Summarizing
o = a
d log[1-r] + a
d log[r]
o = a d loldr dr
o + a
d log[r] + a
d log[1-r]
dr dr
o = -a +
a
+ a
+ -a
= 0
1-r lr2
+ r3 + 1-r
This one equation with one unknown. We know a,
a,
a,
and a
because these are the observed numbers of each class.
Now can substitute the actual numbers and solve for
r.
Example
| class |
PpTt |
Pptt |
ppTT |
pptt |
| observed |
191 |
37 |
36 |
203 |
| symbol |
a1 |
a2 |
a3 |
a4 |
o = -191 + 37
+ 36 - 203
= dL
o = 1-r
+ 3r + r -
1-r = dr
o = -394 + 73
1-r r
o = 1-r
+ r
o = -394 73
1-r = r
394 73
394r = 73 -73r
(394 + 73)r = 73
467r = 73
467r = 73/467
467r = 0.1563 or 15.63%
|
We could use the short formula for testcross
data with the maximum likelihood method:
r = a
+ a
for coupling
n
r = a
+ a
for repulsion
n
For Example:
r = a
+ a
= 37
+ 36 = 73
= 0.1563
n a3
= 3 467 6 =
467
|
Summary
We derived the expected number of each class of progeny
for the testcross with coupling linkage as:
n(1-r), nr,
nr, n(1-r)
2 2
2 2
We used the method of maximum likelihood to find the
most precise estimate of r.
The method of maximum likelihood uses calculus and
sets the first derivative equal to zero.
The method of maximum likelihood finds the best estimate
of r that fits the observed data.
