What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under the following conditions?
Condition: 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.
2. 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?
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 calculate the maximum amount of ethanol that could theoretically be produced in this condition, we can use the stoichiometry of the reaction that converts glucose to ethanol. This reaction can be represented as:
Glucose + 2 ADP + 2 Pi + 2 NADH ⟶ 2 Ethanol + 2 ATP + 2 NAD+
Given the amounts of glucose, ADP, Pi, ATP, NAD, and NADH, we can determine the limiting reactant and calculate the maximum amount of ethanol produced.
1. Calculate the maximum amount of ethanol produced:
The stoichiometric ratio between glucose and ethanol is 1:2, meaning that for every mole of glucose, 2 moles of ethanol are produced. To convert the given amounts into moles, divide by their respective molecular weights:
Glucose: 3.50 × 102 mmol ÷ 180.16 g/mol = 1.942 moles
The limiting reactant is the reactant that is completely consumed in the reaction. To determine the limiting reactant, compare the moles of glucose to the moles of ATP (since both are involved in a 1:1 ratio in the reaction):
ATP: 0.60 mmol ÷ 507.18 g/mol = 0.001182 moles
Since the moles of glucose (1.942) are much greater than the moles of ATP (0.001182), ATP is the limiting reactant.
The stoichiometric ratio between ATP and ethanol is 2:2, meaning that for every mole of ATP, 1 mole of ethanol is produced.
Therefore, the maximum amount of ethanol that could theoretically be produced is equal to the moles of ATP:
Ethanol: 0.60 mmol ÷ 507.18 g/mol = 0.001182 moles
Therefore, the maximum amount of ethanol that could theoretically be produced under these conditions is 0.001182 moles.
2. Calculate the theoretical minimum amount of glucose required to form the maximum amount of ethanol:
The stoichiometric ratio between glucose and ethanol is 1:2, meaning that for every mole of glucose, 2 moles of ethanol are produced.
Therefore, to calculate the minimum amount of glucose required, divide the moles of ethanol by 2:
Minimum glucose required: 0.001182 moles ÷ 2 = 0.000591 moles
To convert this to millimoles, multiply by 1000:
Minimum glucose required: 0.000591 moles × 1000 = 0.591 mmol
Therefore, the minimum amount of glucose (in millimoles) required in the solution to form the maximum amount of ethanol is 0.591 mmol.
To determine the maximum amount of ethanol that can be produced under the given conditions, we need to consider the stoichiometry of the reaction and the available reactants.
The balanced equation for ethanol fermentation in yeast is as follows:
Glucose + 2 ADP + 2 Pi + 2 NADH -> 2 Ethanol + 2 ATP + 2 NAD+
From the given information, we have:
Glucose: 3.50 × 10^2 mmol
ADP: 0.30 mmol
Pi: 0.30 mmol
ATP: 0.60 mmol
NAD: 0.20 mmol
NADH: 0.20 mmol
Comparing the stoichiometric coefficients in the balanced equation, we can see that for every 1 mol of glucose, 2 mol of ethanol is produced. This means that the conversion of glucose to ethanol is a 1:2 ratio.
To calculate the maximum amount of ethanol that can be produced, we need to determine the limiting reactant among the available reactants. The limiting reactant is the one that will be completely consumed first and will determine the maximum amount of product that can be formed.
First, let's consider ADP, Pi, ATP, NAD, and NADH, since their quantities are much lower compared to glucose.
For the conversion of 1 mol of glucose to 2 mol of ethanol, we need:
- 2 mol of ADP
- 2 mol of Pi
- 2 mol of ATP
- 2 mol of NAD
- 2 mol of NADH
Since the quantities of these reactants are higher than what is needed for the conversion of 1 mol of glucose, none of them is the limiting reactant.
Therefore, the limiting reactant is glucose.
Considering the 1:2 ratio, the maximum amount of ethanol that can be produced is half the amount of glucose used:
Maximum amount of ethanol = (3.50 × 10^2 mmol glucose) / 2 = 1.75 × 10^2 mmol ethanol.
Now, to determine the theoretical minimum amount of glucose required to produce the maximum amount of ethanol, we need to reverse the calculation.
If 1.75 × 10^2 mmol of ethanol is produced, then the minimum amount of glucose needed is:
Minimum amount of glucose = 2 × (1.75 × 10^2 mmol ethanol) = 3.50 × 10^2 mmol glucose.
Therefore, the theoretical minimum amount of glucose required in the solution to form the maximum amount of ethanol is 3.50 × 10^2 mmol.