Given A1A1= 35, A1A2==56, A2A2= 24, A1A3= 84, A2A3= 76, A3A3= 31 estimate allele
ID: 33703 • Letter: G
Question
Given A1A1= 35, A1A2==56, A2A2= 24, A1A3= 84, A2A3= 76, A3A3= 31 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 = 306
A1A1= 35
A1A2 = 56
A2A2= 24
A1A3 = 84
A2A3 = 76
A3A3= 31
The genotype frequencies are:
A1A1= 35 / 306 = 0.114 ------- (p2)
A1A2 = 56 / 306 = 0.183
A2A2= 24/ 306 = 0.078 -------- (q2)
A1A3 = 84/ 306 = 0.233
A2A3 = 76/ 306 = 0.248
A3A3= 31/ 306 = 0.101
The allelic frequency of heterozygosity:
Frequency of A1= p = p2 + 1/2 (2pq) = 0.114 + 1/2 (0.183) = 0.112
Frequency of A2= q = 1-p = 1 - 0. 112 = 0.88
The predicted genotype frequencies or this population if you assume Hardy-Weinberg conditions:
A1A1 (p2) = (0.112)2 = 0.012
A1A2 (2pq) = 2 (0.098) = 0.19
A2A2 (q2) = (0.88)2 = 0.22
Expected number of individuals of each genotype:
A1A1= 0.012 × 306= 4
A1A2 = 0.19 × 306= 59
A2A2 = 0.22 × 306=68
CHI - SQUARE (X2):
X2 = (O - E)2 / E
X2 = (35-4)2 /4 + (56-59)2 /59 + (24-68)2 /68
= 240 + 0.15 + 28.47
= 268.62
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