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  • IBseq
    replied
    Hi PRinlgler,
    did you come up with a solution?
    I'm also quite new and tried to understand what is the mathematical process from cufflinks to cuffdiff...why are the fpkm in cufflinks differnt from those given in cuffdiff??

    thanks,
    ib

    Leave a comment:


  • Portah
    replied
    Does some one has an answer for the second part of the question ?

    Leave a comment:


  • PRingler
    started a topic Cuffdiff FPKM and test statistic calculations

    Cuffdiff FPKM and test statistic calculations

    Hello, I have been lurking here a bit, but haven't posted yet, so bear with me.

    I am trying to recreate the calculations of FPKM and the test statistic performed by uffdiff on two sample genes in order to better understand how my RNA-seq data has been tested (and thus better explain my data). I have been reading through the supplemental information from the paper linked on the Cufflinks website (Transcript assembly and quantification by RNA-Seq reveals unannotated transcripts and isoform switching during cell differentiation; Nature Biotechnology 28,511–515(2010)doi:10.1038/nbt.1621), and I have been able to get through part of the calculations, but am still stumped.

    To calculate FPKM, I am using the following formula:

    (10^9)*Xg*(gamma t)
    ~l(t)*M

    Where:
    Xg is the number of fragments aligned to the gene locus (g),
    gamma t is the maximum likelihood estimate of the probability of selecting a fragment from a transcript (t) coming from that gene locus
    ~l(t) is the adjusted transcript length
    Σ from i=1 to t(i)) [F(i)(l(t)-i+1)]
    F(i) = probability that the fragment has a length i
    l(t) = the transcript length
    M is the total number of fragments mapped in that sample

    My sample is run with single-end reads 50 bp long, so I simplified the ~l(t) to the transcript length - 49, and we ran cuffdiff using a GTF file that contains only the loci for one transcript for each gene, so I think gamma should be 1. One of my lab members wrote a program that assigns raw reads to individual genes, so I am using the reads from this program for the raw reads in the FPKM calculation. Using this data I get pretty close to the FPKM calculated:

    Code:
    gene   Reads A   Reads B    Cuff A	Cuff B             Calc A        Calc B
    A	2	81        0.021956	0.823028        0.02116         0.795515
    B	1	40        0.007541	0.279201        0.007374        0.273778
    The difference is most likely from my assumption that the ~l(t) is only looking at the 50 bp read length, or the assumption that gamma is 1.

    From this, I get into much more complicated math when trying to recreate the test statistic. The formula for the test statistic, from the supplemental data, is:

    [log(Xa)+log(gamma a)+ log(Mb) - log(Xb) - log(gamma b) - log (Ma)]
    SQRT[ (psi a*(1+Xa)*(gamma a)^2)/(Xa*(gamma a)^2)+ (psi b*(1+Xb)*(gamma b)^2)/(Xb*(gamma b)^2)]

    Where psi is a variance-covariance matrix that estimates the covariance between gamma (tk) and gamma (tl). As far as I can tell, tk and tl come from Tophat, which splits a read of length l into two smaller reads of length k in order to align across splice junctions. (This may be completely off base)

    When I run the equations using these assumptions, though, I get a value far off from the test statistic calculated by Cuffdiff:

    Code:
    Gene      T stat	     Cufflinks t-stat
    A          0.99128963	-3.42295
    B          0.898803344	-3.07139

    So, I suppose my question is, is there a way for me to calculate (or even estimate) the psi function? Am I making faulty assumptions?

    I apologize if this question has been addressed in a previous thread. I did try to find the answer in the archives before asking here.
    Thank you

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