A particular recessive genetic disorder is fatal before birth, so there are no homozygous recessive individuals. In a particular population, one in 15 individuals is a carrier for this disorder. What are the allele frequencies of the dominant (p) and recessive (q) alleles in this population?

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example,

What are the allele frequencies of the dominant (p) and recessive (q) alleles in this population?

If 1 of 15 individuals are carriers for this disorder, this means that 1/15 are heterozygous, 2pq. So, 2pq = 1/15 = 0.066

Now we must calculate the allelic frequencies.

We know that 1 in 15 individuals are heterozygous, and we also know that there are no recessive homozygous, q², because they can not survive, so of the 15 individuals only one is heterozygous and the rest 14 individuals must be dominant homozygous, p².

The dominant homozygous genotypic frequency is

p²= 14/15 = 0.933

And by clearing the next equation we can get the allelic frequency for p

p²= 0.933

p = √0.933

p = 0.967

So now we know that the allelic frequency for p is 0.967  

This means that the allelic frequency for q or p is 0.033, which we deduce by clearing the equation p + q = 1

                          0.967 + q = 1

                         q = 1 - 0.967

                          q = 0.033

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