set9 mapping

# 11 finding map distances a a a a a a a a b c dco b

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Unformatted text preview: tances a+ a+ a+ a+ a a a a b+ c DCO b+ c+ P b c+ SCO b c SCO b+ c+ SCO bc P b+ c SCO b c+ DCO 9 -pick any 2 genes to start with e.g., a and b 1654 -sum all the offspring with recombinant 131 phenotype of a+ b or a b+ (ignore c) 252 =131 + 252 + 241 + 118 = 742 241 1779 -divide by total number of offspring and 118 multiply by 100 13 = 742 x 100 = 17.7 m.u. a b 4197 4197 -repeat for each gene pair a and c = 9+252+241+13/4197*100 = 12.3 b and c = 9+131+118+13/ 4197 *100 = 6.4 12 Construct the map = 17.7 m.u. a = 12.3 m.u. a = 6.4 m.u. b b c c a c 12.3 b 6.4 17.7 BUT….. 12.3 + 6.4 = 18.7 m.u. not 17.7 as observed WHY? due to double cross overs. 13 Correction for distances between a and b -calculation of m.u. between a and b does not take into consideration double crossovers -does not reflect all of the recombination events that produce a b+ or a+ b a c a+ c+ b+ c+ a a+ b + b c b+ 14 Correction for distances between a and b a c+ b a+ + c b+ -look like parentals for a and b, but are not -these were not included in original calculation -go back to original calculation for a and b and include these numbers 15 Finding map distances a+ a+ a+ a+ a a a a b+ c DCO b+ c+ P b c+ SCO b c SCO b+ c+ SCO bc P b+ c SCO b c+ DCO 9 -pick any 2 genes to start with e.g., a and b 1654 -sum all the offspring with recom...
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