power calculation for qtl association
DESCRIPTION
Power Calculation for QTL Association. Pak Sham, Shaun Purcell Twin Workshop 2001. Biometrical model. GenotypeAAAaaa Frequency(1-p) 2 2p(1-p)p 2 Trait mean-ada Trait variance 2 2 2 Overall meana(2p-1)+2dp(1-p). P ( X ) = G P ( X | G ) P ( G ). - PowerPoint PPT PresentationTRANSCRIPT
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Power Calculation for QTL Association
Pak Sham, Shaun Purcell
Twin Workshop 2001
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Biometrical model
Genotype AA Aa aa
Frequency (1-p) 2 2p(1-p) p2
Trait mean -a d a
Trait variance 2 2 2
Overall mean a(2p-1)+2dp(1-p)
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P(X) = GP(X|G)P(G)
P(X)
X
AA
Aa
aa
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Equal allele frequencies
A
0
0.2
0.4
0.6
0.8
1
-5 -3 -1 1 3 5
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Rare increaser allele
A
0
0.2
0.4
0.6
0.8
1
-5 -3 -1 1 3 5
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Linear regression analysis
-2
-1
0
1
2
3
4
aa Aa AA
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Power of QTL association - regression analysis
N = [z - z1-] 2 / A2
z : standard normal deviate for significance z1- : standard normal deviate for power 1-A
2 : proportion of variance due to additive QTL
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Required Sample Sizes
QTLvariance10%
0
50
100
150
200
250
300
0 0.05 0.1
Significance level
Sa
mp
le s
ize
80% power
95% power
50% power
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Power of likelihood ratio testsFor chi-squared tests on large samples, power is
determined by non-centrality parameter () and degrees of freedom (df)
= E(2lnL1 - 2lnL0)
= E(2lnL1 ) - E(2lnL0)
where expectations are taken at asymptotic values of maximum likelihood estimates (MLE) under an assumed true model
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Between and within sibships components of means
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Variance/Covariance explained
The better the fit of a means model:
- the greater the explained variances and covariances
- the smaller the residual variances and covariances
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Variance of b- component
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Variance of w- component
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Covariance between b- and w- components
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Null model
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Between model
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Within model
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Full model
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NCPs for component tests
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Determinant of a uniform covariance matrix
])1([)( 1 bsabaA sS
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Determinants of residual covariance matrices
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NCPs of b- and w- tests
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Definitions of LD parametersB1 B2
A1 pr + D ps - D p
A2 qr - D qs + D q
r s
pr + D < min(p, r)
D < min(p, r) - pr DMAX = min(ps, rq)
= min(p-pr, r-pr) D’ = D / DMAX
= min(ps, rq) R2 = D2 / pqrs
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Apparent variance components at marker locus
N/22 where
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Exercise: Genetic Power CalculatorUse Genetic Power Calculator, Association Analysis option
Investigate the sample size requirement for the between and within sibship tests under a range of assumptions
Vary
sibship size
additive QTL variance
sibling correlation
QTL allele frequencies
marker allele frequencies
D’
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N for 90% powerIndividuals
0 - 10% QTL variance
QTL, Marker allele freqs = 0.50
D-prime = 1
No dominance
Type I error rate = 0.05
Test for total association
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QTL variance
0
200
400
600
800
1000
1200
0 0.02 0.04 0.06 0.08 0.1
QTL variance
N
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QTL variance
0
20
40
60
80
100
120
0 0.02 0.04 0.06 0.08 0.1
QTL variance
NC
P p
er
ind
ivid
ua
l
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Effect of sibship size
Sibship size 1 - 5
Sib correlation = 0.25 , 0.75
5% QTL variance
QTL, Marker allele freqs = 0.50
D-prime = 1
No dominance
Type I error rate = 0.05
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Total
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
1 2 3 4 5
Sibship size
NC
P p
er
ind
ivid
ua
l
T r = 0.25
T r = 0.75
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Within
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
1 2 3 4 5
Sibship size
NC
P p
er
ind
ivid
ua
l
W r = 0.25
W r = 0.75
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Between
0
0.01
0.02
0.03
0.04
0.05
0.06
1 2 3 4 5
Sibship size
NC
P p
er
ind
ivid
ua
l
B r = 0.25
B r = 0.75
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Exercises1. What effect does the QTL allele frequency have
on power if the test is at the QTL ?
2. What effect does D’ have?
3. What is the effect of differences between QTL and marker allele frequency?
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Allele frequency & LDQTL allele freq = 0.05, no dominance
Sample sizes for 90% power :
Marker allele freq 0.1 0.25 0.5
D’ 1 1 1
N 205 625 1886
Marker allele freq 0.1 0.25 0.5
D’ 0.5 0.5 0.5
N 835 2517 7560