Question

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.

Answer 1

Hardy Weinberg Equilibrium

Explanation:

• Hardy Weinberg Equilibrium formula for allelic frequency is: $p+q=1$

• Hardy Weinberg Equilibrium formula for genotypic frequency is: $p^2+q^2+2pq=1$

• Here,$p$=frequency of dominant allele

                 $q$=frequency of recessive allele

                 $p^2$=frequency of dominant genotype

                 $q^2$=frequency of recessive genotype

                 $2pq$=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:     $q^2$=0.0004

• If $q^2$=0.0004 $q$ will be 0.02;frequency of recessive allele will be 0.02

• Using the formula $p+q=1$

                                       $p+0.02=1\\p=1-0.02=0.98$

• hence,frequency of dominant allele will be 0.98

• Frequency of dominant genotype will be $p^2=(0.98)^2=0.9604$
• Frequency of heterozygous genotype will be $2pq=2*0.98*0.02=0.0392$