Example I
m = n = 0.2, mn = 0.04
| Event |
Gametes |
Proportion |
Example |
| #
of crossovers |
ABC,
abc |
(1-m)(1-n) |
0.64 |
| A-B
single crossovers |
aBC,
Abc |
m(1-n) |
0.16 |
| B-C
single crossovers |
ABc,
abC |
(1-m)n |
0.16 |
| 2-Strand
double crossovers |
AbC,
aBc |
mn |
0.04 |
| Total |
|
1.0 |
1.0 |
The observed recombination fraction between the A and
C loci is m(1-n)+(1-m)n = m+n-2mn = 0.32. If the recombination
fraction was additive, then we would expect the recombination
fraction between A and C loci to be 0.2 + 0.2 = 0.4.
r
= observed recombination between A and B loci.
r
= observed recombination between B and C loci.
r
= observed recombination between A and C loci.
2r
= prob. of a single crossover event between A and B
loci.
2r
= prob. of a single crossover event between B and C
loci.
2r
= prob. of a single crossover event between A and C
loci.
1-2rAC = prob. of no of single crossovers between A
and C loci.
1-2rAB = prob. of no of single crossovers between A
and B loci.
1-2rAC = prob. of no of single crossovers between B
and C loci.
(1-2r)
= (1-2r)(1-2r)
1-2r
= 1-2r
- 2r
+ 4r
r
-2r
= -2r
- 2r
+ 4r
r
r
= r
+ r
- 2r
r
m =
r
= 0.2; n = r
= 0.2; mn = 0.04 = r
r
r
= 0.2 + 0.2 - 2(0.04) = 0.32
=
m + n - 2mn = 0.32
r
does not equal r
+ r
Now we know that r
= 0.16 + 0.04 because we have to add the double crossovers
to the single crossovers. Likewise r
= 0.2
r
= m = 0.2
r
= n = 0.2
Then we would expect that rAC = rAB
+ rBC
Then we would expect that rAC
= 0.2 + 0.2
Then we would expect that rAC
= 0.4
However, the observed recombination fraction between
the A and C loci is 0.32. This is because two-strand
double crossover between A and C result in no recombination
between the A and C loci.
The resulting gametes are ABC, abc, AbC, and aBC. None
of these four types result in observable recombination
between the A and C loci.
If recombination fractions are additive, then r
should equal 0.2+0.2=0.4. It is clear that the observed
recombination between A and C loci is:
| Gametes |
Proportion |
Example |
| aBC,
Abc |
m(1-n) |
0.16
= 0.2(0.8) |
| ABc,
abC |
(1-m)n |
0.16
= 0.2(0.8) |
| Total |
m+n-2mn |
0.32 |
