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 3.50 × 102 mmol glucose, 0.30 mmol ADP, 0.30 mmol Pi, 0.60 mmol ATP, 0.20 mmol NAD , and 0.20 mmol NADH. It is kept under anaerobic conditions.
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?
To determine the maximum amount of ethanol that could theoretically be produced, we need to consider the stoichiometry of the reaction and the available reactants.
The balanced equation for the conversion of glucose to ethanol in yeast is:
C6H12O6 → 2 C2H5OH + 2 CO2
From this equation, we can see that for every molecule of glucose consumed, two molecules of ethanol are produced. Therefore, to find the maximum amount of ethanol, we need to calculate how many moles of glucose can be converted.
Given that the solution contains 3.50 × 102 mmol glucose, we can convert this to moles by dividing by Avogadro's number (6.022 × 10^23).
3.50 × 102 mmol = 3.50 × 10^-4 moles
Since two moles of ethanol are produced from every mole of glucose, the maximum amount of ethanol that can be produced is:
2 × 3.50 × 10^-4 moles = 7.00 × 10^-4 moles
To determine the minimal amount of glucose required to form the maximum amount of ethanol, we need to consider the stoichiometry of the reaction in the opposite direction.
From the balanced equation, we can see that 2 moles of ethanol are produced from 1 mole of glucose. Therefore, to produce the maximum amount of ethanol (7.00 × 10^-4 moles), we would need:
1/2 × 7.00 × 10^-4 moles = 3.50 × 10^-4 moles of glucose
So, the theoretical minimum amount of glucose required to form the maximum amount of ethanol is 3.50 × 10^-4 millimoles.