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Ashdes03 Ashdes03
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10 years ago
 :nerd:In the Caucasian population of North America one out of every 10,000 babies is born with a recessive condition know. As phenylketonuria (PKU) this condition is controlled by a single pair of alleles. People who are homozygous recessive for the PKU gene completely lack the enzyme that is necessary to metabolize the amino acid phenylalanine into harmless by products. The presence of this amino acid in a baby's diet can slow the development of the baby's brain. What percentage would you expect to be heterozygous for the PKU allele ?
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Donated
Valued Member
10 years ago
1/10000 = 0.0001 = frequency of affected individuals (q2)
To get the frequency of the recessive allele, we take sqrt(q2) = sqrt(0.0001)  = 0.01 = q

p = 1-q = 1-.01 = 0.99

From Hardy-Weinberg, the heterozygotes are 2pq:
2(0.01)(0.99) = 0.0198 = 1.98%
Pretty fly for a SciGuy
Ashdes03 Author
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10 years ago
Thanks
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Donated
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10 years ago
Welcome
Pretty fly for a SciGuy
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