Example II
Kramer & Burnham. 1947. Genetics 32:379-390.
| Observed
F2 |
A_B_
a |
A_bb
b |
aaB_
c |
aabb
d |
Total |
| Repulsion |
753 |
292 |
351 |
19 |
1415 |
| Coupling |
1064 |
223 |
259 |
218 |
1764 |
The formula for F2 family score is:
We now substitute a, b, c, and d to find to find the
score for each family.
Repulsion score:
= 4(83.667 - 214.33 + 19)
= 4(-111.667)
= -446.66
Coupling Score:
= 4 [118.22 - 160.67 + 218]
= 4 [175.55]
= 702.22
The scoring formula was developed for repulsion data.
*We change the sign of the score
for coupling data.
Now we will find the total amount of information for
each F
family. We have shown that when p = 0.5, then i = 16/9
and I =n16/9.
For the repulsion family n = 1415, then:
I = n16/9 = 1415(16/9) =2515.56
For the repulsion family n = 1764, then:
I = n16/9 = 1764(16/9) = 3136
The X2 test for deviation from p = 0.5 is
c2/I.
Repulsion family: c = -446.66, I = 2525.56 then:
X2 = c2/I = (-446.66)2/2515.56
= 79.31
Coupling family: c = -702.22, I = 3136 then:
X2 = (-702.22)2/3136
= 157.24
| We
can test for heterogeneity of estimates of deviation
from p = 0.5 among these two families. |
| |
| Source |
score(c) |
I |
X2=c2/I |
df |
| Repulsion |
-446.66 |
2515.6 |
79.31 |
1 |
| Coupling |
-702.22 |
3136 |
157.24 |
1 |
| Sum |
-1148.88 |
|
236.55 |
|
| Total
X2 |
(-1148.88)2/5651.6
= |
233.55 |
2 |
| Heterogeneity
X2 |
|
3.0 |
1 |
Correction -1148.88/5651.6 = -0.2033 = 
Estimate of correction p = 0.5 - 0.2033 = 0.2967
Conclusions:
With 1df a X2
for heterogeneity = 3.0 is not significant at a
= 0.05. Negative signs for scores show that p < 0.5.
Individual X2 for repulsion(79.31) and coupling
(157.24) show significant deviation from p = 0.5 for
both types of families.
We know that p < 0.5 because of the large X2
values of 79.31 and 157.24 for repulsion and coupling
respectively. These significant X2's
are evidence that when p is set to 0.5 the observed
data is a poor fit to the expected values. We can use
the estimate of p = 0.2967 and fit the model a second
time.
To recalculate the X2=
c2/I
for p = 0.2967, we need to recalculate I. Previously
we calculated I with the assumption that p = 0.5. In
general, I = ni.
p = 0.2967, then i = 2[1+2(0.2967)2]
p = 0.2967, then i = [1-(0.2967)2][2+(0.2967)2]
= 2.352
=
2.352
= (0.912)(2.088) 1.904
= 1.235
I = ni = 1415(1.235) = 1747.5 for repulsion data.
We also must recalculate the score for p = 0.2967.
For F2
families in repulsion,
Now we set p = 0.2967, for the repulsion F2
data
= -76.26
For the F2
coupling family data, we use
p = 1-0.2967 = 0.7033
= -123.22
We change to positive 123.22 for coupling data. Now
we calculate i for F2
coupling family data with p = 0.2967 and use 1-p = 0.7033
in place of p, because this is coupling linkage.
= 3.158
I = ni = 1764(3.158) = 5570.6
