Re-edited! Enzyme Kinetic--Good conceptual &graph question

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SaintJude

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Triosephosphate isomerase is a maximally efficient enzyme, which means that it catalyzes its reaction almost as fast as the enzyme and substrate encounter each other. When glutamate-165 is replaced by aspartate, the reaction is 1,000 times slower. The free energy profile for the wild-type enzyme is shown below (1st picture) . What would the profile for the mutant enzyme look like?

EhvD1.jpg


A & B
CB0UT.jpg


C & D

sBrBK.jpg

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Rate and Thermodynamics have got jack **** to do with each other. Delta G dun depend on Energy Activation.

Same graph of Delta G IMO

Before you post it tell me if I'm right.

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Okay, so all the graphs you now added have the same thing. Since it's slower, the energy to overcome must be larger. So I'm gonna guess B....? Could be D, but not sure.
 
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NOICEEEEEE.

So it was just the complex as I said in my first guess. Since it's much slower, it takes more Eact to get up there in the first place. Wasn't sure if that was B or D, but A is right out, and so is C. The complex and stuff is what gets raised in energy, not the end products.
 
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Way 1: Not sure if this is right, but it lead me to the right answer. You have a kinetic product and a thermodynamic product, right? And the kinetic product is faster and has a lower energy transition state, while the thermodynamic product is slower and higher energy. If the mutated enzyme makes the reaction proceed slower than the wild-type enzyme, then the thermodynamic product will probably be favored since it is more stable and a slower reaction means more time for the thermodynamic product to be formed (rather than the kinetic product continuously being formed). Thus, you would expect all of the free energy values to be higher.

Way 2: The shortcut way to this question though is to realize that (since they didn't give you any additional information) the change in free energy should be consistent for all E-S complexes in the graph. In other words, if one increases, they should all increase and if one decreases they should all decrease. Only graph B shows a consistent change so it must be the correct answer.

tl&dr:

Way 1: Slower rate leads to thermodynamic product which has higher energy. Graph B shows an increase in energy.

Way 2: Change in enzyme will be consistent for all the E-S complexes shown, only Graph B shows consistency across all 4 E-S complexes.
 
:thumbup: MedPr, makes a lot of sense for two reasons

1.) Assuming that the 1st and last hump are that of the enzyme-substrate and product-enzyme complexes respectively, then graph B rightly shows no change in the free energy. That's expected, isn't it?

2.) The mutant enzyme, moving slower, will indeed provide more time to allow for the formation of the thermodynamic, higher energy complex (that was the brilliant suggestion I had not seen myself--thanks MedPr)
 
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