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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.

Given the table below determine if the population is in Hardy Weinberg Equilibrium (HWE) and the approximate p-value using thWhat was the chi square value you used from the last question? 9.75 8.29 C 3.46 2.133 None of the above

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Answer #1

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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