Upload
mildred-patrick
View
214
Download
1
Tags:
Embed Size (px)
Citation preview
Continuously moderated effects of A,C, and E in the twin design
Conor V Dolan & Sanja Franić (BioPsy VU)
Boulder Twin WorkshopMarch 4, 2014
Based on PPTs by Sophie van der Sluis & Marleen de Moor
Thanks to Neale family, Sarah Medland, & Brad Verhulst
-4 -3 -2 -1 0 1 2 3
-4-2
02
46
x
yPhenotypicscores
Genetic level (score on A)
EnvironmentalDispersion
Variance ofE given G score
G x E as “genetic control” of sensitivity to different environments: heteroskedasticity
Not all heteroskedasticity is GxE!
Not all heteroskedasticity is GxE!
-4 -3 -2 -1 0 1 2 3
-4-2
02
46
x
yPhenotypicscores
Environmental level (score on E)
GeneticDispersion
ConditionalA variance
G x E as “environmental control” of genetic effects: heteroskedasticity.
Not all heteroskedasticity is moderation!
-4 -3 -2 -1 0 1 2 3
-4-2
02
46
x
yPhenotypicscores
Moderator level (score on moderator)
GeneticDispersion
and / or
Environmental dispersion
Moderation of effects (A,C,E) by measured moderator M: heteroskedasticity.
• Is the magnitude of genetic influences on ADHD the same in boys and girls?
• Do different genetic factors influence ADHD in boys and girls? (DZOS)
Moderation of A effects by binary variable with the bonus of the information provided by DZ opposite sex twins
Sex X A interaction:Moderation of A by sex
Other examples binary moderators
“A” effects moderated by marital status:Unmarried women show greater levels of genetic influence on depression (Heath et al., 1998).
“A” effects moderated by religious upbringing:A religious upbringing diminishes A effects on the personality trait of disinhibition (Boomsma et al., 1999).
Binary moderator: multigroup approach
Continuous Moderators not amenable to multigroup approach
…
treat the moderator as continuous(OpenMx)
Standard ACE model
phenoTwin 1
A C E
phenoTwin 2
A C E
a ce
cea
1
b0
1
b0
11 or .5
Regression on unit: just a way to estimate the mean.
Standard ACE model + Main effect on Means
PhenoTwin 1
A C E
PhenoTwin 2
A C E
a ce
cea
1
b0+ βM M1
1
b0+βMM2
phenoTwin 11
b0+ βM M1
phenoTwin 11
βM
b0
M1
Actually: regression of Phenotype on M ...What is left is the residual, subject to ACE modeling.
Why a triangle? Fixed regressors
res1 res1
equivalent
Summary stats
• Means vector
• Covariance matrix (r = 1 or r=1/2)
22222
222
* ecacar
eca
mean1 mean2
Allowing for a main effect of X
• Means vector
• Covariance matrix (r = 1 or r=1/2)
iMi XmXm 21M
22222
222
* ecacar
eca
Standard ACE model + Effect on Means and a path
Twin 1
A C E
Twin 2
A C E
a+XM1c e c ea+XM2
Twin 1 Twin 21
m+ βM M1
1
m+ βM M2
M has main effect on mean + moderation of A effect
Twin 1
A C E
a+XM1c e
Twin 11
m+ βM M1
M1=0 -> mean= m & A effect = aM1=1 -> mean= m+ βM M1 & A effect = a+XM1
Standard ACE model + Effect on Means and a,c, & e paths
Twin 1
A C E
Twin 2
A C E
a+XM1
c+YM1
e+ZM1a+XM2
c+YM2
e+ZM2
Twin 1 Twin 21
m+MM1
1
m+MM2
• Effect on means: Main effects (regression of phenol on M)
• Effect on a/c/e path loadings: Moderation effects (A x M, C x M, E x M interaction)
Twin 1
A C E
Twin 2
A C E
a+XM1
c+YM1
e+ZM1a+XM2
c+YM2
e+ZM2
Twin 1 Twin 21
m+MM1
1
m+MM2
Expected variances
Standard Twin Model: Var (P) = a2 + c2 + e2
Moderation Model: Var (P|M) = (a + βXM)2 + (c + βYM)2 + (e + βZM)2
Expected MZ / DZ covariances
Cov(P1,P2|M)MZ =
(a + βXM)2 + (c + βYM)2
Cov(P1,P2|M)DZ =
0.5*(a + βXM)2 + (c + βYM)2
Var (P|M) = (a + βXM)2 + (c + βYM)2 + (e + βZM)2
h2 |M = (a + βXM)2 / Var (P|M)
c2 |M = (c + βyM)2 / Var (P|M)
e2 |M = (e + βyM)2 / Var (P|M)
(h2|M + c2|M + e2|M ) = 1 Standardized conditional on value of M
Turkheimer study SESModeration of unstandardized variance components (top)
Moderation of standardized variance components (bottom)
Twin 1
A C E
Twin 2
A C E
a+XM1
c+YM1
e+ZM1a+XM2
c+YM2
e+ZM2
Twin 1 Twin 21
m+MM1
1
m+MM2
1/.5 1
But what have assumed concerning M?
Bivariate distribution of X and Y
Assumption: (y,x)~N(m, S)
x
A C E
y
A C E
Regression of Y on X in terms of latent variables
M1,M2 (co)variance is not included in the model. M is a “fixed regressor”.
Phenotype = BMIModerator = Age .... A fixed regressor
Phenotype = BMIModerator = Intelligence ....Can we treat Intelligence as a fixed fixedIn this context? Depends... on ...
Ec
Eu
T1
au + bau*M1
Cc
M1
am
ac
AcCu
Au
cc
ec
cmem
cu + bcu*M1
eu + beu*M1
T2
Au
Cu
Eu
eu + beu*M2
cu + bcu*M2
au + bau*M2
M2
Cc
Ec
Ac
am em
cm
e c
c c
a c
MZ=1 / DZ=.5
MZ=1 DZ=.5
MZ = DZ = 1
MZ = DZ = 1
M as a random regressor, with its own ACE + ACE cross loadings.
Can we treat M as a fixed effect in this manner?
Twin 1
A C E
Twin 2
A C E
a+XM1
c+YM1
e+ZM1a+XM2
c+YM2
e+ZM2
Twin 1 Twin 21
m+MM1
1
m+MM2
1/.5 1
Not generally
Eu
T1
au + bau*M1
Cc
M1
am
ac
Ec
AcCu
Au
cc
ec
cm
em
cu + bcu*M1
eu + beu*M1
T2
Au
Cu
Eu
eu + beu*M2
cu + bcu*M2
au + bau*M2
M2
Cc
Ec
Ac
am em
cm
e c
c c
a c
MZ=1 / DZ=.5
MZ=1 DZ=.5
MZ = DZ = 1
MZ = DZ = 1
OK if #1: ac/am = cc /cm = ec / em
Eu
T1
au + bau*M1
M1
Ec
Cu
Au
ec em
cu + bcu*M1
eu + beu*M1
T2
Au
Cu
Eu
eu + beu*M2
cu + bcu*M2
au + bau*M2
M2
Ec
eme c
MZ=1 DZ=.5
MZ = DZ = 1
OK if #2: r(M1,M2) = 0
Eu
T1
au + bau*M1
Cc
M1
Cu
Au
cccm
cu + bcu*M1
eu + beu*M1
T2
Au
Cu
Eu
eu + beu*M2
cu + bcu*M2
au + bau*M2
M2
Cc
cm
c c
MZ=1 DZ=.5
MZ = DZ = 1
MZ = DZ = 1
Ok if #3: r(M1,M2) = 1
Eu
T1
au + bau*M1
Cc
M1
am
ac
Ec
AcCu
Au
cc
ec
cm
em
cu + bcu*M1
eu + beu*M1
T2
Au
Cu
Eu
eu + beu*M2
cu + bcu*M2
au + bau*M2
M2
Cc
Ec
Ac
am em
cme c
c c
a c
MZ=1 / DZ=.5
MZ=1 DZ=.5
MZ = DZ = 1
MZ = DZ = 1
What to do otherwise?
MZ: T1=β0,MZ + β1,MZ*M1 + β2,MZ*M2
T2=β0,MZ + β1,MZ*M2 + β2,MZ*M1
DZ: T1=β0,DZ + β1,DZ*M1 + β2,DZ*M2
T2=β0,DZ + β1,DZ*M2 + β2,DZ*M1
E C A
T1
e + βe*M1a + βa*M1
c + βc*M1
A C E
T2
a + βa*M2 e + βe*M2
c + βc*M2
1 / .5
1
1
β0k + β1k*M1 + β2k*M2
1
β0k + β1k*M2 + β2k*M1
But what have assumed concerning the covariance between M and T?
E C A
T1
e + βe*M1a + βa*M1
c + βc*M1
A C E
T2
a + βa*M2 e + βe*M2
c + βc*M2
1 / .5
1
1
β0k + β1k*M1 + β2k*M2
1
β0k + β1k*M2 + β2k*M1
Eu
T1
au + bau*M1
Cc
M1
am
ac
Ec
AcCu
Au
cc
ec
cm
em
cu + bcu*M1
eu + beu*M1
T2
Au
Cu
Eu
eu + beu*M2
cu + bcu*M2
au + bau*M2
M2
Cc
Ec
Ac
am em
cme c
c c
a c
MZ=1 / DZ=.5
MZ=1 DZ=.5
MZ = DZ = 1
MZ = DZ = 1
Eu
T1
au + bau*M1
Cc
M1
am
ac + b
ac *M1
Ec
AcCu
Au
cc + b
cc *M1ec + b
ec*M1cm
em
cu + bcu*M1
eu + beu*M1
T2
Au
Cu
Eu
eu + beu*M2
cu + bcu*M2
au + bau*M2
M2
Cc
Ec
Ac
am em
cm
e c + b ec
*M2c c + b cc
*M2
a c +
b ac*M
2
MZ=1 / DZ=.5
MZ=1 DZ=.5
MZ = DZ = 1
MZ = DZ = 1
Eu
T1
au + bau*M1
Cc
M1
am
ac + b
ac *M1
Ec
AcCu
Au
cc + b
cc *M1ec + b
ec*M1cm
em
cu + bcu*M1
eu + beu*M1
T2
Au
Cu
Eu
eu + beu*M2
cu + bcu*M2
au + bau*M2
M2
Cc
Ec
Ac
am em
cm
e c + b ec
*M2c c + b cc
*M2
a c +
b ac*M
2
MZ=1 / DZ=.5
MZ=1 DZ=.5
MZ = DZ = 1
MZ = DZ = 1
M has to be treated as random and modeled appropriately (as shown)!
Conclusion: Use standard fixed regression approach ifCov(M,T) equally due to A,C,ECov(M1,M2) = zeroCov(M1,M2) = one Use extended fixed regression approach ifCross paths are not moderated Otherwise use full model.
categorical data• Continuous data
– Moderation of means and variances• Ordinal data
– Moderation of thresholds and variances
– See Medland et al. 2009
What about >1 number of moderators and interaction Effects (sex X age)Extend the model accordingly...
ec + bec1*SES + bec2*AGE +
bec2*(AGE *SES)
What about power given such extensions?
Do power calculation.... Exactly, by simulation, by exact simulation ?
Practical
• Replicate findings from Turkheimer et al. with twin data from NTR
• Phenotype: FSIQ• Moderator: SES in children • (cor(M1,M2) = 1
• Data: 205 MZ and 225 DZ twin pairs
• 5 years old
Gene-Environment Correlation
rGE:• Genetic control of exposure to the
environment
Examples: Active rGE: Children with high IQ read more books Passive rGE: Parents of children with high IQ take their
children more often to the library