F
Progeny
Assume Coupling Linkage
If A and B loci are independent, then p = 1/2 and the probability of
each F
genotype = 1/16.
Example:
From: Statistical Genomics:Linkage, Mapping, and QTL Analysis. -
B.H. Liu. pg 174
P(AABB) = 1/4(1-p)2 = 1/4(1-1/2)2 = 1/4(1/4) = 1/16.
| Genotype |
Expected
Proportion |
| A_B_ |
3-2p+p
4 |
| A_bb |
p(2-p)
4 |
| aaB_ |
p(2-p)
4 |
| aabb |
(1-p)
4 |
Example:
There are nine ways to observe A_B_ progeny in the F2.
| Type |
Probability |
*These are
all mutally exclusive types. Compare AABb to AABb. In one case
AB came from the male and Ab came from the female. In the other
case AB came from the female and Ab came from the male.
|
| AABB |
(1-p)2
4 |
| AABb* |
p(1-p)
4 |
| AaBB |
p(1-p)
4 |
| AaBb |
(1-p)2
4 |
| AABb* |
p(1-p)
4 |
| AaBb |
p2
4 |
| AaBB |
p(1-p)
4 |
| AaBb |
p2
4 |
| AaBb |
(1-p)2
4 |
| Sum |
(3-2p+p2)
4 |
We can add probabilities of mutually exclusive types.
For coupling:
| P(A_B_) |
=
1/4[(1-p)2+p(1-p)+p(1-p)+(1-p)2
= +p(1-p)+p2+p(1-p)+p2+(1-p)2] |
| |
=
1/4[3-6p+4p+5p2-4p2] |
| |
=
1/4[3-2p+p2] |
This agrees with the expression on the top of the table on pg. 4 of
those notes.
