Question

A population has the following data:

Genotype Number

AA 65

Aa 5

aa 25

A) Calculate the allelic frequency for the A allele.

B) Calculate the allelic frequency for the a allele.

C) Calculate the expected Hardy-Weinberg genotypic frequency for AA.

D) Calculate the expected Hardy-Weinberg genotypic frequency for Aa.

E) Calculate the expected Hardy-Weinberg genotypic frequency for aa.

F)Convert the expected frequency of AA into total number of people with AA.

G)Convert the expected frequency of Aa into total number of people with Aa.

H)Convert the frequency the expected frequency of aa into total number of people with aa.

I) Do a Chi-square test to determine if the population is at Hardy-Weinberg equilibrium. What is your hypothesis?

J)Do a Chi-square test to determine if the population is at Hardy-Weinberg equilibrium. What is your X² value?

K) Degree of Freedom Value?

L) What is the probability?

M) What is your conclusion?

Answer 1

Answer:

65 A/A = 2 * 65 = 130 “A” alleles

5 A/a = 5 “A” and 5 “a” alleles

25 aa = 50 “a” alleles

Total alleles = 190

A). Frequency of “A” allele = 135/190= 0.71

B). Frequency of “a” allele = 55/190 = 0.29

Three expected genotypes are produced as follows.

Total population = 95

C). Frequency of genotype, A/A = 0.71*0.71 = 5041

D). Frequency of genotype, A/a = 2* 0.71 * 0.29 = 0.4118

E). Frequency of genotype, a/a = 0.29*0.29 = 0.0841

F). A/A individuals = 0.5041 * 570 = 48

G). A/a individuals = 0.4118*95 = 39

H). a/a individuals = 0.0841 * 95 = 8

I & J).

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
A/A	65	48	17	289.00	6.0208
A/a	5	39	-34	1156.00	29.6410
a/a	25	8	17	289.00	36.1250
	95	95			71.7869

Chi-square value = 71.79

K). Degrees of freedom = number of categories – 1

Df = 3-1 = 2

L). Probability = 0.05

Critical value = 5.99

M). The chi-square value of 71.79 is greater than the critical value of 5.99. So we can reject the null hypothesis. Hence the population is not in Hardy-Weinberg equilibrium.