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 #5

Homework
Answers
Assignment #5

F Repulsion Progeny Part II

The A_B_ class has the proportion (p +2)/4 for repulsion linkage. Now let us find the proportion for the other three classes.

class proportion A_bb phenotype
AAbb (1-p)
4
Aabb p(1-p)
4
Aabb p(1-p)
4

(1-p) + p(1-p) + p(1-p) = 1 - 2p + p + p - p + p - p
  4           4          4                         4

= -p + 1 = 1 - p
        4         4

class proportion
aaBB (1-p)
4
aaBb p(1-p)
4
aaBb p(1-p)
4

(1-p) + p(1-p) + p(1-p) =  1-p 
              4                        4

Finally there is only one combination that results in the aabb genotype with probability p/4.

Example

We can sum the probability of all genotypes that are phenotypically A_B_ or A_bb or aaB_, or aabb.

class A_B_ A_bb aaB_ aabb Total
observed a a a a  
observed 753 292 351 19 1415
expectation p2 + 2
4
1-p2
4
1-p2
4
p2
4
 

r = n, s = -a + 2a + 2a + 2a, t = -2a

r = 1415, s = -753 + 2(292) + 2(351) + 2(19) = 571, t = -2(19) = -38

Copyright 2000©, Ted Helms

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