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

Example II

r = 0.2, r = 0.2, r = 0.32

r = r + r - 2r r

m = -1/2ln(1-2r)

m = -1/2 ln (1-2r) = -1/2 ln (1-0.4) = 0.255
m = -1/2 ln (1-2r) = -1/2 ln (1-0.4) = 0.255
m = -1/2 ln (1-2r) = -1/2 ln (1-0.64) = 0.511

Haldane's function converts recombination fractions to map distances which are now additive.

m + m = m

0.255 + 0.255 = 0.511

However note that r + r does not equal r

Why was r = 0.2 converted to m = 0.255?

Because there are double crossovers between the A and B loci. These double crossovers do not result in observable recombination between the A and B loci when the double crossover is a 2-strand type.

However, we have shown that the average recombination for 2, 3, and 4-strand double crossovers is 50% recombination.

Double crossovers should be counted as two crossover events, but the recombination fraction appears the same as a single crossover event. Thus, the relative frequency of crossover events is underestimated by the observable recombination fraction.

Copyright 2000©, Ted Helms

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