Question 6%0D%0AIn a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote%0D%0Agenotype of a certain trait is 0.13. Calculate the percentage of individuals homozygous for the dominant%0D%0Aallele.%0D%0A%0D%0A%0D%0A%0D%0A a%0D%0Ap^2= 29%%0D%0A b%0D%0Ap^2= 40%%0D%0A c%0D%0Aq^2= 57%%0D%0A d%0D%0Aq^2= 23%

Question

Question 6%0D%0AIn a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote%0D%0Agenotype of a certain trait is 0.13. Calculate the percentage of individuals homozygous for the dominant%0D%0Aallele.%0D%0A%0D%0A%0D%0A%0D%0A a%0D%0Ap^2= 29%%0D%0A b%0D%0Ap^2= 40%%0D%0A c%0D%0Aq^2= 57%%0D%0A d%0D%0Aq^2= 23%

Answer 1

b. p^2= 40%

To calculate the percentage of individuals homozygous for the dominant allele, you can use the equation p^2 + 2pq + q^2 = 1, where p and q represent the frequencies of the alleles.

Given that the frequency of the recessive homozygote genotype is 0.13, this corresponds to q^2 = 0.13. Therefore, q = sqrt(0.13) ≈ 0.36.

Since p + q = 1, p = 1 - q = 1 - 0.36 = 0.64.

To find the percentage of individuals homozygous for the dominant allele (p^2), you can square the frequency of the dominant allele (p), which is 0.64, to get 0.64^2 = 0.4096, or 40.96%. Rounded to the nearest whole number, this becomes 40%.