On Thu, Feb 21, 2019 at 8:41 PM Cruzeiro,Vinicius Wilian D <
vwcruzeiro.ufl.edu> wrote:
> Hello Rodrigo,
>
>
> Interesting question.
>
>
> In Amber, HID and HIE are "equally" deprotonated states. When the fraction
> of protonated species is computed in cphstats, it makes no distinction
> between the HID and HIE populations as they both count as "deprotonated"
> (even though you could easily extract the populations for HID and HIE
> yourself from the cpout file).
>
>
> Now, explicitly addressing your question: when the reference energies for
> the HIP residue are computed, the usual approach is to first look at only
> the HID --> HIP transition and compute a reference energy associated only
> to this transition in such a way that the yielded pKa value in the
> simulation is 7.1. Then, the same procedure is repeated considering only
> the HID --> HIE transition but aiming at a pKa of 6.5. In the production
> simulations, we then consider all the 3 protonation states (HIP, HIE and
> HID) and use the two reference energies obtained previously. For
> completeness it is perhaps important to mention that the MC algorithm may
> attempt a protonation state change between the HIE and HID states (Note: as
> the target protonation states are chosen randomly, once you are at the HIE
> state the probability of attempting to go to either HID or HIP is 50%).
> When a protonation state change attempt between HIE and HID is performed,
> the reference energy associated with this transition is the difference
> between the two reference energies previously computed. You can check in
> details all the numbers used in Amber by executing: cpinutils.py --describe
> HIP .
>
>
> Because all 3 protonation states are considered at the same time, at pH
> values 6.5 or 7.1 the fraction of protonated species is not going to be
> 50%. The effective pKa value is going to be between 6.5 and 7.1. This value
> depends also on the charge distributions across all 3 states.
>
Just a minor addendum -- the effective pKa should actually be a bit *under*
6.5. Both states HID and HIE are "deprotonated", so there are more
deprotonated states than there are protonated ones (higher entropy). If
you want to prove this to yourself qualitatively, assume the effective pKa
of the epsilon position is 20 (effectively infinite -- another way of
saying it will *never* deprotonate). In this case, you assume there is
only one titratable site, and the residue would have a pKa of 6.5 (not
somewhere between 6.5 and 20). Lowering the pKa of the epsilon site can
only decrease the pKa of the whole histidine side-chain from the pKa of the
more acidic site, since the only effect it could possibly have is to
increase the total population of deprotonated species at any given pH.
Let's look at what happens at pH 6.5, to a first order approximation. Half
of the histidine concentration will be protonated at the delta position and
half deprotonated. Of the half that is protonated, some fraction of that
will be deprotonated at the epsilon position as well (when it's
deprotonated, the effective pKa of the epsilon site gets very large, so
it'll never deprotonate).
I re-derived these reference energies years ago, but the way they were
originally done (in 2004 and again by myself ~6 years later) was to simply
compute the deprotonation free energy using TI and taking that value
directly. I tried optimizing each of the sites' reference energies by
defining a *new* titratable residue where only the delta position titrated
(only MC steps HID->HIP and HIP->HID could be attempted) or only the
epsilon position titrated (only MC steps HIE->HIP and HIP->HIE could be
attempted), and tweaking the reference energies until I got 50%
protonation/deprotonation at the pH=pKa for that particular site. Turns
out I really couldn't do any better than the plain TI energy (which is good
confirmation of the overall approach as well as the implementations of both
the TI and CpHMD codes).
All the best,
Jason
--
Jason M. Swails
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Received on Wed Feb 27 2019 - 22:30:03 PST
