wild type and knockout gene's CT value always different (1-2CT)and this may effect the experiment result ,it's not? thanks
Yes because you are normalizing to it. You will need a gene that is stably expressed across your wild-type and KO.
Remember each PCR cycling is a doubling process where 1 cycle= 2 fold change, i.e. the difference between 500ng RNA input cDNA will be 1 ct later than 1000ng RNA cDNA input. So a 1 ct difference is a 2-fold change whilst a 2 cycle is 4-fold change (2^X).
Such variation will introduce errors in fold change calculation and is not uncommon.
http://www.gene-quantification.com/hkg.html
http://www.ncbi.nlm.nih.gov/pubmed/24423493
I agree with Can. you will need to perform some tests to identify a set of genes (ideally, more than one) that are not altered by the knockout of your gene of interest. This is key to obtaining reliable results. Some genes you can look at are described in the attached article (these are for human, but I assume a reasonable degree of similarity with the mouse). For my study, I found that RAB7 was suitable as a control for qPCR. Of course you will first want to search the literature to see if anyone has already identified a good control gene for the mice you are using.
Best.
Article Human housekeeping Genes, revisited
I agree with everyone you have to consider one good housekeeping gene, but have you considered to verify your previous procedures? RNA from both cell samples should have good purity ratios and similar values in order to have good results in RT-qPCR. good luck!
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