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) ArmX VectorY VectorZ Vector Alpha 1010 Alpha 200 Alpha 300 Alpha 400 Beta 1-1/21/2-sqrt(2)/2 Beta 2-1/21/2sqrt(2)/2 Beta 31/2-1/2sqrt(2)/2 Beta 41/2-1/2sqrt(2)/2 Gamma 11/2 -sqrt(2)/2 Gamma 21/2 sqrt(2)/2 Gamma 3-1/2 sqrt(2)/2 Gamma 4-1/2 -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 D Tile A Tile B c ĉ.

Orientation Theorem Let T 1 and T 2 be two tiles and let Ɛ 1, Ɛ 2 be the elements of {α 1,...,α 4, β 1,...,β 4, γ 1,...γ 4 }, representing the arms where T 1 and T 2 join. Let let σ i be the lexicographical minimal arm of T i, omitting Ɛ i and its antipodal arm. Write P( Ɛ i,σ i ) for the plane through Ɛ i and σ i. Then the bond angle of (T 1, Ɛ 1 ), (T 2, Ɛ 2 ) is the angle formed between P( Ɛ 1, σ 1 ) and P( Ɛ 2, σ 2 ).

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 H v 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 r 1 = quantity of the base tile we start with r 2 = the quantity of its partner tile A 11 = number of ”a” ends on r 1 A 11 * = number of of ”a*” ends on r 1 A 12 and A 12 * = number of ”a” and ”a*” on r 2 respectively

Example Tile 1 has 3 a’s and 0 a*’s [listed (3,0)]r1(3)+r2A 12 =r1(0)+r2A 12 *r1(3)=r2A12*-r 2 A12 If r1=1 and r2=1, then A 12 * =3 and A12 =0 so the complimentary pair is (0,3) If r1=1 and r2=3, then A 12 * =2 and A12 =1 so the complimentary pair is (1,2)

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*)

Tile Type Arm Partnerships for 3 arm tiles Tile 1Tile 2 a3a*3 a3a,a*2 a2,a*a*3 a2,a*a,a*2 a2,ba*2,b* a,a*,b b*2,b a,a*,bb*3 a,b,ca*,b*,c*

Cross Check r1a3 a2,a* a2,ba,a*,b a,b,c r2a*3a,a*2a*3a,a*2a*2,b*a,a*,b*b*2,bb*3a,*,b*,c* α2, β1xxxxxxxxx α2, β2xxxxxxxxx α4, β1xxxxxxxxx α4, β3TrOct α2, α3PPPPPPPPP α3, β1PPPPPPPPP β1, γ1oTetra ooo o β1, γ3oTrTetra TrTtraTrTetrao β3, γ4PPPPPPPPP β1, γ2xxxxxxxxx

Four Arm Tile Types Non-ConvexConvex α2, γ3, γ4α3, β1, β3α2, β1, γ4 α2, α3, β1α2, α3, β2α2, β2, γ3 α2, α3, β3α2, α3, β4α2, β1, γ1 α3, β1, γ4α2, β2, γ2 α2, β1, β3α2, β2, β4α2, β1, γ2 α2, α3, α4α2, β1, γ3α2, β2, γ2 α2,β2, γ4α3, β1, γ2α2, β1, β2 α3, β1, γ3α3, β2, γ4α2, 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:

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