[klasnikov]

A question from my text book reads:

"To undergo a reaction it is necessary to have 0.25ml of an enzyme at a concentration of 0.2 mgml^-1 in the presence of 0.3 mM enzyme and 200 mM NaCl all in the same buffer. The stock solutions of enzyme, substrate and NaCl are 4mgml^-1, 40mM and 1M all in the same buffer. What volumes of buffer and stock are required for each assay?"

Any ideas how to tackle this?

[klasnikov]

I think I have calculated the dilution factors.

For the enzyme the factor is 20
For the substrate the factor is 100
For the NaCl the factor is 5

So the volume of enzyme required is 0.25/20 = 0.0125 ml

But how about the last part of the question? This is where I am stuck.

[Borek]

Sounds like a simple dilution problem:

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

Note that you have listed enzyme twice, but no substrate.

[klasnikov]

I have read through that page and still don't understand what to do.

If someone could break it down step by step that would be great.

[klasnikov]

Sounds like a simple dilution problem:

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

Note that you have listed enzyme twice, but no substrate.

I didn't see the ability to edit my original post so this is the correct version:

"To undergo a reaction it is necessary to have 0.25ml of an enzyme at a concentration of 0.2 mgml-1 in the presence of 0.3 mM substrate and 200 mM NaCl all in the same buffer. The stock solutions of enzyme, substrate and NaCl are 4mgml-1, 40mM and 1M all in the same buffer. What volumes of buffer and stock are required for each assay?"

[Borek]

It is all about mass conservation. Whatever number of moles you need in the final sample, you have to provide it from the stock solution. Start calculating numbers of moles of all substances in the reaction mixture.