Question

I am having a very hard time with allele frequencies unfortunately even though they seem to be relatively straightforward.

this is the data set I am working with: #1: Adeyemo 2006. HbAA= 0.70, HbAS= 0.26, HbSS=0.04; population size = 5982

and I need to find the fitness of AA, fitness of Aa, and fitness of aa. AA is homozygous with wild type of allele for Sickle-cell anemia, Aa is heterozygous for the Sickle-Cell allele, and aa is homozygous for the sickle-cell allele

just very confused and I appreciate the help

Answer 1

The popultaion size = 5982

HbAA=0.70,

thus number of HbAA individual

=5982x0.7=4187.4 (4187~)

HbAS=0.26

the number of HbAS individual

=5982x0.26=1555.32 (1555~)

HbSS=0.04

the number of HbSS individual

=5982x0.04= 239.28 (239~).

As number of homozygous wild type AA individuals are greatest in number so it is considered to be the fittest among all genotypes.

Fitness of AA = 4187.4/4187.4=1

Fitness of Aa = 1555.32/4187.4= 0.37

Fitness of aa= 239.28/4187.4= 0.057

If you consider the numbers of individual genotypes in whole the answers are same.