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Dilution confusion

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#1 samita

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Posted 17 October 2013 - 07:02 AM

How to make a 16 ug/ml protein dilution from a stock of 1.48 mg/ml.

#2 phage434

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Posted 17 October 2013 - 07:20 AM

Arrgh. Algebra.

(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)

#3 samita

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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

#4 bob1

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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 = ?

#5 samita

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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?/

#6 Missle

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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

#7 samita

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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??

#8 mdfenko

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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).

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#9 samita

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Posted 18 October 2013 - 06:39 AM

actually i need mass  but i was using formula for concentration