NEET 2019 Biology Paper

Frequency of dominant allele (say p) = 0.4
Frequency of recessive allele (say q) = 1 – 0.4 = 0.6
∴‌‌‌ Frequency of homozygous dominantindividuals (AA)
‌‌‌‌‌‌= p‌2= (0.4)‌2 = 0.16
‌‌‌‌‌‌Frequency of heterozygous individuals (Aa)
‌‌‌‌‌‌= 2pq = 2(0.4)(0.6) = 0.48
‌‌‌‌‌‌Frequency of homozygous recessiveindividuals (aa)
‌‌‌‌‌‌= q‌2 = (0.6)‌2= 0.36