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The End-to-End Distance of RNA
as a Randomly Self-Paired Polymer


               Li Tai Fang
      Department of Chemistry & Biochemistry
                      UCLA
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'
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             ●   Efficient translation
    of viral RNA                          of mRNA


                complementary                         RNA binding
                sequence                              protein




    e.g.,
    HIV-1, Influenza, Sindbis, etc.
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, 2010
Circle diagram
Circle diagram
Circle Diagram


●   60% of bases are paired
●   duplex length ≈ 4
●   Inspired the “randomly
    self-paired polymer”
    model
randomly self-paired polymer




    e.g.,
    NT = 1000       NT,eff = 550
    Np = 600        Np,eff = 150
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 R ee is X

X = X (X) · X
preview of the results:




                  Fang, L. T., J. Theor. Biol., 2011
Let's start the grunt work
Reminder:
RNA:               Model:
NT = 1000          NT,eff = 550
Np = 600           Np,eff = 150

            st
Now, the 1 challenge:
probability of a particular set of pairs



     i         j    k          l             m                       n




 p(i)   = 150/550
 p(i­j) =  1 /549       = p (this partial set)
 p(k)   = 148/548
                        = p(i) p(i – j) p(k) p(k – l) p(m) p(m – n)
 p(k­l) =  1 /547
 p(m)   = 146/546
 p(m­n) =  1 /545         depends on NT,eff, Np,eff, and B
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
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!
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
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
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
( NT,eff , X, B )
                     X


= 2   (B – 1)
                         X+B
                            B
                                 NT,eff – X – B – 1
                                         B–1
                    Xi=0
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)
Probability distribution of end-to-end distances




                            <X> = 14.4




                                  Fang, L. T., J. Theor. Biol., 2011
end-to-end distance vs. sequence length




               X =  X (X) · X




                              Fang, L. T., J. Theor. Biol., 2011
1/4
Scaling law: <X> ~ N




                Fang, L. T., J. Theor. Biol., 2011
Once again:
●   The ends of a self-paired polymer, such as
    RNA, are always in close proximity.
    ●   This is a generic feature.

●   Comparison of end-to-end distances:
    ●   random or worm-like polymers: X N1/2
    ●
        randomly branching polymers: X N1/4
    ●   randomly self-paired polymers: X N1/8
Acknowledgment
●   Thesis advisors
    ●   Professors Bill Gelbart and Chuck Knobler

●   Thanks to
    ●   Professor Avinoam Ben-Shaul @ Hebrew University of
        Jerusalem

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The End-to-End Distance of RNA as a Randomly Self-Paired Polymer

  • 1. The End-to-End Distance of RNA as a Randomly Self-Paired Polymer Li Tai Fang Department of Chemistry & Biochemistry UCLA
  • 2. 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'
  • 3. 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
  • 4. 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.
  • 5. 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, 2010
  • 8. Circle Diagram ● 60% of bases are paired ● duplex length ≈ 4 ● Inspired the “randomly self-paired polymer” model
  • 9. randomly self-paired polymer e.g., NT = 1000 NT,eff = 550 Np = 600 Np,eff = 150
  • 10. 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 R ee is X X = X (X) · X
  • 11. preview of the results: Fang, L. T., J. Theor. Biol., 2011
  • 12. Let's start the grunt work Reminder: RNA: Model: NT = 1000 NT,eff = 550 Np = 600 Np,eff = 150 st Now, the 1 challenge:
  • 13. probability of a particular set of pairs i j k l m n p(i)   = 150/550 p(i­j) =  1 /549 = p (this partial set) p(k)   = 148/548 = p(i) p(i – j) p(k) p(k – l) p(m) p(m – n) p(k­l) =  1 /547 p(m)   = 146/546 p(m­n) =  1 /545 depends on NT,eff, Np,eff, and B
  • 14. 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
  • 15. 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!
  • 16. 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
  • 17. 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
  • 18. 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
  • 19. ( NT,eff , X, B ) X = 2 (B – 1)   X+B B NT,eff – X – B – 1 B–1 Xi=0
  • 20. 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)
  • 21. Probability distribution of end-to-end distances <X> = 14.4 Fang, L. T., J. Theor. Biol., 2011
  • 22. end-to-end distance vs. sequence length X =  X (X) · X Fang, L. T., J. Theor. Biol., 2011
  • 23. 1/4 Scaling law: <X> ~ N Fang, L. T., J. Theor. Biol., 2011
  • 24. Once again: ● The ends of a self-paired polymer, such as RNA, are always in close proximity. ● This is a generic feature. ● Comparison of end-to-end distances: ● random or worm-like polymers: X N1/2 ● randomly branching polymers: X N1/4 ● randomly self-paired polymers: X N1/8
  • 25. Acknowledgment ● Thesis advisors ● Professors Bill Gelbart and Chuck Knobler ● Thanks to ● Professor Avinoam Ben-Shaul @ Hebrew University of Jerusalem