Re: [ccp4bb] -RTlnK = delH - TdelS

[Roger Rowlett]

This is absolutely correct. For many enzymes, Km is a conflation of many rate constants in the chemical mechanism. You would have to know for certain, based on prior mechanistic work, that all steps subsequent to substrate binding are slow enough for the experimentally measured Km to approximate Kd for substrate. An additional, and frequently overlooked complication is that the ionization states of active site groups important for binding and/or catalysis may change significantly with temperature because of T-dependent pKa shifts. The way to avoid this is to choose a pH for the experiment that ensures that T-dependent pKa shifts do not change the enzyme ionization state significantly. It is somewhat easier to measure thermodynamic properties for inhibitors, where Ki is typically closer to a true equilibrium constant than Km. Cheers. William G. Scott wrote: Dear James: The problem is that the equation is for an equilibrium constant, and Km is not one. Kd is one. If you can measure Kd, what you describe I think is fine. If you are measuring Km, you have to find a way of justifying the Km ~ Kd approximation, which usually is only safe for enzymes that turn over very slowly. I've never done these kinds of experiments, so someone else needs to answer this. Sorry. I've always been content to push the deltaG button in COOT. All the best, Bill On Nov 18, 2009, at 10:18 AM, james09 pruza wrote: Dear Sir, At equilibrium, how can one determine delH and delS, if temp. and km are known. I have km value and temperatures. I have to calculate delH and delS in equilibrium. Should I plot *lnKm vs 1/T* to have a slope of -delH/R and intercept of delS/R. Multyplying this value with R, will give the value of delH and delS , respectively. If delG= delH - TdelS delG= -RTlnK So, -RTlnK = delH -TdelS lnK = -delH/RT + delS/R Please suggest If I am wrong. I have planned to do ITC with the mutants as well to get the perfect parameters. Thanks for help. On Wed, Nov 18, 2009 at 7:44 PM, William G. Scott < wgsc...@chemistry.ucsc.edu> wrote: The K is an equilibrium constant. Km is not an equilibrium constant. It is a collection of rate constants. Kd is an equilibrium constant, but you have to be careful to make sure that the enthalpy and entropy terms correspond directly to this as well. The equation you quote implies standard state. Km is not the reciprocal of Kd. It is not an equilibrium constant. On Wed, Nov 18, 2009 at 8:24 AM, james09 pruza <james09x...@gmail.com>wrote: Dear All, Sorry for the non-ccp4 query. I have solved a crystal structure of an enzyme and woring on its biochemical aspect. We have a mutant of this enzyme and we are comparing some thermodynamic parameters of this enzyme with mutant( lke delH and delS, delG). we have done the expt at different temp. and know the km value at these temp. Now the question is how to get the value of delH and delS. The relation of these parameters is:- -RTlnK = delH - TdelS Is the K in this relation is km or kd. If it can be treated as kd, so it should be 1/km. All suggestions are welcome. Thanks James -- Roger S. Rowlett Professor Department of Chemistry Colgate University 13 Oak Drive Hamilton, NY 13346 tel: (315)-228-7245 ofc: (315)-228-7395 fax: (315)-228-7935 email: rrowl...@colgate.edu