Molecular Forces in Biological Media:

1. Molecular Forces in Biological Media: • Reference: Nossal, R. & Lecar, H. (1991): ” Molecular & Cell Biophysics” • Addison-Sesley Pub., Redwood City CA. • I. Forces within Macromolecules: • a). Covalent Bonds: Vbond  100 kcal/mole • Compare to thermal energy RT  0.6 kcal/mole • probability for a chemical bond to break • p  exp(-E/RT) = exp(-100/0.6)  0 • Vb = kb(b-bo)2 + kθ(θ-θ0)2 + [lkφl - kφcosnφ] + kω(ω-ω0)2 • Where k’s are force constants which can be obtained from vibrational spectra and • are given in various force field packages. b’s are the bond length, θ’s, φ’s and • ω’s are fluctuating angles. Sub-zeros refer to the equilibrium values. • b). Electrostatic interaction: Vcoul = q1q2/εr • (ε = 1 in vacuuo & ε = 80 in water) • Q: What is ε inside a macromolecule ? Ans: No body knows for sure. • So ? 1. Set ε= 2; 2.ε= distance dependent; 3. Explicit water calculation. (Time consuming calculation) (Sharing electrons) (Boltzmann Distribution) (Stretching, bending backbone torsional ) side chain torsional (Comparable to covalent bond if e = 1)

2. c). Hydrogen bond: • VH = 0.5 – 4 kcal/mole • Comparable to RT (labile protons) • When several H-bonds are formed the additive effect makes the structure quite stable. VH = [(A/rαDA - B/r6DA)cosm(θD.H.A)cosn(θH.A.AA)] • The above equation showed that VH depends on bond length and bond angles between the donor and acceptor. • d). Dispersive Forces (London dispersive forces, van der Waals) • VVDW = -a/r6 (Attractive force) • Lennard-Jones Potential: V(r)LJ = -(a/r6 – b/rm) rDA A H D θD.H.A If m = 12 VLJ(r) is also called 6-12 potential Total Internal Energy: V(r) = Vb + Vc + Ve + VH + VLJ

3. II. Water as a Solvent: • “Bound waters” are found to strongly associate with macromolecules, are therefore affect the stability and conformation of macromolecules strongly. • “Hydrophobic interaction”, the tendency of nonpolar (aliphatic) groups to aggregate in solution, is a major factor in determining the conformation of macromolecules. • Charges groups preferred to be located on the surface, whilst the hydrophobic groups tends to be buried inside. • How do solvent affect macromolecular structure is still not fully understood. III. Electrostatic Forces in Ionic Solution Q: How to calculate the electrostatic interaction in an electrolytic solution? Solution:The Debye-Hückel Theory The electrostatic potential at a point r, Ψ(r), is given by the Poisson’s equation: 2Ψ(r) = -4(r)/ Where (r) is the charge density of the electrolyte in the solution and  is the dielectric constant.

4. The charge distribution of a given mobile ion is given by: ni(r) = ni0exp[-Wi(r)/kBT] (Boltzmann Distribution) Where Wi(r) is the potential energy of such an ion arising from the charge on the stationary surface acting in conjunction. kB is the Boltzmann constant (1.38 x 10-23J/K); ni0 is the concentration of i at Wi(r) = 0. Thus, the total charge density in the solution at r is given by (r) = 1(Zie)ni0exp[-Wi(r)/kBT] (Poisson-Boltzmnn equation) Where Zi is the valance of the ith species and e is the elementary charge (1.6x10-19C). The potential energy of ith ion can be written as: Wi(r) = (Zie)Ψ(r) Thus, 2Ψ = -4-11(Zie)ni0exp[-(Zie)Ψ/kBT] For eΨ/kBT << 1 exp[-(Zie)Ψ/ kBT]  1 - ZieΨ/kBT + … Thus, 2Ψ  -4-11(Zie)ni0[1-ZieΨ/kBT] 2Ψ The first term vannishes due to charge neutrality and  is the “inverse Debye length” and is given by: 2 = 8ne2/ kBT; Debye length RDB = (8ne2/ kBT)-1 where n = 1/21ni0 Zi2 is the “ionic strength” of the solution.

5. Example 1: Under physiological condition, [NaCl] = 140 mM = 0.14M • n = n(Na+) + n(Cl-) = [(0.14)x12+(0.14)x12]/2 = 0.14 • (Z = 1 for both Na+ and Cl-) • and -1 = 8 Å (check it out) • Example 2:Calculate the potential near a planer surface (one dimensional) such as a linear lipid bilayer. • d2Ψ/dx2 = 2Ψ,  Ψ =Ψsexp(-x) • In 100 mM NaCl -1 = 8 Å. Thus, when x = -1 = 8 Å , Ψ =Ψs/2.7 • The effect of ions in solution (“counterions”) is to decrease the range of the interaction between surface charge and any charge in the medium. • Example 3: Potential between two charged sphere of radius a. • Ψ(r) = Z2e2[(1+a/2)2]-1e-(r-a)/r • Where r is the center-to-center distance between the two particles. In comparison, Ψ(r)  1/r between two unshielded charged particles. • Homework: (Due March 17) Problem 1-3

6. Homework 1. (a) Assuming that all other factors are equal, in which ionic solution is the Debye length greater: 0.12M CaCl2 or 0.2 M NaCl ? If the Debye length in a 0.14 M NaCl is 8.2 Å , what is the Debye length in a 0.2 M NaCl solution ? (b) What would happen to the Debye screening if one adds ethanol to an aqueous solution ? 2. (a) The attractive can der Waals energy between two neutral nonpolar small molecules, arising from coupled charge fluctuations, varies as -/R6, where  is the order of 10-58 erg-cm6 (in vacuum, where the dielectric constant is  = 1). Estimate the energy of attraction between two such molecules that are approximately 1 Å apart. At what separation does energy become comparable with kBT, assuming T = 293K ? What would be the effect on these forces if the particles were immersed in a medium with higher dielectric constants (e.g., that of water ?). (b) To first approximation, the interaction between two identical large particles can be calculated by pairwise addition of the charge fluctuation forces between individual atoms located in each of the particles, that is: W  - N2VAdraVBdrb/r6 , where r is the separation between a mass point in particle A and a mass point in particle B, N represents the number of atoms in each particle. If the particles have radius ra are separted by a center-to-center distance R, the interaction energy can be shown to be: W = - (2N2/6)[2a2/(R2 –4a2) + 2a2/R2] + ln(1 – 4a2/R2)] show that when the spheres are very close so that R  2a, W  (2N2)(a/R – 2a)/12 Similarly, show that if R >> 2a , W  (162N2)(a/R)6/9 Although the indicated interaction seems to vary with the square of the number of atoms when the spheres get too big the intra atomic oscillation (induced dipole) no longer in phase and these simple expressions no longer valid.

7. 3: The energy level of a quantized oscillator are given by En = (n + ½)h, where n = 1, 2 …. And h is the Planck’s constant. In thermal equilibrium these levels are populated according to the Boltzmann distribution pn = [exp(-En/kBT)]/[n exp(-En/kBT)] where kB is the Boltzmann’s constant and T is the absolute temperature. The sum in the denominator is the partition function, Z. (a) For an oscillator whose energy spacing is 0.5 ev, determine which levels are occupied by more than 10% of the oscillator when T = 300 K. Explain why the quantity kBT is said to gauge the strength of interaction. (b) The average internal energy per molecule, U, is given by the sum: U = pnEn. Show that U is given by: U = h{1/2 + [exp(h/kBT)–1]-1}. What is the average internal energy when h is much smaller than kBT ? (Hint: Evaluate U by noticing that ne-nx = -d(e-nx)/dx and use this relation to evaluate U from the partition function Z, recall that n=0 xn = (1-x)-1).