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11142011, 02:04 PM  #1 
Junior Member
Location: Nevada Join Date: Mar 2011
Posts: 1

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 RNAseq 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 RNASeq 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 singleend 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 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 variancecovariance 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 tstat 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 
07032012, 10:29 AM  #2 
Member
Location: Cincinnati Join Date: Jan 2012
Posts: 14

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

10162012, 02:47 AM  #3 
Member
Location: uk Join Date: Jul 2012
Posts: 56

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 
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