DCts Vs Folds: statistical analysis? - (Oct/26/2009 )

Hi,
i've an unusual question to ask. 99% of my job is real time PCR. I have my Cts, my normalizer and so on. i calculate DCt by subtraction of Ct sample - Ct normalizer. i then compare groups using DDCt method (DCt of sample-DCt control) and with the 2^-DDCt i have my fold induction.
Question is how do u calculate statistics? i mean do you calculate significativity among groups (treated Vs control) using the DCt s or with folds? same thing for SEM or SD?
i always do my statistics on DCts but now somebody tells me i have to do it on FOLDs. I'm getting confused....
thanks

Do you get different answers/conclusions using folds?

Fizban on Oct 26 2009, 10:21 AM said:

I do it on the final calculation, which would be the fold increase. That's how you end up reporting it in the data, and by virtue of applying the same calculation to each ddCt value, there should be no difference in the stats between ddCt and folds. The means will be different, but the variance of those means will be the same.

fishdoc on Oct 26 2009, 11:54 PM said:

mmmmmm that's interesting. i can agree with u but how do u manage DDCt? i mean i have control and treated samples. i can do my stats on DCts because i have for example 12 controls and 25 treated smpls. in this case i can calculate SD, mean and so on. DDCt is something i decide how to calculate which means that DDCt mean of controls is zero (in fold it will be 1) and then i calculate fold of treated samples. how do u calculated SD of controls if u fix it to zero?

Fizban on Oct 27 2009, 02:03 AM said:

Take each individual Ct value for your control and subtract the mean of the control Cts. For example, if you had a Ct of 12.0, 12.1, and 12.2, you'd average 12.1. Then you'd calculate the variance by 12.0-12.1, 12.1-12.1, and 12.2-12.1. You do end up with a mean difference of zero, but you have the variation about the mean to work with.