Notes BB101 Ranjith

7/23/2019 Notes BB101 Ranjith

http://slidepdf.com/reader/full/notes-bb101-ranjith 1/31

Contents

1 Introduction 4

2 The puzzle of DNA packaging 52.1 Worm-like chain model . . . . . . . . . . . . . . . . . . . . . . 5

2.1.1 Problems for you to solve: . . . . . . . . . . . . . . . . 62.2 Thermal energy vs Elastic energy . . . . . . . . . . . . . . . . 6

3 How proteins fold into a nearly unique structure in 3D? 7

4 Statistical mechanics (or statistical thermodynamics) in bi-ology: Prediction of number of proteins bound to DNA 84.1 If we know thermodynamics, we can predict many things: . . 84.2 Predicting number of proteins bound onto DNA . . . . . . . . 10

4.3 Entropy of the DNA-protein system . . . . . . . . . . . . . . 104.4 Free energy DNA-protein system . . . . . . . . . . . . . . . . 12

5 Diffusion, Einstein relation and Nernst equation 155.1 Diffusion: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.2 Einstein relation . . . . . . . . . . . . . . . . . . . . . . . . . 165.3 Diffusion of charged particles across a membrane channel can

create a potential difference: Nernst equation . . . . . . . . . 18

6 Life in salty water and at low Reynold’s number 206.1 Inertia is negligible for objects in the cellular world . . . . . . 20

6.2 Electrostatic interaction gets “screened” due to the presence of oppositely charged ions . . . . . . . . . . . . . . . . . . . . . . 21

7 Force generation and movement in biology 247.1 A simple polymer . . . . . . . . . . . . . . . . . . . . . . . . . 24

1



7.2 Critical concentration . . . . . . . . . . . . . . . . . . . . . . . 26

7.3 Force generation by a simple polymer . . . . . . . . . . . . . . 277.4 Actin: treadmilling . . . . . . . . . . . . . . . . . . . . . . . . 29

2



This is my quick attempt to prepare a soft copy of my lecture notes! I have

not proof read this, yet. So, read it carefully, as there could be many typo-graphical errors. Don’t trust any statement/equation until you understandthat the statement/equation is physically sensible!

3



Chapter 1

Introduction

Conventionally, biology is being thought (and taught?) as a descriptive subject. The aim of this part of the course is to demonstrate how one canapproach biological problems using physics and mathematics, in a quanti-tative way. The hope is that this will convince you that biology is equallychallenging and exciting as any other subject you have known, if not moreexciting and challenging!

One thing that puts off many of you, while learning biology, is the jargons– myriads of names that one has to remember. Here we will try to use minimalnumber of jargons and names. Instead we will discuss a number of excitingpuzzles in biology. I start with assuming that you know one thing: All living

organisms are made of cells(well, assuming viruses are not living organisms!)

4



Chapter 2

The puzzle of DNA packaging

The size of a typical human cell is roughly 100µm. A human being is madeof trillions of cells. Each and every cell in your body has a very long DNA– in fact every cell (with some exceptions) in your body contains DNA thatis two meters long ; yes, every cell contains two meters long DNA molecule.What is amazing is that this 2-meter long DNA is packaged into a tiny space– a region called nucleus which is only (1/10)th of a cell, roughly (dimensions≈ 10µm). This leads to an immediate puzzle – how is it possible to pack atwo meter long macromolecule

To understand how DNA is packaged, the first thing one needs is to realisethat, to pack in this small space, DNA need to bent and folded. So the first

quantitative thing one need to figure out is the energy needed to bend DNA.

2.1 Worm-like chain model

In a very simple model, one can imagine DNA as a long and thin filament (or,simply, a curve) in 3D space. Any curve can be represented by a parametercalled “arc length" (s) – this is the length along the contour of the filament.Let r(s) represent the position vector at any point s

r(s) = xi + yˆ j + z k

The local bending energy of the elastic filament, at any location s has tobe proportional to the local curvature. Given that curvature is

C = d2r(s)

ds2

5



the bending energy is E b ∝ | C |2. You need to have the square of the curvature

because, symmetry demands that whether you make positive curvature ornegative curvature, the energy to curve is the same. Then, the total bendingenergy of the filament, having any configuration specified by a set of vectorsr(s) is

E b = kb

2

L

0ds

d2r(s)

ds2

2

kb is the bending stiffness of the filament, and it reflects the material propertyof the DNA in a given environment. For DNA, under physiological conditionskb ≈ 200 × 10−30Jm.

2.1.1 Problems for you to solve:

• What is the magnitude of curvature | C | of a circle of radius R?

• What is the bending energy needed bend a DNA of length L to a perfectcircle of radius R = L/2π?

• For what length L, will the above-mentioned energy be comparable tothermal energy kBT ?

2.2 Thermal energy vs Elastic energy

(Please see the pdf file of the lecture uploaded)

6



Chapter 3

How proteins fold into a nearlyunique structure in 3D?

(Please see the pdf file of the lecture uploaded)

7



Chapter 4

Statistical mechanics (orstatistical thermodynamics) inbiology: Prediction of number of proteins bound to DNA

4.1 If we know thermodynamics, we can pre-

dict many things:In your thermodynamics course, you might have studied about Free energy,

entropy, enthalpy and so on. You might have also studied many differentformulas like F = E − T S , G = F + P V and so on. What is the use of studying all these ?

The use is that, if we know free energy of a system, we can predictmany properties of the system1. Yes! sitting at your home, you can makemany predictions. Consider the following example:

Imagine a DNA that has N protein binding sites. We now add large num-ber of proteins into the solution containing this DNA. How many proteins,

on an average, will be bound onto this DNA ? Will all the N binding sites beoccupied by proteins ? Can we predict the average number of proteins that

1Of course, conditions apply! Knowing the free energy, we can predict only the “equi-

librium” properties of the system

8



will be bound onto the DNA ? Yes, we can; if we know the free energy of the

DNA-protein system. How do we calculate the free energy ? Simple. If wesubtract the energy of thermal motion from the total energy of the system,we get the free energy. As the formula says, free energy F is given by

F = E − T S (4.1)

where E is the total interaction energy of the system, T is the temperature,and S is the entropy of the system. Product of temperature and entropy (T S )gives the energy of thermal motion. Remember that at any nonzero temper-ature, all molecules are constantly jiggling in random directions (Brownianmotion). Entropy (S) is a measure of this random, disorderly, jiggling.

How do we calculate E ? If our system consists of charged objects, the

only interaction is electrostatic interaction, and hence E is the electrostaticinteraction energy. If we are dealing with a mechanical system (say, bendinga rod) E is the bending energy. If we are pulling a spring with a force f , E isthe sum of the spring energy ( 1

2kx2) and the energy due to the force applied

(f × x). Soon we will see how to get E for the protein DNA system.Now, how do we calculate entropy, S ? Very easy. Just count the total

number of states that the system can have. For example, if the system is acoin, it can have only two states – head or tail (Let us denote the number of states by Ω. For coin, Ω = 2). If the system is a protein, count all possibleshapes (3D configurations) that the protein can have. And if Ω is the total

number of shapes (or “states"), entropy is defined asS = kB ln(Ω) (4.2)

Why do we say that entropy of a protein is the logarithm of the number of all possible shapes ? As we know, protein molecules are getting constantlyhit by the water molecules doing the Brownian motion. As a result of thishit, protein monomers will start moving in random directions, and this canlead to all possible shapes of a protein. And entropy is the measure of this.

Let us now get back to the example of proteins binding to DNA and tryto the prediction. As stated earlier, If you are given a DNA with N proteinbinding sites, and if we add large number of proteins into the solution con-

taining this DNA, how many proteins, on an average, will be bound ontothis DNA ? In other words, what will be the average protein density on theDNA, when the system reaches equilibrium ?.

9



4.2 Predicting number of proteins bound onto

DNA

DNA is negatively charged. When positively charged proteins bind ontoDNA, the electrostatic interaction energy is at its minimum (You could con-vince yourself by taking an example: consider two oppositely charged parti-cles; write down the Coulomb energy of these two particles. Convince yourself that the energy is minimum when they are the closest.)

The only interaction between DNA and protein is the binding interaction.Let - kBT be the binding energy of each protein. When all the N bindingsites are occupied, the energy is at its minimum (E= -N kBT).

Now, how do we calculate the average number of sites that will be oc-cupied by proteins ? For that, it is not enough to know the energy; but wehave to know the free energy, F = E − T S ; here S is the entropy of thesytem.

4.3 Entropy of the DNA-protein system

: What is the entropy, when m binding sites, out of those N sites, are occu-pied by proteins ?

As we saw earlier, if we can count the total number of ways one can arrangeproteins on DNA, we can calculate entropy. Look at figure 4.1 to see anexample when there are two proteins (m=2), and three binding sites (N=3).

Now let us ask: how many different ways we can rearrange m proteins onthe N sites. The answers is:

Ω = N !

m!(N − m)! (4.3)

Entropy is given by,

S = kB ln Ω = kB ln N !

m!(N − m)! (4.4)

If we use the approximation that ln N ! ≈ N ln N − N (this is called theStirling approximation), we can simplify this as

S = −kBN [ρ ln ρ + (1 − ρ)ln(1 − ρ)] (4.5)

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Figure 4.1: DNA with 3 binding sites; this figure shows the three differentways one can arrange 2 proteins on the 3 sites.

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where ρ = m/N is the density of bound proteins.

4.4 Free energy DNA-protein system

: Now that we know entropy, let us ask, what is the free energy when mbinding sites are occupied by proteins ?

When there are m proteins bound, the energy is given by

E = −m kBT = −Nρ kBT (4.6)

The free energy is given by

F = E − T S (4.7)

= −Nρ kBT + kBT N [ρ ln ρ + (1 − ρ)ln(1 − ρ)] (4.8)

Free energy per binding site, in units of kBT , is given by

F

NkBT = −ρ + ρ ln ρ + (1 − ρ)ln(1 − ρ) (4.9)

where −ρ is the contribution from the energy, and ρ ln ρ + (1 − ρ)ln(1 − ρ)is the cotribution from the entropy. In figure 4.2 we have plotted these

contributions separately. Energy (red curve) is minum when ρ = 1. But freeenergy (blue curve) has a minimum for a different value of ρ, which is lessthan 1.

Let us calculate the minimum of the free energy; the free energy is min-mum when

∂F

∂ρ = 0. (4.10)

⇒ = ln ρ

1 − ρ (4.11)

⇒ ρ = e

1 + e (4.12)

This is the average equilibrium density. In other words, given a value of ,we can predict a value of the average density.

For = 2, the equilibrium density is 0.88; see Fig. 4.3. The energy is thelowest when ρ = 1. The entropy is maximum when ρ = 0.. The interaction

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-2.5

-2

-1.5

-1

-0.5

0

0.5

1

0 0.2 0.4 0.6 0.8 1

S ,

E ,

F ( i n u n i t s o f N

k B T )

Protein density ρ

S

E

F=E-TS

Figure 4.2: The green curve is S N kBT

= −[ρ ln ρ + (1 − ρ)ln(1 − ρ)]; red curve

is E N kBT

= −ρ, with = 2. The blue curve is the free energy ( F N kBT

) givenby eq.4.9. For what density is the free energy minumum ?

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-2.5

-2

-1.5

-1

-0.5

0

0.5 1

0 0.2 0.4 0.6 0.8 1

S ,

E , F

( i n u n i t s o f N k B T )

Protein densit

S

E

F=E-TS

Figure 4.3: Same as fig.4.2; but see the vertical lines : the pink line shows

the density at which entropy is maximum (ρ = 0.5); the black line shows thedensity at which free energy is minum ( ρ = 0.88). The energy is minmumwhen ρ = 1.

energy tries to take the system to ρ = 1, while thermal fluctuations tries tomaximize the entropy – a competition between energy and entropy. In thiscompetition, depending on the value of , the system reaches a density whichis minimum of the free energy.

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Chapter 5

Diffusion, Einstein relation andNernst equation

5.1 Diffusion:

A small bead in water will undergo random Brownian movements. How fardoes the bead move within a time t seconds? To answer this let us do thefollowing thought experiment. Leave the bead at position A in water andmeasure the displacement (r(t)) of the bead after a time of t seconds. Repeat

this experiment many times. Each measurement will give us a r(t), and onewould find that the average displacement

r(t) = 0, (5.1)

and the averaged displacement square is

|r(t)|2 ∝ t. (5.2)

Here the symbol ... means average over many measurements. The propor-tionality constant is related to the diffusion coefficient. It turns out that, in

3 dimensions, |r(t)|2 = 6Dt (5.3)

where D is the diffusion coefficient. (in d dimensions, the relation is |r(t)|2 =2dDt). The dimension of the diffusion coefficient is [D]=L2/T. To quickly

15



show that the mean square displacement (|r(t)|2) is proportional to time

linearly, let us do the following calculation.Now, let us consider a bunch of particles (say, protein molecules) diffusing.Here by "protein molecules", I mean completely folded, globule-like proteins– they can be imagined as "spheres" for all practical purposes! Let C (x, t) bethe concentration of the protein molecules at location x at any time t. Hereconcentration is defined as the number in unit volume (or length in 1D).The protein molecules will diffuse from higher concentration region to lowerconcentration region, and this “flow” of proteins (or the current, or flux) canbe written as

J D = −D∂C

∂x x

(Draw some concentration profiles, and convince yourself that this is true).This is called the Fick’s law of diffusion. Here the current, J D, is defined asthe number of molecules passing through unit area per unit time.

Concentration of molecules at any location will only change, with time,if and only if the flow is space-dependent. If the flow is same everywherein space, at location x, the number of particles flowing in will be equal tonumber of particles flowing out. Therefore, if you want C to change withtime, you should have J D chaining in space. This leads to

∂C

∂t = −

∂J D∂x

[In 3D this equation can be written as ∂C ∂t

= − ∇ · J D]. The above twoequations can be combined to get the diffusion equation

∂C

∂t = D

∂ 2C

∂x2 .

This is called the diffusion equation. Solving this equation one can getC (x, t). Next question is, what exact is D? This question was answeredby Albert Einstein.

5.2 Einstein relation

In one of the most celebrated papers in physics, Einstein, in 1905, presentedthe famous relation between Diffusion coefficient and viscous drag. And hisarguments were essentially the following.

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Consider n particles suspended in a volume V containing a fluid hav-

ing temperature T . Let there be an external force f = −f ez acting on eachparticle (e.g., in case of gravity f = −mgez; for simplicity we shall do the cal-

culation in 1 dimension). In thermal equilibrium particles will be distributedsuch that the concentration is given by the Maxwell-Boltzmann distribution

C (z ) ∝ exp

−f z

kBT

. (5.4)

The diffusion of these particle will generate a flux, J D as defined above. Onthe other hand, the external force f , will lead to movement of particles inthe direction of the force, with a velocity

vf = f ζ

. (5.5)

where ζ is the viscous drag. For sphere-like particles ζ = 6πηa where η isthe viscosity of the medium, and a is the radius of the sphere (or size of thespherical molecule). The corresponding flux, is given by

J f = Cvf = f C

ζ . (5.6)

The equilibrium is obtained when the fluxes are equal in magnitude but

opposite in direction. That is

J f = −J D, (5.7)

giving,

f C

ζ =

DCf

kBT .

⇒ D = kBT

ζ . (5.8)

Problem 2 : Calculate the diffusion coefficient D of a spherical particle of

1 µm diameter, in water.

D = kBT

6πηa =

4.1 × 10−21J

6 × 3.14 × 0.5 × 10−6m × 10−3 J sm3

≈ 0.6 (µm)2/s. (5.9)

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5.3 Diffusion of charged particles across a mem-

brane channel can create a potential differ-ence: Nernst equation

Consider a box partitioned into two equal parts by a membrane. At t = 0,the first part contains C 01 concentration of K + and Cl− ions (equal number of positive and negative ions, so that the partition is charge neutral). The sec-ond partition contains C 02 concentration of K + and C l− ions (equal numberof positive and negative ions, so that the partition is charge neutral). AssumeC 01 > C 02 . Then as the clock starts, the membrane is made permeable onlyto K + ions. What will happen?

The + ions will from the first part will diffuse to part 2, because theconcentration in part 2 is less. The magnitude of this diffusive flow is givenby the equation

J D = −D∂C

∂x .

However, since only + ions are flowing from part 1 to part 2, the numberof - ions in part 1 will be more than the + ions and will lead to a chargeimbalance. The part 1 will be more negatively charged, and part 2 will bemore positively charged. So the + charges in the part 2 will now be attractedback to part 1 (remember only + charges can move through the channel).This will lead to a flow of charges from 2 to 1. This flow is due to theelectrostatic created by the charge imbalance. This flow will be given by

J f = Cvf = f C

ζ

Noting that the electrostatic force is f = −q dφdx

, we get

J f = Cvf = −q dφ

dx

ζ

where φ is the electrostatic potential and ζ is the frictional drag (see above).

As discussed above, the equilibrium is obtained when the fluxes are equal inmagnitude but opposite in direction. That is

J f = −J D, (5.10)

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giving,

−q

dφ

dx

ζ = D ∂C

∂x

Integrating on both sides with x, we have

x2

x1

dx−q dφ

dx

ζ =

x2

x1

dxD∂C

∂x

where x1 and x2 are two nearby points on either sides of the membrane (say,x1 in part 1, and x2 in part 2). Substituting for D from the Einstein’s relation(D = (kBT )/ζ ), we have

φ(x2) − φ(x1) = kBT q

ln C 1C 2

(Check for errors in minus sign!). Here C 1 and C 2 are K + ion concentrationsat equilibrium in part 1 and part 2 respectively. This gives us an equation

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Chapter 6

Life in salty water and at lowReynold’s number

6.1 Inertia is negligible for objects in the cellu-

lar world

Imagine a dead bacterium is put into water with an initial velocity v0. Howfar will it move? For simplicity, let us imagine a bacterium as a sphere of radius a = 1µm. The mass of the sphere essentially the mass of the waterinside (assume no gravity). The velocity will obey the equation

mdv

dt = −ζv

This has the solutionv = v0e−

ζm

t

This tells us that the velocity will be nearly zero when t mζ . Noting

that m =density.volume, and ζ = 6πηa, we can substitute the density andviscosity of water, to get m

ζ ≈ 10−7s. When time is larger than the 100

nanosecond, the velocity will be nearing zero.The total distance travelled by the bacterium is d =

∞

0 vdt = v0mζ

Con-

sidering a reasonable number for initial velocity v0 = 1µm/s, we find thatthe total distance travelled by bacterium, d ≈ 10−13m. The total distancetravelled by the bacterium is less than the size of an atom!! Thismeans that, until you keep pushing, the bacterium will not move. To sustaina nonzero velocity, you need to apply force!

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Now, let us consider the bacterium under an external force f ext (gravity

or any other kind of force). The corresponding Newton’s equation isdv

dt +

ζ

mv =

f ext

m

For typical parameters, since ζ

m very large, we will have situation where the

first term dvdt

negligible when compared to the second term ζ m

v. This wouldmean

v = f ext

ζ

This implies that, we have a situation where force is proportional to velocity!In the cellular world, we are in such a regime where force is proportional to

velocity!This regime, where inertial forces are negligible compared to the viscous

forces, is known as the low Reynolds number regime. Reynolds number isdefined as:

Re = Inertial forces

Viscous forces.

Life, in the cellular level, is typically at very low Reynolds number. If youare enthusiastic in knowing more, google search for an article by E M Purcellnamed "Life at low Reynolds number". Both the suggested books discussmore details.

6.2 Electrostatic interaction gets “screened” due

to the presence of oppositely charged ions

When an electrically neutral big molecule is put in water, many small ions willcome out of the molecule and will wander in solution such that the entropy of the system increases. For example, the chemical group COOH, in water, willexist as COO− and H+, so that the H+ ion can wander around in solution(reason: Brownian forces, entropy). Such small charges are typically called

“counter ions”. In physiological conditions, there is also salt: NaCl existing

as Na+

and Cl−

; similarly K+

and Cl−

. But, remember, overall, the systemis charge neutral. That is, q i = 0.

Imagine a big positively charged macro-ion (eg. a protein) – given its charge,many small negative ions (counter-ions) will be hanging around this big ion

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– the resulting system can be imagined as a big positive ion with cloud of

negative ion around it. How will this protein–counter-ion system will interactwith another such protein–counter-ion system?Given that there are charges all around, what is the probability density of

finding a charge q i at any location? Given that this is a statistical mechanicalsystem at a finite temperature T , the probability density is given by theBoltzmann equation

P i = A exp

−

q iφ

kBT

where φ is the electrostatic potential of that location and A is a constant.Now, to find out the potential at any location, one has to solve the Poisson

equation

∇2φ = −C

0r

where the charge density C is nothing but

C =

i

q iP i = A

i

q i exp

−

q iφ

kBT

where we substitute for the probability from the Boltzmann equation. Usingthe expression for charge density in the Poisson equation, we get

∇2φ = −A

0r

i

q i exp

− q iφ

kBT

The above equation is known as the Poisson Boltzmann equation. The so-lution of this equation will give you the potential φ due to the the chargedsystem, at any location, r distance apart. However, solving this equation isvery difficult.

When qφkBT

1, one can expand the exponential function as

exp

−

q iφ

kBT

≈ 1 −

q iφ

kBT

This gives us

∇2φ = −A

0r

i

q i

1 −

q iφ

kBT

22



Noting that

i q i = 0, we have

∇2φ = A

0r

i

q 2i φ

kBT

Rewriting the above equation, we get

∇2φ =

1

λ2D

φ

where

λD =

i

kBT 0r

q 2i

A

. You can immediately see that such an equation is likely to have a solution

φ ∝ exp

−r

λD

when one demands that the potential at r → ∞ is zero. It turns out that,this equation, when solved appropriately (taking spherical symmetry, etc),one gets that the electrostatic potential due to the protein-counterion system,at distance r apart is

φ(r) = Bexp

−rλD

rwhere B is a constant. This is called the Debye-Huckel potential. This tellsus that, unlike in the normal electrostatic system, the potential decreasesexponentially fast, and when r λD the potential is zero. This length (λD)beyond which electrostatic potential dies out is known as the Debye lengthor the Debye screening length. In physiological conditions, the Debye lengthis λD ≈ 1nm. This implies that beyond one nanometer distance, in biology,electrostatic interaction will die out. Because effect of any charge will be

“screened" by oppositely charged ions that are wandering around! Therefore

the potential φ(r) = Bexp

−rλDr

is also known as the “screened-Coulomb"

potential.

23



Chapter 7

Force generation and movementin biology

In typical cells, force generation and cell movement are driven by certainkind of polymers known as actin.

7.1 A simple polymer

First, let us consider a simple filament-like polymer (see Fig. 7.1). Assumeone end of the polymer is inert, and the other end can polymerise and depoly-

merise with a rate kon and koff . For example kon = 5s−1

means the polymergrows by 5 subunits every second. Similarly koff = 2s−1 means the polymershrinks by 2 subunits every second. The net growth speed, on an average isgiven by,

v = dl

dt = kon − koff

where l is the average length of the filament (number of subunits on thefilament) and t is time. The unit of speed is subunits/second. In a sim-ple model, we can imagine that the polymerisation rate will depend on theavailability of free monomers to polymerise such that

kon = k0C

where k0 is the intrinsic rate constant and C is the free subunit (monomer)concentration available in the solution (I will be using the term subunit and

24



Figure 7.1: A simple polymer, polymerising with rate kon and depolymerisingwith rate koff at one end. The other end is assumed to be inert. The poly-merisation rate could depend on the free monomer concentration C , floatingin the solution, such that kon = k0C , where k0 is the intrinsic rate constant.

25



monomer interchangeably). This gives

v = k0C − koff (7.1)

7.2 Critical concentration

Critical concentration is the concentration of free monomers below whichthe filament would not grow. Just above the critical concentration, filamentgrows with a nonzero speed. In other words, critical concentration is theconcentration at which the average velocity v = 0. From Eq. 7.1, we get

C crit = koff

k0

.

Now, if we start with C > C crit, the filament will polymerise; as the filamentpolymerises and grows, the free monomer concentration will reduce. Finallythe growth will happen until the free monomer concentration is equal to thecritical concentration.

Problems:

1. Free monomer concentration can be defined C = N f /V where N f is thenumber of free monomers and V is the volume. As filament polymerises,the free monomer concentration decreases such that

dN f (t)dt

= −k0C (t) + koff

Taking V as a constant, using the relation C = N f /V , solve the aboveequation and obtain C (t). Substitute this result in the equation forvelocity and plot the velocity vs time. Note the velocity as t → ∞.Can you integrate the velocity equation to obtain l as a function of time?

2. Now assume a more general case where both ends can polymerise anddepolymerise with rates k+

on, k+off , k−on and k−off respectively. Here plus

signs represent rates for one end and the minus signs represent ratesfor the other end. Since both ends are symmetric, thermodynamicsimposes a constraint that

k+off

k+0

= k−off

k−0.

26



In other words C +crit = C −crit. If you think about it physically, you

will realise that, since these intrinsic rates depend only on the interac-tion between two monomers (how two monomers are bonded together),these ratios have to be the same at both ends as long as the monomersare of the same kind.

Now, write down the equation for the change in length (or averagevelocity) of the whole filament

dl

dt = v =?

What is the critical concentration (C crit) for the whole filament?

7.3 Force generation by a simple polymer

The polymerisation – chemical energy of binding – can be converted to me-chanical energy – force and work, by designing the system appropriately. Forexample, consider the system shown in Fig. 7.2. A polymer is polymerisingand depolymerising against a movable rigid wall on the right side. At theback (left side), it is supported by a fixed immovable rigid wall. Imaginethat you are applying an external force f on the movable wall on the right(see figure) against the direction of growth. As the filament polymerises, the

movable wall will keep moving to the right. As the filament depolymerises,the wall will keep moving to the left.

See the Fig. 7.2. To add a new subunit on the polymer, the wall needs tobe pushed by a distance d, where d is the size of a single subunit. In otherwords, to insert a single new subunit, the work needed to be done is f d.Since the force is against the growth, the polymerisation rate will decreaseas force increases. The polymerisation rate in the presence of the force canbe assumed to be

kon = k0C exp

−f d

kBT

.

When we put f = 0 we get back the old rate. For simplicity, let us assumekoff is independent of force. This gives us the growth velocity as

v = k0C exp

−f d

kBT

− koff (7.2)

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External forcef

rigid immovable

wall supportWall is free to move

Figure 7.2: A polymer is polymerising and depolymerising against a movablerigid wall on the right side. At the back (left side), it is supported by afixed immovable rigid wall. An external force f is being applied on themovable wall on the right, against the direction of growth. As the filamentpolymerises, the movable wall will keep moving to the right. As the filamentdepolymerises, the wall will keep moving to the left.

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Maximum force that can be generated: As you increase the external

force, the velocity of the polymer will decrease. At some force value (whenf = f s), the polymer will not grow anymore. This is the maximum force thepolymer can generate. This is also known as the stall force. This is given bytaking v = 0

k0C exp

−f sd

kBT

= koff (7.3)

f s = −kBT

d ln

koff

k0C (7.4)

f s is the maximum force such a polymer can generate. Cells use this kind of force to polymerise and depolymerise.

Problems:

1. Plot force vs velocity according to equation 7.2

2. Equate v = 0 in eq. 7.2. Plot the resulting equation as f versus C .Plot it with C in x-axis and and f s in the y-axis. The curve you getdivides the space into two. In one portion, for any value of f and C ,you will find that the polymer will grow. The in the other portion, forany value of f and C , the polymer will shrink. This is like a “phasediagram” with two phases. Shrinking phase and growing phase.

3. Assume that kon is independent of force, and the whole force is used toremove (disassemble) the monomer. What would be the expression forthe force-dependent off-rate ? What would be the resulting stall force?

4. For the above case, plot force vs velocity.

7.4 Actin: treadmilling

Actin is a protein. Actin in solution typically binds to ATP (ATP is thefuel for biological machines to work!) ATP-bound actin can polymerise and

become actin filament (often known as F-actin). Once polymerised onto thefilament, ATP-bound actin can change its chemical state (it will go undergo achemical reaction known as ATP hydrolysis) and become ADP-bound actin.In this process, actin monomer also undergoes some structural change. This

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ATPADP

ATP

ATP ATP

ATP

ATP

ADP ADP

Figure 7.3: At the right (plus) end: ATP-bound actin can polymerise withrate kT

on and depolymerise with rate kT off . At the left(minus) end, ADP-

bound actin can only depolymerise. Let kD−off be the rate of ADP-bound

actin depolymerisation from left end (minus end). Assume that all otherpossible rates are zero.

hydrolysis process happens only on the filament. Typically, it does not hap-pen in solution. Therefore, you will have an asymmetric polymer whereone end has newly polymerised ATP-bound monomer and the other end hasADP-bound monomer. It also turns out that ADP-bound actin is unstableand prefers to depolymerise quickly. All these information can be summarisedas shown in Fig.7.3. ATP-bound actin can polymerise with rate k T

on and de-polymerise with rate kT

off from the plus end (right end) (Please note that plusand minus are notations to distinguish both the ends. It has nothing to withelectric charges). Let kD−

off be the rate of ADP-bound actin depolymerisationrate from left end (minus end). (see figure). Assume that all other possible

rates are zero.Based on the what we did for a simple polymer, at the plus end, we canwrite,

kT on = kT

0 C

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where C is the concentration of the free ATP-bound actin. The net rate of

growth at the plus end is v+ = kT 0 C − kT

off

When

C = kT

off

kT 0

= C +crit

the growth velocity v+ will be zero. For any concentration C > kT

off

kT 0

, the right

end (plus end) will keep polymerising and make the polymer grow on anaverage. However, since the left (minus) end has only depolymerisation, theleft end will always contribute to the polymer shrinking. The left (minus)end will never grow. In other words, the velocity of the minus end v− =

−kD−off (always negative). By varying the concentration, we can achieve the

condition where the plus end growth is equal to the minus end shrinking

kT 0 C − kT

off = kD−off

The phenomenon of plus end growth and minus end shrinkage (such thatfilament length is constant on an average) is called treadmilling. The con-centration at which this happens is known as the tread milling concentration.

C tread = kD−

off + kT off

kT 0

At this concentration, filament, on an average will not grow. But note thatC tread > C +crit. That means, at this concentration, the plus end will keepgrowing, and the minus end will keep shrinking, on an average. (that willgive some idea on why we called it a plus end and a minus end!). However,we have no net filament growth, but the centre of mass of the polymer keepsmoving. This is like a simple machine that can move!

Other parts we discussed during lecture 10 (that is not coveredhere) will not be asked for the mid semester test

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Notes BB101 Ranjith

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