Haldane Map Function And Genetic Linkage

Question: Haldane Map Function And Genetic Linkage

8.6 years ago by

User 9334 • 60 wrote:

if i have two markers that are 100 Mb apart on the same chromosome, and i know that for my chromosome, centiMorgan to megaBase ratio is .9, then how many offspring in a generation can i expect to recombine between A and B?

since 1 centiMorgan is roughly 1 megaBase in this case, and 1 centiMorgan is 1% of the generation, then it would not make sense to say that 100% of the offspring would have recombined. The recombination frequency is bounded at 0.5.

so how can this be estimated? must i use haldane's function? what is a good way to think of this? thanks.

linkage genetics snp • 7.3k views

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written 8.6 years ago by User 9334 • 60

8.6 years ago by

Haibao Tang ♦ 3.0k

Mountain View, CA

Haibao Tang ♦ 3.0k wrote:

You are right about using a map function. See a derivation here. The relationship between map unit and recombination frequency is:

alt text

When r is small, r and m is roughly equal - so you are right about 1cM ~ 1% of the generation. It does not hold when r is approaching 0.5.

alt text

ADD COMMENT • link written 8.6 years ago by Haibao Tang ♦ 3.0k

if I plug in 100 cM into the equation for r from http://statgen.ncsu.edu/qtlcart/manual/node46.html i still get that r = 0.5, which cannot be right. if snp A and snp B are 100 Mb apart on the same chromosome, they will not segregate independently

ADD REPLY • link written 8.6 years ago by User 9334 • 60

@unknown (google): see the plot above (x-axis is r, y-axis is m), remember 100cM = 1 morgan, and when you look at m = 1, r is somewhere between 0.4 and 0.5. Sean gave an exact calculation.

ADD REPLY • link written 8.6 years ago by Haibao Tang ♦ 3.0k

@unknown (google): it is not clear to me where your intuition comes from. 100Mb is a very large distance, it could be the two ends of a chromosome in human. I wouldn't expect strong linkage across that distance. Unless you really meant 100Kb rather than 100Mb - for example if snp A is 100Kb away from snp B, then that's 0.1cM and roughly 0.1% recombination frequency.

ADD REPLY • link written 8.6 years ago by Haibao Tang ♦ 3.0k

so what would it be in this case? i am not sure i see the answer

ADD REPLY • link written 8.6 years ago by User 9334 • 60

0.5(1-exp(-21)) = 0.4323324

ADD REPLY • link written 8.6 years ago by Sean Davis ♦ 26k

This makes no sense, since if you have something that's roughly 100 Mb away, it can't be that the recombination frequency is around 40%. It should be far less. if snp A is 100 Mb away from snp B, what is their predicted recombination frequency?

ADD REPLY • link written 8.6 years ago by User 9334 • 60

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