Use the following information to answer the next three questions: 1) Jean Mannin

Question

Use the following information to answer the next three questions:

1) Jean Manning, Charles Kerfoot, and Edward Berger studied the frequencies at the phosphoglucose isomerase (GPI) locus in the cladoceran Bosmina longirostris. At one location, they collected 650 animals from Union Bay in Seattle, Washington, and determined their GPI genotypes by using electrophoresis. They counted the following # for each genotype: S1S1 = 121; S1S2 = 205; S2S2 = 324. What is the frequency of S1? Give your answer to 2 decimal places. For example, 0.12345 = .12

2) Using the information in the previous problem (to 2 decimal places), what is your calculated Chi-squared value to determine if this population is in Hardy-Weinberg equilibrium or not?  Give your answer to 2 decimal places.

3) Is the population in the previous problem in Hardy-Weinberg equilibrium?

Explanation / Answer

(1)

Frequency of S1 = sqrt( Frequency of S1S1)

Frequency of S1S1 = 121 / ( 121 + 205 + 324) = 0.186

Frequency of S1 = sqrt( 0.186) = 0.43

(2)

For S1S1

Chi square value = Sum of [ ( Observed - Expected)^2 / Observed ]

Observed = 121 , Expected = 0.43 * 121 = 52.03

( Observed - Expected)^2 / Observed = (121 - 52.03)^2 / 52.03 = 91.43

For S1S2

Observed = 205 , Expected = sqrt( 205 / ( 121 + 205 + 324) ) * 205 = 115.13

( Observed - Expected)^2 / Observed = ( 205 - 115.13)^2 / 115.13 = 70.15

For S2S2

Observed = 324 , Expected = sqrt( 324 / ( 121 + 205 + 324) ) * 324 = 228.75

( Observed - Expected)^2 / Observed = (324 - 228.75)^2 / 228.75 = 39.66

Chi Square value = 91.43 + 70.15 + 39.66 = 201.24

(c)

If (Frequency of S1S1 + Frequency of S1S2 + Frequency of S2S2) = 1 then the population is is Hardy Weinberg equilibrium

Frequency of S1S1 = 121 / ( 121 + 205 + 324) = 0.19

Frequency of S1S2 = 205 / ( 121 + 205 + 324) = 0.31

Frequency of S2S2 = 324 / ( 121 + 205 + 324) = 0.50

So,

0.19 + 0.31 + 0.50 = 1

Hence population is in Hardy Weinberg equilibrium

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