In E. coli, fatty acids (CH 3 -(CH 2 ) n -COOH) with an even number of carbons a
ID: 206503 • Letter: I
Question
In E. coli, fatty acids (CH3-(CH2)n-COOH) with an even number of carbons are degraded to acetate by the removal of one 2 carbon unit (acetate) at a time according to the reaction:
(CH3-(CH2)n-COOH) + FAD + NAD+ <--> (CH3-(CH2)n-2-COOH) +FADH2 + NADH + acetate (CH3COOH)
The acetate is then broken down to CO2 and H2O by the usual pathway of energy metabolism. (For this problem consider acetate to be equivalent to acetyl CoA. (This is a simplification.)
IF E. coli can live on a medium containing fatty acids as the only energy source then the presence of O2 must be ______.
necessary
unnecessary
harmful
xx only 3 choices for this problem xxx
necessary
unnecessary
harmful
xx only 3 choices for this problem xxx
Explanation / Answer
Ans. Note the following points-
I. As shown in the reaction, there is reduction of FAD and NAD+ into FADH2 and NADH during removal of one 2C units.
II. The cell has a limited pool of FAD and NAD+. So, they must be re-generated/ recycled by oxidation of FADH2 and NADH into FAD and NAD+ respectively.
III. To oxidize NADH and FADH2, there must be some oxidizing agents present in the system. O2 is one of the most common oxidizing agents – terminal electron acceptor.
IV. As mentioned in the question, the end products are CO2 and H2O. So, O2 must be present as the oxidizing agent in the system because O- form H2O by acting as terminal electron acceptor while oxidizing NADH and FADH2 for regeneration.
#V. If there were some other oxidising agent, say, NO3, the end product would have been NH3 instead of H2O.
# Therefore, presence of O2 must be necessary.
So, correct option is- A. Necessary.