Presentation on theme: "Discussion topic for week 3 : Entropic forces"— Presentation transcript:
1 Discussion topic for week 3 : Entropic forces In the previous lectures we have stressed that random motion is the dominant feature of cellular dynamics.Why don't cells exploit the long-range Coulomb interactions to accelerate enzyme reactions, binding of ligands to proteins, etc?
2 Thermodynamics in cells (Nelson, chap. 6) See the Statistical Physics notes for detailed descriptions of Entropy,Temperature and Free Energy.Entropy is a measure of disorder in a closed system. It is defined aswhere Ω is the number of available states in the phase spaceStatistical postulate: An isolated system evolves to thermal equilibrium.2nd law: DS 0, equilibrium is attained when the entropy is at maximum.Entropy of an ideal gas:
3 Example:Temperature is defined from entropy asFor the ideal gas:
4 DxTo create order, we have to dowork on a systemvolume changework doneTo push the piston f > FP, then DS > 0. At the equilibrium DS = 0
5 aOpen systems:Consider the same system but nowimmersed in a heat bath whichkeeps the temp. constant at T.As the spring expandsConsider the total entropy of the system a+Bremains const.B
6 Helmholtz free energy of a system in contact with a thermal bath at T System comes to an equilibrium when the total entropy is maximumor when its free energy of is minimum.How much work can we extract from such a system?[This suggests that we can define an entropic force from free energy]Maximum possible work:
7 Assuming the initial length is Li, it will expand until Internal kinetic energy remains the sameThe free energy changes byIntroduceWork done on the load is
8 The difference between free energy and work done is wasted as heat! In order to extract maximum possible workQuasi-static processes provide the most efficient way for extracting workIf we bring this system intocontact with a heat bath atT2 < T1 and recycle this processwe obtain a heat engine.T2T1
9 Microscopic systems:Consider a microscopic system in contactwith a heat bath which keeps the temp.constant at T.According to the statistical postulateall the allowed states in the joint systemhave the same probability P0Probability of a particular state in “a”Boltzmann distribution
10 Two-state system:Consider a system, which is in thermal equilibrium with a reservoir andhas only two allowed states with energies E1 and E2The probabilities of being in these states follow from the Boltzmann dist.
11 Activation barrier:In a more typical situation, there is also an energy barrier in the forwarddirection (i.e. E2 E1), which is called the activation barrier, DE‡Rate constant: (probability of transition per unit time)For N particles, the rate of flow is given by N.k
12 The forward and backward rate constants and flux are given by At equilibrium,Approach to equilibrium(cf. Nernst pot.)
13 At equilibrium,To solve the DE, introduceHere t is called the relaxation (or decay) time to equilibrium.It critically depends on the activation barrier,If DE >> kT, k- 0, and transition probability becomes
14 Free energy in microscopic systems: In a microscopic system, fluctuations in energy can be large, so we needuse the average energy. Given the probability of each state as PiAverage energyEntropy (from Shannon’s formula)By analogy, we define the free energy of a microscopic system asWe need to show that this free energy is minimum at the equilibriumand leads to the Boltzmann distributionTo find the minimum of Fa, set
15 where Z is the partition function Substitute in Fa
16 Complex two-state systems: Because biological macromolecules are flexible, they do not have uniquestructures. Rather in a given configuration, there are an ensemble ofstates whose energies are very close.Consider two such configurations I and II, with multiplicities NI and NIIIf all the states in I had the same energy EI and in II, EII, thenMore generally, they are different, so we need to use a Boltzmann dist.Using
17 The relative probability becomes Similarly, the transition rates between two complex states becomesNote: when the volume changes in a reaction, the quantity to consider isthe Gibbs free energyThis is important for reactions involving the gas phase but can beignored in biological systems, where reactions occur in water
18 Example: RNA folding as a two-state system (Bustamante et al, 2001) Single molecule experiment using optical tweezersIncreasing the force on the bead triggers unfolding of RNA at ~14.5 pNDNA handle
19 Reversible folding of RNA Black line: stretchingGray line: relaxingDz = b-a ~ 22 nmWhen f = 14.1 pN, half of the RNAin the sample is folded and halfunfoldedf.Dz = 14.1 x 22 = 310 pN.nm = 76 kTDF = 79 kT
20 Entropic Forces (Nelson, chap. 7) Entropic forces are in action when a system in a heat bath does worke.g., expansion of gas molecules in a box of volume VIdeal gas law applies equally to solutions
21 Osmotic pressure:Sugar solution rises in the vessel until it reaches an equilibrium height zfm
23 Calculate the work done by osmosis when the piston moves from Li Lf The sugar concentration changes asFor maximum efficiency, assume that we operate near equilibriumThis is the same expression we calculated for the free energy changein an expanding volume of gas molecules. Dilute solutes in water behave the same way as gas molecules.
24 Osmotic pressure in cells: Cells are full of solutes; proteins occupy roughly 30% of the cell volume.Volume of a protein:Protein concentration:Note: c(mole/liter)=c(SI)/(103 NA), so c = 0.12 mMFrom van’t Hoff relation(cf. 1 atm = 105 Pa)Laplace’s formulafor surface tensionThus S 0.5 x 10-5 x 300 = 1.5 x 10-3 N/m = 1.5 pN/nm, rupture tension!
25 Osmotic pressure and depletion force When small and large molecules coexist, large ones tend to aggregate.The short range force driving this process is called the depletion force.When d < 2R, depletion forces act to reduce the volume inaccessibleto smaller molecules.d > 2Rd < 2R
26 Electrostatic forces in solutions Review of electrostatics:Coulomb’s law:Electric field:Gauss’ law:Differential form:Central field:Electric potential:& energy
27 Dielectrics:In a dielectric medium charges are screened: q q/e, r r/ewhere e is the dielectric constant of the medium.Electric forces, fields and potentials are screened by the same amount.Water has a large dipole moment and hence a large e (e = 80)Proteins and lipids have a much lower e (e ≈ 2 - 5)Boundary between two media with different e forms a polarizable interfaceEexe1 Induced surface charge densitye2
28 e1 = 80 e2 = 2 Examples: 1. Charge near a membrane image charge q d Estimates of reaction force for an ion near a membrane (q’ ≈ q/84)Useful quantities to remembervacuum waterimage chargeqde1 = 80dq’e2 = 2
29 e1 = 80 e2 = 2 2. Charge near a protein (spherical) q r Note that the monopole term (l = 0) vanishes.The dipole term givesqrNo simple image charge solutionSeries solution gives:e1 = 80e2 = 2a
30 e1 = 80 e2 = 2 3. Charge inside a channel (infinite cylinder) Series solution gives (too complicated!)The cylinder results are very different because, unlike the other cases,the charge is inside a closed boundary (cf. induced charge is 40 timeslarger).qe1 = 80e2 = 2a
31 Poisson-Boltzmann (PB) equation: Mobile ions in water respond to changes in the potential so as tominimize the potential energy of the system.Instantaneous potential is given by the Poisson eq’n.At equilibrium, the mobile ions assume a Boltzmann distributionSubstituting this density in the Poisson eq’n gives the PB eq’nBecause of the exp dependence, the PB equation is very difficult to solve.
32 Gouy-Chapman solution of the PB eq’n for a charged surface: 1-1 electrolyte solution (e.g. NaCl)surface charge density, sno fixed charges for x > 0Boundary cond.:Solution (see the Appendix)where k is the inverse Debye length,xs
33 Linearized PB equation: Expand the Boltzmann factor in the ion density asThe first term vanishes from electroneutrality; substituting the secondterm in the PB eq’n, we obtain the linearized PB eq’nThis is much easier to solve, e.g. for the previous problem we haveBoundary condition:
34 Charge density:Debye length controls the thickness of the charge cloud near a chargedsurface (at x = 1/k, it drops to about 1/3 of its value at x = 0)For a 1-1 salt solution at room temperatureRule of thumb: for a 90 mM solution, 1/k = 10 ÅAlthough the approximate solution is obtained forit remains valid up toFor nonlinear effects take over and the potential decaysfaster than the exponential.where c0 is in moles
35 Energy stored in a diffuse layer: Potential energy due to an external field:Energy per unit area for a charged surfaceEstimate the energy for a unit charge per lipid head groupFor a cell of radius 104 nm,Ion clouds are permanent fixtures of charged surfaces.
36 Debye-Huckel solution of the linearized PB eq’n for a sphere: Ion clouds modify the Coulomb interaction between macromolecules.We can solve the PB equation for a sphere to describe this effect.Represent the total charge q of the molecule with a point charge at r = 0The coefficients c1 and c2 are determined from the boundary conditionsat r = a, namely, j and eE are continuous across the boundaryaqe=80e=1
37 Using the second boundary condition gives Again we have an exponential “Debye” screening of the Coulomb potent.To find the screening charge usemaximum at r = 1/k
38 Total screening charge Energy stored in the diffuse layerForThis is too small, but charges on a protein are on the surface and q >> e Coulomb interaction is mediated by the ion clouds around proteins.
39 Why is water special?Large dipole moment (p = 1.8 Debye) large dielectric const. (e = 80)Tetrahedral structure of watermolecules in iceMolecular structure of H2OCovalent O-H bond: 1 ÅH-bond distance: 1.7 ÅO-O distance: 2.7 Åq ~ 0.4 eqq104o1 Å1 Å-2q
40 Energy scales involved in binding of molecules: ~ 100 kT single covalent bonds, e.g. C-C, C-N, C-O, O-H~10 kT H-bonds, e.g. O---H, N---H (recall a-helix and DNA)~1 kT van der Waals attraction between neutral moleculesNote that the above values are for two molecules in vacuum.In water, as with all other interactions, the H-bond energy is also reducedto about 1-2 kT. Thus, H-bond energy is comparable to the average K.E.of 3kT/2, and it can be relatively easily broken.In ice, there are 4 H-bonds per water molecule.In water at room temperature, the average number of H-bonds per watermolecule drops to ~3.5. That is, H-bond network is maintained to ~90% !*** Dynamics of water involves making and breaking of H-bonds ***
41 Solvation of molecules in water: Ions gain enough energy from solvation so that salt crystals dissolve inwater to separate ions. Born energy for solvationEnergy gain when a monovalent ion goes from vacuum into water:This energy is larger than the binding energy of an ion in crystal form.In water, ions have a tightly bound hydration shell around them.Typically there are 6 water molecules in the first hydration shell.Solventberg model: conductance of ions in water increases with temp.for most ions
42 Solvation of nonpolar molecules and hydrophobic interactions: When nonpolar molecules such as hydrocarbon chains are mixed inwater, they disrupt the H-bond network, costing free energy.For small molecules, water molecules can form a ‘clathrate cage’ aroundthe intruder, which almost maintains the H-bond in bulk water.But such an ordered structure costs entropy, i.e. free energy still goes up.Size dependence of solubilityof small nonpol. molecules:Butanol, C4H9OH (top)Pentanol, C5H11OHHexanol, C6H13OHHeptanol, C7H15OH (bottom)
43 Appendix: Gouy-Chapman solution of PB eq. Solve the diff. eq.Subject to the boundary condition:First introduce the dimensionless potential,and the inverse Debye length,The diff. eq. becomes,Multiply both sides by and integrate from ∞ to x
44 Take the square root of the diff. eq. using Integrate [0, x]Substitute(reject the + solution)
45 Using the identityWe can writeBoundary conditions:
46 Approximate solution for Boundary condition:Charge dist.: