agridat (version 1.12)

butron.maize: Maize yields in a multi-environment trial. Pedigree included.

Description

Maize yields in a multi-environment trial. Pedigree included.

Format

A data frame with 245 observations on the following 5 variables.

gen

genotype

male

male parent

female

female parnet

env

environment

yield

yield, Mg/ha

Details

Ten inbreds were crossed to produce a diallel without reciprocals. The 45 F1 crosses were evaluated along with 4 checks in a triple-lattice 7x7 design. Pink stem borer infestation was natural.

Experiments were performed in 1995 and 1996 at three sites in northwestern Spain: Pontevedra (42 deg 24 min N, 8 deg 38 min W, 20 m over sea), Pontecaldelas (42 deg 23 N, 8 min 32 W, 300 m above sea), Ribadumia (42 deg 30 N, 8 min 46 W, 50 m above sea).

A two-letter location code and the year are concatenated to define the environment.

The average number of larvae per plant in each environment:

Env	Larvae
pc95	0.54
pc96	0.91
ri96	1.78
pv95	2.62
pv96	3.35

Examples

# NOT RUN {
data(butron.maize)
dat <- butron.maize

# require(lattice)
# densityplot(~yield|env, dat, layout=c(1,5))

require(reshape2)
mat <- acast(dat, gen~env, value.var='yield')
mat <- sweep(mat, 2, colMeans(mat))
mat.svd <- svd(mat)
# Calculate PC1 and PC2 scores as in Table 4 of Butron
# Comment out to keep Rcmd check from choking on '%*%'
# round(mat.svd$u[,1:2] %*% diag(sqrt(mat.svd$d[1:2])) %*% diag(c(-1,1)),3)

biplot(princomp(mat), main="butron.maize", cex=.7) # Figure 1 of Butron

# ----------------------------------------------------------------------------

# }
# NOT RUN {
# Here we see if including pedigree information is helpful for a multi-environment model

# Create the pedigree
ped <- dat[, c('gen','male','female')]
ped <- ped[!duplicated(ped),] # remove duplicates
unip <- unique(c(ped$male, ped$female)) # Unique parents
unip <- unip[!is.na(unip)]
# We have to define parents at the TOP of the pedigree
ped <- rbind(data.frame(gen=c("Dent","Flint"), # genetic groups
                        male=c(0,0),
                        female=c(0,0)),
             data.frame(gen=c("A509","A637","A661","CM105","EP28",
                          "EP31","EP42","F7","PB60","Z77016"),
                        male=rep(c('Dent','Flint'),each=5),
                        female=rep(c('Dent','Flint'),each=5)),
             ped)
ped[is.na(ped$male),'male'] <- 0
ped[is.na(ped$female),'female'] <- 0

# View the pedigree.  Can't use kinship2 ... plants can be both male/female
# Not the best view...too much overplotting
require(synbreed)
pe <- with(ped[1:57,], create.pedigree(gen, male, female, gener=NULL))
windows(9,6)
synbreed::plot.pedigree(pe, vertex.size=10, vertex.label.cex=.5, asp=.5) #

require(asreml)
ped.ainv <- asreml.Ainverse(ped)$ginv

m0 <- asreml(yield ~ 1+env, random = ~ gen, data=dat)
m1 <- asreml(yield ~ 1+env, random = ~ ped(gen), ginverse=list(gen=ped.ainv), data=dat)
m2 <- update(m1, random = ~ id(env):ped(gen))
m3 <- update(m2, random = ~ diag(env):ped(gen))
m4 <- update(m3, random = ~ fa(env,1):ped(gen))
AIC(m0,m1,m2,m3,m4)
##    df      AIC
## m0  2 229.4037
## m1  2 213.2487
## m2  2 290.6156
## m3  6 296.8061
## m4 11 218.1568

p0 <- predict(m0, classify="gen")$pred$pvals
p4 <- predict(m4, classify="gen")$pred$pvals
p4par <- p4[1:12,]   # parents
p4 <- p4[-c(1:12),]  # hybrids
# Careful!  Need to manually sort the predictions
p0 <- p0[order(as.character(p0$gen)),]
p4 <- p4[order(as.character(p4$gen)),]

# lims <- range(c(p0$pred, p4$pred)) * c(.95,1.05)
lims <- c(6,8.25) # zoom in on the higher-yielding hybrids
plot(p0$predicted.value, p4$predicted.value,
     pch="", xlim=lims, ylim=lims, main="butron.maize",
     xlab="BLUP w/o pedigree", ylab="BLUP with pedigree")
abline(0,1,col="lightgray")
text(x=p0$predicted.value, y=p4$predicted.value, p0$gen, cex=.5, srt=-45)
text(x=min(lims), y=p4par$predicted.value, p4par$gen, cex=.5)

# Including the pedigree provided little benefit

# }
# NOT RUN {
 # dontrun

# }