Re: [ccp4bb] -RTlnK = delH - TdelS
James, I think you need to be a little more specific about what you want to calculate. Keq for a reaction A <=> B will not change with an enzyme mutation as the thermodynamic relationships between the reactants and products do not change. As a catalyst, the enzyme impacts only the free energy of activation delG "double-dagger" and not the reaction thermodynamics. However, a mutation can markedly impact the Keq of ligand binding, but Km here is of no use. As mentioned by others, Km is an amalgam of many intermediate equilibria in a reaction up to the rate limiting step. Moreover, Km equals Kd in only the most simplest of enzymes, which sadly few of us work on. Thus, you need to determine actual Kds for all your ligands, which in turn requires you to determine the reaction characteristics of your enzyme (random sequential, ordered, bi-bi, etc.). It is not easy and real enzymologists are rapidly dying off (or at least retiring). The old Cantor and Schimmel book "Biophysical Chemistry Part III" (1971) should help you here. I actually cannot think of a recent monograph on pure enzymology published in the last decade or so. If you want the change in activation energy caused by a mutation (delE "double-dagger"), the the "k" in RTlnk is the kcat and an Arrhenius plot is just the thing to use. But as Roger said, you MUST carefully control for temperature dependent pH changes. Good luck and fire up the spec for some long days, Michael R. Michael Garavito, Ph.D. Professor of Biochemistry & Molecular Biology 513 Biochemistry Bldg. Michigan State University East Lansing, MI 48824-1319 Office: (517) 355-9724 Lab: (517) 353-9125 FAX: (517) 353-9334Email: garav...@msu.edu On Wed, Nov 18, 2009 at 8:24 AM, james09 pruza 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
Re: [ccp4bb] -RTlnK = delH - TdelS
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 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
Re: [ccp4bb] -RTlnK = delH - TdelS
Km is the amount of substrate needed to get the enzyme to half-maximum velocity. So if the brain enzyme needs 10 times less substrate to reach half-maximum velocity, you would see the apparent Km go down from 25, not up. Any undergrad biochemistry text will have all this stuff in it. Or am I doing your homework assignments for you? On Nov 18, 2009, at 10:54 AM, james09 pruza wrote: > Dear Sir, > I need one more help. > If there are two isozymes at two different compartment, say one in brain and > one in serum. If the serum one has Km of 25 and that of brain has Km of 2.5. > If by chance, the brain one comes to the serum, What should be the Km of > the serum sample of the person --- Will it be close to 2.5 or will it be > 27.5 ? > Please help. > Thanks. > James > > On Wed, Nov 18, 2009 at 8:38 PM, William G. Scott < > wgsc...@chemistry.ucsc.edu> 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 >> 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 > >> >>
Re: [ccp4bb] -RTlnK = delH - TdelS
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 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 >>> >> >> >> >>
Re: [ccp4bb] -RTlnK = delH - TdelS
James- It's: lnKd = deltaH- RT(deltaS) As for Km ~ Kd, here be dragons...Km values cannot be treated as the Kd of a complex because dissociation of a catalytic complex has two potential fates, one that is denoted by a rate constant that is correlated to turnover (so, the forward reaction leading to product, generically called k2) and one that represents the degeneracy or breakdown of the enzyme/substrate complex (the reverse reaction or k-1). The Km will most likely be very similar to the Kd but it is only an apparent Kd (Kdapp). Since you have measured enzyme activities at multiple temperatures, you could graph an Arrhenius plot and determine the activation free energies for the wt and mutant enzymes by a statistical fit to the plotted data. This would provide a comparative thermodynamic analysis of the enzyme. If you have a good substrate analog (inhibitor), you could determine Ki with your normal activity assay and compare between wt and mutant. Of course, there's always calorimetry, too... HTH- Brad On Wed, Nov 18, 2009 at 11:24 AM, james09 pruza 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 >
[ccp4bb] -RTlnK = delH - TdelS
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
