Question 14: Genetics: Given the table below determine if the population is in Hardy Weinberg Equilibrium (HWE) and the approximate p-value using the critical values table here.
Homework Answers
Answer:
None of the above
Not in HWE, p-value is <0.05
Chi-square value = 8.29
Explanation:
A/A 17 = 17*2=34 A alleles
A/a 55 = 55 A & 55 a alleles
a/a 12 = 24 a alleles
Total alleles = 168
Total of A alleles = 34+55 = 89
Frequency of A allele = 89/168 = 0.53
Total of a alleles = 24+55 = 79
Frequency of a allele = 79/168 = 0.47
Expected number of each genotype:
Total population = 84
A/A = 0.53*0.53*84 = 23.6
A/a = 2*0.53*0.47*84 = 41.85
a/a = 0.47*0.47*84 = 18.55
|
Phenotype |
Observed(O) |
Expected (E) |
O-E |
(O-E)2 |
(O-E)2/E |
|
A/A |
17 |
23.6 |
-6.6 |
43.56 |
1.8458 |
|
A/a |
55 |
41.85 |
13.15 |
172.92 |
4.1320 |
|
a/a |
12 |
18.55 |
-6.55 |
42.90 |
2.3128 |
|
Total |
84 |
84 |
8.2905 |
Chi-square value = 8.29
DF = number of categories – 1
DF = 3-1=2
P value =5.99 at 0.05 significance.
The chi-square value of 8.29 is greater than the critical value of 5.99. Hence we can reject the null hypothesis and the population is not in Hardy-Weinberg equilibrium
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