency 0.2 = h0m0zygous recessive frequency ( q )

∴ Frequency of the h0m0zygous dominant frequency ( p ) = 0.8.

Back to equation ( 1 )

0.8² + 2Pq + 0.2² =  1

2Pq = 1 - ( 0.8 )² - (0.2)²

       = 0.32  = 32%

Hence In Conclusion the percentage of the population that is heterozygous is 32%

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