Under anaerobic conditions in yeast, glucose is converted into ethanol using the following reactions and stoichiometry:
1) Glycolysis: Glucose + 2 NAD + 2 ADP + 2 Pi → 2 Pyruvate + 2 NADH + 2 H+ + 2 ATP.
2) Alcoholic fermentation: Pyruvate + NADH + H+ → Ethanol + CO2 + NAD+
To calculate the maximum amount of ethanol that could theoretically be produced under the given conditions, we need to identify the limiting reactants in these reactions. Since we have 2.50 × 10² mmol glucose, ATP, ADP, Pi, NAD, and NADH are the main limiting reactants.
Reaction 1:
0.40 mmol ADP + 0.40 mmol Pi can produce 0.40 mmol ATP; since 2 ATP are produced from 1 glucose, 0.40 mmol ATP / 2 = 0.20 mmol glucose can be used in glycolysis.
Reaction 2:
0.20 mmol NAD + 0.20 mmol NADH can drive the formation of 0.20 mmol ethanol since 2 pyruvates will be consumed per glucose to make ethanol (1:1 stoichiometry between glucose and ethanol in this case).
Given that the conditions are anaerobic, we can consider that all the available glycolytic ATP will be used to produce ethanol. In this case, the maximum amount of ethanol that will be produced is the same as the limiting amount of glucose utilized, which is 0.20 mmol.
For the second part of the question, the theoretical minimum amount of glucose required to form the maximum amount of ethanol is also 0.20 mmol, as it is the limiting factor in the anaerobic fermentation process.
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 2.50 × 102 mmol glucose, 0.40 mmol ADP, 0.40 mmol Pi, 0.80 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?
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