[BioC] HTqPCR

[Heidi Dvinge]

Hi Deborah,

> Good morning Heidi,
>
> Yes, the order of my samples is different in my qPCRset compared to my
> files_essai$Treatment. Do I have to order them in the same way ?
>
The order doesn't have to be the same, as long as you make sure you always
remember re-order one of them when you use them both in the same function,
such as the limmaCtData examples you provided last time.

I don't trust myself to always remember such things (or to do it
correctly!), so I always order the samples in the same way ;). But that's
not a requirement.
>
<snip>
>
> And with normalizeCtData I obtained :
>> deltaCtnorm <- normalizeCtData(essai.cat, norm =
>> "deltaCt",deltaCt.genes=  c("gene85", "gene86", "gene87", "gene88",
>> "gene89", "gene90","gene91"))
> Calculating deltaCt values
>         Using control gene(s): gene85 gene86 gene87 gene88 gene89 gene90
> gene91
>         Card 1: Mean=20.93      Stdev=2.74
>         Card 2: Mean=21.58      Stdev=2.73
>         Card 3: Mean=21.73      Stdev=2.81
>         Card 4: Mean=20.96      Stdev=2.73
>         Card 5: Mean=21.69      Stdev=2.69
>         Card 6: Mean=21.73      Stdev=2.83
> I chose this method because my director gave me a file where he has chosen
> only seven housekeeper genes on the eight (one of them has different Cp
> results in almost each sample) so I did the same thing.
>
In this case the values of all these housekeepers do look robust across
your samples, although the Ct values are higher than for typical
housekeepers such as b-actin. (Which BTW isn't necessarily a bad thing).
So as long as you/your boss is happy with it, I guess that's fine.

> And about the other normalization methods, I tried them only for seeing
> the difference between them.
> I did :
>>par(mfrow=c(3,2))
>>plot(exprs(essai),exprs(essai_g.mean),pch=20,main="Normalisation avec
>> geometric.mean",col=rep(brewer.pal(6,"Spectral"),each=96))
>>plot(exprs(essai),exprs(essai_scale.rank),pch=20,main="Normalisation avec
>> scale.rankinvariant", col=rep(brewer.pal(6,"Spectral"), each=96))
>>plot(exprs(essai),exprs(essai_deltaCt),pch=20,main="Normalisation avec
>> deltaCt",col=rep(brewer.pal(6,"Spectral"),each=96))
>>plot(exprs(essai),exprs(essai_q.norm),pch=20,main="Normalisation avec
>> quantile",col=rep(brewer.pal(6,"Spectral"),each=96))
>>plot(exprs(essai),exprs(essai_norm.rank),pch=20,main="Normalisation avec
>> norm.rankinvariant", col=rep(brewer.pal(6,"Spectral"), each=96))
>
In your case you have relatively few genes (96), which may not be quite
enough for some of the methods. If the deltaCt-normalised data doesn't
look too discrepant from all the other methods you're probably fine.

> And then I wanted to compare only one of the sample as your example and I
> used abline() but it didn't work.
>>plot(exprs(DU145)[,3],exprs(essai_g.mean)[,3],pch=20,col="magenta")
>>abline(exprs(DU145)[,3],exprs(essai_scale.rank)[,3],pch=20,col="blue")
>>abline(exprs(DU145)[,3],exprs(essai_deltaCt)[,3],pch=20,col="purple")
> I did'nt get error message but there was only one plot. I also changed the
> xlim value...
>
Well, you only use one plot() command, hence only one plot gets produced.
Do you perhaps want to add data from multiple objects using points()?

plot(exprs(DU145)[,3],exprs(essai_g.mean)[,3],pch=20,col="magenta")
points(exprs(DU145)[,3],exprs(essai_norm.rank)[,3],pch=20,col="blue")
...etc..

> About the p.value, I tried to plot them as you had suggest me to do.
> Here are the p.values of each test.
>> MWTEST<-read.csv("essai_3h_CT_mwest1.csv", sep = ";",
>> dec=",",header=TRUE)
>> MWTEST$p.value
>  [1] 0.2452781 0.2452781 0.2452781 1.0000000 0.2452781 1.0000000 0.2452781
> 0.2452781 0.2452781 0.2452781 0.2452781 0.2452781 0.2452781 0.2452781
> 0.2452781
> [16] 0.2452781 1.0000000 0.2452781 0.2452781 0.6985354 0.2452781 0.2452781
> 0.2452781 0.6985354 0.2452781 0.6985354 0.2452781 0.6985354 0.2452781
> 0.2452781
> [31] 0.2452781 0.2452781 0.2452781 0.2452781 0.2452781 0.2452781 0.2452781
> 0.2452781 0.2452781 1.0000000 0.2452781 1.0000000 0.2452781 0.2452781
> 0.2452781
> [46] 0.2452781 0.2452781 0.2452781 1.0000000 0.2452781 0.6985354 0.2452781
> 0.6985354 0.2452781 0.2452781 0.2452781 0.2452781 0.2452781 1.0000000
> 0.2452781
> [61] 0.2452781 0.2452781 0.2452781 0.2452781 0.2452781 1.0000000 0.2452781
> 0.2452781 0.2452781 0.6985354 0.2452781 0.2452781 0.2452781 0.2452781
> 0.2452781
> [76] 0.2452781 0.2452781 0.2452781 0.2452781 1.0000000 0.2452781 0.2452781
> 0.2452781 0.2452781
>
>> TTEST<-read.csv("essai_3h_CT_ttestFINAL.csv",sep=";",dec=",")
>> TTEST$p.value
>  [1] 0.0000445916 0.0007203397 0.0011614062 0.0013499661 0.0021628924
> 0.0026335148 0.0028568178 0.0031900227 0.0047685877 0.0048604855
> 0.0058520334
> [12] 0.0061487450 0.0101863324 0.0101863324 0.0101863324 0.0101863324
> 0.0101863324 0.0101863324 0.0101863324 0.0101863324 0.0118721483
> 0.0130066108
> [23] 0.0166399589 0.0169334479 0.0170523949 0.0209596603 0.0364914159
> 0.0411074450 0.0427607922 0.0448349199 0.0476675408 0.0494563227
> 0.0512518277
> [34] 0.0514256601 0.0572253657 0.0625769920 0.0656449529 0.0797613537
> 0.0804359345 0.0819791697 0.0821601769 0.0909080090 0.0918788852
> 0.0986962901
> [45] 0.0993682993 0.1208709609 0.1261907225 0.1331849915 0.1338834931
> 0.1590798074 0.1611312123 0.1657960803 0.1718536210 0.1844114022
> 0.2035267116
> [56] 0.2092967748 0.2111576859 0.2192894000 0.2223339619 0.2393817321
> 0.2416781885 0.2479843103 0.2570206534 0.2570800840 0.2610404909
> 0.2755461365
> [67] 0.2886380998 0.3133822666 0.4574691996 0.4790123864 0.4963391483
> 0.5780428714 0.5827604076 0.6029711831 0.6738622120 0.6905548966
> 0.7800699292
> [78] 0.8045384637 0.8399336418 0.9347460142 0.9531859762 0.9719743053
> 0.9774759886 0.9934469629
>
>> LIMMATEST<-read.csv("essai_limmaFINAL.csv",sep=";",dec=",")
>> LIMMATEST$X3h.CT.p.value
>  [1] 4.538818e-01 9.722424e-01 9.681357e-01 1.478327e-02 9.765722e-01
> 1.224899e-01 4.579647e-01 1.035570e-03 6.190137e-02 6.862192e-03
> 1.828032e-02
> [12] 2.354413e-02 2.027634e-01 3.924912e-03 1.245438e-03 9.538478e-07
> 5.915158e-04 2.714424e-01 2.449646e-04 9.943747e-02 8.115928e-02
> 1.014429e-01
> [23] 1.730959e-04 2.283943e-01 4.106429e-02 8.292733e-01 7.384857e-01
> 9.053543e-04 3.031922e-05 4.381594e-02 8.697809e-05 1.730959e-04
> 9.949012e-01
> [34] 3.584419e-03 3.713434e-04 7.691588e-01 1.336464e-01 3.141131e-01
> 3.500428e-02 5.853026e-06 6.234777e-02 1.096195e-01 5.065608e-01
> 1.425943e-02
> [45] 7.720779e-01 2.074906e-05 2.596116e-04 6.080595e-02 6.472036e-01
> 1.730959e-04 6.924510e-05 8.243564e-03 2.010885e-01 9.367344e-01
> 2.535135e-01
> [56] 9.788777e-01 1.730959e-04 6.717992e-02 1.041109e-01 3.951307e-04
> 1.152792e-01 2.552804e-04 8.276034e-01 6.578508e-03 3.226937e-02
> 1.730959e-04
> [67] 1.730959e-04 1.052211e-01 6.826300e-05 1.730959e-04 2.939883e-01
> 1.116254e-02 2.997326e-01 5.701757e-02 2.319393e-03 3.023084e-02
> 8.304573e-01
> [78] 4.892519e-01 6.178556e-01 4.863336e-01 8.506124e-02 1.730959e-04
> 1.380221e-01 7.850957e-03
>
> So I did :
> par(mfrow=c(1,3))
> plot(LIMMATEST$X3h.CT.p.value,col="green",main="p-value LIMMA")
> plot(TTEST$p.value,col="red",main="p-value Student")
> plot(MWTEST$p.value,col="blue",main="p-value Mann-Whitney")
> Considering the graphs that I obtained, I can say that the p.values don't
> follow a general trend... So there is a real problem somewhere... Is that
> alright ?
>
I'm sorry for not being clear, I meant plot them against each other. For
example plot(TTEST$p.value, MWTEST$p.value).

> Actually, I have a question about the "Summary" component of the
> limmaCtData : how do it do to calculate if it is up- or down-regulation ?
> Because when I calculated the expression (2^ddCt) of the gene18, I
> obtained expression = 11.41.
> So the gene18 is up-regulated whereas in the "Summary" there is no change.
> Is there a link between the "Summary" and the expression ?
>
In Summary, -1/0/1 should correspond to down-regulation/no
difference/up-regulation respectively.

The summary is linked to the expression, but it requires that the change
in expression is statistically significant at p<0.05. Otherwise it's just
"0" in the output.

Best,
\Heidi

> Thank you again for your help and your advice,
>
> Deborah.
>