In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote genotype of a certain trait is 0.13. Calculate the percentage of individuals homozygous for the dominant
allele.
In Hardy-Weinberg equilibrium, the frequencies of the genotypes in a population can be calculated using the following equations:
p^2 + 2pq + q^2 = 1
where p is the frequency of the dominant allele and q is the frequency of the recessive allele. Since q^2 represents the frequency of the recessive homozygote genotype, we know that q^2 = 0.13.
Therefore, q = √0.13 = 0.36
Since p + q = 1, we can calculate that p = 1 - 0.36 = 0.64
To find the percentage of individuals homozygous for the dominant allele (p^2), we multiply p^2 by 100:
0.64^2 * 100 = 40.96%
Therefore, in a population in Hardy-Weinberg equilibrium where the recessive homozygote genotype frequency is 0.13, approximately 40.96% of the individuals are homozygous for the dominant allele.