In non-cyclic photophosphorylation 1 ATP and 2 $NADPH_2$ molecules are produced.

In cyclic photophosphorylation 2 ATP molecules are produced.

For production of one molecule of Glucose 18 ATP and 12 $NADPH_2$ molecules are reqiured.

According to my understanding,6 turns of cyclic and non-cyclic photophosphorylation is required to produce one molecule of glucose.

Because 6 turns of non cyclic photophosphorylation will produce 12 $NADPH_2$ and 6 ATP molecules.

And cyclic photophosphorylation will produce 12 ATP molecules.

Total=12 $NADPH_2$ and 18 ATP.

I think Iam missing something , Please answer me if I am correct or not.

  • 1
    $\begingroup$ You seem to have mixed up cyclic and noncyclic, as you're first saying that noncyclic produces NADPH (correct) but then that cyclic produces NADPH (wrong). But the overall stoichiometry seems right, assuming the numbers you give at the top. What is it exactly that you have trouble with? $\endgroup$ – Roland Mar 2 '17 at 19:00
  • $\begingroup$ Yeah!!A mistake while writing but now I have edited.I just want to know if I have said it correct $\endgroup$ – RuBisCO Mar 3 '17 at 4:35
  • $\begingroup$ Well, your conclusion is just plain addition, so not much to say about that. The ATP:NADPH ratio of photorespiration varies though -- since it is a chemoosmotic process, the stoichiometry is not fixed, and it might differ between species, conditions etc. Where did you get your numbers from? $\endgroup$ – Roland Mar 3 '17 at 7:16
  • $\begingroup$ The cyclic and non cyclic part was given in the textbook.but the 6 turns part was a conclusion from myself. So that means there can't be a fixed ratio. $\endgroup$ – RuBisCO Mar 3 '17 at 8:30

In non-cyclic photophosphorylation the stoichiometry of ATP:NADPH is predicted as 3:2, i.e, for every two NADPH formed, a total of three ATP are produced.

Source: Cyclic, pseudocyclic and noncyclic photophosphorylation: New links in the chain. John F. Allen. TRENDS in Plant Science Vol.8 No.1 January 2003.

Cyclic photophosporylation, does not occur under normal conditions in higher plants. It is believed to occur only when either light intensity is low (lower frequency so as only PS1 can function) or when carbon fixation is not at par with light irradiance (i.e, too much light).

And hence it is not advisable to come to this generalization "Six Turns of non cyclic photophosphorylation will produce 12 NADPH2 and six ATP molecule"

  • $\begingroup$ In non-cyclic photophosphorylation the stoichiometry of ATP:NADPH is predicted as 3:2, i.e, for every two NADPH formed, a total of three ATP are produced.I don't understand what u said . Can u elaborate a little more or provide a link. $\endgroup$ – RuBisCO Mar 3 '17 at 16:48
  • $\begingroup$ Kindly read this: jfallen.org/~john/webstar/john/pdf/32775.pdf $\endgroup$ – Tushar Mar 4 '17 at 6:15
  • $\begingroup$ I will add the link to the answer $\endgroup$ – RuBisCO Mar 5 '17 at 18:19

The numbering of 1ATP from noncyclic photophosphorylation and 2ATP from cyclic is little too much to be right. However, the formation of 2NADPH + H+ out of noncyclic photophosphorylation is correct.

Because according to the Chemiosmotic Hypothesis (Mitchell), the number of protons accumulating in the lumen of the granal thylakoid during noncyclic transport of 4e- from P680 towards Ctyb6f complex is 12 (8H+ transferring from stroma by Q-cycle + 4H+ from photolysis of 2H2O at Oxygen-evolving Complex). And at every 5th e- transport, the cyclic electron transport occurs and helps in transporting 2H+(protons) from stroma region towards the lumen of granum. So, a total of 14 H+ accumulates per an O2 release by photolysis of water. These 14H+ find its way to stroma region through CF particle and 3ATP are synthesized.

Therefore, H+/ATP = 4.67, e-/NADPH + H+ = 2, Total ATP release in both photphosphorylation during evolution of an oxygen molecule = 3, total NADPH + H+ during evolution of an oxygen molecule at PS-II = 2
Therefore, ATP/NADPH formation from cyclic photophosrylation at P680 = 7/9 (instead of the expected 3/2) and ATP alone = 1.29 (instead of 3).


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