Hello all,
this is a question I have on a practice exam.
If the glucose is labeled with
C-14 at the 1-position, what fraction of the isotope will not have
been released after each two-carbon fragment has gone through four complete turns of the cycle?
The answer: Glucose labeled in the 1-position will produce methyl-labeled acetyl CoA. After four complete
turns, one fourth of the input-labeled carbon will be present in oxaloacetate.
I have read the Lehninger "Principles of BIochemistry" textbook many times on how to go about reaching this answer but I still do not understand. please help!
