Saturday, April 21, 2012

Calculate the average distance between a given DNA motif within DNA sequences in R

Suppose that we want to calculate the expected distance of a DNA motif within a DNA target sequence, if we know the composition bias or the probability distribution (multinomial model) we can compute it just fine.

Download the R code <- here

FIRST PART

For example, supose that we want to compute the expected distance of the motif "GATC" in a sequence composed of 10,000 bases given that the whole sequence follows a probability distribution of p(A) = 0.3, p(C) = 0.2, p(G) = 0.2, p(T) = 0.3.

So open an R prompt and load the functions:

source("motifOccurrence.R")

Then, enter the initial values:

lengthSeq <- 10000
motif <- c("G","A","T","C")
probDistr <- c(0.3,0.2,0.2,0.3)

And finally, compute the average expected distance of every occurrence of the motif inside the target sequence. Using the following formula (this computation corresponds to the "computeExpectedDistance" function of the R script "motifOccurrence.R"):

    expectedDistance = lengthDNAseq/(lengthDNAseq*p))
 
where "p" stands for the joint probability of the motif, in other words: p = p(GATC) = p(G)*p(A)*p(T)*p(C) = 0.2*0.3*0.3*0.2 = 0.0036

To easily compute the expected distance in R, type:
 
expectedDist <- computeExpectedDistance(probDistr,lengthSeq,motif)

As you see, it returns the value of "277", this number means that, for the "GATC" motif inside a sequence of 10,000 bases with a composition bias of p(A) = 0.3, p(C) = 0.2, p(G) = 0.2, p(T) = 0.3 we may expect a distance of 277 bases between each "GATC".

Or, a little more graphic:

277 bases | GATC | 277 bases | GATC | .....

SECOND PART

Another way to compute this (even though it involves more computations) we can simulate X number of DNA sequences with a fixed length with an equal probability distribution per sequence and extract the coordinates of the motif within each sequence to finally compute the average distance of the motif.

There is a function titled "iterateComputeDistance" to do the calculations for you. Add the next parameter to the R environment:

iterations <- 100

Compute the average distance of the "GATC" motif within 100 DNA sequences (the other parameters remain equal)

expectedDistWithSimSeqs <- iterateComputeDistance(probDistr,lengthSeq,motif,iterations)

As we expected, the results of the two approaches are highly similar (ouuu yeah!)

THIRD PART

But, what happens when we already have a sequence and want to know the expected distance of that motif inside of it?.

Just like "Hey dude, I have an E.coli plasmid DNA sequence and want to know the average distance of the GATC motif".

Lets test using the sequence "Escherichia coli 2078 plasmid pQNR2078 complete sequence" <- http://www.ncbi.nlm.nih.gov/nuccore/HE613857.1

Ok, use the "ape" library to import the sequence to the R environment (if this library is not installed, type: install.packages("ape"))

NOTE: the GenBank sequences are in lowecase, so it will be needed to use a motif in lowercase to do the right computations.

Import the library:
require("ape")

Import the sequence:
plasmid <- read.GenBank("HE613857.1")
plasmidDNA <- as.character.DNAbin(plasmid)
plasmidDNA <- plasmidDNA[[1]]
motifEcoli <- c("g","a","t","c")

Get the coordinates:
plasmidDNAcoord <- coordMotif(plasmidDNA,motifEcoli)
Get the average distance between the motif occurrences.
plasmidDNAmotifDistance <- computeDistance(plasmidDNAcoord)

   > plasmidDNAmotifDistance
   [1] 270

Compute the number of occurrences of the motif among the plasmid (the result is 151 occurrences):

The number of occurrences of the motif in R is:
(length(plasmidDNAcoord)-1)


So, the main distance between the motif "gatc" finally is 270 bases and we are done :D

CODE:
#
#    Script: motifOccurrence.R
#    Author: Benjamin Tovar
#    Date: 21/April/2012
#
################################################################################

#                       ############
#                        FUNCTIONS:
#                       ############

##############################################################################
iterateComputeDistance <- function(multinomialDNAmodel,
                                   lengthDNAseq,
                                   motif,
                                   numberOfIterations){

    # This function returns the mean distance 
    # of a given motif given X number of DNA sequences given a multinomial model
    # (probability distribution of each base).
    
    # So, it will generate X number of DNA sequences using a given 
    # probability distribution and then it will compute the distance among 
    # that mofit within the total set of sequences to finally returns 
    # the average distance of the motif.
    
    result <- rep(NA,numberOfIterations)
    for(i in 1:numberOfIterations){
        currentGenome <- NA
        currentCoordinatesOfMotif <- NA        
        currentSequence <- sample(c("A","C","G","T"),
                                    lengthDNAseq,rep=T,
                                    prob=multinomialDNAmodel)
        currentCoordinatesOfMotif <- coordMotif(currentSequence,motif)
        result[i] <- computeDistance(currentCoordinatesOfMotif)
        cat(" *** Iteration number: ",i," completed *** | average distance = "
            ,result[i],"\n")  
    }
    result <- trunc(mean(result))
    cat(" \n*** Computation status: DONE ***\n\n")
    return(result)  
}
##############################################################################
coordMotif <- function(targetSequence,motif){

    # This function returns the coordinates of the motif of study in a target 
    # DNA sequence. In other words, if I found the motif, tell me exactly in
    # which position of the DNA sequence is.
    
    lengthMotif <- length(motif)
    lengthTargetSeq <- (length(targetSequence)-lengthMotif)
    motif <- toString(motif)
    motif <- gsub(", ","",motif)
    res <- 1    
    for(i in 1:lengthTargetSeq){
        currentTargetSeq <- targetSequence[i:(i+(lengthMotif)-1)]
        currentTargetSeq <- toString(currentTargetSeq)
        currentTargetSeq <- gsub(", ","",currentTargetSeq)
        if(currentTargetSeq == motif){
            res[(length(res)+1)] <- i
        }
    }
    return(res)
}
##############################################################################
computeDistance <- function(coordinatesOfMotif){
    
    # This function returns the mean distance 
    # of a given motif given its coordinates within a target DNA sequence.
    # In other words, If I already got a list with the coordinates where the 
    # motif is inside a DNA sequence, tell me the average distance between
    # this coordinates to get the expected distance of that motif.

    currentDistance <- rep(NA,(length(coordinatesOfMotif)-1))
    lengthCoord <- length(currentDistance)
    for(i in 1:lengthCoord){
        currentDistance[i] <- coordinatesOfMotif[i+1]-coordinatesOfMotif[i]
    }
    res <- trunc(mean(currentDistance))
    return(res)   
}
##############################################################################
computeExpectedDistance <- function(multinomialModel,
                                    lengthDNAseq,
                                    motif){
                                    
    # This function computes the expected distance of a given motif in a DNA
    # sequence given its multinomial model (probability distribution of 
    # each base)
    
    # Convert the motif into an index                                       
    motifIndex <- gsub("A",1,motif); motifIndex <- gsub("C",2,motifIndex)
    motifIndex <- gsub("G",3,motifIndex); motifIndex <- gsub("T",4,motifIndex)
    motifIndex <- as.numeric(motifIndex)
    # Compute p value of the motif given the multinomial model
    p <- rep(NA,length(motif))
    for(i in 1:length(motifIndex)){
        p[i] <- multinomialModel[motifIndex[i]]
    }    
    p <- prod(p)
    result <- trunc(lengthDNAseq/(lengthDNAseq*p))
    return(result)
}                                   
##############################################################################

# Benjamin

Benjamin

Friday, April 13, 2012

Introduction to Markov Chains and modeling DNA sequences in R

Markov chains are probabilistic models which can be used for the modeling of sequences given a probability distribution and then, they are also very useful for the characterization of certain parts of a DNA or protein string given for example, a bias towards the AT or GC content.

For now, I am going to introduce how to build our own Markov Chain of zero order and first order in R programming language. The definition of a zero order Markov Chain relies in that, the current state (or nucleotide) does not depends on the previous state, so there's no "memory" and every state is untied.

For the first order Markov Chain the case is different because the current state actually depends only on the previous state. Given that points clear, a second order Markov Model will be a model that reflects that the current state only depends on the previous two states before it (This model will be useful for the study of codons, given that they are substrings of 3 nucleotides long).

Download the scripts (R script and graphviz scripts) here

Example of a Markov Chain of zero order (the current nucleotide is totally independent of the previous nucleotide).

The multinomial model is:
p(A)+p(C)+p(G)+p(T) = 1.0
0.4 +0.1 +0.1 +0.4 = 1.0


Example of the structure of the zero order model:




Example of a Markov Chain of first order (the current nucleotide only depends on the previous nucleotide).

    The multinomial model per base is:
    p(A|A)+p(C|A)+p(G|A)+p(T|A) = 1.0
    p(A|C)+p(C|C)+p(G|C)+p(T|C) = 1.0
    p(A|G)+p(C|G)+p(G|G)+p(T|G) = 1.0
    p(A|T)+p(C|T)+p(G|T)+p(T|T) = 1.0      

    So:
    0.6 + 0.1 + 0.1 + 0.2  = 1.0
    0.1 + 0.5 + 0.3 + 0.1  = 1.0
    0.05+ 0.2 + 0.7 + 0.05 = 1.0
    0.4 + 0.05+0.05 + 0.5  = 1.0


Example of the structure of the first order model:
Code: 
#    Author: Benjamin Tovar
#    Date: 13/April/2012
#
#    Example of a Markov Chain of zero order (the current nucleotide is
#    totally independent of the previous nucleotide).

#    The multinomial model is:
#    p(A)+p(C)+p(G)+p(T) = 1.0
#    0.4 +0.1 +0.1 +0.4  = 1.0

# Define the DNA alphabet that will be used to put names to the objects
alp <- c("A","C","G","T")
# Create the vector that represents the probability distribution of the model
zeroOrder <- c(0.4,0.1,0.1,0.4)
# Put the name of reference of each base 
names(zeroOrder) <- alp 
# Create a sequence of 1000 bases using this model.
zeroOrderSeq <- sample(alp,1000,rep=T,prob=zeroOrder)

# ***** Study the composition bias of the sequence *****
# We wil use the "seqinr" package.
# For the installation of the package, type:
# install.packages("seqinr")
# Then, load the package:
require("seqinr")

# Count the frequency of each base 
# in the sequence using the "count" function
zeroOrderFreq <- count(zeroOrderSeq,1,alphabet=alp,freq=TRUE)

# Count the frequency of dinucleotides 
# in the sequence using the "count" function
zeroOrderFreqDin <- count(zeroOrderSeq,2,alphabet=alp,freq=TRUE) 
 
# Now, plot the results in the same plot:
layout(1:2)
barplot(zeroOrderFreq,col=1:4,main="Compositional bias of each nucleotide",
        xlab="Base",
        ylab="Base proportion")
barplot(zeroOrderFreqDin,col=rainbow(16),
        main="Compositional bias of each dinucleotide",
        xlab="Base",
        ylab="Base proportion")

# &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

#    Example of a Markov Chain of first order (the current nucleotide only
#    depends on the previous nucleotide).

#    The multinomial model per base is:
#    p(A|A)+p(C|A)+p(G|A)+p(T|A) = 1.0
#    p(A|C)+p(C|C)+p(G|C)+p(T|C) = 1.0
#    p(A|G)+p(C|G)+p(G|G)+p(T|G) = 1.0
#    p(A|T)+p(C|T)+p(G|T)+p(T|T) = 1.0        

#    So:
#    0.6 + 0.1 + 0.1 + 0.2  = 1.0
#    0.1 + 0.5 + 0.3 + 0.1  = 1.0
#    0.05+ 0.2 + 0.7 + 0.05 = 1.0
#    0.4 + 0.05+0.05 + 0.5  = 1.0
    
    
# Create the matrix that will store the probability distribution given 
# a certain nucleotide:
firstOrderMat <- matrix(NA,nr=4,nc=4)
# Put names to the 2 dimensions of the matrix 
colnames(firstOrderMat) <- rownames(firstOrderMat) <- alp 
# Add the probability distribution per base:
firstOrderMat[1,] <- c(0.6,0.1,0.1,0.2)  # Given an A in the 1st position
firstOrderMat[2,] <- c(0.1,0.5,0.3,0.1)  # Given a  C in the 1st position
firstOrderMat[3,] <- c(0.05,0.2,0.7,0.05)# Given a  G in the 1st position
firstOrderMat[4,] <- c(0.4,0.05,0.05,0.5)# Given a  T in the 1st position

# Now we got a matrix
#    > firstOrderMat
#         A    C    G    T
#    A 0.60 0.10 0.10 0.20
#    C 0.10 0.50 0.30 0.10
#    G 0.05 0.20 0.70 0.05
#    T 0.40 0.05 0.05 0.50

# In order to continue, we need an initial probability distribution to know
# which base is the most probable to start up the sequence.
inProb <- c(0.4,0.1,0.1,0.4); names(inProb) <- alp
# So, the sequence will have a 40% to start with an A or a T and 10% with C or G

# Create a function to generate the sequence.
# NOTE: To load the function to the current environment, just copy
# the entire function and paste it inside the R prompt.

generateFirstOrderSeq <- function(lengthSeq,
                                  alphabet,  
                                  initialProb,
                                  firstOrderMatrix){
#    lengthSeq = length of the sequence
#    alphabet = alphabet that compounds the sequence 
#    initialProb   = initial probability distribution
#    firstOrderMatrix = matrix that stores the probability distribution of a 
#                       first order Markov Chain

    # Construct the object that stores the sequence
    outputSeq <- rep(NA,lengthSeq)
    # Which is the first base:
    outputSeq[1]  <- sample(alphabet,1,prob=initialProb) 
    # Let the computer decide:
    for(i in 2:length(outputSeq)){
        prevNuc <- outputSeq[i-1]    
        currentProb <- firstOrderMatrix[prevNuc,]
        outputSeq[i] <- sample(alp,1,prob=currentProb)
    } 
    cat("** DONE: Sequence computation is complete **\n")
    return(outputSeq)
}

# Use the generateFirstOrderSeq function to generate a sequence of 1000 bases 
# long
firstOrderSeq <- generateFirstOrderSeq(1000,alp,inProb,firstOrderMat)

# ***** Study the composition bias of the sequence *****
# We wil use the "seqinr" package.
# For the installation of the package, type:
# install.packages("seqinr")
# Then, load the package:
require("seqinr")

# Count the frequency of each base 
# in the sequence using the "count" function
firstOrderFreq <- count(firstOrderSeq,1,alphabet=alp,freq=TRUE)

# Count the frequency of dinucleotides 
# in the sequence using the "count" function
firstOrderFreqDin <- count(firstOrderSeq,2,alphabet=alp,freq=TRUE) 
 
# Now, plot the results in the same plot:
layout(1:2)
barplot(firstOrderFreq,col=1:4,main="Compositional bias of each nucleotide",
        xlab="Base",
        ylab="Base proportion")
barplot(firstOrderFreqDin,col=rainbow(16),
        main="Compositional bias of each dinucleotide",
        xlab="Base",
        ylab="Base proportion")
        
## Lets plot the 4 plots in one window
    layout(matrix(1:4,nr=2,nc=2))
    # Results from the zero order 
    barplot(zeroOrderFreq,col=1:4,
            main="Compositional bias of each nucleotide\nZero Order Markov Chain",
            xlab="Base",
            ylab="Base proportion")
    barplot(zeroOrderFreqDin,col=rainbow(16),
            main="Compositional bias of each dinucleotide\nZero Order Markov Chain",
            xlab="Base",
            ylab="Base proportion")
    # Results from the first order 
    barplot(firstOrderFreq,col=1:4,
            main="Compositional bias of each nucleotide\nFirst Order Markov Chain",
            xlab="Base",
            ylab="Base proportion")
    barplot(firstOrderFreqDin,col=rainbow(16),
            main="Compositional bias of each dinucleotide\nFirst Order Markov Chain",
            xlab="Base",
            ylab="Base proportion")
        
        
# end.   

     



Plot of the zero order model:


Plot of the first order model



All plots


Benjamin

Wednesday, April 11, 2012

Generate artificial DNA or protein sequences in R in a single line of code.

To generate an artificial DNA sequence of  "n" bases long with a fixed composition bias in just one line of code, just open your R prompt and type:

seqX <- sample(c("A","C","G","T"),10000,rep=TRUE,prob=c(0.4,0.1,0.1,0.4))

As you see, the alphabet is the four letter alphabet of the DNA (so, if desired, you can replace that alphabet with any other alphabet that you might need, like the amino acids 20 letter code), the next parameter is the length of the desired output sequence. rep=TRUE means that each letter of the alphabet could be repeated and finally, I think that the most useful parameter of the function sample is the prob parameter because it allows you to select the multinomial model (the proportion or "bias" of each base). For example our simulated sequence is 80% AT rich and 20% GC rich given that for the "A" base we got a probability of 0.4, for the "C" base 0.2 and so on.

Now, to check it out that our artificial sequence follows that bias, a simple plot will tell us more than thousand words:

Lets extract the proportion of each base along the sequence using the table and the length function:

freqX <- table(seqX)/length(seqX)

Now, lets plot it doing:

barplot(freqX,col=1:4,main="Compositional bias of seqX",xlab="Base",ylab="Base proportion")

And finally we got this awesome plot.



So, the barplot shows that effectively each base is well represented by the multinomial model and our artificial sequence is loyal to it. 

Benjamin

Epic R is Epic <- Beginners command reference card

This reference card has been written by Tom Short.

Click the image to see the R magic:


Thank you for your support Tom.
Benjamin


Friday, April 6, 2012

Using the Blogger app for Android..

Well, the application has a simple and elegant UI.

Let's check it out and see how it works :P

Install R 2.15 and further versions in Debian Squeeze

The last Friday, March 30th, the last stable version of R, the version 2.15.0 was released.

So, to install it in Debian Squeeze, or in another Distro powered by Debian (I actually use CrunchBang Linux), just follow the same instructions described here for the installation of R 2.14.2 by clicking here.

Do not forget to upgrade the packages too.
Benjamin.

Install octave 3.6.1 in Debian Squeeze

Following the nice instructions found it here <- http://verahill.blogspot.mx/2012/02/debian-testing-wheezy-64-compiling.html , I have successfully installed Octave 3.6.1 in Squeeze (I had to rename some packages names to fit with the Squeeze packages).

FIRST STEP: Install the required libraries and dependencies.

sudo apt-get install gfortran build-essential
sudo apt-get install libqhull-dev libpcre++-dev libblas-dev liblapack-dev libreadline-dev
sudo apt-get install libcurl4-openssl-dev
sudo apt-get install libfltk1.1-dev
sudo apt-get install libgraphicsmagick++-dev
sudo apt-get install libhdf5-serial-dev
sudo apt-get install libqrupdate-dev
sudo apt-get install libsuitesparse-dev
sudo apt-get install  glpk gperf flex bison libfontconfig1-dev


SECOND STEP: Download Octave and extract it.

wget ftp://ftp.gnu.org/gnu/octave/octave-3.6.1.tar.gz
tar -xvf octave-3.6.1.tar.gz


THIRD STEP: Compile it.

cd octave-3.6.1/
./configure
make -j3

where 3 is the number of cores +1 (for me 2 cores)

FOURTH STEP: Validate the compiled version:

make check


FIFTH STEP: Install Octave

sudo make install


That's all. Thanks to Lindqvist - a blog about Linux and Science. Mostly.
Benjamin