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

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In a population in Hardy-Weinberg equilibrium, the frequency of homozygous recessive individuals can be calculated as 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