Ethanol + NAD+ acetaldehyde + NADH + H+
You have a 1 ml solution containing 100 uM Ethanol, and 1 mM NAD+. You then add 1 mg of alcohol dehydrogenase. Assume all of the ethanol gets converted to acetaldehyde.
a) What will the final concentration of NADH be in the solution?
Ethanol + NAD+ ----> acetaldehyde + NADH + H+
To determine the final concentration of NADH in the solution, we need to calculate the amount of NADH produced when all the ethanol is converted to acetaldehyde.
First, let's convert the volume of the solution to liters:
1 ml = 0.001 L
Next, let's calculate the moles of ethanol in the solution:
Ethanol concentration = 100 uM = 100 x 10^-6 mol/L
Volume of solution = 0.001 L
Moles of ethanol = Ethanol concentration x Volume of solution
Moles of ethanol = (100 x 10^-6 mol/L) x (0.001 L)
Moles of ethanol = 0.0001 mol
According to the balanced equation, 1 mole of ethanol is converted to 1 mole of acetaldehyde. Therefore, the moles of acetaldehyde produced will be the same as the moles of ethanol:
Moles of acetaldehyde = 0.0001 mol
Now, let's calculate the moles of NADH produced. According to the balanced equation, 1 mole of ethanol is converted to 1 mole of NADH:
Moles of NADH = 0.0001 mol
To calculate the concentration of NADH in the final solution, we divide the moles of NADH by the final volume of the solution:
Final volume of solution = 0.001 L
Concentration of NADH = Moles of NADH / Final volume of solution
Concentration of NADH = 0.0001 mol / 0.001 L
Concentration of NADH = 0.1 mol/L
Finally, we can convert the concentration of NADH to micromoles per liter (uM):
Concentration of NADH = 0.1 mol/L = 0.1 x 10^6 uM
Concentration of NADH = 100 uM
Therefore, the final concentration of NADH in the solution will be 100 uM.