If allele frequency for a dominant allele is 0.4. What will be the number of heterozygous individuals if the population is of 100 individuals with diploid traits?

Explanation

As p+q =1, where p is frequency of dominant allele and q is frequency of recessive allele. 0.4 + q =1, q = 1-0.4 = 0.6. The Hardy-Weinberg equation is p2 +2pq + q2=1. Over here, p is 0.4 while q is 0.6, as we can subtract 0.4 from 1 as p + q = 1. Using the equation we will find 2pq and multiply it by 100 to get the number of heterogeneous individuals.The frequency of homozygous dominant individuals will be p2 =(0.4)^2 = 0.16. The frequency of homozygous recessive individuals will be q2=(0.6)^2 = 0.36. The frequency of heterozygous individuals will be 1 - (0.16 + 0.36) = 0.48. To find the number of heterozygous individuals in a population of 100 individuals, we multiply the frequency of heterozygous individuals by the total population size and then round to the nearest whole number: 0.48 x 100 = 48.Therefore, the answer is B. 48.