DNA Strand-displacement Gate

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

Worked example — DNA Strand-displacement Gate

Boolean output	1
Effective trigger	100 nM
Pseudo-first-order rate	1.00 × 10⁻³ s⁻¹
Time to 90% output	38.4 min

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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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

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.
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.
  
  $y = 1 - e^{- k C t} = 0.9 \Rightarrow k C t = ln 10 = 2.3026$
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.
  
  $t = \frac{ln 10}{k C}$
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.
  
  $k C = 1 0^{4} \times 1 0^{- 7} = 1 0^{- 3} s^{- 1} \Rightarrow t = 2, 303 s = 38.4 min$
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.
  
  $t_{10} = \frac{10 ln 10}{k C} = 23, 026 s = 6.4 h$
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.
  
  $\frac{86 , 400}{2 , 303} = 37.5 gates deep per day$
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.
  
  $k C = 1 0^{6} \times 1 0^{- 6} = 1 s^{- 1} \Rightarrow t = 2.30 s, 37, 523 deep$
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

Leads to Energy per operation seconds per operation.

References (2)

Experimental characterization and kinetic modeling of toehold-controlled strand displacement: D. Y. Zhang and E. Winfree, “Control of DNA Strand Displacement Kinetics Using Toehold Exchange.” Journal of the American Chemical Society 131(47), 17303–17314, 2009.
The scalable Boolean DNA circuit architecture behind the gate examples: L. Qian and E. Winfree, “Scaling up digital circuit computation with DNA strand displacement cascades.” Science 332(6034), 1196–1201, 2011.

By the 42Math team · Updated 2026-08-22

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

medium AND gate - A medium toehold at 100 nM reaches 90% output in 38.4 min.
weak OR gate - A hundredfold weaker toehold takes a hundred times longer: 64 h to the same 90%.
strong AND gate - A strong toehold at the same concentration gets there in 23 s.
missing AND input - One input missing, and in this idealised model the AND gate never produces output at all.