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DNA Strand-displacement Gate

Toeholds turn molecular recognition into logic, but the clock runs in seconds to hours.

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DNA gates compute by chemistry, not voltage

An exposed toehold recruits an input strand; branch migration then displaces an output strand. Sequence design chooses who can react, while concentration and toehold binding choose how quickly.

Two sliders, one number

The response is y = 1 − exp(−kCt), so the time to any fraction depends only on the product of rate constant and trigger concentration. Double k and halve C and the curve is identical, both reaching 90% at 2.3026/kC — the model has one degree of freedom. A slow toehold can be paid for with concentration, until the concentration runs into the cost of DNA.

Thirty-eight minutes to switch

At the default 10⁴ M⁻¹s⁻¹ and 100 nM the gate needs 38 minutes to reach 90%, about 2 × 10¹⁴ times a 10 ps silicon gate. Push both to a plausible ceiling, 10⁶ M⁻¹s⁻¹ and 1 µM, and it is 2.3 seconds — still 2 × 10¹¹ times slower. Molecular logic wins on energy per event by roughly 10⁴ and loses on time by 10¹¹.

Problem solved in full

  1. How deep a DNA circuit can get in one day, at 38 minutes a gate 6 steps

    A strand-displacement gate reaches 90% output at a rate constant of 10⁴ M⁻¹s⁻¹ and a trigger at 100 nM. Derive the time. Then work out how many of these can be chained in a day, and say what that limits.

    Open this problem in the tool

    1. The response is first order in the trigger: y = 1 − e^(−kCt). Set y = 0.9 and rearrange, and the 0.9 becomes ln 10 = 2.3026, because 1 − 0.9 is one tenth.

      y=1−e−kCt=0.9  ⇒  kCt=ln⁡10=2.3026y = 1 - e^{-kCt} = 0.9 \;\Rightarrow\; kCt = \ln 10 = 2.3026

    2. So t = ln 10 ÷ (kC), and everything about the gate is in that one product kC. Nothing else in the expression survives the rearrangement.

      t=ln⁡10kCt = \frac{\ln 10}{kC}

    3. Put the numbers in. 100 nM is 1 × 10⁻⁷ M, so kC = 10⁴ × 10⁻⁷ = 10⁻³ s⁻¹, and t = 2.3026 ÷ 10⁻³ = 2,303 s, which is 38.4 minutes.

      kC=104×10−7=10−3 s−1  ⇒  t=2,303 s=38.4 minkC = 10^{4} \times 10^{-7} = 10^{-3}\ \mathrm{s^{-1}} \;\Rightarrow\; t = 2{,}303\ \mathrm{s} = 38.4\ \mathrm{min}

    4. Now the part the panel does not model: it shows one gate, and a circuit is gates in series. Ten of them, each waiting for the last, take 23,026 s, which is 6.4 hours.

      t10=10ln⁡10kC=23,026 s=6.4 ht_{10} = \frac{10 \ln 10}{kC} = 23{,}026\ \mathrm{s} = 6.4\ \mathrm{h}

    5. Turn that round and ask what a day buys. 86,400 ÷ 2,303 = 37.5 gates deep. A day of chemistry gets you a circuit thirty-seven layers deep, and that is the whole computation, not thirty-seven operations per second.

      86,4002,303=37.5 gates deep per day\frac{86{,}400}{2{,}303} = 37.5\ \text{gates deep per day}

    6. Push both knobs to the plausible ceiling, 10⁶ M⁻¹s⁻¹ and 1 µM: kC becomes 1 s⁻¹ and a gate takes 2.30 s. A day is then 37,523 gates deep — a thousand times better, and a 10 ps silicon gate still manages 8.64 × 10¹⁵ in the same day.

      kC=106×10−6=1 s−1  ⇒  t=2.30 s,37,523 deepkC = 10^{6} \times 10^{-6} = 1\ \mathrm{s^{-1}} \;\Rightarrow\; t = 2.30\ \mathrm{s}, \quad 37{,}523\ \text{deep}

    Answer

    38.4 minutes a gate, so 37.5 gates deep in a day, or 37,523 at the best rate anyone can plausibly reach. The number that matters is the DEPTH, not the count, because a strand-displacement circuit is enormously parallel and hopelessly serial at the same time: every gate in one layer runs at once in the same tube, and the next layer cannot start until this one has finished. So the right comparison is not against a chip's operations per second — it is against a chip's critical path, and there the gap is 2.3 × 10¹⁴ per layer. That is why published DNA circuits are shallow: the square-root circuit Qian and Winfree built in 2011 ran a handful of layers and took hours, and no amount of scaling the tube up changes the depth. Chemistry buys width for free and pays for depth in wall-clock time.

Learning path

Computing with molecules

References (2)

· Reviewed against published sources. Corrections welcome.