Example 1
Testcross AaBb x aabb
| |
AaBb |
Aabb |
aaBb |
aabb |
Total |
| observed |
140 |
38 |
32 |
150 |
360 |
| expected |
90 |
90 |
90 |
90 |
360 |
| = |
(140-90)2 |
+ |
(38-90)2 |
+ |
(32-90)2 |
+ |
(150-90)2 |
= 135.19 |
| 90 |
90 |
90 |
90 |
This X2 has 3df and is highly significant.
Now we will break this overall X2 test into
three parts, each with 1df.
First test for a ½Aa:½aa ratio for the A-a locus.
| |
Aa |
aa |
Total |
| observed |
140+38 |
32+150 |
360 |
| expected |
180 |
180 |
360 |
| X2 = |
S(O-E)2 |
= |
(178-180)2 |
+ |
(182-180)2 |
= 0.025 |
| E |
180 |
180 |
This X2 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.
| |
Bb |
bb |
Total |
| observed |
172 |
188 |
360 |
| expected |
180 |
180 |
360 |
| X2 = |
S(O-E)2 |
= |
(172-180)2 |
+ |
(188-180)2 |
= 0.62 |
| E |
180 |
180 |
This X2 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 = 70.
| |
Parental |
Recombinant |
Total |
| observed |
290 |
70 |
360 |
| expected |
180 |
180 |
360 |
| 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.
