substrate concentration= 3.5mM (c2)

Question

buffer=0.4 ml
10mM PNPP stock= ?ml (V1)
water=?ml
enzyme= 0.2ml
total volume =1ml (V2)

I need help with a dilution, I understand that you would be using c1v1=c2v2 but im confused as to what we need to find the amount of water and PNPP stock needed all for a total volume of 1ml, when we aren't given C1? My teacher said that substrate concentration was c2, 10mM of PNPP was V1, and the total volume is 1 ml which is v2. We need to find amount of PNPP and Water to get a total volume of 1ml at the substrate concentration 3.5mM...So how do I find the ml of both water and PNPP without a c1?

Answer 1

I"m not a biochemist so all of this substrate business is not my cup of tea; however, my educated guess is to do this.
(I think C1 is 10 mM PNPP.)
10*V1 = 3.5*1 mL
V1 = 0.35 mL.
Then buffer + 0.4 mL
enzyme = 0.2 mL
Total so far is 0.95 mL, then add water to make a total of 1 mL.
Since this is my best educated guess you should confirm with another source.