Serial Dilution BSA Bradford measurement - What to do? (May/06/2010 )
Dear all,
I am experiencing real difficulties with something very simple such as serial dilutions and preparation of appropriate stock solutions.
For my Bradford protein measurement I need to prepare 0.1, 0.25, 0.5, 1, 1.5 mg/ml protein. I would also like to prepare stock solution in order to use ready dilutions in the future. However, I don`t have any idea how to do that. Neither do I know how much powder of BSA to dillute?
I understand that this is something very basic and everyone knows it and this makes me even more insecure and embarrassed. I really hope I can find some help here...please!!! :oops:
Thank you so much!
you can prepare a few ml of 1.5 mg/ml bsa (3 mg makes 2 ml, 4.5 mg makes 3 ml,...) in diluent (water, buffer, ...).
take 1 ml of 1.5 mg/ml and add 0.5 ml diluent=1 mg/ml.
take 0.5 ml 1 mg/ml and add 0.5 ml diluent=0.5 mg/ml.
take 0.5 ml 0.5 mg/ml and add 0.5 ml diluent=0.25 mg/ml.
take 0.1 ml 1 mg/ml and add 0.9 ml diluent=0.1 mg/ml.
simple.
Thanks a lot, but how did you calculate the serial dilutions.
I tried with the C1xV1=C2xV2 but I do not know if this is correct?
So when I calculate:
10x stock solution: 15mg/ml (I put 0.0015g in 10ml)
then
for 1mg/ml I have 15mg/ml x V1 = 1ml x 1mg/ml => V1 = 1/15=0.067ml
Thus I take 0.067ml from 10x stock solution and add 0.023ml diluent?
Do you think that could work? 
vigri on May 7 2010, 02:40 PM said:
I tried with the C1xV1=C2xV2 but I do not know if this is correct?
So when I calculate:
10x stock solution: 15mg/ml (I put 0.0015g in 10ml)
then
for 1mg/ml I have 15mg/ml x V1 = 1ml x 1mg/ml => V1 = 1/15=0.067ml
Thus I take 0.067ml from 10x stock solution and add 0.023ml diluent?
Do you think that could work?
What you really need is a math tutor, not a biology one...
0.0015g=1.5mg, so if you put 1.5mg in 10ml, your stock solution will be 1.5mg/10ml=0.15mg/ml
To make 10ml of 15mg/ml stock solution, you need to weigh
15mg/ml x 10ml = 150mg = 0.15g of BSA
The equation C1xV1=C2xV2 is correct, but 0.067ml+0.023ml=0.1ml, and that's not your V2.
To make 1ml of 1mg/ml working solution from 15mg/ml stock solution, you need to measure
(1mg/ml x 1ml)/(15mg/ml)=0.067ml of stock solution (You got this part right)
Then to make it a full 1ml, you need to add
1ml - 0.067ml = 0.923ml of diluent.
check your decimal places and your arithmatic.
0.0015g is 1.5mg not 15mg.
.067+.023=.09
c1*v1=c2*v2 works quite well but sometimes intuition and experience work well, too.
i wrote as though the high concentration stock is the highest concentration standard so that not all standards will be dilutions. this way inaccuracies of the pipette will not factor as much in one standard but it is not necessary, you can make a 10x stock, if you wish.
then i just figured easy dilutions, eg: 1ml*1.5mg/ml=1.5ml*1mg/ml, 0.5ml*1mg/ml=1ml*0.5mg/ml, and so on.
so, if you start with a 15mg/ml stock then: 0.1ml*15mg/ml=1ml*1.5mg/ml, etc.
edit: proteinwork and i wrote pretty much the same thing at the same time but for 1 error, 0.067+0.023=.09 and 1-0.067=0.933
mdfenko on May 7 2010, 03:29 PM said:
0.0015g is 1.5mg not 15mg.
.067+.023=.09
c1*v1=c2*v2 works quite well but sometimes intuition and experience work well, too.
i wrote as though the high concentration stock is the highest concentration standard so that not all standards will be dilutions. this way inaccuracies of the pipette will not factor as much in one standard but it is not necessary, you can make a 10x stock, if you wish.
then i just figured easy dilutions, eg: 1ml*1.5mg/ml=1.5ml*1mg/ml, 0.5ml*1mg/ml=1ml*0.5mg/ml, and so on.
so, if you start with a 15mg/ml stock then: 0.1ml*15mg/ml=1ml*1.5mg/ml, etc.
edit: proteinwork and i wrote pretty much the same thing at the same time but for 1 error, 0.067+0.023=.09 and 1-0.067=0.933
Yes, you are right. I thought he was just missing one decimal place so I did not really do the math.
Thank you both,
I really appreciate your help and yes, I hate math.
Have a great weekend!