The enzyme alcohol dehydrogenase catalyzes the following reaction:
Ethanol + NAD+ acetaldehyde + NADH + H+
You have a 1 ml solution containing 100 µM 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?
b) What will the final absorbance of the solution be at 340 nm? (the molar extinction coefficient of NADH at 340 nm is 6.22 x 103 l/mol cm
Some of the texts are showing funny. So, I do not want to tackle this problem.
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To answer this question, we need to calculate the final concentration of NADH and the final absorbance of the solution using the given information and the stoichiometry of the reaction.
a) To calculate the final concentration of NADH, we need to determine the number of moles of NADH produced. From the reaction, we can see that for every mole of ethanol converted to acetaldehyde, one mole of NADH is produced.
Given:
Initial concentration of NAD+ (NAD) = 1 mM = 1 mmol/L
Volume of solution = 1 mL = 1 cm^3 (since 1 mL = 1 cm^3)
Added alcohol dehydrogenase (ADH) = 1 mg
First, let's calculate the number of moles of NAD+:
Number of moles of NAD+ = (Initial concentration of NAD+) x (Volume of solution)
= (1 mmol/L) x (1 cm^3 / 1000 µL) x (1000 µL / 1 mL)
= 1 µmol
Since alcohol dehydrogenase is not consumed in the reaction, we don't need to consider it in the calculations.
Now, since all of the ethanol (100 µM) is converted to acetaldehyde, the concentration of NADH produced will be the same as the concentration of ethanol (100 µM).
Therefore, the final concentration of NADH = 100 µM.
b) To calculate the final absorbance of the solution at 340 nm, we can use the Beer-Lambert Law, which states that absorbance is proportional to the concentration of the absorbing species and the path length of the light through the solution.
Given:
Molar extinction coefficient of NADH at 340 nm (ε) = 6.22 x 10^3 L/mol cm
Concentration of NADH = 100 µM = 100 x 10^-6 mol/L
Path length (l) = 1 cm
Final absorbance (A) = ε x c x l
= (6.22 x 10^3 L/mol cm) x (100 x 10^-6 mol/L) x (1 cm)
= 6.22 x 10^-4
Therefore, the final absorbance of the solution at 340 nm will be 6.22 x 10^-4.