agridat (version 1.16)

butron.maize: Multi-environment trial of maize with pedigrees

Description

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

Arguments

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

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

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

if(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
  # Including the pedigree provided little benefit

  # 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) #
# }
# NOT RUN {
# ----------------------------------------------------------------------------

# }
# NOT RUN {
  # asreml3
  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, data=dat, classify="gen")$pred$pvals
  p4 <- predict(m4, data=dat, 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)
# }
# NOT RUN {
# ----------------------------------------------------------------------------

# }
# NOT RUN {
##   require(asreml4)
##   ped.ainv <- ainverse(ped)

## if(FALSE){
##   # asreml4 does not use ginverse
##   m0 <- asreml(yield ~ 1+env, data=dat, random = ~ gen)
##   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, data=dat, classify="gen")$pred$pvals
##   p4 <- predict(m4, data=dat, 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
## } # if(FALSE)

# }
# NOT RUN {
 # dontrun

# }

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