| |
= | Nr |
| N |
For testcross data we know the number of recombinant progeny directly. Let Nr equal the number of recombinant genotypes and N equal the total numbr of progeny.
F1 AaBb x aabb is the testcross.
Example:
We do a X2 for Ho:1:1:1:1 ratio and fail to accept Ho. Also we test Ho:1:1 ratio for A locus and B locus and fail to reject Ho. Next we test for Ho: independence between A and B loci. We fail to accept Ho for independence. Now let us estimate p.
| AaBb | Aabb | aaBb | aabb | Total | |
| observed | 140 | 38 | 32 | 150 | 360 |
| expected | 90 | 90 | 90 | 90 | 360 |
This is clearly coupling linkage, from the shortage of Aabb and aaBb classes.
= 70 = 0.194 or 19.4%
360