Question

Cystic fibrosis is a genetic disorder in homozygous recessives that causes death during the teenage years. 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 model? f(A1) = 0.9800, f(A2) = 0.0200 f(A1) = 0.9700, f(A2) = 0.0300 f(A1) = 0.9997, f(A2) = 0.0003 f(A1) = 0.9604, f(A2) = 0.0392

Answer 1

The correct answer is: f(A2) = 0.0300 f(A1) = 0.9997

If 9 in 10,000 newborn babies have the disease, that means that the frequency of recessive homozygous is 9/10000=0.0009.

According to Hardy-Weinberg equilibrium p2+2pq+q2=1 (p+q=1) where p2 is frequency of dominant homozygous (only p is dominant allele), 2pq is frequency of heterozygous and q2 is the frequency of recessive homozygous (only q is the frequency of recessive allele). This means that q2=0.0009 and (A2) q=0.03. p (A1)=1-0.03=0.97