Lecture 5: Electrostatic Interactions & Screening Lecturer: Prof. Brigita Urbanc (brigita@drexel.edu) 10/06/2009 PHYS 461 & 561, Fall 2009-2010 1 A charged particle (q=+1) in water, at the interface between water (=80) and protein (=3) Find the electric field produced by the charge q at an arbitrary point 2: 1 q=+1 2 ⨯ water //////////////////////////// protein // How do we solve for the electric field and why does the presence of water—protein interface matter? in the absence of the interface, electric field E = q/4 0r12 ➔ 10/06/2009 PHYS 461 & 561, Fall 2009-2010 2 Method of Image Charges Griffiths, David J. (1998). Introduction to Electrodynamics (3rd ed.). Prentice Hall. ISBN 0­13­805326­X. problem solution Coulomb's law: … normal distance from the y=0 plane: = (x2+z2)1/2 10/06/2009 PHYS 461 & 561, Fall 2009-2010 3 A similar problem of an interface between vaccum (=0) and a metal (=ꝏ): electric field locally perpendicular to the interface ➔ NO field in a metal (or else current) ➔ 2⨯ r12 r02 SOLUTION (above the interface): E = q/4 0r12 + (­q)/4 0r02 ➔ ➔ +q 1 vaccum (=1) ­q 0 metal (=ꝏ) reflection effect: +q induces the shift of electrons on the high metal side 10/06/2009 PHYS 461 & 561, Fall 2009-2010 4 Water—protein interface: mirror charge approach inverse problem—the original charge in water (high permittivity 1=80 side of the interface) ➔ force lines are repelled from the low =3 (protein) side ➔ a charged atom on the water side gets surrounded by electronegative parts of polar water molecules ➔ positive mirror charge q' = q ( 1 – 2)/( 1 + 2) ➔ ~ q (if 1 » 2 ) 10/06/2009 E = q/4 0r12 + q'/4 0r02 PHYS 461 & 561, Fall 2009-2010 5 We can use the generalized form of the field around the charge q: E = q/4 eff 0r12 = q/4 1 0r12 (1 + r12/r02) With effective permittivity eff depending on the position r: eff = 80 for the point 2 close to the charge 1 ➔ eff = 40 everywhere else in water ➔ eff = ( 1 + )/2 ~ 40 for any point 2 ➔ below the surface ➔ = ( 1 + )/2 ~ 40 for point 1 eff inside the protein except if r12 « a, then eff = 10/06/2009 ­ ­ ­ +1 ­­ ­­ ­ + + + + + water ///////////////////////////// protein PHYS 461 & 561, Fall 2009-2010 6 Effective permittivity across the protein: 1 + + + + + water protein ­ ­ ­ ­ ­ water 2 eff = 200 due to water electric dipole and induced polarization 10/06/2009 PHYS 461 & 561, Fall 2009-2010 7 Values of effective permittivity eff around and inside a protein: +1 40 80 40 80 80 protein 40 water 80 80 60 80 +1 3 protein 80 200 Except inside the protein very close to the charge, the electric field is strongly reduced due to screening by polar water molecules 10/06/2009 PHYS 461 & 561, Fall 2009-2010 8 The medium of high permittivity (water) attracts the charge: a charge on the protein surface is repelled from the protein ➔ a charge inside the protein is attracted to water ➔ Why does water have a high permittivity? permittivity determined by atomic structure of water ➔ permittivity proportional to the polarization induced in the medium by an external electric field → polarity of H2O ➔ Induced polarization produces effective internal field of the opposite sign, thereby diminishing the total field (relative to vaccum) ➔ 10/06/2009 PHYS 461 & 561, Fall 2009-2010 9 Electrostatic Interaction Between Two Oppositely Charged Atoms consider distances 3—4 Å → no water molecules in­ between possible! What is the value of effat such small distances? ➔ example: Na+Cl­ in water dissolves (Van de Waals distance between Na and Cl ~ 3 Å) ➔ EI potential energy: ­1.5 kcal/mol ( eff=80) ­3.0 kcal/mol ( eff=40) ­6.0 kcal/mol ( eff=20) > EHB 10/06/2009 PHYS 461 & 561, Fall 2009-2010 10 Free energy change associated with dissociation: oxalic acid (diprotic acid – 2 H­atoms per molecule): Step 1: H2C2O4 ⇄ HC2O4− + H+ ➔ − 2− + Step 2: HC O ⇄ C O + H 2 4 2 4 ➔ Step 1 occurs at pH ~ 2 & Step 2 at pH ~ 4.5 → pH ~ 2.5 pH value associated with H+ concentration, [H+]: [H+] = 10pH = exp(2.3 ⨯ pH) ➔ pH → [H+] → change in the Gibbs free energy G G = RT {ln([H+]b) ­ ln([H+]a)} = RT (2.3 ⨯ pH) ➔ For oxalic acid pH = 2.5 → G~ 3.5 kcal/mol ( eff~40) ➔ similar result for dissociation of the carbonic acid ➔ 10/06/2009 PHYS 461 & 561, Fall 2009-2010 11 Physical Interpretation: Neighboring water molecules “pulling charges” from sides - +1 - + - - + 10/06/2009 + - + + + - + - + ­1 + + - PHYS 461 & 561, Fall 2009-2010 12 Experimental Determination of the Effective Permittivity (A. Fersht – protein engineering) enzymes exhibit optimal activity at a pH­optimum ➔ mutation of the active site amino acid to a charged amino acid shifts the pH­optimum (mutation at the protein surface) ➔ pH­optimum happens because the active site needs a fixed concentration of protons, [H+] = 10­pH (by definition) [OH­] = 10­14+pH (by definition) ➔ active site (AS) accepts a proton H+, AS + H+ = ASH+: ([ASH+] = [AS][H+] ⨯ probability for H+ binding) [ASH+]/[AS] = exp(­FASH+/RT) ⨯ [H+] ➔ 10/06/2009 PHYS 461 & 561, Fall 2009-2010 13 FASH+ = Free Energy of H+ binding [ASH+]/[AS] = exp(­FASH+/RT) ⨯ 10­pH = exp(­FASH+/RT) ⨯ exp (­2.3 ⨯ pH) = exp{­(FASH+/RT + 2.3 ⨯ pH)} Mutation—induced charge induces the potential eU (e=+1, the charge of H+): eU = FASH+ (M) – FASH+(0) M ­­­ with mutation 0 ­­­ without mutation ↓ (a) FASH+(M)/RT + 2.3 ⨯ pHM = FASH+(0)/RT + 2.3 ⨯ pH0 (b) eU = FASH+(M) – FASH+(0) = ­ 2.3 ⨯ RT (pHM ­ pH0) (c) eU = eq/4 eff 0r, q ­­­ mutation introduced charge 10/06/2009 PHYS 461 & 561, Fall 2009-2010 14 Equation for eff : eq/4 eff 0r = ­ 2.3 RT pH Results of Fersht's experiments: eff ~40 to ~120 Protein Engineering (Fersht, the “father”): ­ changing a codon on the protein gene induces the mutation at an exact site of the protein globule ­ structural changes monitored by X­ray & NMR ­ protein as microscopic electrometer NEGLECTED: INTs between dipoles & quadrupoles (smaller than INTs between charges, decrease faster with r) 10/06/2009 PHYS 461 & 561, Fall 2009-2010 15 Electrostatic Interaction Between Two Free Charges in Water U(r) = q1 q2 / 4 eff 0r ⨯ exp(­r/D) (exponential decay) eff and D depend on the properties of the medium (water) D ­­­ Debye­Hückel radius: D = 3/I1/2Å also on ionic strength I I = ½ cizi2 ci ­­­ concentration of the ion type i zi ­­­ charge of the ion type I I = 0.1—0.15 [mol/l] (physiological conditions) ↓ For water at physiological conditions D ~ 8 Å 10/06/2009 PHYS 461 & 561, Fall 2009-2010 16 The origin of the screened electrostatic interaction temperature dependence of electrostatic effects: ➔ eff (T=273) = 88 & eff (T=373) = 55 88/55 ~ 1.6 & 373/273 ~ 1.4 The electrostatic interactions in water decreases with absolute temperature almost linearly → entropic effect! Electrostatic INTs in water are caused by the ordering of water molecules around the charges & variation of this ordering with distance (similar to the hydrophobic effect). 10/06/2009 PHYS 461 & 561, Fall 2009-2010 17 Disulfide (S—S) bonds formed between the side chains of two cysteines (Cys) ➔ two Cys side chains ( –C H2–SH ) release two H­atoms: ➔ –C H2–SH + –C H2–SH → –C H2–S–S–C H2– + H2 during formation of a disulfide bond formation and breakdown of S—S bond in cells is catalyzed by an enzyme called disulfide isomerase (only to accelerate the processes) & reversible ➔ absence of disulfide isomerase “freezes” the formed S—S bonds, S—S bonds typical of secreted proteins ➔ 10/06/2009 PHYS 461 & 561, Fall 2009-2010 18 Coordinate bonds ➔ formed by N, O, S­atoms of the protein & O­atom of water to di­ and trivalent ions of Fe, Zn, Co, Ca, Mg (metals) metal ions characterized by vacant orbits of low energy, capable of bonding an electron pair ➔ N, O, S­atoms are electron donors (radius ~1.5 Å) : their electrons occupy the vacant orbits of the metal ion (radius ~0.7 Å), forming a coordinate bond (only 1 bonded atom), several kcal/mol (similar to hydrogen bonding) ➔ if the donor atoms in the protein conformation are in a proper position for coordinate bonding, the ion gets released from water, bonds to protein (S of water increases!) → chelate complex ➔ 10/06/2009 PHYS 461 & 561, Fall 2009-2010 19