12050 VLSBDad phase unknownA1A20.5(total # meioses)Odds = 1/2[(1-r)n • rk] + 1/2[(1-r)n • rk]odds ratioWhat single r value best explains the data?A1 DA2 dorA1 dA2 DFor this, you need to search r’s.maximum likelihoodr = 0.13Oops: a numerical mistake (thanksto Jonathan for detective work)In real life this correction does matter…best r = 0.2873best r = 0.2771Accounting for both phasesUsing only one phasefamily 1: 10 meioses, 1 (or 9) apparent recombinantsfamily 2: 10 meioses, 4 (or 6) apparent recombinantsfamily 3: 10 meioses, 3 (or 7) apparent recombinantsfamily 4: 10 meioses, 3 (or 7) apparent recombinantstotal LOD = LOD(family 1) + LOD(family 2) + LOD(family 3) + LOD(family 4)2Locus heterogeneityage of onsetCoinsOdds = P(your flips | r)P(your flips | r = 0.5)r = intrinsic probability of coming up heads (bias)Odds = (1-r)n • rk 0.5(total # flips)Odds ratio ofmodel that coinis biased,relative to nullCoins3 heads2 heads1 heads010.01440.90.10240.80.30240.70.61440.610.51.38240.41.64640.31.63840.21.16640.100oddsr0 heads010.00160.90.02560.80.12960.70.40960.610.52.07360.43.84160.36.55360.210.4980.1160oddsr4 heads16110.4980.96.55360.83.84160.72.07360.610.50.40960.40.12960.30.02560.20.00160.100oddsr010.12960.90.40960.80.70560.70.92160.610.50.921 60.40.70560.30.40960.20.12960.100oddsr011.16640.91.63840.81.64640.71.38240.610.50.61440.40.30240.30.10240.20.01440.100oddsrr = intrinsic probability of coming up heads (bias)Significance cutoff(single family)3The analogy againTesting lots of markers for linkage to a trait isanalogous to having lots of students, eachflipping a coin.The search for the coin’s bias parameter isanalogous to the search for recombinationdistance between markers and disease locus.Testing lots of markers for linkage to a trait isanalogous to having lots of students, eachflipping a coin.The search for the coin’s bias parameter isanalogous to the search for recombinationdistance between markers and disease locus.Each student is analogous to a marker.The analogy againTesting lots of markers for linkage to a trait isanalogous to having lots of students, eachflipping a coin.The search for the coin’s bias parameter isanalogous to the search for recombinationdistance between markers and disease locus.Each student is analogous to a marker.Each coin flip is analogous to a family memberin pedigree.The analogy againMultiple testing, shownanother way1. Simulate thousands of markers, inheritedfrom parents to progeny.2. Assign some family members to have adisease, others not.3. Test for linkage between disease andmarkers, knowing there is none.E. Lander and L. Kruglyak, Nature Genetics 11:241, 19954SimulationSimulationEverymarker isanalogousto a studentflippingA real world scenarioYou have invested a bolus of research money in a linkage mappingstudy of a genetic disease segregating in families. For each familymember, you do genotyping at a bunch of markers.When you finally run the linkage calculation, the strongest marker givesa LOD of 2. You desperately want to believe this is significant.You simulate a fake trait with no genetic control 1000 times.You find that in 433 of these simulations, the fake trait had a LOD > 2.This means that in your real data, the probability of your precious linkagepeak being a false positive is 433/1000 = 0.433.If you spent more money and time to follow this up, it could be acomplete waste. Essential to know.Simulation/theory5Simulation/theorySimulate 1000times, ask howfrequently youget a peak overa certainthreshold.Simulation/theoryWith modestmarker spacingin a humanstudy, LOD of3 is 9% likely tobe a falsepositive.Simulation/theoryBut this wouldchange in adifferentorganism, withdifferentnumber ofmarkers, etc.Simulation/theoryBut this wouldchange in adifferentorganism, withdifferentnumber ofmarkers, etc.So in practice,everyone doestheir ownsimulationspecific to theirown study.6More markers = more tests =more chance for spurious highlinkage score.More markers = more tests =more chance for spurious highlinkage score.Not true when you add individuals(patients)! Always improves results.Multiple testing in geneticsMultiplemarkersnotnecessaryMarker density matters?But if the only marker you testis >50 cM away, will get nolinkage.7Marker density matters?But if the only marker you testis >50 cM away, will get nolinkage.So a mapping experiment is adelicate balance between toomuch testing and notenough…Candidate gene approach:apple pigmenthttp://waynesword.palomar.edu/images/apple3b.jpgCandidate gene approach:apple pigmentCandidate gene approachHypothesize that causal variant will be inknown pigment gene or regulator. NOTrandomly chosen markers genome-wide.8Candidate gene approachCandidate gene approachRed progeny haveRFLP pattern likered parentCandidate gene approachUnpigmented progenyhave RFLP pattern likeunpigmented parentBut if you can beatmultiple testing, why notdo the whole genome…9Testing for linkage doesn’talways mean countingrecombinants.Back to week 4Fig. 3.12Qualitative but polygenicFig. 3.12Two loci.Need one dominant allele at eachlocus to get phenotype.A simulated cross: test one locusAAbbaaBBAaBbGenotype atmarker close to AlocusFlower colorinter-mateTwo loci.Need onedominant alleleat each locus toget phenotype.10A simulated cross: test one locusAAbbaaBBAaBbFlower colorinter-mateTwo loci.Need onedominant alleleat each locus toget phenotype.AABb AaBb aaBb AaBB aaBB AabbGenotype atmarker close to AlocusA simulated cross: test one locusAAbbaaBBAaBbFlower colorinter-mateTwo loci.Need onedominant alleleat each locus toget phenotype.AAB b AaBb aaBb AaBB aaBB AabbGenotype atmarker close to AlocusA simulated cross: test one locusAAbbaaBBAaBbFlower colorinter-mateTwo loci.Need onedominant alleleat each locus toget phenotype.AAB b AaBb aaBb AaBB aaBB AabbGenotype atmarker close to AlocusPurpleflowersresult fromAA or Aa.No need to count recombinantsAAbbaaBBAaBbFlower colorinter-mateTwo loci.Need onedominant alleleat each locus toget phenotype.AAB b AaBb aaBb AaBB aaBB AabbGenotype atmarker close to Alocus32Bottomallele13Topallelewhitepurple11No need to count recombinantsAAbbaaBBAaBbFlower colorinter-mateTwo loci.Need onedominant alleleat each locus toget phenotype.AAB b AaBb aaBb AaBB aaBB AabbGenotype atmarker close to Alocus32Bottomallele13Topallelewhitepurpleχ2 = Σ(O - E)2ENo need to count recombinantsAAbbaaBBAaBbFlower colorinter-mateTwo loci.Need onedominant alleleat each locus toget
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