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  • jaymiRNA
    Junior Member
    • Nov 2011
    • 5

    #1

    Differential expression DESeq using GLM for timecourse experiment

    Dear Members
    I am currently working on NGS data and performed some differential expression analysis using DESeq. At the moment I am trying to find differentially expressed miRNAs using GLM functionality.
    My experimental design consists of 4 samples (4stages of developing flower) with 2 biological replicates each. I would like to identify the differentially expressed miRNA during different developmental stages(Stage 1,2,3 and 4). However, I have some doubts regarding my final data interpretation and here is the code I used and the output.

    My Dataset:
    >Str(SSSA)
    Dataframe: 662 obs. of 8 variables:
    $ Flower1 : int 1 1 6 0 0 0 6709 0 3 0 ...
    $ Flower1.1: int 1 1 1 0 0 0 2261 2 1 2 ...
    $ Flower2 : int 0 4 49 0 0 0 7311 0 4 0 ...
    $ Flower2.1: int 1 14 69 0 0 0 11211 0 1 0 ...
    $ Flower3 : int 0 8 89 1 0 0 1978 0 0 1 ...
    $ Flower3.1: int 0 4 72 0 0 0 5392 0 1 1 ...
    $ Flower4 : int 0 1 67 0 0 0 4675 0 0 0 ...
    $ Flower4.1: int 0 0 21 0 0 0 4629 0 1 5 ...


    Code:
    My code
    groups <- c( rep("F1",2), rep("F2",2), rep("F3",2), rep("F4",2) )
    cds <- newCountDataSet(SSSA , groups )
    cds <- estimateSizeFactors( cds )
    cds <- estimateDispersions( cds, "pooled" )
    fit0 <- fitNbinomGLMs ( cds, count ~ 1 )
    fit1 <- fitNbinomGLMs ( cds, count ~ groups )
    pvals <- nbinomGLMTest( fit1, fit0 )
    padjGLM <- p.adjust( pvals, method="BH" )
    fit1.pval <- cbind(fit1, padjGLM)
    My Result
    > str(fit1.pval)
    'data.frame': 662 obs. of 7 variables:
    $ (Intercept): num 1.23 1.16 2.77 -32.46 NA ...
    $ groupsF2 : num -3.17 1.07 2.17 -2.1 NA ...
    $ groupsF3 : num -32.569 0.859 2.973 30.973 NA ...
    $ groupsF4 : num -31.675 -2.008 2.835 -0.875 NA ...
    $ deviance : num 0.746 1.895 2.041 0.234 NA ...
    $ converged : logi TRUE TRUE TRUE TRUE NA NA ...
    $ pval : num 0.28395 0.44052 0.00622 0.86603 NA ...

    My doubts.....
    1.In my case, is it better to set the conditions as a data frame instead of a vector- “groups <- c( rep("F1",2), rep("F2",2), rep("F3",2), rep("F4",2) )”?
    2.In the code, I calculated the object fit0 setting the value 1 after the tilde -“fit0 <- fitNbinomGLMs ( cds, count ~ 1 )” but I am not very clear.. what count ~ 1 stand for/refers to ?

    3.My final results( above) shows 7 variables(including the adjusted pvalue) in the data.frame and I am not clear in interpreting this data. In particular , by using ‘groups’ in calling fit 1 i.e. fit1 <- fitNbinomGLMs ( cds, count ~ groups ),I can’t understand if I really compared F1 against all other stages F2,F3,F4, as I intend or If I made any other comparisions.

    Could anyone help me on this?

    Jay
  • jaymiRNA
    Junior Member
    • Nov 2011
    • 5

    #2
    nobody knows?

    Comment

    • Simon Anders
      Senior Member
      • Feb 2010
      • 995

      #3
      re 1.: If you pass a data frame, you can name you factor. You used "groups". If you pass just a vector, it will always be "condition".

      re 2.: You are testing whether the differences between samples from different groups are significantly stronger than the difference between replicates.

      Is this what you want to test?

      Comment

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