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

Kosambi's Map Function

Haldane's mapping function assumes that there is no interference which would increase or decrease the proportion of double crossovers. Kosambi's mapping function is based on empirical data regarding the proportion of double crossovers as the physical distance varies. Kosambi's function adjusts the map distance based on interference which changes the proportion of double crossovers. Where 2m = ; is the probability of a crossover event; and m is the map distance.

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

- see Liu text pgs. 322-324
Haldane's map function is:
rAC = rAB + rBC - 2rAB rBC

and did not adjust for crossover interference.

Kosambi's map function is:
rAC = rAB + rBC -2CrAB rBC

where C is the coefficient of coincidence

and m = 1/4log [(1+2r)/(1-2r) for 0 <= r < 0.5

We can calculate m or map distance when we know r.
(recombination fraction).

How do we estimate r from observed numbers of crossover and non-crossover progeny? There are several methods which include the product method and maximum likelihood.

See Lui pg 320-321

Let d = map distance in centiMorgans,
Let p = observed recombination fraction.

Let d = -1/2ln(1 - 2p) for 0 <= p <= 0.5
Let d = infinity for p >= 0.5.

Also, p = 1/2(1 - e)

Example:

p = 0.392, find d.

d = -1/2ln(1 - 2(0.392))
d = -1/2ln(1 - 0.784)
d = -1/2ln(0.216)
d = -1/2(-1.5325)
d = 0.766M = 76.6cM

Now let d = 50cM = 0.5M, find p.

p = 1/2(1 - e-2d)
p = 1/2(1 - e-1) = 1/2(1 - 1/e)
p = 1/2(1 - 0.36)
p = 0.32 = 32%

Therefore, 50cM distance does not equate to independent assortment between two loci.

Copyright 2000©, Ted Helms

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