Lecture 6: Entropy Rules! (contd.) Lecturer: Brigita Urbanc Office: 12-909 (E-mail:
[email protected]) Course website: www.physics.drexel.edu/~brigita/COURSES/BIOPHYS_2011-2012/ 10/11/2011
PHYS 461 & 561, Fall 2011-2012
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Boltzmann distribution (by counting: Textbook, pages 230-231) Information theory: based on the constraints (for example the total energy) derive the least biased probability distribution Shannon entropy:
S(p1, p2, p3 ..., pN) ∑i=1
N
pi ln pi
where p is the probability of the system to be in the i-th microstate. i
For example, if nothing is known about the system except that there is N microstates, then for all p = 1/N and S = ln N (the maximal value).
i
How do we formally derive this result? 10/11/2011
PHYS 461 & 561, Fall 2011-2012
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Maximize Shannon entropy S' using Lagrange multiplier method for each constraint:
S' ∑i=1 N pi ln pi [ ∑i=1 N pi 1] Maximization equations:
∂S'/∂ → ∑i=1 N pi 1 ∂S'/∂pi → lnpi 1 pi exp(1 Note that the probabilities p do not depend on i! i ∑i=1 N pi 1 → ∑i=1 N exp(1 1 → exp(1
pi 10/11/2011
PHYS 461 & 561, Fall 2011-2012
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Boltzmann distribution is a maximum entropy distribution with a fixed average energy:
S' ∑i pi ln pi [ ∑i pi –1] [ ∑i pi Ei – ‹E›] lnpi1 – Ei→pi exp(1exp(Ei ∑i pi 1 →exp(1+∑i exp(EiZ pi exp(Ei∑i exp(Ei
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PHYS 461 & 561, Fall 2011-2012
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Ideal gas approximation: the interaction range is small in comparison to the mean spacing between molecules For a system with N independent variables (x the probability distribution can be factorized:
, x , … x ) 1 2 N
P(x1, x2, … xN) = P(x1) P(x2) … P(xN)
– uniform spatial distribution of ideal gas molecules.
Microscopic state of the system is described by (x, p ) so the key x
information that we need to derive is P(p ) knowing the kinetic x
energy of E 10/11/2011
2 x
= p /(2m). PHYS 461 & 561, Fall 2011-2012
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Probability distribution for a system with average energy is:
exp[-px2/(2m)] P(px) = ∑states exp[-px2/(2m)] Instead of a sum, we use an integral over a continuous variable p
:
x
∑states → ∫-∞ ∞ dpx where we will use the integral:
∫-∞ ∞ exp(-px2) dpx = √ 10/11/2011
PHYS 461 & 561, Fall 2011-2012
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According to the equipartition theorem, each degree of freedom is associated with :
‹E› = ½ kBT Calculate ‹E› using the Boltzmann distribution: ∞
2 x
Z = ∫-∞ exp[-p /(2m)] dpx = √2m ‹› ∫-∞ ∞ px2/(2m) exp[-px2/(2m)] dpx
We use the following trick:
‹› ∂Z∂½ so we showed that the Lagrange multiplier (k T) . B 10/11/2011
PHYS 461 & 561, Fall 2011-2012
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Free energy and chemical potential of a dilute solution: Application of a lattice model (and ideal gas approx.)
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PHYS 461 & 561, Fall 2011-2012
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Configurational entropy of N objects placed into W available spots:
!
W(N,) =
S = kB lnW N! (– N)!
How to calculate the chemical potential of a dilute solutions?
SOLUTE
= (∂GTOT /∂NS)T,p
GTOT = NW 0W + NS S – T SMIX
water G + solute energy + mixing entropy contribution Mixing entropy contribution: -independent solute molecules (ideal gas) -lattice model: N + N is a total number of lattice sites W
S
W(NW,NS) = (NW + NS)! / (NW! NS!) 10/11/2011
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S = kB lnW = ln[(NW + NS)! / (NW! NS!)] ≈ kB {NW ln[NW/(NW + NS)] + NS ln[NS/(NW + NS)]} = kB [NW ln(1 - NS/NW) + NS ln(NS/NW)] Taking into account the Taylor expansion of ln(1+x) ≈ x, we get:
SMIX = kB [NS ln(NS/NW) – NS]
GTOT (T,p,NW,NS) = NW W0 + NS S(T,p) + kBT(NS ln(NS/NW) – NS) S(T,p) = S(T,p) + kBT ln(c/c0)
or in a general form expressed in concentration c=N/V:
i = i0 + kBT ln(ci/ci0 ) 10/11/2011
PHYS 461 & 561, Fall 2011-2012
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Osmotic pressure Is an Entropic Effect ➔
➔
consider a cell in an aqueous environment exchanging material with a solution (intake of food, excretion of waste) chemical potential difference proportional to G:
G = ( 1 2) dN ≤ 0 (spontaneous) ➔
cell with a crowded environment of biomolecules: tendency of almost all components to move out causes a mechanical pressure called osmotic pressure
➔
lipid membranes with ion channels to regulate ion concentration
➔
calculate osmotic pressure due to a dilute solution of NS molecules
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PHYS 461 & 561, Fall 2011-2012
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Osmotic pressure on a semipermeable membrane, which only allows water molecules through
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PHYS 461 & 561, Fall 2011-2012
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W(T,p) = ∂G ∂ W p W (T,p) kBT NS/NW 0
For both sides of the membrane in equilibrium:
W (T,p1) W (T,p2) kBT NS/NW 0
0
Expand the chemical potential at p around the p value: 2
1
W (T,p2) ≈ W (T,p1) (∂ W /∂p) (p2 – p1) 0
and consider that
0
0
(∂ W0/∂p) = v = NW/V is volume per
water molecule so that
(p2 – p1) = kBT NS/V 10/11/2011
PHYS 461 & 561, Fall 2011-2012
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Measuring interstrand interactions in DNA using osmotic pressure
p(dS) = F0 exp(-dS/c) dS … interstrand spacing; F0 depends on the ionic solution pressure:
10/11/2011
PHYS 461 & 561, Fall 2011-2012
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Law of Mass Action and Equilibrium Constants (Chemical Reactions) ➔
chemical equilibrium between A, B and their complex AB:
A + B AB
➔
final equilibrium independent of whether we start with only A and B, or with a high concentration of AB and no A or B
➔
NA, NB, NAB
➔
In equilibrium: dG = 0
… number of A, B, and AB molecules
0 = (∂G/∂NA) dNA + (∂G/∂NB) dNB + (∂G/∂NAB ) dNAB ➔
A more convenient and general expression:
∑i=1 N i dNi 0
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PHYS 461 & 561, Fall 2011-2012
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Stoichiometric coefficients for each of the reactants are defined as:
i ±1
depending on whether the number of particles of the i-th type increases or decreases during the reaction:
∑i=1 or or
N
i i 0
∑i=1 N i0 i kBT ∑i=1 N ln(ci/ci0 )i ∑i=1 N i0 i ln[∏i=1 N (ci/ci0 )i]
∏i=1 N cii = (∏i=1 N ci0 i) exp(∑i=1 N i0 i) where we define the equilibrium constant K : eq Keq = (∏i=1 N ci0 i) exp(∑i=1 N i0 i) 10/11/2011
PHYS 461 & 561, Fall 2011-2012
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Kd = 1/Keq … dissociation constant In our case of the reaction A
+ B AB, we can express Kd as:
Kd = ∏i=1 N ci i = cA cB / cAB
Example: total concentration: 50 M
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PHYS 461 & 561, Fall 2011-2012
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Application to Ligand-Receptor Binding:
L + R LR Kd = [L][R]/[LR] or [LR] = [L] [R]/Kd Binding probability:
[LR] [L]/Kd pbound = --------------- = -------------[LR] + [R] 1 + [L]/Kd A natural interpretation of K : K is the concentration at which d d the receptor has a probability of ½ of being occupied by a ligand. Based on out prior result, we can express it in terms of lattice model parameters as: K 10/11/2011
= exp() d
PHYS 461 & 561, Fall 2011-2012
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Important: K depends on the concentration of free ligands not d
their total concentration!
(1) (2)
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(C)
(D)
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Cooperative Ligand-Receptor Binding: The Hill Function Biological function: an on-off switch behavior triggered by Binding of a ligand to a receptor involves a cooperative (all-or-none) mechanism:
L + L + R L2R 2 2 Kd = [L] [R]/[L2R] pbound
10/11/2011
([L]/Kd)2 ([L]/Kd)n = ---------------- = -------------------2 n 1 + ([L]/Kd) 1 + ([L]/Kd)
PHYS 461 & 561, Fall 2011-2012
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The larger the n, the sharper the binding curve (probability of binding versus ligand concentration)
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PHYS 461 & 561, Fall 2011-2012
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