Answer: 2pq = 0.0207 (heterozygous frequency)
Explanation:
The Hardy-Weinberg Principle states that <u>the genetic composition of a population remains in equilibrium as long as no natural selection or other factors are active and no mutations occur.</u> 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
