Linkage Studies - Part II
Example B:
The closer the linkage between two loci, the greater
the amount of information per testcross individual.
Let p = 0.04, n = 1000 with repulsion linkage.
testcross I = ni; i = 1 / [p(1 - p)] = 1 / (0.04 x 0.96)
= 26.04
I = 1000(26.04) = 26042
| F2 I =ni; i = |
2(1 + 2p2) |
= |
2(1 + 2(0.0016)) |
| (1 - p2)(2 + p2) |
(0.9984)(2.0016) |
| |
|
|
| |
|
= |
2.0064 |
| |
|
1.9984 |
| |
|
|
| |
|
= |
1.004 |
I = ni = 1000(1.004) = 1004
The ratio 1004/26042 = 0.04. This shows that with close linkage in
repulsion, the testcross data provides 25 times the
amount of information from F2.
We can compare testcross data for n = 1000 with p =
0.40 versus p = 0.04 for repulsion linkage. We showed
that when p = 0.04, then I = 4167. The conclusion is
that the closer the linkage the greater the amount of
information and precision of the estimate of p.
Now let's compare repulsion vs coupling F2 data when
n = 1000 and p = 0.4.
F2 repulsion: We showed I = 1455.3
For coupling F2 data use 1-p = 1-0.4 = 0.6
for coupling linkage in this formula:
| i = |
2(1 + 2p2) |
= |
2(1 + 2(0.36)) |
| (1 - p2)(2 + p2) |
|
(0.64)(2.36) |
| |
|
|
|
| |
|
= |
3.44 |
| |
|
1.51 |
| |
|
|
|
| |
|
= |
2.278 |
I = ni = 1000(2.278) = 2278
With F2 repulsion, I = 1455. With F2 coupling I = 2278.
This show that for the same value of p, F2 coupling
data has 1455/2278 or 1.56 times the amount of information
as repulsion data.
