From: <arthur.gilmour_at_DPI.NSW.GOV.AU>

Date: Tue, 29 May 2007 10:16:37 +1000

Date: Tue, 29 May 2007 10:16:37 +1000

Dear Felicity,

re:

Could someone please help me in how I might calculate a pooled residual

variance from the output below as I need it to calculate heritability? I

don't suppose it's as simple as an arithmetic mean, harmonic mean or

quadratic mean, or is it a bit more complicated than that?

Source Model terms Gamma Component Comp/SE %

C

Residual 1260 1252

at(Trial,2).Rep 4 4 0.842880E-02 0.842880E-02 0.51 0

P

Female 281 281 0.211808 0.211808 5.99 0

P

Trial.Female 1124 1124 0.103475 0.103475 4.87 0

P

Variance[ 1] 492 0 0.700046 0.700046 12.54 0

P

Residual AR=AutoR 12 0.118128 0.118128 2.06 0

U

Residual AR=AutoR 41 0.167493 0.167493 3.18 0

U

Variance[ 2] 492 0 0.472797 0.472797 12.62 0

P

Residual AR=AutoR 12 0.195265E-01 0.195265E-01 0.32 0

U

Residual AR=AutoR 41 0.237822 0.237822 4.58 0

U

Variance[ 3] 120 0 0.520579 0.520579 6.11 0

P

Residual AR=AutoR 12 -0.345290E-01 -0.345290E-01 -0.25 0

U

Residual AR=AutoR 10 0.193473 0.193473 1.50 0

U

Variance[ 4] 156 0 0.662102 0.662102 6.74 0

P

Residual AR=AutoR 12 0.121112E-01 0.121112E-01 0.11 0

U

Residual AR=AutoR 13 0.300947 0.300947 2.98 0

U

Since I have not noticed anyone venturing a response, here is my attempt.

The usual way to pool variances is weighted by their individual degrees of

freedom.

That is a bit difficult in this case as the degrees of freedom are not

easily obtained, being affected

by the correlated model as well as by the fixed terms in the model.

So, it may be possible to extract the residual stratum variance degrees of

freedom from

separate analyses of each trial, fitting 'female' as random, and use that.

Otherwise, weighting just be number of plots (less fixed effects), would

not be too bad in this instance

given the two largest trials have the most diverse variances (.70 and .47)

which are not extremely different.

You have constrained the genetic variances to be common across trials, so

you could also

constrain the residual variance to be common and see what the 'common'

value is.

This of course highlights the general problem that 'heritability' was

inverted in the context of homogeneous variances

but now we want to use it with heterogeneous variances.

So, an alternative would be to fit

Trial.Female

with

Trial 0 CORUH 0.7

.31 .31 .31 .31

Female

instead of 'Female Trait.Female'

and then calculate 4 heritabilites (each trial) since this would be more

informative.

It seems to me that heritability as a concept in plot trials is a bit

vague in any case because

it will be affected by plot size. I think Brian has an alternative way to

do the calculation for plot trials.

May you know Jesus Christ and His blessing in 2007,

Arthur Gilmour, His servant .

Mixed model regression mapping for QTL detection in experimental crosses.

Computational Statistics and Data Analysis 51:3749-3764 now available at

http://dx.doi.org/10.1016/j.csda.2006.12.031

Profile: http://www.dpi.nsw.gov.au/reader/17263

Personal website: http://www.cargovale.com.au/

mailto:Arthur.Gilmour_at_dpi.nsw.gov.au, arthur_at_cargovale.com.au

Principal Research Scientist (Biometrics)

NSW Department of Primary Industries

Orange Agricultural Institute, Forest Rd, ORANGE, 2800, AUSTRALIA

fax: 02 6391 3899; 02 6391 3922 Australia +61

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Received on Sat May 29 2007 - 10:16:37 EST

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