Etchevers:Notebook/STRA6 in eye development/2009/05/07
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How to express ChIP results
I will fill in the details later on what we have done when I can go through my desk papers. Meanwhile, yesterday and today Sadaf was analyzing her data and writing her presentation of what she's carried out. We can all be pretty proud.
For each immunoprecipitation, from the concentration of DNA in the "total input" sample, we can calculate how much total input DNA we started with in a given IP.
In our case, out of the 1.1 mL or so of chromatin, 50 μL were removed for TI isolation. This was purified, resuspended in 60 μL, and the concentration determined to be 68 ng/μL. That means a yield of 4080 ng for the original 50 μL, or 81.6 ng/μL for the 1 mL of chromatin.
Since we took 150 μL of that same preparation per IP (3 IP's first round on the 30th April, 4 IP's second round on 5/5), each one of these * 81.6 ng/μL = 12.24 μg DNA per IP (the REAL "total input").
For each amplicon, therefore, it is possible to compare the Ct of the TI well with those of each amplicon if you remember that in the TI well you put a fraction of the TI (2.5 ng/μL instead of 68 ng/μL, so divide by 27.2) and you had put 3x as much DNA total input into the IP than into what you extracted from the TI fraction (150 μL vs 50 μL). Therefore, the equation to calculate the % of TI in pg/μL is:
IP (amplicon) = 2.5ng/ul / 27.2 / 3 / 2^(Ct IP - Ct TI) * 1000