Hi Abhilash,

On 26/06/2008, at 1:59 AM, Abhilash Venu wrote:

> Hi Mark,
>
> I a have analyzed a set of data, with the following target file.
> FileName                                                           
> Target
> 251485013291_S01_GE1-v5_95_Feb07_1_1.txt    test
> 51485013291_S01_GE1-v5_95_Feb07_1_2.txt     test
> 251485013291_S01_GE1-v5_95_Feb07_1_3.txt    test
> 251485013291_S01_GE1-v5_95_Feb07_1_4.txt    test
> 251485013285_S01_GE1-v5_95_Feb07_1_1.txt    norm
> 251485013285_S01_GE1-v5_95_Feb07_1_2.txt    norm
> 251485013285_S01_GE1-v5_95_Feb07_1_3.txt    norm
> 251485013285_S01_GE1-v5_95_Feb07_1_4.txt    norm
>
> During normalization I am getting the following warning message? is  
> it going to create any trouble for me?
> MA<-normalizeBetweenArrays(log2(Rgene$G), method="quantile")
> Warning message:
> In log(c(65230.44761, 175.96476, 170.2672, 186.6677, 176.24364,  :
>   NaNs produced
The NaN warning means that some of your numbers cannot be logged.
Check which values have been converted to NaN's, and determine if they  
are because of malformed numbers in the text files (unlikely) or due  
to negative raw values. If you leave the NaN's in, then the ratio for  
that probe will also be NaN, and you'll loose info about those genes  
that may go from off to highly on. Thus adding a small offset to make  
them positive (BEFORE logging them) is quite commonly used.

>
> Created the following design file,
> > design=cbind(norm=c(0,0,0,0,1,1,1,1),test=c(1,1,1,1,0,0,0,0))
> > design
>      norm test
> [1,]    0    1
> [2,]    0    1
> [3,]    0    1
> [4,]    0    1
> [5,]    1    0
> [6,]    1    0
> [7,]    1    0
> [8,]    1    0
> >
>  fit<- lmFit(MA,design
>  cont.matrix=makeContrasts(normvstest=test-norm,levels=design))
> fit2=contrasts.fit(fit,cont.matrix)
> fit3=eBayes(fit2)
>  toptable(fit3,genelist=RG$genes,adjust="fdr")
>
> Whether my analysis is fine? As you mentioned I am getting the  
> adjusted p values above 0.05, and the data is showing high variance  
> among the samples (within the test and in normal also).
Your code looks good, however i'm not 100% sure that your use of MA is  
correct, since MA objects usually contain 2 colour data. Generally  
stats on MA objects operates on the MA$M data.frame, that is, the log2  
ratios of red vs green channel.
A foolproof approach would be to provide the matrix of log2, bg  
subtracted single channel data to a single channel normalisation  
method, such as normalizeQuantiles, then provide this single channel  
object to the lmFit.

> In this scenario should I try other normalization methods like  
> 'normexp' or 'vsn'.
Well, you would need to look into the assumptions behind these  
methods, and whether their application is appropriate. I haven't used  
normexp, but for vsn, i usually start by plotting the average  
expression level vs the stdev of each probe to see if there is an  
intensity dependent bias that could be fixed by vsn. Perhaps someone  
else can comment here.
>
>
> One more question I would like to ask is Can I get the fold value  
> for each test sample by comparing with all the normal? or is there  
> any other better approach for single color (Agilent) analysis. This  
> will help me to create the heatmap for genes across the samples.

- you can get the logged values for each array by MA$A
you can get the averaged values for test and norm from fit 
$coefficients, and the log ratio for test vs norm by  
fit2$coefficients, or from the logFC column in the topTable(fit3). If  
you want a heatmap of all 8 columns of data, then you'll need to use  
the single channel, normalised object that i talked about above.

cheers,

Mark
>
>
> Thank you in advance.
>
> Best
> Abhilash
>
> On Wed, Jun 18, 2008 at 8:43 PM, Abhilash Venu <abhivenu@gmail.com>  
> wrote:
> Hi
>
> Thank you Mark
> Thank you Francois.
> Currently I am I analyzing the entire data, once it got over I will  
> get back to you.
>
> Regards,
> Abhilash
>
>
> On Wed, Jun 18, 2008 at 11:21 AM, Mark Cowley <m.cowley0@gmail.com>  
> wrote:
> Hi Abhilash,
> In addition to Francois' comments, for me the biggest indicator is  
> that your adjusted P-values are all greater than 0.05.
> My interpretation of this is that after multiple testing correction,  
> none of your genes are statistically significantly differentially  
> expressed.
> This probably implies that either the differences between your two  
> groups are not large, or that there is higher inter-sample variance;  
> again, plotting some of these DE genes will help inform you as to  
> which is the case.
>
> cheers,
> Mark
>
>
> On 18/06/2008, at 2:48 AM, Francois Pepin wrote:
>
> Hi Abilash,
>
> you might want to read up a bit on those statistics and how they are  
> generated. It definitely helps when it comes to interpreting their  
> results properly.
>
> The B statistic is a log odds ratio. It simply means that the actual  
> log odds are between 0 and 1: less than 50% probability of  
> differential expression according to this test.
>
> The t statistics can be positive or negative. A negative t-statistic  
> simply means that the mean of the second group ("test" in this case)  
> is higher. A high negative t-statistic would have the same evidence  
> of differential expression as a high positive t-statistic.
>
> Francois
>
> Abhilash Venu wrote:
> Hi Mark,
> I think it was a great suggestion and I am providing the result,  
> which I got
> after the command, topTable(fit, coef="normvstest", adjust="fdr").
> I believe this preliminary results solved the problem of very high  
> fold
> value, which I was getting earlier. I will be looking at the entire  
> data and
> proved a better view at the earliest. But I have some doubts in this  
> table
> itself. Here I am getting negative odds ratio and in some cases  
> negative t
> value. What should I do in these scenario?
> topTable(fit, coef="normvstest", adjust="fdr")
>          logFC   AveExpr         t      P.Value adj.P.Val         B
> 35726 -2.1103554 11.936825 -9.602581 8.028641e-06 0.3614093 -3.912633
> 1968  -1.3413791  9.974470 -6.960746 9.138620e-05 0.7305539 -3.971277
> 5558  -1.6566417 10.885625 -6.506724 1.487751e-04 0.7305539 -3.987395
> 34497  1.0445013 10.251047  6.185219 2.132063e-04 0.7305539 -4.000460
> 33195 -1.3603874 13.373106 -6.116817 2.305481e-04 0.7305539 -4.003438
> 44662  0.9528248 11.180259  6.045345 2.503347e-04 0.7305539 -4.006630
> 24980 -1.5689151 10.824414 -5.932376 2.855094e-04 0.7305539 -4.011846
> 30206  2.2991372 13.647875  5.926758 2.873946e-04 0.7305539 -4.012112
> 26046 -1.1709614  9.505652 -5.746545 3.557246e-04 0.7305539 -4.020911
> 27210  1.4815342  9.416698  5.656415 3.964086e-04 0.7305539 -4.025537
> Thanks in advance
> Best
> Abhilash
> On Tue, Jun 17, 2008 at 4:38 AM, Mark Cowley <m.cowley0@gmail.com>  
> wrote:
> Hi Abhilash,
> Your code looks good, except that usually you will want to normalise  
> log
> transformed data. thus try:
>
> MA<-normalizeBetweenArrays( log2(Rgene$G), method="quantile")
>
> If your logFC ratios still look very high, then try convincing  
> yourself of
> their accuracy by looking at the raw data (RG$R) for some of the most
> differentially expressed genes, and also plot the expression values  
> for some
> of these DE genes.
>
> good luck,
> Mark
> Peter Wills Bioinformatics Centre
> Garvan Institute of Medical Research
>
>
>
> On 17/06/2008, at 1:17 AM, Abhilash Venu wrote:
>
> Hi list,
> I am still wonder about the data, which I analyzed by the limma. I  
> accept
> that  I am a biology graduate student, and in the learning stage. I am
> analyzing the single color data, which had been generated by Agilent  
> 4x44k
> platform. With the help of mailing list and limma users guide, I  
> have done
> the following analysis. But logFC gives very high values like 320,  
> 1320
> etc.
> I don't know how really the fitting is happening. Can I rely on this
> result.
> How should I go about it.
> #Reading the data.
>
> RG<-read.maimages(txt_files, columns = list(G = "gMeanSignal", Gb =
> "gBGMeanSignal",
> R="gMedianSignal",Rb="gBGMedian
>
> Signal"),
> annotation= c("Row", "Col",
> "ProbeUID","ProbeName", "GeneName",))
>
>
> Rgene<-backgroundCorrect(RG,method='subtract')
>
> #Considering only G as it is single color experiment.
> MA<-normalizeBetweenArrays(Rgene$G,method="quantile")
>
> design <- cbind(norm=1,normvstest=c(1,1,1,1,0,0,0,0))
> fit <- lmFit(MA, design)
> fit <- eBayes(fit)
> topTable(fit, coef="normvstest", adjust="fdr")
> --
>
> Regards,
> Abhilash
>
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>
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>
> ----------------------------------------------------------------------
> Mark Cowley, BSc (Bioinformatics)(Hons)
>
> Peter Wills Bioinformatics Centre
> Garvan Institute of Medical Research
> 384 Victoria St                                 Tel:  +61 2 9295 8542
> Darlinghurst, NSW 2010          Fax:  +61 2 9295 8538
> Australia                                       email:
> m.cowley@garvan.org.au
> www.garvan.org.au
> ----------------------------------------------------------------------
>
>
>
>
>
>
>
> -- 
>
> Regards,
> Abhilash
>
>
>
> -- 
>
> Regards,
> Abhilash


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