Contingency X2
P(aaB_) = P(aa) P(B_) = 0.25(0.857)
= 0.21425.
The expected number of aaB_ progeny is 0.21425(280)
= 59.99.
P(aabb) = P(a) P(b) = 0.25(0.143) = 0.03575.
The expected number of aabb progeny is 0.03575(280)
= 10.5
| |
B_ |
bb |
| A_ |
180
(179.97) |
30
(30.03) |
| aa |
60
(59.99) |
10
(10.01) |
We can now test Ho:A and B are independent factors
using a Contingency X2 test. We will use
Yate’s correction, due to small numbers in some classes.
If O-E is less than 0.5, then we set O-E = O.
| = |
(|180 - 179.97| - 1/2)2 |
+ |
(|30 - 30.03| - 1/2)2 |
| 179.97 |
30.03 |
| + |
(|60 - 59.99| - 1/2)2 |
+ |
(|10 - 10.5| - 1/2)2 |
| 59.99 |
10.5 |
= 0.00 + 0.00 + 0.00 + 0 = 0.0
A contingency X2
test of 0.0 with1df is not significant. When α
= 0.05 the critical X2
value must be 3.84 or larger. We could use the formula
for a 2 x 2 contingency X2
test with Yate’s correction term.
| X2 = |
(ad-bc - 1/2N)2N |
|
|
| (a+b)(a+c)(c+d)(b+d) |
|
|
| |
B_ |
bb |
|
| A_ |
180(a) |
30(b) |
a+b |
| aa |
60(C)
a+c |
10(d)
b+d |
c+d |
| X2 = |
[180(10) - 30(60) - 1/2(280)]2280 |
|
|
| (210)(240)(70)(40) |
|
|
| = |
(1800-1800-140)2280 |
|
|
| 141120000 |
|
|
= 0.039 ~ 0
Which is a non-significant X2
when testing Ho:Factors A and B are independent. The
explanation for the significant deviation from the 9:3:3:1
ratio is that there is a distorted segregation ratio
at the B locus.
The
rows show how the proportions vary for the B locus for
the same level of the A factor. The columns show how
the proportions vary for the A locus and a fixed level
of factor B.
