Question

2. Suppose 7 glucose molecules enter glycolysis. Calculate the number of inorganic phosphate molecules required as well as the number of pyruvate molecules produced.

In yeast, ethanol is produced from glucose under anaerobic conditions. A cell-free yeast extract is placed in a solution that contains 2.50 x 10 mmol glucose, 0.35 mmol ADP, 0.35 mmol Pi, 0.70 mmol ATP, 0.20 mmol NAD', and 0.20 mmol NADH. It is kept under anaerobic conditions. What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under these conditions? maximum amount of ethanol: 0.4 mmol 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? minimum amount of glucose required: mmol

Pi=

Answer 1

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

1                 2            2               2  

250           0.35          0.35          

limiting reagent ADP and Pi

250 mmol glucose = 500 mmol Ethanol

0.35 mmol ADP = 0.35 mmol Ethanol

0.35 mmol Pi = 0.35 mmol Ethanol

answer : 0.35 mmol ethanol

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

x mole glucose --------------> 0.35 mol ethanol

x = 0.35 / 2 = 0.20

glucose millimoles = 0.18