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In fruit flies, vestigial wings are recessive to normal wings, and ebony body is recessive to normal body. A cross was made between flies (all true-breeding) with normal wings and normal bodies to flies with vestigial wings and ebony bodies. The F1 offspring were then self-crossed to produce an F2 generation. The following data were obtained (observed):

Category Predicted ratio Observed numbers
normal wings, normal body 9 452
normal wings, ebony body 3 126
vestigial wings, normal body 3 101
vestigial wings, ebony body 1 41
720

Given that we would expect equal proportions of each class of flies after a testcross, test the goodness of fit using a chi-square test. What is the chi-squared value for this set of data?

Respuesta :

Oseni

Answer:

See the answer below

Explanation:

One will expect the offspring of the F2 generation to phenotypically segregate in 9:3:3:1 according to the Mendelian ratio.

Using the formula for Chi-square:

    = [tex]\frac{(O - E)^2}{E}[/tex] where O = observed frequency and E = expected frequency

Phenotype                                    O             E                          Chi-square

normal wings, normal body       452     9/16 x 720= 405      [tex]\frac{(452 - 405)^2}{405}[/tex]= 5.45

normal wings, ebony body         126      3/16 x 720= 135       [tex]\frac{(126 - 135)^2}{135}[/tex]= 0.6

vestigial wings, normal body      101       3/16 x 720= 135        [tex]\frac{(101 - 135)^2}{135}[/tex]= 8.56

vestigial wings, ebony body        41         1/16 x 720 = 45         [tex]\frac{(41 - 45)^2}{45}[/tex]= 0.36

Total Chi-square = 5.45+0.6+8.56+0.36

                       = 14.97

Degree of freedom = 4 - 1 = 3

Critical Chi-square value (alpha = 95%)  = 7.81

The calculated Chi-square value (14.97) is more than the critical value (7.81), hence, the outcome of the cross does not fit-in to Mendel's ratio.

The Chi-square value for the data set is 14.97.

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