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What is the number of ATP molecules formed during the photosynthetic processes which consume 8 molecules of $\text{H}_2\text{O}$ due to noncyclic electron transport and subsequent photophosphorylation?

Assume that quinone cycle facilitates the transfer 4H+ to the lumen of thylakoid membrane for every two electrons passing through electron transport system and one ATP is formed for every 3H+ moving down the proton gradient by the mediation of F0F1-ATPase

MY ATTEMPT: In the lumen, due to hydrolysis 16 H+ are formed. Due to quinone cycle 32 H+ are taken from stroma and transferred to the lumen and NADP+ takes 48 H+ so in stroma there is deficiency of 80 and there are only 48 H+ in the lumen. If I assume that if there were initially hundred H+ on both sides then after non cyclic transport inside the lumen there are 148 H+ and in stroma there are 20 H+ so to have equilibrium then there should be 84 H+ on both side so total 64 H+ should travel then I am getting 64/3 ATP which is wrong. Where I am going wrong?

Screenshot of question paper enter image description here

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    $\begingroup$ I had attempted but iam getting 64/3 ATP as answer which is wrong. $\endgroup$ – JM97 Jan 18 '16 at 6:11
  • $\begingroup$ I have no knowledge in photophosphorylation, but isn't ATP synthase creates 3 ATP for every 10 H+? and so maybe it's 64/3.3 ATP $\endgroup$ – A. Steiner Jan 25 '16 at 14:44
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diagram for non-cyclic photophosphorylation from "Pradeep's: A textbook of Biology for Class XI

Sorry for the poor quality of the image, but its just as a reference for my answer.

You have made the question too complex. From the figure, we find that 1 H2O gives 2 H+ and 2 e-. 2 e-, through quinone cycle, provide 4 H+ i.e. total 6 H+ which form 6/3 = 2 ATP i.e.

1 H2O => 6 H+ => 2 ATP

Multiply this equation by 8 and you get:

8 H2O => 48 H+ => 16 ATP

So, 16 ATP should be the answer.

Reference:

Pradeep's: A textbook of Biology for Class XI, 2015, Vol. II, page IV/106

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  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – JM97 Feb 10 '16 at 15:24
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    $\begingroup$ I threw an answer up, but according to my calculation, and ignoring ATP from NADPH, 16 ATP should be correct. I upvoted this. $\endgroup$ – CKM Feb 10 '16 at 17:19
  • $\begingroup$ @JM97 is it okay now? $\endgroup$ – another 'Homo sapien' Feb 10 '16 at 17:22
  • $\begingroup$ @aayush Srivastav in your answer you are not considering the fact that reduction of NADP to NADPH is causing a deficiency of protons which effects the proton gradient $\endgroup$ – JM97 Feb 11 '16 at 16:45
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    $\begingroup$ That is the real question which I was really asking, WHY DONT WE APPLY PRICIPLES OF CHEMISTRY IN THIS PROCESS? $\endgroup$ – JM97 Feb 12 '16 at 12:50
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I think we can look at it like this enter image description here

Your noncyclic pathway liberates 4H+ from two water molecules. We're doing that because to get one O2, you've got to split 2H2O. This happens in photosystem II. So in the next step (you know, really broadly) the electrons are passed to cytochrome b6-f complex, where each electron allows it to pump 2H+ into the thylakoid space. Since we got 4e- from splitting 2H2O, from that we can get 8H+. So net H+ = 8 + 4 or 12H+/O2, or we can rewrite it as 4ATP/O2.

For 2H2O you also got 2NADPH, which can also be used to get ATP. I think it's 3ATP/NADPH, so you get (6+4)ATP/2H2O and repeat four times:

8H2O = 40ATP

Ignoring NADPH oxidation:

8H2O = 16ATP

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  • $\begingroup$ please see the z scheme given at Wikipedia yours is different , according to Wikipedia only 2e are needed for reduction of NADP to NADPH but according to yours 4e are reducing 2NADP, SO WHICH ONE IS CORRECT? $\endgroup$ – JM97 Feb 11 '16 at 17:13
  • $\begingroup$ The reduction of NADP to NADPH requires two electrons per NADPH produced, diagram. So, either representation is correct, but I utilize 4e > 2NADPH because the light reactions don't just release radical oxygen, they release O2, hence I always begin with 2H2O. @JM97 $\endgroup$ – CKM Feb 11 '16 at 18:35
  • $\begingroup$ why don't you consider the fact that reduction of NADP to NADPH also contributes to proton gradient $\endgroup$ – JM97 Feb 12 '16 at 1:25
  • $\begingroup$ Because you specified noncyclic electron transport. $\endgroup$ – CKM Feb 12 '16 at 1:47
  • $\begingroup$ reconsidering it again from starting then 8 water molecules gives 16 protons and 16e then first inside thylakoid lumen there is an increase of 16H+ second quinone cycle facilitates transfer of 32H+ from stroma to lumen so there is decrease of proton of stroma and increase of protons in lumen. Third 16e will reduce only 8NADP to 8NADPH so there is decrease of only 8e on stroma. Overall:inside lumen +48 , in stroma -40 so to get equilibrium as per chemistry principles only 44H+ should pass therefore we get 44/3 ATP. $\endgroup$ – JM97 Feb 12 '16 at 1:47
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As we know that for 4 H²O molecules one non cyclic electron transport occur which gives us 4H+ ions ..this 4H+ ions gives 2 ATP so 8H+ ions will gives us 4 ATP....so Ans= 4ATP molecules /8H+ ions

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ACCORDING TO ME THE ANSWER IS 24 ATP THAT IS BECAUSE 8H+ IS PRODUCE DE TO PHOTOLYSIS. 16 H+ WILL BE TRANSFERRED INTO LUMEN. SO 8 ATP WILL BE PRODUCE BY ATPASE {TOTAL 24 H+] 4 NADPH2 WILL PRODUCE 12 ATP 4 ATP WILL BE RELEASED FROM ELECTRON WHILE MOVING IN PS1. SO TOTAL 8+12+4 = 24 ATP.

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    $\begingroup$ First: Please don't write in all caps as this is seen as yelling. And then it would be nice to show some references. $\endgroup$ – Chris Aug 26 '17 at 6:27

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