How to make a 16 ug/ml protein dilution from a stock of 1.48 mg/ml.
Posted 17 October 2013 - 07:20 AM
(1.48 mg/ml) / x = 16 ug/ml
x = (1480 ug/ml) / 16 ug/ml = 92.5
So, you need to dilute by 92.5 times.
Add 10 ul of your stock to 915 ul of water (more likely buffer)
Posted 17 October 2013 - 07:33 AM
that I never understand that, if I need to make a 50 or 75 ul of protein dilution
Posted 17 October 2013 - 01:10 PM
For a defined volume use:
C1V1=C2V2 where C= concentration (before and after) and V=volumes before and after.
In your case C1=1480 ug/ml, C2 = 16 ug/ml, V2=75 (or 50)
therefore V1= (16 x 75)\1480 = ?
Posted 18 October 2013 - 02:56 AM
I am still wrong in my question I think. What you told me I know. But I dont know how to have a 32 ug proteiin from 1.48 mg/ml stock solution in 50 ul for bradford assay?/
Posted 18 October 2013 - 04:01 AM
It's simplistic but think of it this way: 1.48mg/ml = 1.48ug/ul
you want 32 ug = 32/1.48 = 21.6ul
you want that 32ug in a total of 50ul, so 50-21.6 = 28.4ul diluent
Posted 18 October 2013 - 04:06 AM
even if I want 32 ug in 75 ul I have to take 21.6 ul of protein??
Posted 18 October 2013 - 04:23 AM
do you want 32ug (mass) or 32ug/ml (concentration)?
if mass then you need 21.6ul of 1.48mg/ml (as missle said).
if concentration then you need (16 x 75)\1480 = 0.81ul protein + 74.19ul buffer (or water) (as bob1 said).
genius does what it must
i do what i get paid to do
Posted 18 October 2013 - 06:39 AM
actually i need mass but i was using formula for concentration