Physics of Macromolecules-STATISTICAL MECHANICS

8/8/2019 Physics of Macromolecules-STATISTICAL MECHANICS

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Physics of Macromolecules

Navin Singh

November 2, 2010

1 STATISTICAL MECHANICS

1.1 A review

The foundation of statistical mechanics lies in the thermodynamics of ideal gases. This is the reason whythis subject in the beginning called it as Statistical Thermodynamics . Later on people realized thepower of this subject and its application in solving the equation of states and hence termed it as Statistical

Mechanics . With its diversity in the other elds now people often termed this subject as StatisticalPhysics .

Starting with thermodynamics of ideal gas we will discuss more issues related with this subject.In thermodynamics, we study the properties of the system at macroscopic level. For e.g. the pressure of

the gas is nothing but the collision of molecules with the surface. Temperature is nothing but the averagekinetic energy of all the molecules.

In statistical mechanics, we study properties of system at microscopic level. At microscopic level, onehas to deal with particles or system of particles. Problem with these kind of systems is their largeness. Onecan not solve the dynamics of these particles with the help of Newtonian classical mechanics or quantummechanical equation. In those subjects, we treat system as a single particle and then solve its equation of state.

In one mole of an ideal gas, there are 10 23 number of particles. To solve the equation of state, one hasto solve 1023 number of equations !!. This is nearly impossible job even with most powerful computer in theworld. Here comes the power of statistical mechanics.

This subject is based on the theory of probability. According to this theory, we do not predict the exactstate of any occurrence, but ask the probability of occurring that event.

Let me take an example. An ideal gas having N number of particles occupying a space of V . If the available energy to the system is E , the macroscopic behaviour of the system will depend on how atmicroscopic level, this energy is distributed among the N number of particles. The distribution, as always,

have to obey certain rules. What are those rules?i= N

i =1

n i = N i= N

i =1

n i i = E (1)

which tells us that there are total N number of particles and i th particle is having energy i . Any distributionwhich obey the above equation is possible. Others are discarded. Thus, there will be a large number of waysin which this distribution is possible. Each of these ways is called as microstate of the system. The statecorresponding to any set of N , V and E is known as macrostate of the system. The number of ways isdenoted by a symbol which is a function of N , V and E .

Interestingly, at this point, famous scientist, Boltzmann discovered with the help of many mathematicalformulations that the entropy of the system is somehow related with this microscopic quantity ( N,V,E ).

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He stated,S = kB ln(N,V,E ) (2)

His argument was based on many probabilistic calculations which is beyond the scope of this discussion. Letus see, whether this equation works or not? For a system in which there is a unique way to distribute theabove mentioned energy, the entropy of the system will be, S = kB ln = 0, as = 1. Any other form of relation between entropy and is not possible. As disorderness is more, randomness and hence the numberof ways in which the energy can be distributed is more.

What a simple and beautiful expression this is? This was a major breakthrough in the development of this subject. This not only connect the thermodynamics with probability but also the macroscopic worldto microscopic world. This leads to formulation of ensemble theory which made this subject more vital anddiverse.

1.2 Phase Space

Let us introduce the concept of phase space which is required to calculate the quantities explaining thebehaviour of the system. Since in statistical mechanics we dealt with energy and its distribution, we need todene a space which is more convenient to locate a microstate of the system. A space consists of momentaand coordinate of the particle (they are also known as canonical conjugate quantities) is known as Phasespace . Just like to locate a particle in a Euclidean space you need space coordinates, to locate a particlein phase space you need the information of its momenta and coordinates. A denite point in this phasespace exactly corresponds to one microscopic state of motion of the whole system. The temporal ( i.e. time)evolution of the system corresponds to one curve ( q, p) in the phase space which is called the phase spacetrajectory. This can be determined by Hamiltons equation of motion as,

qi =H p i

pi = H qi

(3)

Thus the energy of the system can be expressed in terms of Hamiltonian.

An example: Harmonic OscillatorThe Hamiltonian of a harmonic oscillator in one dimension is,

H (q, p) =p2

2m+

12

kq2 (4)

where m is the mass of the particle and k the oscillator constant (spring constant). The phase space of thissystem is nothing but an ellipse as, H (q, p) = E = constant , whose half-axes are a = 2mE and b = 2e/k .Mathematically the phase space is denoted as, and an element in phase space is denoted as d =d3N pd3N q.

1.3 Ensemble Theory

The aim of any theoretical attempts or equation is to validate the experimental observations. Experimentallywhatever is being observed it is a function of time. How one can emulates those experimental things usinga theoretical tool. In ensemble theory, we generate M number of systems which are mental copies of eachother and corresponds to each of the macrostate stated by ( N,V,E ). We dene an expectation value of anyobservable as,

f = the ensemble average of f the time average of (ensemble average of ) f (5)Since, the time and ensemble averages are completely independent things, one can restate the above equationas,

f = the ensemble average of (time average of ) f (6)

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For a long period of time, the time average means we passed through all the members of ensemble. Thus,taking ensemble average has no meaning. Hence,

f = long time average of f f exp (7)

1.3.1 Micro Canonical Ensemble

If a state of the system is dened by x number of particles, volume and energy, the proposed ensemble is

called as Micro Canonical Ensemble . One can calculate the energy surface using equation discussed aboveH (q, p) = E to nd out the number of ways one can distribute the energy E among the particles N . Thenone can calculate the thermodynamical quantities from the relation S = kB ln.

1.3.2 Canonical Ensemble

It is difficult to control or measure the energy E and hence to satisfy the relation H (q, p) = E some whatmore tedious. In stead of energy we x the temperature which is a controllable quantity and dene themacrostate of the system by N,V,T . This is the canonical ensemble . In this ensemble, energy can take anyvalue from 0 to and we ask the probability of nding a particle in an energy state E r . This probability,P r is found to be proportional to exp( E r ), where = 1 /k B T . The normalized probability is,

P r =exp(E r )r exp(E r )

(8)

The denominator in the above equation is called as partition function which means sum over all possibleenergy states and denoted by symbol, Z . The thermodynamical relation of this quantity is,

A(N,V,T ) = kB T ln Z (N,V,T ) (9)

1.3.3 Grand Canonical Ensemble

If we do not impose the restriction over the number of particles, i.e. it can take any value between 0 to ,the proposed ensemble is known as grand canonical ensemble . The relation between canonical partitionfunction and grand canonical partition function is

(z ,V,T ) = zN Z (N,V,T ) (10)

where z is known as the weight factor or fugacity corresponding to exchange of particles and z = exp( )where is the chemical potential.

1.4 Some applications

With these preliminaries knowledge of the subject one can explain various interesting phenomena around us.

Bose Einstein condensation

The specic heat of the solids

The paramagnetism

Black body radiation

The conformation of polymers

The helix coil transition in proteins and DNA

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1.5 Problems

1. Consider a system A having N number of particles, occupying V space and having energy E . Calculatethe number of microstates available to the system and hence the entropy of the system.

Sol: Let us consider M identical systems in contact with each other. Let the number of systems inthe quantum state i be n i out of M . There are n1 systems in 1 state, n2 systems in 2 state and soon. The number of ways in which one can arrange n1 systems in 1 states etc. can be written as,

= M !n1!n2!n3!n4!n5!..

= M !i = M i =1 n i !

(11)

The entropy of the system is,

S M = kB ln = kB ln M !i

ln n i ! kB M ln M i

n i ln n i

= kBi

n i M i

ln n i = kBi

n i lnn iM

=

kB M

i

n i

M ln

n i

M =

kB M

i

pi ln( pi ) (12)

Entropy per particle can be written as,

S =S M M

= kBi

pi ln( pi ) (13)

If there are quantum states available to the system, the probability to be in one of these states is pi = 1 . Thus, the entropy per particle will be,

S = kBi

1

ln1

= kB ln (14)

2. For an arbitrary system, the number of microstates is given as, = a exp( bEV ), where a and b aresome constants. By working out the steps, nd out the relation between the energy and the temperature

of the system. What will be the value of the energy for which the temperature of the system will be

zero?

3. A system at 300 K temperature is releasing a heat of 106 J and is adding the same to a system at 299 K. How much is the total entropy changed during this conversion? By what factor does the number of

accessible states increase?

4. Find out the temperature of a system whose microstates are given as = A exp[ (EV )1/ 2], with as

some constant. For which value of energy, the temperature of the system wil l be zero?

5. The entropy of black body radiation is given by the formula, S = 43 V 1/ 4E 3/ 4 where is some constant.

Determine the temperature of the radiation and show that P V = 13 E .

6. Three energy levels having energies, E 1 = 0 J, E 2 = 1 .4 1023 J, E 3 = 2 .8 1023 J.

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bit simple as l i l j = 0. Why? Since each bond vector is free to rotate in the free space with the sameangle, the average of their scalar product will be equal to average of cos( ) which is zero.

Thus,r 2 = nl 2 (18)

The ratio r2

nl 2 is known as characterstic ratio which measures the stiffness of the chain.

2.3 Radius of GyrationThe radius of gyration is dened as the root mean square distance of an array of atoms or group of atomsor groups from their common center of gravity. For a chain of n + 1 identical atoms joined by n bonds, if

i j

rri j

rij

Centre of mass

Figure 2: The radius of gyration.

RGi is the distance of skeltal element i from the centre of gravity, then,

R2G =1

n + 1

n

i=0

R2Gi (19)

Let us consider a polymer composed of ( n + 1) monomers. The i th monomer will be having coordinate(x i , yi , zi ) and has mass m i . The centre of gravity is a point where

N i=1 m i r i = 0, where, r i is a vector from

the centre of mass to the unit i. For idential monomers, ir

i = 0. Now,

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l

lcos

lsin

ld

x

y

z

Figure 3: The bond vector and its projection along x-axis.

angle between the bond vector and the z-axis, the mean square projection length l2x will be l2 cos2() . Thusto nd out the average projection we have to nd out the canonical average of cos 2(). Since distributionof cos() is equally probable in all direction, the number of vectors lying in the region and + d willbe determined by the area of the annular ring which is obtained by rotating the bond vector about thex-axis with the xed angle between the vector & x-axis. This area is 2 l 2 sin()d. Thus, the mean squareprojection length will be,

l2x =

0 l2 cos2 (2l 2)sin d

0 2l2 sin d

=l2

3(22)

Thus, the average root-mean square projection along a given axis is l/ (3) Now the average steps l/ 3along the chosen axis occur with random statistics in negative and positive directions. If there are N bonds,i.e. N steps, the distance traversed along x-axis will be,

x = (N 1 N 2)l3 (23)where, N 1 is the step in positive direction, while N 2 is the step in -ve direction. Also N = N 1 + N 2 . Theprobability W (N 1 , N 2 , N ) of N 1 & N 2 steps out of total steps N will be,

W (N 1 , N 2 , N ) =1

2N N !

N 1!N 2!(24)

where, 12N is the probability of generating a particular sequence of positive and negative steps whileN !

N 1 !N 2 !

represents the number of particular sequence of steps.Using Stirlings approximation which states that,

ln N ! N ln N N +12 ln(2) +

12 ln N (25)

and under the condition that N is large and dening M = N 1 + N 2 , we can solve the equation for probability(the detailed derivation is left for you to work out). Thus,

ln W (N 1 , N 2 , N ) =12

ln2

n (1 M 2 /N 2) N 2

ln 1 M 2

N 2 M 2

ln1 + M/N 1 M/N

(26)

For large x, we have,

ln1 + x1 x

2x & ln(1 x) x (27)Using this, one can argue for large N (N >> M ) the eq.(26) can be re-written as,

ln W = ln 2N + N 2 M 2

N 2 M 2 2M N = ln 2N M 2

2N (28)

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Hence,

W = 2N exp M 22N (29)Gaussian distribution.

Let us calculate the probability in terms of distance travelled. Dening the distance variable as

x =ml3M

2 =3x2

l2(30)

In terms of distance variable one can ask the probability per unit interval of distance that x takes betweenx and x + dx. For xed N , the value of M changes in units of two as, for M = N 1 + N 2 . This means that xchanges by 2l3 , the relation W (x, N ) =

W (M,N )interval requires an interval equivalent to two units of M , i.e.

2l3 .

W (x, N ) =32l

W (M, N ) = 32N 1l exp 3x22Nl 2 (31)With similar aruguement we can write for W (y, N ) & W (z, N ). Thus,

W (x,y,z )dxdydz =B

3

exp(B 2R2)dxdydz (32)

where, B = 32Nl 21/ 2 & R2 = x2 + y2 + z2

2.5 Statistical Segment

We have dened the characterstic ratio as,

C N =r 2

Nl 2(33)

for FJC C N = 1. For real chain, for large N , C N approaches as C .For random walk or FJC model, l i .l j = 0 for all i = j . In case of real chains, l i .l j 0 for sufficiently

large

| j

i

|which means that bonds separated by sufficiently large distance are randomly oriented.

This gives the concept of statisical segment for real chains. This means that large polymeric chains canbe approximated as FJC where bond lengths are replaced by statisical segments. In the limiting case,

C =r 2

Nl 2 r2 = C Nl

2 (34)

in terms of statistical segment,r 2 = N e l2e (35)

where, N e is the number of statistical segments while le is the length of that segment. Also, the maximallength of the chain will be, r m = N e le = fNl , where, f is some number between 0 & 1 (equal to cos( / 2)for even number N ). Solving all these equations will give us,

N e =f 2

C N & le =

C f

l (36)

This means that N e 1

C and le C . Remember, the C signies the stiffness in the chain.An example

For polymethylene, f = 0 .83, C = 6 .7. This implies that N e 0.1N N N e = 10. This means thatstatistical segment spans approximately 10 bonds.

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r

l1

a

Figure 4: The average projection of r on the rst bond vector.

2.6 Persistence Length

The average projection of end-to-end distance vector r on the rst bond of the chain l 1 .

a =l 1

l1.

N

i=1

l i (37)

It is a measure of the length over which the chain persists in the same direction, as the rst bond. In general,

a =l i

li.

N

j =1

l j (38)

For chain composed of identical bond length, l, the persistence length is closely related to the limitingcharacterstic ratio C .

r 2 = Nl 2 + 2N

j = i+1

N

i=1( l i .l j ) = Nl 2 + 2

N

j =2(l 1 .l j ) + 2

N

j = i+1

N

i=2(l i .l j ) (39)

As per denition of persistence length, as dene in eq. (37), the second term in eq. (39) can be written as,

N

j =2( l 1 .l j ) = al l 1 .l 1 = al l2 as N (40)

Similarly we can partitioned the third term in eq. (39) for i = 3 and so-on. This simply means that thethird term in eq. (39) can be replaced by 2( n 1)(al l2). Thus,

r 2 = Nl 22al 1 = C

a =l(C + 1)

2(41)

This ratio gives us a physical feeling of limiting characterstic ratio.

2.7 Porod Kratky (Worm Like Chain) model

From various studies it has been found that long DNA chain is not like a rod but behaves like worm. It isrod like at shorter length but as the length of the chain increases its properties are more closer to a worm.

To study the elastic response of dsDNA chains, a simple model was proposed by Porod and Kratkywhich is also known as Worm like chain (WLC) model. This model assumes the chain as a continuouslycurving chain. The direction of curvature at any point is assumed to be random. In this model, there arex hypothetical bonds of length lb are assumed. Each bond makes an angle with its neighbour and free torotate in the space. The average projection, , of the end-to-end distance vector r on the rst bond is,

= lbx1

k=0(cos )k (42)

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b

Figure 5: The worm like chain model.

For large x, a and the summation conveges to [1 cos ]1 . Thus,a =

lb1 cos

(43)

If we reduce the lb in such a way that a and the contour length Lc are held constant at their pre-determinedvalues. In that case, lb 0 and 0 which means that 1 cos 0 and x . Hence,

lb1

cos

=lb

ln[cos ]

= a (44)

This will give, cos = exp( lb/a ). Here we have used an assumption that ln x x 1 for x 1. In thislimiting case, the summation (discrete) can be replaced by integration (continuous) as,

= lbx1

k=0

exp(klb/a ) = lb x

0exp(klb/a )dk =

L c

0exp(L/a )dL = a[1exp(Lc /a )] (45)

where, the summation is converted into integration for small lb and x x 1 for x >> 1; L = lbk.Let us try to establsih a relation between end-to-end distance vector and the persistence length. From

gure one can infer that r .dr = dLc . Taking differential of both sides,

dc = 2 r .dr = 2 dL c = 2 a[1

exp(

Lc /a )]dLc (46)

Integrating both sides for the complete chain, as,

r 2 = 2 a L c

0[1exp(Lc /a )]dLc = 2 aL c 1

aLc {1 exp(Lc /a )} (47)

In the limit of large Lc ,r 2

Lc= 2 a (48)

while for Lc /a < 1, we have

r 2 = L2c 1 13

Lca

+112

Lca

2

+ .... (49)

One can draw the following conclusions on looking on these two limiting case.

One can argue that for Lc


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Figure 6: The schematic of bonding for the formation of -helix. The NCC model.

3 Helix Coil transition

It is proved through X-ray crystallographic, infrared spectra, optical rotational viscosity etc studies thatpolypeptide can form -helical in the crystalline as well as in solution. During helix formation thehydrogen bonding between oxygen of carboxyl group at i + 4 & hydrogen of amide group at i if formed. Therotation angle and [between C-C and C-N] remains xed at +50 (left handed) and -50 (right handed).

In solution at some values of physical parameters, the polypeptides are in helical form but with the

change in these physical parameters (temperature, solvent composition, pH etc.) this form disrupts. Poly-i-glumate have ionizable carboxyl group and is a function of pH of the solution. Thus conformation of thispolymer varies with variation in pH of the solution.

These plots show that as the value of pH is increased there is a sharp change in the optical rotation,intrinsic viscosity and continuous change in ionization. All these changes reveals that the conformation of helix form of poly-Lglumate changes to some other form, may be random coil. From ooptical rotation andintrinsic viscosity curve, signature of random coil appears as thier values at higher pH is smaller. Secondiportant information is that they are showing the transition to be co-operative, i.e. all or none kind. Thetransition from helix to coil can be made by changing the temperature of the solution. The formation of helix depends on the chain length.

3.1 Molecular Mechanism of the transition

From experimental obsevation it is clear that transition from coil to helix and vice versa is a sharp transition.There is a co-operativity in the transition. What is the reason behind this co-operativity?

Let us try to understand the mechanism involve at molecular level.Hydrogen bonding is between C = O H N . To make a hydrogen bond (the rst one), six dihedral

angles have to move. When chain is in random coil conguration all set of , conguration are possible.To make this bond the change in entropy is large. However, making the second and successive bonds changein entropy is low. Also to minimise the end effects the length of helix should be large.

3.2 Thermodynamics of the transition

The idea discussed here was proposed by John Schellman in 1955. He made several assumptions to explainthe thermodynamical behaviour of this transition. The assumptions are:

Each chain is assumed to be either all helix or all coil, no intermediate structure.

The free energy change for converting coil to helix is Gr = H r T S r (51)

where, H r is the change in enthaply in forming a hydrogen bond while S r is the loss in entropy forthe same.

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Figure 7: Six dihedral angles have to rotate for rst bond, while only two bonds are required to rotate.

To calculate the net change in the Gibbs free energy of the system one can argue that it should be GT =N Gr , which is not correct for polypeptide chain. In this case we have to look at the microscopic levelwhich will give us, that net change in the free energy will be,

At the ends four bonds (4 of carboxyl and 4 of amide) are still free which will reduce the net enthaplyas (n 4) H r .

The entropy contribution from the two ends will increase the net entropy as ( N 2) S rThus, the net change in Gibbs free energy will be

GT = ( N 4) H r (N 2) S r (52)The equilibrium constant K for the reaction is,

K = cbN =[Helix ][Coil ]

GT = RT ln K = RT ln c NRT ln b

ln c N ln b = GT RT

= ( N 4) H rRT (N 2)

T S rRT

= N H rRT

S rR 4

H rRT 2

S rR

(53)

Equating RHS & LHS will give,

b = exp H rRT

+ S r

R(54)

c = exp 4 H rRT 2

S rR

(55)

Point to note here is that as predicted by eq (53) for large N , small change in b can cause sharp change inK . The parameter c reects the decit of hydrogen bonds at the ends while b for bonding.

During the folding of polypeptides, is there a critical number below which the polypeptide can not fold?Let us try to nd out the answer. If we believe that there is a temperature T = T m at which the unfoldedstate changes to folded state (or vice-versa) at which the Gibbs free energies of helix & coil states are equal.At this temperature GT = 0 and K = 1. Putting these in eq.(52) will give

GT

= 0 = ( N

4) H

r (N

2)T

m S

r

T m

=(N

4)

(N 2) H r S r

(56)

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This is reason why the minimum number of residues should be 5, otherwise T m will be negative which isunphysical.

At N = 5T m =

13

H r S r

for N T m H r S r

(57)

i.e. variation of T m is independent of N . Check out the value of N for which II part of eq. (57) holds.Let us observe the sharpness of the transition. If we plot (fraction of helicity) as a function of temper-

ature, the slope at T m will give the information about the sharpness of the transition. Since, equilibriumconstant is

K =[Helix ][Coil ]

=

1 ln K = ln ln(1 ) = (N 4)

H rRT

(58)

Differentiating both sides with respect to temperature, we have

1

ddT T = T m

+1

1 ddT T = T m

= ( N 4) H rRT 2

=1

(1 )ddT

(59)

At T = T m , = 12 . Substituting these along with the value of T m from eq.(56), we have,

d

dT

=1

4

(N 2)2(N 4)

S 2r

H r(60)

which means slope of the vs temperature varies with N . However this model oversimplify the actualphenomenon. The assumption of all or none is litte bit supercial as it is found that there are intermediatestates during helix to coil (or vice versa) transition.

3.3 Partition function for the Simple Chains

The partition function is useful tool to estimate the energy of helix-coil transition. In this method we calculateall possible states in which system has nite probability to exist. A partition function Z for molecule can bedened as,

Z =i

gi exp( i ) where =1

kB T (61)

Here, gi is the statisical weight (number of states that have energy i ), i is the energy of i th state. gi is alsoknown as degeneracy of the i th state.

Let us consider a hypothetical chain consisting of four elements A in which there are two states A+ andA are possible. The possible conformation if the chain will have, A+ AAA+ , A+ A+ AA, and so-on.Taking care of directionality in the chain A+ A+ AA and AAA+ A+ will be treated as two states.

The possible conformation

All AAAA states will contribute as Z 0 .

All A+ AAA states will contribute as Z 1 . There are 4!3!1! = 4 number of ways to realise it.

All A+ A+ AA states will contribute as Z 2 . There are 4!2!2! = 6 number of ways to realise it.

All A+ A+ A+ A states will contribute as Z 3 . There are 4!3!1! = 4 number of ways to realise it.

All A+ A+ A+ A+ state will contribute as Z 4 . There is a unique way to realise this state.Thus, the partition function will be,

Z = Z 0 + 4 Z 1 + 6 Z 2 + 4 Z 3 + Z 4 = Z 0[1 + 4K 1 + 6 K 2 + 4 K 3 + K 4] (62)

where, K i =Z iZ 0 is known as equilibrium constant.

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For most of the calculations we do not require the value of Z 0 . This will be more clear with an example.Let us nd out the fraction of molecules with a particular number of A+ elements. If 2+ is the fraction of two A+ molecules,

2+ =6Z 2Z

=6K 2Z 0

Z 0[1 + 4K 1 + 6 K 2 + 4 K 3 + K 4](63)

=6K 2

1 + 4 K 1 + 6 K 2 + 4 K 3 + K 4(64)

One can assume that Z 0 = 1 as reference state and the change in free energy can be calculated for convertingA to A+ . Thus, Gi = RT ln K i . To convert an A state to A+ , we need,

K 1 = exp GRT

= s and K 1 = exp 2 GRT

= s2 and so on (65)where, s is the statisical weight. Thus,

Z = 1 + 4 s + 6 s2 + 4 s3 + s4 =4

i=0i s i = (1 + s)4 (66)

where, i is the number of ways of putting iA+ elements in a chain of four unit. Using bionomial expansionmethod, one can argue that it is nothing but i = 4!(4i )! . Thus, the partition function is now expressed as asimple sum of terms involving powers of a single statistical weight parameters s. which is assigned to everyA+ element, with a factor of unity assigned to each unit in the A+ state.

Rules for constructing the partition function

In general the partition function for these kinds of systems are generated with the help of following steps:

Consider all possible conformations of the chain.

Establish a reference state for a chain element, and assign this state a statistical weight of unity.

For each conformation of the chain, assign an appropriate statistical weight for each element that isnot in the reference state.

Construct the partition function by summing over all conformations, with appropriate products of statistical weighting factors assigned to each conformation.

3.4 The Zipper model

In this model, each residue has two possible states, helix (h) or coil (c). Their existence in a given confor-mation however has to follow some restrictions. Initiation of helix state can occur at point, but all the helixstate should be in one stretch. That is, hhhcccc , ccchhhhhccc like states are allowed but hhcchhhhccc kind

of states are not allowed.In an -helix, out of m residues, only m 2 residues adopt a unique value of and . There will bem 4 number of hydrogen bonds are possible in the complete chain. In this model, we consider n unitswhich can exist in either helical or coil state. One can construct the partition function for this chain by themethod discussed below.

Let us put all coil conguration as the reference state and assign it a statisical weight of unity. If s is thestatisical weight or equilibrium constant in converting one coil state into a helix state at the end of helicalsegment, then,

s = cccchhhhh h cccc cccchhhhh c cccc

(67)

This is the propagation step. The inititation of a helical section with an all coil chain is much more difficultthan the propagation. Growth or propagation is relatively easier that nucleation or inititation.

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Assume that s be the equilibrium constant (

n=

1n

n

k=1

kp(k) =1

nZ

n

k=1

k(n k + 1) s k =s

nZ Z s

=1n

ln Z ln s

(75)

=s

n(s 1)3ns n +2 (n + 2) sn +1 + ( n + 2) s n1 + s(s1) 2 (s

n +1 (n + 1) s + n)(76)

Thus, for s = 0 = 0 while for s = = 1. For long chains, s 1, 0.5. s can be changed bytemperature or pH of the solution.

This model is good for short chain as the basic assumption of this model upto some limit, is valid. Forlonger chain, this model fails to explain the qualitative behaviour of the transition.

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Physics of Macromolecules-STATISTICAL MECHANICS

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