The document summarizes a seminar on the physics of DNA, RNA, and RNA-like polymers. It discusses how DNA drives the initial infection process in bacteriophage due to its stiff and self-repelling properties. It also describes how RNA forms secondary structures through base pairing between nucleotides and how the 5' and 3' ends spontaneously associate in long RNA molecules, bringing them into close proximity. Finally, it compares models of RNA secondary structure, such as the randomly self-paired polymer model and successive folding model, and how they predict different scaling laws for end-to-end distance with polymer length.

1. Exit Seminar:
Physics of
DNA, RNA, and RNA-like Polymers
Li Tai Fang
Department of Chemistry & Biochemistry
UCLA

2. Bacteriophage: DNA as genome
● DNA is a stiff,
self-repelling polymer
measure length:
● Capsid is highly gel electrophoresis
pressurized LamB
● DNA is released from
capsid upon binding
LamB
DNase

3. Entropic Pulling Force
no pulling force:
unaware of an
outside world
entropic
pulling
force
DNase

4. 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

5. RNA
a biopolymer
consisting of 4
different species of
monomers (bases):
G, C, A, U
G–C
A –U secondary
G–U structure
3'
5'

6. generic vs. sequence-specific properties
● Regardless of sequence or length, we can
predict
●
Pairing fraction:  60%
●
Average loop size:  8
● Average duplex length:  4

7. 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

8. Association of 5' – 3' required for:
● Efficient replication ● Efficient translation
of viral RNA of mRNA
complementary RNA binding
sequence protein
e.g.,
HIV-1, Influenza, Sindbis, etc.

9. 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

10. Circle Diagram

11. Circle Diagram

12. Circle Diagram
● 60% of bases are paired
● duplex length ≈ 5
● Inspired the “randomly
self-paired polymer”
model

13. randomly self-paired polymer
e.g.,
NT = 1000 NT,eff = 520
Np = 600 Np,eff = 120

14. 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 – pj)·(1 – pk)· … ·(1 – pz)
(X) = P(X) – P(X–1) = probability Ree is X
X = X (X) · X

15. preview of the results:
X

16. End-to-End Distances:
● flexible or worm-like polymers: X N1/2
– dsDNA, denatured RNA
●
randomly self-paired polymers: X N 1/4
– RNA-like polymer
● SuccessiveFold/MFold/Vienna: X N 0
– RNA folding algorithms (no pseudoknot)

17. Let's start the grunt work
Reminder:
RNA: Model:
NT = 1000 NT,eff = 520
Np = 600 Np,eff = 120
st
Now, the 1 challenge:

18. probability of a particular set of pairs
i j k l m n
p(i) = 120/520
p(i­j) = 1 /519 = p (this partial set)
p(k) = 118/518
= p(i)  p(i – j)  p(k)  p(k – l)  p(m)  p(m – n)
     
p(k­l) = 1 /517
p(m) = 116/516
p(m­n) = 1 /515 depends on NT,eff, Np,eff, and B

19. Next challenge:
● We have pi = p(NT,eff, Np,eff, B)
● We want P(X) = 1 – (1 – pi)·(1 – pj)·(1 – pk)· … ·(1 – pz)
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
x1 x2 x3 x4
B = 3:
i j k l m n

20. Task: find (B)
● 1st, find the number of sets {x1, x2, …, xB+1},
such that X = x1+ x2+ … + xB+1
● for B = 3, X = 10: # of ways to arrange these:
X+B (X+B)!
=
B X!  B!

21. For each {xi}, how many ways to move the
middle regions?
vs.
i j k l i j k l
Navailable NT,eff – X – B – 1
=
B–1 B–1

22. Consider all X's
X
 X+B
B
NT,eff – X – B – 1
B–1
Xi=0
Missing something...... base-pairing “crossovers:”
(a) (b) (c) vs. (a) (b) (c)
i j k l i j k l

23. Crossovers are also known as pseudoknots
● X = xa + xb + 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

24. ( NT,eff , X, B )
X
(B – 1)
= 2   X+B
B
NT,eff – X – B – 1
B–1
Xi=0

25. Once again, the general approach
where end-to-end distance  X
P(X) = at least one of these pairs will occur
P(X) = 1 – (1 – pi)·(1 – pj)·(1 – pk)· … ·(1 – pz)
P(X) = 1 – (1 – pB=1) B=1 · (1 – pB=2) B=2 · … · (1 – pBmax) Bmax
●
(X) = P(X) – P(X–1)

26. (X)

27. X
X =  X (X) · X

28. X

29. Problems:
● Pseudoknots are rare in RNA
● Not held in check in the self-paired
polymer model
● Successive RNA Folding Model:
● Pseudoknots completely prohibited

30. Successive Folding Model:
– created by Prof. Avi Ben-Shaul

31. randomly self-paired polymer Successive Fold

32. End-to-End Distances:
– generic feature
● flexible or worm-like polymers: X N1/2
– dsDNA, denatured RNA
●
randomly self-paired polymers: X N 1/4
– RNA-like polymer
● SuccessiveFold/MFold/Vienna: X N 0
– RNA folding algorithms (no pseudoknot)

33. 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

34. For an RNA of N = 1000, pairing fraction = 0.6
Probability that the ends will be no more than 20 unpaired bases apart?
B = 1:
p = (120/520) (1/519) = 1/2249 = 4.45x10-4
 = 231
P(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
Prob (X 20) = 1 – (0.902  0.706  0.635  0.725 …  1) = 0.81