Locus EST Locus ICD Locus LA Locus PAP Locus ME SS SF FF SS SF FF SS SF FF SS SF
ID: 202099 • Letter: L
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)
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
Explanation / Answer
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
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