dissertation defense: the physics of dna, rna, and rna-like polymers
DESCRIPTION
Oral Defense of my DissertationTRANSCRIPT
Exit Seminar:
Physics ofDNA, RNA, and RNA-like Polymers
Li Tai Fang
Department of Chemistry & BiochemistryUCLA
● DNA is a stiff,
self-repelling polymer● Capsid is highly
pressurized● DNA is released from
capsid upon binding LamB
Bacteriophage: DNA as genome
DNase
LamB
measure length:gel electrophoresis
Entropic Pulling Force
DNase
no pulling force:unaware of anoutside world
entropicpullingforce
Generic properties of DNA- independent of sequence
● Stiff and self-repelling● persistence length radius of capsid
● contour length diameter of capsid
● Confinement● entropy outside entropy inside
● Physical properties of DNA drive the initial infection process
RNA
a biopolymer consisting of 4 different species of monomers (bases): G, C, A, U
GAG
secondarystructure
–––
CUU
5'3'
● Regardless of sequence or length, we can predict● Pairing fraction: 60%
● Average loop size: 8
● Average duplex length: 4
generic vs. sequence-specific properties
generic vs. sequence-specific properties
● Regardless of sequence or length, we can predict● Pairing fraction: 60%
● Average loop size: 8
● Average duplex length: 4
● 5' – 3' distance
Association of 5' – 3' required for:
● Efficient replication of viral RNA
● Efficient translation of mRNA
e.g., HIV-1, Influenza, Sindbis, etc.
complementary sequence
RNA bindingprotein
Question:How do the 5' and 3' ends of long RNAs find each other?Answer:The ends of RNA are always in close proximity, regardless of sequence or length !
Yoffe A. et al, 2009
Circle Diagram
Circle Diagram
Circle Diagram
● 60% of bases are paired
● duplex length ≈ 5
● Inspired the “randomly self-paired polymer” model
randomly self-paired polymer
e.g., N
T = 1000
Np = 600
NT,eff
= 520
Np,eff
= 120
general approach
1) pi = probability that the ith set of “base-pair(s)”
-------will bring the ends to less than/equal to X
2) P(X) = at least one of those sets will occur
= 1 – (1 – pi)·(1 – p
j)·(1 – p
k)· … ·(1 – p
z)
(X) = P(X) – P(X–1) = probability Ree
is X
X = X (X) · X
preview of the results:
X
End-to-End Distances:
● flexible or worm-like polymers: X N1/2
– dsDNA, denatured RNA
● randomly self-paired polymers: X N1/4
– RNA-like polymer
● SuccessiveFold/MFold/Vienna: X N0
– RNA folding algorithms (no pseudoknot)
● flexible or worm-like polymers: X N1/2
– dsDNA, denatured RNA
● randomly self-paired polymers: X N1/4
– RNA-like polymer
● SuccessiveFold/MFold/Vienna: X N0
– RNA folding algorithms (no pseudoknot)
Let's start the grunt work
RNA:N
T = 1000
Np = 600
Model:N
T,eff = 520
Np,eff
= 120
Reminder:
Now, the 1st challenge:
probability of a particular set of pairs
i j k l m n
p(i) = 120/520p(ij) = 1 /519p(k) = 118/518p(kl) = 1 /517p(m) = 116/516p(mn) = 1 /515
= p (this partial set)
= p(i)p(i – j) p(k) p(k – l) p(m) p(m – n)
depends on NT,eff
, Np,eff
, and B
Next challenge:
● We have pi = p(N
T,eff, N
p,eff, B)
● We want P(X) = 1 – (1 – pi)·(1 – p
j)·(1 – p
k)· … ·(1 – p
z)
Let (B) = number of ways to make a set of pairs
Then, P(X) = 1 – (1 – pB=1
)B=1 · (1 – pB=2
)B=2 · … · (1 – pBmax
)Bmax
i j k l m n
B = 3: x
1 x
2 x
3 x
4
Task: find (B)
● 1st, find the number of sets {x1, x
2, …, x
B+1},
such that X = x1+ x
2+ … + x
B+1
● for B = 3, X = 10: # of ways to arrange these:
X + B ( X + B ) !
B X! B!=
For each {xi}, how many ways to move the
middle regions?
i j k l i j k l
vs.
Navailable
B – 1N
T,eff – X – B – 1
B – 1=
Consider all X's
X + B B
NT,eff
– X – B – 1
B – 1
X
X
i=0
Missing something...... base-pairing “crossovers:”
vs.
i j k l i j k l
(a) (b) (c) (a) (b) (c)
Crossovers are also known as pseudoknots
● X = xa + x
b + x
c
as long as xb j – i
____ and xb l – k
● 2 ways to connect each middle region
● undercount by 2(B – 1)
Now, let's put it all together
X + B B
NT,eff
– X – B – 1
B – 1
X
X
i=0
= 2(B – 1)
( NT,eff
, X, B )
Once again, the general approach
● pi = probability that the ith set of “base-pair(s)”
bringing together the ends to Ree
X will
occur● P = none of those pair(s) actually occurs
= (1 – pi)·(1 – p
j)·(1 – p
k)· … ·(1 – p
z)
● Prob(X) = [1 – P(X)] – [1 – P(X–1)]
● X = X Prob(X) · X
where end-to-end distance X
P(X) = at least one of these pairs will occur
P(X) = 1 – (1 – pi)·(1 – p
j)·(1 – p
k)· … ·(1 – p
z)
P(X) = 1 – (1 – pB=1
)B=1 · (1 – pB=2
)B=2 · … · (1 – pBmax
)Bmax
● (X) = P(X) – P(X–1)
(X)
X
X = X (X) · X
X
Problems:
● Pseudoknots are rare in RNA● Not held in check in the self-paired
polymer model
● Successive RNA Folding Model:● Pseudoknots completely prohibited
Successive Folding Model:– created by Prof. Avi Ben-Shaul
randomly self-paired polymer Successive Fold
End-to-End Distances:– generic feature
● flexible or worm-like polymers: X N1/2
– dsDNA, denatured RNA
● randomly self-paired polymers: X N1/4
– RNA-like polymer
● SuccessiveFold/MFold/Vienna: X N0
– RNA folding algorithms (no pseudoknot)
Acknowledgment● Thesis advisors
Professors Bill Gelbart and Chuck Knobler
● Special thanks to
Professor Avi Ben-Shaul
● Thesis committee
Professors Joseph Loo, Giovanni Zocchi, Tom Chou
● Group members and former group members:
Aron Yoffe, Ajay Gopal, Odisse Azizgolshani, Peter Prinsen, Ruben Cadena, Cathy Jin, Maurico Comas-Garcia, Rees Garmann, Peter Stavros, Vivian Chiu Glover, Venus Vakhshori, Yufang Hu, Roya Zandi
B = 1:p = (120/520) (1/519) = 1/2249 = 4.45x10-4
= 231P(1) = (1 – 4.45x10-4)231 = 0.902
B = 2:p = (120 x 118) / (520x519x518x517) = 1.96x10-7
= 1.78 x 106
P(2) = (1 – 1.96x10-7)1.78E6 = 0.706
B = 3:p = 8.55x10-11
= 5.301x109
P(3) = (1 – 8.55x10-11)5.301E9 = 0.635
B = 4:p = 3.70x10-14
= 8.72x1012
P(4) = (1 – 3.70x10-14)8.72E12 = 0.725
For an RNA of N = 1000, pairing fraction = 0.6Probability that the ends will be no more than 20 unpaired bases apart?
Prob (X 20) = 1 – (0.902 0.706 0.635 0.725 … 1) = 0.81