Mixed Inhibition: Finding Ki & Ks Relation

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The discussion focuses on understanding the relationship between the equilibrium constants (Ki and Ks) in a mixed inhibition scenario within a thermodynamic cycle. Participants clarify that the equilibrium constant between enzyme (E) and enzyme-substrate complex (ES) can be expressed as Ks = [E][S]/[ES]. The conversation emphasizes that while these constants can be related, they are generally not equal. The need to simplify the representation of the reaction scheme is highlighted to facilitate understanding. Overall, the thread aims to clarify the mathematical relationships governing mixed inhibition.
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Homework Statement


I have this mixed inhibition, and knowing that it is a thermodynamic cycle i am asked to find the relation between the different Ki and Ks.
https://scontent.xx.fbcdn.net/hphotos-xfp1/v/t35.0-12/12699252_1024216030972374_1605707906_o.jpg?oh=8b60ba8c953ea368fe8fc09af6fe5eaa&oe=56C77997

Homework Equations

The Attempt at a Solution


I am not even sure to understand the question...
 
Last edited:
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Dassinia said:

Homework Statement


I have this mixed inhibition, and knowing that it is a thermodynamic cycle i am asked to find the relation between the different Ki and Ks.
https://scontent.xx.fbcdn.net/hphotos-xfp1/v/t35.0-12/12699252_1024216030972374_1605707906_o.jpg?oh=8b60ba8c953ea368fe8fc09af6fe5eaa&oe=56C77997
2. Relevant equation

The Attempt at a Solution


I am not even sure to understand the question...

Look at the top line only. What is the equilibrium constant between E and ES?
Well okay since you say you don't understand the question Ks = [E][ S]/[ES] = ?

Now rub out to the arrows of the top line. Leave the others. What is the above-defined equilibrium constant equal to?

Can these equilibrium constants be different?
 
epenguin said:


Look at the top line only. What is the equilibrium constant between E and ES?
Well okay since you say you don't understand the question Ks = [E][ S]/[ES] = ?

Now rub out to the arrows of the top line. Leave the others. What is the above-defined equilibrium constant equal to?

Can these equilibrium constants be different?
Between E and ES Ks=k-1/k1 Is it equal to k'-1/k'1=Ks' ?
 
Dassinia said:
Between E and ES Ks=k-1/k1 Is it equal to k'-1/k'1=Ks' ?

In general no.

Rub out to the arrows of the top line. Leave the others. In fact just write out the scheme again without the k1, k-1 steps. Can't you work out what that gives you for [E][ S]/[ES] ?
 

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