Around 6 molecules of G3P is produced at the end of the Calvin Cycle (light independent reactions of photosynthesis), and 5 of which are reverted back to RuBp.

The general equation that I read is 5 G3P -----> 3 RuBp using 3 ATP.

It however doesn't seem to add up (regarding balancing of the atoms) as 5 phosphates would be converted to 6.

I would appreciate a link to any resource explaining this process in detail.

Thank you and have a nice day ^^


1 Answer 1


You need to account for free phosphates (Pi) that derive from ATP and are released in phosphatase reactions. The regeneration of 3 ribulose-1,5-2P has the overall reaction

5 glyceraldehyde-3P + 3 ATP $\rightarrow$ 3 ribulose-1,5-2P + 3 ADP + 2 Pi

So in total eight phosphates (here counting ATP as 1) are redistributed, 6 of which end up in ribulose-1,5-2P, and two are released as free phosphates. (These Pi are then recaptured during ATP synthesis in the light-dependent reactions.)

The above is a summary reaction of course; the complete scheme involves 10 enzymatic reactions that rearrange carbons in various ways to form 5-carbon sugars from 3-carbon ones (which takes a bit of juggling of atoms). Release of free Pi occurs in the fructose biphosphatase and seduheptulose biphosphatase steps.

A detailed account can be found in any major biochemistry textbook, like Stryer's Biochemistry. Online resources like MetaCyc can also be helpful.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .