Question

For each populations, calculate the expected genotype frequencies to determine which of these populations is NOT in Hardy-Weinberg equilibrium.

1 AA, 18 Aa, 81 aa

20 AA, 20 Aa, 5 aa

25 AA, 10 Aa, 1 aa

50 AA, 20 Aa, 30 aa

Calculate the Chi-square statistic (rounded to 2 significant figures) to show that your answer choice in the previous question is not in Hardy-Weinberg equilibrium.

3.5   

26

34

4.9

Answer 1

The answers are

For each populations, calculate the expected genotype frequencies to determine which of these populations is NOT in Hardy-Weinberg equilibrium.

The answer is 50 AA, 20 Aa, 30 aa

Calculate the Chi-square statistic (rounded to 2 significant figures) to show that your answer choice in the previous question is not in Hardy-Weinberg equilibrium.

The answer is 34

Find the below explanation

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For 1 AA, 18 Aa, 81 aa

The steps involved here

Step 1: Calculation number of alleles

Genotype	Freequency	Allele A	Allele a	Total
AA	1	2	0	2
Aa	18	18	18	36
aa	81	0	162	162
Total	100	20	180	200

Step 2: Calculation of allele frequencies

Frequency = No of a alleles/Total no of alleles

Allele A (Dominant allele)	= 20/200	= 0.1
Allele a (Recessive allele)	= 180/200	= 0.9

Step 3: Expected genotype frequencies estimation

Genotype	Genotype frequencies	For 1000 population the freqeunecies
AA	= 0. 1*0.1= 0.010	*100	=1
Aa	= 2*0.1*0.9= 0.180	*100	=18
aa	= 0.9*0.9= 0.810	*100	=81

As the same value of expected and observed indicates that the population is in HW equilibrium.

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For 20 AA, 20 Aa, 5 aa:

The steps involved here

Step 1: Calculation number of alleles

Genotype	Freequency	Allele A	Allele a	Total
AA	20	40	0	40
Aa	20	20	20	40
aa	5	0	10	10
Total	45	60	30	90

Step 2: Calculation of allele frequencies

Frequency = No of a alleles/Total no of alleles

Allele A (Dominant allele)	= 60/90	= 0.67
Allele a (Recessive allele)	= 30/90	= 0.33

Step 3: Expected genotype frequencies estimation

Genotype	Genotype frequencies	For 1000 population the freqeunecies
AA	= 0.67*0.67= 0.444	*45	=20
Aa	= 2*0.67*0.33= 0.444	*45	=50
aa	= 0.33*0.33= 0.111	*45	=5

As the same value of expected and observed indicates that the population is in HW equilibrium.

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For 25 AA, 10 Aa, 1 aa:

The steps involved here

Step 1: Calculation number of alleles

Genotype	Freequency	Allele A	Allele a	Total
AA	25	50	0	50
Aa	10	10	10	20
aa	1	0	2	2
Total	36	60	12	72

Step 2: Calculation of allele frequencies

Frequency = No of a alleles/Total no of alleles

Allele A (Dominant allele)	= 60/72	= 0.83
Allele a (Recessive allele)	= 12/72	= 0.17

Step 3: Expected genotype frequencies estimation

Genotype	Genotype frequencies	For 1000 population the freqeunecies
AA	= 0.83*0.83= 0.444	*36	=25
Aa	= 2*0.83*0.17= 0.444	*36	=10
aa	= 0.17*0.17= 0.111	*36	=1

As the same value of expected and observed indicates that the population is in HW equilibrium.

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For 50 AA, 20 Aa, 30 aa:

The steps involved here

Step 1: Calculation number of alleles

Genotype	Freequency	Allele A	Allele a	Total
AA	50	100	0	100
Aa	20	20	20	40
aa	30	0	60	60
Total	100	120	80	200

Step 2: Calculation of allele frequencies

Frequency = No of a alleles/Total no of alleles

Allele A (Dominant allele)	= 120/200	= 0.60
Allele a (Recessive allele)	= 80/200	= 0.40

Step 3: Expected genotype frequencies estimation

Genotype	Genotype frequencies	For 1000 population the freqeunecies
AA	= 0.60*0.60= 0.360	*100	=36
Aa	= 2*0.60*0.40= 0.480	*100	=48
aa	= 0.40*0.40= 0.160	*100	=16

                               

Step 4: Chi-square test

Null hypothesis: The observed values are not deviating from the expected values.

Category	AA	Aa	aa
Observed values	50	20	30
Exprected Values	36	48	16
Deviation	14	-28	14
D^2	196	784	196
D^2/E	5.44	16.33	12.25	34.03
X^2	34.03
Degrees of freedom	1

Inference: The calculated chisquare value i.e. 34.03 is greater than the table value i.e.3.84 at 1 DF and 0.05 probability, hence the null hypothesis is rejected, which means the population is not in HW equilibrium.