Question

2. Hardy-Weinberg Equilibrium; chi-square test

Sickle cell anemia is a recessive disorder caused by a recessive mutation (S) in the b-hemoglobin gene. 80% of affected SS individuals die before reproducing.   Heterozygotes (AS) and homozygous dominant (AA) individuals do not have sickle cell anemia.

The table below shows the number of people of each genotype in a population of 100 people in population of Cameroon.

Observed # individuals in a Cameroon population	AA	AS	SS
	62	32	6

1. What are the genotype, allele, and phenotype frequencies in this population?

Show your work (that is, explain how you arrived at your answer).

b)   Is this population in Hardy-Weinberg Equilibrium? Show your work, including equations, and make sure at the end that you explain why you answered “yes” or “no” to this question.

Answer 1

Answer:

Observed genotype frequencies:

AA = 62/100 = 0.62

AS = 32/100 = 0.32

SS = 6/100 = 0.06

Allele frequencies:

Total number of A alleles = 2*AA+AS = 2*62+32 = 156

Total number of S alleles = 2*SS +AS = 2*6+32 = 44

Total alleles = 200

Fr(A) = 156/200 = 0.78

Fr(S) = 44/200 = 0.22

Expected genotype frequencies:

AA = 0.78*0.78 = 0.6084

Number of AA individuals = 0.6084 *100 = 60.84

SS = 0.22*0.22 = 0.0484

Number of SS individuals = 0.0484 *100 = 4.84

AS = 2*0.78*0.22 = 0.3432

Number of AS individuals = 0.3432 * 100 = 34.32

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
AA	62	60.84	1.16	1.35	0.022
AS	32	34.32	-2.32	5.38	0.157
SS	6	4.84	1.16	1.35	0.278
	100	100			0.457

Chi-square value = 0.457

Degrees of freedom = Number of phenotypes – 1

Df = 3-1=2

Chi-square value of 0.457 is lesser than the critical value of 5.99. We can accept the null hypothesis and the population is in Hardy-Weinberg equilibrium.