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Please note that all the questions are part of one question. So, please answer a

ID: 205169 • Letter: P

Question

Please note that all the questions are part of one question. So, please answer all of them and show your work as well. Thanks in advance!

Question 1

Part 1

You survey a population of cotton rats at the Coastal Center for variation at a SNP locus. You find 30 individuals that are A/A, 71 are A/G and 67 are G/G. what is the frequency of the A allele in this population?

Part 2

For the population in the previous question, what will the frequency of the A allele be when the population is in Hardy Weinberg equilibrium?

Part 3

When the population in Question 1 is in Hardy Weinberg Equilibrium, what will be the frequency of heterozygous individuals?

Part 4

Is the population in Question 1 in Hardy-Weinberg equilibrium? For your answer, also give the value of the chi-square statistic, the degrees of freedom, the p-value (use the table in your text).

Part 5

In the current population, as described in Question 1, what would be the frequency of mating between A/A and A/G individuals?

Explanation / Answer

Part 1:
The freequency of A allele : 0.39

Part 2
The freequency of A allele : 0.39

Part 3
The vfrequency of heterozygous individuals = 80

Part 4

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

Test static:

Category

AA

AG

GG

Observed values

30

71

67

168

Exprected Values

25

80

63

168

Deviation

5

-9

4

0

D^2

25

81

16

122

D^2/E

1

1.0125

0.253968

2.266468

X^2

2.266468

Degrees of freedom

1

Inference: The calculated chisquare value i.e. 2.266 is less than the table value i.e. 3.84 at 1 DF and 0.05 probability, hence the null hypothesis accepted.

For answers find the below calculations

Genotype

Freequency

Allele A

Allele G

Total

AA

30

60

0

60

AG

71

71

71

142

GG

67

0

134

134

Total

168

131

205

336

Allele A

= 131/ 336

0.39

Allele G

= 205 / 336

0.61

After HW equilibrium

AA

= 0.39 * 0.39 = 0.152

X 168 = 25

AG

= 2 * 0.39 *0.61 = 0.476

X 168 = 80

GG

= 0.61 *0.61 = 0.372

X 168 = 63

Genotype

Freequency

Allele A

Allele G

Total

AA

25

50

0

50

AG

80

80

80

160

GG

63

0

126

126

Total

168

130

206

336

Allele A

= 130 / 336

0.39

Allele G

= 206 / 336

0.61

Category

AA

AG

GG

Observed values

30

71

67

168

Exprected Values

25

80

63

168

Deviation

5

-9

4

0

D^2

25

81

16

122

D^2/E

1

1.0125

0.253968

2.266468

X^2

2.266468

Degrees of freedom

1

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