For the production of one glucose molecule in the Calvin cycle a plant uses 18 molecules ATP, but when the same glucose molecule is oxidised — first in the cytoplasm and then in the mitochondrion — it can obtain approximately 36-38 molecules ATP. How is the theory of conservation of energy maintained here?

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    $\begingroup$ Please add references for the numbers you cite. Wikipedia or Berg et al. (NCBI Bookshelf online) would be suitable. $\endgroup$
    – David
    Commented Feb 26, 2017 at 20:56
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    $\begingroup$ @sreekara plant uses 18atp to fix light energy which is more than energy released during respiration. $\endgroup$
    – JM97
    Commented Feb 27, 2017 at 1:23
  • $\begingroup$ @JM97 — When you open a comment box to a question you see the following: "Use comments to ask for more information or suggest improvements. Avoid answering questions in comments." Please do what it says, otherwise you are breaking the Stack Exchange system. In the present case your answer is contentious (I contend that it is incorrect), but unless it is formulated into an answer I cannot vote it down or comment on it (without compounding your behaviour). If the poster thinks it correct, he cannot accept it as an answer and the question would appear unanswered. $\endgroup$
    – David
    Commented Feb 27, 2017 at 10:02
  • $\begingroup$ @David, I studied it in our text book and even our teachers taught the same. $\endgroup$
    – sreekara
    Commented Feb 27, 2017 at 14:48
  • $\begingroup$ @David: But if you put a simple & obvious answer into an actual 'answer', you will get numerous complaints about it not being supported by references &c. As to 'breaking' the SE system, the only thing that breaks it are the over-officious moderators who insist on short-circuiting good discussions. $\endgroup$
    – jamesqf
    Commented Feb 27, 2017 at 19:27

2 Answers 2


The Fallacies in the argument

The question contains two main fallacies (some would say sleights of hand) in the energetic comparison of glucose synthesis from CO2 in the Calvin cycle and glucose oxidation via glycolysis, the tricarboxylic acid cycle and the electron transport chain:

  1. The descriptions of the two reactions are incomplete — important co-substrates are ignored.
  2. The formulation of the question seems to assume that the energetics of a biochemical reaction series is reflected solely in the interconversion of ATP and ADP, rather than the free energy changes that occur in all associated reactions.

Detailed explanation

To make a valid comparison of the thermodynamics of glucose oxidation to CO2 with its synthesis form CO2 we have to consider the single reversible reaction: Glucose to carbon dioxide (The ‘6H’ may appear rather odd, but it is accounted for in the reduction of cofactors etc.. We cannot include oxygen in the equation because it is not involved chemically in the synthesis of glucose. Therefore for this treatment the electron transport chain is not included, although it is discussed later.)

The reaction from left to right is associated with a certain decrease in Gibbs free energy (ΔG) and that from right to left in a corresponding increase of the same value. In a non-biological context this might involve the evolution and use of heat energy, but in the cell it involves the chemical energy of bonds between atoms. We therefore have to consider all the chemical reactions to which the reaction above is coupled, i.e. which receive or transfer free energy. The additional reactions and the free energy changes they involve can be found in the chapters on glycolysis, the tricarboxylic acid cycle and the Calvin cycle in Berg et al. and are (omitting water, hydrogen ions and inorganic phosphate): Energetics of glucose oxidation and synthesis From which it can be seen that the chemical energy input required for synthesis is greater than that obtained, contrary to what is asserted in the question

What about the ATP?

Yes, the cell oxidizes NADH and FADH2 and uses the free energy change to build an electrochemical gradient, the dissipation of which generates ATP (30 molecules per glucose molecule is the current estimate). However that is separate process, with no counterpart in the Calvin Cycle, where the NADPH is generated from photosynthetic reduction and not from ATP. However, if you perform a naive ‘currency conversion’ at the rate of 3ATP per NAD(P)H and 2ATP per FADH2, the balance is: input for synthesis 54 ATP, output from oxidation 38 ATP, i.e. the same general result as above.

Popularizing science is not easy, and those who make the effort can be excused for equating energy with ATP (often as clip-art lightening flashes). However if you wish to study metabolism you need to think scientifically about chemical thermodynamics and free energy. The fact that the hydrolysis of ATP to ADP is accompanied by a decrease in Gibbs free energy is not particularly chemically remarkable in the context of the free energy changes of other chemical conversions in the cell (including NAD(P) oxido-reductions). What is remarkable is that the cell has evolved enzyme that catalyse reactions in which the free energy change from this conversion is not lost as heat, but can be used to offset the +ve ΔG of a reaction to which it is coupled. Reactions involving NAD(P)H do the same thing, but they are limited to reductions, and the free energy change is inconveniently large.

Why the accepted answer to this question is a red herring

The accepted answer addresses a different question to that posed. It cites a calculation of the photon energy required for converting carbon dioxide to glucose. This is irrelevant to the thermodynamics of the Calvin Cycle, which can occur in the dark. It is relevant only to the efficiency of the use of photon energy in phosphorylating ADP and reducing NADP+ (incidentally in an open system) — a distinct reaction. The thermodynamics of the reactions the question refers to — glucose / carbon dioxide interconversion processes — do not involve the generation of their co-substrates.

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    $\begingroup$ Thank you@David, I wasn't considering the NADPH if we consider them and then equate 12 NADPH produces approximately 12×3= 36 ATPs, and adding 18ATPs it gives 54 ATPs ,which are more than that produced in oxidising it that is 36ATPs. $\endgroup$
    – sreekara
    Commented Feb 27, 2017 at 15:03
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    $\begingroup$ @ David I have another doubt, the mechanism of ATP synthesis in mitochondria and chloroplast are almost the same, but there is a big difference in mitochondria the f1 particle uses 2 H+ ions to synthesis one ATP and whereas in chloroplast it uses 3H+ ions for the same, shall I post it as a separate question? $\endgroup$
    – sreekara
    Commented Mar 11, 2017 at 13:33
  • $\begingroup$ @sreekara — I'd definitely post that as a separate question. (And a small linguistic point. I realise that 'doubt' is used in Indian English to introduce a question. However in British and US English it is never used that way. 'Doubt' only conveys the idea of disbelief or uncertainty.) $\endgroup$
    – David
    Commented Mar 11, 2017 at 14:11
  • $\begingroup$ I will post the question tomorrow because I have posted one today. $\endgroup$
    – sreekara
    Commented Mar 11, 2017 at 14:14
  • $\begingroup$ I have now revised my answer as I thought it important to address the misaprehensions about ATP and "energy" clearly with a table of free energy changes. The gist is the same as before. $\endgroup$
    – David
    Commented Apr 6, 2017 at 14:49

Background: Law of conservation of energy states that energy in a closed system remains constant. So the amount of energy in a molecule remains same when its annihilated( also consider law of energy mass equivalence).

Before answering this I would like to point that in the formation of glucose not only energy from ATP is utilised to fix carbon but also light energy is also utilised, which is about :

It takes eight (or perhaps 10 or more) photons to utilize one molecule of CO2. The Gibbs free energy for converting a mole of CO2 to glucose is 114 kcal, whereas eight moles of photons of wavelength 600 nm contains 381 kcal, giving a nominal efficiency of 30%. $^{(1)}$

So, how is the light energy utilised in Calvin cycle?

This light energy is used for formation of NADPH which used in Calvin cycle. Therefore not only ATP but also NADPH are required for generation of energy. So after considering the above facts one can argue quantitatively that law conservation of energy is not violated in case of oxidation of glucose by mitochondria.

  • $\begingroup$ I do not think this answers the question posed, as it relates to the energetics of the generation of ATP and NADPH by light in a thermodynamically open system. The Calvin cycle does not involve light. I believe my own answer is correct in addressing the failure to consider the energetics of anything other than the hydrolysis of ATP. $\endgroup$
    – David
    Commented Apr 6, 2017 at 14:52
  • $\begingroup$ @David I never said that Calvin cycle directly utilises light energy. The OP asked broadly about violation of law of energy conservation between photosynthesis and respiration. My answer too says that the NADPH is the entity which the OP has forgotten. Just to have a broad perspective I highlighted the fact that a lot more of light energy is used to form glucose than the energy released on oxidation. How is my answer misleading? $\endgroup$
    – JM97
    Commented Apr 6, 2017 at 15:09
  • $\begingroup$ I am sorry, but the poster did not ask broadly about the violation of energy conservation, he argued specifically from figures for ATP involvement in glucose synthesis and oxidation. By answering in terms of the energetics of a different reaction (the light reaction) you were implying that it was relevant to the thermodynamics of the dark reaction. It isn't. $\endgroup$
    – David
    Commented Apr 6, 2017 at 16:07
  • $\begingroup$ I can't force OP to accept your answer as accepted. $\endgroup$
    – JM97
    Commented Apr 6, 2017 at 16:11
  • $\begingroup$ That's not the point. The accepted answer need not be the best answer. It is valid to comment with diagreement, however. I am just copying the practice common of the original SE StackOverflow, which I was familiar with before joining this SE. $\endgroup$
    – David
    Commented Apr 6, 2017 at 16:57

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