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weqwewe [10]
3 years ago
5

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.
Biology
1 answer:
Simora [160]3 years ago
7 0

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

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