FINAL POWERPOINT

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```					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:
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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