Conditional Probability
| |
(1-r)/2
AQ |
r/2
Aq |
r/2
aQ |
(1-r)/2aq |
| (1-r)/2
AQ |
(1-r)2/4 |
r(1-r)/4 |
r(1-r)/4 |
(1-r)2/4 |
| r/2
Aq |
r(1-r)/4 |
r2/4 |
r2/4 |
r(1-r)/4 |
| r/2
aQ |
r(1-r)/4 |
r2/4 |
r2/4 |
r(1-r)/4 |
| (1-r)/2
aq |
(1-r)2/4 |
r(1-r)/4 |
r(1-r)/4 |
(1-r)2/4 |
| |
QQ |
Qq |
qq |
|
| AA |
(1-r)2/4 |
r(1-r)/4
+ r(1-r)/4 |
r2/4 |
1/4 |
| Aa |
r(1-r)/4
+ r(1-r)/4 |
(1-r)2/4
+ r2/4 + r2/4 + (1-r)2/4 |
r(1-r)/4
+ r(1-r)/4 |
1/2 |
| aa |
r2/4 |
r(1-r)/4
+ r(1-r)/4 |
(1-r)2/4 |
1/4 |
| |
1/4 |
1/2 |
1/4 |
1.0 |
Prob.(QQ/AA) = Prob(QQ and AA) = (1-r)2/4
= (1-r)2
Prob.(QQ/AA) = Prob(AA) 4
Prob.(Qq/Aa) = Prob(Qq and Aa)
Prob.(Qq/Aa) = Prob(Aa)
Prob.(Qq/Aa) = [(1-r)2/2 + r2/2]
= (1-r)2 + r2
Prob.(Qq/Aa) = 1/2
Marker
Genotype |
P(Qj/M) |
| QQ |
Qq |
qq |
| AA |
(1-r)2 |
2r(1-r) |
r2 |
| Aa |
r(1-r) |
(1-r)2
+ r2 |
r(1-r) |
| aa |
r2 |
2r(1-r) |
(1-r)2 |
Suppose we select individuals that have the AA marker genotype. What
is the probability that the QTL genotype will be QQ?
P(Qj/M) = P(QQ/AA) = (1-r)2
The proportion of QQ selected.
This result shows that as r increases for coupling linkage between
A and Q, selection for QQ will be less effective.
Let us select only individuals with the AA marker genotype. Now we
have reduced the sample space.
| |
QQ |
Qq |
qq |
|
| AA |
(1-r)2/4 |
r(1-r)/2 |
r2/4 |
1/4 |
| Aa |
r(1-r)/2 |
(1-r)2/4
+ r2/4 |
r(1-r)/2 |
1/2 |
| aa |
r2/4 |
r(1-r)/2 |
(1-r)2/4 |
1/4 |
The reduced sample space is now:
| |
QQ |
Qq |
qq |
|
| AA |
(1-r)2 |
2r(1-r) |
r2 |
1.0 |
P(QQ/AA) + P(Qq/AA) + P(qq/AA)
= (1-2r+r2) + (2r-2r2) + r2 = 1
