Say we have an enzyme that has multiple structural conformations, say A, B, C and D. Additionally, we say that the enzyme must go through all conformations, to get to one end to the other, i.e.

$$\large \ce{A <=>[k_{A-B}][k_{B-A}] B <=>[k_{B-C}][k_{C-B}] C <=>[k_{C-D}][k_{D-C}] D}$$

So there's no skipping any of these steps to go from A to D.

Given all the rate constants derived experimentally, and thus knowing the equilibrium constants for every step, is it possible to derive an overall equilibrium constant that describes which conformation the enzyme is most likely to be found, and perhaps more importantly, the relative fractions for every conformation?

  • $\begingroup$ Of course it's possible, are you asking how to do it in general or specific cases? $\endgroup$ – Mithoron Mar 14 '18 at 23:07

If we use the equilibrium constants:

$$\ce{A <=>[$K_1$] B <=>[$K_2$] C <=>[$K_3$] D}$$

And we call $[\text{E}]$ the total enzyme concentration, we have:

$$ [\text{A}] + [\text{B}] + [\text{C}] + [\text{D}] = [\text{E}]\label{a}\tag{1}$$

It's not hard to see that $\displaystyle K_1 K_2 K_3 = \frac{[\text{D}]}{[\text{A}]}$. From here, we have that $\displaystyle [\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3}$. We also get:

$$[\text{B}] = K_1[\text{A}] = \frac{[\text{D}]}{K_2 K_3}$$ $$[\text{C}] = K_2[\text{B}] = \frac{[\text{D}]}{K_3}$$

Hence, substituting these last three equalities in $\ref{a}$, we get:

$$ \frac{[\text{D}]}{K_1 K_2 K_3} + \frac{[\text{D}]}{K_2 K_3} + \frac{[\text{D}]}{K_3} + [\text{D}] = [\text{E}]$$


$$[\text{D}] = \frac{K_1 K_2 K_3 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$

From here we can also express the concentrations of the other species in terms of $[\text{E}]$:

$$[\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3} = \frac{[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$ $$[\text{B}] = K_1 [\text{A}] = \frac{K_1 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$ $$[\text{C}] = K_2 [\text{B}] = \frac{K_1 K_2[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$

With these equations we may compare the concentrations of each species at equilibrium, thus we are able to determine which one is the highest based on the values for the equilibrium constants; and we can also obtain their mole fractions (divide each by $[\text{E}]$, which will eliminate this term from all the numerators).

  • $\begingroup$ Just to be clear, $[B] = K_1 [A]$ is just an algebraic construct, right? Because physically it wouldn't make sense to claim that $[B]$ is only dependent on this small part of the reaction. $\endgroup$ – komodovaran_ Mar 15 '18 at 10:11
  • $\begingroup$ Yes and no. In principle, at equilibrium $[A]$ can only convert to $[B]$, so its concentration at equilibrium is only dependent on $[B]$ at equilibrium. They are all related, you may also rewrite all the equations in term of any component, meaning that they all depend on each other. $\endgroup$ – ralk912 Mar 15 '18 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.