Question

Locus EST			Locus ICD			Locus LA			Locus PAP			Locus ME
	SS	SF	FF	SS	SF	FF	SS	SF	FF	SS	SF	FF	SS	SF	FF
California	36	20	7	48	4	4	20	11	2	16	7	10	16	11	5
Alaska	3	3	2	7	2	2	3	2	3	1	3	2	1	2	1

For which loci and populations are the genotypes not in Hardy-Weinberg equilibrium? Confirm your conclusions by performing a chi-square test for each of the five loci in both the California and Alaska populations. You must show your work! (you will require more space then provided below) (note: n is different for each locus investigated)

Answer 1

California:

Locus EST: total no of samples= 36+20+7=63. x²= 36/63=0.571 . so x= 0.755. similarly y²= 7/63=0.11, so y= 0.33.

since x+y is not equal to 1, they are not in equilibrium.

Locus ICD: total no of samples = 56. x=0.92 & y=0.26. Hence , not in equilibrium.

Locus LA: total no of samples=33. x=0.77 & y=0.39. Not in equilibrium.

Locus PAP:total no of samples =33. x=0.69 & y=0.55. Not in equilibrium.

Locus ME: total no of samples= 32. x=0.70 & y= 0.39. Not in equilibrium.

Alaska:

Locus EST: total no of samples:8. x=0.61 & y=0.5. not in equilibrium

Locus ICD: total no of samples:11. x=0.79 & y=0.42 not in equilibrium

Locus LA: total no of samples:7. x=0.53 & y = 0.65 not in equilibrium

Locus PAP: total no of samples=7. x=0.65 & y=0.36. since x+y =1 . in equilibrium.

Locus ME: total no of samples= 4. x=0.5 & y= 0.5. in equilibrium