Part I

Part II

Part III

Summarizing

F2 Data

F2 Progeny Part I

F2 Progeny Part II

Standard Error Of 'p', The Recombination Fraction

Summary

The Amount Of Information And Its Uses

Homework
Questions Assignment #6

 

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F2 Repulsion Data

classes
A_B_
A_bb
aaB_
aabb
observed # a1 a2 a3 a4
expected frequency
2+p2
4
1-p2
4
1-p2
4
p2
4

The above table is for repulsion linkage from the cross

to get the F2 progeny.

L=
n!
(
2 + p2
)
a2
(
1 - p2
)
a2
(
1 - p2
)
a4
(
p2
)
a4
a1!a2!a3!a4!
4
4
4
4

The equation to be maximized with respect to r is:

log(L) = C+a1 log
[
(2 + p2)
]
+ a2 log
[
(1 - p2)
]
 
4
4
 

  + a3 log
[
(1 - p2)
]
+ a4 log
[
 p2
]
 
4
4

dL
= 0 +
a1 (2p)
+ a2
 -2p 
+ a3
 (-2p) 
+ a4
(2p)
dp
2 + p2
1 - p2
1 - p2
p2

Using the chain rule that:

d log Q
=
d log Q
dQ
       
dp
dQ
dp
       

  =
 1 
dQ
       
 
Q
dp
       

 
 dL 
= 2p
[
   a1   
-
(a2 + a3)
+
 a4 
]
= 0
 
dp
2 + p2
(1 - p2)
p2

 
  a1  
-
(a2 + a3)
+
 a4 
= 0  
 
2 + p2
1 - p2
p2
 

 
=
   a1(1 - p2)   
+
(2 + p2)(-a2 - a3)
+
 a4 
= 0
 
(2 + p2)(1 - p2)
(2 + p2)(1 - p2)
p2

=
a1(1-p2)p2
+
(2+p2)(-a2-a3)p2
+
(2+p2)(1-p2)a4
= 0
(2+p2)(1-p2)p2
(2+p2)(1-p2)p2
(2+p2)(1-p2)p2

= a1(p2 - p4) + 2p2(-a2 - a3) + p4(-a2 - a3)
     + 2a4 - 2p2a4 + p2a4 - p4a4 = 0

= -a1p4 - a2p4 - a3p4 - a4p4 + a1p2 - 2a2p2 - 2a3p2
     - 2a4p2 + 2a4 = 0

= p4(-a1 - a2 - a3 - a4) + p2(a1 - 2a2 - 2a3 - 2a4) + 2a4 = 0

= -p4n + p2(a1 - 2a2 - 2a3 -2a4) + 2a4 = 0

= p4n + p2(-a1 + 2a2 + 2a3 + 2a4) -2a4 = 0

We can solve for p using the quadratic equation in the form of rx2 + sx + t = 0. r = +n, s = -a1 + 2a2 + 2a3 + 2a4,
t = -2a4

Copyright 2000©, Ted Helms

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