If the frequency of recessive allele in the population is 40%. Find the number of dominant individuals out of a population of 5000 individuals.

The correct option is B 4200
Frequency of recessive allele (q) = 0.4
Frequency of dominant allele (p) = 1-.4=0.6
Hence, dominant individuals would be = homozygous + heterozygous $=2pq+p^2$
$=(2\times 0.4\times 0.6)+(0.6\times 0.6)$
= 0.48 + 0.36
= 0.84 = 84%
So, dominant individuals in a population of 5000 = 0.84 X 5000 = 4200.