Multi-environment trial of maize with pedigrees

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 parent

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

Used with permission of Ana Butron.

Source

Butron, A and Velasco, P and Ord\'as, A and Malvar, RA (2004). Yield evaluation of maize cultivars across environments with different levels of pink stem borer infestation. Crop Science, 44, 741-747. https://doi.org/10.2135/cropsci2004.7410

Examples

# \dontrun{

  library(agridat)
  data(butron.maize)
  dat <- butron.maize

  libs(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] 

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


  if(0){
    # Fixme: This section is broken because synbreed has been removed from CRAN
    # 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
    
  }


  # asreml 4
  if(0){
    libs(asreml)
    ped.ainv <- ainverse(ped)
      
    m0 <- asreml(yield ~ 1+env, data=dat, random = ~ gen)
    m1 <- asreml(yield ~ 1+env, random = ~ vm(gen, ped.ainv), data=dat)
    m2 <- update(m1, random = ~ idv(env):vm(gen, ped.ainv))
    m3 <- update(m2, random = ~ diag(env):vm(gen, ped.ainv))
    m4 <- update(m3, random = ~ fa(env,1):vm(gen, ped.ainv))
    #summary(m0)$aic
    #summary(m4)$aic
    ##    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")$pvals
    p1 <- predict(m1, data=dat, classify="gen")$pvals
    p1par <- p1[1:12,]   # parents
    p1 <- p1[-c(1:12),]  # remove parents
    # Careful!  Need to manually sort the predictions
    p0 <- p0[order(as.character(p0$gen)),]
    p1 <- p1[order(as.character(p1$gen)),]
    
    # lims <- range(c(p0$pred, p1$pred)) * c(.95,1.05)
    lims <- c(6,8.25) # zoom in on the higher-yielding hybrids
    plot(p0$predicted.value, p1$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$predicted.value, y=p1$predicted.value,
         p0$gen, cex=.5, srt=-45)
    text(x=min(lims), y=p1par$predicted.value, p1par$gen, cex=.5, col="red")
    round( cor(p0$predicted.value, p1$predicted.value), 3)
    # Including the pedigree provided very little change
    
  }
# }