Statistical mechanics of biological...

Dispense del Corso di Biofisica, Dipartimento di Fisica, Università di Cagliari. A.A.: 2015/2016 Docente: Dott. Attilio Vittorio Vargiu NB: Queste dispense non sostituiscono il materiale didattico suggerito a piè del programma! 1

Statistical mechanics of biological processes

Dispense del Corso di Biofisica, Dipartimento di Fisica, Università di Cagliari. A.A.: 2015/2016 Docente: Dott. Attilio Vittorio Vargiu NB: Queste dispense non sostituiscono il materiale didattico suggerito a piè del programma!

Modeling biological processes •  Describing biological processes requires models.

•  If a “reaction” occurs on timescales much faster than that of connected processes quasi-equilibrium situation: 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.

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Modeling biological processes •  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.

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Microstates in biology Lattice model very useful to describe statistical mechanics of molecular

recognition events (concentration arises naturally as key variable).

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•  Example with ligand-receptor

binding.

•  Macrostates: bound vs. unbound.


Microstates in biology DNA (and any polymer) in solution

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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.

Definition of microstates not absolute, but depends on problem of interest!


Modeling biological processes Boltzmann’s distribution: key equations

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p Ei( ) = 1Ze−Ei kBT

Z = e−Ei kBTi∑

F = − 1βlnZ

E =1Z

Eie−Ei kBT = −

∂∂βlnZ

i∑

...


Modeling biological processes Boltzmann’s distribution: examples of two-states systems

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Transport through membrane channel



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Binding of ligands to a rigid receptor (no d.o.f.)


Ligand-receptor binding •  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:

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pbound =1bound

states∑

1boundstates∑ + Lunbound

states∑


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

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

Ligand-receptor binding

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1boundstates∑ = Nmicrostates

1bound = e−βεb × L −1( )unboundstates∑

L −1( )unbound = e−β L−1( )εsol

states∑ =

Ω!L −1( )! Ω− L −1( )%& '(!states

∑ e−β L−1( )εsol

Lunbound = e−βLεsolstates∑ =

Ω!L! Ω− L( )!states

∑ e−βLεsol


•  If Ω >> L as often happens, one can approximate:

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

•  Probability can be written after rearrangement of equation as:

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


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Ω!Ω− L( )!

≈ ΩL

Δε = εb −εsol c = L ΩVbox

pbound =c c0( )e−βΔε

1+ c c0( )e−βΔεLangmuir adsorption

isotherm or Hill function with coefficient 1

c0 =1 Vbox


Langmuir adsorption isotherm


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Estimate of parameters by choosing

Vbox = 1 nm3 c0 = 1 molecule/nm3 =

= 1024 molecules/l =

= 1024/NA M ~ 1.66 M


Langmuir adsorption isotherm


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Estimate of parameters by choosing

Vbox = 1 nm3 c0 = 1 molecule/nm3 =

= 1024 molecules/l =

= 1024/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.



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Expression of specific protein (transcription)


StatMech of gene expression…

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.

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StatMech of gene expression… Focusing 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 to bind to specific promoter region of DNA.

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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.



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•  Problem similar to that seen for ligand binding receptor.

•  pb given by calculating relative weight of 1 polymerase among P binding to promoter with energy vs. binding to NNS non specific sites with

energy .

pSb =1Sb

states∑1Sb

states∑ + PNS

bstates∑

εpdS

εpdNS



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•  Numerator statistical weight of having 1 polymerase bound specifically.

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

1Sbstates∑ = e−βε

Sb × P −1( )NSb

states∑

P −1( )NS = e−β P−1( )εNSb

states∑ =

NNS !P −1( )! NNS − P −1( )$% &'!states

∑ e−β P−1( )εNSb

PbNS = e−βPε

NSb

states∑ =

NNS !P! NNS −P( )!states

∑ e−βPεNSb


•  If NNS >> P as often happens, one can approximate:

•  Introducing energy difference:

•  Probability can be written after rearrangement of equation as:

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


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NNS !NNS −P( )!

≈ NNS( )P

Δε = ε Sb −εNSb

pbound =PNNS

e−βΔε

1+ P NNSe−βΔε

=1

1+ NNSP e

βΔε


•  Difference between strong and weak promoter can be attributed to


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Δε

Δε = −2.9kBT

Δε = −8.1kBT


•  Chemical potential Gibbs free energy per unit mass change.

•  µ of solute can be calculate from total Gibbs free energy Gtot.

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

•  Dilute solution No interaction between various solute molecules.

Energy of solute molecules given by sum of per-molecule contribution

(solvation cost εs) multiplied by Ns.

Chemical potential of dilute solution

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µsolute =∂Gtot

∂Ns

"

#$

%

&'T ,p

Gtot = NH2Oµ 0H2O + Nsεs −TSmix


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


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Ns NH2O


Entropy of mixing Smix given by number of ways of arranging Ns molecules within Ns + NH2O lattice points:

Exploiting Stirling approximation and low concentration c = Ns/NH2O gives:


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Smix = kB lnW NH2O,Ns( ) = kB ln

NH2O+ NS( )!

NH2O!Ns!

Smix ≈ −kB Ns lnNs

NH2O

− Ns

#

$%%

&

'((

Gtot T, p,NH2O,Ns( ) = NH2O

µN2O0 T, p( )+ Nsεs T, p( )+ kBT Ns ln

Ns

NH2O

− Ns

"

#$$

%

&''


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

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


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µs =∂Gtot

∂Ns

= εs + kBT lncc0

µi = µi0 + kBT lncici0


Considering a binary spontaneous reaction between two species 1 and 2, second law of thermodynamics states that ΔG must be non positive:

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

Since, at least for dilute solutions:

This force can be seen arising in part from different concentrations of two species.

Same 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.

Osmotic pressure as entropic effect

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dG = µ1dN1 +µ2dN2 = µ1 −µ2( )dN1 ≤ 0

Δµ = Δµ0 + kBT lnc1c2


Consider a cell in water solution:

•  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…


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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).


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

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

µH2O=∂Gtot

∂NH2O



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µH2O= ∂Gtot ∂NH2O

At equilibrium chemical potentials must coincide, and this will generate difference in pressure in two compartments:

Gtot T, p,NH2O,Ns( ) = NH2O

µH2O0 T, p( )+ Nsεs T, p( )+ kBT Ns ln

Ns

NH2O

− Ns

"

#$$

%

&''

µH2OT, p( ) = µH2O

0 T, p( )− kBTNs

NH2O

µH2O

0 T, p1( ) = µH2O0 T, p2( )− kBT

Ns

NH2O



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If Δp << p1 we can expand chemical potential on right-hand side of equation:

Since molecular volume is given by derivative of µ with respect to p:

Equilibrium condition becomes:

µH2O0 T, p2( ) ≈ µH2O

0 T, p1( )+∂µH2O

0

∂pp1

p2 − p1( )

∂µH2O0

∂pp1

= vH2Omol =V1 NH2O≈V2 NH2O

≡V NH2O

p2 − p1( ) = kBTNs

V→Δp = kBTcs

van’t Hoff formula

gives osmotic pressure as function of concentration of solute


Osmotic pressure in E.coli

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

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

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

Since kBT ~ 4.1 pN⋅m, Δp amounts to:

0.1M = 0.1NA 1L ≈ 61022 1015 fL→ 6107molecules

c ≈ 6107 109nm3 = 0.06molecules nm3

Δp ≈ 0.064.1molecules nm3 ≈ 0.25 pN nm2 ≈ 2.5atm


Law of Mass Action

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

reactants and products.

•  Can be derived by application of 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.


Law of Mass Action

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Consider simple reaction with two reagents and one product, in which total number of molecules is fixed:

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 = 0↓

∂G ∂NA( )B,AB dNA + ∂G ∂NB( )A,AB dNB + ∂G ∂NAB( )A,B dNAB = 0

↓

µAdNA +µBdNB +µABdNAB = 0


Law of Mass Action

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

•  Equilibrium condition becomes:

•  For dilute solutions this condition can be written:

dG = 0→ µidNii=1

N

∑ = µiν idN = 0i=1

N

∑ → µiν ii=1

N

∑ = 0

dNi =ν idN

µiν ii=1

N

∑ = µi0 + kBT lncici0

"

#$

%

&'ν i

i=1

N

∑ = 0→ µi0ν ii=1

N

∑ = −kBT ln cici0

"

#$

%

&'

νi

i=1

N

∑

↓

−1kBT

µi0ν ii=1

N

∑ = ln cici0

"

#$

%

&'

νi

i=1

N

∏


Law of Mass Action

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•  Exponentiating formula and putting all constants on right-hand side gives:

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

•  Inverse of dissociation constant Kd:

ciνi

i=1

N

∏ = ci0νi

i=1

N

∏ e−1kBT

µi0νii=1

N

∑

Keq T( ) ≡ ci0νi

i=1

N

∏ e−1kBT

µi0νii=1

N

∑

Law of Mass Action

Explains and predicts dynamic equilibrium by relating concentrations of reactants and products at a given

temperature and pressure.

Kd =1Keq


Law of Mass Action

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•  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:

ciνi

i=1

N

∏ = cA−1cB

−1cAB1 =

cABcAcB

= Keq T( ) = 1Kd


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

•  Reconcile views by expressing pbound as function of Kd:

•  Kd cast in terms of microscopic parameters of lattice model as:

Ligand-receptor binding revisited

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Kd =L[ ] R[ ]LR[ ]

pbound =c c0( )e−βΔε

1+ c c0( )e−βΔε

Chemical " dissociation constant Kd: Statistical " binding energy Δε:

pbound =LR[ ]

R[ ]+ LR[ ]=

L[ ] Kd

1+ L[ ] Kd

Former result: dissociation constant Kd corresponds to concentration of

ligands for which pbound = 1/2

Kd =1Vbox

eβΔεLink between languages of chemistry and of 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.

Ligand-receptor binding revisited

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•  Energy released upon reaction ATP ADP + Pi depends on concentration of reactants and products.

•  Consider change ΔG of single reaction upon hydrolysis:

•  For dilute solutions:

•  Considering change with respect to reference concentration (ΔGref):

Thermodynamics of ATP hydrolysis

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dG = µidNii=1

N

∑ = µiν idNi=1

N

∑ dN=1" →"" ΔG = µiν ii=1

N

∑

ΔG = µi0 + kBT lncici0

"

#$

%

&'ν i

i=1

N

∑

ΔG = ΔGref + kBT ln cνiii=1

N

∏ cνii,refi=1

N

∏#

$%

&

'(


Choosing equilibrium state as reference implies:

Thus previous equation referred to equilibrium state becomes:

Since standard state free energy (at c = 1M) can be expressed as ΔG0 = -kBT ln Keq, by adding and subtracting ΔGref0 one obtains (expressing Keq in molar units!):


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cνiii=1

N

∏ = KeqΔGref = 0

ΔG = kBT ln cνiii=1

N

∏ Keq

#

$%

&

'(

ΔG = ΔG0 + kBT lnci1M"

#$

%

&'νi

i=1

N

∏


Considering hydrolysis of ATP and expressing concentrations in molar units:

Considering typical experimental values:

•  ΔG0 -12.5 kBT

•  All concentrations near millimolar range:

–  [ATP] ~ 5⋅10-3

–  [ADP] = 0.5⋅10-3

–  [Pi] = 10⋅10-3

•  Gives result introduced earlier:


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ΔG = ΔG0 + kBT ln ADP[ ] Pi[ ] ATP[ ]

ΔGhATP ≈ −12.5kBT + ln

5 ⋅10−410−2

5 ⋅10−3≈ −12.5kBT − 6.9kBT ≈ −20kBT


•  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.

Cooperativity and Hill function

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•  This behavior is seen in many biological processes and arises from cooperativity.

•  Can be understood considering a simple process with two ligands.

•  For sake of simplicity we consider ideal cooperative behavior, that means no intermediate with one ligand only bound to receptor:

L + L + R L2R


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Kd2 =

L[ ]2 R[ ]L2R[ ]

pbound =L2R[ ]

R[ ]+ L2R[ ]

pbound =L[ ] Kd( )

2

1+ L[ ] Kd( )2

Hill function with coefficient n=2


More generally:


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pbound =L[ ] Kd( )

n

1+ L[ ] Kd( )n

Hill function with Hill coefficient n

•  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.


References •  Books and other sources

•  Physical Biology of the Cell (2nd ed.), Phillips et al., Chap. 6

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