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

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  

i believe the answer is c

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