Hi
I am new to luciferase assays. I am finding it difficult how to normalize data.
I have psicheck2 vector which has both renilla and firefly. I ligated my gene of interest into 3'UTR of renilla gene.I need to see the reduction in renilla level, While firefly is working as control.
After performing luc assay i got readings using 20/20 glomax luminometer. First reading was of firefly and second was of renilla.
Can any one please help me out that how i can i normalize my data or how can i calculate the reduction in renilla level.
Thanks in advance
normalization of dual luciferase assay
Started by ariaz, Oct 22 2010 01:04 PM
4 replies to this topic
#1
Posted 22 October 2010 - 01:04 PM
#2
Posted 27 October 2010 - 07:30 AM
In your case, for each well I will calculate the ration Renilla/Firefly and then compare those ratios among your experimental groups. You probably get lost because it is usally the Firefly plasmid being modified and the Renilla one used as normalizator, so the glomax luminometer would give the calculations as Firefly/Renilla. In your case, you have to do Renilla/Firefly
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#3
Posted 27 October 2010 - 10:16 AM
Thanks alot:
I was going to do in the same way but some one said that i should first calculate ratio Renilla/Firefly and then devide the renilla values with the averages of these ratios (Renilla/renilla firefly reatio average). I dont know it is right or not. What do you say.
Thanks again
I was going to do in the same way but some one said that i should first calculate ratio Renilla/Firefly and then devide the renilla values with the averages of these ratios (Renilla/renilla firefly reatio average). I dont know it is right or not. What do you say.
Thanks again
96well, on 27 October 2010 - 07:30 AM, said:
In your case, for each well I will calculate the ration Renilla/Firefly and then compare those ratios among your experimental groups. You probably get lost because it is usally the Firefly plasmid being modified and the Renilla one used as normalizator, so the glomax luminometer would give the calculations as Firefly/Renilla. In your case, you have to do Renilla/Firefly
#4
Posted 28 October 2010 - 05:16 AM
Imagine you have a single well in your experiment, so a single end-point
Renilla = 150
Firefly = 8000
R/F = 0.01875
average R/F = 0.01875
R/(averageR/F) = 8000
Is like if you divide Renilla/Renilla/Firefly: you erase Renilla values and obtain just the Firefly, I don't see how this could be useful.
Renilla = 150
Firefly = 8000
R/F = 0.01875
average R/F = 0.01875
R/(averageR/F) = 8000
Is like if you divide Renilla/Renilla/Firefly: you erase Renilla values and obtain just the Firefly, I don't see how this could be useful.
Nature magazine. Do you qualify for a free subscription?
#5
Posted 28 October 2010 - 06:48 AM
96 well,
Yes you are right.
I understand it now.I dont know y it seemed to complicated to me.
Thank you very much.
Yes you are right.
I understand it now.I dont know y it seemed to complicated to me.
Thank you very much.













