attac 2. (2.5 pts) Based on an example of fitness in sickle cell genotypes, use
ID: 197655 • Letter: A
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attac 2. (2.5 pts) Based on an example of fitness in sickle cell genotypes, use the Hardy/We natural selection has led to evolution in the population below. nber g formula to determine if Step 2 Step 1 Observed o aileles Observed Genotypes # of 1 genotype 45 40 15 100 people FrequenciesA AS sS Total 1.00 Step 3 allele frequencies Step 4 Plug the allele frequencies into the Hardy/Weinberg formula (p2+2pq+-1) to get the expected genotype frequencies for AA, AS, and SS Step 5 Compare the observed genotype frequencies from step #1 to the expected genotype frequencies from ste # 4, Are they all the same? genotype frequencies that 2 Do you observe any occur more or less than expected, as predicted by Hardy/Weinberg? Step 1 Step 5 types obs ASExplanation / Answer
Allele frequencies estimation:
Genotype
Freequency
Allele A
Allele S
Total
AA
45
90
0
90
AS
40
40
40
80
SS
15
0
30
30
Total
100
130
70
200
Allele A
= 130 / 200
= 0.65
Allele S
= 70 / 200
= 0.35
Expected Genotype Frequencies
AA
= 0.65* 0.65 = 0.423
*100 = 42
AS
= 2 * 0.65 *0.35 = 0.455
*100 = 46
SS
= 0.35* 0.35 = 0.123
*100 = 12
Chisquare test:
Null hypothesis: The observed values are not deviating from the expected values.
Test static:
Category
AA
AS
SS
Observed values
45
40
15
Exprected Values
42
46
12
Deviation
3
-5
3
D^2
7.5625
30.25
7.5625
D^2/E
0.18
0.66
0.62
1.46
X^2
1.46
Degrees of freedom
1
Inference: The calculate chisquare value i.e. 1.46 is less than the table value i.e. 3.84 at 3 DF and 0.05 probability, hence the null hypothesis is accepted. Which means the population is in HW equilibrium.
Genotype
Freequency
Allele A
Allele S
Total
AA
45
90
0
90
AS
40
40
40
80
SS
15
0
30
30
Total
100
130
70
200
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