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In the common Calvin cycle diagram, it is commonly stated that "three cycles are combined to show the production of 1 molecule of G3P".

How does the G3P molecule escape from this cycle in one of these turns?

Is one carbon atom "released" from this cycle at each turn and the G3P forms in another reaction?

Or alternatively, is the G3P released at the end of the three "turns"? If this is the case, then shouldn't this diagram represent only one "cycle" of the Calvin cycle? [1]: https://i.stack.imgur.com/mzizw.jpg

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  • $\begingroup$ Your question is unintelligible without reference to a diagram. Please revise your question accordingly, and phrase it so that it addresses a biological problem, rather than the interpretation of a particular mode of presentation in a specific text. $\endgroup$
    – David
    Jan 14 at 19:34
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    $\begingroup$ Not quite sure I understand what you are asking; the cycle is clearly not closed, there is a reactant entering and a product leaving, as well as interactions with other cycling molecules as ATP/ADP and NADPH/NADP+. $\endgroup$
    – Bryan Krause
    Jan 14 at 21:26
  • $\begingroup$ This is a great question - the secret in a nutshell is that five three-carbon G3Ps are turned into three five-carbon RuBPs at the left. This involves a lot of making change with sugars - see the Wikipedia article section on the regeneration step. You might also compare the pentose phosphate pathway. I should submit an answer but I'm not feeling energetic enough to write out all those steps right now ... which is presumably how this textbook author felt! $\endgroup$ Jan 15 at 1:21

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