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.