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Saturated vs ACE model- different correlations

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No user picture. JuanJMV Joined: 07/20/2016
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Behavioral Genetics Models
Hi all,

I am trying to fit a univariate model.

I fitted the saturated model first and I got correlations of 0.64 and 0.34 for MZ and DZ respectively.

However, when I fitted the ACE model, I got these results:
A=0.04
C=0.52
E=0.44

So, I decided to check the correlations from the ACE model and I got 0.56 and 0.54 for MZ and DZ respectively.
I know that correlations may change from the ACE to the saturated model. However, these differences are huge.

I have tried with different scripts, umx and with/without covariates and I get the same results.

My sample comprise 100 pairs and there are no missing data.

Do you know why there are so big differences between the saturated and the ACE model?

Thank you so much in advance.

Replied on Tue, 08/27/2019 - 13:45
Picture of user. AdminRobK Joined: 01/24/2014

I agree that the discrepancy between the saturated-model and ACE-model results seems too big to be real. I notice that the phenotypic correlations suggest a substantial a² and a near-zero c², but the ACE results are the opposite of that. Have you perhaps made an error such that some of your MZ twins are being treated as DZ twins, and vice versa?

You might as well post your full script, preferably as an attachment. I would also like to see the model-expected MZ and DZ covariance matrices from the saturated model.

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Replied on Tue, 08/27/2019 - 15:19
No user picture. JuanJMV Joined: 07/20/2016

Hi Rob,

Thank you so much for your prompt response.

Please find attached the saturated and ACE scripts.

Here the model-expected covariance matrices from the saturated model.
> fit$MZ.covMZ
SymmMatrix 'covMZ'

$labels
[,1] [,2]
[1,] "vMZ1" "cMZ21"
[2,] "cMZ21" "vMZ2"

$values
[,1] [,2]
[1,] 0.057890691 0.043141926
[2,] 0.043141926 0.078377175

$free
[,1] [,2]
[1,] TRUE TRUE
[2,] TRUE TRUE

$lbound
[,1] [,2]
[1,] 1e-24 0e+00
[2,] 0e+00 1e-24

> fit$DZ.covDZ
SymmMatrix 'covDZ'

$labels
[,1] [,2]
[1,] "vDZ1" "cDZ21"
[2,] "cDZ21" "vDZ2"

$values
[,1] [,2]
[1,] 0.0333631015 0.0099068022
[2,] 0.0099068022 0.0260063531

$free
[,1] [,2]
[1,] TRUE TRUE
[2,] TRUE TRUE

$lbound
[,1] [,2]
[1,] 1e-24 0e+00
[2,] 0e+00 1e-24

I am using the same file for both scripts so I do not think some MZ twins are being treated as DZ twins.

Let me know if you need something else.

Thank you so much for your help.

File attachments
SCRIPTS.txt
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Replied on Tue, 08/27/2019 - 15:51
Picture of user. AdminRobK Joined: 01/24/2014

The phenotypic variance among MZ twins is about twice that among DZ twins. Is there a reason for that? Anyhow, that's a pretty important piece of information. Keep in mind that the ACE model imposes equal phenotypic variances for both zygosity groups. I don't think you should try to interpret standardized parameter estimates from the ACE model with this dataset. In fact, I'm not sure the ACE model should be interpreted at all in this case. How does it compare to the non-biometrical models in terms of fit? What do the raw (i.e., unstandardized) variance components look like?
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Replied on Wed, 08/28/2019 - 06:18
No user picture. JuanJMV Joined: 07/20/2016

Hi Rob,

Thank you so much for your response. I do not know why the variance is bigger in MZ twins. The sample is not too big so maybe that is the problem.

Here you can find the outputs. Please let me know if you need something else.

1-Comparison

> mxCompare( fit, fitACE )
base comparison ep minus2LL df AIC diffLL diffdf p
1 oneSATca 12 -82.097179 188 -458.09718 NA NA NA
2 oneSATca oneACEca 6 -65.081461 194 -453.08146 17.015718 6 0.0092256499
>

2-Saturado

> sum
Summary of oneSATca

free parameters:
name matrix row col Estimate Std.Error A lbound ubound
1 b11 b1 1 1 -0.0090484542 0.0067378309
2 b12 b2 1 1 0.0743644333 0.0382335972
3 mMZ1 MZ.meanMZ 1 1 -0.0754166735 0.1652120793
4 mMZ2 MZ.meanMZ 1 2 -0.0670462403 0.1663562850
5 vMZ1 MZ.covMZ MidPeriphery1 MidPeriphery1 0.0578906907 0.0111658450 1e-24
6 cMZ21 MZ.covMZ MidPeriphery1 MidPeriphery2 0.0431419265 0.0108939007 ! 0!
7 vMZ2 MZ.covMZ MidPeriphery2 MidPeriphery2 0.0783771747 0.0151543292 1e-24
8 mDZ1 DZ.meanDZ 1 1 -0.0414547261 0.1548536765
9 mDZ2 DZ.meanDZ 1 2 -0.0225095071 0.1543364226
10 vDZ1 DZ.covDZ MidPeriphery1 MidPeriphery1 0.0333631015 0.0070266021 0!
11 cDZ21 DZ.covDZ MidPeriphery1 MidPeriphery2 0.0099068022 0.0045876895 ! 0!
12 vDZ2 DZ.covDZ MidPeriphery2 MidPeriphery2 0.0260063531 0.0054515852 0!

3-ACE

> fitEsts(fitACE)
b11 b12 xbmi a11 c11 e11
-0.0106 0.0524 0.0182 0.0424 0.1625 0.1497
A C E SA SC SE
VC 0.0018 0.0264 0.0224 0.0355 0.5219 0.4426

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Replied on Wed, 08/28/2019 - 10:31
Picture of user. AdminRobK Joined: 01/24/2014

I don't think it means very much for the saturated model to fit better than the ACE model. The saturated model is usually kind of "silly", at least in cases where the order of twins in a pair is arbitrary. How do your `fitEMVO` and `fitEMVZ` compare to the ACE model?
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Replied on Wed, 08/28/2019 - 08:07
Picture of user. AdminNeale Joined: 03/01/2013

In small sample sizes, variance differences sometimes occur due to outliers. Are there perhaps outlier observations in the MZ data set? However, the variances of both MZ twin 1 and MZ twin 2 are both about twice those of DZ twins, so one might expect both members of the pair to be outliers. What is the phenotype? For some traits, we observe contrast effects - parents ratings of their children’s activity for example.

There are models for sibling interaction that predict different variances. Here the phenotypes of the twins directly influence each other. If there is genetic variation, the sibling interaction generates greater variance in MZ pairs than DZ. But I would inspect the data for outliers first.

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Replied on Fri, 08/30/2019 - 04:21
No user picture. JuanJMV Joined: 07/20/2016

Thank you Rob and Mike,

Here the comparisons:

> mxCompare( fitEMVO, fitACE )
base comparison ep minus2LL df AIC diffLL diffdf p
1 eqMVarsTwin 8 -78.770636 192 -462.77064 NA NA NA
2 eqMVarsTwin oneACEca 6 -65.081461 194 -453.08146 13.689175 2 0.0010652053

> mxCompare( fitEMVZ, fitACE )
base comparison ep minus2LL df AIC diffLL diffdf p
1 eqMVarsZyg 6 -65.081461 194 -453.08146 NA NA NA
2 eqMVarsZyg oneACEca 6 -65.081461 194 -453.08146 9.6851238e-10 0 NA

I have removed 3 twin pairs that were outliers (all MZ and one of them with high means in both members of the twin pair)

Here the results without outliers:

> mxCompare( fit, subs <- list(fitCov, fitEMO, fitEMVO, fitEMVZ) )
base comparison ep minus2LL df AIC diffLL diffdf p
1 oneSATca 12 -118.89221 182 -482.89221 NA NA NA
2 oneSATca testCov 10 -113.78681 184 -481.78681 5.10539909 2 0.077871165
3 oneSATca eqMeansTwin 10 -118.31624 184 -486.31624 0.57596443 2 0.749774926
4 oneSATca eqMVarsTwin 8 -116.08046 186 -488.08046 2.81174413 4 0.589807172
5 oneSATca eqMVarsZyg 6 -114.07218 188 -490.07218 4.82002346 6 0.567095201

$covDZ
SymmMatrix 'covDZ'

$labels
[,1] [,2]
[1,] "vDZ1" "cDZ21"
[2,] "cDZ21" "vDZ2"

$values
[,1] [,2]
[1,] 0.033687101 0.010184924
[2,] 0.010184924 0.026238597

$covMZ
SymmMatrix 'covMZ'

$labels
[,1] [,2]
[1,] "vMZ1" "cMZ21"
[2,] "cMZ21" "vMZ2"

$values
[,1] [,2]
[1,] 0.034779181 0.019073070
[2,] 0.019073070 0.046862384

And now the correlations are:

SATURATED: MZ=0.47 y DZ=0.34

ACE: MZ=0.43 y DZ=0.39

Comparison between ACE and fitEMVO/fitEMVZ

> mxCompare( fitEMVO, fitACE )
base comparison ep minus2LL df AIC diffLL diffdf p
1 eqMVarsTwin 8 -116.08046 186 -488.08046 NA NA NA
2 eqMVarsTwin oneACEca 6 -114.07218 188 -490.07218 2.0082793 2 0.36635969

> mxCompare( fitEMVZ, fitACE )
base comparison ep minus2LL df AIC diffLL diffdf p
1 eqMVarsZyg 6 -114.07218 188 -490.07218 NA NA NA
2 eqMVarsZyg oneACEca 6 -114.07218 188 -490.07218 0 0 NA

*The phenotype is an objective measure of the eye.

Thank you so much for your helpful comments.

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