From chemistry-request@www.ccl.net Fri Feb 5 02:55:37 1999 Received: from mailbox2.ucsd.edu (mailbox2.ucsd.edu [132.239.1.54]) by www.ccl.net (8.8.3/8.8.6/OSC/CCL 1.0) with ESMTP id CAA22380 Fri, 5 Feb 1999 02:55:36 -0500 (EST) Received: from physics.ucsd.edu (physics.ucsd.edu [132.239.69.26]) by mailbox2.ucsd.edu (8.9.1a/8.9.1) with SMTP id XAA28182; Thu, 4 Feb 1999 23:55:33 -0800 (PST) Received: from localhost by physics.ucsd.edu (SMI-8.6/SMI-SVR4) id XAA08386; Thu, 4 Feb 1999 23:53:48 -0800 Date: Thu, 4 Feb 1999 23:53:48 -0800 (PST) From: Margaret Cheung X-Sender: cheung@physics To: kynn@panix.com cc: chemistry@www.ccl.net Subject: Re: CCL:G:Histogram method, redux In-Reply-To: <199902042041.PAA20610@panix3.panix.com> Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Hi, KJ, Thank you for raising such a fun topic to discuss about. I guess there are some ideas needed to be carefully defined. 1. If the folding is "perfectly" two-state like behavior, and both the unfolded and folded states are locally stable on the energy landscape, then the free energy diagram should have a double well like feature as a function of some order parameters, such as radius of gyration, contact numbers, or number of hydrogen bonds formation etc. Such a feature which follows distinct two-state behaviors should be considered as the first order phase transition. The transition temperature, also the folding temperature (Tf) in this case, will be very sharp and smoothed out by the finite size effect. The signature of Tf could be also obtained by observing the variation of fluctuations in the energys, specific heat, from the simulation. So, by plotting Cv vs T, you can still get Tf from the energy output from the simulation. 2. Your argument 4 can not be led to argument 5. What you observed in the simulation was related to the stability between the folded and the unfolded states under various temperature. It is the signature of the first order phase transition to have such a free energy profile--when temperature is low, configuration space was trapped around the folded state on the energy landscape and the MFPT (mean first passage time) for such a configuration to escape from this local minumum is indeed very long. The result you had will be more related to the identification of where transition states are along the reaction coordinates, instead of the determination of thermodynamic value Tf. More comments are welcomed. Regards, Margaret S. Cheung Physics/Biophysics Department 0350 University of California, San Diego 9500 Gilman Drive, La Jolla, CA 92093-0350 On Thu, 4 Feb 1999 kynn@panix.com wrote: > > > > Thanks for the replies to my earlier query about the histogram method. > I have attempted to read some of the references suggested (given > below, for those interested, along with other references requested), > but I must confess that I can barely understand them. Before spending > any more time trying to make sense of them, it would be wise to > determine whether these methods would actually serve my purposes. So > here I give a schematic description of the problem I'm dealing with, > in the hope that someone may be able to tell me whether the "multiple > histogram method" would be of any use. > > Consider a model of protein folding dynamics, with the following > features: > > 1. The folding is perfectly two-state, the two states being designated > U (unfolded) and F (folded). Moreover, the two states are well > separated in terms of energy; i.e., if we were able to produce an > energy histogram at the folding transition temperature (Tf), the > distribution would be distinctly bimodal, with a largely unpopulated > region separating the peaks corresponding to F and U. > > 2. Let t_F(T) and t_U(T) be the mean passage times for the folding > (U->F) and the unfolding (F->U) transitions, respectively, as > functions of the temperature T. > > 3. We distinguish three temperature regimes I < II < III. In regime > I, for all practical purposes, t_U(T) is infinite, while t_F(T) is > "reasonably small". Conversely, in regime III, t_F(T) is practically > infinite, while t_U(T) is small. > > 4. In regime II, a narrow neighborhood of the folding transition > temperature Tf, both t_F(T) and t_U(T) are roughly equal, and both are > extremely large (of the order of 2-3 CPU-weeks on our workstations). > > (Of course, no sharp boundaries separate these three temperature > regimes; t_F(T) and t_U(T) grow rapidly, as one approaches regime II > from regime I and III, respectively.) > > 5. Because of the slowing down described in 4, it is extremely > difficult to determine Tf. > > The single histogram method (which I described in my original post) > would require a single run near Tf, long enough to sample both the F > and U states in a way that actually reflects their relative > probabilities. For the reason mentioned in (4) and (5) this is > impractical, which seems to rule out this method. > > But it was suggested by a few responders to my earlier post that the > *multiple* histogram method may be useful in this case. The only way > I can see to apply this method is by producing two histograms, one at > temperature T1 in regime I to sample F, and one at temperature T3 in > regime III to sample U. Then, if we "absorb" the temperature into the > definition of the Hamiltonian (as Ferrenberg and Swendsen do, see > reference below), then this could be construed a case of having two > "Hamiltonians", H/(k*T1) and H/(k*T2), (where H is the actual > Hamiltonian, in the standard nomeclature). Then I would use the > method of Bennett (see reference below) to obtain a free energy > difference between these two states (as a function of T). Is this > correct? > > (The method of Ferrenberg and Swendsen is a generalization of > Bennett's method to more than two histograms, if I've understood them > correctly). > > These multiple histogram methods require a certain amount of overlap > between the histograms, which leads, for the system described above, > to a somewhat paradoxical situation: the greater the overlap between > the two states (meaning that neither P(U;T)/P(F;T) nor P(F;T)/P(U;T) > is too small), the closer the simulation temperatures have to be to > Tf, and therefore the longer it would take to adequately sample the > phase space (as remarked in (4) above). This suggests that the > multiple histogram method would not be very useful either. Am I > right? > > Thanks for your patience. Below are the references mentioned above. > Regards, > > KJ > >