Question

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 3.00 × 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?

Answer 1

Each glucose molecule is broken down into two pyruvate molecules

the process is known as Glycolysis

the equation of Glycolysis:

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

then..specifically:

2 CH3COCOO- --->2CH3CHO + 2CO2

2CH3CHO + 2NADH ---> 2 C2H5OH

It is easier if we sytnehtise all:

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

From the stoichiometric relationships:

300 mmol glucose = 600 mmol Ethanol (2x ratio)

0.30 mmol ADP = 0.30 mmol Ethanol (1x ratio)

0.30 mmol Pi = 0.30 mmol Ethanol (1x ratio)

since all are limiting reactants... the max mol of ethanol produces --> 0.30 mmol of ethanol

Q2.

For 0.30 mmol of ethanol, reacall that

1 mole of Glucose gives 2 moles Ethanol.

0.30/2 = 0.15 mmol --->

0.15 mmol Glucose.