In a population in Hardy-Weinberg equilibrium,75 percent of the individuals have a dominant allele for a particular gene (p=0.75)and 25 percent have a recessive allele (q=0.25).The proportion of homozygous recessive individuals in the F₁ generation will be

Question

Quizplus · Accepted Answer

According to the Hardy-Weinberg equilibrium, the proportion of homozygous recessive individuals is given by q^2. 
Therefore, q^2 = (0.25)^2 = 0.0625 or 6.25%.

In a Population in Hardy-Weinberg Equilibrium,75 Percent of the Individuals