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Not Telling
Sort of, people do use it, and it can work, but there is still no selection for the other plasmid, so it might get degraded or methylated, while the selection plasmid is maintained.
#108839 is there a method to set up stable transfectants without drug selection
Posted
on 04 May 2011 - 02:09 PM
#104979 why phosphorylated protein relative to total protein?
Posted
on 28 March 2011 - 06:18 AM
bettan, on 28 March 2011 - 05:54 AM, said:
I was wondering, why do i have to calculate the ratio between p-protein relative to total protein? many i have talked to just say: thats how you do wheen you present you data. The reason i ask, wheen i try to get my manuscript published someone in the editorial board points out, i must present my data in this way. But i dont see the advantage to present my data like this. Can you please give me you reflections on this
/ Bettan
/ Bettan
I'm going to assume you are trying to show that a certain Treatment/Condition induces phosphorylation.
The reason to calculate the ratio between p-protein and total protein is to show that your treatment is specifically inducing phosphorylation and that the increase in p-protein is not due to differences in total protein (which could be due to induction of protein expression rather than phosphorylation).
#100299 Molecular Cell Biology
Posted
on 11 February 2011 - 05:52 AM
Um, you should probably do your own homework!
#98771 Percentage dilution
Posted
on 27 January 2011 - 08:15 AM
The general concentration equation is:
C1*V1=C2*V2
Where C is the concentration and V is the volume, 1 refers to starting conc/volume and 2 refers to the finishing volume. Basically it means that the concentration of the reagent multiplied by it's volume is always the same.
So for your question the starting concentration is 10% (so that is 0.1) and the end concentration is 0.3% (so that is 0.003) The end volume is 10ml
So in the equation we get:
0.1 x V1 = 0.003 x 10ml = 0.03
V1 = 0.03/0.1 = 0.3ml = 300ul
So you'd need to add 300ul of your 10% solution to 9700ul (about 10ml) of your diluent.
C1*V1=C2*V2
Where C is the concentration and V is the volume, 1 refers to starting conc/volume and 2 refers to the finishing volume. Basically it means that the concentration of the reagent multiplied by it's volume is always the same.
So for your question the starting concentration is 10% (so that is 0.1) and the end concentration is 0.3% (so that is 0.003) The end volume is 10ml
So in the equation we get:
0.1 x V1 = 0.003 x 10ml = 0.03
V1 = 0.03/0.1 = 0.3ml = 300ul
So you'd need to add 300ul of your 10% solution to 9700ul (about 10ml) of your diluent.
#96786 dilution question
Posted
on 06 January 2011 - 05:48 AM
if you add 3 µl of water, you still have 49 ng of your compound, this doesn't change no matter how much water you add
the only thing that changes is the concentration, in the beginning you had 7 ng/µl, now you have 49 ng in 10 µl ==> 4,9 ng/µl
the only thing that changes is the concentration, in the beginning you had 7 ng/µl, now you have 49 ng in 10 µl ==> 4,9 ng/µl
