Question

A.) In yeast, ethanol is produced from glucose under anaerobic conditions. What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under the following conditions? A cell-free yeast extract is placed in a solution that contains 325 mmol glucose, 0.35 mmol ADP, 0.35 mmol Pi, 0.70 mmol ATP, 0.20 mmol NAD , and 0.20 mmol NADH. It is kept under anaerobic conditions. (answer in mmol)

B.) Under the same conditions, what is the theoretical minimum amount of glucose (in millimoles) required in the solution to form the maximum amount of ethanol? (answer in mmol)

Answer 1

Each glucose molecule is broken down into two pyruvate molecules in a process known as glycolysis.Glycolysis is summarized by the equation:

C₆H₁₂O₆ + 2 ADP + 2 P_i + 2 NAD⁺ ---> 2 CH₃COCOO^- + 2 ATP + 2 NADH + 2 H₂O + 2 H⁺

Again this CH₃COCOO^- reacts to form ethanol according to the following reaction-

2 CH₃COCOO^- --->2CH₃CHO + 2CO₂

2CH₃CHO + 2NADH ---> 2 C₂H₅OH

Combining all, we get,

C₆H₁₂O₆ + 2 ADP + 2 P_i ---> 2 C₂H₅OH + 2 ATP + 2 CO₂

The whole reaction can be summarized using following image-

From the coefficients in the reactions, we obtain that

325 mmol glucose = 650 mmol Ethanol

0.35 mmol ADP = 0.35 mmol Ethanol

0.35 mmol Pi = 0.35 mmol Ethanol

Hence, of all the reactants, ADP or Pi forms least amount of ethanol and hence, they are the limiting reagent. So, Ethanol obtained is 0.35 mmol

Part 2

So, here 0.35 mmol of Ethanol is being formed. 1 mole of Glucose gives 2 moles Ethanol. So, for 0.20 mmol Ethanol we need only 0.175 mmol Glucose.