Question

Among 40 individuals, 4 are heterozygous for a particular locus with two alleles (A1 and A2), 16 are homozygous for allele A1, and 20 are homozygous for allele A2.

1. What is the frequency of allele A1?

2.What is chi-squared for your observed/expected genotypes?

3.Does the population depart significantly from Hardy-Weinberg expectations? (For 1 degree of freedom χ,> 3.84 is significant)

Answer 1

Given:

Out of 40 individuals,

No. of A1A1 individuals = 16
No. of A1A2 individuals = 4
No. of A2A2 individuals = 20

Total Number of Alleles = 40 * 2 = 80

Number of A1 allele = 16 * 2 + 4 = 36
Number of A2 allele = 20 * 2 + 4 = 44

1)

Frequency of A1 allele = p = 36/80 = 0.45
Frequency of A2 allele = q = 44/80 = 0.55

Expected Number of A1A1 indiviuals = N * p² = 40 * 0.45 * 0.45 = 8.1
Expected Number of A1A2 indiviuals = N * 2pq = 40 * 2 * 0.45 * 0.55 = 19.8
Expected Number of A2A2 indiviuals = N * q² = 40 * 0.55 * 0.55 = 12.1

2)

Null Hypothesis; H₀ : The population is in Hardy-Weinberg Equibrium
Alternate Hypothesis; H_A : The population is not in Hardy-Weinberg Equibrium

Calculation of Chi-Squared Statistic for given data

Genotype	Observed (O)	Expected (E)	O - E	(O - E)2	(O - E)2/E
A1A1	16	8.1	7.9	62.41	7.705
A1A2	4	19.8	-15.8	249.64	12.61
A2A2	20	12.1	7.9	62.41	5.158
					= 25.473

The statistic for the given data is 25.473.

3)

The population does depart significantly from Hardy-Weinberg Expectations, as the statistic for the data; 25.473 > 3.84. Therefore, the population is not in Hardy-Weinberg Equilibrium.