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:
"c" = 4 |
[ |
a |
- |
b + c |
+ |
d |
] |
9 |
3 |
1 |
We now substitute a, b, c, and d to find to find the
score for each family.
Repulsion score:
"c" = 4 |
[ |
a |
- |
b + c |
+ |
d |
] |
= 4 |
[ |
753 |
- |
643 |
+ |
19 |
] |
9 |
3 |
1 |
9 |
3 |
1 |
= |
4(83.667 - 214.33 + 19) |
= |
4(-111.667) |
= |
-446.66 |
Coupling Score:
"c" = 4 |
[ |
a |
- |
b + c |
+ |
d |
] |
= 4 |
[ |
1064 |
- |
482 |
+ |
218 |
] |
9 |
3 |
1 |
9 |
3 |
1 |
= 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 F2 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 = 2515.56 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.
i = |
2(1 + 2p2) |
(1 - p2)(2 + p2) |
| p = 0.2967, then i = |
2[1+2(0.2967)2] |
|
|
|
[1-(0.2967)2][2+(0.2967)2] |
|
|
|
= |
2.352 |
= |
2.352 |
|
|
(0.912)(2.088) |
1.904 |
|
|
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,
c = |
[ |
a |
- |
b + c |
+ |
d |
] |
2p |
(2 + p2) |
(1 - p2) |
p2 |
Now we set p = 0.2967, for the repulsion F2
data
c = |
[ |
753 |
- |
643 |
+ |
19 |
] |
(0.5934) = -76.26 |
2.088 |
0.912 |
0.088 |
For the F2
coupling family data, we use
p = 1-0.2967 = 0.7033
c = |
[ |
1064 |
- |
(223 + 259) |
+ |
218 |
] |
(1.4067) = -123.22 |
2.495 |
0.505 |
0.495 |
We change to positive 123.22 for coupling data. Now
we calculate 
for F2
coupling family data with p = 0.2967 and use 1-p = 0.7033
in place of p, because this is coupling linkage.
i = |
2(1 + 2p2) |
= |
2[1 + 2(0.7033)2] |
(1 - p2)(2 + p2) |
[1 - (0.495)][2 + (0.495)] |
= |
3.979 |
|
(0.505)(2.495) |
|
|
|
= |
3.979 |
|
|
1.26 |
|
|
|
|
= |
3.158 |
I = ni = 1764(3.158) = 5570.6
