Question

make a hardy weinbery equilbrium question: where gene A has 4 alleles , gene B has 3 alleles and gene C has 2 alleles.
provide allele frequencies for each allele. frequency must have 6 decimal places.

Answer 1

According to the hardy weinbery equation for larger population that allelic and genotypic frequency remains constant.

So for more than two alleles the genotype = n* n+1/2

Then,

Gene A has 4 alleles therefore genotype = 4*5/2 = 10

In the 10 genotype the 4 genotype is homozygous.

Allelic frequency is 4/10 = 0.4 = 40%

Gene B has 3 alleles therefore genotype = 12/2 = 6

Alleles frequency is 3/6*100 =50%

Gene C has 2 allele

Genotype = 6/2=3

Allele frequency 2/3*100 = 66%