design in mixed ref and dye-swap experiment
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@silvano-piazza-986
Last seen 6.8 years ago
Hello to everyones, the experiments that I have to consider is very simple: I want to find significant genes between 2 conditions A and B, but I have only few experiment so I have to collect both ref versus conditions (A or B) either dye swap experiment (A versus B and B versus A) so targets is SlideNumber Cy3 Cy5 array1 ref A array2 ref B array3 ref B array4 ref B array5 A B array6 B A of course array5 and array6 are the dye-swap. So to design the procedure, I follow the LIMMA user guide (by Gordon Smith), Chapter 14.5 Weaver Mutant Data. so >design <- modelMatrix(targets, ref = "ref") Found unique target names: B A ref >design A B array1 0 1 array2 1 0 array3 1 0 array4 1 0 array5 -1 1 array6 1 -1 >fit <- lmFit(MA,design) >cont.matrix <- makeContrasts(A.B=A-B,levels=design,weight=MA$weights) >fit2 <- contrasts.fit(fit, cont.matrix) > fit2 <- eBayes(fit2) >topTable(fit2,adjust.method="fdr") ....omissis... M A t P.Value B 209 3.801460 6.538782 8.315672 1.0000000 -4.209468 2328 1.184194 7.343676 6.717978 1.0000000 -4.228492 7877 1.904360 6.504330 6.114349 1.0000000 -4.239110 27187 -4.0759493.771499 -5.783558 1.0000000 -4.246099 3709 3.434542 3.467492 5.639159 1.0000000 -4.249459 7561 2.002753 5.159913 5.616194 1.0000000 -4.250013 7130 2.580527 3.863867 5.600047 1.0000000 -4.250405 19983 -2.1176246.836539 -5.567882 1.0000000 -4.251194 So all genes have P.Value equal to 1!!!!!! in previous posts I read that this happen when you have to consider multivariate test, which i don't known how to manage..., but anyway 1) Am I doing something wrong in the design? 2) Am I doing something wrong in the subsequent evaluation steps? Any ideas Thank you to all Silvano Dr.Silvano Piazza LNCIB, Area Science Park, Padriciano 99 Trieste, ITALY Tel. +39040398992 Fax +39040398990
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Naomi Altman ★ 6.0k
@naomi-altman-380
Last seen 8 weeks ago
United States
Dear Silvano, As I have indicated elsewhere on this list, the "p-values" reported by TopTable are actually "q-values". Hence, if you have fewer "significant" genes than expected by chance under the null hypothesis, the reported p-value is 1.0. e.g. Suppose you have 1000 genes. Then if the number of genes significant at alpha% is less than 1000*alpha for each alpha, your TopTable p-value will be 1.0 (i.e. all of the significant genes are estimated to be false positives). Your experiment design is needlessly complex and also wasteful. If you have only 2 conditions, you should do one of the following: hybridize both conditions to every array (in dye-swap pairs) with no technical replicates (This is most efficient) use a reference design with the reference sample always in the same channel. (This is simplest, but has 1/2 the efficiency.) Mixing these 2 designs, especially with a mix of biological and technical replicates needlessly complicates your analysis. It also requires a mixed model ANOVA to take into account the different levels of replication. --Naomi At 10:28 AM 3/25/2005, Silvano Piazza wrote: >Hello to everyones, >the experiments that I have to consider is very simple: > >I want to find significant genes between 2 conditions A and B, but I have >only few experiment so I have to collect both ref versus conditions (A or >B) either dye swap experiment (A versus B and B versus A) > >so targets is >SlideNumber Cy3 Cy5 >array1 ref A >array2 ref B >array3 ref B >array4 ref B >array5 A B >array6 B A > >of course array5 and array6 are the dye-swap. > >So to design the procedure, I follow the LIMMA user guide (by Gordon >Smith), Chapter 14.5 Weaver Mutant Data. > >so > >design <- modelMatrix(targets, ref = "ref") > Found unique target names: > B A ref > >design > A B > array1 0 1 > array2 1 0 > array3 1 0 > array4 1 0 > array5 -1 1 > array6 1 -1 > >fit <- lmFit(MA,design) > >cont.matrix <- makeContrasts(A.B=A-B,levels=design,weight=MA$weights) > >fit2 <- contrasts.fit(fit, cont.matrix) > > fit2 <- eBayes(fit2) > >topTable(fit2,adjust.method="fdr") > ....omissis... > M A t > P.Value B > 209 3.801460 6.538782 8.315672 1.0000000 > -4.209468 > 2328 1.184194 7.343676 6.717978 1.0000000 -4.228492 > 7877 1.904360 6.504330 6.114349 1.0000000 -4.239110 > 27187 -4.0759493.771499 -5.783558 1.0000000 -4.246099 > 3709 3.434542 3.467492 5.639159 1.0000000 -4.249459 > 7561 2.002753 5.159913 5.616194 1.0000000 -4.250013 > 7130 2.580527 3.863867 5.600047 1.0000000 -4.250405 > 19983 -2.1176246.836539 -5.567882 1.0000000 -4.251194 >So all genes have P.Value equal to 1!!!!!! >in previous posts I read that this happen when you have to consider >multivariate test, which i don't known how to manage..., but anyway > >1) Am I doing something wrong in the design? >2) Am I doing something wrong in the subsequent evaluation steps? >Any ideas > > > >Thank you to all > >Silvano > > > > > > > > > > > >Dr.Silvano Piazza >LNCIB, >Area Science Park, >Padriciano 99 >Trieste, ITALY >Tel. +39040398992 >Fax +39040398990 > >_______________________________________________ >Bioconductor mailing list >Bioconductor@stat.math.ethz.ch >https://stat.ethz.ch/mailman/listinfo/bioconductor Naomi S. Altman 814-865-3791 (voice) Associate Professor Bioinformatics Consulting Center Dept. of Statistics 814-863-7114 (fax) Penn State University 814-865-1348 (Statistics) University Park, PA 16802-2111
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Dear Naomi First of all, thank you very much for your answer. > > > As I have indicated elsewhere on this list, the "p-values" reported by > TopTable are actually "q-values". Hence, if you have fewer > "significant" genes than expected by chance under the null hypothesis, > the reported p-value is 1.0. > > e.g. Suppose you have 1000 genes. Then if the number of genes > significant at alpha% is less than 1000*alpha for each alpha, your > TopTable p-value will be 1.0 (i.e. all of the significant genes are > estimated to be false positives). > that's very clear now, thanks again. > Your experiment design is needlessly complex and also wasteful. If > you have only 2 conditions, you should do one of the following: > > hybridize both conditions to every array (in dye-swap pairs) with no > technical replicates (This is most efficient) > use a reference design with the reference sample always in the same > channel. (This is simplest, but has 1/2 the efficiency.) > > Mixing these 2 designs, especially with a mix of biological and > technical replicates needlessly complicates your analysis. It also > requires a mixed model ANOVA to take into account the different levels > of replication. > Yes, I know I know.... but unfortunately I could not decide, in this case, how to make the experiments, so my situation is: these experiments are available at the moment and I have to find out DE genes, and only for this reason I was wondering if there is any correct methods to work in "mixed" (exp vs ref and dye-swap) design, thats means to extract more information that it is possible. Thank you Silvano > --Naomi > > At 10:28 AM 3/25/2005, Silvano Piazza wrote: >> Hello to everyones, >> the experiments that I have to consider is very simple: >> >> I want to find significant genes between 2 conditions A and B, but I >> have only few experiment so I have to collect both ref versus >> conditions (A or B) either dye swap experiment (A versus B and B >> versus A) >> >> so targets is >> SlideNumber Cy3 Cy5 >> array1 ref A >> array2 ref B >> array3 ref B >> array4 ref B >> array5 A B >> array6 B A >> >> of course array5 and array6 are the dye-swap. >> >> So to design the procedure, I follow the LIMMA user guide (by Gordon >> Smith), Chapter 14.5 Weaver Mutant Data. >> >> so >> >design <- modelMatrix(targets, ref = "ref") >> Found unique target names: >> B A ref >> >design >> A B >> array1 0 1 >> array2 1 0 >> array3 1 0 >> array4 1 0 >> array5 -1 1 >> array6 1 -1 >> >fit <- lmFit(MA,design) >> >cont.matrix <- makeContrasts(A.B=A-B,levels=design,weight=MA$weights) >> >fit2 <- contrasts.fit(fit, cont.matrix) >> > fit2 <- eBayes(fit2) >> >topTable(fit2,adjust.method="fdr") >> ....omissis... >> M A t >> P.Value B >> 209 3.801460 6.538782 8.315672 1.0000000 >> -4.209468 >> 2328 1.184194 7.343676 6.717978 1.0000000 -4.228492 >> 7877 1.904360 6.504330 6.114349 1.0000000 -4.239110 >> 27187 -4.0759493.771499 -5.783558 1.0000000 -4.246099 >> 3709 3.434542 3.467492 5.639159 1.0000000 -4.249459 >> 7561 2.002753 5.159913 5.616194 1.0000000 -4.250013 >> 7130 2.580527 3.863867 5.600047 1.0000000 -4.250405 >> 19983 -2.1176246.836539 -5.567882 1.0000000 -4.251194 >> So all genes have P.Value equal to 1!!!!!! >> in previous posts I read that this happen when you have to consider >> multivariate test, which i don't known how to manage..., but anyway >> >> 1) Am I doing something wrong in the design? >> 2) Am I doing something wrong in the subsequent evaluation steps? >> Any ideas >> >> >> >> Thank you to all >> >> Silvano >> >> >> >> >> >> >> >> >> >> >> >> Dr.Silvano Piazza >> LNCIB, >> Area Science Park, >> Padriciano 99 >> Trieste, ITALY >> Tel. +39040398992 >> Fax +39040398990 >> >> _______________________________________________ >> Bioconductor mailing list >> Bioconductor@stat.math.ethz.ch >> https://stat.ethz.ch/mailman/listinfo/bioconductor > > Naomi S. Altman 814-865-3791 (voice) > Associate Professor > Bioinformatics Consulting Center > Dept. of Statistics 814-863-7114 (fax) > Penn State University 814-865-1348 (Statistics) > University Park, PA 16802-2111 > > > Dr.Silvano Piazza LNCIB, Area Science Park, Padriciano 99 Trieste, ITALY Tel. +39040398992 Fax +39040398990
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@gordon-smyth
Last seen 8 minutes ago
WEHI, Melbourne, Australia
> Date: Tue, 29 Mar 2005 11:22:22 +0200 > From: Silvano Piazza <piazza@lncib.it> > Subject: Re: [BioC] design in mixed ref and dye-swap experiment > To: Naomi Altman <naomi@stat.psu.edu> > Cc: bioconductor@stat.math.ethz.ch > Message-ID: <107fbaba4c9ce4b7c1ecc6051aac92ff@lncib.it> > Content-Type: text/plain; charset=US-ASCII; format=flowed > > Dear Naomi > > First of all, thank you very much for your answer. >> >> >> As I have indicated elsewhere on this list, the "p-values" reported by >> TopTable are actually "q-values". Hence, if you have fewer >> "significant" genes than expected by chance under the null hypothesis, >> the reported p-value is 1.0. >> >> e.g. Suppose you have 1000 genes. Then if the number of genes >> significant at alpha% is less than 1000*alpha for each alpha, your >> TopTable p-value will be 1.0 (i.e. all of the significant genes are >> estimated to be false positives). >> > > that's very clear now, thanks again. > > >> Your experiment design is needlessly complex and also wasteful. If >> you have only 2 conditions, you should do one of the following: >> >> hybridize both conditions to every array (in dye-swap pairs) with no >> technical replicates (This is most efficient) >> use a reference design with the reference sample always in the same >> channel. (This is simplest, but has 1/2 the efficiency.) >> >> Mixing these 2 designs, especially with a mix of biological and >> technical replicates needlessly complicates your analysis. It also >> requires a mixed model ANOVA to take into account the different levels >> of replication. >> > > Yes, I know I know.... > but unfortunately I could not decide, in this case, how to make the > experiments, so my situation is: these experiments are available at the > moment and I have to find out DE genes, and only for this reason I was > wondering if there is any correct methods to work in "mixed" (exp vs > ref and dye-swap) design, thats means to extract more information that > it is possible. You analysis is already correct, given the arrays that you have. If you are expecting to see differential expression here but aren't, you might revisit the pre-processing and QC steps for this data. Good pre-processing can make a spectactular difference to differential expression results. Gordon > Thank you > > Silvano > >> --Naomi >> >> At 10:28 AM 3/25/2005, Silvano Piazza wrote: >>> Hello to everyones, >>> the experiments that I have to consider is very simple: >>> >>> I want to find significant genes between 2 conditions A and B, but I >>> have only few experiment so I have to collect both ref versus >>> conditions (A or B) either dye swap experiment (A versus B and B >>> versus A) >>> >>> so targets is >>> SlideNumber Cy3 Cy5 >>> array1 ref A >>> array2 ref B >>> array3 ref B >>> array4 ref B >>> array5 A B >>> array6 B A >>> >>> of course array5 and array6 are the dye-swap. >>> >>> So to design the procedure, I follow the LIMMA user guide (by Gordon >>> Smith), Chapter 14.5 Weaver Mutant Data. >>> >>> so >>> >design <- modelMatrix(targets, ref = "ref") >>> Found unique target names: >>> B A ref >>> >design >>> A B >>> array1 0 1 >>> array2 1 0 >>> array3 1 0 >>> array4 1 0 >>> array5 -1 1 >>> array6 1 -1 >>> >fit <- lmFit(MA,design) >>> >cont.matrix <- makeContrasts(A.B=A-B,levels=design,weight=MA$weights) Why are you using 'weights='? That is not an argument for makeContrasts(). Gordon >>> >fit2 <- contrasts.fit(fit, cont.matrix) >>> > fit2 <- eBayes(fit2) >>> >topTable(fit2,adjust.method="fdr") >>> ....omissis... >>> M A t >>> P.Value B >>> 209 3.801460 6.538782 8.315672 1.0000000 >>> -4.209468 >>> 2328 1.184194 7.343676 6.717978 1.0000000 -4.228492 >>> 7877 1.904360 6.504330 6.114349 1.0000000 -4.239110 >>> 27187 -4.0759493.771499 -5.783558 1.0000000 -4.246099 >>> 3709 3.434542 3.467492 5.639159 1.0000000 -4.249459 >>> 7561 2.002753 5.159913 5.616194 1.0000000 -4.250013 >>> 7130 2.580527 3.863867 5.600047 1.0000000 -4.250405 >>> 19983 -2.1176246.836539 -5.567882 1.0000000 -4.251194 >>> So all genes have P.Value equal to 1!!!!!! >>> in previous posts I read that this happen when you have to consider >>> multivariate test, which i don't known how to manage..., but anyway >>> >>> 1) Am I doing something wrong in the design? >>> 2) Am I doing something wrong in the subsequent evaluation steps? >>> Any ideas >>> >>> >>> >>> Thank you to all >>> >>> Silvano >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> Dr.Silvano Piazza >>> LNCIB, >>> Area Science Park, >>> Padriciano 99 >>> Trieste, ITALY >>> Tel. +39040398992 >>> Fax +39040398990 >>> >>> _______________________________________________ >>> Bioconductor mailing list >>> Bioconductor@stat.math.ethz.ch >>> https://stat.ethz.ch/mailman/listinfo/bioconductor >> >> Naomi S. Altman 814-865-3791 (voice) >> Associate Professor >> Bioinformatics Consulting Center >> Dept. of Statistics 814-863-7114 (fax) >> Penn State University 814-865-1348 (Statistics) >> University Park, PA 16802-2111 >> >> >> > Dr.Silvano Piazza > LNCIB, > Area Science Park, > Padriciano 99 > Trieste, ITALY > Tel. +39040398992 > Fax +39040398990
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