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During the dark reaction, 9 molecules of ATP are consumed for every 3 carbon dioxide molecules, yet only 6 molecules of inorganic phosphate ultimately leave the cycle. Does that mean that there is a greater concentration of ADP than inorganic phosphate returning to the light reactions?

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Not 6, but 9 molecules of Pi are formed in the Calvin cycle, from the conversion of 9 ATP to 9 ADP:

Calvin Cycle
Source: Columbia University

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  • $\begingroup$ I thought the inorganic phosphates were incorporated into phosphoglycerate to form DIphosphoglycerate, but then given off again in the formation of G3P. Then it is once again incorporated into glyceraldehyde phosphate to form ribulose BIsphosphate, but is not released anywhere else. $\endgroup$ – lightweaver Apr 25 '15 at 13:23
  • $\begingroup$ There's definitely some confusion. In the C3 cycle reactions, the conversion of 3-phospho-D-glycerate to diphosphoglycerate (DPG) actually affixes the inorganic phosphate from ATP to the DPG. The reaction that follows releases NADP+ and inorganic phosphate at a ratio of 2 per CO2 in the cycle, so for 3CO2 used in making one half a glucose, thats 6 inorganic phosphate liberated. If you're also considering the regeneration of rubisco, two inorganic phopsphate are liberated there for a total of 8 inorganic phosphate in the products per 3CO2. $\endgroup$ – CKM Apr 25 '15 at 13:44
  • $\begingroup$ And then, here's a link to a more detailed reaction schematic: chem.qmul.ac.uk/iubmb/enzyme/reaction/polysacc/polygif/… $\endgroup$ – CKM Apr 25 '15 at 13:47

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