biol139-lecture20-2011

7 mu events once have accurate map distance for

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Unformatted text preview: from vg/b and have map distance between pr and b Why are double crossover events so rare? Why Product of separate independent probabilities: Probability of recombinants from crossover in region 1 alone = 10% or 10 m.u. 10 Probability of recombinants from crossover in region 2 alone = 20% or 20 m.u. 20 The probability of a double crossover is 0.10 x 0.20 = 2% 0.10 1.56 m.u y w 4.06 m.u ec 5.5 m.u For our working example: probability of single crossover in region 1 = 0.0156 probability of single crossover in region 2 = 0.0406 Expected probability double crossover is .0156 x .0406 = .063 .063 Observed double crossovers are 3+3 = 6/10,000 = .06 .06 Interference Interference The number of observed double crossovers may The observed be less than expected be expected Why ? Because sometimes a crossover in one region of the chromosome will reduce the likelihood of a crossover in an adjacent part of the chromosome A smaller observed value than expected would indicate interference Interference Interference It is not uniform, and may vary for different regions of the It chromosome chromosome A quantitative measure of the amount of interference in a quantitative particular chromosomal region is first obtained by calculating the coefficient of coincidence coefficient Coefficient of coincidence = Frequency observed (DCO) Frequency expected (DCO) Interference = 1- coefficient of coincidence Interference If there is no interference (interference = 0), then the observed frequency of double crossovers is equal to the expected frequency. If interference is complete (interference = 1), then there are no double crossovers observed. Interference Interference Sample problem: The probability of a crossover in Region 1 is 20%, and in Region 2 is 6%. in Region Region The observed rate of double crossovers for the observed area encompassing Regions 1 and 2 is 0.9%. area 0.9%. Calculate the interference in this region. Expected double crossover frequency = (0.2)(0.06) = 0.012 = 1.2% 1.2% 0.9 0.9 Coeffic...
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This note was uploaded on 10/04/2011 for the course BIOL 139 taught by Professor Christinedupont during the Spring '10 term at Waterloo.

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