Hi Hyungwon, They do, and you could probably use that package as well. However, in this case with only two time points I am not sure you need anything more than a two factor ANOVA. Best, Jim Hyungwon Choi wrote: > Hi James, > > Just a guess, but doesn't Tai and Speed (Ann Statist, 34(5), 2006) have > timecourse package for this? > > Hyungwon > > James W. MacDonald wrote: >> Hi Sanjat, >> >> Sanjat Kanjilal wrote: >> >>> James W. MacDonald <jmacdon at="" ...=""> writes: >>> >>> >>>>> I have two strains (B and W) and two different time points (2 and >>>>> 1 >>> hours). I am trying to find genes which >>> >>>> respond differently: >>>> >>>>> 1. in B vs W, when comparing time points 1 and 2 (i.e. I compare >>>>> BvsW in >>> time point 1 to BvsW in time point 2) >>> >>>>> 2. in time in different strains (i.e. I compare B1vsB2 versus >>>>> W1vsW2) I am getting the same results, or they should be the >>>>> same? >>>>> Could you comment on whether this is a right way to analyse time >>>>> series? >>>>> >>>> It appears you want to test for the interaction between time and >>>> strain. Both of your points (1 and 2 above) are essentially >>>> identical. You want to know what genes react differently over time >>>> in the two strains (and worded differently, but meaning the same >>>> thing - what genes react differently between the strains at >>>> different times). These two things are verbo-algebraically the same >>>> (yup, new word. My next move is to start a new Wikipedia entry >>>> describing exactly what it means ;-D) >>>> >>>> >>>>> My code is below. >>>>> Thank you, >>>>> Lev. >>>>> > temp<-rma(data) >>>>> > targets <- readTargets("Targets.txt") >>>>> >>>>> >>>>>> lev <- c("W.1","B.1","W.2","B.2") >>>>>> f <- factor(targets$Target, levels=lev) >>>>>> design <- model.matrix(~0+f) >>>>>> colnames(design) <- lev >>>>>> fit <- lmFit(temp, design) >>>>>> cont.dif <- makeContrasts(Diff.Time=(B.2-W.2)-(B.1-W.1), >>>>>> Diff.Strain=(B.2- >>>>>> >>> B.1)-(W.2-W.1), levels=design) >>> >>>> As the statements above are verbo-algebraically the same, these two >>>> terms are algebraically the same. >>>> >>>> (B.2-W.2)-(B.1-W.1) = (B.2-B.1)-(W.2-W.1) >>>> >>>> Hence you should get the same results from each contrast. And yes, >>>> this contrast does give you the interaction. >>>> >>>> Best, >>>> >>>> Jim >>>> >>>> >>> Thanks for your helpful explanation Jim. >>> >>> As a follow-up question: >>> >>> Can the (B.2-W.2) - (B.1-W.1) contrast be thought of as comparing B @ >>> time 2 to B @ time 1 while 'normalizing' to the effects of the wild >>> type (ie I'd like to subtract out the effects of the control group)? >>> If not, is there a term that would do that? >>> >> >> Yes, you can think of it that way. That's the same idea behind a >> paired t-test. Each individual might have an inherently different >> baseline (the wt in this case), and you are just interested in seeing >> if the treatment has the same relative effect in both samples. Since >> the difference in baseline between samples is probably not >> interesting, you subtract it out. >> >> Best, >> >> Jim >> >> >> >>> Thanks, >>> Sanjat >>> >>> _______________________________________________ >>> Bioconductor mailing list >>> Bioconductor at stat.math.ethz.ch >>> https://stat.ethz.ch/mailman/listinfo/bioconductor >>> Search the archives: >>> http://news.gmane.org/gmane.science.biology.informatics.conductor >>> >> >> > -- James W. MacDonald, M.S. Biostatistician Affymetrix and cDNA Microarray Core University of Michigan Cancer Center 1500 E. Medical Center Drive 7410 CCGC Ann Arbor MI 48109 734-647-5623