substrate concentration= 3.5mM (c2)

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?

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.

To find the amounts of water and PNPP stock required to achieve a total volume of 1 ml at a substrate concentration of 3.5 mM, you will need to determine the necessary dilution factor.

Here's how you can approach the problem:

1. Determine the dilution factor (DF):
DF = c1 / c2
Since you are not given the value of c1, you can solve for DF using the information given. In this case, DF = (V1 + water + enzyme) / total volume (V2) = (V1 + water + 0.2 ml) / 1 ml.

2. Substitute the values and rearrange the equation to solve for V1:
DF = V1 / (c2 * V2)
V1 = DF * c2 * V2

Note: In this case, c2 is given as 3.5 mM, V2 is given as 1 ml, and we have the dilution factor (DF) equation from step 1.

3. Substitute the values into the equation and solve for V1:
V1 = DF * c2 * V2
V1 = (V1 + water + 0.2 ml) / 1 ml * 3.5 mM * 1 ml
V1 = V1 + water + 0.2 ml
water = V1 - 0.2 ml

Now you have the value of V1, which represents the amount of PNPP stock required. Subtracting 0.2 ml (the volume of the enzyme) from V1 will give you the amount of water (in ml) needed to achieve a total volume of 1 ml. Remember to use the dilution factor equation you obtained earlier!

I hope this helps clarify the process of determining the amounts of water and PNPP stock needed for the dilution.