If you can provide some sound reasoning that touches on tertiary structures of proteins and does not use a lot of advanced chemistry jargon that might be really helpful, especially for an intro biology student. I think my problem is that activation energy is lowered by an enzyme and I know how its like on the Gibbs Free Energy diagrams. However I think I've been taking this as an axiom and I'm not really sure why.
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1$\begingroup$ Have you read the wiki article? What specifically don't you understand? How should we provide a detailed chemical explanation without using advanced chemistry jargon? You need to edit your question and explain exactly what you know and where exactly you're stuck. Otherwise, this question may be closed as violating our homework policy. "Homework" is interpreted to mean any academic or other assignment, test preparation, or task given in relation to a class, educational setting, or self-learning. $\endgroup$– MattDMoCommented Sep 19, 2015 at 2:28
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2$\begingroup$ Take a piece of wood and set it down on a slippery surface. Now take a saw and without holding the piece of wood try sawing through the wood. Much of the energy will likely be dissipated by the board sliding around and you will expend a lot of your own energy trying to cut the wood without much success. Now take the wood and securely fasten it between two clamps, like on a work bench. Now use the saw and cut through the wood. All of the energy that you are expending is transferred to the cutting action and isn't dissipated because the wood cannot move around. The clamps are holding it in place $\endgroup$– AMRCommented Sep 19, 2015 at 3:10
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2$\begingroup$ Nice description @AMR :) As an add-on - enzymes with multiple substrates often bind these substrates and bring them close to each other such that it is a near-perfect 3D confirmation to establish the bonding / breakage of chemical bonds. $\endgroup$– AliceD ♦Commented Sep 19, 2015 at 3:55
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1$\begingroup$ @AliceD Thanks... If the OP had said that they were taking Biochemistry then I would agree that enzymes would not be specific enough, but they did say Intro Biology, which means at this point of first semester they may be around Chapter 8 of Campbell and they likely are just struggling with the basic understanding, although Campbell, goes into a little more detail, they are taking very broad brush terms. We aren't Academia SE where it is geared towards graduate students and professionals only, so I don't think it needs to be closed. You are right though that there are many types of enzymes. $\endgroup$– AMRCommented Sep 19, 2015 at 4:49
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1$\begingroup$ @MattDMo the wikipedia article does not say how enzymes lower activation energy. Actually it is not easy to find answer to this question by a simple google search and most basic level biochem textbooks also do not explain this issue properly. $\endgroup$– WYSIWYGCommented Sep 20, 2015 at 10:05
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1 Answer
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BACKGROUND
We should first understand what activation energy really is and what does Arrhenius equation mean. For any chemical reaction to happen you need to break some existing bonds in order to form new bonds. Breaking existing bonds consumes some energy; in other words you have to convert the reactant to an active form which undergoes a reaction readily. The energy that is required to do so is called activation energy.
It is essential to note that activation energy is an empirical term because it does not actually consider the reaction mechanism. The above example of bond-breaking and re-formation can happen in many ways. Some mechanisms may require less activation energy than the others.
The Arrhenius equation denotes the relationship between reaction rate and activation energy:
$$k=Ae^{-\frac{E_a}{RT}}$$
where $k$ is the reaction rate constant, $A$ is the Arrhenius constant (an arbitrary proportionality constant), $E_a$ is the activation energy, $R$ is the universal gas constant and $T$ is the absolute temperature.
This equation tells you that the rate of reaction decreases with increase in activation energy and increases with increase in temperature.
The temperature indicates the internal energy of the system i.e. the movement of molecules. Collision between molecules is essential for chemical reaction to happen and increased temperature increases the chance of collisions (due to higher molecular movement).
How do enzymes lower activation energy of a reaction?
They either do that by increasing the chance of interaction between the reactants or diverting the reaction to follow a different mechanism that has a lower activation energy.
As you know, the enzyme binds to the reactants and forms a complex. This effectively brings the reactants closer thereby increasing the probability of their "collision" which in turn increases the reaction rate. This can be further extended to the level of molecular orientation. The analogy presented by AMR in their comment is great. It is one of the mechanisms by which the free energy of the system would increase — by reduction of entropy. Clamping the substrate in place can expose its reactive groups to the co-reactants, thereby allowing the reaction to proceed faster.
In most other mechanisms of enzyme catalysis, the functional groups in the enzyme active site react with the substrate to form intermediates and these intermediates which have lower activation energy give rise to product faster than the substrate in absence of the enzyme.
It is to be noted that this entire process of catalysis adds a few extra steps to the actual reaction (binding, formation of intermediates etc). The catalyst would be effective only if these combined rates are still faster than that of the uncatalyzed reaction.
