Answer to Question #212401 in Genetics for MOHAMAD NORASREY B

Question

a.      A certain form of sickle cell anemia is caused by an autosomal recessive allele. Assume that the mutation rate is 1 x 10-5 and that persons having this condition can only produce about 80% of the normal offspring.

 

i.        At equilibrium between mutation and selection, what will the frequency of the allele for sickle cell anemia be? Show your calculation.

Answer 1

The Hardy-Weinberg equilibrium $p2+2pq+q2=1$  can be applied when the gene variation is at equilibrium, where 

p = frequency of the dominant allele in the population

q = frequency of the recessive allele in the population

p² = percentage of homozygous dominant individuals

q² = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals

Sickle cell anemia is caused by an autosomal recessive allele (homozygous). 80% or 0.8 of the normal offspring includes the homozygous dominant and heterozygous dominant, who are normal as well as carriers who will not express the condition. This accounts for the p²+2pq.

Thus, $q2=1-p2+2pq$  can help in finding the percentage of the recessive allele.

$q2 =1−(p2+2pq)\\ q2=1−0.8\\= 0.2$

The frequency of the recessive allele$q =\sqrt{\smash[b]{0.2}}=0.447$

Thus the frequency of the allele for sickle cell anemia is 0.447