Recessive traits: If the frequency of a recessive trait such as cystic fibrosis or PKU is known, it is possible to calculate allele frequencies and genotype frequencies using the Hardy Weinberg equation and its assumptions are as follows: 

(i) say 1 in 1, 2500 Indian newborns have cystic fibrosis which means that the frequency of homozygotes for this recessive trait is 

 q² = 1/2,500 = 0.0004 

(ii) The square root of the frequency of recessives is equal to the allele frequency of the cystic fibrosis allele q = (0.0004)0.5 = 0.02 

(iii) The frequency of the normal allele is equal to 1 - the frequency of the cystic fibrosis allele p = 1- q = 1 - 0.02 = 0.98105 

(iv) The frequency of carriers (heterozygotes) for the cystic fibrosis allele is 2pq = 2 (0.98)(0.02) = 0.04 or 1/25 69 v. The frequency of homozygotes for the normal allele is 

p² = (0.98)² = 0.96

(vi) Thus the population is composed of three genotypes at the calculated frequencies of homozygous normal = 0.96, heterozygous carriers = 0.04, homozygous affected = 0.0004