Question

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:

a) 0.0625
b) 0.25
c) 0.375
d) 0.5

Answer 1

The proportion of homozygous recessive individuals in the F1 generation, given a p value of 0.75 and q value of 0.25 in a Hardy-Weinberg equilibrium population, is 0.0625, calculated as q^2.

Step-by-step explanation:

The question asks to determine the proportion of homozygous recessive individuals in the F1 generation for a population in Hardy-Weinberg equilibrium, given that 75 percent have a dominant allele (p=0.75) and 25 percent have a recessive allele (q=0.25). According to the Hardy-Weinberg principle, which states that genotype frequencies in a population tend to remain constant from generation to generation in the absence of disturbing factors, the proportion of homozygous recessive individuals can be calculated using the equation p2 + 2pq + q2 = 1. Therefore, to find the frequency of the homozygous recessive genotype (aa), we need to calculate q2, which is (0.25)2 = 0.0625. This means the correct answer is a) 0.0625, representing the frequency of homozygous recessive individuals in the population.