Question

Phi-value analysis - Free energy of activation contribution

Answer 1

So as people know the phi-value in given the mathematical definition of
\[\Large \phi = \frac{ \Delta \Delta G_{\ddagger-D} }{ \Delta \Delta G_{N-D} }\]
In this post I will try go over theory and considerations for the contributions to the term \(\Delta \Delta G_{\ddagger-D}\) in terms of statistical mechanics and reaction kinetics.

My wish is for the users to consider if the approximations seems legit and useable. Please keep the standard high for replying.

Answer 2

So lets look at a folding reaction. We start by setting up the reaction as:
\[\Large \sf E_u \rightleftharpoons E^{\ddagger} \rightarrow E_f\]
Where \(\sf E_u\) is the unfolded enzyme and \(\sf E_{f}\) is the folded enzyme. As the reaction suggest we assume a rapid equilibrium between the unfolded state and transition state.
\[\Large K^{\ddagger}=\frac{ \left[ \sf E^{\ddagger} \right] }{ \left[ \sf E_u \right] } \rightarrow \left[ \sf E^{\ddagger} \right]=K^{\ddagger} \left[ \sf E_u \right] \]

Answer 3

If we assume that the last step is the rate determining step: and so write:
\[\Large v=k^{\ddagger}\left[ \sf E^{\ddagger} \right]=K^{\ddagger}k^{\ddagger}\left[ \sf E_u \right]=k_r\left[ \sf E_u \right]\]
That gives the reaction rate constant the following term:
\[\Large k_r=K^{\ddagger}k^{\dagger}\]

Answer 4

The rate constant has been handled as by Atkins et al. QMC and write
\[\Large k^{\ddagger}=\kappa \nu^{\ddagger}\]
Here is \(\kappa\) the transmission coefficient and \(\nu^{\ddagger}\) the frequencies of transition. By convention it normal to set \(\kappa\) to 1, if unknown.

In order to handle the equilibrium constant of transition \(K^{\ddagger}\), I've used statistical mechanics and specifically the partition function for the vibrational contribution.
We can express the equilibrium constant in terms of
\[\Large K^{\ddagger}=\frac{ q_{E^{\ddagger}} }{ q_{E_{u}} } \exp \left( \frac{ -\Delta E_0 }{ RT } \right)\]
This is done from the relationship between the partitions and equilibrium constant:
\[\Large K=\left\{ \prod_{j}^{}\left( \frac{ q_{j,m}^{\Theta} }{ N_A } \right)^{v_j} \right\} \exp \left( \frac{ - \Delta E_0 }{ RT } \right)\]

Answer 5

The partition is for vibration is considered very low during transition state and much lower than in their folded or unfolded state. I there for look at the first considerable vibration mode:
\[\Large q_{E^{\ddagger}}=\frac{ 1 }{ 1-(1-\frac{ h\nu^{\ddagger} }{ kT }+...) }\]
By the assumptions above and that \(\large \frac{ h\nu^{\ddagger} }{ kT }<<1\) we get
\[\Large q_{E^{\ddagger}}=\frac{ kT }{ h \nu^{\ddagger} } \bar{q}_{E^{\dagger}}\]
Where \(\large \bar{q}_{E^{\dagger}}\) is the partition function for vibration minus 1 mode of vibration.

Answer 6

Using this we get the equilibrium constant at transition as
\[\Large K^{\ddagger}=\frac{ kT }{ h \nu^{\ddagger} }\frac{ \bar{q}_{E^{\ddagger}} }{ q_{E_{u}} } \exp \left( \frac{ -\Delta E_0 }{ RT } \right)=\frac{ kT }{ h \nu^{\ddagger} }\bar{K}^{\ddagger}\]
Where \(\large \bar{K}^{\ddagger}\) is the transition equilibrium constant minus 1 mode of vibration in the contribution to transition state partition.

When putting it together we get the reaction rate constant as:
\[\Large k_r=K^{\ddagger}k^{\ddagger}=\frac{ kT }{ h \nu^{\ddagger} } \bar{K}^{\ddagger} \kappa \nu^{\ddagger}=\frac{ kT \kappa }{ h } \bar{K}^{\ddagger}\]
Using this method we get rid of the frequencies of transition which are usually unknown and can relate the "equilibrium constant" to the rate constant of folding.

My question is. Is it a fair approximation to treat \(\large \bar{K}^{\ddagger}\) as a equilibrium constant for which we can substitute the relation between \(\Delta G\) and the equilibrium constant:
\[\Large K=\exp \left( \frac{ -\Delta G }{ RT } \right)\]

@aaronq

Answer 7

the relationship between \( \Delta G^{o} \) and \(K \) is \( \Delta G^{o} = -RT ~Ln~K \), which allows you to calculate any \( K\) from standard-state free energy of reaction

when the reaction is at equilibrium, \(Q=K \) the driving force behind a chemical reaction is zero, \(\Delta G = 0 \)

the relationship between free energy of reaction at any moment, \(\Delta G \), and standard-state free energy reaction is \( \Delta G = \Delta G^{o}~ RT ~Ln~Q\)


for summary \(\href{ http://finedrafts.com/files/CUNY/chemistry/General/Brown-Holme/Chemistry_Engineering_Students_2nd_txtbk.pdf }{chapters~11,12,13} \)

or if you have time \(\href{ http://finedrafts.com/files/CUNY/chemistry/General/Denbeigh/The%20Principles%20of%20Chemical%20Equilibrium,%204th%20edition%20by%20K.%20Denbigh.pdf }{principles~of~chem~equilibrium} \)

Answer 8

well that link is a bummer
it was previewing correctly

Answer 9

standard-state free energy of reaction is \(\Delta G^{o} \)

Answer 10

which I know that you already know

Answer 11

Yeah it is super annoying never works my self these days with href...

Indeed the relations you bring up can be used. And was also the fundament of my first thinking. The problem for me was to lay the mathematical fundament for data collection. And to use the relation:
\[\Large \Delta_{\ddagger} G=-RTln(K^{\ddagger})\]
The problem is we cannot stop the reaction to study the transition.

Answer 12

no, we cannot

Answer 13

we can only observe the initial and final

Answer 14

But it seems like somebody found a way around it, they suggest reaction kinetics and look at the initial and final points as you along with the rate constant during folding and unfolding, and from there say something about the transition state.
This relationship is what I kinda I try figure out, if this kinda of approximation is even valid.

Answer 15

My biggest concern was to get rid of the frequencies of transition as it is unknown in any experimental setup I can think of. But I think this kind of analysis mostly belong to transition state theory (a subject I didn't study that much to be honest)