Expected Frequency Example II
Let r
= r
= 0.2
| |
Expected
Frequency |
| Genotype |
with
double crossover |
without
double crossover |
| AaBbCc |
0.32 |
1/2(0.6)
= 0.3 |
| AaBbcc |
0.08 |
1/2(0.2)
= 0.1 |
| AabbCc |
0.02 |
0 |
| Aabbcc |
0.08 |
0.1 |
| aaBbCc |
0.08 |
0.1 |
| aaBbcc |
0.02 |
0 |
| aabbCc |
0.08 |
0.1 |
| aabbcc |
0.32 |
0.3 |
The above frequencies assume that double crossovers
consist of only 2-strand double crossovers that result
in no recombinant progeny. In the absence of double
crossovers,
rAC = 0.1
+ 0.1 + 0.1 + 0.1 = 0.4
1.0
When double crossovers are considered, then the observed
rAB = 0.08 + 0.08 + 0.08 + 0.08
= 0.32
1.0
However, the actual r must include 2-strand double
crossover events that fail to show observed recombination.
rAC = 0.08 + 0.08 + 0.08 + 0.08
+2(0.02+0.02) = 0.40
1.0
