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Please answer questions 4 and 5 using the provided genotypes and answers from qu

ID: 271836 • Letter: P

Question

Please answer questions 4 and 5 using the provided genotypes and answers from questions 1-3... THANK YOU!!

Genotypes:

A1/A1: 80

A1/A2: 20

A2/A2: 90

Total: 190

1. Allele frequency estimation

Genotype

Freequency

Allele A1

Allele A2

Total

A1 A1

80

160

0

160

A1 A2

20

20

20

40

A2 A2

90

0

180

180

Total

190

180

200

380

Allele A1

= 180/380

0.47

Allele A2

= 200/380

0.53

2. Expected frequency estimation:

A1 A1

= 0.47*0.47 = 0.224

A1 A2

= 2*0.47*0.53= 0.499

A2 A2

= 0.53*0.53 = 0.277

3. Expected number estimation:

A1 A1

= 0.47*0.47 = 0.224 * 190

= 43

A1 A2

= 2*0.47*0.53= 0.499* 190

= 95

A2 A2

= 0.53*0.53 = 0.277*190

= 53

4. Use a X2 test to determine if your population sample conforms to the expected Hardy-Weinberg proportions. Provide a clear statement of how you would interpret the results of this test

5. If the population is in Hardy-Weinberg proportions, explain what that means. If the population does not appear to be in Hardy-Weinberg proportions, propose a mechanism that might account for the observed pattern and briefly explain your answer.

Genotype

Freequency

Allele A1

Allele A2

Total

A1 A1

80

160

0

160

A1 A2

20

20

20

40

A2 A2

90

0

180

180

Total

190

180

200

380

Explanation / Answer

4. Chi-square test:

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

Category

A1 A1

A1 A2

A2 A2

Observed values

80

20

90

Exprected Values

43

95

53

Deviation

37

-75

37

D^2

1396.399

5585.596

1396.399

D^2/E

32.76

58.96

26.53

118.25

X^2

118.25

Degrees of freedom

1

Inference: The calculated chisquare value i.e. 118.25 is greater than the table value i.e. 3.84 at 1 DF and 0.05 probability, hence the null hypothesis is rejected.

5. If the population is in Hardy-Weinberg proportions which means the population is in steady state and not evolving. If the population does not appear to be in Hardy-Weinberg proportions which means the population is not in steady state and the population is evolving .

Category

A1 A1

A1 A2

A2 A2

Observed values

80

20

90

Exprected Values

43

95

53

Deviation

37

-75

37

D^2

1396.399

5585.596

1396.399

D^2/E

32.76

58.96

26.53

118.25

X^2

118.25

Degrees of freedom

1