Question

the master mix now contains 0.16 ml of 10 μm ethanol, 0.2 ml of 10 μm nad⁺, and 0.54 ml buffer. we will add 0.1 ml of 1 mg/ml adh so that the total volume will be 1 ml. what is the substrate concentration in this master mix? a) 0.0016 mm b) 0.016 mm c) 0.16 mm d) 1.6 mm

Explanation:

Step1: Use the dilution formula $C_1V_1 = C_2V_2$.

We know the initial volume $V_1$ and concentration $C_1$ of the substrate - containing solution, and the final volume $V_2$ of the master - mix. We want to find the final concentration $C_2$.

Step2: Identify the values.

The volume of the substrate - containing solution added $V_1=0.16\ mL$. Let's assume the concentration of the substrate in the added solution is $C_1$ (not given in full context, but we can calculate based on the fact that we are using dilution principles). The final volume of the master - mix $V_2 = 1\ mL$.
If we assume the substrate is added in a pure form (for the sake of calculating concentration in the final mix), and we use the formula $C_2=\frac{C_1V_1}{V_2}$.
Let's assume the initial concentration of the substrate in the added solution is such that when we calculate the concentration in the final $1\ mL$ mix.
The volume of the added solution $V_1 = 0.16\ mL$ and final volume $V_2=1\ mL$.
If we assume the initial concentration of the substrate in the added solution is $10\ \mu M$ (from the problem description), then $C_2=\frac{10\ \mu M\times0.16\ mL}{1\ mL}=1.6\ \mu M$. Since $1\ mM = 1000\ \mu M$, then $1.6\ \mu M=0.0016\ mM$.

Answer:

a) 0.0016 mM