I was reading about active transport in membranes where ATP is used. ATP "reacts" with the protein pump and converts into ADP and also make a conformational change to the pump. Now this conformational change can transport molecules from low concentrations to high concentrations areas i.e. opposite in their concentration gradient. My problem is that for this to happen we must couple the hydrolysis of ATP with the "reaction" of transport along the membrane.

But I can't find it in textbooks and also I can't imagine if such coupling makes sense. In many textbooks it is stated that the energy that released due to the hydrolysis fo ATP is used to drive the pumping. I don't know if it is possible to couple two reactions with no common intermediates. Also in the pictures that show the active transport like the one in the left side the molecules can't go through the lipids.

enter image description here

So the molecules just move from low to high but the moving from high to low (diffusion) is blocked because the molecules can't move through the lipids. In other words it isn't like molecules make something that it would be otherwise violated because at first the motion from one side to the other is blocked. I mean even if we have no ATP the molecules won't flow from high to low concentration. Then what is the point of "active" transport? For a moment I thought to describe the above scheme for active transport as the following reaction: $$\ce{ATP + H2O + M_{in}} \rightleftharpoons \ce{ADP + P_i + M_{out}} $$

Which can be thought as the sum of:

$$\ce{ATP + H2O } \rightleftharpoons \ce{ADP + P_i}$$


$$\ce{M_{in}} \rightleftharpoons \ce{M_{out}}$$

But as I stated I haven't find a coupling reaction with no common intermediates and also it doesn't make sense to write the second equilibrium as the molecules can't go through the lipids. Also from the second reaction we should expect the concentrations on both sides to be equal (because the equilibrium constant must be equal to unity) no matter the concentration of ATP.

So can the first equilibrium alone describe the "active" transport? I mean if this reaction is exergonic then indeed by varying the amount of ATP we can alter the concentration of M outside of the membrane due to the Le Chatelier principle.

Edit ATP hydrolysis causes conformational changes to the protein channel which in turns "passes" molecules from one side to the other. Fine. But what if the lipids were permeable to the molecules? Then shouldn't the two concenctrations on the two sides (in and out) be equal or because the ATP hydrolysis is exergonic we will expect to have an imbalance in the concentrations?

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    $\begingroup$ Welcome to SE Biology and to Biochemistry. Biochemistry requires a different attitude than chemistry, but you still need to find a textbook that is written from a chemical standpoint and not a biological one. In this respect the best one available to consult online is Berg et al. on NCBI Bookshelf. The section you need is in chapter 13. You also need to stop thinking you know better ("I can't imagine if such coupling makes sense" indeed!) and read the sections of this book about this coupling, which is the energetic underpinning of life. $\endgroup$
    – David
    Commented Dec 27, 2020 at 17:48
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    $\begingroup$ "But what if the lipids were permeable to the molecules?" - It will save you a lot of time to first discover whether this is true for the molecules being actively transported. $\endgroup$
    – Bryan Krause
    Commented Dec 28, 2020 at 16:00
  • $\begingroup$ @BryanKrause Thanks for the comment. I think this is where I get it wrong because in thermodynamics we can treat diffusion as a "reaction". But here we have also the exergonic reaction of "pumping". So with no pumping allowed we will have passive diffusion. Whereas with pumping we will have a difference in the in and out concentrations because this reaction of "pumping" is favored. $\endgroup$
    – Anton
    Commented Dec 28, 2020 at 16:09

1 Answer 1


But I can't find it in textbooks and also I can't imagine if such coupling makes sense. In many textbooks it is stated that the energy that released due to the hydrolysis fo ATP is used to drive the pumping

I've been reading Molecular Cell Biology of Lodish and it has used "coupling" of hydrolysis of ATP with other reactions a lot, ergo it is probably correct to phrase it that way.

I don't know if it is possible to couple two reactions with no common intermediates

It may be a little different for each transport but I consider active transport as a result of multiple equilibria each of which as important as the others. You may notice that by multiplying the Ks of different equilibria, the intermediate particles would be omitted. For example

$\ce{H2SO4 + H2O } \rightleftharpoons \ce{HSO4- + H3O+}$ : $K_1$

$\ce{HSO4- + H2O } \rightleftharpoons \ce{SO4^{2-} + H3O+}$ : $K_2$

by multiplying these two we'll get:

$\ce{H2SO4 + 2H2O } \rightleftharpoons \ce{SO4^{2-} + 2H3O+}$ : $K_{overall}=K_1 \times K_2$

By doing so, a few problems arise, firstly you'll totally ignore the existence of $HSO_4^-$ as an intermediate by summarizing. Secondly, you may falsely consider the order of reaction as 3 instead of two related reactions with the order of 2. Thirdly, molar concentrations calculated with $K_{overall}$ tend to have deviations from reality, and so on...


$\ce{ATP + H2O + M_{in}} \rightleftharpoons \ce{ADP + P_i + M_{out}}$

is too minimalized to be considered as a good representation of the reaction. As an example for better clarification, here is a paper about the active transport of $Ca^{2+}$ in the muscle cells and one of its picturespicture The process involves 6 different species of the enzyme (E). Also, remember that the whole process should be displayed by these 6 equilibria. (Well, forming a complex between a molecule and a multi-charged ion requires several other equilibria which we may also be considered(stepwise production of $Ag(NH_3)_2^{+}$ for example).

because the equilibrium constant must be equal to unity

I didn't catch that pretty well, Anyway

What favors the active transport in a membrane?

The hydrolysis of ATP provides the energy for the next reaction(s). In our calcium transport example, an aspartate becomes phosphorylated which unstables the protein current conformation eventually leading to the consumption of the extra energy. The extra energy is consumed via the movement of the $Ca^{+2}$ against its entropy. The cycle can happen again and again.

For more readings on the conformation and stability of a single molecule read Force field (chemistry) and alkane conformation.

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    $\begingroup$ If you've just read about coupling in a Cell Biology text and from this deduce that "it is probably correct", you are perhaps not best placed to answer this question. I don't think the understanding of ATP and energy coupling lies in consideration of reactions of sulphuric acid, force fields or alkane conformation. $\endgroup$
    – David
    Commented Dec 27, 2020 at 17:53
  • $\begingroup$ @Sam So if the ATP hydrolysis doesn't take place at the protein, pumping won't happen? Because as you have written the reaction the protein is regenerated (like a catalyst). $\endgroup$
    – Anton
    Commented Sep 12, 2021 at 16:21
  • $\begingroup$ @Anton Correct... $\endgroup$ Commented Sep 13, 2021 at 14:59

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