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cystic fibrosis is a genetic disorder in humans where the person has two homozygous recessive alleles for the gene. if the disease is left untreated, it causes severe health problems in the individual. if 9 in 10,000 newborn babies have the disease, what are the expected frequencies of the dominant (a1) and recessive (a2) alleles according to the hardy-weinberg equation?

Sagot :

The expected frequencies of the dominant (A1) and recessive (A2) allele according to the Hardy-Weinberg equation are F(A1) = 0.97 and F(A2) = 0.03

According to Hardy Weinberg's law,

  • p+q = 1
  • (p+q)² = 1
  • p² + q² + 2pq =1

with:

p2 = dominant homozygous allele genotype frequency.

2pq = heterozygous genotype frequency

q2 = homozygous recessive allele genotype frequency

Given that cystic fibrosis is a recessive genetic disorder, the frequency of the disease is 9/10,000.

q2 = 9/10000

q = √(9/10000)

q = 3/100

q = 0.03.

So, the genotype frequency of the recessive homozygous allele is (A2) = 0.03

And the dominant homozygous allele frequency is

f(A1) = 1- f(A2)

f(A1) = 1.00 - 0.03

f(A1) =0.97.

Learn more about Hardy Weinberg's law at https://brainly.com/question/16823644

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