Theoretical and Experimental DNA Computation (Natural ...

86 4 Complexity Issues
At level D(S) we carry out steps 1 through 7, as described above. However,
at steps 4 and 7 we retain all strands of length ≥ 3l. We are now then ready
to implement the final read-out phase. This involves a simple interpretation
of the final gel visualisation. Since we know the unique length uj of each
z j
D(S) section of the strand for each output gate tj, the presence or absence
of a strand of length uj +2l in the gel signifies that tj has the value 1 or 0
respectively.
We now illustrate this simulation with a small worked example. We
describe the DNA simulation of the small NAND-gate circuit depicted in
Fig. 4.6a. The sequences chosen to represent each gate strand are represented
in Fig. 4.6b, the shape of each component diagrammatically representing its
sequence. It is clear from the diagram that the input and output components
of connected gates fit each other in a “lock-key” fashion. Given the inputs
shown in Fig. 4.6a, the simulation proceeds as follows. Gate g1 has input values
of 0 and 1, so we add to tube T0 a strand representing the complement of
the second input component of the strand representing g1. Gateg2 has both
inputs equal to 1, so we add to tube T0 strands representing the complement
of both input components of g2. We then add tube T0 to tube T1, allow annealing,
and ligate. We next add the appropriate restriction enzyme. The strands
representing g1 are not cleaved, since the first restriction site is not made
fully double stranded. The strands representing g2 are cleaved, since the first
restriction site is fully double stranded. We then sort T1 on length using a gel
following PCR, “snip” off the output components of full-length strands using
a second blunt ended restriction enzyme, and remove the output components.
These retained strands are then added as input to T2, and we repeat the above
procedure. After sorting on length, the presence of intact, full-length strands
indicates that the circuit has the output value 1 (Fig. 4.6c).
However, there is a potential problem in the restriction of side-by-side annealed
oligonucleotides in that there is a nick in the double stranded DNA.
This problem can be potentially solved by using DNA ligase. The problem
is that it is difficult to ensure 100% ligation. Failure of ligation yields false
positives, i.e., strands that should have been restricted escape into the next
stage. Therefore the restriction enzyme used will have to be indifferent to
the presence of a nick. One potential group of such nick resistant restriction
enzymes is the type III restriction enzymes. The type III enzymes cut at a
specific distance 3’ from the recognition site; e.g., restriction enzyme Eco57I
recognizes CTGAAG, and then cuts the top 16 nucleotides and the bottom
14 nucleotides of the recognition site. We may engineer Eco57I sites in all
the input oligonucleotides such that the restriction site extends across adjacent
sequences. Digestion of the DNA should be possible only if both input
oligonucleotides are double stranded. We can easily test nick resistance and
dependence on double strandedness of both input oligonucleotides by creation
of artificial substrates with and without nicks and substrates that are fully
double stranded, half double stranded, and fully single stranded.