Power calculation for QTL association (discrete and quantitative traits) - PowerPoint PPT Presentation

About This Presentation
Title:

Power calculation for QTL association (discrete and quantitative traits)

Description:

1. Calculate critical value (Inverse central 2) Alpha. 0 (under ... Family-based VC association: between and within tests. what is the impact of sibship size? ... – PowerPoint PPT presentation

Number of Views:408
Avg rating:3.0/5.0
Slides: 42
Provided by: shaunp2
Category:

less

Transcript and Presenter's Notes

Title: Power calculation for QTL association (discrete and quantitative traits)


1
Power calculation for QTL association(discrete
and quantitative traits)
  • Shaun Purcell Pak Sham
  • Advanced Workshop
  • Boulder, CO, 2003

2
  • Quantitative

Threshold
Discrete
Variance components
Case-control
Case-control
TDT
TDT
3
Discrete trait calculation
  • p Frequency of high-risk allele
  • K Prevalence of disease
  • RAA Genotypic relative risk for AA genotype
  • RAa Genotypic relative risk for Aa genotype
  • N, ?, ? Sample size, Type I II error rate

4
Risk is P(DG)
  • gAA RAA gaa gAa RAa gaa
  • K p2 gAA 2pq gAa q2 gaa
  • gaa K / ( p2 RAA 2pq RAa q2 )
  • Odds ratios (e.g. for AA genotype) gAA / (1-
    gAA )
  • gaa / (1- gaa )

5
Need to calculate P(GD)
  • Expected proportion d of genotypes in cases
  • dAA gAA p2 / (gAAp2 gAa2pq gaaq2 )
  • dAa gAa 2pq / (gAAp2 gAa2pq gaaq2 )
  • daa gaa q2 / (gAAp2 gAa2pq gaaq2 )
  • Expected number of A alleles for cases
  • 2NCase ( dAA dAa / 2 )
  • Expected proportion c of genotypes in controls
  • cAA (1-gAA) p2 / ( (1-gAA) p2 (1-gAa) 2pq
    (1-gaa) q2 )

6
Full contingency table
  • A allele a allele
  • Case 2NCase ( dAA dAa / 2 ) 2NCase ( daa
    dAa / 2 )
  • Control 2NControl ( cAA cAa / 2 ) 2NControl (
    caa cAa / 2 )

7
Threshold selection
  • Genotype AA Aa aa
  • Frequency q2 2pq p2
  • Trait mean -a d a
  • Trait variance ?2 ?2 ?2

8
P(X) ?GP(XG)P(G)
P(X)
Aa
AA
aa
X
9
P(GXltT) P(XltTG)P(G) / P(XltT)
P(X)
Nb. the cumulative standard normal distribution
gives the area under the curve, P(X lt T)
Aa
T
AA
X
10
Incomplete LD
  • Effect of incomplete LD between QTL and marker

Note that linkage disequilibrium will depend on
both D and QTL marker allele frequencies
11
Incomplete LD
  • Consider genotypic risks at marker
  • P(DMM) (pm1 d)2 P(DAA)
  • 2(pm1 d)(qm1- d) P(DAa)
  • (qm1- d)2 P(Daa)
  • / m12
  • Calculation proceeds as before, but at the marker

Haplo.
Geno.
AM/AM
AAMM
AM/aM or aM/AM
AaMM
aM/aM
aaMM
MM
12
Discrete TDT calculation
  1. Calculate probability of parental mating type
    given affected offspring
  2. Calculate probability of offspring genotype given
    parental mating type and affected
  3. Calculate overall probability of heterozygous
    parents transmitting allele A as opposed to a
  4. Calculate TDT test statistic, power

13
Fulker association model
The genotypic score (1,0,-1) for sibling i is
decomposed into between and within components
deviation from sibship genotypic mean
sibship genotypic mean
14
NCPs of B and W tests
Approximation for between test
Approximation for within test
Sham et al (2000) AJHG 66
15
Practical Exercise
  • Calculation of power for simple case-control
    study.
  • DATA frequency of risk factor in 30 cases and
    30 controls
  • TEST 2-by-2 contingency table chi-squared
  • (1 degree of freedom)

16
Step 1 determine expected chi-squared
  • Hypothetical risk factor frequencies
  • Case Control
  • A allele present 20 10
  • A allele absent 10 20

Chi-squared statistic 6.666
17
Step 2. Determine the critical value for a given
type I error rate, ?
- inverse central chi-squared distribution
P(T)
Critical value
T
18
Step 3. Determine the power for a given critical
value and non-centrality parameter
- non-central chi-squared distribution
P(T)
Critical value
T
19
Calculating Power
1. Calculate critical value (Inverse central ?2)
Alpha
0 (under the null)
20
  • http//workshop.colorado.edu/pshaun/gpc/pdf.html
  • df 1 , NCP 0
  • ? X
  • 0.05
  • 0.01
  • 0.001

21
Determining power
  • df 1 , NCP 6.666
  • ? X Power
  • 0.05 3.84146
  • 0.01 6.6349
  • 0.001 10.827

22
1. Planning a study
  • Candidate gene study
  • A disease occurs in 2 of the population
  • Assume multiplicative model
  • genotype risk ratio Aa 2
  • genotype risk ratio AA 4
  • 100 cases, 100 controls
  • What if the risk allele is rare vs common?

23
2. Interpreting a negative result
  • Negative candidate gene TDT study,
  • 82 affected offspring trios
  • affection scoring gt2 SD above mean
  • candidate gene SNP allele frequency 0.25
  • Desired 80 power, 5 type I error rate
  • What is the minimum detectable QTL variance
    (assume additivity)?

24
Planning a study
  • p N cases (N controls)
  • 0.01 1144
  • 0.05 247
  • 0.2 83
  • 0.5 66
  • 0.8 126
  • 0.95 465
  • 0.99 2286

25
Interpreting a negative result
  • QTL Power
  • 0.00 0.05
  • 0.01 0.34
  • 0.02 0.60
  • 0.03 0.78
  • 0.04 0.88
  • 0.05 0.94

26
Exploring power of association using GPC
  • Linkage versus association
  • difference in required sample sizes for specific
    QTL size
  • TDT versus case-control
  • difference in efficiency?
  • Quantitative versus binary traits
  • loss of power from artificial dichotomisation?

27
Linkage versus association
QTL linkage 500 sib pairs, r0.5 QTL
association 1000 individuals
28
Case-control versus TDT
p 0.1 RAA RAa 2
29
Quantitative versus discrete
K0.5
K0.2
K0.05
To investigate use threshold-based
association Fixed QTL effect (additive, 5,
p0.5) 500 individuals For prevalence K Group
1 has N and T Group 2 has N
and T
30
Quantitative versus discrete
  • K T (SD)
  • 0.01 2.326
  • 0.05 1.645
  • 0.10 1.282
  • 0.20 0.842
  • 0.25 0.674
  • 0.50 0.000

31
Quantitative versus discrete
32
  • Incomplete LD
  • what is the impact of D values less than 1?
  • does allele frequency affect the power of the
    test?
  • (using discrete case-control calculator)
  • Family-based VC association between and within
    tests
  • what is the impact of sibship size? sibling
    correlation?
  • (using QTL VC association calculator)

33
Incomplete LD
  • Case-control for discrete traits
  • Disease K 0.1
  • QTL RAA RAa 2 p 0.05
  • Marker1 m 0.05 D 1, 0.8, 0.6, 0.4, 0.2,
    0
  • Marker2 m 0.25 D 1, 0.8, 0.6, 0.4, 0.2,
    0
  • Sample 250 cases, 250 controls

34
Incomplete LD
  • Genotypic risk at marker1 (left) and marker2
    (right) as a function of D

35
Incomplete LD
  • Expected likelihood ratio test as a function of D

36
Family-based association
  • Sibship type
  • 1200 individuals, 600 pairs, 400 trios, 300
    quads
  • Sibling correlation
  • r 0.2, 0.5, 0.8
  • QTL (diallelic, equal allele frequency)
  • 2, 10 of trait variance

37
Family-based association
  • NCP proportional to variance explained
  • Between test
  • ? with ? sibship size and ? sibling correlation
  • Within test
  • 0 for s1, ? with ? sibship size and ? sibling
    correlation

38
Between-sibship association
39
Within-sibship association
40
Total association
41
GPC
  • Usual URL for GPC
  • http//statgen.iop.kcl.ac.uk/gpc/

Purcell S, Cherny SS, Sham PC. (2003) Genetic
Power Calculator design of linkage and
association genetic mapping studies of complex
traits. Bioinformatics, 19(1)149-50 
Write a Comment
User Comments (0)
About PowerShow.com