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

Tile A

Tile DTile A

Tile Bc ĉ.

c ĉ.

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 radiansπ/2 radians2π/3 radiansπ 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

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

* *1 11 2 12 1 11 2 12rA +r A =rA +r A

Example

Tiles with more than just ‘a’’s

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

Solve all three with each other in mind Only possible result is (a,b,c) and

(a*,b*,c*)

* *1 11 2 12 1 11 2 12rB +r B =rB +r B

* *1 11 2 12 1 11 2 12rC +r C =rC +r 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 Checkr1 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 TypesNon-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