Question

Inbreeding in a population causes a deviation from Hardy–Weinberg expectations such that there are more homozygotes than expected. For a locus with a rare deleterious allele at a frequency of 0.04, what would be the frequency of homozygotes for the delete- rious allele in populations with inbreeding coefficients of F = 0.0 and F = 0.125?

Answer 1

Answer:

For population with F= 0.0, aa=0.0016. For population with F=0.125, aa= 0.0064.

Explanation:

In this population, a deleterious allele (a) has a frequency of 0.04, so:

$q=0.004\\p+q=1$

And:

$p= 1-0.04\\p= 0.96$

The question asks for the frequency of homozygotes or aa in a population with two possible inbreeding coefficients. This coefficient (F)  measures the strength of the inbreeding or the probability that two alleles in an individual are identical by descent (IBD).

So, it is known:

$aa=(1-F)q_{outbred} ^{2} +Fq_{inbred} \\aa= q^{2}+F(q-q^{2})\\aa=q^{2}+Fq(1-q)\\aa=q^{2}+Fpq$

Therefore, for F= 0.0, aa will be:

$aa=q^{2}+Fpq\\ aa=0.04^{2}+(0.0)(0.96)(0.04)\\  aa=0.04^{2}+0\\ aa=0.0016$

And, when F=0.125, aa will be:

$aa=q^{2}+Fpq\\ aa=0.04^{2}+(0.125)(0.96)(0.04)\\aa=0.0064$