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Title: A recessive trait appears in 81% of the individuals in a population that is in Hardy-Weinberg equilibrium.? Post by: smag on Sep 13, 2012 A recessive trait appears in 81% of the individuals in a population that is in Hardy-Weinberg equilibrium. What percent of the population in the next generation is expected to be homozygous dominant?
Title: A recessive trait appears in 81% of the individuals in a population that is in Hardy-Weinberg equilibrium.? Post by: Sm4450 on Sep 13, 2012 In Hardy-Weinberg equilibrium, (p + q)^2 = p^2 + 2pq + q^2 = 1; and p + q = 1
The frequency of recessive trait, q^2(aa) = 0.81 The frequency of recessive allele, q(a) = SQRT 0.81 = 0.9 The frequency of dominant allele, p(A) = 1 - 0.9 = 0.1 Percentage of homozygous dominant in the population in the next generation would be p^2 (AA) = 0.1^2 = 0.01 |