Lecture 4: Mechanical and Chemical Equilibrium In the Living Cell (Contd.)

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1 Lecture 4: Mechanical and Chemical Equilibrium In the Living Cell (Contd.) Lecturer: Brigita Urbanc Office: ( Course website: 1

2 How do we find the equilibrium state or configuration? Are thermal fluctuations important? NO YES minimize potential energy minimize free energy 2

3 Finding displacement/charge density that minimizes the total potential energy 3

4 Potential energy function: f(u1, u2) 4

5 Find a minimum: f(u1, u2)/ u1 = 0 f(u1, u2)/ u2 = 0 For example: f(u1, u2) = ½ (A11 u12 + A22 u22 + 2A12 u1 u2) results in: A11 u1 + A12 u2 = 0 A12 u1 + A22 u2 = 0 5

6 Expansion in a Taylor series around local/global minima 6

7 Taylor expansion for cos(x) as a function of order n 7

8 Elastic stretching model: Young modulus E define strain: = L/L; note that in general, = (x,y,z) F = - k a or F/A = E L/L (A cross section area) E measures the stiffness of the beam; F/A stress 2 n = A/a, n=l/a ; F = n k a = A E L/L = AE a/a A 0 0 A 0 E = k/a0 A na n 8

9 Elastic deformation energy: a quadratic function of the strain Estrain = ½ EA ( L/L)2 dxestrain = ½ EA [du(x)/dx]2 dx F-actin filament lipid bilayer 9

10 Finding the macromolecule configuration that minimizes the total Gibbs free energy F = E TS; S = kb ln W; W # of microstates or multiplicity Example: possible arrangements of Np proteins on a DNA with N binding sites 10

11 S = kb ln W(Np; N) W(Np; N) = N! / [Np! (N Np)!] For example: 10 copies of Lac repressor protein for 5 x 106 DNA binding sites within E. coli genome N = 5 x 106; Np = 10 W ~ 3 x 1060 S = kb ln{n! / [Np! (N Np)!]} Using Stirling's approximation, we get: S/kB N ln(n) N [Np ln(np) Np] - [(N - Np) ln(n - Np) (N Np)] S/kB N [c ln(c) (1-c) ln(1-c)], where c = Np/N 11

12 Entropy is maximal at c = ½ 12

13 Hydrophobic Effect is Related to Entropy of Water Molecules Nonpolar (hydrophobic) molecules in solution deprive water molecules of the capacity to form hydrogen bonds and consequently take away part of their orientational entropy. 13

14 Local tetrahedral arrangement of water molecules: 6 possible orientations of central H2O molecule 14

15 When one of the four H2O neighbors is populated by a nonpolar (hydrophobic) molecule: 3 of 6 configurations are forbidden, thus: Shydrophobic Ghydrophobic = kb ln3 kb ln6 = - kb ln2 = n kbt ln2 n # of water molecules adjacent to the nonpolar molecule Ghydrophobic Examples: = hydrophobic Ahydrophobic -oxygen molecule O2 in water: ~kbt -octane in water: ~ 15 kbt 15

16 Isolated system: - can do no work on the environment and vice versa - no heat can flow from it or vice versa - no external fields - no particle flow Macroscopic equilibrium of an isolated system a system with a maximal entropy, that is, the largest number of microscopic realizations. Consider three different isolated two-compartment systems such That the barrier separating the two compartments (1,2) permits: (A) energy flow: E1 + E2 = ETOT = const. (B) volume change: V1 + V2 = VTOT = const. (C) particle flow: N1 + N2 = NTOT = const. Examine the total entropy: STOT = S1(E1, V1, N1) + S2(E2,V2, N2) 16

17 (A) Energy flow: Maximal entropy principle: ds = ( S/ E1) de1 + ( S/ E2) de2 = [ ( S/ E1) - ( S/ E2)] de1 = 0 where we considered: de2 = - de1 (ds/de1,2 ) = 1/T1,2 thermodynamic definition of temperature T1 = T2 17

18 (B) Volume change: Maximal entropy principle: ds = ( S/ V1) dv1 + ( S/ V2) dv2 = [ ( S/ V1) - ( S/ V2)] dv1 = 0 where we considered: dv2 = - dv1 ( S/ V1,2 )E,N = p1,2 /T TD identity p1 = p2 18

19 (C) Particle flow: Maximal entropy principle: ds = ( S/ N1) dn1 + ( S/ N2) dn2 = [ ( S/ N1) - ( S/ N2)] dn1 = 0 where we considered: dn2 = - dn1 - ( S/ N1,2 )E,V = 1,2 /T TD definition of a chemical potential : 1 = 2 19

20 Isolated versus closed system: Maximizing entropy versus Minimizing free energy Closed system exchanges energy and work (volume changes) with environment. closed system + environment = isolated system Open system exchanges energy, work (volume changes), and matter with environment. 20

21 maximizing the entropy of a closed system and environment minimizing the free energy of a closed system only: dstot = dsr + dss 0 der = T dsr p dvr (first law of TD) (heat added work done) by reservoir dsr = (der + p dvr )/T dstot = dss + der/t + p dvr/t 0 der = des & dvr = dvs dss des/t p dvs/t 0 dg = d(es + p Vs T Ss) 0 At fixed T,p; the free energy G of a closed system is minimized! 21

Lecture 5: Entropy Rules!

Lecture 5: Entropy Rules! Lecturer: Brigita Urbanc Office: 12-909 (E-mail: brigita@drexel.edu) Course website: www.physics.drexel.edu/~brigita/courses/biophys_2011-2012/ 1 DEFINITION OF A MICROSTATE Example: