PBG 650 Advanced Plant Breeding and Quantitative Genetics
(HW 4 from 2011 class)
Analysis of Half-sib Families of Meadowfoam; 1 location, 2 replications, 214 families
The data set is from a meadowfoam half-sib progeny trial. Data for the check entries were removed so that we could obtain estimates
of genetic variances from the 214 families representing the parent population MF187. We will be analyzing the variable ‘thousand
seed weight’ (TSW) which is measured in grams. For the purposes of this assignment, we will ignore the incomplete block structure of
the original lattice design and just consider the two complete blocks (=Reps) as an RCBD.
Use the annotated SAS output to respond to the questions on the right of the table.
#
1
SAS PROGRAM
PROC GLM data=ONE plots (unpack)=diagnostics;
Title 'Fixed Effect Model using PROC GLM';
class Rep Entry;
model TSW = Rep Entry;
random rep entry;
lsmeans Entry/stderr;
RUN;
QUESTIONS
Use the output to calculate
VP
- VG
- VX
- VA
- h2
MSF 0.5218469

 0.26092345
r
2
VG  VF   MSF  MSE  / r  0.5218469  0.2249413 / 2  0.14845
VP 
MSE 0.224913

 0.11247065
r
2
VA  4 * VF  4 *0.14845  0.5938
V
VG
0.14845
h2  G 

 0.5689
VP VG  VX 0.26092345
VX 
1
#
2
SAS PROGRAM
PROC VARCOMP METHOD = REML MAXITER = 100
data=ONE;
Title 'PROC VARCOMP for a random model - REML
estimation';
CLASS Rep Entry;
MODEL TSW = Rep Entry;
RUN;
QUESTIONS
PROC VARCOMP provides estimates of variance components
when all effects in the model are random. Compare the REML
estimates of variance components to your calculations in Part 1.
Use the output to calculate
- VA
- h2
VA  4 * VF  4 * Var(Entry)  4 *0.14845  0.5938
h2 
VG
V(entry)
0.1485


 0.5691
VP V(entry)  V(Residual) 0.1485  0.2249
2
2
Use the output from the asymptotic covariance matrix to calculate a
standard error for the estimate of VG.
V( G2 )  0.000758
3
PROC GLIMMIX data=one;
Title 'GLIMMIX Procedure';
class Rep Entry;
model TSW = /solution;
Random Rep Entry/solution;
/* I am not completely sure about this,
but I think this will give us a likelihood
ratio test for the variance among families*/
covtest general 0 1;
run;
se  0.000758  0.0275318
Compare the estimates of VG and its standard error to previous
results.
They are the same.
The following calculation illustrates the meaning of BLUP and the
relationship between the eblups and heritability for a balanced
experiment using unrelated parents:
Pick an eblup estimate for one of the entries from the solutions
output for random effects. Calculate the difference between the
observed mean for that entry (from the lsmeans in #1) and the grand
mean of all entries. Is the magnitude of the observed deviation larger
or smaller than the eblup? Calculate the ratio eblup/observed
deviation. What does that tell you about the expected response to
selection?
2
4
5
PROC MIXED data=one;
Title 'Mixed Procedure - Full Model';
class Rep Entry;
model TSW =;
Random Rep Entry;
run;
PROC MIXED data=one;
Title 'Mixed Procedure - Reduced Model';
class Rep;
model TSW =;
Random Rep;
run;
DATA prob;
chiprob=0.5*(1-probchi(LR, 1));
PROC PRINT;
RUN;
QUIT;
Entry 1 8.960-9.777=-0.817
Larger; eblup=-0.4649
Eblup/observed deviation = -0.4649/-0.817=0.5690h2
Use the values for -2 Res Log Likelihood
from the output for the full and reduced models to calculate the
Likelihood Ratio for the hypothesis that VG = 0.
L r z

LR  2ln   2 L r z  L  z 
L z





 

LR=799.7-763.0=36.7
In class you will substitute your LR value in the chiprob equation
shown on the left. The resulting probability value can be divided by
two because we are really only considering a one-sided test for
VG>0. What do your results indicate?
chiprob=0.5*(1-probchi(36.7, 1))=6.88897E-10
The small probability value indicates that the estimated variance
among entries is much greater than zero.
3