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Discussion topic for week 3 : Entropic forces In the previous lectures we have stressed that random motion is the dominant feature of cellular dynamics.

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Presentation on theme: "Discussion topic for week 3 : Entropic forces In the previous lectures we have stressed that random motion is the dominant feature of cellular dynamics."— 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 as where Ω is the number of available states in the phase space Statistical postulate: An isolated system evolves to thermal equilibrium. 2 nd law:  S  0, equilibrium is attained when the entropy is at maximum. Entropy of an ideal gas:

3 Example: Temperature is defined from entropy as For the ideal gas:

4 To create order, we have to do work on a system volume change work done xx To push the piston f > F P, then  S > 0. At the equilibrium  S = 0

5 Open systems: Consider the same system but now immersed in a heat bath which keeps the temp. constant at T. As the spring expands Consider the total entropy of the system a+B a B remains const.

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 maximum or 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 L i, it will expand until Internal kinetic energy remains the same The free energy changes by Introduce Work done on the load is

8 The difference between free energy and work done is wasted as heat! In order to extract maximum possible work Quasi-static processes provide the most efficient way for extracting work If we bring this system into contact with a heat bath at T 2 < T 1 and recycle this process we obtain a heat engine. T2T2 T1T1

9 Microscopic systems: Consider a microscopic system in contact with a heat bath which keeps the temp. constant at T. According to the statistical postulate all the allowed states in the joint system have the same probability P 0 Probability of a particular state in “a” Boltzmann distribution

10 Two-state system: Consider a system, which is in thermal equilibrium with a reservoir and has only two allowed states with energies E 1 and E 2 The 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 forward direction (i.e. E2  E1), which is called the activation barrier,  E ‡ 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, introduce Here  is called the relaxation (or decay) time to equilibrium. It critically depends on the activation barrier, If  E >> 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 need use the average energy. Given the probability of each state as P i Average energy Entropy (from Shannon’s formula) By analogy, we define the free energy of a microscopic system as We need to show that this free energy is minimum at the equilibrium and leads to the Boltzmann distribution To find the minimum of F a, set

15 where Z is the partition function Substitute in F a

16 Complex two-state systems: Because biological macromolecules are flexible, they do not have unique structures. Rather in a given configuration, there are an ensemble of states whose energies are very close. Consider two such configurations I and II, with multiplicities N I and N II If all the states in I had the same energy E I and in II, E II, then More 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 becomes Note: when the volume changes in a reaction, the quantity to consider is the Gibbs free energy This is important for reactions involving the gas phase but can be ignored 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 tweezers Increasing the force on the bead triggers unfolding of RNA at ~14.5 pN DNA handle

19 Reversible folding of RNA Black line: stretching Gray line: relaxing  z = b-a ~ 22 nm When f = 14.1 pN, half of the RNA in the sample is folded and half unfolded f.  z = 14.1 x 22 = 310 pN.nm = 76 kT  F = 79 kT

20 Entropic Forces (Nelson, chap. 7) Entropic forces are in action when a system in a heat bath does work e.g., expansion of gas molecules in a box of volume V Ideal gas law applies equally to solutions

21 Osmotic pressure: Sugar solution rises in the vessel until it reaches an equilibrium height z f m

22 Osmotic flow Reverse osmosis mg van’t Hoff relation

23 Calculate the work done by osmosis when the piston moves from L i  L f The sugar concentration changes as For maximum efficiency, assume that we operate near equilibrium This is the same expression we calculated for the free energy change in 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)/(10 3 N A ), so c = 0.12 mM From van’t Hoff relation (cf. 1 atm = 10 5 Pa) Laplace’s formula for surface tension Thus  0.5 x x 300 = 1.5 x 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 inaccessible to smaller molecules. d < 2R d > 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/   /  where  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  (  = 80  Proteins and lipids have a much lower  (  ≈  Boundary between two media with different  forms a polarizable interface E ex   Induced surface charge density  

28 Examples: 1. Charge near a membrane Estimates of reaction force for an ion near a membrane (q’ ≈ q/84) Useful quantities to remember vacuum water   = 80   = 2 q d q’ image charge

29 2. Charge near a protein (spherical) Note that the monopole term (l = 0) vanishes. The dipole term gives   = 80   = 2 qrqr No simple image charge solution Series solution gives: a

30 q 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 times larger).   = 80   = 2 a

31 Poisson-Boltzmann (PB) equation: Mobile ions in water respond to changes in the potential so as to minimize the potential energy of the system. Instantaneous potential is given by the Poisson eq’n. At equilibrium, the mobile ions assume a Boltzmann distribution Substituting this density in the Poisson eq’n gives the PB eq’n Because 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,   no fixed charges for x > 0 Boundary cond.: Solution (see the Appendix) where  is the inverse Debye length,  x 

33 Linearized PB equation: Expand the Boltzmann factor in the ion density as The first term vanishes from electroneutrality; substituting the second term in the PB eq’n, we obtain the linearized PB eq’n This is much easier to solve, e.g. for the previous problem we have Boundary condition:

34 Charge density: Debye length controls the thickness of the charge cloud near a charged surface (at x = 1/ , it drops to about 1/3 of its value at x = 0) For a 1-1 salt solution at room temperature Rule of thumb: for a 90 mM solution, 1/  = 10 Å Although the approximate solution is obtained for it remains valid up to For nonlinear effects take over and the potential decays faster than the exponential. where c 0 is in moles

35 Energy stored in a diffuse layer: Potential energy due to an external field: Energy per unit area for a charged surface Estimate the energy for a unit charge per lipid head group For a cell of radius 10 4 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 = 0 The coefficients c 1 and c 2 are determined from the boundary conditions at r = a, namely,  and  E are continuous across the boundary a q 

37 Using the second boundary condition gives Again we have an exponential “Debye” screening of the Coulomb potent. To find the screening charge use maximum at r = 1/ 

38 Total screening charge Energy stored in the diffuse layer For This 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. (  = 80) Tetrahedral structure of water molecules in ice Molecular structure of H 2 O Covalent O-H bond: 1 Å H-bond distance: 1.7 Å O-O distance: 2.7 Å q ~ 0.4 e q -2q q 1 Å 104  1 Å

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  -helix and DNA) ~1 kT van der Waals attraction between neutral molecules Note that the above values are for two molecules in vacuum. In water, as with all other interactions, the H-bond energy is also reduced to 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 water molecule 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 in water to separate ions. Born energy for solvation Energy 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 in water, they disrupt the H-bond network, costing free energy. For small molecules, water molecules can form a ‘clathrate cage’ around the 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 solubility of small nonpol. molecules: Butanol, C 4 H 9 OH (top) Pentanol, C 5 H 11 OH Hexanol, C 6 H 13 OH Heptanol, C 7 H 15 OH (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 byand integrate from ∞ to x

44 Take the square root of the diff. eq. using Integrate [0, x] Substitute (reject the + solution)

45 Using the identity We can write Boundary conditions:

46 Approximate solution for Boundary condition: Charge dist.:


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