Question

In yeast, ethanol is produced from glucose under anaerobic conditions. A cell‑free yeast extract is placed in a solution that contains 3.50 × 10 2 mmol glucose, 0.35 mmol ADP , 0.35 mmol P i , 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?

Answer 1

0.35 milli moles of ethanol can be theoretically be produced under these conditions.

Step-by-step explanation:

`C_6H_(12)O_6 + 2 ADP + 2 Pi\rightarrow 2 C_2H_5OH + 2 ATP + 2 CO_2`

Moles of glucose =
`3.50* 10^2` milli mole

Moles of ADP = 0.35 milli mole

Moles of Pi = 0.35 milli mole

Moles of ATP = 0.70 milli mole

As we can see that ADP and Pi are in limiting amount which means tat they are limiting reagent. So, the moles of ethanol produced will depend upon the moles of ADP and Pi.

According to reaction, 2 moles of ADP gives 2 moles of glucose.

Then 0.35 milli moles of ADp will give :

`(2)/(2)* 0.35 mmol=0.35 mmol` of ethanol

0.35 milli moles of ethanol can be theoretically be produced under these conditions.