Cystic fibrosis is an autosomal recessive disorder. In one population, the frequency of affected individuals (A 2 A 2) is 0.0004. Assuming that this population is under Hardy-Weinberg equilibrium, calculate all allele frequencies and genotype frequencies. Enter your numerical answers in the boxes below. Express each answer to four decimal places.

Hardy Weinberg Equilibrium

Explanation:

Hardy Weinberg Equilibrium formula for allelic frequency is: Hardy Weinberg Equilibrium formula for genotypic frequency is: Here,=frequency of dominant allele

                 =frequency of recessive allele

                 =frequency of dominant genotype

                 =frequency of recessive genotype

                 =frequency of heterozygous genotype

frequency of affected individuals=0.0004(given)

here,affected individuals means the one's having Cystic Fibrosis and since it is recessive disease so we can write frequency of recessive genotype as:     =0.0004

If =0.0004 will be 0.02;frequency of recessive allele will be 0.02Using the formula

                                       

hence,frequency of dominant allele will be 0.98Frequency of dominant genotype will be Frequency of heterozygous genotype will be

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