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

Explanation:

Step1: Use dilution formula

$C_1V_1 = C_2V_2$, where $C_1$ is initial concentration, $V_1$ is initial volume, $C_2$ is final concentration and $V_2$ is final volume.
We know $V_1=0.16\ mL$, $V_2 = 1\ mL$. Let the initial concentration of substrate be $C_1$ and final concentration be $C_2$.

Step2: Rearrange for $C_2$

$C_2=\frac{C_1V_1}{V_2}$. Given the initial concentration of ethanol (sub - strate) is $10\ \mu M=10\times10^{- 6}\ M$.
$C_2=\frac{10\times10^{-6}\ M\times0.16\ mL}{1\ mL}=1.6\times10^{-6}\ M = 1.6\ mM$

Answer:

c) $1.6\ mM$