PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Figure 4-17Slide 32Slide 33Slide 34Slide 35Slide 36A. Review: Mapping with molecular markersB. The relationship between RF and map distance(Haldane's mapping function)'Haplotypes'C. Measuring linkage with LOD scoresLOD= 'Log of the odds'D. Fine mapping and map-based gene cloningGenetic map [cM] versus Physical map [bp]Map-based cloning = Positional cloningE. Recombinant Inbred (RI) Lines Lecture 9Monday September 15, 2014Linkage, Recombination, and Mapping Genes - IIIMapping genes with molecular markersMolecular (DNA-based) markers versus visible markers- Mendelian - Codominant- Robust - AbundantTypesSNPs (single nucleotide polymorphisms)SSLPs (short sequence length polymorphisms)VNTRs (variable number tandem repeat)RFLPs (restriction fragment length polymorphisms)Scoring (=‘genotyping’) alleles of molecular markersPCR and electrophoretic separation of DNA for RFLP: add Restriction digestion stepLater: By hybridizationStrain 1AACGTTAACATT Strain 21/1 2/2 1/21/12/2 2/2ForwardReverseAAGTCGTGACAGTACGTAC-----AACGTT----- TTCTAACGACCCGTAGACAGTAG12Restriction Fragment Length Polymorphism (RFLP)1. PCR3. Gel2. Restriction enzymeCo-dominanceHindIIIAACATTLinkage between a visible trait and a molecular markeror VNTR 4 distinct alleles of marker M Pp ppto synthesize DNA from each alleleElectrophoresisEvidence for Linkage?= DNA fragmentsPp ppCo-dominanceLinkage between a visible trait and a molecular markerAn SSLP marker is located 2 map units away from the X-linked locus responsible for red-green color blindness. Three alleles of this SSLP are known: “4 repeats,” “5 repeats,” and “7 repeats.” A woman with normal color vision, but whose father is color blind, marries a man with normal vision, and they have a son. Family member SSLP genotypeWoman’s father “5 repeats”Woman’s husband “3 repeats”Woman “5 and 7 repeats”Couple’s child “5 repeats” Given the SSLP genotypes of the various family members shown above, what is the probability that the couple’s son is color blind? A) 2%B) 50%C) 98%D) 99%E) 100%Linkage problem IIIRF = 2 cM = 2% rg / + 5 / 7+ / Y. / rg / Y 5+ / .7 / . rg / Y 5 /Alternative: + / Y 5 /Gametes of rg 5 Par. 49/100 + 7 Par. 49/100 rg 7 Rec. 1/100 + 5 Rec. 1/100 rg + 5 798%2%Linkage problem IIIIF sun carries allele “5”, then the Probability (rg) P (rg) = 98%GametesA. Review: Mapping with molecular markersB. The relationship between RF and map distance(Haldane's mapping function)'Haplotypes'C. Measuring linkage with LOD scoresLOD= 'Log of the odds'D. Fine mapping and map-based gene cloningGenetic map [cM] versus Physical map [bp]Map-based cloning = Positional cloningE. Recombinant Inbred (RI) Lines Lecture 9Monday September 15, 2014Linkage, Recombination, and Mapping Genes - IIIThe relationship between recombination frequencies and map distanceRF: 1.0% 2.0%3.0%RF: 15.0% 21.0%32.0%Recombination frequencies are ‘not additive’ over long distances. But the units for map distances should be additive (for convenience). map units: 1cM 2cM 3cMmap units: 20.0 cM 30.0 cM50 cMCase I: Short map distance Case II: Long map distance empirical dataderived from empirical dataPeter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.A mapping function to estimate the map distance from the recombination frequency (Haldane)because crossovers are randomly distributed[cM]RealitySimplisticRF [%]Is the discrepancy between (empirical) RF and (additive) map units due to double crossovers ?Scenario: Long map distance between A and B AVERAGE crossover frequency = 1But in reality, ...Scenario: Long map distance between A and B AVERAGE crossover frequency = 1But in reality, crossovers are randomly distributed...... some meioses will have no crossover...Scenario: Long map distance between A and B.AVERAGE crossover frequency = 1... and even meioses with 2 crossovers...... have RF=50%Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.A mapping function to estimate the map distance from the recombination frequency (Haldane)Probability of 0 crossovers (crossovers are randomly distributed!)[cM]RealitySimplisticRF [%] Q: What is the probability of 0 crossover?p(0)p (>0)Recombination Frequency -> Map units = crossover frequencies- Empirical - Derived- Not quite additive - Perfectly additiveGiven: Average number of c/o= m Question: What is the probability of i= 0, 1, 2, 3… c/o between A and B?Poisson distribution: p(i) = [ e -m m i ] / i! A BabA Ba bHaldane’s Mapping FunctionSolution: Reverse the Problem. Trick: What is the probability of at least 1 crossover? p(>0) = 1-p(0) = 1- e -m * im=1410Recombination Frequency -> Map units = crossover frequencies- Empirical - Derived- Not quite additive - Perfectly additiveGiven: Average number of c/o= m Question: What is the probability of i= 0, 1, 2, 3… c/o between A and B?Poisson distribution: p(i) = [ e -m m i ] / i! 1 23 4 5<= *A BabA Ba bHaldane’s Mapping FunctionMapunitsAverage Number ofcrossovers / meiosisp(0)p(1)p(2)1-p(0)[1-p(0)]/ 2Recombinationfrequency [cM]=m [%] 5 cM0.10.900.090.0050.100.05 5 10 cM0.20.820.160.020.180.09 9 50 cM1.00.370.370.180.630.32326Solution: Reverse the Problem. Trick: What is the probability of at least 1 crossover? p(>0) = 1-p(0) = 1- e -m RF [%] = 50 x [1 - e -0.02m.u. ]remember: i! = i x (i-1) x (i-2) x … x 1*RF = 0.5 x [1 - e -m ]<= *50 cM1.001.0001.000.5050 Ideal case im=1410Haplotype: A stretch of DNA with linked genes (alleles)A section of a chromosome cM RF 1-RF in [generations] in [years]50cM 50% 50% 1 (1-0.5) 1 = 0.5 2512cM 12% 88% 5 (1-0.12) 5 ~ 0.5 125 1cM 1% 99% 70 (1-0.01) 70 ~ 0.5 1750Map Distance Half life of the genetic interval (humans)Inheritance of ‘haplotypes’:Short segments of a chromosome are inherited as intactpieces for many generations and long periods of time.A. Review: Mapping with molecular markersB. The relationship between RF and map distance(Haldane's