AMPHIPATHIC ALPHA HELICES

1 Proteins

1.1 Amphipathic Alpha Helices

Alpha helices are often embedded in a protein so that one side faces the surface and the other

side faces the interior. Such helices are often termed amphipathic because the surface side is

hydrophilic and the interior side is hydrophobic. A simple way to decide whether a sequence

of aminoacids might form an amphipathic helix is to arrange the aminoacids around what is

known as âhelix-wheel projectionâ, which is a 2d projection of the aminoacid positions (see

Fig. 1). If the hydrophobic and hydrophilic aminoacids are segregated on opposite sides of

the wheel, the helix is amphipathic. Using the helix-wheel projection, decide which of the

three peptides here might form an amphipathic helix

- S L I K S V I E M V D E W F R T F L
- F L I R V L R K V F R V L T R I L S
- R L F R S R V L K I A V I R F L L I

(underlined aminoacids are hydrophobic one, they can be memorized by the mnemonic

“FAMILY VW”).

1.2 Amphipatic beta sheets

Like alpha helices, beta sheets often have a one side facing the surface of the protein and one

side facing the interior, giving rise to an amphipathic sheet with one hydrophobic surface

and one hydrophilic surface. From the sequences listed below pick the one that could form

a strand in an amphipathic beta sheet. - A L S C D V E T Y W L I
- D K L V T S I A R E F M
- D S E T K N A V F L I L
- T L N I S F Q M E L D V
- V L E F M D I A S V L D

1.3 Buried Aminoacids

Small proteins may have only one or two amino acid side chains that are totally inaccessible

to solvent. Even in large proteins, only about 15% of the amino acids are fully buried. A list

Figure 1: You can use this figure to decide which of the three sequences might form an

amphipatic helix (problem 1.1).

of buried side chains from a study of twelve proteins is shown in the table below. The list is

ordered by the proportion of amino acids of each type that are fully buried. What types of

amino acids are most commonly buried? Least commonly buried? Are there any surprises

on this list?

Aa prop. buried Aa prop. buried

I 0.60 S 0.22

V 0.54 E 0.18

C 0.50 P 0.18

F 0.50 H 0.17

L 0.45 D 0.15

M 0.40 Y 0.15

A 0.38 N 0.12

G 0.36 Q 0.07

W 0.27 K 0.03

T 0.23 R 0.01

2 Ideal Flexible Polymers

2.1 Radius of gyration

For an ideal polymer of length N we have shown that the average end-to-end distance R~ is:

hR~ 2

i = a2

N (1)

where a is the average bond length. We define now the radius of gyration as follows:

Rg =

vuut

1

N + 1

X

N

i=0

(~ri ~rcm)

2 (2)

where ~ri is the position of the monomer i (i = 0, 1,â¦N) and where the center of mass

position is given by:

~rcm = 1

N + 1

X

N

i=0

~ri (3)

a) Show that

R2

g = 1

2(N + 1)2

X

N

i=0

X

N

j=0

(~ri ~rj )

2 (4)

b) Use the previous formula to calculate the average1 hR2

gi

c) Show that in the limit N ! 1 one has

hR2

gi â a2N

6 (5)

so that the asymptotic N dependence behavior is similar to that of the end-to-end

distance of Eq. (72).

1You will need to use the following relations

X

N

k=1

k = N(N + 1)

2

X

N

k=1

k2 = N(N + 1)(2N + 1)

6

2.2 The Gaussian Chain

We have shown that in general the end-to-end distance for an ideal polymer at large N is a

gaussian distribution

P(R, N ~ ) = â 3

2â¡N a2

â3/2

exp

3R~ 2

2N a2

!

(6)

irrespectively of the type of model considered. In the calculation we have expanded the

Fourier transform of g(~u) for small k. This approximation is not necessary for a Gaussian

chain, which is defined by the following bond distribution:

g(~u) = â 3

2â¡a2

â3/2

e

3~u2

2a2 (7)

Show that for a Gaussian chain the probability distribution function can be computed exactly

for any values of N, not necessarily large, and show that the result of the calculation gives

Eq. (6).

2.3 One dimensional polymer

We consider the following one dimensional model of a lattice polymer

g(u) = 1

2 ((u + a) + (u a))

Let us suppose the first monomer is placed in x = 0. Since this is a lattice polymer, the

monomers can only occupy discrete positions x = na with n an integer. The end point will

be in R = ma. It is then more convenient to write the end distribution as P(m, N).

a) Calculate the distribution P(m, N) for every value of N.

b) Show that in the limit N m and N and m both large P(m, N) reduces to a gaussian

which agrees with the general expression derived in the course.

3 Ideal Semi-flexible Polymers

3.1 Inextensible Continuous Polymers

Inextensible discrete polymers have bonds with fixed length. Continuous polymers are described by a curve ~r(s), where 0 ï£¿ s ï£¿ L is the curvilinear coordinate. The condition of

inextensivity can be expressed as follows

|d~r(s)| â |~r(s + ds) ~r(s)| = ds (8)

which implies that the tangent d~r(s)/ds has unit length. Using Eq. (8) find ~r(s) for

a) A circle of radius R in the xy plane.

b) An helix of radius R and pitch l, where the helical axis is along the z direction.

3.2 The Worm Like Chain in two dimensions

During the lecture we found that a 3d WLC the correlation between two tangent vectors

decays exponentially as

hub(0) Â· ub(s)i = es/lp (9)

where lp = Ka/kBT is the persistence length. We want to consider here configurations of

the polymer in 2d. The polymer as DNA can be absorbed to a planar surface orthogonal to

the axis z. The DNA is bound to the surface along the z-direction, but can still fluctuate

along the x and y directions.

a) Calculate the partition function of a discrete WLC in 2d in the limit K 1.

b) Following the same procedure as in 3d calculate the tangent-tangent correlator hub(0) Â·

ub(s)i in 2d.

c) Calculate finally hR2i, where the end-to-end distance is given by

R~ =

Z L

0

ds ub(s) (10)

emphasizing the diâµerences with the 3d case.

3.3 Bending a polymer

The continuum WLC model is described by the following energy:

E = lp

2

Z L

0

âdub

ds â2

ds (11)

where = 1/kBT is the inverse temperature, lp the persistence length and ub(s) is the unit

vector tangent to the polymer in s. Let ~r(s) be a continuous curve in the space and s the

arclength. For an infinitesimal variation ds we must have |d~r| = ds. Hence the derivative:

d~r

ds = ub (12)

defines a unit vector tangent to the curve.

In this problem we estimate the energies of some specific conformations of a polymer.

Ï

DNA

(b) protein (a)

Figure 2: (a) Some proteins bind at two sides of a DNA molecule and induce a loop formation

with a bend (teardrop shape). (b) The teardrop can be approximated as a combination of

an arc of a circle and two straight segments.

a) Using (11) estimate Ecircle(L), the energy of a circle of length L. Express your result

in units of kBT.

b) Calculate Ehelix(R0, l0) the energy for a full turn of an helix of radius R0 and pitch l0

(the pitch is the length covered along the helix axis after one turn).

Some proteins bind to two sites of a DNA molecule and form loops as shown in Fig. 8.

One can calculate the shape that minimizes (11) which resembles a teardrop. The minimal

energy for a teardrop of length L can be computed using elliptic integrals and the result is

2 :

Eteardrop = 14.055

kBTlp

L (13)

c) Show that this is lower than the energy of a circle of the same length L.

d) Approximate the teardrop shape using an arc of a circle and two straight segments

tangent to the circle. Find the minimal energy value for this configuration keeping a

fixed total length L. Compare the result to Eteardrop. (Note: perform the minimization

numerically. You will have to find the energy function E(), where is the angle

shown in Fig. 8(b). Plot this function and find the numerical value of its minimum).

3.4 DNA persistence length

The persistence length of a double stranded DNA molecule measured at room temperature

and 10 mM of NaCl is of lp = 50 nm. Using the WLC model estimate the persistence length

at 70 C, assuming that the molecule remains double stranded. (Note: Suciently high

temperature induces separations of the two strands of the double helix, this transition from

double stranded to single stranded DNA is known as DNA melting)

2See:H. Yamakawa, W.H. Stockmayer, J. Chem. Phys. 57, 2843 (1972).

4 Real-polymers

4.1 DNA package in the cell nucleus

We can use random walks, or better self-avoiding walks, to describe the configuration of a

polymer in the space. As mentioned before the double stranded DNA can be considered as

composed of segments of length a = 50nm and the distance between base pairs along the

DNA is 0.34nm. The total number of bases in the human DNA is that of 3.3 â¥ 109. This

is divided into several chromosomes, but we assume that the whole DNA is a single long

polymer.

a) Compute the typical radius of the human DNA using random walk statistics and selfavoiding walk statistics3.

b) We know that the DNA is contained in the nucleus of the cell and that the nucleus has

approximately the radius of 1Âµm. Conclude from the value computed in a) that DNA

is much more tightly packed in the nucleus than what expected from the random and

self-avoiding walk analysis.

4.2 Random and self-avoiding walks

We want to estimate, in d-dimensional space, the typical number of contacts of a random

walk of length N, as N ! 1.

a) How does the radius of the sphere Rd containing the random walk scales with N? How

does the volume Vd scales with N?

b) Consider now the monomers as independent particles confined in Vd. How does the

probability that a given pair of monomers meet in the same point scales with N? How

does the total probability of any given pair to be found at the same point scales with

N?

4.3 Relevance of excluded volume interactions

In this problem we analyze the onset of excluded volume eâµects. We expect that if the

polymer is âthinâ the self-avoidance will produce small corrections and the polymer will

show an ideal behavior R â¡ aN1/2. If the polymer is âthickâ self-avoidance is relevant and

R â¡ aNâ«, with â« = 3/5 in 3d.

Show that the equation which gives the minimal R in 3d is

â R

a

p

N

â3

+

â«cN1/2

12a3 = 3

2

â R

a

p

N

â5

(14)

(this is the same equation discussed in the course where we added the correct prefactors).

We apply the previous equation to double stranded DNA. We can consider it as a polymer

composed by N segments. Each segment is approximated as a cylinder of length a = 50 nm

(the dsDNA persistence length) and diameter d = 2 nm. One segment corresponds to 150

base pairs. The parameter â«c is the volume occupied by one segment, which you can estimate

from the volume of the cylinder.

Solve Eq. (14) and find R for a number of segments ranging from N = 10 to N = 107.

Plot your results of R vs. N in a log-log scale. A scaling law of the type R â¡ aNâ« would

3Recall that for self-avoiding walks in three dimensions R â N0.6.

appear as a straight line in a log-log plot. Do you obtain such a line? Explain your results.

How does the graph R vs. N change if our DNA would be have an hypothetical thickness

of 15 nm?

5 Stretching Polymers

5.1 Stretching a polymer: the low force limit

In this problem we will show that a generic polymer stretched with a very weak force at the

two extremities will extend as a spring

f â¡ KhXi (15)

where the force is applied along the Ëx direction and hXi is the average extension along

the direction of the force. The previous relation, is valid at low forces f, i.e. in the limit

fhXi â§ kBT. We say then that a polymer at low forces behaves as an entropic spring.

Our starting point is the partition function of the most generic polymer model stretched

by a force f at its two ends:

ZN (f) = Z Y

N

i=1

d~uieE(~u1, ~u2,â¦ ~uN ) e ~f Â· (~rN ~r0) (16)

where E(~u1, ~u2,â¦ ~uN ) is the energy of a configuration identified by segments ~u1, ~u2,â¦ ~uN .

We do not require any specific property for the enrgy E.

Note that the extension is the average end-to-end distance along the direction of the

applied force:

X â h(~rN ~r0) Â· xËi (17)

a) Show that at low forces:

ZN (f) = ZN (0) â

1 + 2f 2

2 hX2

if=0 + â¦â

(18)

where the dots denote higher order terms and hX2if=0 is the average of the squared

x-component of the end-to-end vector in absence of applied force.

b) Using the previous equation show that Eq. (15) holds and express the spring costant

K as a function of the temperature and hX2if=0.

c) How does the spring constant of a polymer with excluded volume interaction scales as

a function of the polymer length N?

5.2 Stretching a gaussian polymer

We consider the following energy for a gaussian polymer to which a constant force is applied

to the two ends:

E = K

X

N

i=1

(~ri ~ri1)

2 ~f Â· (~rN ~r0) (19)

Show that

â¦

(~rm+n ~rn)

2âµ

= ma2 +

m2a2

9

â f a

kBT

â2

(20)

F

F

Î¾b

Figure 3: Blobs formed in a polymer under tension as discussed in Problem 5.2. The force introduces a

new length scale in the system â b = kBT /F, which is the size of a blob.

where a2 â h(~ri ~ri1)2if=0 is the averaged squared extension of a bond at zero force.

Note that the previous equation tells us that at weak forces and short separation between

monomers the polymer behaves as a random walk:

â¦

(~rm+n ~rn)

2âµ

â ma2 (21)

However for larger m and forces stretching dominates

â¦

(~rm+n ~rn)

2âµ

â m2a2

9

â f a

kBT

â2

(22)

The boundary between the two regimes is obtained by equating the two terms in the lhs of

Eq. (20) which yields:

mb =

â3kBT

f a â2

(23)

The polymer can be viewed as being made of blobs each containing mb monomers and of

size â b = a

pmb = 3kBT/f (see Fig. 3). Monomers within a blob, i.e. separated by distances

m<mb, are unperturbed by the force.

Hint: To derive Eq. (20) write the partition function in term of the bond variables ~ui â

~ri ~ri1. With this change of variables the partition function factorizes ZN = (Z1)N and Z1

can be integrated easily. Eq. (20) becomes:

â¦

(~rm+n ~rn)

2âµ

= h

m

X

+n

k=n+1

~uk

!2

i = mh~u2

i + m(m 1)h~ui

2 (24)

where the last equality follows from the fact that diâµerent segments are independent hence if

i 6= j we have h~ui Â· ~uj i = h~uii Â· h~uj i = h~ui2.

Answer:

AMPHIPATHIC ALPHA HELICES