FINAL POWERPOINT

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FINAL POWERPOINT Powered By Docstoc
					Design Optimization for
DNA Nanostructures III:
Realizable Graphs

   Talk By: Brian Goodhue, Daniel Koch
   Saint Michael’s College
   Collaborators: Jacob Girard, Andrew Parent, Mary Spuches, Thomas
   Dickerson, Andrew Gilbert, Dan Lewis
Outline
 Review Problem
 Define Octet Truss
 Two Arm Tile Constructions
 Three Arm Tile Constructions
 Four Arm Tile Constructions
 Five and Six Arm Tile Constructions
 Seven and Greater Arm Tile Constructions
 Conclusion
Problem Statement
 Finding structures within the octet truss
 Purpose : optimize the design so that DNA
  nanostructures will be more efficient to
  produce
 Approach : Minimize cost of producing
  DNA complexes by minimizing the number
  of different molecular building blocks
Summary of Design Constraints
 Arms are straight and rigid.
 The positions of the arms are fixed.
 The arms do not experience twist strain.
 No molecule has more than twelve arms
  or less than two arms.
 Final DNA structures must be complete.
 No design may allow structures smaller
  than the target structure to form.
A vertex and its neighborhood in
the Octet Truss
 Composed of three intersecting planes
 Each plane contains 4 arms w/ 90 degree
  angles between consecutive arms
 Planes identified as alpha, beta and
  gamma
 We will try to arrange the tiles in
  lexicographical minimal form starting with
  the alpha plane
A Vertex in the Octet Truss
Coordinates (if vertex is at origin)
Arm       X Vector   Y Vector   Z Vector
Alpha 1   0          1          0
Alpha 2   -1         0          0
Alpha 3   0          -1         0
Alpha 4   -1         0          0
Beta 1    -1/2       1/2        -sqrt(2)/2
Beta 2    -1/2       1/2        sqrt(2)/2
Beta 3    1/2        -1/2       sqrt(2)/2
Beta 4    1/2        -1/2       sqrt(2)/2
Gamma 1   1/2        1/2        -sqrt(2)/2
Gamma 2   1/2        1/2        sqrt(2)/2
Gamma 3   -1/2       -1/2       sqrt(2)/2
Gamma 4   -1/2       -1/2       -sqrt(2)/2
Orientation
 We need to be able to describe the
  orientation at which two tiles connect
 We want to calculate the bond angle
  between the two tiles
                  c      ĉ.
    Tile A                           Tile B



                  c       ĉ.
    Tile A                             Tile D
Orientation Theorem
Let T1 and T2 be two tiles and let Ɛ1, Ɛ2 be
  the elements of {α1,...,α4, β1,...,β4,
  γ1,...γ4}, representing the arms where T1
  and T2 join. Let let σi be the
  lexicographical minimal arm of Ti,
  omitting Ɛi and its antipodal arm. Write
  P(Ɛi,σi) for the plane through Ɛi and σi.
  Then the bond angle of (T1, Ɛ1), (T2, Ɛ2)
  is the angle formed between P(Ɛ1, σ1)
  and P(Ɛ2, σ2).
Orientation
   This theorem allows us to make sure tiles
    are oriented properly with respect to one
    another in the final construct.
Convex Theorem
Theorem: If G is a complete complex
 constructed from rigid tiles types, then at
 least one of the tiles must have a
 geometric configuration such that the
 convex hull Hv formed by the vertex v and
 the end points of the half edges has v as a
 corner point.
What does this mean to us?
Two Armed Tile Type
Configurations
 There are only four possible unique two
  armed tile types.
 The four are two arms with the angles:
   π/3 radians
   π/2 radians
   2π/3 radians
   π radians
Structures
 Interior Angle Formula: (a)(n) = (n-2) π,
   where n = # of sides, a = angle measure
 π/3 radians: triangle
 π/2 radians: square
 2π/3 radians: hexagon
 π radians: never forms a complete
  structure
 Any structures that leave the plane would
  create a spiral and never form a complete
  structure
Three Armed Tile Type
Configurations
   10 different tile types given by program
   We will go through each tile individually to see if
    a complete structure can be created
   We will create a list of all the possible tile type
    partnerships of sticky and cohesive ends
              Parity Theorem
   If the tile has an odd number of arms, then
    there needs to be two tile types in order to
    create a complete structure.

    Example: If the tile type has 2 ‘a’ ends and
    1 compliment ‘a*’ end, then it would never
    complete. There needs to be a tile with 1
    ‘a’ and 2 ‘a*’ or something similar.
Three Armed Tile Type
Configurations
Tiles with only ‘a’ and ‘a*’ sticky ends
  r1A 11 + r2 A 12 = r1A + r2 A
                      *
                      11
                                  *
                                  12

 r1 = quantity of the base tile we start with
 r2 = the quantity of its partner tile
 A11 = number of ”a” ends on r1
 A11* = number of of ”a*” ends on r1
 A12 and A12* = number of ”a” and ”a*” on r2
  respectively
Example
Tiles with more than just ‘a’’s
      r1B11 + r2B12 = r B + r2B
                        *
                      1 11
                                *
                                12

   We solve for this equation simultaneously
    with the first the equation with ‘a’ ends.

      r1C11 + r2C12 = r1C + r2C
                          *
                          11
                                     *
                                     12

 Solve all three with each other in mind
 Only possible result is (a,b,c) and
  (a*,b*,c*)
Tile Type Arm Partnerships
       for 3 arm tiles
     Tile 1      Tile 2
       a3          a*3
       a3         a,a*2
     a2,a*         a*3
     a2,a*        a,a*2
      a2,b       a*2,b*
     a,a*,b      a,a*,b
     a,a*,b       b*2,b
     a,a*,b        b*3
      a,b,c      a*,b*,c*
                              Cross Check
r1        a3       a3       a2,a*     a2,a*     a2,b     a,a*,b    a,a*,b   a,a*,b     a,b,c

r2        a*3     a,a*2     a*3       a,a*2    a*2,b*    a,a*,b*   b*2,b     b*3      a,*,b*,c*

α2, β1    x        x         x         x         x         x         x        x          x

α2, β2    x        x         x         x         x         x         x        x          x

α4, β1    x        x         x         x         x         x         x        x          x

α4, β3   TrOct   TrOct     TrOct     TrOct     TrOct     TrOct     TrOct    TrOct      TrOct

α2, α3    P        P         P         P         P         P         P        P          P

α3, β1    P        P         P         P         P         P         P        P          P

β1, γ1    o      Tetra     Tetra      Tetra      o         o         o       Tetra       o

β1, γ3    o      TrTetra   TrTetra   TrTetra   TrTetra   TrTetra   TrTtra   TrTetra      o

β3, γ4    P        P         P         P         P         P         P        P          P

β1, γ2    x        x         x         x         x         x         x        x          x
Four Arm Tile Types
            Non-Convex            Convex
    α2, γ3, γ4      α3, β1, β3   α2, β1, γ4
    α2, α3, β1      α2, α3, β2   α2, β2, γ3
    α2, α3, β3      α2, α3, β4   α2, β1, γ1
    α2, β1, γ1      α3, β1, γ4   α2, β2, γ2
    α2, β1, β3      α2, β2, β4   α2, β1, γ2
    α2, α3, α4      α2, β1, γ3   α2, β2, γ2
    α2,β2, γ4       α3, β1, γ2   α2, β1, β2
    α3, β1, γ3      α3, β2, γ4   α2, β1, β4
                                 α2, γ1, γ2
Tile Type Arm Partnership
 Best case scenario we will only need one
  tile type.
 Two options:
     (a2,a*2)
     (a,a*,b,b*)
Four Arm Constructions
 Cuboctahedron:
 •   α 1,α 2, β 1, β 4

 Octahedron:
 • α 1,α 2, γ 1, γ 2
    Cuboctahedron, Octahedron




Source: http://www.gardendome.com/cuboctahedron.GIF
Greater Arm Constructions
 5 and 6 arms may have less convex tile
  types.
 We can look at these possibilities.
 7 and greater will have no convex tile
  types.
Conclusions
 We have proved all the structures possible
  up to 4 arm tile types
 With the convex theorem and exhaustive
  proofing, we have the tools and
  techniques to look at the rest
Acknowledgements

				
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