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If the recessive disease phenylketonuria (pku) occurs in a genetically constant population with a frequency of 1 in 1o000, the frequency of the carrier genotype
If the recessive disease phenylketonuria (PKU) occurs in a genetically constant population with a frequency of 1 in 1o000, the frequency of the carrier genotype
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Abhijeet Gaurav Best Answer
We that from Hardy-Weinburg principle that : p+q = 1 Here p = Dominant q = Recessive Now here, q^2 = 1/10000 so, q = 1/100 Now, p = 1-q = 1- 1/100 = 99/100 so, p^2 = 9801/10000 Now according to Hardy Wrinburg equation : (p+q)^2 = p^2+ 2pq+ q^2 Now, Carrier genotype is 2pq So, 2pq = 1- (9801/10000 +1/10000) = 198/10000 = 1.99%