Linkage Studies - Part II

Example B:

The closer the linkage between two loic, 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.