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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.

Respuesta :

Hardy Weinberg Equilibrium

Explanation:

  • Hardy Weinberg Equilibrium formula for allelic frequency is: [tex]p+q=1[/tex]
  • Hardy Weinberg Equilibrium formula for genotypic frequency is: [tex]p^2+q^2+2pq=1[/tex]
  • Here,[tex]p[/tex]=frequency of dominant allele

                 [tex]q[/tex]=frequency of recessive allele

                 [tex]p^2[/tex]=frequency of dominant genotype

                 [tex]q^2[/tex]=frequency of recessive genotype

                 [tex]2pq[/tex]=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:     [tex]q^2[/tex]=0.0004

  • If [tex]q^2[/tex]=0.0004 [tex]q[/tex] will be 0.02;frequency of recessive allele will be 0.02
  • Using the formula [tex]p+q=1[/tex]

                                       [tex]p+0.02=1\\p=1-0.02=0.98[/tex]

  • hence,frequency of dominant allele will be 0.98
  • Frequency of dominant genotype will be [tex]p^2=(0.98)^2=0.9604[/tex]
  • Frequency of heterozygous genotype will be [tex]2pq=2*0.98*0.02=0.0392[/tex]

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