Yeast, ethanol is produced from glucose under anaerobic conditions. What is the

Question

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 275 mmol glucose, 0.30 mmol ADP, 0.30 mmol P_i, 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?

Explanation / Answer

Part A)

Gylcolysis is summarized by the equation given below,

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

Again this CH3COCOO- reacts to form ethanol according to the following reaction-

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

2CH3CHO + 2NADH ---> 2 C2H5OH

Combining all above, we have,

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

from the above reaction,

1 mol of glucose produce 2 mole ethanol.

Therefore,

275 mmol glucose= 550 mmol ethanol.

0.30 mmol ADP=0.30 mmol ethanol

0.30 mmol Pi=0.30 mmol ethanol

As, in given reaction it is concluded that, Pi and ADP are limiting reactants because they produce

least amount of ethanol. Therefore,

Ethanol obtained will be 0.30 mmol

Part B)

1 mole glucose -----------------> 2 mol ethanol

y mmol glucose-------------> 0.30 mmol ethanol

Thereofore,

mmol of glucose required = 0.15 mmol

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