Question

Researchers determined the genotypes of a number of individuals from each of the same five sites on the island of Guinea. They examines a common locus known to have only two alleles (X741 and C443). Their data are summarized below:

Genotype	site 1	site 2	site 3	site 4	site 5
X741/X741	21	18	6	12	17
X741/C443	44	32	10	25	30
C443/C443	17	15	5	15	13
Total	82	65	21	52	60

a) What are the frequencies of the X741 and C433 alleles among the population on the island?

b) Give the number of each genotype expected if the population is in Hardy Weinberg equilibrium.

c) Use chi square test to determine whether this population is in equilibrium. Be sure to show the chi-square statistic, the number of degrees of freedom, the critical value, and your conclusion.

Answer 1

Genotype	1	2	3	4	5	Sum
X741/X741	21	18	6	12	17	74
X741/C443	44	32	10	25	30	141
C443/C443	17	15	5	15	13	65

Total Population = 280

frequency of X741/X741(aa) = 74/280 = 0.26

frequency of C443/C443 (bb) = 65/280 = 0.23

frequency of X741/C443 (2ab) = 141/280 = 0.50

if the population is in Hardy Weinberg Equilbrium

a2 + b2 + 2ab = 1

0.26+0.23+2ab=1

2ab = 0.51

Expected Heterozygote population if it is in a Hardy Weinberg equilbrium = 0.51

observed heterozygote population = 0.50

Sum of population observed for expressing for Phenotype

with  X741 allele =Homozygote +Hetrozygote = 0.26+0.50 = 0.76

Sum of population expected for population in Hardy Weinberg equilbrium  for expressing Phenotype

with  X741 allele =Homozygote +Hetrozygote = 0.26+0.51 = 0.77

χ² = (expected-observed)2/ expected = (0.77-0.76)2/ 0.77 =1.29*10-4  

Degree of freedom = no of categories of expected phenotype - 1 = 2-1=1

p =0.05 for significance .

using the two values we get χ2 critical value is got from chi square table.(given below) =3.841

On comparing both the values we find that  χ2 calculated value < χ2 critical value .So the two population are similar and follow Hardy Weinberg Equilbrium.