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) .0.016 mm
b) 0.0016 mm
c) 0.16 mm
d) 1.6 mm

Explanation:

Step1: Identify substrate and dilution law

The substrate is ethanol. Use dilution formula $C_1V_1 = C_2V_2$, where $C_1$ = initial concentration, $V_1$ = initial volume, $C_2$ = final concentration, $V_2$ = final total volume.

Step2: List known values

$C_1 = 10\ \mu\text{M}$, $V_1 = 0.16\ \text{mL}$, $V_2 = 1\ \text{mL}$

Step3: Solve for final concentration

Rearrange formula: $C_2 = \frac{C_1V_1}{V_2}$

$$\begin{align*} C_2 &= \frac{10\ \mu\text{M} \times 0.16\ \text{mL}}{1\ \text{mL}} \\ &= 1.6\ \mu\text{M} \end{align*}$$

Step4: Convert units to mM

Since $1\ \text{mM} = 1000\ \mu\text{M}$, $C_2 = \frac{1.6}{1000}\ \text{mM} = 0.0016\ \text{mM}$

Answer:

b) 0.0016 mM