Estimation of Recombinant Faction
|
= |
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.
F
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
