Question

A presumed dihybrid in Drosophila B/b ; F/f is testcrossed to b/b ; f/f. (B = black body; b = brown body; F = forked bristles; f = unforked bristles.)

The results were:

Black, forked 230

Black, unforked 210

Brown, forked 240

Brown, unforked 250

Use the chi-square test to determine if these results fit the results expected from testcrossing the hypothesized dihybrid.

df 0.5 0.01 0.001 1 Probability level (alpha) 0.10 0.05 0.02 2.706 3.841 5.412 4.605 5.991 7.824 6.251 7.815 9.837 7.779 9.48

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Answer #1

Null hypothesis - B and F genes show independent assortment.

BbFf × bbff

Gametes bf
BF BbFf
Bf Bbff
bF bbFf
bf bbff

Expected ratio = 1 : 1 : 1 : 1

Total # progenies = 230 + 210 + 240 + 250 = 930

Expected # of each type of progeny = 232.5

Chi square test -

Genotype #observed #expected (O-E)^2/E
BbFf 230 232.5 0.026
Bbff 210 232.5 2.177
bbFf 240 232.5 0.241
bbff 250 232.5 1.317

Chi square value = 0.026 + 2.177 + 0.241 + 1.317 = 3.761

DOF = 4 - 1 = 3

p value = 0.1 to 0.9

Null hypothesis accepted.

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