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 prove this. X^2 = S [(observed - expected)^2/expected] X_2, 0.05^2 = 6.00 Observed Genotypes AA = 125 A_a = 105 aa = 25 N = 255 Observed Genotypes AA = 800 A_a = 150 aa = 200 N= 1150

Solution

Genotype frequency using the given data (total population size = 225

AA (p²) = 125/225 = 0.555;
Aa (2pq) = 105/225 = 0.466;
aa (q²) = 25/225 = 0.111

Allele frequencies:

Frequency of A = p = p² + 1/2 (2pq) = 0.555 + 1/2 (0.466) = 0.555 + 0.233 = 0.788

Frequency of N = q = 1-p = 1 - 0.788 = 0.212

According to Hardy-Weinberg law, expected genotype frequency:

AA (p²) = (0.788)² = 0.620; Aa (2pq) = 2 x 0.788 x 0.212 = 0.334; aa (q²) = (0.212)² = 0.044

Expected number of genotypes:

AA = 0.620 X 225 = 139.5

Aa = 0.334 X 225 = 75.1

aa = 0.044 X 225 =9.9

Since X² = (O - E)² / E; putting the values:

X² = (125-139.5)² /139.5 + (105-75.1)² /75.1 + (25-9.9)² /9.9

(125-139.5)2 /139.5 + (105-75.1)2 /75.1 + (25-9.9)2 /9.9

= 1.50+ 11.9+ 23.03= 36.43

X²_(calculated) > X²_(table)

Since there is statistically significant difference between observed and expected values, the null hypothesis is rejected and the population is NOT in Hardy-Weinberg equilibrium.

Genotype frequency using the given data (total population size = 1150

AA (p²) = 800/1150 = 0.695
Aa (2pq) = 150/1150 = 0.130
aa (q²) = 200/1150 = 0.173

Allele frequencies:

Frequency of A = p = p² + 1/2 (2pq) = 0.695 + 1/2 (0.130) = 0.695 + 0.065 = 0.76

Frequency of N = q = 1-p = 1 - 0.76 = 0.24

According to Hardy-Weinberg law, expected genotype frequency:

AA (p²) = (0.76)² = 0.577; Aa (2pq) = 2 x 0.76 x 0.24 = 0.364; aa (q²) = (0.24)² = 0.057

Expected number of genotypes:

AA = 0.577 X 1150 = 663.55

Aa = 0.364 X 1150 = 418.6

aa = 0.057 X 1150 =65.5

Since X² = (O - E)² / E; putting the values:

X² = (800-663.55)² /663.55 + (150-418.6)² /418.6 + (200-65.5)² /65.5

800-663.55)2 /663.55 + (150-418.6)2 /418.6 + (200-65.5)2 /65.5

= 28.05+ 172.35+ 276.18= 476.58

X²_(calculated) > X²_(table)

Since there is statistically significant difference between observed and expected values, the null hypothesis is rejected and the population is NOT in Hardy-Weinberg equilibrium.