Below are two populations One Is in HardyWeinberg equilibriu

Below are two populations. One Is in Hardy-Weinberg equilibrium and the other is not. Go through the process, including the chi square test to prose this. X^2 = sigma [(observed - expected)^2/expected] Population one Observed Genotypes AA = 125 Aa = 105 aa = 25 N = 255 Population two Observed Genotypes AA = 800 Aa = 150 aa = 200 N = 1150

Solution

In population 1^st

AA= 125

Aa=105

aa=25

Total= 255

There are two alleles A and a. according the HWE

(A+a)² = 1

AA + 2Aa + aa = 1;

HWE rule follows the law of independent assortment and tells that addition of allele frequencies are always 1. According the HWE rule, we assume that each genotype will get equal number of progenies and if so the allele frequencies will be:

Number of genotypes / total genotypes

For A allele = 1 / 4 = 0.25 which is alos for ‘a’ allele

For Aa the frequency would be = 2/ 4 = 0.50

Therefore, the crosses which follow the above rule and frequencies will show HWE.

Now in the population 1^st

Allele frequency of ‘A’ allele = number of progenies/ total progenies

= 125/ 255 = 0.49

Repeating same process

For ‘a’ allele = 25/ 255 =0.1

The results show do not follow the rule of HWE.

Now calculate chi square:

Ideally, the cross should follow the rule of HWE and then the progenies would be approximately:

AA genotype = frequency x total progenies = 0.25 x 255 = 64 =aa genotype

Aa genotype = 0.5 x255 =127

Genotypes

Observed

Expected

(O-E)²

X²

AA

125

64

3721

58.14

aa

25

64

1521

23.77

Aa

105

127

484

3.81

Total X²

85.72

X² = 85.72

X = 9.26

Genotypes	Observed	Expected	(O-E)2	X2
AA	125	64	3721	58.14
aa	25	64	1521	23.77
Aa	105	127	484	3.81
Total X2				85.72