Columbia BIOL W3034 - Biotech Topic 5

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Slide 1Slide 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 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 53Slide 541Balanced Translocation detected by FISH2Red-Chrom. 5probeGreen-Chrom. 8probe32D Protein Gels45MS-peptide size signature: match to all predicted proteins61. Follow the mutation2. Follow which regions of DNA are co-inherited (linked)Positional Cloning by Recombination Mapping71. Follow the mutationTo determine disease genepresence or absence (genotype) from phenotype you must first establishDominant / recessiveAurosomal / sex-linkedPositional Cloning by Recombination Mapping8SINGLE GENE DEFECTSModes of InheritanceTo deduce who (likely) has one or two copies of mutant geneAffected FemaleUnaffected Male9AUTOSOMAL DOMINANT+/+D/+D/++/+10RECESSIVEAUTOSOMALX-LINKEDRECESSIVEa/+a/aa/+x/++/Yx/+x/Y+/Y11122. Follow which DNAs are co-inherited (linked)Use DNA sequences that differamong individuals within a family-Polymorphisms.Positional Cloning by Recombination MappingT GC A13VNTR / STRP DETECTION14A1A1A3A1A3A4A2A1A2A2 A4A4 A3A315XXX2 3ParentGameteChildA1A2B1B2C1C2B1B1C1C1A2A216Recombination MappingMeasures distance between 2 sites on a chromosome according to frequency of recombinationDistance between 2 DNA markersorDistance between a “disease gene”and a DNA marker17No fixed proportionalConversion betweenGenetic distance (cM)andPhysical distance (kb, Mb)18FAMILY AA1 DA2 +NR NR NR NR NR RD D D D+ +19FAMILY BA1 DA2 +NR NR NR NR NR RA1 +A2 DR R R R R NR20INFORMATIVE MEIOSISIdeally:- unambiguous inheritance of mutation and markers(requires heterozygosity for each in parent)knowledge of which alleles linked in parent (phase)21Assign numbers to results of linkage analysisto deal with non-ideal meiosesto sum data from many meioses in a familyto sum data from several families22Likelihood of RLikelihood of NRIf linked and RF = If unlinked:-1 -1/21/2Family A has 1 recombinant and 5 Non-RecombinantsLikelihood, given linkage ofL ( )= . (1- )5L (1/2) = (1/2)6Z = Lod = log { L ( ) / L (1/2)}Or given unlinked:-Z 0.58 0.62 0.51 0.3 0 0.1 0.2 0.3 0.4 0.523Z = 3Lod24FAMILY BA1 DA2 +NR NR NR NR NR RA1 +A2 DR R R R R NR25Family B:- Disease gene may be linked to A1 or A2Consider equally likely50% chance Family B has 1 R and 5 NR50% chance Family B has 5 R and 1 NRL ( )L (1/2) = (1/2)6Z = Lod = log { L ( ) / L (1/2)}Z 0.28 0.32 0.22 0.08 0 0.1 0.2 0.3 0.4 0.5= . (1- ) 51/2 { }+ . (1- ) 51/2 { }26Z 0.28 0.32 0.22 0.08 0 0.1 0.2 0.3 0.4 0.5Phase unknownZ 0.58 0.62 0.51 0.3 0 0.1 0.2 0.3 0.4 0.5Phase known27Z = Z1 + Z2 + Z3 + Z4 +…..Z = Z(A) + Z(B) + Z(C) + Z(D) + ….For family “A” with meioses 1, 2, 3, 4 …..For multiple families, “A”, “B”, “C”, “D”…..Assumption: same gene responsible for disease in all familiesProblem: locus heterogeneity28Z = 3Lod2930LINKAGE DISEQUILIBRIUMManygenerations31PCR test DNA segments3233Testing for specific mutations34ARMS 3’ mis-match of primer3536OLA37383940Aa BB CC DD Ee FF Gg HH II JJ AA BB CC Dd Ee FF GG HH II JjAA BB CC Dd Ee FF Gg HH II JJa B C D e/E F G H I JA B C d E/e F G H I JA B C D E/e F g H I JA B C D e/E F G H I jMother FatherSon/DaughterFamily Trio SNP genotypes reveal haplotypesDeduced haplotypes- ignoring recombinationCreation of variant sequencesRearrangement of sequence variants by recombinationFirst, consider just the creation of variant sequences within a short stretch of DNA where there is no significant rearrangementdue to recombination (an assumption that turns out to be valid)ABCDEFGHIJKLMNOPQRSTAbCDEFGHIJKLMNOPQRSTABCDEFgHIJKLMNOPQRSTAbCDEFGHIJKLMNOPqRSTaBCDEFgHIJKLMNOPQRSTAbCDEFGHIJkLMNOPqRSTAbCDEFGhIJkLMNOPqRSTABCDEfGHIJKLMNOPQRSTaBCDEFgHIJKLMNOPQrSTaBCDEFgHIJKLMnOPQrSTbbqbqkbqkhggagargarnfHistoryABCDEFGHIJKLMNOPQRSTAbCDEFGHIJKLMNOPQRSTABCDEFgHIJKLMNOPQRSTAbCDEFGHIJKLMNOPqRSTaBCDEFgHIJKLMNOPQRSTAbCDEFGHIJkLMNOPqRSTAbCDEFGhIJkLMNOPqRSTABCDEfGHIJKLMNOPQRSTaBCDEFgHIJKLMNOPQrSTaBCDEFgHIJKLMnOPQrSTbbqbqkbqkhggagargarnfRetention & amplification of only a few haplotypesFor any short region of DNA typically only 4-6 haplotypesare found in a sampling of present day humans (of the manymillions that must have existed in at least one copy en route).These local haplotypes provide some information about ancestry.Now consider how the major haplotypes of each short region ofDNA are associated with neighboring haplotypes to see whererecombination events took place.aBCDEFgHIJKLMnOPQrSTUVwXyZaBCDEFgHIJKLMnOPQrSTUVwXyZ aBCDEFgHIJKLMnOPQrSTUVwXyZaBCDEFgHIJKLMnOPQrSTUVwXyZ aBCDEFgHIJKLMnOPQrSTUVwXyZ aBCDEFgHIJKLMnOPQrSTUVwXyZaBCDEFgHIJKLMnOPQrSTUVwXyZ   High LD regions?aBCDEFgHIJKLMnOPQrSTUVwXyZaBCDEFgHIJKLMnOPQrSTUVwXyZ aBCDEFgHIJKLMnOPQrSTUVwXyZaBCDEFgHIJKLMnOPQrSTUVwXyZ  aBCDEFgHIJKLMnOPQrSTUVwXyZ aBCDEFgHIJKLMnOPQrSTUVwXyZaBCDEFgHIJKLMnOPQrSTUVwXyZHigh LD segment High LD segmentRecombination hot-spot85% of genome made up of 5-20kb high LD blocksOnly 4-5 different major haplotypes per block in the world!Haplotype blocks100 kb1 2 3 4 5 6 7 8Disease No disease/2,000 /3,000Minor allele frequencySNP-2a 93 130SNP-2b 21 27SNP-3a 140 62SNP-3b 24 35SNP-3c 140 260SNP-3d 87 120……. …. ….……. …. ….……. ….


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Columbia BIOL W3034 - Biotech Topic 5

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