in a population that is in hardy Weinberg equilibrium the frequency of the recessive homozygous genotype of a certain tree is 0.13 calculate the percentage of individuals homozygous for the dominant allele
In Hardy-Weinberg equilibrium, the frequency of homozygous recessive individuals (q^2) can be determined by taking the square root of the frequency of the recessive genotype. In this case, the frequency of the recessive homozygous genotype is 0.13, so q^2 = 0.13.
Taking the square root of 0.13, we find that the frequency of the recessive allele (q) is approximately 0.36.
The frequency of the dominant allele (p) can be calculated as 1 - q, so p ≈ 1 - 0.36 = 0.64.
To calculate the percentage of individuals homozygous for the dominant allele (p^2), we square the frequency of the dominant allele. p^2 = (0.64)^2 = 0.4096.
Converting this into a percentage, we get 40.96% of individuals homozygous for the dominant allele in the population.