0920 Calculation of Load Score

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Lecture Outline 9/20 ã Mapping genes in human pedigrees ã Mapping the centromere using tetrads Mapping Human Genes ã Here is a pedigree that shows blood types and “nail patella syndome” – NPS is a rare genetic disease occurring in only about 2/100,000 births, characterized by underdeveloped nails and kneecaps. Announcements I have posted a new homework assignment on the web page First Exam is next Wednesday (9/28). Review session tomorrow at 4:30 NPS is a dominant trait Lod scores ã Test to c
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  1 Lecture Outline 9/20 ãMapping genes in human pedigreesãMapping the centromere using tetradsAnnouncements I have posted a new homework assignment on theweb pageFirst Exam is next Wednesday (9/28).Review session tomorrow at 4:30 Mapping Human Genes ãHere is a pedigree that shows blood types and “nailpatella syndome”  –NPS is a rare genetic disease occurring in only about 2/100,000births, characterized by underdeveloped nails and kneecaps. NPS is a dominant trait Lod scores ãTest to compare the likelihood that two lociare linked, vs the likelihood that the two lociare unlinked.LOD = log 10 likelihood if linkedlikelihood if the loci are unlinked  LOD = “logarithm of the odds” Testcross BN/bn x bn/bn GenotypeExpected Expectedif unlinkedif “r” cM apartBN/bnBn/bnbN/bnbn/bn0. Prob(no recombination) = 1-r Split that among two parental classesProb(recombination) = r Split that among two recombinant classes Lod score calculation ãAssume you have a testcross family with 8 non-recombinantoffspring and zero recombinants between genes B and NãThen, start with a hypothesis about linkage:  –maybe they are 10 cM apart ãWhat is the probability of no recombinantion for each gamete? 1-r = prob(no recombination) = 0.90 Remember there are two different parental genotypes.If you keep track of individual genotypes, then each has probability(1-r)/2 so, (1-r)/2 = 0.45 ãProbability of 8 non-recombinants if genes are linked ( at 10 cM )is: 0.45 8 ãProbability if not linked is 0.25 8 remember, r=0.5 when loci are not linked ãLod score = log 10 (0.45 8 /0.25 8 )= log 10 (110.19)= 2.04 The pattern of 8 non-recombinants is 110times more likelyif the genes are linked at 10cMthan if theywere unlinked (LOD must be >3.0 to be “significant”)   Human pedigree mapping 1.Determine genotypes in the pedigree2.Find informative families (e.g. test crosses)3.What fraction of offspring are recombinant?O/O n/n x B/O N/n  2 Calculate LOD score ãFam 1 : 2/13 recombinantãFam 2:0/2ãFam 31/1 OOBOBOBO Log Odds Ratio =Log (L(r=x) / L(r=0.5)) Both offspring are non-recombinant Prob (seeing this non-recombinant genotype)= (1-r)2If r=0.5, then Prob = 0.25If r=0.1, then Prob = 0.45 B/O N/nB/O N/n Calculate LOD score ãFam 1 : 2/13 recombinantãFam 2:0/2ãFam 31/1 Family 2Likelihood if independent? 0.25*0.25=0.0625Likelihood if linked at r=0.1?0.45*0.45   =0.2025Ratio = 3.24LOD = 0.51 OOBOBOBO Log Odds Ratio =Log (L(r=x) / L(r=0.5)) Prob(getting 2 non-recombinant) Calculate LOD score ãFam 1 : 2/13 recombinantãFam 2:0/2ãFam 31/1 Family 1Likelihood if independent? 0.25 11 * 0.25 2 =1.49 x10 -8 Likelihood if linked at r=0.1?0.45 11 * 0.05 2 =3.83 x10 -7 Ratio = 25.7LOD = 1.41 11 non-recombinants2 recombinants .45*.45*.45*.45* . . . *.45.05*.05 Calculate LOD score ãFam 1 : 2/13 recombinantãFam 2:0/2ãFam 31/1 Family 3Likelihood if independent? 0.25=0.25Likelihood if linked at r=0.1?0.05=0.05Ratio = 0.20LOD = -0.69 BOBOOO 1 recombinant Calculate LOD score ãFam 1: 2/13 LOD = 1.41ãFam 2:0/2LOD = 0.51ãFam 31/1LOD = -0.69Add the LOD scores from all 3 families: 1.41 + 0.51 + -0.69 =1.22Now, do that whole exercise for other values of r! Should be bigger than3.0 for significant linkage Calculate LOD score RLOD0.01- LOD is maximumaround r=0.2, so that isthe best estimate of linkage  3 Human pedigree mapping ãIt is hard to get big enough pedigrees,with enough informative families Mapping a hair loss gene ãAllopecia universalisãFound a large family in Pakistanwith many affected individualsãLooked for linkage betweenmolecular markers and the trait  –300 highly polymorphic markers –Searched for markers that werehomozygous in affected individualsand heterozygous otherwise –One such marker was found onchromosome 8p12 Source: Ahmed et al., 1998. Science 279:720 Finding the actual gene ãIn other work, they had also mapped the“hairless” gene ( hr  ) using somatic cell hybridmapping  –Used the mouse hr  gene to amplify the humangene sequence –They fused bits of human chromosomes tohamster cells, to find the fragments thatcorresponded to the mouse hairless gene –It also mapped to8p12 Finding the amino acid ãThe hr  gene was a goodcandidate, so they sequencedthe gene from individuals in thispedigree.ãAll affected individuals had anallele of  hr  that differed in asingle base, A to G  –Changes threonine to alanine But, it all started with this pedigree . . . Another example ãRenal Fanconi syndrome ãLichter-Konecki et al, 2001 Am J Hum Gen ãRFS is a dominant disorder ãUsed a big pedigree from Wisconsin and alot of polymorphic markersãFound a set of markers on Chromosome 15that were associated with the disorder   4 Note the recombinationin II-1 and III-11Either green or red must cause disease Must be somewherein this region Tetrad analysis ãIn some haploid fungi and algae, the meioticproducts stay together in ‘tetrads’ ã Why are there four?  ãBecause you see all products of meiosis, you can domore detailed analyses ãlike mapping the centromereãdistinguishing various kinds of crossovers ãIn many ways, everything is the same exceptMeiotic tetrads are counted, not individuals Tetrads can be either ordered or unordered ãThe bread mold Neurospora crassa produces orderedtetrads. Mapping the centromere ãCentromeresalwayssegregate in Meiosis I  –AAaa or aaAA Draw that out to convince yourself . . . ãIf alleles segregate in meiosis II, then theremust have been a crossover between thatgene and the centromere.  – AaAa or AaaA Again, draw this out to convince yourself . . . http://www.stanford.edu/group/neurospora/RajuNcrassaWeb/Fig49.WTxcys3.Raju781308.gif  First division patternSecond division pattern Tetrad example ãTetrad types from a heterozygous Aa individual.ãWhat is the distance to the centromere? PatternNumber of asciiAAaa 36aaAA44AaAa4aAaA6AaaA3aAAa7Total: 100  – In Neurospora, each of these gametes would be doubled, producing ascii with 8 spores. Here I’m just showing the four meiotic products. How many show second-division segregation?What proportion of  gametes will be recombinant?
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