Question

cystic fibrous is a genetic disease characterized by the production of mucus that is abnormally viscous. Normal watery mucus is associated with the dominant allele for CFTR ( cystic fibrous tramsmembrane conductance regulator) gene; abnormally thick mucus is associated with the recessive allele for that gene. in the human population of the U.S , 2499 out of every 2500 people produce normal secretions, and so are free of cystic fibrous. if we assume the human population of the U.S. to be at Weinberg equilibrium with respect to the CFTR gene, then what is the frequency of individuals heterozygous for the cystic-fibrous allele?

Answer 1

Given that the cystic fibrosis allele is in Hardy-Weinberg equilibrium so allele frequency will be 1 so will be the genotype frequency.

Given that the out of 2500 only 1 individual will secret which have cystic fibrosis.

So genotype frequency of cystic fibrosis individual will be 1/2500 = 0.004

Also p2 +2pq + q2 =1

q2 = genotype freuqerncy of cycsitc fibrosis.

So allelic frequency will be 0.0004^1/2

0.02

So q= 0.02

also p+q= 1

p = 1-0.02 = 0.98

Frequency of heterozygous individaul will be

2*p*q =

2*0.98*0.02 = 0.0392.

Number of heterozygous individuals will be 98