From: <Jeremy.Brawner_at_CSIRO.AU>

Date: Mon, 1 Feb 2010 12:05:40 +1100

Date: Mon, 1 Feb 2010 12:05:40 +1100

Dear all,

I have been able to run a similar bivariate-normal model as to what is described below in ASREML.

Is this possible to run this in ASREML-R; if yes how do I specify the different distributions?

Kind regards,

Jeremy

-----Original Message-----

From: ASReml users discussion group [mailto:ASREML-L_at_DPI.NSW.GOV.AU] On Behalf Of Arthur Gilmour

Sent: Friday, 8 May 2009 3:23 PM

To: ASREML-L_at_DPI.NSW.GOV.AU

Subject: Re: bivariate binomial-normal analysis

Dear Anita,

see below.

On Thu, 2009-05-07 at 09:37 -0300, Anita Schmidek wrote:

*> Dear all,
*

*>
*

*>
*

*>
*

*> I’m working with pre-weaning calf mortality (1=death; 0=alive).
*

*>
*

*> Calf birth weight seems to be an important characteristic regarding
*

*> mortality (average optimum), and show possibility to be used as an
*

*> indicator trait.
*

*>
*

*> For these reason, I want to fit a bivariate analysis, using mortality
*

*> and birth weight as traits.
*

*>
*

*> Initially, I’ve done univariate analysis, in order of get heritability
*

*> estimates, as well as starting values for bivariate analysis.
*

*>
*

*> The model I used for the bivariate analysis, regarding sire model, is
*

*> as follows:
*

*>
*

*>
*

*>
*

*> Analise bivariada da mortalidade ONM210 PN sire model
*

*>
*

*> anim !P # calf number
*

*>
*

*> pai !P # sire number
*

*>
*

*> mae !P # dam number
*

*>
*

*> nmae 3576 !I # dam number with number of dams in file
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*>
*

*> idv 4 !I # dam age class (4)
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*>
*

*> sx !I # calf sex
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*>
*

*> gc 68 !I # contemporary group
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*>
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*> onm210 !M -9 #missing = -9 # pre-weaning mortality
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*>
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*> pn !M -9 #missing = -9 # calf birth weight
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*>
*

*> ped_mtger.dat # pedigree file
*

*>
*

*> mt_to.dat !TYPEIIISS !EXTRA 50 !REPORT !ASUV #!CONTINUE #data file
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*>
*

*> onm210 pn !BIN !LOGIT ~ Trait Trait.idv Trait.sx ,
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*> !r Trait.pai !f Trait.gc
*

*>
*

*>
*

*>
*

*> 1 2 1
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*> 0
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*>
*

*> Trait 0 US 3.29 3.32 13.40 !GPFP
*

*> # in fact, the value in .asr for “3.29” was “1”, but considering
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*> # I used a logit link, I put the error as 3.29 (is that correct?)
*

** NO , assuming no over/under dispersion, it should read

Trait 0 US 1 2 13.40 !GFPP

** But with over/under dispersion, !GP

*>
*

*> Trait.pai 2
*

*>
*

*> Trait 0 US 0.04 0.15 1.96 !GP
*

*>
*

*> pai 0 AINV
*

*>
*

*>
*

The variance parameter 1 is the dispersion parameter for the GLMM

model, which uses weights derived from the fitted value usng BINOMIAL

assumption and logit transformation. Since a logisitc variable has a

variance of 3.29, we can pretend/presume/assume/whatever that on the

underlying scale (on which the effects are estimated and the genetic

variance (V4) applies) has a residual variance of 3,29 (assuming no

over/under dispersion. So to derive genetic parameters we might use

F phenvar1 1 * 3.29 + 4

F phenvar23 2:3 + 5:6

F addvar 4:6 * 4

H heritA 10 7

H heritB 12 9

etc

*>
*

*> and the .pin file:
*

*>
*

*> F phenvar 1:3 + 4:6
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*>
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*> F addvar 4:6 * 4
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*>
*

*> H heritA 10 7
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*>
*

*> H heritB 12 9
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*>
*

*> R phencorr 7 8 9
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*>
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*> R gencor 4:6
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*>
*

*>
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*>
*

*> .pvc file
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*>
*

*> 7 phenvar 1 1.131 0.2906E-01
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*>
*

*> 8 phenvar 2 3.226 0.5354E-01
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*>
*

*> 9 phenvar 3 95.39 0.5863
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*>
*

*> 10 addvar 4 0.2804E-01 0.1074
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*>
*

*> 11 addvar 5 -0.3762 0.2141
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*>
*

*> 12 addvar 6 5.337 0.8535
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*>
*

*> heritA = addvar 10/phenvar 7= 0.0248 0.0944
*

*>
*

*> heritB = addvar 12/phenvar 9= 0.0560 0.0088
*

*>
*

*> phencorr = phenvar /SQR[phenvar *phenvar ]= 0.3106 0.0068
*

*>
*

*> gencor 2 1 = Trait 5/SQR[Trait 4*Trait 6]= -0.9725 1.8858
*

*>
*

*>
*

*>
*

*> The program converged, but I don’t know if that’s the right (or
*

*> better) model…
*

*>
*

*> Should I have similar heritability values between uni and bivariate
*

*> analysis? (because they are quite different)…
*

*>
*

** They should indeed be quite similar, assuming the model is the same

(except for the addition of the correlation). You could take the above

model and fix the covariances at 0.0 and you should get the univariate

results.

*> Is there a reason (and a way of dealing with) the high SE found in
*

*> genetic correlation?
*

*>
*

** In this model, it will depend on the number of sires.

** In this case, the genetic variance (.028) has a high SE (.10) so

the genetic correlation will also have a high SE implying that the

estimate is hardly useful.

*>
*

*>
*

*> I thank all in advance for the attention and help.
*

*> Regards
*

*>
*

*>
*

*>
*

*> Anita Schmidek
*

*>
*

*>
*

*>
*

*>
*

*>
*

*>
*

*>
*

*>
*

*> This message is intended for the addressee named and may contain confidential information. If you are not the intended recipient, please delete it and notify the sender. Views expressed in this message are those of the individual sender, and are not necessarily the views of their organisation.
*

--

May the God and Father of the Lord Jesus Christ guide and bless you.

Arthur Gilmour

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Received on Mon Feb 01 2010 - 12:05:40 EST

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