352SLecture09_S11 - Gene$cs and Evolu$on: Lecture #9...

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Unformatted text preview: Gene$cs and Evolu$on: Lecture #9 Today’s Topics: 1.  Biases from mul$ple crossovers 2.  Three ­point testcross 3.  Review? •  Readings: Pierce, pgs. 175 ­179; 183 Linkage and Recombination I( Biases in 2 ­Point Crosses: 1.  Distance between linked genes may result in the appearance of them being unlinked. 2.  Double crossovers. a)  Second crossover undoes the effect of the first – results in more nonrecombinant gametes. b)  result in distances being underes$mated. •  Figure 7.12 1 Double Crossovers: Example: 1.  2 genes that are 10 cM apart 2.  250 meio$c events that produce 1000 chroma$ds 3.  50 crossovers, but 5 are double crossovers 7.  The data: #crossovers # meioDc events (parental cells) 0 205 1 40 2 5 Double Crossovers: Example: #nonrecombinant chromosomes = (205*4) + (40*2) + (5*2) = 910 #recombinant chromosomes = (40*2) + (5*2) = 90 Recombina$on freq. = 90/1000 = 0.90 Map distance = 9.0 cM Double Crossovers: Double crossovers: 1.  More common for genes that are far apart. 2.  Can be detected by examining a 3rd gene between the 2 crossovers. 3 ­Point Testcross: 1.  Can establish gene order in one genera$on 2.  Can detect double crossovers – provides more accurate map. 3.  Inves$gate 3 genes simultaneously. 3 ­Point Testcross: An Example •  Figure 7.14: Drosophila 1. Eye color: scarlet (st) recessive to red (st+) 3. Body color: ebony (e) recessive to gray (e+) 3. Bristle size: spineless (ss) recessive to normal (ss+) 4.  All three loci are on the 3rd chromosome 5.  Triple heterozygotes (TH) crossed with triple recessive parent. •  Note: Gene posi$on is arbitrarily assigned at this point. 2 3 ­Point Testcross: An Example •  Note: 1.  Dominant phenotype progeny inherit dominant from the TH. 2.  X ­over doesn’t affect testcross parent gametes. 3.  X ­over will affect TH gametes 4.  23 = 8 phenotypic classes Determining Gene Order Here’s how to determine gene order: 1.  Iden$fy nonrecombinant progeny – 2 most numerous phenotypic classes. 2.  Iden$fy double X ­over progeny – 2 least numerous phenotypic classes a)  Spineless b)  Scarlet and ebony 3.  Draw the 3 possible TH chromosome arrangements. 3 Determining Gene Order 4.  Carry out double X ­over 5.  Which arrangement gives the double X ­over phenotypes? Mapping 3 ­Point Cross •  Re ­write data as seen in Figure 7.15: •  Nonrecombinant classes •  Double X ­over classes •  Single X ­over classes 4 Mapping 3 ­Point Cross •  Recombina$on frequency: 1.  Add up all recombinants (single AND double X ­overs) 2.  Divide by total number of progeny 3.  Mul$ply by 100% •  Calculate distance between st and ss: [(50 + 52 + 5 + 3)/755] = 14.6% = 14.6 cM •  Calculate distance between ss and e: [(43 + 41 + 5 + 3)/755] = 12.2% = 14.6 cM The Map: B B The Map: Next Time: st 14.6 cM ss 12.2 cM e EXAM 1 5 ...
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