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Relative abundances of soil bacteria from 27 samples collected in nine forest and 18 grassland sites in Germany. The data set includes abundances of 18 bacterial phyla (including three candidate phyla) and five proteobacterial classes.
data(Bacteria)
A data frame with 27 observations on the following 24 variables.
Land use type
a factor with levels forest
grassland
Acidobacteria
a numeric vector
Actinobacteria
a numeric vector
Bacteroidetes
a numeric vector
Chloroflexi
a numeric vector
Cyanobacteria
a numeric vector
Deinococcus-Thermus
a numeric vector
Fibrobacteres
a numeric vector
Firmicutes
a numeric vector
Fusobacteria
a numeric vector
Gemmatimonadetes
a numeric vector
Nitrospira
a numeric vector
OP11
a numeric vector
Planctomycetes
a numeric vector
Spirochaetes
a numeric vector
Tenericutes
a numeric vector
TM7
a numeric vector
Verrucomicrobia
a numeric vector
WS3
a numeric vector
Alphaproteobacteria
a numeric vector
Betaproteobacteria
a numeric vector
Deltaproteobacteria
a numeric vector
Gammaproteobacteria
a numeric vector
Epsilonproteobacteria
a numeric vector
Relative abundances of 18 bacterial phyla (including three candidate phyla) and five proteobacterial classes (alpha, beta, gamma, delta and epsilon) from two ecological metagenomics studies (Will et al. 2010, Nacke et al. 2011). There are 27 observations altogether, nine of which stem from forest and 18 from grassland plots in Germany.
One goal of these investigations was to unravel differences in bacterial diversity and community composition between the land use types forest and grassland.
The bacteria's relative abundances were determined by analyzing the V2-V3 region of the 16S rRNA gene via pyrosequencing-based DNA techniques.
data(Bacteria)
str(Bacteria)
### Assess whether there is a difference in biodiversity and
### community composition species richness (Shannon index,
### Simpson index) between grassland and forest.
### Bootstrap times set to 50 due to example time settings
library(simboot)
mcpHill(dataf=Bacteria[,2:24], fact=Bacteria[,1], boots=50, qval=c(0,1,2))
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