tag:blogger.com,1999:blog-9633767.post4591685408678498812..comments2021-01-12T12:51:03.444-08:00Comments on The Curious Wavefunction: ∆G, ∆G†† and All That: Implications for NMRWavefunctionhttp://www.blogger.com/profile/14993805391653267639noreply@blogger.comBlogger22125tag:blogger.com,1999:blog-9633767.post-30243149914380147462008-12-22T10:07:00.000-08:002008-12-22T10:07:00.000-08:00This might help:http://people.clarkson.edu/~wilcox...This might help:<BR/>http://people.clarkson.edu/~wilcox/Design/<BR/>gibbreac.doc<BR/><BR/>Keep em coming...and let us know MJ when you hit the blogstreet.Wavefunctionhttps://www.blogger.com/profile/14993805391653267639noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-62766111546671377772008-12-21T15:22:00.000-08:002008-12-21T15:22:00.000-08:00GMC2007"At equilibrium, increasing the proportion ...GMC2007<BR/><BR/>"At equilibrium, increasing the proportion of either products or reactants leads to an increase in in free energy. Both of these processes will be associated with a positive delta-G. Discuss.<BR/><BR/>1:34 PM, December 18, 2008"<BR/><BR/>Quite true. But the equilibrium constant is all about proportions, and you can't increase the proportion of products (or reactants) without leaving equilibrium, which is why deltaG is positive in these cases. <BR/><BR/>Think about it a bit more.<BR/><BR/>RetreadAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-51625037126022610882008-12-21T13:38:00.000-08:002008-12-21T13:38:00.000-08:00MJ,At last a mention of standard states! The delt...MJ,<BR/><BR/>At last a mention of standard states! The delta-G is the standard free energy change and this should be indicated with a small superscripted circle (like a degree symbol). If you work thru the math you find the concentration associated with the equilibrium constant corresponds to the concentration that defines the standard state. <BR/><BR/>It is correct to state that<BR/><BR/>delta-G = -RTlogKd<BR/><BR/>provided that you define delta-G to be the standard free energy change for the process and that the concentrations are in units of the standard concentration. But this is seldom done. I prefer to write (assume the 'o' is a 'degree symbol'):<BR/><BR/>delta-Go = -RTlog(Kd/Co)<BR/><BR/>Let Wavefunction and I know when you do start blogging so we drop by for some more entertainment.Georg-Martin Krapperhttps://www.blogger.com/profile/15416686863175197568noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-82933094516723323362008-12-21T08:48:00.000-08:002008-12-21T08:48:00.000-08:00"Mr. Boswell,"True enough. But, perhaps, through ..."Mr. Boswell,"<BR/><BR/>True enough. But, perhaps, through elaboration, correction, and extension of earlier arguments - as well as the introduction of new ones - some degree of understanding can be found. I naturally mean "argument" here less in a contentious way and more in a stepwise process thinking through a question to find some sort of answer.<BR/><BR/>Curiously enough, the original reason I bothered to comment could be of interest - many of those experiments I was thinking of in my first comment obtain data in various combinations of force, position/distance, and velocity. I recall that people have attempted to map energy surfaces from such measurements. It seems that there's a certain conceptual simplicity there - we all learn about the relationships between energy, force, and work when beginning in physics. Most of the systems I've heard/read about have been protein-protein interactions, or particular cases like streptavidin and biotin, but I think it could be applied more broadly. If it's possible to extend the "molecular wire" concept (made famous by Harry Gray and his group, mostly in relation to metalloproteins that I recall) to any sort of ligand-receptor system, you could put your binding moiety on the tail that would bind to the receptor, the long wire that leads to the receptor surface, and an appropriate linker that you can attach to run these sorts of force measurement studies, it would be pretty cool. If anyone has heard of someone already doing it, citations would be greatly appreciated. I'd love to see it.<BR/><BR/>I really should get my own blog. One of these days...<BR/><BR/>- MJAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-30356398795070993652008-12-21T07:49:00.000-08:002008-12-21T07:49:00.000-08:00I have found you an argument; I am not obliged to ...I have found you an argument; I am not obliged to find you an understanding. <BR/>James BoswellAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-27968762036429539062008-12-20T15:51:00.000-08:002008-12-20T15:51:00.000-08:00While doing some holiday shopping earlier today, I...While doing some holiday shopping earlier today, I realized that my earlier argument, while the intent and spirit behind it were valid, the details are prone to laborious nitpicking that might be staved off. Let me repackage it so it hangs together slightly more coherently, I hope.<BR/><BR/>We have a receptor-ligand binding situation where<BR/><BR/>P + L < - > PL.<BR/><BR/>The < - > notation is used here to indicate that it is a reversible process (it's a limitation of my ability to enter symbols here). We can define the equilibrium constant, Keq, as<BR/><BR/>Keq = [PL]/[P][L],<BR/><BR/>the dissociation constant, Kd, as<BR/><BR/>Kd = [P][L]/[PL],<BR/><BR/>and reaction quotient, Q, as<BR/><BR/>Q = [PL]/[P][L].<BR/><BR/>Here, I am substituting in the concentrations for activities, so when using concentrations in lieu of activities is no longer is a good approximation, this falls apart. Note here that since it's a case of 1 ligand binding to 1 receptor, the superscript for each concentration term in Q is 1. (#) We know the following expressions for delta-G, the first for a system under standard (equilibrium) conditions,<BR/><BR/>delta-G = -RT*ln(Keq)<BR/><BR/>and the second for a system under non-standard conditions,<BR/><BR/>delta-G = delta-G(0) + RT*ln(Q).<BR/><BR/>So we follow the same idea as above, substituting -RT*ln[Keq(0)] in for the delta-G(0) term, which represents the change in standard Gibbs energy for the process, and do the same manipulations until we get<BR/><BR/>delta-G = RT*ln[Q/Keq(0)]<BR/><BR/>In this case, as Q has the same form dimensionally as Keq for the receptor-ligand binding event (reciprocal concentration, M^(-1)), and that the reaction quotient, like the equilibrium constant, is a ratio of product to reactants (or of the product to starting species, if we're not talking about a reaction per se, as in this case, but rather some other physical process), consider the following. An equilibrium constant is a measure of the extent of a process under standard (equilibrium) conditions – the reaction quotient is, in a coarse-grained fashion, a measure of the extent of a process under non-standard conditions. Given that nature has the temerity to not do everything under standard conditions (from the pressurized depths of the ocean to the dry heat of the desert), and that it works out neatly with regard to the math, we could substitute Keq for Q in the above equation so we now have<BR/><BR/>delta-G = RT*ln[Keq/Keq(0)]<BR/><BR/>or, if we were so inclined, <BR/><BR/>delta-G = RT*ln[Kd/Kd(0)],<BR/><BR/>where we exchanged Kd for Keq. It's not the most rigorous argument out there, but as the thought of including all sorts of correction factors for activities of solutes (that are also electrolytes – what's the old saying again, “Debye-Huckel is good for slightly salted water”?) makes me quiver and not in a good way (although if I were to do it, I'd rather do it properly and submit it to, I don't know, J. Chem. Ed. for the heck of it), I'll leave it at that. <BR/><BR/>Of course, GMC2007, if you happen to have a nice reference for the proper way to calculate free energies of binding for a receptor-ligand complex from binding and/or equilibrium constants, I would be interested in seeing it after having thought through all of this.<BR/><BR/>- MJ<BR/><BR/>#: I think all of this would also apply in principle to a more complicated situation, such as a protein interacting with a substrate and, say, a ligand that modulates the protein's function, but I haven't thought through how you'd work through all of the mathematics.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-59702236414258844812008-12-20T08:35:00.000-08:002008-12-20T08:35:00.000-08:00GMC2007, Duly noted, I didn't catch that bit of di...GMC2007, <BR/><BR/>Duly noted, I didn't catch that bit of dimensional analysis. Thinking about it more, here might be a cute and simple way to avoid that bit of mathematical heresy and work from the basics. <BR/><BR/>We start out with the following, I figure, rather general statement:<BR/><BR/>delta-G = delta-G(0) + RT*ln Q<BR/><BR/>where delta-G is the change in Gibbs energy for the process under investigation, delta-G(0) is the change in Gibbs energy under standard conditions (insert laundry list of requirements for 'standard conditions' here), and RT we know, and Q is the reaction quotient. Substitute in -RT ln Q(0) for the delta-G(0) term, where Q(0) is the reaction quotient for the process under standard conditions, and we have<BR/><BR/>delta-G = -RT*ln Q(0) + RT*ln Q .<BR/><BR/>Some algebra gives us<BR/><BR/>delta-G = RT [ln Q - ln Q(0)],<BR/><BR/>and a tiny bit more gives us<BR/><BR/>delta-G = RT*ln [Q/Q(0)].<BR/><BR/>Now, at equilibrium, the reaction quotient is equal to the equilibrium constant, so we can now write the following expression:<BR/><BR/>delta-G = RT*ln [Kd/Kd(0)].<BR/><BR/>Here, since Kd and Kd(0) have the same units, the units cancel out and we have a unitless number of which the logarithm can be taken without issue, I would figure. FWIW, this equation looks familiar from p.chem., so it isn't anything new. <BR/><BR/>For completeness's sake, the ratio of Kd to Kd(0) determines whether the process under investigation is spontaneous, as when Kd = Kd(0), the delta-G is zero and the system is at equilibrium. <BR/><BR/>If this fails to pass muster, I look forward to see what will be asked next....<BR/><BR/>- MJ<BR/><BR/>P.S. - Minor digression here, but the earlier comment about NMR and paramagnetic systems compels me. Utilizing paramagnetism in NMR is becoming a somewhat trendy - but, IMO, extremely interesting and worthwile - thing to do in relation to structural biology. Whether it's a naturally paramagnetic system, a metalloprotein that has had its naturally occurring metal substituted for a paramagnetic one, or utilizing a paramagnetic tag to introduce paramagnetism into a diamagnetic protein, it's making inroads in both solution and solid state NMR communities. Maybe this is one of things I could use a blog for to ramble on about for a while....Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-17201098036988933022008-12-20T02:17:00.000-08:002008-12-20T02:17:00.000-08:00MJ,This is not even close to what I was getting at...MJ,<BR/><BR/>This is not even close to what I was getting at. Kd is simply a the name that protein folk give to the equilibrium constant for protein ligand binding equilibria. It has units of concentration and there lies the problem. You only calculate a logarithm of a number and this is the case whatever the base of the logarithm.Georg-Martin Krapperhttps://www.blogger.com/profile/15416686863175197568noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-56363503959429357022008-12-19T19:12:00.000-08:002008-12-19T19:12:00.000-08:00Since I am a little short of time today, two comme...Since I am a little short of time today, two comments for now:<BR/>1. I am not Retread<BR/>2. Please continue the discussion. We are looking for particularly itchy-mouse-click-fingered humans here. Digressers will be rewarded.Wavefunctionhttps://www.blogger.com/profile/14993805391653267639noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-87428980056110272632008-12-19T17:38:00.000-08:002008-12-19T17:38:00.000-08:00GMC2007,The first thing that appears to be wrong w...GMC2007,<BR/><BR/>The first thing that appears to be wrong with the equation you gave is that Kd is equal to the product of the reactant concentrations (protein and ligand) divided by the product concentration (the protein-ligand complex), while Keq - as is typically meant to be plugged into that equation - is equal to the product of the product concentrations divided by the product of the reactant concentrations. The equation needs to be rewritten to account for this fact, as well as include a correction term of some sort, the form of which can depend on your preferences. The other thing that could be problematic, IMO, is whether you meant to use the natural logarithm, ln, or some other logarithm of undefined base. But that's a simple matter of algebra to straighten that out, if it applies to your question.<BR/><BR/>I'm quite sure I just went off on a rather tedious divergence in my earlier comments on this post about "equilibrium free energy differences" since I have seen that phrase in the literature - see http://www.ncbi.nlm.nih.gov/pubmed/14528008 just to show that I'm not just making it up - regarding particular types of experiments which, while interesting to me, aren't really under discussion here and are only tangentially of interest at best. I should be taking my leave right about now.....<BR/><BR/>- MJAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-27412236264819252282008-12-19T15:38:00.000-08:002008-12-19T15:38:00.000-08:00My dear MJ,It's not particularly meaningful to tal...My dear MJ,<BR/><BR/>It's not particularly meaningful to talk about two species having the same Gibbs free energy since the Gibbs free energy is an extensive variable. However the Gibbs free energy of a system can be minimised, which is the case at equilibrium. This is why displacing the system from equilibrium is leads to an increase in the Gibbs free energy of the system. <BR/><BR/>Now here's my challenge. Can you tell me what might be wrong with writing:<BR/><BR/>delta-G = -RTlogKd<BR/><BR/>for the binding of ligand to a protein?Georg-Martin Krapperhttps://www.blogger.com/profile/15416686863175197568noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-65857568643826462762008-12-19T08:17:00.000-08:002008-12-19T08:17:00.000-08:00Addendum -The delta-G that I was thinking of, the ...Addendum -<BR/><BR/>The delta-G that I was thinking of, the difference in Gibbs energies (or Helmholtz energies) between the original state (P,L,C) and the new state (P', L', C'), is what is referred to as equilibrium free energy differences, at least from what I've been able to gather. <BR/><BR/>My apologies to our host for this post, I should have not had such an itchy mouse click finger....<BR/><BR/>- MJAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-82361324029965615722008-12-19T08:13:00.000-08:002008-12-19T08:13:00.000-08:00I really should get my own blog account, if for no...I really should get my own blog account, if for no other reason than to not be mistaken for someone I'm not.<BR/><BR/>This is how I read the blog post given the earlier comments that were discussed on the blog with regard to NMR, protein-ligand complexes, and the like, which is completely up my alley. <BR/><BR/>What we have is some starting system of a protein (let's call it P) and a ligand (call it L) that is bound in some averaged conformation (call it C, and it's composed of, just to make life simple, a two-site hop between c1 and c2). This starting system is in thermodynamic equilibrium, which means that the Gibbs energy of the protein-ligand complex in conformation 1 (call it G1) is equal to the Gibbs energy of protein-ligand complex is conformation 2 (call it G2). To no great surprise, delta-G is zero here. This is not what I was getting at in my earlier post, and I should probably be smacked for not being clearer about this issue. <BR/><BR/>What I was mentioning in the earlier comment is that one perturbs the system, such that P goes to P', L goes to L', and C goes to C' (and c1 and c2 go to c1' and c2'), and that this new system, composed of P' and L', where L' is bound in its new averaged conformation C', is also at equilibrium. The delta-G I was mostly focused on is the delta-G between the Gibbs energy of the new system (where G1’ = G2’ = G’) and the original system (where G1 = G2 = G). This perturbation could be anything of experimental interest (temperature, oxidation state if the protein is a metalloprotein, pressure, whatever you can vary in a reasonable manner). <BR/><BR/>That’s what I was trying to get at in my earlier comment, although I hope now in a less confused manner.<BR/><BR/>- MJ, who is definitely not a retired physicianAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-28643605220369155922008-12-18T13:34:00.000-08:002008-12-18T13:34:00.000-08:00Retread,Your vapid and confused response does not ...Retread,<BR/><BR/>Your vapid and confused response does not suggest that your are in a position to instruct me to 'think about it a bit more'. <BR/><BR/>At equilibrium, increasing the proportion of either products or reactants leads to an increase in in free energy. Both of these processes will be associated with a positive delta-G. Discuss.Georg-Martin Krapperhttps://www.blogger.com/profile/15416686863175197568noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-84146750893081000302008-12-18T08:58:00.000-08:002008-12-18T08:58:00.000-08:00That's what your blogger was talking about, even i...That's what your blogger was talking about, even if not explicitly...and even if in retrospect. I have to admit that the term "equilibrium free energy" is a little confusing. In this particular case I think the word "equilibrium" is used only because the energy relates to the "equilibrium constant". But as MJ says, I have also seen the term used in some places, including textbooks. The way I see it, the reason ∆G in this equation is not zero is simply because it's related to K, which as Retread noted, is related to unequal concentrations of reactants and products. The ∆G which <I>is</I> zero is a different one as far as I understand; it's the free energy change for driving the reaction one way or the other. I want to relate that to chemical potential but I think I should think a little more about that.Wavefunctionhttps://www.blogger.com/profile/14993805391653267639noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-48749847222398094612008-12-17T17:40:00.000-08:002008-12-17T17:40:00.000-08:00While not the blogger, I might be able to clarify ...While not the blogger, I might be able to clarify the issue of "equilibrium free energy differences" up. Then again, maybe not....<BR/><BR/>The Gibbs energy is typically defined as G = H - TS (the Helmholtz energy is analogous with F = U - TS) for a state in equilibrium. So if you have state "A" which then goes to state "B" then (delta)G = G(state B) - G(state A). [Sorry I can't do the neat Greek symbols here.] Unfortunately, since nomenclature and phrasing can get a little sloppy, G and F can get described as the Gibbs (or Helmholtz) free energy when we should just drop the "free" from our vocabulary, but that's not going to happen any time soon. Speaking as someone who has traveled enough in thermo and stat mech over the years, I can vouch for the phrase "equilibrium free energy difference" cropping up pretty regularly, esp. in relation to the Jarzynski equality (in fact, Jarzynski uses it himself in his first paper on the topic, I believe).<BR/><BR/>If that's not what our blogger was talking about, I digress. <BR/><BR/>- MJAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-48656794176584926012008-12-17T12:08:00.000-08:002008-12-17T12:08:00.000-08:00Retread,Perhaps you can humour me by explaining ex...Retread,<BR/><BR/>Perhaps you can humour me by explaining exactly how the free energy difference between reactant and products determines the equilibrium constant. I'm keen to see whether you're really in a postion to instruct me to 'think about it a bit more'. I also challenge you to define 'equilibrium free energy difference'. It's not a term that one encounters a lot in thermodynamics text books.Georg-Martin Krapperhttps://www.blogger.com/profile/15416686863175197568noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-83367987857856126882008-12-16T13:51:00.000-08:002008-12-16T13:51:00.000-08:00The simplest way to remember the distinction betwe...The simplest way to remember the distinction between thermodynamic stability and kinetics is to think about the hydrogen gas + oxygen gas system. An explosion results when a spark is thrown into the brew, everything is converted to the much stabler product (water). This never happens (at room temperature) without the spark (which supplies the activation energy for the first few reactions) because it is so high. The number of molecules at room temperature moving fast enough (with enough energy) to get over the activation energy barrier is effectively zero. After that, the heat released keeps things going.<BR/><BR/>GMC2007: You are forgetting concentrations. The free energy difference between products and reactants determines the equilibrium constant (which involves concentrations). At equilibrium as many prodcts are going back to reactants and reactants are going forward to products. Concentrations adjust themselves accordingly. Think about it a bit more.<BR/><BR/>I've got a new post up today on "The Skeptical Chymist" today which protein aficionadoes might find interesting. <BR/><BR/>RetreadAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9633767.post-81404240275419926202008-12-16T13:00:00.000-08:002008-12-16T13:00:00.000-08:00Can you explain the term equilibrium free energy d...Can you explain the term equilibrium free energy difference? I would have thought that at equilibrium the free energy of the reactants equals that of the products and that the difference would be zero.Georg-Martin Krapperhttps://www.blogger.com/profile/15416686863175197568noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-62024550064988176232008-12-14T09:51:00.000-08:002008-12-14T09:51:00.000-08:00Well, unless you're at high field and have a param...Well, unless you're at high field and have a paramagnetic species around. Of course, in that case you have a whole other set of problems.Sparkyhttps://www.blogger.com/profile/10444352252473400411noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-4468128198200225242008-12-14T08:27:00.000-08:002008-12-14T08:27:00.000-08:00Quite true; I should have stated that. But I guess...Quite true; I should have stated that. But I guess a rate of say 10*5/s will almost always be fast compared to the difference in the chemical shifts, won't it?Wavefunctionhttps://www.blogger.com/profile/14993805391653267639noreply@blogger.comtag:blogger.com,1999:blog-9633767.post-19585832671678270882008-12-13T08:31:00.000-08:002008-12-13T08:31:00.000-08:00That's a pretty good discussion. I think it's impo...That's a pretty good discussion. I think it's important to clarify that the low resolution time is a property of the <I>chemical shift</I>. This is important because when we talk about whether something is "fast", "slow" or "intermediate" on the "NMR timescale" we are not talking about the raw process speed. These terms indicate the relationship between the magnitude of Δ<I>ω</I> and <I>k</I>ex. Thus, a rate of 500 /s can be slow <I>or</I> fast on the NMR timescale depending on the nucleus in question and the chemical shift differences involved.Sparkyhttps://www.blogger.com/profile/10444352252473400411noreply@blogger.com