Osmotic pressure and protein binding

Transcription

1 Osmotc pressure and proten bndng Igor R. Kuznetsov, KochLab Symposum talk 5/15/09 Today we take a closer look at one of the soluton thermodynamcs key ponts from Steve s presentaton. Here t s: d[ln(k off )] d[osmolal] = N w 55.6 the change n the proten-dna unbndng rate wth respect to the water actvty (n terms of solute osmolal concentraton) s proportonal to the change n the number of the "excludng" water molecules durng dssocaton. Some of us may see how ths result comes about rght away, but f you are lke me, understandng t takes some tme. What I am gong to do today s show how we can arrve at ths mportant formula, one baby step at a tme. Soluton thermodynamcs Frst we wll revew the bascs of the soluton thermodynamcs brefly, especally the quanttes that are mportant for ths talk: chemcal potental µ, Gbbs free energy G, and osmotc pressure P osm. Then we wll look at how water actvty changes wth osmotc pressure. Fnally, we ll arrve at the aforementoned dssocaton water actvty dependence. Let us start wth conservaton of energy. We postulate that the energy s conserved or, more specfcally, that the nternal energy of the system E s conserved unless heat s added (changng entropy), work s done (changng volume), or more matter s ntroduced nto the system (changng the number of partcles). Thus, we wrte E = E(S, V, N ). The change n E s gven by chan rule as de = ( ) ds + S V,{N } ( ) ( ) dv + dn. V S,{N } N V,S We call varous partal dervatves n the above equaton: T, -P, µ: ( ) ( ) ( ) T, P, µ S V N V,{N } S,{N } n other words, the frst law of thermodynamcs becomes V,S de = TdS pdv + µ dn. Snce the state varables S, V, and N are all extensve, we can use Euler s homogeneous functon theorem and wrte explct defnton of nternal energy as E = TS pv + µ N. 1

2 In context of expermental observaton, we are nterested n thngs lke: what reactons are gong to take place; n what drecton and to what extent; what wll be the fnal equlbrum state. Examnng the nternal energy potental dfferental, we see that when all the state varables are kept constant, the change n E s zero. In other words, the thermodynamc potental nternal energy has an absolute mnmum at equlbrum: when S, V, and N of a closed system are held constant, the nternal energy E decreases and reaches a mnmum value. Ths, n theory, should determne the drecton of all chemcal reactons that take place under gven condtons, as well as ther extent. In practce however, we rarely deal wth sentropc processes: for example, dlutng salt n water s drven by entropy change. The nternal energy potental therefore s not very convenent when lookng at chemcal reactons that take place n a lab, where the constant parameters are pressure and temperature. Fortunately, we can use a more sutable thermodynamc potental, whch has P and T as state varables. Ths potental s defned as G = E + PV TS (= µ N ). Takng a dfferental, we obtan dg = du + p dv + V dp TdS SdT, dg = TdS p dv + µ dn + p dv + V dp TdS SdT, dg = V dp SdT + µ dn, along wth a useful dentty (Gbs-Duhem corollary) µ dn = V dp SdT. The Gbbs free energy G s mnmzed when a system reaches equlbrum at constant pressure and temperature. As such, t s a convenent crteron of spontanety for processes wth constant pressure and temperature. Also, note that the chemcal potental s now convenently defned as a partal dervatve of G wth respect to N : ( ) G µ. N Transtonal entropy of soluton, free energy, chemcal potental, osmotc pressure Consder a bnary soluton - a mxture of 2 knds of molecules. Assume that before mxng both substances had equal temperature and pressure. Then the T,P 2

3 spontaneous process of mxng s drven purely by entropy: the only state varable that changes s entropy S, and so the change n G s gong to come from ncrease n entropy of mxng alone (rearrangement of molecules). Before mxng, the free energy of consttuents s gven by whle after the mxng t s G 0 = (E + PV TS) 0 = µ 0 N 0, G = µ 0 N0 T S. To fgure out the change n Gbbs free energy we need to know the change n S due to rearrangement. If we start from two substances (say water "w" and solute "s") perfectly separated, and end up wth a perfectly mxed soluton at equlbrum, the change wll be gven by Boltsman s entropy formula S = k B lnw, where W s the number of ways N w molecules of water and N s molecules of solvent can be rearranged on a lattce of the sze N = N w + N s. The number of such permutatons, takng nto account that N w molecules and N s molecules are dentcal among themselves, s gven by W = N! N w!n s!. Usng Strlng s formula for large N ln N! NlnN + N 1, we obtan ( S = k B N w ln( N w N ) + N sln( N ) s N ). The free energy of a bnary mxture s gven by ( G = G 0 T S = µ 0 w N0 w + µ0 s N0 s + k BT N w ln( N w N ) + N sln( N ) s N ). From here we calculate the chemcal potentals of solvent by takng partal dervatves of G wth respect to N w and N s : µ s = µ 0 s + k BTln(x s ), µ w = µ 0 w + k B Tln(1 x s ), where x s s a mole fracton of the solvent, and x s + x w = 1. Let us examne these results. The equlbraton of the chemcal potental s the drvng force behnd the molecular dffuson, just lke the temperature equlbraton s the drvng force behnd the heat transfer. If we try to apply the above formulaton to the case when two solutons of dfferent solvent concentraton are brought n contact across a sem-permeable membrane, we seem to be 3

4 n trouble". Imagne two tanks, one wth salt soluton and the other wth pure water, separated by a membrane that s only permeable to water (just lke RBC membrane). Chemcal potental of water n the soluton s lower than that n pure water, so more water molecules are gong to be transported n the drecton of the lower µ than n the opposte drecton. But no matter how long ths contnues, the solt concentraton n soluton s never gong to go to zero, and as a result the entropc parts of the chemcal potental on the two sdes of the membrane wll never balance. Is the osmoses gong to drve all of the avalable pure water nto the soluton, desperately tryng to equlbrate µ? Not really, because the enthalpc part of the chemcal potental µ 0 depends on pressure: µ w (T, P) = µ 0 w (T, P) P osmv w = µ 0 w (T, P 0) + Pv w P osm v w. The hydrostatc pressure n the tank wth water and solt rses to counter the osmoss across the sem-permable membrane (the enthropc logarthmc term). Hence we can talk of the "Osmotc pressure", defnng t as P osm = (µ µ o )/v w. Proten-DNA bndng In smple terms, a proten-dna assocaton reacton can be wrtten as The chemcal potental s gven by At chemcal equlbrum, we get P + D < > PD. µ = µ 0 + k BTln(x ). µ PD = µ P + µ D, so the assocaton constant can be calculated as K = k off = [x ( PD] µ 0 k on [x P ][x D ] = exp PD µ 0 P ) ( ) µ0 D G = exp, kt KT where G s change n free energy gong from PD to P and D. Imagne we ntroduce an osmotc agent (say PEG) n the water. Followng the argument of [2], we observe that the unbndng state dffers from bndng state n the amount of molecules of water that are excluded,.e. the amount of molecules unable to solvate PEG. It s analogous to the case of the sempermeable membrane: n order to unbnd, the system PD has to draw addtonal water molecules aganst the chemcal potental gradent (aganst the osmotc pressure). The addtonal work to overcome the osmotc pressure resstance s equal to the dfference n chemcal potentals multpled by the number of addtonal water molecules. Ths work can also be expressed as osmotc stress multpled by the volume of addtonal water molecules that need to be drawn. 4

5 The addtonal work requrement wll make the dssocaton events less frequent. In other words, the assocaton constant depends on osmotc stress K = K(P osm ). Snce G = k B TlnK (from above), and G has an addtonal osmotc pressure contrbuton P osm V w, we have d(lnk) dp osm = V w k B T. Osmotc pressure s usually measured n osmolal unts, so that P osm = (k B T/ν w )(osmolalty/55.6), where ν w s the molecular volume of water, and 55.6 s the molarty of pure water. Taken together, these numbers gve an osmotc pressure of 24 atm per osmolal at 20. We can rewrte the above expresson as d(lnk) d[osmolarty] = N w I don t see how we can expand Parsegan s dervaton to case when the dssocaton reacton s not at equlbrum. In order to arrve at the result from [1] that Steve used, we wll need to resort to the phenomenologcal Arrhenus type reacton rate for proten unbndng (ths s from talkng to Evan). Thus we would postulate that the off-rate s gven by ( ) G k off = k 0 exp, kt where G s the potental energy barrer that needs to be overcome n order for the proten to unbnd. If we were to pull on the molecule (say wth optcal tweezers), the potental energy barrer heght G would reduce, and the reacton off-rate would ncrease accordngly. The osmotc stress on the other hand leads to the ncrease of the barrer heght commensurate wth the product of the addtonal excludng water volume and the osmotc pressure. Dfferentatng the Arrhenus law above, we obtan the desred References d[ln(k off )] d[osmolal] = N w [1] Sdorova, N.Y. and D. C. Rau, Bopolymers, 53(5), (2000) [2] Parsegan,V.A., Rand, P.R. and D.C. Rau, Meth. Enz., 259, (1995) 5