Question

If the frequency of homozygous dominant is 60%, the frequency of heterozygous is 20%, and the frequency of homozygous recessive is 20%, what is the frequency of the dominant allele, and the frequency of the recessive allele

Answer 1

When a genetic population follows Hardy-Weinberg Equilibrium (HW), it states that certain biological tenets or requirements must be met. Given so, then HW states that the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimal. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
So the problem states that homozygous dominant individuals (p^2) account for 60%, or 0.60. Thus the square root (sr) of p^2 = p or the dominant allele frequency in the population. So sr(p^2) = sr(0.60) -->
p = 0.775 or 77.5%
Homozygous recessive individuals (q^2) account for 20%, or 0.20. Thus sr(q^2) = q or the recessive allele frequency in the population. So sr(q^2) = sr(0.20) --> q = 0.447 or 44.7%
But since 44.7% + 77.5% = 122.2%, which is not equal to 1, we have a situation in which the allele frequencies do not match up, therefore this population cannot be determined using the Hardy-Weinberg Equation.

Answer 2

One way of calculating allelic frequencies from genotypic frequencies, is to add half the value of heter0zyg0us frequency to each of the h0m0zyg0us ones. In the exposed example, p = 0.7 and q = 0.3.

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Available data:

• Frequency of the h0m0zyg0us dominant = 60% = 0.6
• Frequency of heter0zyg0us = 20% = 0.2
• Frequency of the h0m0zyg0us recessive = 20% = 0.2

We need to calculate the allelic frequencies.

To get them , we can not just calculate the square root of genotypic h0m0zyg0us frequencies, because we would be missing the alleles included in the heter0zyg0us genotype.

What we need to do is to add half of the heter0zyg0us value, to each of the h0m0zyg0us values. So, following to Hardy-Weinberg theory:

Genotypic frequencies:

• h0m0zyg0us dominant genotype = F(XX) = p² ⇒ X being the dominant allele

p² = 60% = 0.6

• heter0zyg0us genotype = F(Xx) = 2pq

2pq = 20% = 0.2

• h0m0zyg0us recessive genotype = F(xx) = p² ⇒ x being the recessive allele

q² = 20% = 0.2

Allelic frequencies:

• f(X) = p = dominant allelic frequency

p = p² + 1/2 (2pq)

p = 0.6 + 1/2 (0.2)

p = 0.6 + 0.1

p = 0.7

• f(x) = q = recessive allelic frequency

q = q² + 1/2 (2pq)

q = 0.2 + 1/2 (0.2)

q = 0.2 + 0.1

q = 0.3

So, clearing the following equations, the result should equal 1.

                   p + q = 1                                   p² + 2pq + q² = 1

                0.7 + 0.3 = 1                                0.6 + 0.2 + 0.2 = 1

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