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CF is an autosomal recessive condition. A population of 387 individuals has 7 people with CF....

CF is an autosomal recessive condition. A population of 387 individuals has 7 people with CF. If we assume that the CF gene is in Hardy-Weinberg equilibrium

what is the estimated frequency of the CF allele? Give answer in decimal places

Solutions

Expert Solution

Hardy-Weinberg Equilibrium states that a large interbreeding population will have same allelic and genotypic frequencies for several generations if it does not encounter any mutation, migration, or selection.

Hence, the frequency of the alleles can be denoted by p and q (generally p for the dominant type and q for teh recessive type) while the genotypic frequencies of the homozygous genotypes are given by p2 and q2. The heterozygous population is given by 2pq. These terms are obtained by the expantion of the following equation:

According to the Hardy-Weinberg Equilibrium:

p+q=1

Taking square on both the sides:

(p+q)2=1

p2 + q2 + 2pq=1

In the given question:

Total population=387

Individuals with CF or cystic fibrosis gene=7

Rest of the individuals = 380.

Geneotype frequency (q2)= The number of individuals with CF/Total individuals in the population

=7/387

=0.018

The allele frequency for the CF allele (q)= 0.134

According to the Hardy-Weinberg Equilibrium:

p+q=1

p=1-0.134

p=0.866

p2=0.749

According to the Hardy-Weinberg Equilibrium:

p2 + q2 + 2pq=1

0.749 + 0.018 + 2pq = 1

2pq= 1-0.767

= 0.233

Genotype with homozygous dominant individuals = p2 x 387 = 0.749 x 387 = 290

Genotype with homozygous recessive individuals = q2 x 387 = 0.018 x 387= 7

Genotype with heterozygous individuals= 2pq x 387 = 0.233 x 387= 90

Hence the frequency of the recessive allele or the CF allele (q) = 0.134

gladiator answered 1 year ago
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