Selected Solutions to MAM1000W Homework Test 3 Samples – 2011
Linkage in Selected Samples
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Linkage in Linkage in Selected SamplesSelected Samples
Boulder Methodology WorkshopBoulder Methodology Workshop20052005
Michael C. NealeMichael C. NealeVirginia Institute for Virginia Institute for Psychiatric & Behavioral Psychiatric & Behavioral
GeneticsGenetics
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Basic Genetic ModelBasic Genetic Model
Q: QTL Additive Genetic F: Family Environment E: Random Environment3 estimated parameters: q, f and e Every sibship may have different model
Pihat = p(IBD=2) + .5 p(IBD=1)
F1
P1
F2
P2
11
1
1
Q1
1
Q2
1
E2
1
E1
Pihat
e f q q f e
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Mixture distribution modelMixture distribution modelEach sib pair i has different set of WEIGHTS
p(IBD=2) x P(LDL1 & LDL2 | rQ = 1 )
LDL1 LDL2
Q1 F1 E1 Q2F2E2
T1
q f e qfe
1.00
rQ
1.00
1.00
1.00 1.001.001.00
m mLDL1 LDL2
Q1 F1 E1 Q2F2E2
T1
q f e qfe
1.00
rQ
1.00
1.00
1.00 1.001.001.00
m mLDL1 LDL2
Q1 F1 E1 Q2F2E2
T1
q f e qfe
1.00
rQ
1.00
1.00
1.00 1.001.001.00
m m
p(IBD=1) x P(LDL1 & LDL2 | rQ = .5 )p(IBD=0) x P(LDL1 & LDL2 | rQ = 0 )
Total likelihood is sum of weighted likelihoods
rQ=1 rQ=.5 rQ=.0
weightj x Likelihood under model j
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QTL's are factorsQTL's are factors
• Multiple QTL models possible, at different Multiple QTL models possible, at different places on genomeplaces on genome
• A big QTL will introduce non-normalityA big QTL will introduce non-normality• Introduce mixture of means as well as Introduce mixture of means as well as
covariances (27ish component mixture)covariances (27ish component mixture)
• Mixture distribution gets nasty for large Mixture distribution gets nasty for large sibshipssibships
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Biometrical Genetic ModelBiometrical Genetic Model
0
d +a-a
Genotype means
AA
Aa
aa
m + a
m + d
m – a
Courtesy Pak Sham, Boulder 2003
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Mixture of Normal Distributions Mixture of Normal Distributions
Equal variances, Different means and different proportions according to allele frequencies
Aa AAaa
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Implementing the ModelImplementing the Model
• Estimate QTL allele frequency pEstimate QTL allele frequency p
• Estimate distance between homozygotes 2aEstimate distance between homozygotes 2a
• Compute QTL additive genetic variance asCompute QTL additive genetic variance as• 2pq[a+d(q-p)]2pq[a+d(q-p)]22
• Compute likelihood conditional on Compute likelihood conditional on • IBD statusIBD status• QTL allele configuration of sib pair (IBS)QTL allele configuration of sib pair (IBS)
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27 Component Mixture27 Component Mixture
AA
Aa
aa
AA Aa aa
Sib1
Sib 2
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19 Possible Component 19 Possible Component MixtureMixture
AA
Aa
aa
AA Aa aa
Sib1
Sib 2
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Results of QTL SimulationResults of QTL Simulation3 Component vs 19 Component
200 simulations of 600 sib pairs each GASP http://www.nhgri.nih.gov/DIR/IDRB/GASP/
Parameter True 3 Component
19 Component
Q 0.4 0.414 0.395
A 0.08 0.02 0.02
E 0.6 0.56 0.58
Test Q=0 (Chisq) --- 13.98 15.88
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Information in selected samplesInformation in selected samples
• Deviation of pihat from .5Deviation of pihat from .5• Concordant high pairs > .5Concordant high pairs > .5• Concordant low pairs > .5Concordant low pairs > .5• Discordant pairs < .5Discordant pairs < .5
• How come?How come?
Concordant or discordant sib pairs
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Pihat deviates > .5 in ASPPihat deviates > .5 in ASP
tIBD=2IBD=1IBD=0
t
Larger proportion of IBD=2 pairs
Sib 2
Sib 1
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Pihat deviates <.5 in DSP’sPihat deviates <.5 in DSP’s
tIBD=2IBD=1IBD=0
t
Larger proportion of IBD=0 pairs
Sib 1
Sib 2
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Sibship informativeness : sib pairsSibship informativeness : sib pairs
-4 -3 -2 -1 0 1 2 3 4Sib 1 trait -4-3
-2-10
12 3 4
Sib 2 trait
0
0.5
1
1.5
2
Sibship NCP
-4 -3 -2 -1 0 1 2 3 4Sib 1 trait -4
-3-2
-10
12
34
Sib 2 trait
0
0.5
1
1.5
2
Sibship NCP
-4-3
-2-1
01
23
4Sib 1 trait
-4-3
-2-1
01
23
4
Sib 2 trait
0
0.5
1
1.5
2
Sibship NCP
unequal allelefrequencies
dominance
rarerecessive
Courtesy Shaun Purcell, Boulder Workshop 03/03
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Two sources of informationTwo sources of information
• Phenotypic similarityPhenotypic similarity• IBD 2 > IBD 1 > IBD 0IBD 2 > IBD 1 > IBD 0• Even present in selected samplesEven present in selected samples
• Deviation of pihat from .5Deviation of pihat from .5• Concordant high pairs > .5Concordant high pairs > .5• Concordant low pairs > .5Concordant low pairs > .5• Discordant pairs < .5Discordant pairs < .5
• These sources are independentThese sources are independent
Forrest & Feingold 2000
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Implementing F&FImplementing F&F
• Simplest form test mean pihat = .5Simplest form test mean pihat = .5
• Predict amount of pihat deviationPredict amount of pihat deviation
• Expected pihat for region of sib pair Expected pihat for region of sib pair scoresscores
• Expected pihat for observed scoresExpected pihat for observed scores
• Use multiple groups in MxUse multiple groups in Mx
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Predicting Expected Pihat deviationPredicting Expected Pihat deviation
tIBD=2IBD=1IBD=0
t
+Sib 2
Sib 1
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Expected Pihats: TheoryExpected Pihats: Theory• IBD probability conditional on phenotypic scores IBD probability conditional on phenotypic scores
xx11,x,x22
• E(pihat) = p(IBD=2|(xE(pihat) = p(IBD=2|(x11,x,x22))+.5p(IBD=1|(x))+.5p(IBD=1|(x11,x,x22))))
• p(IBD=2|(xp(IBD=2|(x11,x,x22)) = )) = IBD=2IBD=2(x(x11,x,x22) /) /• [[IBD=2IBD=2(x(x11,x,x22) + 2) + 2IBD=1IBD=1(x(x11,x,x22) + ) + IBD=0IBD=0(x(x11,x,x22))]]
p(IBD=2 |(x|(x11,x,x22)) )+p(IBD=1 |(x|(x11,x,x22)) )+p(IBD=0 |(x|(x11,x,x22)) )
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Expected PihatsExpected Pihats• Compute Expected Pihats with pdfnorCompute Expected Pihats with pdfnor
• \pdfnor\pdfnor(X_M_C(X_M_C))• Observed scores X (row vector 1 x nvar)Observed scores X (row vector 1 x nvar)• Means M (row vector)Means M (row vector)• Covariance matrix C (nvar x nvar)Covariance matrix C (nvar x nvar)
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How to measure covariance?How to measure covariance?
tIBD=2IBD=1IBD=0
t
+
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AscertainmentAscertainment• Critical to many QTL analysesCritical to many QTL analyses
• DeliberateDeliberate• Study designStudy design
• AccidentalAccidental• Volunteer biasVolunteer bias• Subjects dyingSubjects dying
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Exploiting likelihoodExploiting likelihood• Correction not always necessaryCorrection not always necessary
• ML MCAR/MARML MCAR/MAR
Simulate bivariate normal data X,YSimulate bivariate normal data X,Y Sigma = 1 .5Sigma = 1 .5 .5 1.5 1 Mu = 0, 0Mu = 0, 0
Make some variables missingMake some variables missing Generate independent random normal variable, Z, if Z>0 then Y missingGenerate independent random normal variable, Z, if Z>0 then Y missing If X>0 then Y missing If X>0 then Y missing If Y>0 then Y missingIf Y>0 then Y missing
Estimate elements of Sigma & MuEstimate elements of Sigma & Mu Constrain elements to population values 1,.5, 0 etcConstrain elements to population values 1,.5, 0 etc Compare fitCompare fit Ideally, repeat multiple times and see if expected 'null' distribution emergesIdeally, repeat multiple times and see if expected 'null' distribution emerges
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ResultsResults of simulationof simulationMissingness mean x mean y var x cov xy var y LR
ChisqMCAR (rand)
MLE -0.0116 -0.1 1.0505 0.4998 0.8769 6.492
sample -0.0116 -0.0919 1.0505 0.8839
MAR (on x) MLE 0.0048 0.0998 1.0084 0.4481 1.1025 5.768
sample 0.0014 0.4437 1.0084 0.9762
NMAR (on y) MLE -0.0204 0.6805 0.9996 0.1356 0.2894 227.262
sample 0.0448 0.7373 0.9996 0.2851
Population covariance 1 .5 1 Means 0, 0Population covariance 1 .5 1 Means 0, 0
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Weighted likelihood Weighted likelihood approachapproach
• Usual nice properties of ML remainUsual nice properties of ML remain
• FlexibleFlexible
• Simple principle Simple principle • Consideration of possible outcomesConsideration of possible outcomes• Re-normalizationRe-normalization
• May be difficult to computeMay be difficult to compute
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Example: Two Coin TossExample: Two Coin Toss3 outcomes
HH HT/TH TTOutcome
0
0.5
1
1.5
2
2.5 Frequency
Probability i = freq i / sum (freqs)
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Example: Two Coin Example: Two Coin TossToss
3 outcomes
HH HT/TH TTOutcome
0
0.5
1
1.5
2
2.5 Frequency
Probability i = freq i / sum (freqs)
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Non-random ascertainmentNon-random ascertainment• Probability of observing TT globallyProbability of observing TT globally
• 1 outcome from 4 = 1/41 outcome from 4 = 1/4
• Probability of observing TT if HH is not Probability of observing TT if HH is not ascertainedascertained• 1 outcome from 3 = 1/31 outcome from 3 = 1/3
• or 1/4 divided by ‘or 1/4 divided by ‘Ascertainment Ascertainment CorrectionCorrection' of 3/4 = 1/3' of 3/4 = 1/3
Example
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Correcting for ascertainmentCorrecting for ascertainmentUnivariate case; only subjects > t ascertained
0 1 2 3 4-1-2-3-40
0.1
0.2
0.3
0.4
0.5
xi
likelihood
t
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AscertainmentAscertainment CorrectionCorrection Be / All you can beBe / All you can be
tx (x) dx
(x)
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Affected Sib PairsAffected Sib Pairs
tx
ty
1
0
0 1
tx ty(x,y) dy dx
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Ascertainment Corrections for Ascertainment Corrections for Sib PairsSib Pairs
tx ty(x,y) dy dx ASP ++
tx ty(x,y) dy dx
-DSP +-
tx ty
(x,y) dy dx-
CUSP +--
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Correcting for ascertainmentCorrecting for ascertainment
• Multivariate selection: multiple integrals Multivariate selection: multiple integrals • double integral for ASPdouble integral for ASP• four double integrals for EDACfour double integrals for EDAC
• Use (or extend) weight formulaUse (or extend) weight formula
• Precompute in a calculation groupPrecompute in a calculation group• unless they vary by subjectunless they vary by subject
Linkage studies
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Initial Results of SimulationsInitial Results of Simulations Null ModelNull Model
50% heritability50% heritability No QTLNo QTL Used to generate Used to generate
null distributionnull distribution .05 empirical .05 empirical
significance level at significance level at approximately 91 approximately 91 Chi-squareChi-square
QTL SimulationsQTL Simulations 37.5% heritability37.5% heritability 12.5% QTL12.5% QTL Mx: 879 significant Mx: 879 significant
at nominal .05 p-at nominal .05 p-value value
Merlin: 556 Merlin: 556 significant at significant at nominal .05 p-value nominal .05 p-value
Some apparent Some apparent increase in powerincrease in power
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1.00
F
S1 S2 S3 Sm
l1 l2 l3lm
e1 e2 e3 e4
Measurement is KEYMeasurement is KEYNeed continuous interval scales
Most complex traits not measured this way
Use latent trait instead
Factor model equivalent toItem response theory model
Can allow for non-normal Factors
Measurement of multiple Sx
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ConclusionConclusion Quantifying QTL variation in selected Quantifying QTL variation in selected
samples can be donesamples can be done
Can be computationally challengingCan be computationally challenging
May provide more powerMay provide more power
Permits multivariate analysis of correlated Permits multivariate analysis of correlated traitstraits