X Contingency Testing

If two events are independent, the probability of both events occurring is the product of the probabilities of each separate event.

Event A occurs with P(A) and event B occurs with P(B), then the probability of both A and B occurring is P(A) P(B). The contingency X test evaluates whether one factor influences the probability of a second factor.

Example

In a di-hybrid F cross with complete dominance gene action at both loci the expected F progeny segregation is 9A_B_:3A_bb:3aaB_:1aabb, provided the two loci segregate independently and the individual loci segregate in a 3:1 ratio.

Strickberger, pgs. 317-318.

The X with Yates correction is 17.14 with 3df and the associated probability is less than 0.001. We are testing Ho:9:3:3:1 with a = 0.05, so we fail to accept Ho.

Significance

There are three possible explanations for the overall X test being significant:

1. distorted segregation at the A locus;
2. distorted segregation at the B locus;
3. the two loci do not segregate independently. We can set up three hypothesis:
1. Ho:3A:1a
2. Ho:3B:1b
3. Ho:A and B loci do not segregate independently.

The X tests show evidence that the A locus segregates 3:1, but the B locus does not segregate 3:1. There is evidence of a distorted segregation ratio at the B locus. Now we will test for independence between the two loci.

P(A) = 210 = 0.75
P(A) = 280

P(a) =  70 = 0.25
P(a) = 280

P(B) = 240 = 0.857
P(B) = 280

P(b) =  40 = 0.143
P(b) = 280

If we assume independence we can find the expected relative frequencies of each class. We then multiply the expected relative frequencies by the total number to get the expected number of each class.

P(AB) = P(A)P(B) = 0.75(0.857) = 0.64275.
The expected number of A_B_ is 0.64275(280)
= 179.97.

P(Ab) = P(A)P(b) = (0.75)(0.143) = 0.10725.
The expected number of A_bb is 0.10725(280)
= 30.03.