Sebastian Zöllner
University of Michigan
Keng-Han Lin
Matthew Zawistowski
Mark Reppell
GWAS have been successful.
Only some heritability is explained by common variants.
Uncommon coding variants (maf 5%-0.5%) explain less.
Rare variants could explain some ‘missing’ heritability.
◦ Better Risk prediction.
◦ Rare variants may identify new genes.
◦ Rare exonic variants may be easier to annotate functionally and
interpret.
Testing individual variants is
unfeasible.
◦ Limited power due to small number of
observations.
◦ Multiple testing correction.
Alternative: Joint test.
◦ Burden test (CMAT, Collapsing, WSS)
◦ Dispersion test (SKAT, C-alpha)
Gene-based tests have low power.
◦ Nelson at al (2010) estimated that 10,000 cases &
10,000 controls are required for 80% power in
half of the genes.
Large sample size required
More heterogeneous sample =>Danger of
stratification
Stratification may differ from common
variants in magnitude and pattern.
(202 genes, n=900/900,
MAF < 1%, Nonsense/nonsynonymous
variants)
Expected Number of variants
per kb
African-American
Southern Asia
South-Eastern Europe
South-Western Europe
Western Europe
Central Europe
North-Western Europe
Eastern Europe
Northern Europe
Finland
A gradient in diversity
from Southern to Northern
Europe
Sample Size
• Measure of rare variant diversity.
• Probability of two carriers of the minor alleles being
from different populations (normalized).
Median EU-EU: 0.71
Median EU-EU: 0.86
Median EU-EU: 0.98
1.
Select 2 populations.
2.
Select mixing parameter r.
3.
4.
Sample 30 variants from
the 202 genes.
Calculate inflation based
on observed frequency
differences.
Zawistowski et al.
2014
If multiple affected family members are collected, it may
be more powerful to sequence all family members.
Family-based tests can be robust against stratification.
TDT-Type tests are potentially inefficient.
How to leverage low frequency?
◦ Low frequency risk variants should me more common in cases.
◦ And even more common on chromosomes shared among
many cases.
S=0
•
Consider affected sibpairs.
•
Estimate IBD sharing.
Compare the number of
rare variants on shared
(solid) and non-shared
chromosomes (blank).
•
Any aggregate test can be
applied.
S=1
S=2
Twice as many non-shared as shared
chromosomes.
Null hypothesis determines test:
Shared alleles : Non-shared alleles=1:2
Test for linkage or association
Shared alleles : Non-shared alleles=
Shared chromosomes : Non-shared chromosomes
Test for association only
IBD sharing is known.
Individuals don’t need phase to identify shared variants.
Except one configuration: IBD 1 and both sibs are heterozygous
Configuration 1
Configuration 1
+1 shared
+2 non-shared
Under null, probability of configuration 2 is allele frequency.
Under the alternative, we need to use multiple imputation.
Assume chromosome sharing status
S=0
is known for each sibpair.
Count rare variants; impute sharing
status for double-heterozygotes.
S=1
Compare number of rare variants
between shared and non-shared
chromosomes with chi-squared test
(Burden Style).
S=2
Classic CaseControl
Internal
Control
S=0
S=1
S=2
Selected
Cases
Consider 2 populations.
p=0.01 in pop1, p=0.05 in pop2.
1000 sibpairs for internal control design.
1000 cases, 1000 controls for selected cases.
1000 cases and 1000 controls for case-control.
Sample cases from pop1 with proportion .
Test for association with α=0.05.
0.8
0.4
0.0
Type I Error Rate
Internal Control
Selected Cases
Conventional
0.0
0.2
0.4
0.6
Proportion
0.8
1.0
Realistic rare variant models are unknown
◦
◦
◦
◦
◦
Typical allele frequency
Number of risk variants/gene
Typical effect size
Distribution of effect sizes
Identifiabillity of risk variants
Goal: Create a model that summarizes these
unknowns into
◦ Summed allele frequency
◦ Mean effect size
◦ Variance of effect size
Assume many loci carrying risk variants.
Risk alleles at multiple loci each increase the
risk by a factor independently.
Frequency of risk variant: A
◦ Independent cases
P( R | A) P(A|R)P(R)
◦ On shared chromosome
P( R | AA) P( AA | R) P( R)
Affected
AA Affected relative
pair
R Risk locus
genotype
Relative risk is sampled from
distribution f with mean ,
variance σ2.
Simplifications:
◦ Each risk variant occurs only
once in the population.
◦ Each risk variant on its own
haplotype.
Then the risk in a random
case is
A
Affected
r1,r Carrier status of
2
chromosome 1,2
m1, Relative risk of
m2 risk variants on
1,2
Mean effect size
σ2 Variance of effect
size
P( A | r1 , r2 ) m1r1 m2r2 f (m1 ) f (m2 ) r1 r2
To calculate the probability of
having an affected sib-pair we
condition on sharing S.
For S>0, the probability
depends on σ2. E.g. (S=2):
P( AA | r1 , r2 , S 2)
m12 r1 m22 r2 f (m1 ) f (m2 )
AA Affected rel pair
ri
Carrier stat
chrom i
mi Relative risk of
variant on i
f
Distribution of
RR
Mean RR
σ2 Variance of RR
S
Sharing status
E( f 2 )r1 E( f 2 )r2 ( 2 2 )r1 r2
Select μ, σ2 and cumulative frequency f
Calculate allele frequency in cases/controls
P(R|A).
Calculate allele frequency in shared/nonshared chromosomes.
=> Non-centrality parameter of χ2
distribution.
0.6
Conventional
Case-Control
Internal Control
Selected Cases
0.0
sMAF
0.2 0.4
f=0.2
f=0.01
1
2
3
4
5
1 2 3 4 5 1
Mean Relative Risk
2
3
4
5
1.0
2.5
function(x) power.sas(mu = x, sigma2 = sigma2, f = 0
n_sb = n1))
0.0
Power
0.4
0.8
f=0.01
4.0 1.0
2.5
4.0 1.0
Mean Relative Risk
, function(x) power.sas(mu = x, sigma2 = sigma2, f =
n_sb = n1))
f=0.05
f=0.2
Internal Control
Selected Cases
Conventional
2.5
4.0
0
1
2
y(x, function(x) power.sas(mu = mu, sigma2 = x, f = 0.05,
n_sb = n1))
0.0
Power
0.4
0.8
f=0.01
3
40
1
2
3
40
Variance of Relative Risk
ly(x, function(x) power.sas(mu = mu, sigma2 = x, f = 0.2,
n_sb = n1))
f=0.05
f=0.2
Internal Control
Selected Cases
Conventional
1
2
3
4
Gene-gene interaction affects power in families.
For broad range of interaction models, consider
two-locus model.
G now has alleles g1,g2. The joint effect is
P( A | r1, r2 , g1, g2 ) Lr1 r2 Gg1 g2 (r1 r2 )( g1 g2 )
We compare the effect of while adjusting L and
G to maintain marginal risk.
0.8
Power
0.4
0.0
IC SRR=2
IC SRR=8
Conventional
0.2 0.4 0.6 0.8
Interaction Coefficient
1.0
0.8
Power
0.4
0.0
IC SRR=2
IC SRR=8
Conventional
1.0
1.2 1.4 1.6 1.8
Interaction Coefficient
2.0
Stratification is a strong confounder for rare variant
tests.
Family-based association methods are robust to
stratification.
Comparing rare variants between shared and nonshared chromosomes is substantially more powerful
than case-control designs.
All family based methods/samples depend on the
model of gene-gene interaction. Under antagonistic
interaction power can be lower than a population
sample.
Thank you for your attention
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