If the frequency of the recessive allele that causes phenylketonuria (PKU)when homozygote is about 0.01 in European and American populations,what is the estimated heterozygote frequency in these populations? (Assume that people mate randomly with respect to PKU,and that the mutant allele does not have any effects on fitness.)

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The frequency of homozygous recessive individuals can be estimated through the Hardy-Weinberg equation: p² + 2pq + q² = 1, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. Given that the frequency of the recessive allele is 0.01, q² = 0.01, and q = 0.1. The heterozygote frequency can be estimated by 2pq, which is 2(0.1)(0.9) = 0.18. However, this is the frequency of carriers, not necessarily heterozygous individuals with one affected allele. Therefore, we divide by 2 to get an estimated heterozygote frequency of 0.09 or about 0.0198 when rounded to the nearest hundredth.

If the Frequency of the Recessive Allele That Causes Phenylketonuria