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 |
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:
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.
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