Jump to content

  • Log in with Facebook Log in with Twitter Log in with Windows Live Log In with Google      Sign In   
  • Create Account

Submit your paper to J Biol Methods today!
- - - - -

Calculating Standard Errors and Normalising to Transgene

standard errorr transgene normalise

  • Please log in to reply
1 reply to this topic

#1 Final year phd at wits end

Final year phd at wits end


  • Active Members
  • Pip
  • 7 posts

Posted 03 February 2013 - 06:56 AM


Maths has never been a particular strong point and I need someone to explain this to me in the most simple way possible.....and I'm relatively new to RT-PCR.

There are three different problems relating to two different data sets in this post:

Problem 1. I have measured the Ct values for a number of genes of interest and a reference gene in the left atria and right atria of both mutant and wildtype samples (i.e. 4 groups). I have then compared left and right atria between wild type; left and right atria between mutant; left atria between mutant and wildtype and finally, right atria between mutant and wildtype.

I have calculated the fold change for my genes of interest using both the Delta delta Ct method (i.e. not correcting for PCR efficiency, which is what the previous 'expert' in the lab did) and correcting for PCR efficiency using REST2009 (which provides me with a whisker-box plot and errors). But my boss insists on wanting to see standard errors, the errors on the whisker plots are huge so I need to show him something that is comparable to the previous 'expert'. So I have taken the DeltaDelta Ct values (i.e. % of Ref Gene), calculated the average and the standard error using these values and plotted them. Is this correct?

He said that he wanted to see the standard error for the fold change, but as not all my samples are properly paired (i.e. mutant and wildtype are different samples) I have calculated the fold change using the mean Ct values for the group. So as far as I can see there is no correct way of applying errors to this value?

Problem 2: I have measured Ct values for a number of genes of interest and a reference gene in mutant and wildtype samples (i.e. 2 groups). Having calculated the fold change for my genes of interest using both the Delta delta Ct method and a method that factors in PCR efficiency (REST2009), my bosses want me to correct for differential transgene expression. Has anyone done this? If so, how? Again, because the Ct values are logarithmic I'm not sure how to go about it. Should I inverse log the DdCt for the GOI, and divide by the inverse log of the DdCt for the transgene, then re-log the value?

I'm not sure if I'm being asked to do the impossible? And the previous 'experts' methods are questionable at best!

Problem 3: more generic. The 'expert' also normalises the DdCt of all samples to a calibrator sample, which is a wt sample selected at random. Is this normal practise? To do this, they calculate 2^(DdCt sample-DdCt calibrator)?? I could understand using the average of the WT group but not one sample?

Any advice would be gratefully received. I'm going around in circles!


#2 doxorubicin



  • Moderators
  • PipPipPipPipPipPipPipPipPipPip
  • 193 posts

Posted 04 February 2013 - 08:40 AM

1) Generally, these calculations can be carried out by the qPCR program software if you have a reasonably modern machine. The standard errors associated with a ratio are usually larger than most bosses like to see, though. By hand, you'll have to do something like a student's t-test to calculate 95% confidence intervals of these ratios...I think graphpad should be useful for this. I personally like to do quantitative pcr with a standard curve and report quantities (ng) for each gene, each with its own error bars to avoid having to plot these ratios....but sometimes it is avoidable.

2) For the transgene bit, I don't exactly follow....so if there is one sample with 2X transgene, you just divide the final quantities by 2?

3) Using the average is probably going to make your figure look nicer, since the wild-type control will be at 100% rather than at 95% or whatever.

Home - About - Terms of Service - Privacy - Contact Us

©1999-2013 Protocol Online, All rights reserved.