To calculate the fraction of the population that is heterozygous in a population in Hardy-Weinberg equilibrium, we need to use the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
Where p represents the frequency of allele A and q represents the frequency of allele a.
Given that the frequency of allele A is 0.4, we can assign p = 0.4. Therefore, q = 1 - p = 1 - 0.4 = 0.6.
The term 2pq represents the frequency of heterozygotes (Aa) in the population. Plugging in the values, we have:
2(0.4)(0.6) = 0.48
Therefore, the fraction of the population that is heterozygous is 0.48.
In a population that is in Hardy-Weinberg equilibrium, there are two possible alleles for a certain gene, A and a. If the frequency of allele A is 0.4, what fraction of the population is heterozygous?
0.40
0.60
0.16
0.48
1 answer
1 answer