Statistical Mechanics of Biological Processes

Statistical mechanics of biological processes

• Describing biological processes requires models. • If “reaction” occurs on timescales much faster than that of connected processes  quasi-equilibrium: laws of thermodynamics can be used.

• Biological systems/processes involve large number of interacting molecules.

 • Deterministic description impossible, resort to probabilistic description with evaluation of average properties.

• Statistical mechanics theoretical framework appropriate to quantitatively describe thermodynamics of processes at molecular level. • Thermodynamic state functions interpreted through concepts of microstates of systems compatible with a given macrostate.

• Microstate  particular realization of microscopic arrangement of constituents of system/process of interest.

• Macrostate  identified by a particular set of macroscopic independent parameters (e.g. E, N, V for an isolated thermodynamic system) which affect dynamics of constituents.

Microstates compatible with a given macrostate are different possible ways system can achieve that macrostate.

 • Statistical mechanics allows to calculate probability of each microstate under a set of constraints acting on the system.

• Boltzmann distribution  probability determined by energy of microstate.

Boltzmann’s distribution: key equations

1 −Ei kBT p(Ei ) = e Z Z = ∑e−Ei kBT i 1 F = − ln Z β

1 −Ei kBT ∂ E = ∑Eie = − ln Z Z i ∂β ...

Lattice model very useful to describe statistical mechanics of molecular recognition events (concentration arises naturally as key variable).

• Example with ligand-receptor binding. • Macrostates: bound vs. unbound.

DNA (or any other polymer) in solution

Shape of polymer (microstate) can be characterized in different ways: • By using function r(s) to characterize positions of points of molecule (s distance of point along molecule). • By reporting coordinates of all atoms of DNA.

DNA (or any other polymer) in solution

Definition of microstates not absolute, but depends on problemShape of ofpolymer interest! (microstate) can be characterized in different ways: • By using function r(s) to characterize positions of points of molecule (s distance of point along molecule). • By reporting coordinates of all atoms of DNA.

Boltzmann’s distribution: examples of two-states systems Transport through membrane channel: electrophysiology experiments

Boltzmann’s distribution: examples of two-states systems Binding of ligands to a rigid receptor (no internal d.o.f.)

• Lattice model to describe thermodynamics of molecular recognition useful because concentration arises naturally as a key parameter. Consider:

• L ligands, Ω boxes each with volume Vbox. • Only two classes of states:

1. No ligand bound to receptor, all compatible microstates have same energy εsol.

2. One ligand bound to receptor, all compatible microstates have energy εb.

• pbound given by:

∑ 1bound states pbound = ∑ 1bound + ∑ Lunbound states states

• Numerator  statistical weight of having 1 ligand bound and L-1 in solution

1 W 1bound e−βεb L 1 ∑ bound = microstates = × ∑ ( − )unbound states states

−β L−1 ε Ω! −β L−1 ε L −1 = e ( ) sol = e ( ) sol ∑ ( )unbound ∑ % ' states states (L −1)!&Ω−(L −1)(!

• Denominator  configurational partition function for L ligands swimming in solution with no binding to receptor:

−βLεsol Ω! −βLεsol ∑ Lunbound = ∑ e = e states states L!(Ω− L)!

• If Ω  L as often happens, approximation holds:

Ω! L ≈ Ω (Ω− L)!

• Introducing energy difference, concentration c and concentration standard c0 (L = Ω ):

Δε = εb −εsol c = L ΩVbox c0 =1 Vbox

• Probability can be written after rearrangement of equation as:

−βΔε Langmuir adsorption isotherm (c c0 )e or pbound = −βΔε 1+(c c0 )e Hill function with coefficient 1

-βΔε  Weights to have zero or one ligands bound are 1 and c/c0 e respectively.

Estimate of parameters by choosing

3 3 Vbox = 1 nm  c0 = 1 molecule/nm = = 1024 molecules/l =

24 = 10 /NA M ~ 1.66 M

Estimate of parameters by choosing

3 3 Vbox = 1 nm  c0 = 1 molecule/nm = = 1024 molecules/l =

24 = 10 /NA M ~ 1.66 M

Value of dissociation constant Kd corresponds to concentration of

ligands for which pbound = 1/2  Perfect balance between entropic and energetic terms of free energy of binding.

Boltzmann’s distribution: examples of two-states systems Expression of specific protein (transcription)

Cells express different genes in different amounts at different times. Regulation of gene expression is complex and requires several control mechanisms as well as degradation of both mRNA and proteins.  Amount of proteins present at any time depends on all processes involved.

Keep it simple: focus on first step, reduce complexity of gene expression by: 1. Considering only amount of mRNA produced by RNA polymerase. 2. Considering only binding of transcription factors (namely activators) to promoter region: if TF bound, then polymerase starts transcription.

 Reduce problem to calculation of probability of polymerase binding to specific promoter region of DNA.

Further assumption (corroborated by experiments): all RNA polymerases are bound to DNA. 

Probability of expression of gene calculated considering competition between binding to specific promoter site and non specific binding to all remaining ones along 1D lattice.

• Problem similar to that seen for ligand binding receptor.

• pb given by calculating relative weight of 1 S polymerase among P binding to promoter with energy ε pd NS vs. binding to NNS non specific sites with energy ε p d .

S ∑ 1b S states pb = S NS ∑ 1b + ∑ Pb states states

• Numerator  statistical weight of having 1 polymerase bound specifically.

S NS 1S e−βεb P 1 ∑ b = × ∑ ( − )b states states

NS P 1 NS N ! P 1 NS P −1 = e−β( − )εb = NS e−β( − )εb ∑ ( ) ∑ $ & states states (P −1)!%N NS −(P −1)'!

• Denominator  total partition function, need to calculate weight of P polymerases binding nonspecifically:

NS NS NS −βPεb N NS ! −βPεb ∑ Pb = ∑ e = e states states P!(N NS − P)!

• If NNS  P, approximate:

N NS ! P ≈ (N NS ) (N NS − P)! • Introducing energy difference:

S NS Δε = ε b −ε b

• Probability can be written after rearrangement of equation as: P e−βΔε N NS 1 pbound = = P −βΔε N βΔε 1+ e 1+ NS e N NS P

-βΔε • Weights to have zero or one specific bindings 1 and P/NNSe respectively.

Difference between strong and weak promoter can be attributed to Δε

Δε = −8.1kBT

Δε = −2.9kBT

• Chemical potential  Gibbs free energy per unit mass change.

• µ of solute can be calculate from total Gibbs free energy Gtot: " % ∂Gtot µsolute = $ ' # ∂Ns &T, p

• Total free energy can be written as sum of solvent free energy, solute energy and entropy of mixing:

G = N µ 0 + N ε −TS tot H2O H 2O s s mix

• Dilute solution  No interaction among solute molecules.  Energy of solute molecules given by sum of

per-molecule contribution (εs) multiplied by Ns. Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 23 Chemical potential of dilute solution

Lattice model allows to derive equation for chemical potential of dilute solutions

Ns NH2O



Entropy of mixing Smix given by number of ways of arranging Ns molecules within Ns + NH2O lattice points: N + N ! ( H2O S ) S = k lnW N + N , N = k ln mix B ( H2O s s ) B N !N ! H2O s

Exploiting Stirling approximation and low concentration c = Ns/NH2O gives: # & Ns S ≈ −k %N ln − N ( mix B % s N s ( $ H2O '  " N % G T, p, N , N N 0 T, p N T, p k T $N ln s N ' tot H2O s = H2OµN2O ( ) + sεs ( ) + B s − s ( ) $ N ' # H2O &

Deriving with respect to Ns after introduction of solute and reference concentrations c = Ns/Vbox and c0 = NH2O/Vbox gives chemical potential:

∂Gtot c µs = = εs + kBT ln

∂Ns c0

which can be generalized to any dilute solution by writing chemical potential with respect to a standard reference state indicated by suffix 0:

ci µi = µi0 + kBT ln ci0

Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 26 Osmotic pressure as entropic effect Considering a binary spontaneous reaction between two species 1 and 2, second law of thermodynamics states that ΔG must be non positive:

dG = µ1dN1 + µ2dN2 = µ1 − µ2 dN1 ≤ 0 ( ) 

Δµ = µ1 – µ2 driving force for mass transport: if µ1 ≥ µ2  dN1 ≤ 0 and viceversa. For dilute solutions:

c1 Δµ = Δµ0 + kBT ln c2  Force arising in part from different concentrations of two species. Identical reasoning applies to same species in two compartments 1 and 2. If concentration is higher in 1 and particles are allowed to cross boundary between 1 and 2, they will flow from 1 to 2.

• Interior of cell very crowded  concentration of proteins, DNA and many other molecules much higher than outside, while concentration of water much lower.

 • Force will tend to move components outside and water inside cell.

• Mechanical force acting on membrane induces osmotic pressure.

How to cope with constant stress induced by osmotic pressure?

• Bacteria endowed with rigid cell wall outside plasma membrane.

• Pumps actively drying cells’ interior by expelling water and modulating concentration of ions and sugars, etc…

Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 28 Osmotic pressure as entropic effect Osmotic pressure is a purely entropic effect. • Rationalized using thermodynamics for dilute solutions.

• Consider two compartments, one with water and another with dilute solution, separated by semipermeable membrane (only water can cross).

• At equilibrium chemical potentials of water identical in two compartments.

• Chemical potential of water on side of dilute solution can be derived as:

∂G µ = tot H2O N ∂ H2O

G N µH2O = ∂ tot ∂ H2O " N % G T, p, N , N N 0 T, p N T, p k T $N ln s N ' tot H2O s = H2OµH2O ( ) + sεs ( ) + B s − s ( ) $ N ' # H2O &  N T, p 0 T, p k T s µH2O ( ) = µH2O ( ) − B N H2O

At equilibrium chemical potentials in compartments 1 and 2 must coincide:

N 0 T, p 0 T, p k T s µH2O ( 1 ) = µH2O ( 2 ) − B N H2O

Difference in pressure in two compartments needed to maintain equilibrium.

If Δp  p1 expand chemical potential on right-hand side of equation: ∂µ 0 0 T, p 0 T, p H2O p p µH2O ( 2 ) ≈ µH2O ( 1) + ( 2 − 1) ∂p p1 Since molecular volume is given by derivative of µ with respect to p: ∂µ 0 H2O = vH2O= V N ≈ V N ≡ V N ∂p mol 1 H2O 2 H2O H2O p1

Equilibrium condition becomes: van’t Hoff formula Ns (p2 − p1) = kBT → Δp = kBTcs gives osmotic pressure as function V of concentration of solute

Estimate of osmotic pressure in E. coli obtained by considering following assumptions:

• Concentration of inorganic ions ~ 100 mM, which means, since V ~ 1fL, a number of ions within bacterium of about:

22 15 7 0.1M = 0.1N A 1L ≈ 6 10 10 fL → 6 10 molecules

Concentration as number of solute molecules Ns/V can be calculated considering that 1 fL = 109 nm3:

c ≈ 6 107 109 nm3 = 0.06molecules nm3

Since kBT ~ 4.1 pN ⋅ nm, Δp amounts to: Δp ≈ 0.06 4.1molecules nm3 pN nm ≈ 0.25 pN nm2 ≈ 2.5atm

Chemical reactions at equilibrium obey Law of Mass Action, which introduces equilibrium constants setting ratio between concentrations of reactants and products.

• Can be derived from statistical mechanics, which also gives microscopic description of equilibrium in terms of stoichiometric coefficients and concentrations of species involved in reaction. • Entropy maximization is a way to obtain equilibrium constants.

Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 33 Law of Mass Action Consider simple reaction with two reagents and one product: A + B  AB

• Stoichiometric coefficients νi must be introduced to count changes in number of molecules during reaction.

• In case above if A and B decrease by one unit,

AB must increase by same quantity  νA = νB = -νAB. • At equilibrium differential of Gibbs free energy must be zero since G is minimum:

dG(N A, NC, NC ) = 0 ↓ G N dN G N dN G N dN 0 (∂ ∂ A )B,AB A +(∂ ∂ B )A,AB B +(∂ ∂ AB )A,B AB = ↓

µAdN A + µBdNB + µABdN AB = 0 Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 34 Law of Mass Action

In a general case, expressing all dNi through stoichiometric coefficients:

dNi = νidN equilibrium condition becomes:

N N N dG = 0 → ∑µidNi = ∑µiνidN = 0 → ∑µiνi = 0 i=1 i=1 i=1 • For dilute solutions this condition can be written: N N " % N N " %νi ci ci ∑µiνi = ∑$µi0 + kBT ln 'νi = 0 → ∑µi0νi = −kBT∑ln$ ' i=1 i=1 # ci0 & i=1 i=1 # ci0 & ↓ N N " %νi 1 ci − ∑µi0νi = ln∏$ ' kBT i=1 i=1 # ci0 & Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 35 Law of Mass Action Exponentiation of formula and reordering of all constants on right-hand side gives:

Law of Mass Action N 1 N N − µ ν k T ∑ i 0 i Explains and predicts dynamic νi νi B i=1 ∏ci = ∏ci0e equilibrium by relating concentrations i=1 i=1 of reactants and products at a given temperature and pressure.

Constant on right hand side defined as equilibrium constant of reaction Keq(T):

N 1 N − µ ν k T ∑ i 0 i K T cνi e B i=1 eq ( ) ≡ ∏ i0 i=1

Inverse of dissociation constant Kd: 1 Kd = Keq

Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 36 Law of Mass Action • Equilibrium constant can be measured experimentally, but previous formula allows to set microscopic interpretation of chemical equilibrium.

• For instance, considering again simple reaction A + B  AB one gets: N νi −1 −1 1 cAB 1 ∏ci = cA cB cAB = = Keq (T) = i=1 cAcB Kd Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 37 Ligand-receptor binding revisited

Example to highlight link between two ways of describing of molecular processes:

Chemical " dissociation constant Kd: Statistical " binding energy Δε: −βΔε [L][R] (c c0 )e Kd = pbound = −βΔε [LR] 1+(c c0 )e

• Reconcile views by expressing pbound as function of Kd: Former result: dissociation constant [LR] [L] Kd pbound = =  Kd corresponds to concentration of R LR 1 L K [ ] + [ ] + [ ] d ligands for which pbound = 1/2

• Kd cast in terms of microscopic parameters of lattice model as: Link between languages 1 βΔε Kd = e of chemistry and of Vbox statistical physics

Note about experimental settings

• pbound depends on [L], that is concentration of free ligands and not total one. • Important subtlety since in typical experiment we pipette in total concentration, while free concentration is determined by molecular properties of L/R interaction. 

• Often to determine Kd convenient to work at concentrations where ligands are significantly more than receptors (excess of ligands). 

• Free ligand concentration nearly equal to total ligand concentration.

• Energy released upon reaction ATP  ADP + Pi depends on concentration of reactants and products. • Considering change ΔG upon hydrolysis of one molecule:

N N N dN=1 dG = ∑µidNi = ∑µiνidN """→ΔG = ∑µiνi i=1 i=1 i=1

• For dilute solutions: N " % N N " %νi ci ci ΔG = ∑$µi0 + kBT ln 'νi = ∑µi0 + kBT ln∏$ ' i=1 # ci0 & i=1 i=1 # ci0 &

• Considering change with respect to reference concentration (ΔGref): # N N & νi νi ΔG = ΔGref + kBT ln%∏ci ∏ci,ref ( $ i=1 i=1 '

Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 40 Thermodynamics of ATP hydrolysis Choosing equilibrium state as reference implies: N ν ΔG = 0 c i = K ref ∏ i,ref eq i=1 Thus previous equation referred to equilibrium state becomes: # N & νi ΔG = kBT ln%∏ci Keq ( $ i=1 '

Since standard state free energy (at c = 1M) can be expressed as:

ΔG = −k T ln K 0 B eq by adding and subtracting ΔG0 one obtains (expressing Keq in molar units!):

N νi " ci % ΔG = ΔG0 + kBT ln∏$ ' i=1 #1M &

Considering hydrolysis of ATP and expressing concentrations in molar units:

ΔG = ΔG0 + kBT ln [ADP][Pi ] [ATP] Considering typical experimental values:

• ΔG0  -12.5 kBT • All concentrations around mM range: – [ATP] ~ 5⋅10-3 – [ADP] = 0.5⋅10-3

-3 – [Pi] = 10⋅10 • Gives result introduced earlier:

−4 −2 ATP 5⋅10 10 ΔG ≈ −12.5k T + ln ≈ −12.5k T − 6.9k T ≈ −20k T h B 5⋅10−3 B B B

• Many cellular processes are based on two allowed states only. • Molecule or cell needs to be either on or off as a function of concentration of signal received  turn analog signal into digital output.  • “Response” curves such as Langmuir adsorption isotherm not appropriate to describe these processes. • Binding curve must be switch-like, step or sigmoidal function.

• Behavior seen in many biological processes, arises from cooperativity. • Can be understood considering a simple process with two ligands. • For sake of simplicity we consider ideal cooperative behavior: no intermediate with one ligand only bound to receptor:

L + L + R  L2R  2 2 [L] [R] [L2 R] Kd = pbound = [L2 R] [R] + [L2 R] 

2 [L] Kd p = ( ) Hill function with bound 2 coefficient n=2 1+([L] Kd )

More generally:

n [L] Kd p = ( ) Hill function with bound n Hill coefficient n 1+([L] Kd )

• Increasing n increases cooperativity and thus sharpness of transition between states. • Usually n is found by fitting binding data to Hill curves directly without reference to underlying origins of a given Hill coefficient. • In derivation assumed negligible [LR]  binding is either all or nothing. Not strictly true  measured value of Hill coefficient may not be an integer.

Biophysics Course held at Physics Department, University of Cagliari, Italy. Academic Year: 2017/2018. Dr. Attilio Vittorio Vargiu PLEASE NOTE! This material is meant just as a guide, it does not substitute the books suggested for the Course. 45 References • Books and other sources • Physical Biology of the Cell (2nd ed.), Phillips et al., Chap. 6