Given the following genotypes of A1A1= 188, A1A2= 240 and A2A2= 93 estimate alle

Question

Given the following genotypes of A1A1= 188, A1A2= 240 and A2A2= 93 estimate allele frequencies and heterozygosity. what are the predicted genotype frequencies or this population if you assume Hardy-Weinburg conditions? Des this population deviate from Hardy weingburg equilibrium (test by chi square)? The critical value for comparison in a chi- square test is 3.85 when there is one degree of freedoms as is the case here (number of alleles-1). Does it appear that this sample comes from a population in Hardy-Weinburg equilibrium?

Explanation / Answer

Based on the given data,

The total number of population = 521

A1A1= 188

A1A2= 240

A2A2= 93

The genotype frequencies are:

A1A1= 188 / 521 = 0.360---- (p2)

A1A2= 240 / 521 = 0.460--- (2pq)

A2A2= 93 / 521 = 0.178------ (q2)

The allelic frequency of heterozygosity:

Frequency of A1= p = p2 + 1/2 (2pq) = 0.360 + 1/2 (0.460) = 0. 360 + 0.23 = 0.59

Frequency of A2= q = 1-p = 1 - 0. 59 = 0.47

The predicted genotype frequencies or this population if you assume Hardy-Weinburg conditions:

A1A1 (p2) = (0.59)2 = 0.34

A1A2 (2pq) = 2 (0.59) (0.47) = 0.55

A2A2 (q2) = (0.47)2 = 0.22

Expected number of individuals of each genotype:

A1A1= 0.34 × 521= 177

A1A2 = 0.55 × 521= 287

A2A2 = 0.22 × 521=115

CHI - SQUARE (X2):

X2 = (O - E)2 / E

X2 = (188-177)2 /177 + (240-287)2 /287 + (93-115)2 /115

= (11)2 /177 + (-47)2 /287 + (-22)2 /115

= 0.683 + 7.696 + 4.2

= 12.5

X2(calculated) < X2(table) [3.841, 1 df, 0.05 ls]

Therefore, conclude that there is no statistically significant difference between what you observed and what you expected under Hardy-Weinberg. That is, you fail to reject the null hypothesis and conclude that the population is in Hardy weingburg equilibrium.

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