Percent Recombinant Genotypes versus Map Distance

Haldane's Mapping Function

Example I

Summarizing

Example II

Expected Frequency Example I

Expected Frequency Example II

Poisson Distribution

An Example

Kosambi's Map Function

Summary Of Map Distance Versus Observed Recombination Fraction

Homework Assignment #7 Questions

Homework Assignment #7 Answers

An Example

r estimated from observed recombination = 0.30.

r = 1/2(1 - e) = 0.30
1 - e = 0.60
0.4 = e-2d
ln(0.4) =-2d
-ln(0.4) = 2d
-(-0.916) = 2d
d = 0.458

l is the mean number of crossover events. Each crossover results in 50% recombination. Then l/2 = m.
m = recombination fraction

** The observed recombination percent was 0.3, the actual relative frequency of recombinant types is 0.458. The adjusted recombination percent is multiplied by two to equal the crossover percent. This is because each crossover event results in 50% non-parental gametes. The map is 45.8cM.

Haldane's mapping function converts the relationship of the sum fo two intervals to the sum of the total interval into a linear relationship.

r does not equal r + r

m = m + m

pg 20. Liu, B.H. Statistical Genomics.

Copyright 2000©, Ted Helms

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