Introduction to Genetic Analysis 136

Introduction to Genetic Analysis 136 - 44200_04_p115-150...

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135 4.6 Accounting for unseen multiple crossovers and this formula is the mapping function we have been seeking. Let’s look at an example of how it works. Assume that in one testcross we obtain an RF value of 27.5 per- cent (0.275). Plugging this into the function allows us to solve for m: so e 2 m 5 1 2 (2 3 0.275) 5 0.45 Using a calculator we can deduce that m 5 0.8. That is, on average there are 0.8 crossovers per meiosis in that chromosomal region. The ±nal step is to convert this measure of physical map distance to “corrected” map units. The following thought experiment reveals how: “In very small genetic regions, RF is expected to be an accurate measure of physical distance because there aren’t any multiple crossovers. In fact meioses will show either no crossovers or one crossover. The frequency of crossovers (m) will then be translatable into a ‘correct’ recombinant frac- tion of m/2 because the recombinants will be 1/2 of the chromatids arising from the single-crossover class. This de- ±nes a
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This note was uploaded on 01/10/2011 for the course BIOL BIOL taught by Professor Johnson during the Spring '08 term at Aberystwyth University.

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