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Cystic fibrosis is an autosomal recessive disease characterized by two copies of a mutated CFTR gene....

Cystic fibrosis is an autosomal recessive disease characterized by two copies of a mutated CFTR gene. If one in 100 people in the United States have cystic fibrosis and one in 5.0505050505 people are carriers for cystic fibrosis, calculate the number of individuals that are homozygous dominant. In other words, how many people would have two copies of the normal (non-mutated) CFTR gene. Use the Hardy-Weinberg equation and explain how you determined this.

Expert Solution

Answer :- As cystic fibrosis is autosomal recessive that means if only both the alleles are mutated then the only disease will only happen. That means patients have to be homozygous.

According to Hardy-Weinberg equation -

p = frequency of wild type allele

q = frequency of mutated alleles = 1% = 1/100 = .01

so, p^2 = homozygous population for wild type allele .

q^2 = homozygous population for mutated type allele which patients are affected by cytric fibrosis .

p+q = 1

so, p= 99% (as q = 1%)

Carrier is those who are heterozygous for mutated alleles; that means, they have one normal allele and one mutated allele.

2pq = 1/5.0505050505 = 0.198

(p+q)^2 = 1

or, p^2 + 2pq + q^2 = 1

or, p^2 = 1 - q^2 - 2pq = (1- 0.198- .01) = 0.792 = 79.2%

1000(79.2/100) = 792 peoples

Hence 79.2 % (792 people out of 1000 people) people are homozygous for dominant.

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gladiator answered 5 months ago

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