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?
I'm not a biochemistry but if you know the limiting reagent I can show you what to do. First you need to know the equation and which of those materials is the LR.
To determine the maximum amount of ethanol that can be produced from glucose in yeast, we need to consider the stoichiometry of the reaction and the limiting reactant. The balanced equation for this reaction is:
Glucose + 2 ADP + 2 Pi + 2 NADH ↔ 2 Ethanol + 2 ATP + 2 NAD+
From the given information, we have:
Glucose: 2.50 × 10^2 mmol
ADP: 0.40 mmol
Pi: 0.40 mmol
ATP: 0.80 mmol
NAD: 0.20 mmol
NADH: 0.20 mmol
First, we need to determine the limiting reactant. To do this, we compare the ratios of the reactants to their stoichiometric coefficients:
For glucose:
Glucose / 1 = 2.50 × 10^2 mmol / 1 = 2.50 × 10^2 mmol
For ADP:
ADP / 2 = 0.40 mmol / 2 = 0.20 mmol
For Pi:
Pi / 2 = 0.40 mmol / 2 = 0.20 mmol
For NADH:
NADH / 2 = 0.20 mmol / 2 = 0.10 mmol
We can see that glucose is the limiting reactant because it has the smallest ratio to its stoichiometric coefficient.
Since 1 mole of glucose produces 2 moles of ethanol, the maximum amount of ethanol that can be produced is:
2.50 × 10^2 mmol glucose * (2 mmol ethanol / 1 mmol glucose) = 5.00 × 10^2 mmol ethanol
So, the maximum amount of ethanol that could theoretically be produced is 5.00 × 10^2 mmol.
Now, to determine the theoretical minimum amount of glucose required to form this maximum amount of ethanol, we can use the same stoichiometry ratio:
Glucose / Ethanol = (2.50 × 10^2 mmol glucose) / (5.00 × 10^2 mmol ethanol) = 0.50 mmol glucose/mmol ethanol
The minimum amount of glucose required would be the reciprocal of this ratio:
1 mmol ethanol / 0.50 mmol glucose = 2 mmol glucose
Therefore, under the given conditions, the theoretical minimum amount of glucose required in the solution to form the maximum amount of ethanol is 2 mmol.
To determine the maximum amount of ethanol that could theoretically be produced, we need to consider the stoichiometry of the reaction.
The balanced chemical equation for the production of ethanol from glucose is:
C6H12O6 → 2 C2H5OH + 2 CO2
This means that for every molecule of glucose, two molecules of ethanol and two molecules of carbon dioxide are produced.
Given that there are 2.50 × 10^2 mmol of glucose, we can calculate the maximum amount of ethanol produced as follows:
2.50 × 10^2 mmol glucose × (2 mmol ethanol / 1 mmol glucose) = 5.00 × 10^2 mmol ethanol
Therefore, the maximum amount of ethanol that could theoretically be produced is 5.00 × 10^2 mmol.
To determine the theoretical minimum amount of glucose required to form this maximum amount of ethanol, we reverse the stoichiometric ratio:
1 mmol glucose / (2 mmol ethanol) = 0.5 mmol glucose / (1 mmol ethanol)
Therefore, the theoretical minimum amount of glucose required is 0.5 times the maximum amount of ethanol:
0.5 × 5.00 × 10^2 mmol ethanol = 2.50 × 10^2 mmol glucose
Therefore, the theoretical minimum amount of glucose required is 2.50 × 10^2 mmol.