The frequency of two alleles in a gene pool is 0.17 (A) and 0.83(a). Assume that the population is in Hardy-Weinberg equilibrium.
Calculate the percentage of heterozygous individuals in the population.
To calculate the percentage of heterozygous individuals in the population, we can use the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
Where:
p = frequency of allele A
q = frequency of allele a
p^2 = frequency of homozygous dominant individuals (AA)
2pq = frequency of heterozygous individuals (Aa)
q^2 = frequency of homozygous recessive individuals (aa)
Given:
p = 0.17
q = 0.83
Using this information, we can substitute the values into the equation:
(0.17)^2 + 2(0.17)(0.83) + (0.83)^2 = 1
0.0289 + 0.2822 + 0.6889 = 1
0.999 = 1
This shows that the allele frequencies in the population add up to 1, which means the population is in Hardy-Weinberg equilibrium.
Now, the percentage of heterozygous individuals can be calculated by multiplying the frequency of heterozygous individuals (2pq) by 100:
2pq = 2(0.17)(0.83) = 0.2822
Percentage of heterozygous individuals = 0.2822 * 100 = 28.22%
Therefore, the percentage of heterozygous individuals in the population is 28.22%.