Enzyme Thermodynamic/Kinetics Question

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If it is thermodynamically spontaneous for substrate and product to bind to the enzyme, why does the enzyme ever release the product? For example the K association of a product to the enzyme might be 5,000 1/M, and so is obviously favored. Why does the product eventually leave the enzyme's active site?
 

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  • #2
chemisttree
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Does the product bind with the same affinity as the starting material? Has the structure of the substrate undergone a change? Does the enzyme change it's structure (shape) when bound to the product vs. the substrate?
 
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The substrate has higher binding.
 
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chemisttree
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Which enzyme system are you referring to?
 
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I'm not referring to a specific one. It is more of a thought experiment. I guess what I need help with is balancing kinetics with thermodynamics, and how all of it inter-relates.
 
  • #6
Andy Resnick
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There's a lot of material out there you can read; I am looking at "Biological Thermodynamics", Donald Haynie, ch.8 (reaction kinetics).

Basically we can allow there to be a rate k_on for the complex E*S and a rate k_off for the decomposition to E S. They do not have to be the same. Also, there is a rate k_2 for the formation of product P from E*S. Combining them, we get to the Michaelis-Menten equation for enzyme kinetics.

Thermodynamically, the various states E S, E*S, and P are all metastable states. None of them are global minima in the energy landscape, and so there is both a probability associated with the existence of a particular state and a rate constant connecting the various local minima.

In terms of the Gibbs free energy, it is true that some of the processes occur spontaneously- the dissociation of a G-protein, for example, in response to a binding event. The cell then must use GTP (in this case, more often it's ATP) to drive the reaction the other way. Kinases and phosphatases often use ATP for similar reactions. Na-K-ATPase uses ATP to move ions against the electrochemical potential.
 

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