Example 1

Testcross AaBb x aabb

overall X = S(O-E)
                    E

= (140-90) + (38-90) + (32-90) + (150-90) = 135.19
        90            90            90             90

This X has 3df and is highly significant.

Now we will break this overall X test into three parts, each with 1df.

First test for a ½Aa:½aa ratio for the A-a locus.

X = S(O-E) = (178-180) + (182-180) = 0.025
            E            180             180

This X with 1df is not significant. The A-a locus segregates 1:1 as expected, so there is no problem with different viability for either allele.

Now let us test for distorted segregation ratio in the B-b locus. First we will sum across the AaBb and aaBb classes to derive the total number in the Bb class. Next we will sum the Aabb and aabb classes to determine the total number of the bb class.

X = S(O-E) = (172-180) + (188-180) = 0.62
            E            180             180

This X w/1df is not significant so we fail to reject Ho:½Bb:½bb ratio.

We have an overall X test w/3df. We have used 1df to test Ho:½Aa:½aa and a second df to test Ho:½Bb:½bb.

If we fail to reject this third hypothesis, the evidence supports independent assortment or no linkage.

Ho: Independent Assortment

We will sum the coupling classes together and then sum the repulsion class together to test for linkage: Coupling class are 140AaBb + 150aabb = 290. Repulsion classes are 38Aabb + 32aaBb = 60.

X² = S(O-E)² = (290-180)² + (70-180)² = 134.44
            E            180             180

This X² w/1df is highly significant. We rejection Ho and conclude that the two loci are linked.