04BIS1012012RecombinLect4b

1 rf is equivalent to 1 map unit mu or 1 centimorgan

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Unformatted text preview: ates of map distances are obtained from the sum of the distances calculated from shorter subintervals (10 m.u.). However, a formula developed by JSB Haldane (1919) called the "mapping function" can be used to correct for the effects of multiple crossovers and provide more accurate map distances. e-m mi (i) = i! A a March 19, 2012 or RF = 1/2(1- e-m ) B b C c 43 BIS101001, Spring 2012--Genes and Gene Expression, R.L. Rodriguez 2012 Mapping Function: Solving for m By solving for e-m (the natural log of the mean crossovers per meiosis) you can find m, using a calculator or natural log table (see next slide). e-m = 1 - RF(2) m/2 = RF RF = 1/2(1- e-m ) Since it has been shown empirically that m/2=RF, you can easily convert the original RF values into accurate map distances. For examples, if RF = 27.5% or 27.5 m.u. then... e-m = 1 - 0.275(2) m = 0.8 March 19, 2012 e-m = 1 - 0.55 e-m = 0.45 0.8/2 = 0.4 or 40 m.u. 44 BIS101001, Spring 2012--Genes and Gene Expression, R.L. Rodriguez 2012 M...
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This note was uploaded on 03/18/2012 for the course BIS 101 taught by Professor Simonchan during the Winter '08 term at UC Davis.

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