How to Calculate Allele Frequencies from Homozygote Frequency in Hardy-Weinberg Equilibrium

The Hardy-Weinberg equilibrium is a fundamental principle in population genetics, providing a mathematical model for predicting genotype and allele frequencies in a population that is not evolving. If you know the total frequency of homozygotes in a diploid population, you can use the Hardy-Weinberg equation to determine the frequencies of each allele.

Problem Overview

• Given: The frequency of homozygotes (both AA and aa) in a diploid population is 0.68.

• Task: Find the frequencies of the two alleles (A and a) in the population.

Step 1: Recall the Hardy-Weinberg Equation

For two alleles, A and a:

p2+2pq+q2=1p2+2pq+q2=1

• p2p2 = frequency of homozygous dominant (AA)

• $2pq$ = frequency of heterozygotes (Aa)

• q2q2 = frequency of homozygous recessive (aa)

The total frequency of homozygotes is:

p2+q2=0.68p2+q2=0.68

Step 2: Use the Relationship Between p and q

Since $p+q=1$, let’s express $q$ in terms of $p$:

p2+(1−p)2=0.68p2+(1−p)2=0.68

Expand (1−p)2(1−p)2:

p2+(1−2p+p2)=0.68p2+1−2p+p2=0.682p2−2p+1=0.682p2−2p+0.32=0p2+(1−2p+p2)=0.68p2+1−2p+p2=0.682p2−2p+1=0.682p2−2p+0.32=0

Divide both sides by 2:

p2−p+0.16=0p2−p+0.16=0

Step 3: Solve the Quadratic Equation

p2−p+0.16=0p2−p+0.16=0

Use the quadratic formula:

p=1±1−4×0.162p=21±1−4×0.16

Calculate the discriminant:

1−4×0.16=1−0.64=0.360.36=0.61−4×0.16=1−0.64=0.360.36=0.6

So,

p=1±0.62p=21±0.6

This gives two solutions:

• p=1+0.62=1.62=0.8p=21+0.6=21.6=0.8

• p=1−0.62=0.42=0.2p=21−0.6=20.4=0.2

Since p and q are interchangeable (A and a), the frequencies are 0.8 and 0.2.

Step 4: Match with the Provided Options

• (1) 0.1 and 0.9

• (2) 0.2 and 0.8

• (3) 0.4 and 0.6

• (4) 0.5 and 0.5

The correct answer is (2) 0.2 and 0.8.

Conclusion

If the frequency of homozygotes in a diploid population is 0.68, the frequencies of the two alleles are 0.2 and 0.8 under Hardy-Weinberg equilibrium