Example I
For F2
families
L = n!
(m1)a(m2)b(m3)c(m4)d
a!b!c!d!
| class |
A_B_ |
A_bb |
aaB_ |
aabb |
Total |
| observed |
a |
b |
c |
d |
n |
| expected |
m1 |
m2 |
m3 |
m4 |
|
d log L |
= 2p |
[ |
a |
- |
b + c |
+ |
d |
] |
= 0 |
dp |
(2 + p2) |
(1 - p2) |
p2 |
|
|
|
|
|
|
|
|
|
|
|
= |
[ |
a |
- |
b + c |
+ |
d |
] |
2p |
|
(2 + p2) |
(1 - p2) |
p2 |
2p = 0.5(2) = 1, so we can leave out the 2p above. Now we substitute
p=1/2 in the above equation.
a |
- |
(b + c) |
+ |
d |
= |
a |
- |
b + c |
+ |
d |
(2 + 0.25) |
(1 - 0.25) |
0.25 |
2.25 |
0.75 |
0.25 |
= |
a |
- |
b + c |
+ |
d |
= 4 |
[ |
a |
- |
b + c |
+ |
d |
] |
9/4 |
3/4 |
1/4 |
9 |
3 |
1 |
| |
|
|
|
|
|
|
|
|
|
|
|
|
|
| |
|
|
|
|
|
= Fisher's score for F2 families |
Now the amount of information per individual ( i )
for F2
family is:
i = |
2(1 + 2p2) |
= S |
( |
1 |
( |
dm |
) |
2 |
) |
(1 - p2)(2 + p2) |
m |
dp |
|
We set p = 0.5, then:
| i = |
2(1 + 2p2) |
 |
2(1 + 2(0.25)) |
| (1 - p2)(2 + p2) |
(1 - 0.25)(2 + 0.25) |
| |
|
|
|
= |
2(1.5) |
= |
3 |
|
(0.75)(2.25) |
27/16 |
|
| |
|
|
|
= |
3 |
( |
16 |
) |
= |
16 |
|
1 |
27 |
9 |
|
Now the total amount of information for a specific F family when p=1/2
is ni = I = n16/9.
Fisher showed that c2/I
is distributed as X2
w/1df for a test of significance for deviation from
50% recombination. Each c2/I can be calculated for each
type of family which are combined to obtain the combined
estimate of p.
