Question

In a typical Mendelian population of 100 individuals, those carrying gene 'a' in homozygous conditon suffer form phenylketonurea and those with gene 'A' in homozygous condition are normal. Others who are heterozygotic with 'Aa' are carriers . Calculate the number of
1.normal individuals,
2.carriers,
3.sufferers and
4.phenotypically normal individuals.

Answer 1

Dear Student,

According to Hardy Weinsberg Principle:
p² + 2pq + q² = 1
p² = frequency of the homozygous genotype
2pq = frequency of the heterozygote genotype
q² = frequency of the homozygote genotype

Phenylketonuria (recessive autosomal), of which the deleterious gene has a frequency of 1/100:
--> q = 1/100
therefore, the frequency of this disease is q² = 1/10 000,
and the frequency of heterozygotes (carriers) is 2pq = 2 x 99/100 x 1/100 = 2/100;

Note that there are a lot of heterozygotes: 1/50, two hundred times more than there are individuals suffering from the condition. .

For a rare disease, p is very little different from 1, and the frequency of the heterozygote = 2q.

Regards,