statistical physics of rna folding · statistical physics of rna folding ralf bundschuh...

Statistical Physics of RNA folding

Ralf Bundschuh

Collaborators: Terry Hwa, UCSD, Ulrich Gerland, Universitat zu Koln,Robijn Bruinsma, UCLA

Ohio State University

March 18, 2008

Ralf Bundschuh (Ohio State University) Statistical Physics of RNA folding March 18, 2008 1 / 111

Outline

Outline of all lectures

Part I : Introduction to RNA biology

Part II : Homogeneous RNA

Part III : Sequence disorder

Part IV : Force-extension experiments


Part I

Introduction to RNA biology


Outline of part I

1 RNA structure

2 Making RNA

3 Functions of RNA

4 Summary


RNA structure

Outline of part I

1 RNA structureChemical structureSecondary structureTertiary structure

2 Making RNA

3 Functions of RNA

4 Summary


RNA structure Chemical structure

Chemical structure of RNA

RNA is a heteropolymer

Backbone is phosphate-sugar chain

Directionality

Quite flexible

Negatively charged in solution(The Biology Project, U Arizona)

Four different side chains (“bases”)

HydrophobicTwo groups: purines (A,G) and pyrimidines (C,U)


RNA structure Secondary structure

Secondary structure of RNA

Primary structure: sequence (5’ → 3’)GCGGAUUUAGCUCAGUUGGGAGAGCGCCAGACUGAAAAUCUGGAGGUCCUGUGUUCGAUCCACAGAAUUCGCACCA

Secondary structure: base pairing

Watson-Crick (G–C and A–U)Wobble (G–U)non-canonical (anything else)Allows formation of hydrogen bondsProtects hydrophobic bases from waterCosts configurational entropyBrings like charges close togetherFree energy per (WC) base pair ≈2 kcal

mol ≈4kBTStrongly dependent on temperature andsalt concentration


RNA structure Tertiary structure

Tertiary structure of RNA

Tertiary structure: three-dimensional arrangementWeaker interactions than base pairingDepends on MagnesiumConfers biological function

−→

wikipedia


Making RNA

Outline of part I

1 RNA structure

2 Making RNA

3 Functions of RNA

4 Summary


Making RNA

Transcription

Template for RNA is stored ingenome

RNA is produced duringtranscription

RNA is produced by RNApolymerase

RNA is exact copy of DNAsequence

http://www.dadamo.com/wiki/wiki.pl/

Transcription (DNA transcription)


Functions of RNA

Outline of part I

1 RNA structure

2 Making RNA

3 Functions of RNAMessenger RNATransfer RNARibosomal RNAOther RNA

4 Summary


Functions of RNA Messenger RNA

Translation

Messenger RNA codes forprotein sequence

Translation from messengerRNA occurs in ribosome

Three bases (a codon) in themessenger RNA code for oneamino acid via the genetic code

RNA structure is not veryimportant (nobelprize.org)


Functions of RNA Transfer RNA

Transfer RNA

Adapter molecule that implements genetic code

Binds amino acid at one end

Base pairs with messenger RNA at other end

Very specific clover leaf structure recognized by ribosome

wikipedia http://ghs.gresham.k12.or.us/science/ps/sci/ibbio/chem/nucleic/chpt15/tRNA.gif


Functions of RNA Ribosomal RNA

Ribosomal RNA

Ribosomal RNA is majorcomponent of the ribosome

Ribosomal RNA catalyzes actualextension of protein sequence

Overall structure conservedthroughout all of cellular life

⇒ RNA world hypothesis Center for Molecular Biology of RNA, UC-Santa Cruz


Functions of RNA Other RNA

Other RNA

snoRNA guides chemical modification of other RNAs

microRNA regulates the translation of other genes

Ribozymes catalyze chemical reactions(RNaseP, group I and II intron, Hammerhead ribozyme, . . .)

Guide RNA guides RNA editing

RNA viruses use single or double stranded RNA as their genome

New categories found all the time . . .(ENCODE project finds that very large fraction of human genome istranscribed, but only few percent are protein coding)


Summary

Outline of part I

1 RNA structure

2 Making RNA

3 Functions of RNA

4 Summary


Summary

Summary of part I

RNA is a heteropolymer of four bases

RNA forms a well-defined hierarchy of structures

RNA is involved in nearly every aspect of life


Part II

Homogeneous RNA


Outline of part II

5 Boltzmann partition function

6 Molten RNA

7 Molten-native transition — hairpin

8 Molten-native transition — Cayley tree

9 Summary


Boltzmann partition function

Outline of part II

5 Boltzmann partition functionSecondary structurePartition functionRecursion equation

6 Molten RNA



9 Summary


Boltzmann partition function Secondary structure

Definition of RNA secondary structure

Definition

An RNA secondary structure on an RNAsequence of length N is a set S of basepairs (i , j) with 1 ≤ i < j ≤ N that fulfillthe following conditions:

Each base is involved in at most onebase pair

No pseudoknots, i.e., if (i , j) and (k, l)are base pairs with i < k, eitheri < k < l < j or i < j < k < l


Boltzmann partition function Secondary structure

Diagrammatic representation

An RNA secondary structure can be represented by an arch diagram

Every base is end point to at most one arch

Two arches never cross

One to one correspondance between arch diagrams that fulfil theabove conditions and secondary structures


Boltzmann partition function Partition function

Energy models

Each structure S has a certain (free) energy E [S ]

Different models possible

All base pairs are equal E [S ] = ε0|S | (ε0 < 0)Energy associated with base pairs E [S ] =

∑(i,j)∈S e(i , j)

Turner parameters (nearest neighbor model):energy associated with loops

Base pair stacks

Hairpin loops

Interior loops

Bulges

Multi-loops

interior loop

bulge

hairpin loop

stacking loop

multiloop


Boltzmann partition function Partition function

Partition function

Definition

The partition function of an RNA molecule with energy function E [S ] isgiven by

Z ≡∑S

e−E [S]RT

R = 1.987 calmol K

Temperature T in Kelvin ⇒ RT ≈ 0.6kcalmol

Ensemble free energy F = −RT lnZ

Thermodynamics completely specified if partition function is known


Boltzmann partition function Recursion equation

Calculating partition functions

Definition

Let Zi ,j be the partition function for the RNA molecule starting at base i

and ending at base j . Denote this quantity by i j.

Observation

For the base pairing energy model the Zi ,j obey the recursion equation

Σ+=jj−1kikjj−1ii j k+1k−1

Zi ,j = Zi ,j−1 +

j−1∑k=i

Zi ,k−1e− e(k,j)

RT Zk+1,j−1

Calculate from shortest to longest substrands

O(N3) algorithm for arbitrary sequence


Molten RNA

Outline of part II


6 Molten RNAEnergy modelz transformPropertiesGeometric derivation



9 Summary


Molten RNA Energy model

Energetics in molten phase

Definition

In the molten phase of RNA every base can pair with every other baseequally well, i.e., e(i , j) = ε0.

Properties

Applies to repetitive sequences: AUAUAUAUAUAUAUAUAU,GCGCGCGCGCGCGC, GACGACGACGACGACGACGAC

Applies to arbitrary sequences at temperatures close to denaturationon a coarse-grained level

Minimum energy is always N2 ε0

Structural entropy plays a major role


Molten RNA Energy model

Molten phase partition function

Reminder

Σ+=jj−1kikjj−1ii j k+1k−1

Zi ,j = Zi ,j−1 +

j−1∑k=i


RT Zk+1,j−1

Simplification

In the molten phase the partition function depends only on the length j − iof the substrand, not on i and j individually: Zi ,j ≡ G (j − i + 2)

Consequence

G (N + 1) = G (N) + qN−1∑k=1

G (k)G (N − k) with q ≡ e−ε0RT


Molten RNA z transform

z transform

Definition

For any series Q(N) the z transform Q is defined as

Q(z) ≡∞∑

N=1

Q(N)z−N

Properties

Function of the complex variable z

Analytic outside a radius of convergence

Discrete version of Fourier transform

Back transform: Q(N) = 12πi

∮Q(z)zN−1dz

Convolution property:∑N−1

k=1 Q(k)W (N − k) = Q(z) · W (z)



z transform for molten RNA

G (z) can be calculated:

G (N + 1) = G (N) + qN−1∑k=1

G (k)G (N − k)

G (N + 1)z−N = G (N)z−N + qz−NN−1∑k=1

G (k)G (N − k)

∞∑N=1

G (N + 1)z−N = G (z) + qG 2(z)

zG (z)− G (1) = G (z) + qG 2(z)

zG (z)− 1 = G (z) + qG 2(z)

G (z) =1

2q

[z − 1−

√(z − 1)2 − 4q

]



z transform for molten RNA

G (z) can be calculated:

G (N + 1) = G (N) + qN−1∑k=1

G (k)G (N − k)

G (N + 1)z−N = G (N)z−N + qz−NN−1∑k=1

G (k)G (N − k)

∞∑N=1

G (N + 1)z−N = G (z) + qG 2(z)

zG (z)− G (1) = G (z) + qG 2(z)

zG (z)− 1 = G (z) + qG 2(z)

G (z) =1

2q

[z − 1−

√(z − 1)2 − 4q

]Ralf Bundschuh (Ohio State University) Statistical Physics of RNA folding March 18, 2008 30 / 111


Back transform I

Integral expression for G (N)

G (N) =1

2πi

∮G (z)zN−1dz

=1

4πqi

∮ [z − 1−

√(z − 1)2 − 4q

]zN−1dz

= − 1

4πqi

∮ √(z − 1)2 − 4q zN−1dz

Singularity structure

Branch cut from 1− 2√

q to z0 ≡ 1 + 2√

q

Contour has to go around branch cut

Contour can be contracted to branch cut



Back transform II

Reminder

G (N) = − 1

4πqi

∮ √(z − 1)2 − 4q zN−1dz

Leading behavior

For large N only the singularity with largest real part contributes⇒ G (N) ≈ zN

0 = (1 + 2√

q)N

Prefactor ∫ µc

µ0

(µc − µ)αeµNdµ ≈ Γ(1 + α)N−(1+α)eµcN

⇒ G (N) ∼ N−3/2(1 + 2√

q)N


Molten RNA Properties

Properties of molten RNA

G (N) ≈(

1 + 2√

q

4πq3/2

)1/2

N−3/2 (1 + 2√

q)N

N−3/2 characteristic behavior due to entropy

Can be observed in pairing probability:

Pr{1 and k paired} = qG (k)G (N − k)

G (N + 1)∼ k−3/2 (N − k)−3/2

N−3/2≈ k−3/2

Free energy is F = −RT lnG (N) ≈ −RTN ln(1 + 2√

q)

For q � 1: F ≈ −RTN ln(2√

q) = −RT2 N ln(q) = N ε0

2

For q = 1: G (N) =number of secondary structures∼ N−3/23N .


Molten RNA Geometric derivation

A geometric argument

Mapping

5

10

121314

11

98

7

64

321

151617

18

⇒6 7 91 32 4 5 8 15 1611 1210 13 14 17 18

via

Properties

One to one mapping between structures and mountains

Mountains are random walks

At each point two choices with weight√

q and one with weight 1⇒ (1 + 2

√q)N

Never go below zero but return at the end ⇒ N−3/2


Molten-native transition — hairpin

Outline of part II


6 Molten RNA

7 Molten-native transition — hairpinModelSolutionResults


9 Summary


Molten-native transition — hairpin Model

Model for molten-native transition

Observation

Structural RNAs have to fold into a specific “native” structure⇒ there must be something in the sequence that prefers this structure

Model

Use perfect hairpin as native structure

1 N/2

NAssign binding energy ε1 to all native basepairs

Assign binding energy ε0 to all other basepairs

11

N

N


Molten-native transition — hairpin Solution

Partition function

Definition

Let Z (N; q, q) be the partition function of the Go model for themolten-native transition with 2N − 2 bases, q = e−ε0/RT , and q = e−ε1/RT

Definition

Let W (N; q) ≡ Z (N; q, q = 0) be the partition function of the Go model

for 2N − 2 bases in which native contacts are disallowed .

Observation

Z (N; q, q) =(p.f. with 0 native contacts) + (p.f. with 1 native contacts)+ (p.f. with 2 native contacts) + . . .

=+ +...+ +



Individual terms

0 native contacts

(p.f. with 0 native contacts) = = W (N; q)

⇒ z-transform: W (z)

1 native contact

(p.f. with 1 native contacts) = = q∑N−1

k=1 W (k; q)W (N − k; q)

⇒ z-transform: qW 2(z)

2 native contacts

(p.f. with 2 native contacts) = =q2∑

1≤k1<k2<N W (k1; q)W (k2 − k1; q)W (N − k2; q)

⇒ z-transform: q2W 3(z)



Putting it all together

Summing up

+ +...+ +

Z (z ; q, q) = W (z) + qW 2(z) + q2W 3(z) + . . . =W (z)

1− qW (z)

What is W (z)?

For q = q we have Z (N; q, q = q) = G (2N − 1; q)

⇒ Z (z ; q, q) = E (z ; q) ≡∑∞

N=1 G (2N − 1; q)z−N

⇒ W (z) =E (z)

1 + qE (z)⇒ Z (z ; q, q) =

E (z)

1− (q − q)E (z)




Summing up

+ +...+ +

Z (z ; q, q) = W (z) + qW 2(z) + q2W 3(z) + . . . =

W (z)

1− qW (z)

What is W (z)?


⇒ Z (z ; q, q) = E (z ; q) ≡∑∞

N=1 G (2N − 1; q)z−N

⇒ W (z) =E (z)

1 + qE (z)⇒ Z (z ; q, q) =

E (z)

1− (q − q)E (z)




Summing up

+ +...+ +

Z (z ; q, q) = W (z) + qW 2(z) + q2W 3(z) + . . . =W (z)

1− qW (z)

What is W (z)?


⇒ Z (z ; q, q) = E (z ; q) ≡∑∞

N=1 G (2N − 1; q)z−N

⇒ W (z) =E (z)

1 + qE (z)⇒ Z (z ; q, q) =

E (z)

1− (q − q)E (z)




Summing up

+ +...+ +

Z (z ; q, q) = W (z) + qW 2(z) + q2W 3(z) + . . . =W (z)

1− qW (z)

What is W (z)?


⇒ Z (z ; q, q) = E (z ; q) ≡∑∞

N=1 G (2N − 1; q)z−N

⇒ W (z) =E (z)

1 + qE (z)

⇒ Z (z ; q, q) =E (z)

1− (q − q)E (z)




Summing up

+ +...+ +

Z (z ; q, q) = W (z) + qW 2(z) + q2W 3(z) + . . . =W (z)

1− qW (z)

What is W (z)?


⇒ Z (z ; q, q) = E (z ; q) ≡∑∞

N=1 G (2N − 1; q)z−N

⇒ W (z) =E (z)

1 + qE (z)⇒ Z (z ; q, q) =

E (z)

1− (q − q)E (z)



Behavior of E (z)

Expression for E (z)

E (z) =∑∞

N=1 G (2N − 1; q)z−N can be calculated similarly to G (z).

E (z) =1

2q− 1

4qz

[√(z − 1)2 − 4q +

√(z + 1)2 − 4q

].

Properties

Vanishes as z →∞Square root branch cut at z0 = 1 + 2

√q

Finite limit E (1 + 2√

q) at branch cut

Other branch cuts have smaller real part 0 5 10 15 20z0

0.05

0.1

E(z)

z0

E(z )0



Singularity structure of Z (z)

Candidate singularities

Z (z ; q, q) =E (z)

1− (q − q)E (z)

Square root branch cut at z0 = 1 + 2√

q

Pole at z1(q) given by E (z1) = 1/(q − q) 0 5 10 15 20z0

0.05

0.1

E(z)

q−q~1

z0

E(z )0

z (q)~1

Dominant singularity

Square root branch cut at z0 if 1/(q − q) > E (z0)

Pole at z1(q) if 1/(q − q) < E (z0)

⇒ Phase transition


Molten-native transition — hairpin Results

Properties of molten-native transition

Characterization of phase transition

Critical bias qc = q + 1/E (z0)

If q < qc regular molten behavior Z ∼ N−3/2(1 + 2√

q)N

−→ native base pairs do not play any role

If q > qc native behavior with Z ∼ N0z1(q)N

−→ finite fraction of native base pairs but still many “bubbles”

For q � qc we get Z ∼ N0qN

−→ only native base pairs

Conclusion

It takes a finite amount of sequence bias to enforce a native structure.


Molten-native transition — Cayley tree

Outline of part II


6 Molten RNA


8 Molten-native transition — Cayley treeModelRecursion equationsNumericsAnalytical solutionOpen problems

9 Summary


Molten-native transition — Cayley tree Model

Modeling branchy native states

Problem

Natural RNA structures are branched

Solution

Use Cayley tree as native structure

⇒ Characterize by order k instead of lengthN = 2k+2 − 2


Molten-native transition — Cayley tree Recursion equations

Restricted partition function

Restricted partition function

Define Q(k, n) as the partition function of all structures with 2n bases“outside” the native base pairs and all these bases unpaired

Full partition function

W (n) is still partition function of non-nativebase pairs among 2n bases

⇒ Z (k; q, q) =∑n

W (n)Q(k, n)

W

n=2 n=2

Q


Molten-native transition — Cayley tree Recursion equations

Recursion equations

Recursion equations from level k to k + 1

n > 0

First native base pair is open⇒ both sub-trees contribute to accessible base pairs⇒ Q(k + 1, n) =

∑m

Q(k,m)Q(k, n −m)

n = 0

First native base pair is paired⇒ both sub-trees contribute to first bubble of non-native base pairs⇒ Q(k+1, n=0) = q

∑n1,n2

Q(k, n1)Q(k, n2)W (n1+n2)n=1


Molten-native transition — Cayley tree Numerics

Pinch free energy

Free energy

Q(k, n) −→ Z (k; q, q) −→ F (k; q, q) = −kBT log Z (k; q, q)

Expectation

Small q ⇒ molten phase ⇒ F ≈ N−3/2zN0

Large q ⇒ native phase ⇒ F ≈ zN1

(N = 2k+2 − 2)

Pinch free energy

F (k+1)−2F (k) =

{32k molten

const. native


Molten-native transition — Cayley tree Numerics

Numerics

Numerically iterate recursion equations for Q(k, n) at q = 4

0 10 20k

0

5

10

15

20

25

∆F(k

)/k B

T

~q=4~q=20~q=40~q=50~q=56~q=62~q=68~q=74~q=80~q=90

pinch free energies

0 10 20 30 40 50 60 70 80 90 100~q0

1

2

3

4

θ

N >= 250N >= 30000

fitted exponents

Phase transition at q ≥ 80


Molten-native transition — Cayley tree Analytical solution

Stationary solution

Asymptotics

In native phase: Z (k) ≈ zN1 with N = 2k+2 − 2

⇒ assume Q(k, n)/z2k+2−21 → w(n) for large k

Recursions for w(n)

w(n) =n−1∑m=0

w(m)w(n − 1−m) and

w(0) = (q − q)∞∑

n1=0

∞∑n2=0

w(n1)w(n2)G (2(n1 + n2); q)



n > 0

Recursion

w(n) =n−1∑m=0

w(m)w(n − 1−m)

Solution

convolution ⇒ z-transform ⇒ quadratic equation

⇒ w(z) =z

2−√

z2

4− zw(0)

(w(0) still to be determined)



n = 0

Boundary condition

w(0) = (q − q)∞∑

n1=0

∞∑n2=0

w(n1)w(n2)G (2(n1 + n2); q)

z space

w(0) =q−q

2πi

∮ {1

2q− 1

4q√

z

[√(√

z−1)2−4q +√

(√

z+1)2−4q

]}×

×

(1

2z−√

1

4z2−w(0)

z

)2

dz .

Self-consistency equation for w(0)



Evaluation the self-consistency equation

Molten phase

w(0) = 0 is always a solution

Interpretation: molten phase

However: only know that Q(k, n)/zN1 −→ 0; do not know Θ

Native phase

Non-zero solution only if branch cuts do not overlap

As w(0) → 1/[4(1 + 2√

q)2], integral converges to ≈ 4(1 + 2√

q)2

Non-zero solution only for q > qc ≈ q + 4(1 + 2√

q)2

Interpretation: native phase

There is a phase transition at finite q


Molten-native transition — Cayley tree Open problems

Open problems

Analytical argument that high temperature phase is molten

Position of phase transition (at q = 4 numerics says qc ≈ 80,analytical calculation says qc ≈ 96)

Order of phase transition (numerically appears very weak)

70 80 90~q4×10

-5

5×10-5

6×10-5

7×10-5

d2 (F/N

k BT

)/d

~ q2

k=13k=14k=15k=16k=17k=18k=19

10 20 30~q0

0,001

0,002

0,003

0,004

d2 (F/N

k BT

)/d~ q2

k=8k=9k=10


Summary

Outline of part II


6 Molten RNA



9 Summary


Summary

Summary of part II

The partition function of RNA can be calculated in polynomial time

Asymptotic behavior for homogeneous RNA can be calculated byanalytical methods

The partition function of molten RNA has a characteristic N−3/2

behavior

It takes a finite amount of sequence bias to enforce a native structure.

Branchiness of the native structure strongly affects the molten-nativephase transition


Part III

Sequence disorder


Outline of part III

10 Introduction

11 Numerical evidence

12 Analytical arguments for glass transition

13 Properties of the glass transition

14 Summary


Introduction

Outline of part III

10 Introduction




14 Summary


Introduction

Importance of sequence disorder

Homogeneous models only good at high temperatures

At low temperatures differences between bases matter

Still want to know thermodynamic behavior

Have to average over sequences

Need quenched average

F = −RT log Z


Numerical evidence

Outline of part III

10 Introduction

11 Numerical evidenceNumerical methodLow temperature propertiesEvidence for glass transition



14 Summary


Numerical evidence Numerical method

Zero temperature numerics

Observation

At T = 0 the free energy F equals the ground state energy E .

Definition

Let Ei ,j be the energy of the lowest energy structure for the RNA moleculestarting at base i and ending at base j .

Recursion equation

Ei ,j =j−1

mink=i

{Ei ,j−1,Ei ,k−1 + e(k, j) + Ek+1,j−1}

Calculate from shortest to longest substrands

O(N3) algorithm for arbitrary sequence

Repeat for many randomly chosen sequences


Numerical evidence Low temperature properties

Zero temperature properties

Properties

Return probability pi ,j ≡ Pr{i and j paired}∼ |i − j |−3/2 in molten phase

Size of structure 〈h〉 =∑N/2

i=1

∑Nj=N/2+1 pi ,j

∼ N1/2 in molten phase <h>

Findings−1.34l

return probabilityMuller et al., Europ. Phys. J. E (2003)

100 1000N

1

10

100

<h>

N0.7

size


Numerical evidence Low temperature properties

Is there a glass transition?

Observation

Thermodynamics at T = 0 different from molten phase.

Question

Is random RNA . . .

. . . always molten for T > 0?

. . . always glassy for T < ∞?

. . . glassy below a finite glass transition temperature Tc and moltenabove?


Numerical evidence Evidence for glass transition

Finite temperature numerics

Partition function numerics as a function of temperature

Singularity in second derivative of specific heat

Pagnani et al., Phys. Rev. Lett. (2000)

Very smooth transition at finite Tc


Analytical arguments for glass transition

Outline of part III

10 Introduction


12 Analytical arguments for glass transitionLimit on the molten phaseLimit on the glass phase


14 Summary


Analytical arguments for glass transition Limit on the molten phase

Strategy

Objective

Show that there is a finite temperature glass transition.

First step

Show that molten phase cannot be stable for all T > 0.

Approach

Assume that molten phase is stable for all T > 0 and show that this isinternally inconsistent.



Pinching energy

Object of interest

Pinching energy ∆F as afunction of sequence length N.

Direct calculation

Molten phase stable for all T

⇒ G (T ,N) ≈ A(T )N−3/2 exp[−f0(T )N]

⇒ ∆F = −RT logG (T ,N/2)G (T ,N/2)

G (T ,N)≈ 3

2RT log(N)



Lower bound

Best match

Find match of exactly complementary pieces

Can be found as long as N24−` > 1

Longest one has length ` = log N/ log 2.

Pinch free energy

Each base pair in exact match contributes ε0

Unpinched partition functionZunp ≥ exp[ ε0

RT `] exp[−(N − 2`)f0(T )]

Pinch free energy ∆F ≥ [ε0 + 2RTf0(T )]`



Putting it together

Combining everything

3

2RT ≥ [ε0 + 2RTf0(T )]/ log 2 (∗)

Observation

ε0 + 2R limT→0

[Tf0(T )] > 0

Conclusion

(*) must break down at some temperature T∗⇒ The molten phase cannot be stable below T∗⇒ The glass transition must happen at a temperature above T∗


Analytical arguments for glass transition Limit on the glass phase

Small disorder

Second step

Show that molten phase persists down to some finite critical temperature.

Approach

Understand molten (no disorder) phase⇒ introduce a little bit of disorder⇒ Treat disorder as a perturbation

Model

Simplified model: e(i , j) independent Gaussian random variables withmean ε0 and variance D.



Replica approach

n non-interacting RNAs in the same disorder (same “sequence”)

Partition function of one molecule: Z

Partition function of n molecules: Zn

Disorder average: Zn

Why care?

Expansion of quenched free energy −RT log Z in powers of disorder Dstarts as

−RT log Z = −RT log Z +RT

2

[Z 2 − Z

2]

+ O(D2)

⇒ Small n replica averaged partition functions determine smalldisorder behavior



Replicated partition functions

How do the replicated partition functions look like?

Z 1 is molten phase partition function (homogeneous model)

In general:Zn is partition function of n homogeneous RNA molecules that gainan extra energy D whenever two molecules form a common bond

7D



Calculating replicated partition functions

Single replica

Molten phase ⇒ already calculated

Arbitrary n

Open problem, only approximate calculations exist, see next subsection

Two replicas

Can be calculated exactly along the lines of what we did in part II

Too long to present here ⇒ only rough idea(see me if you want to get the details)



Two replica calculation

Order configurations of 2 replica system by common bonds

7D

Common bonds ( ) form RNA structurethemselves

represents sum over all possible choicesof non-common bonds in the two replicas

1 replica → `−3/2 ⇒ 2 replicas → (`−3/2)2 = `−6/2

effective picture: single RNA with“6-dimensional” loop entropies



Results from two replica calculatrion

Two replica results

phase transition at finite Dc

D < Dc :

Bubbles dominateZ 2 ∼ N−3

2 replicas fluctuate independently (molten)

D > Dc :

Common bonds dominateZ 2 ∼ N−3/2

2 replicas have same configuration (glass)

Consequences for glass transition

Molten phase perturbatively stable


Properties of the glass transition

Outline of part III

10 Introduction



13 Properties of the glass transitionCritical exponentsPhysical picture

14 Summary


Properties of the glass transition Critical exponents

Properties of the glass transition

Field theory of the glass transitionLassig & Wiese, Phys. Rev. Lett. (2006); David & Wiese, Phys. Rev. Lett. (2007)

Approach

Replica partition function

Expansion in strength of the disorder

Renormalization through Operator Product Expansion

Two loop calculation

n → 0 limit

Results

Critical exponents the same at transition and in glass phase

〈h〉 ∼ N0.64, pi ,j ∼ |i − j |−1.36 (very good agreement with numerics)


Properties of the glass transition Physical picture

Picture of the glass transition

Look at pairs of structures (Lassig & Wiese, Phys. Rev. Lett. (2006))

Above glass transition:

Structures are largely different

Bubbles of ground state structureappear

Bubble size diverges as transitionis approached

Below glass transition

Structures are largely locked intoground state structure

Bubbles of unlocked bases appear

Bubble size diverges as transitionis approached


Summary

Outline of part III

10 Introduction




14 Summary


Summary

Summary of part III

Random RNA is in a distinct glass phase at low temperatures

A finite temperature glass transition separates the glassy from themolten phase

The critical properties of the glass transition are perturbatively knownthrough an RG calculation

Exact results on the properties of the glass phase are still outstanding


Part IV

Force-extension experiments


Outline of part IV

15 Force-extension experiments

16 Quantitative modeling

17 Results for simple force-extension experiments

18 Nanopores

19 Summary


Force-extension experiments

Outline of part IV

15 Force-extension experimentsMotivationExperimental setup



18 Nanopores

19 Summary


Force-extension experiments Motivation

Experimental methods to determine RNA structure

Experimental methods

X-ray crystallography

Nuclear Magnetic Resonance

Biochemical evidence: protection essays

Correlated mutation analysis

Problem

RNA is very floppy ⇒ two RNA molecules rarely have the samethree-dimensional structure

Idea

Look at one RNA molecule at a time.


Force-extension experiments Experimental setup

Experimental setup

Attach ends of RNA molecule to two beads

Keep beads at fixed distance R with optical tweezers

Measure force f on beads as function of distance R

r


Force-extension experiments Experimental setup

Measurement

Subject

P5ab hairpin of Tetrahymena thermophila group I intron

Data

(Liphardt, Onoa, Smith, Tinoco, and Bustamante, Science, 2001)


Quantitative modeling

Outline of part IV


16 Quantitative modelingForce from free energySecondary structuresBackbonePutting it back together


18 Nanopores

19 Summary


Quantitative modeling Force from free energy

Calculating the force

r

Basic physics

energy = force · distance

Force from free energy

Need free energy F (r) at fixed end to end distance r ⇒ force f =∂F (r)

∂r



Partition function I

r

Free energy from partition function

F (r) = −RT lnZ (r) ⇒ need partition function Z (r) at fixed distance r

Partition function

Z (r) =∑

secondary structures S

∑polymer configurations P withdistance r and structure S

e−E [S]+E [P]

RT



Partition function II

r

Z (r) =∑


∑polymer configurations P withdistance r and structure S

e−E [S]+E [P]

RT

=∑


∑polymer configurations Pof the “exterior bases” withdistance r and structure S

e−E [S]+E [P]

RT

=N∑

m=0

∑secondary structures Swith m exterior bases


e−E [S]+E [P]

RT



Partition function III

r

Z (r) =N∑

m=0



e−E [S]+E [P]

RT

=N∑

m=0


e−E [S]RT

∑polymer configurations Pof m “exterior bases”with distance r

e−E [P]RT

=N∑

m=0


e−E [S]RT

∑

polymer configurations P of ssRNAwith m bases with distance r

e−E [P]RT



Partition function IV

Z (r) =N∑

m=0

Q(m)W (m, r)

withQ(m) ≡


e−E [S]RT

andW (m, r) ≡

∑polymer configurations P of ssRNAwith m bases with distance r

e−E [P]RT


Quantitative modeling Secondary structures

Recursion equation

Reminder Σ+=jj−1kikjj−1ii j k+1k−1

Zi ,j = Zi ,j−1 +

j−1∑k=i


RT Zk+1,j−1

Definition

Let Qj(m) be the partition function for the first j bases with exactly mexterior bases (

��

1 j).

Generalization

��

��

��Σ+=

1 j 1 j−1 j k 1 k j−1 jk−1 k+1

Qj(m) = Qj−1(m − 1) +

j−1∑k=1

Qk−1(m)e−e(k,j)RT Zk+1,j−1


Quantitative modeling Secondary structures

Properties of recursion equation

��

��

��Σ+=

1 j 1 j−1 j k 1 k j−1 jk−1 k+1

Qj(m) = Qj−1(m − 1) +

j−1∑k=1

Qk−1(m)e−e(k,j)RT Zk+1,j−1

Properties:

O(N3) complexity

Q(m) = QN(m)

Can be easily generalized to Turner parameters

Can be calculated by modifying the Vienna package(Hofacker, Fontana, Stadler, Bonhoeffer, Tacker, and Schuster,Monatshefte f. Chemie, 1994)


Quantitative modeling Backbone

Polymer physics

Need

W (m, r) =∑

polymer configurations P of ssRNAwith m bases with distance r

e−E [P]RT

Model

Elastic freely jointed chain

Persistence length 1.9nm/base distance 0.7nm

Partition function

Calculation of W (m, r) is standard polymer physics.


Quantitative modeling Putting it back together

RNApull

Putting it together

Q(m), W (m, r)

−→ Z (r) =N∑

m=0

Q(m)W (m, r)

−→ F (r) = −RT lnZ (r)

−→ f (r) =∂F (r)

r

Web server

http://bioserv.mps.ohio-state.edu/rna


http://bioserv.mps.ohio-state.edu/rna

Results for simple force-extension experiments

Outline of part IV



17 Results for simple force-extension experimentsHairpinA “real” moleculeWhy the force-extension curve is smooth?

18 Nanopores

19 Summary


Results for simple force-extension experiments Hairpin

Hairpin

Apply to P5ab hairpin of Tetrahymena thermophila group I intron

Quantitative agreement!


Results for simple force-extension experiments A “real” molecule

The full group I intron

The group I intron of Tetrahymena thermophila

Group I intron contains pseudo-knot!

Quantitative modeling ignores pseudo-knot⇒ known inactive conformation(Pan and Woodson, J. Mol. Biol., 1998)

G AGCCUUGC

AAAGGGUA

U AG C

CGU A

AU

AG U

GCU AG CA

U GA UG CA UG CU AU GU GC GA GAA

UUGCC

CGG

GACAACA

A

A

GGG

CUCUCUAAA

AGGUUUCCA

UUUAU

UAG

CAAU A

G CG UU GC G

GCGU AA

AAGCGGCGU

CACGCAG UCCAAG UU AC G

UA

U

AG UU AC GA UA UC GA U

A

A

G CA UUCU

AAA

AUACGGUAUUAGUGCUACGCGAC

GUAC

U AC G

C UA UG CA

AC

GAC

CUGAU

G

UAA

AUGUCG

AUA

AAU

A

GUC

GGAAGA

GCUCUUCU

UAUG

U

G

G

CG

AAC GG UG CA UA UA UAU AU AU AG CG CC A

CAU

AUA

GUUAG

A

CAGCGUCA

A

CUCCU

AG GGGUAAU

CUAGCGG

GGUCGCCGAG

GUCAC

AUGAAGUG

AU

GCAA

GAACUAAUUUGUAUGCGAAAGUAUAUUGAUUAGUUU

UGGAGU

ACUCG

P5b

P5aP5

P4

P6a

P6b

P5c

P2

P1

P7

Alt P3

P2.1

P9

P9.1

P9.1a

P9.2

5’3’

P8

Computational result

No sign of secondary structure!

0 100 200 300<R> [nm]

0

10

20

30

40

<F

> [

pN]


Results for simple force-extension experiments Why the force-extension curve is smooth?

What’s happening?

Intermediate structure

Look at intermediate structure (here r =100nm)

Like “Socks on the clothes line”

CUCUCU

AAAUAG

CAA

UAUUUAC

CUU

UGGA G

GGAA

AAG

UU

AU

CA

GGC

AUGC

A CC

UG

GU

AG

CUA

GU

CU

UU

AA

AC

CA

AU

AGAUU

GC A U CGGUUUAAAAGGCAA@@@@

@@@ A A A U U G C G G G A A A G GGGUCAA

C AG C CGUUCAG

U A C C A AGU

CU

CA

GG

GGA

A AC

UU

UG

AG

AU

G

GC

CU

UGC

AAAGGG

UA UGG

UAA

U AA

GCUG

ACGGAC

AU

GGUCCU

AACCAC

GCAGCCAAG

UCC

UA

A GUCAACAGAU C

UU

CU

GU

UG

AUA

UGG

AU

G

C

AG

UUC A

CAG

AC

UA A A U@@@@@@@GG G G A A G A

U GUAUUCUUCUCA U A A G A U A U AGUCGGACCUC

U C C U UAAU

GGGAGC

UA

GC

GG A U

GAA

GU

G AUG

CAAC

AC

UGGA

GC

CG

CU

GG

GAACUAAUU

UGUAUGC

GAA

AG

UA

UA

UUG

AU

UA

GU

UUU

GGAGUAC

UCG


Results for simple force-extension experiments Why the force-extension curve is smooth?

What’s happening?

Compensation effect

Extension r is increased

One of the “socks” disappears

The other “socks” take up the slag

⇒ No rapid change in force as sock disappears

⇒ smooth force-extension curve


Nanopores

Outline of part IV




18 NanoporesIntroductionForce-extension experiments

19 Summary


Nanopores Introduction

Nanopores

What is a nanopore?

A nanopore

is a little hole

is so small that only single-stranded but no double-stranded RNA canpass through it

can be placed between two chambers such that there is only onenanopore connecting the chambers


Nanopores Introduction

Nanopores

Types of nanopores

natural ion channel (α-hemolysin)(Meller, J. Phys. Cond. Mat., 2003)

solid state pore(Storm et al., Nature Mat., 2003)


Nanopores Force-extension experiments

Combining nanopores with force-extension experiments

Suggestion

Combine a nanopore with a force-extension setup

n nucleotides

r

f

"cis" "trans"

Prediction

The force will rise at every structural element

The structural elements will break in their order along the sequence



Simulation approach

n nucleotides

r

f

"cis" "trans"

Simulation

Fix r(t) to be linear (set by experiment)

Degree of freedom: number n of base in the pore

At each time n can

increase:−→ calculable gain in mechanical energy on the right−→ potentially loss in binding energy calculable from E [S ]decrease:−→ calculable loss in mechanical energy on the right

Monte Carlo simulation (see also part V)



Simulation result

Apply to group I intron of Tetrahymena thermophila

0 50 100 150 200 250 300<R> [nm]

0

10

20

30

40

50

<f>

[pN

]

n=50 84 162 357

P9.1P9

P9.2 P6

200 238 310

P4-P5aP5bc

387

P1P2

P2.1

100

P7

Alt P3

P8

G AGCCUUGC

AAAGGGUA

U AG C

CGU A

AU

AG U

GCU AG CA


UUGCC

CGG

GACAACA

A

A

GGG

CUCUCUAAA

AGGUUUCCA

UUUAU

UAG

CAAU A

G CG UU GC G

GCGU AA

AAGCGGCGU

CACGCAG UCCAAG UU AC G

UA

U

AG UU AC GA UA UC GA U

A

A

G CA UUCU

AAA

AUACGGUAUUAGUGCUACGCGAC

GUAC

U AC G

C UA UG CA

AC

GAC

CUGAU

G

UAA

AUGUCG

AUA

AAU

A

GUC

GGAAGA

GCUCUUCU

UAUG

U

G

G

CG

AAC GG UG CA UA UA UAU AU AU AG CG CC A

CAU

AUA

GUUAG

A

CAGCGUCA

A

CUCCU

AG GGGUAAU

CUAGCGG

GGUCGCCGAG

GUCAC

AUGAAGUG

AU

GCAA

GAACUAAUUUGUAUGCGAAAGUAUAUUGAUUAGUUU

UGGAGU

ACUCG

P5b

P5aP5

P4

P6a

P6b

P5c

P2

P1

P7

Alt P3

P2.1

P9

P9.1

P9.1a

P9.2

5’3’

P8

Signature of every structural element

Can extract sequence position of stalling sites



Structure reconstruction I

Repeat pulling in opposite direction

10

20

30

40

50

<f>

[pN

]

0 50 100 150 200 250 300<R> [nm]

0

10

20

30

40

50

<f>

[pN

]

n=43 62

n=50 84 115 162 200 233

112 169 226 289 317

253 331 389

268

⇒

CUCUCUAAAUAGCAAUAUUUACCUUUGGAGGGAAAAGUUAUCAGGCAUGCACCUGGUAGCUAGUCUUUAA...<.......................>........<......<........>.........<.......

(((((((((((....))......)))))))))...((((((((((......)))))))))).((((((((ccccccccc..............ccccccccc...aaaaaaaaaa......aaaaaaaaaa.gggggg..

ACCAAUAGAUUGCAUCGGUUUAAAAGGCAAGACCGUCAAAUUGCGGGAAAGGGGUCAACAGCCGUUCAGU.................>.........>..............<..<.....<.........<........

(((.((.......)).))))).)))))).............((((((....((((((....(((.(((((......................gggggg.............jjjj......kkkkkk....iiii.....

5’−

ACCAAGUCUCAGGGGAAACUUUGAGAUGGCCUUGCAAAGGGUAUGGUAAUAAGCUGACGGACAUGGUCCU.......................>.....<.........>.....>..........>..>........>.

(((..(((((((((....)))))))))..(((.....)))....)))....).)))))))...))))))..............................llll...llll................iiii...kkkkkk.

AACCACGCAGCCAAGUCCUAAGUCAACAGAUCUUCUGUUGAUAUGGAUGCAGUUCACAGACUAAAUGUCG....>..>.<......<......<................>....>............<...........

..)).))))((...((((...((((((((.....))))))))..))))...))...(.(((((..((((.

.....jjjj.............dddddddd...dddddddd.................hhhhh.......

GUCGGGGAAGAUGUAUUCUUCUCAUAAGAUAUAGUCGGACCUCUCCUUAAUGGGAGCUAGCGGAUGAAGU......<..<..........>..............>.<......<........>.......<......<.

....(((((((.....)))))))....))))))))))((((.(((((....)))))(((((((....(((

....eeeeeee.....eeeeeee........hhhhh......ffff......ffffbbbbbbb.......

GAUGCAACACUGGAGCCGCUGGGAACUAAUUUGUAUGCGAAAGUAUAUUGAUUAGUUUUGGAGUACUCG

...............bbbbbbb...............................................

−3’(......))))....))))))).((((((((.((((((....)))))).))))))))..)).....)).

..................>........................>...........>....>........

Match stalling sites by sequence comparison ⇒ reconstruct structure



Structure reconstruction II

UGUCG

AUA

G

3’

P9

P9.1

P9.2

U

GUAAU

AG

GUCAC

AUGAAGUG

AU

GCAA

U

AGUAUAUGAUUAGUUU UU

GCC GAG

UACUA

G CG

P9.0

P9.1a

P8UAUG

U

AC

GUAC

AAA

AUAUAUAGCGC

AUC

AU

A

AGA

GC

UU

AAGACCGUCAAAU

P2

P2.1

A

AU

U CA

GCA

AUAUU

C

5’

P1

P7AG C

AA

Alt P3A

A

UC

GUA G

P5b

P5aP5

AA

GCGCA

U AGA

A

A

U AG C

CGU A

UGC

U AG C


GACAACA

GAA

AUA

G

UP5c

GACG G

A

P6b

P6a

G

G UU AC GC G

UA

U

A UA

U UC

CC

AA

GC

AGU

U

P4CA

C GU GC GU AC GU GA UA UA U

A UCG

U GAU

A UU G

GCA UG CG C

G CU GC GU AU AU A

GCGUA

CGC

GCC G

GUG U

CGU AG C

G CC G

UA U

G

C

G

C

A

U

A

U

A U

GU AC GA UA UC GA UG C

GCU AG CA UU A

C GU G

GCU AU AC GU A

C

GA

UC

GG

C

C

AA

GCGUAUGUAG ACUAA U

U

GCUGGC

UAGCGG C

G

3’

P9

P9.1

P9.2

CUCCU

GAAU

GAG

GUCAC

AUGAAGUG

AU

GCAA

GAACUAAUUU

UAUAUUGAUUAGUUU UU

GCC GAG

UACUA

G CG

P9.0

P9.1a

P8UAUG

U

AC

GUAC

AAA

AUAUAUAGCGC

AUC

AU

A

AGA

GC

UU

A

P2

P2.1

A

AU

U CA

GCA

AUAUU

C

5’

P1

P7AG C

GAU

A

AA

UG

AGAUA

U

P3

C

A

UAAA

P5b

P5aP5U A

G CG UU GC G

GC

AA

GCGCA

U AGA

A

A

U AG C

CGU A

UGC

U AG CA


UGCC

CGG

GACAACA

U

GAA

AUA

G

UP5c

GACG G

A

P6b

P6a

G

G UAGG

UAAG U

U UC

CC

AA

GC

AGU

U

P4CA

U

C GGGAGGUUU

UCUCUAAA

GGACUAUUGA U

CGA

CCUGGU

GUCUUU

CGGAAA

GACCG

CUGGC

UGCG

CA

GC

G

CC

GA

U

A

U

U AC G

UAA U

GCA U

CGA U

AUC

CU

GACUA

CUGAU

CUCUUCU

GGGAAGA

GGCGAUC

CCGCUGG

AGGGU AAAG

GCGGUAU

Overall structure reconstructed correctly

Can distinguish different structures on the same sequence

Even pseudoknot reconstructed

“Just” needs to be implemented experimentally . . .


Summary

Outline of part IV




18 Nanopores

19 Summary


Summary

Summary of part IV

Quantitative description of single-molecule experiments possible

Force-extension curves do not reveal secondary structure informationdue to compensation effects

Single-molecule experiments reveal structure information with thehelp of a nanopore


statistical physics of rna folding · statistical physics of rna folding ralf bundschuh...

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