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Can biological enzymes catalyze thermodynamically unfavorable reactions? I read that an enzyme lowers the activation energy of a reaction by offering an alternative reaction pathway with a lower activation energy, however the ΔG of the reaction is unchanged. If enzymes can do this, then how exactly does this work?

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    $\begingroup$ I've deleted the comments as the question has an accepted answer and flags started to pour in. Please go to chat to resolve outstanding matters, if any. They served their purpose as a few good quality answers are in place now. Q and A's also seem to be on the same page now in terms of the can and how issue. $\endgroup$ – AliceD Jun 26 '17 at 21:50
  • $\begingroup$ "I read that [...]" - for completeness, could you add where you read this? $\endgroup$ – arboviral Jun 28 '17 at 7:21
  • $\begingroup$ I read this in my school homework question, hence the tag. But there's a link here that mentions something to that effect: rsc.org/Education/Teachers/Resources/cfb/enzymes.htm $\endgroup$ – Jonathan Smith Jun 28 '17 at 11:18
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Enzymes can catalyze a thermodynamically unfavorable reaction by coupling it with a thermodynamically favorable reaction. Most often, enzymes use ATP hydrolysis reaction (energetically favorable) as a source of energy (in simple terms) to drive the unfavorable reaction forward. One important point to keep in mind here is that enzymes don't drive a reaction equilibrium forward or backward; they just help in achieving the reaction equilibrium faster. Also pay attention that enzyme itself does not change the thermodynamics of a reaction (as you say, ∆G of the concerned reaction remains the same, except at equilibrium), coupling of a favorable reaction with an unfavorable one only helps in making the overall reaction favorable. This phenomenon, of one reaction changing the rate of another reaction, is called induced catalysis and has nothing to do with the enzyme itself.

I will talk about how enzymes do this using examples, while the other answer talks about the semantics. An example which will help you understand this concept is ATP synthase. It is an enzyme found in the inner mitochondrial membrane which allows the movement of H+ from the inter-membrane space towards the mitochondrial matrix, using the energy of movement (since this is favorable and exergonic) to generate ATP. The reaction would look like this:

$$\ce{ADP + P_i + 3 H^+_{IMS} \rightarrow ATP + 3 H^+_{matrix}}$$ (remember that exactly 3 H+ are not required, this is a simplified version)

But, ATP synthase does not drive the reaction forward; it just helps in reaching the following equilibrium faster: $$\ce{H^+_{IMS} \leftrightharpoons H^+_{matrix}}$$ i.e. keeping the amount of H+ in inter-membrane space and matrix the same. This is why a constant gradient is required to keep them in working condition. For this, the electron transport chain continuously throws H+ into the inter-membrane space to maintain an electrochemical gradient. Along with this, enzymes ATP-ADP translocase and phosphate carrier continuously take out ATP from the matrix and throw in ADP and Pi. However, this lowers the electrochemical gradient (exchanging ADP3- for ATP4- in inter-membrane space, while phosphate carrier catalyzes electroneutral reaction), making the final reaction: $$\ce{ADP + P_i + 4 H^+_{IMS} \leftrightharpoons ATP + 4 H^+_{matrix}}$$

However, since the enzyme just helps in maintaining equilibrium, it can also work backwards i.e. it can also transport H+ against electrochemical gradient (unfavorable), coupling it with ATP hydrolysis (favorable) to reach equilibrium. The enzyme is now called ATPase and the reaction becomes: $$\ce{ATP + 4 H^+_{matrix} \rightarrow ADP + P_i + 4 H^+_{IMS}}$$

There are many more such enzymes which couple ATP hydrolysis (or other favorable reactions) to catalyze reactions which are either unfavorable or very slow (i.e. require a source of activation energy) under normal cellular conditions. Another factor, as the other answer explains, which helps in this is changing the concentration of products. As seen above, ATP-ADP translocase and phosphate carrier continuously decrease the concentration of products (ATP) and increase concentration of reactants (ADP and Pi) so that the overall equilibrium shifts forward and the enzyme keeps on driving the reaction forward (this, again, has nothing to do with with the enzyme itself, it is called Le Chatelier's Principle). Again, the other answer focuses on this issue.

References:

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    $\begingroup$ @david Well, I reread my answer and don't find any line which says that enzyme has something to do with thermodynamics. In fact, it is this point itself, to disregard which I say again and again that enzymes only help in attaining equilibrium faster. I don't know how you concluded that point from my answer, but I don't think anybody with even basic knowledge of enzymes would get misled by my answer. Also, I don't like giving one-line answers & try to add as many details as possible. Yet, you can post an answer if mine 'bamboozles' you. $\endgroup$ – another 'Homo sapien' Jun 26 '17 at 11:14
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    $\begingroup$ Excellent, clear answer. I don't know what David is on about, I can't see any way this answer would be misleading to the OP. $\endgroup$ – Bryan Krause Jun 26 '17 at 14:46
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    $\begingroup$ @David Is that really the basis of your complaint about this answer? The original question was "can" - the rest of this answer addresses "how". $\endgroup$ – Bryan Krause Jun 26 '17 at 18:56
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    $\begingroup$ Perhaps coupled reactions is what the OP was looking for, but it so that's not clear from the question. Coupling two reactions is an entirely different concept than catalysis, which simply means lowering the activation energy, so I think the wording of this answer is confusing. The correct answer is that catalysis has nothing to do with whether the reaction is favorable or not, as @David points out. (Also, ATP synthase is a rather poor example of a coupled reaction I think, as it is not fully stoichiometric; something like hexokinase would be better.) $\endgroup$ – Roland Jun 26 '17 at 19:17
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    $\begingroup$ @Roland I don't think OP was looking for anything in particular in a correct answer. I suspect, though I don't know for sure, that the OP is trying to come to terms with two obvious facts: 1) Not all reactions in biology are, by themselves, energetically favorable, and 2) Enzymes can't change ΔG, effectively they only alter reaction rates. So how does this work? This answer clearly answers that question. $\endgroup$ – Bryan Krause Jun 26 '17 at 19:33
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Can enzymes catalyze thermodynamically unfavourable reactions?

Enzymes don't change the equilibrium of a reaction, but the fact that an equilibrium exists means that the reaction proceeds in both the forward and reverse directions. Before equilibrium is attained, ΔG for the reaction is not 0. Thus, by definition, one direction is thermodynamically favourable (ΔG<0) while the other is thermodynamically unfavourable (ΔG>0). Since an enzyme lowers the activation energy of a reaction, it is catalysing both the forward and reverse reactions. This is a semantic argument since the reaction will still proceed in the direction which is thermodynamically favourable.


How can enzymes catalyze thermodynamically unfavourable reactions?

Other than in the sense I described above, enzymes do not catalyze thermodynamically unfavourable reactions. However, biological systems can make unfavourable reactions favourable. In addition to the reaction coupling discussed in the other answer (sometimes referred to as pushing) thermodynamically unfavourable reactions can be made favourable by pulling. In this case, for reactions where ΔG°>0 (ie the reaction under standard conditions is unfavourable), the reaction can be made favourable by decreasing the concentration of products relative to reactants, such as by immediately using them in a subsequent, thermodynamically favourable reaction. Mathematically, the relationship between the free energy change of the reaction under standard conditions and the actual reaction is given by:

$$\Delta G=\Delta G^{\circ} +R T\ln\left(\frac{[P]}{[R]}\right)$$

As the ratio of product to reactant concentrations decreases, the actual free energy change of the reaction decreases and can be made thermodynamically favourable.

To be clear, it is not the enzyme that makes an unfavourable reaction favourable (again, it only lowers activation energy). Rather, the enzyme is catalyzing a reaction that is now thermodynamically favourable.

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    $\begingroup$ Sorry for cluttering your answer again, but I just love the dualistic approach. +1 it also points out a fundamental problem on the why versus how questions though. Often they seem to go hand in hand and obscuring the question core. Your approach solves matters nicely :) $\endgroup$ – AliceD Jun 26 '17 at 22:00
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    $\begingroup$ Nice answer, +1! At least now I understand what I was missing in my answer ;) I've edited my answer to make clear the difference between what mine and your answers discuss :) $\endgroup$ – another 'Homo sapien' Jun 27 '17 at 4:55
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    $\begingroup$ @Roland (1) I'm not sure I appreciate this point, yet. (2) The answer is that enzymes don't make unfavourable reactions favourable. Then I go on to explain what can make unfavourable reactions favourable. $\endgroup$ – canadianer Jun 27 '17 at 15:23
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    $\begingroup$ @Roland (this applies to your comment here and on the other answer as well) - I think the discrepancy here is between biochemistry of enzymes as functionally interesting in and of themselves, and the OP and the answers are in a different mode of thinking, where enzymes are a means to an end, part of the rest of biology. Neither approach is wrong, but for the second approach, it is still within the scope of the problem to include other helper mechanisms, because the goal is to understand the whole process, not the action of one enzyme or one specific reaction. $\endgroup$ – Bryan Krause Jun 27 '17 at 16:34
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    $\begingroup$ If we were talking about cars, in answering "how does pressing the accelerator make a car go faster?" someone could answer that "well it lets fuel and air enter the engine faster which increases RPMs and translates to torque that makes the car go fast." Then someone else says no no the accelerator pedal doesn't do any of that, that's the job of the butterfly valve and the transmission and the wheels. The pedal just signals the car, mechanically or electrically, to open the butterfly valve. Both answers are right, but at OP's level of understanding the first answer makes more sense. $\endgroup$ – Bryan Krause Jun 27 '17 at 16:38

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