## Serial Dilution BSA Bradford measurement

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### Serial Dilution BSA Bradford measurement

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.

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!!!

Thank you so much!

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.

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!!!

Thank you so much!

that's fairly simple. Just prepare e.g. 1.5 mg/ml stock solution. Now take e.g. 10 ml of this solution and add 20 ml of water and you will get 3/2 that is 1 mg/ml. Now you can dilute it twice, so take10 ml and add 10 ml of water. Than again. And the last one dilute 2.5 alias 5/2 times, so take 10 ml and add 15 ml of water.

I don't know, what volume will you need exactly, but probably 10-times less than I wrote will be sufficient, but that doesn't matter much

I don't know, what volume will you need exactly, but probably 10-times less than I wrote will be sufficient, but that doesn't matter much

http://www.biolib.cz/en/main/

*Cis*or*trans*? That's what matters.### Re: Serial Dilution BSA Bradford measurement

Great thank you indeed!

Which equation did you use?

I was trying to use the C1V1=C2V2 formula

10x stock solution: 15mg/ml (I put 0.0015g in 10ml)

then according to the equation:

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?

What do you think about that?

Which equation did you use?

I was trying to use the C1V1=C2V2 formula

10x stock solution: 15mg/ml (I put 0.0015g in 10ml)

then according to the equation:

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?

What do you think about that?

### Re: Serial Dilution BSA Bradford measurement

vlad81 wrote:10x stock solution: 15mg/ml (I put 0.0015g in 10ml)

that looks to me as 1.5 mg per 10 ml, that is 0.15 mg/ml

vlad81 wrote: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?

I think, that 1 ml - 0.067 ml is not 0.023, but rather like 0.933 ml?

The equation is perfect, but you can try to think about it just simply.

If you had 15 mg/ml and want 1 mg/ ml, than you want to dilute it 15× and thus the volume must be 1/15 of the resulting volume

That's actually application of the previous equation:

c1V1 = c2V2 => c1/c2 = V2/V1

http://www.biolib.cz/en/main/

*Cis*or*trans*? That's what matters.### Re: Serial Dilution BSA Bradford measurement

Well, that was really very helpful. Thanks

Best Wishes JackBean!

Best Wishes JackBean!

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