If 33 of the 300,000 people in corpus christi, tx, have pku, how many people are carriers (heterozygous) for pku?

2pq = 0.0207 (heterozygous frequency)

Explanation:

The Hardy-Weinberg Principle states that the genetic composition of a population remains in equilibrium as long as no natural selection or other factors are active and no mutations occur. That is, Mendelian inheritance, by itself, does not generate evolutionary change. Under certain conditions, after a generation of random mating, the frequencies of the genotypes of an individual locus will be fixed at a particular equilibrium value. It also specifies that these equilibrium frequencies can be represented as a simple function of the allelic frequencies at that locus.

In the simplest case, with a locus with two alleles A and a, with allele frequencies of p and q respectively, this principle predicts that the genotypic frequency for the dominant homozygous AA is p^2, that of the heterozygous Aa is 2pq and that of the recessive homozygous aa, is q^2.

Phenylketonuria (PKU) is a genetic condition that causes phenylalanine to build up in the body. PKU is inherited as an autosomal recessive pattern

So, n=300,000

33 have PKU, so 33/300,000 is the genotypic frequency = 0.00011 (homozygous recessive, aa = q^2). This is the proportion of people who has both recessive alleles, so people who has the condition.

Then, 0.00011 = aa =q^2

q = 0.0105, which is the allele frequency for the recessive allele.

Knowing that p + q = 1 and p^2 + 2pq + q^2 = 1

p + 0.0105 = 1

p = 0.9895, which is the allele frequency for the dominant allele.

So, p^2 = 0.9895^2 = 0.979 which is the genotypic frequency for homozygous dominant.

And to calculate the genotypic frequency for heterozygous, we do 2pq= 2 x 0.9895 x 0.0105 = 0.0207

Or we can also do p^2 + 2pq + q^2 = 1 ---> 1 - 0.9895^2 - 0.0105^2 = 2pq = 0.0207

So, the genotypic frequency (how many people are heterozygous) is 0.0207