A cell chooses what to remember

Miraia S. Chiou

Survey · Biology

A cell chooses what to remember

Gene switches preserve chosen fates; adapting receptors remember the background to keep noticing change.

2026-09-08 · 17 min read · 2 interactive demos

Imagine giving a cell two instructions. First: remember which door you entered through. Second: keep noticing changes in the room’s smell.

These requests pull memory in different directions. The door should remain in memory after it disappears from view. The room’s steady smell should become familiar, freeing the cell to notice the next change. Both jobs require retaining something about the past. They differ in what should remain visible at the output.

Let’s build two tiny mathematical cells. One will remember a choice through genes that suppress each other. The other will retain a moving reference against which it senses its surroundings. Epistemic Physics gives us a particularly inviting route between them: positive coordinates, persistent directions, and changes of reference frame.

We will eventually connect the two cells. Could a receptor that adapts away a sensation use that very act of adaptation to write a lasting memory elsewhere?

Two genes hold each other quiet

A gene-regulatory network is a collection of interactions that control how much product genes make. In our first network, gene x makes a product that inhibits gene y, while gene y returns the favor. This is mutual repression.

The mechanism has an experimental foothold. Gardner, Cantor, and Collins constructed a synthetic toggle in Escherichia coli with two mutually inhibitory regulatory units. Temporary chemical or temperature interventions switched the circuit between stable expression states. Their 2000 paper established a concrete form of addressable cellular memory.

Choose a symmetric teaching model:

\begin{aligned} \dot{x} & = \frac{a}{1 + y^{2}} - x, \\ \dot{y} & = \frac{a}{1 + x^{2}} - y . \end{aligned}

Here $x, y$ are nonnegative concentrations divided by a repression threshold. The dot means a derivative with respect to time, measured in degradation times. The positive parameter $a$ controls production relative to loss. The squared concentration gives a Hill-2 response, a chosen cooperative repression curve. We give both genes identical parameters to make the symmetry visible.

Read the first equation aloud: x is made at a rate that falls when y rises, and existing x disappears at a rate proportional to its concentration.

Now give x a small advantage. More x suppresses y. Less y releases x from repression. A loop of two negative influences can reinforce an initial difference. Whether that reinforcement defeats degradation depends on $a$ .

Watch one destination split into two

At an equilibrium, concentrations stop changing. An equal-expression equilibrium has $x = y = s$ , where

s + s^{3} = a .

There is one positive solution because the left side increases continuously with $s$ .

What happens to a tiny difference between the genes? Subtract the dynamical equations:

\begin{aligned} \frac{d}{d t} (x - y) & = (x - y) R, \\ R & = \frac{a (x + y)}{(1 + x^{2}) (1 + y^{2})} - 1. \end{aligned}

Here $R$ is the instantaneous growth rate per unit expression difference. Close to the equal equilibrium, it becomes

λ_{difference} = \frac{s^{2} - 1}{s^{2} + 1} .

This growth rate says whether a small difference shrinks or grows. It is negative when $s < 1$ , zero at $s = 1$ , and positive when $s > 1$ . Since $s = 1$ corresponds to $a = 2$ , we have found the threshold.

Below $a = 2$ , small differences fade. Above it, the equal equilibrium becomes unstable to differences and two stable unequal states appear. This splitting is called a pitchfork bifurcation, named after the shape of the equilibrium branches.

There is a delicious control experiment hiding in the factored equation. If $x - y$ is exactly zero, its derivative is exactly zero. A perfectly symmetric deterministic cell stays symmetric even above the threshold. The equations amplify a difference; an initial bias, perturbation, or added noise supplies it.

Predict before touching: place the cell exactly equal at $a = 3$ and release it. Then add just $0.05$ to x and release again. Finally, pulse y. Can a brief intervention change which gene stays high after the intervention ends?

Two genes, two possible memories

Give a cell a lasting nudge

The cell begins with gene x in charge. Choose how long to pulse gene y, then remove the pulse and wait. Can a brief intervention leave y in charge on its own?

Gene x2.59

Gene y0.39

Pulse duration, in model time units

Place the cell, change production, or pulse x

Production a 3.00

Follow the expression path and its cone coordinates

Gene expression

The diagonal is equal expression. Drag the dark point or use its arrow keys, then release.

The same cell in the cone

Gold dashed: A² − P² = 4. Unequal settled states lie here; the moving cell can leave it.

x-dominant fixed pointy-dominant fixed pointReleased pathDuring pulse

Gene x: 2.593
Gene y: 0.392
Cone quantity M = 2√(xy): 2.015
Current expression: x leads

A small initial x advantage has had 24 time units to grow. Two stable fixed points are available. Exact equality persists until a perturbation supplies a difference.

Dimensionless Hill-2 model: dx/dt = a/(1 + y²) − x and dy/dt = a/(1 + x²) − y. A pulse adds 4 to the named production rate for the selected duration. Each path shows the most recent release; all randomness is off.

Open the gene-switch experiment in the playground.

The pulse in this experiment adds production to the chosen gene for $1.5$ time units. It then disappears completely. The remaining trajectory shows what the mutual repression network does by itself.

Two memories on one hyperbola

We can solve the unequal equilibria with a short piece of algebra. At equilibrium,

\begin{aligned} x (1 + y^{2}) & = a, \\ y (1 + x^{2}) & = a . \end{aligned}

Subtracting gives

(x - y) (1 - x y) = 0.

Whenever $x \neq y$ , therefore, $x y = 1$ . Substituting this back gives $x + y = a$ . For $a > 2$ , one unequal state has

\begin{aligned} x & = \frac{a + \sqrt{a^{2} - 4}}{2}, \\ y & = \frac{a - \sqrt{a^{2} - 4}}{2} . \end{aligned}

Swapping x and y gives the other state. Now align the cell with familiar sum-and-difference coordinates:

Gene expression	Cone coordinates
Total concentration $x + y$	$A = x + y$
Expression difference $x - y$	$P = x - y$
Product $x y$	$M^{2} = A^{2} - P^{2} = 4 x y$
Unequal equilibria, $x y = 1$	Hyperbola $A^{2} - P^{2} = 4$

Because concentrations are nonnegative, $A \geq | P |$ : the state lies in a positive cone. The two persistent fates sit at

\begin{matrix} A = a, M = 2, \\ P = \pm \sqrt{a^{2} - 4} . \end{matrix}

The teal and rose fixed points in both plots are the same two cells under this coordinate change. Their local relaxation rates, the eigenvalues describing small disturbances, are $- 1 + 2 / a$ and $- 1 - 2 / a$ . Both are negative for $a > 2$ .

We can even use rapidity, the logarithmic ratio coordinate

ρ = \frac{1}{2} \log \frac{x}{y} .

On the unequal equilibrium branches, $x y = 1$ gives $x = e^{ρ}$ , $y = e^{- ρ}$ , hence

A = 2 \cosh ρ, P = 2 \sinh ρ .

Increasing production opens two equal-and-opposite rapidity directions. The cell’s feedback chooses which direction persists.

The hyperbola describes the settled unequal states. Along a moving trajectory, $M$ can change. For example, at $x = y = 1$ and $a = 3$ , both concentrations initially increase at rate $1 / 2$ , so $M = 2 \sqrt{x y}$ initially increases at rate $1$ . Its eventual fixed-point value supplies no conservation law for the whole motion.

What the cell writes, and what an observer infers

The pulse works by moving the cell into the other basin of attraction, the set of starting states that approach that destination. Once the pulse ends, feedback maintains the chosen expression pattern.

Lowering $a$ below two leaves one stable equal-expression equilibrium. An old expression difference then fades. Waiting a finite time can leave a very small residual difference, however; raising $a$ again may amplify it. The demo’s exactly-equal control deliberately performs a stronger reset than merely waiting.

The resulting memory lives in concentrations and regulatory dynamics. To attach a Beta distribution to an observer’s belief about the cell requires another choice: what is the observer estimating, how are readings generated, and which readings share a lineage? Repeated measurements of one cell could update an estimate of its state without becoming independent examples of cell-fate choice.

A related mathematical puzzle makes the distinction productive. Take a Beta-distributed probability $q$ with positive shapes $α, β$ , and transform it to $ρ_{q} = \frac{1}{2} \log [q / (1 - q)]$ . Define $A_{β} = α + β$ and $P_{β} = α - β$ . Including the change-of-variable factor, its negative log density is

\begin{aligned} V (ρ_{q}) & = A_{β} \log (2 \cosh ρ_{q}) \\ - P_{β} ρ_{q} + C, \\ V^{″} (ρ_{q}) & = A_{β} {sech}^{2} ρ_{q} > 0. \end{aligned}

Here $C$ is a constant and $sech ρ = 1 / \cosh ρ$ . This potential is strictly convex: its slope always increases, leaving a single minimum. If we choose relaxation down this potential as a dynamical law, it has one resting destination. Our gene network supplies additional structure through coupled feedback.

This claim concerns the density in rapidity. A Beta density in $q$ can instead rise toward both endpoints when both shapes are below one. Density peaks depend on the coordinate used to measure their height.

For a population with two persistent fates, a mixture is a natural next construction: one distribution for x-dominant cells, another for y-dominant cells, and weights describing how many occupy each group. Add molecular fluctuations and we can ask how often cells cross between basins. Those weights, crossing rates, and an observer’s uncertainty about them become distinct quantities to investigate. How life prepares for an uncertain world follows the next possibility: allocating a population across several ways to survive.

The second cell remembers its background

Our second cell has a different assignment. A lasting concentration step should eventually become familiar, while a fresh change should remain detectable.

This is adaptation: an output returns toward its previous level while the changed stimulus remains present. Yi and colleagues identified integral feedback within a model of bacterial chemotaxis, the regulation of movement in response to chemicals, and used it to explain robust return to a target output. Their analysis provides the biological starting point for our smaller receptor model.

Let $c > 0$ be an external concentration, and choose a concentration unit $c_{0}$ . Write its natural logarithm as $ℓ = \log (c / c_{0})$ . Give the cell an internal reference $m$ , a target occupancy $0 < p_{*} < 1$ , and an adaptation rate $κ > 0$ :

\begin{aligned} p & = \frac{1}{1 + e^{- (ℓ - m)}}, \\ \dot{m} & = κ (p - p_{*}) . \end{aligned}

The first equation is the logistic function, which turns the difference $ℓ - m$ into a number between zero and one. It also equals $p = c / (c + K)$ with a moving effective binding scale $K = c_{0} e^{m}$ .

The second equation adjusts the reference whenever occupancy misses its target. If $p$ is too high, $m$ rises, reducing $ℓ - m$ and bringing $p$ down. If $p$ is too low, the correction reverses. This is negative feedback.

It is also integral feedback because

m (t) - m (0) = κ \int_{0}^{t} [p (s) - p_{*}] d s .

The internal variable retains accumulated output error. At constant input its equilibrium is

\begin{aligned} m_{\infty} & = ℓ - \log \frac{p_{*}}{1 - p_{*}}, \\ p_{\infty} & = p_{*} . \end{aligned}

The output returns while the internal reference changes. The cell has kept the background in a place where it can subtract it from future sensations.

Predict before touching: compare a concentration jump from one to four with a jump from ten to forty. What must the second cell remember before the jumps for their entire response curves to match? Then freeze its reference.

A cell compares today with its recent past

Help a receptor keep its bearings

1 / 3 · Recover after a sudden jump. After a 4× jump, bring occupancy back to 0.50 (±0.03) by time 30.

Receptor occupancy0.500

Before the input changes

Incoming step at t = 44.0× concentrationChoose a policy, then run to time 30.

Each policy replays the chosen receptor model from time zero. Input begins at t = 4; the checkpoint is t = 30 model units. Times are dimensionless.

Open the adaptive-reference experiment in the playground.

A moving reference has a speed limit

Let concentration grow exponentially, so its logarithm follows a steady ramp: $\dot{ℓ} = v$ . Define the sensed difference $d = ℓ - m$ . Then

\dot{d} = v - κ (p - p_{*}) .

A tracking state keeps $d$ fixed while both input and reference move. Setting $\dot{d} = 0$ gives

\begin{matrix} p_{track} = p_{*} + \frac{v}{κ}, \\ - κ p_{*} < v < κ (1 - p_{*}) . \end{matrix}

The strict bounds follow from $0 < p < 1$ . Within them, the tracking state is stable because increasing $d$ increases $p$ , which pushes $\dot{d}$ back down. Beyond them, the reference cannot keep pace and occupancy approaches saturation. At equality, maintaining the limiting occupancy would require an infinite sensed difference, so there is no finite-offset tracking state there either.

With $p_{*} = 1 / 2$ , $κ = 1$ , and $v = 0.2$ , the tracking occupancy is $0.7$ . The receptor carries a sustained report of the rate of change. Switch to $v = 0.7$ and watch the target become unreachable. Increase $κ$ to two and tracking becomes possible again.

Near the target, the local return time is $1 / [κ p_{*} (1 - p_{*})]$ . At $p_{*} = 1 / 2$ , this is $4 / κ$ . Faster reference adjustment makes a step familiar sooner while allowing steeper log ramps to be followed.

Multiply the world and follow the same response

Multiply every input concentration by a fixed factor $b > 0$ . This shifts $ℓ$ by $\log b$ . If we also shift the initial reference by $\log b$ , the whole system obeys

\begin{aligned} ℓ^{'} (t) & = ℓ (t) + \log b, \\ m^{'} (t) & = m (t) + \log b, \\ p^{'} (t) & = p (t) . \end{aligned}

Primes label the scaled experiment. The difference $ℓ^{'} - m^{'}$ is unchanged, and equal occupancies give equal reference velocities. These relations therefore persist throughout the trajectory.

This is scale equivariance: a transformation of the input has a matching transformation of the internal state. The resulting output invariance is a form of fold-change detection, where the complete response depends on concentration ratios. Shoval and colleagues studied this property and its connection to sensory searches that remain the same when an entire input field is multiplied by a constant. Their 2010 paper distinguishes matching whole response curves from merely returning to the same final output.

Movement adds another twist. If the cell follows position $r (t)$ with velocity $u$ , the log concentration it encounters changes at rate

\frac{d ℓ}{d t} = \frac{\partial ℓ}{\partial t} + u \cdot \nabla ℓ .

The first term is environmental change at a fixed place. The second is change caused by moving through a spatial gradient, the direction and rate at which concentration varies across space. The receptor feels their sum. Distinguishing a strengthening chemical source from swimming toward it becomes a task for additional movement information or an active experiment. How to read a living landscape follows signals and orientation through that larger spatial world.

Let one memory write into the other

Now connect our two cells into one imagined circuit. Let occupancy above target add production to gene x, and occupancy below target add production to gene y:

\begin{aligned} u_{x} & = B [p - p_{*}]_{+}, \\ u_{y} & = B [p_{*} - p]_{+} . \end{aligned}

Here $B > 0$ is a coupling strength and $[z]_{+} = max (z, 0)$ keeps each added production rate nonnegative. Add $u_{x}, u_{y}$ to the corresponding toggle equations.

Something lovely follows immediately:

u_{x} - u_{y} = B (p - p_{*}) = \frac{B}{κ} \dot{m} .

The receptor’s reference adjustment supplies a signed drive to the persistent memory. For a concentration step between fully adapted states, its total signed drive is

\int (u_{x} - u_{y}) d t = \frac{B}{κ} \log \frac{c_{after}}{c_{before}} .

The sensation can fade while its consequences remain in the gene circuit. A tenfold-larger world gives the same writing signal when the references are matched. Changing the time profile can still matter: the nonlinear toggle responds to the trajectory of the drive, even when two inputs deliver the same total signed amount.

Try imagining a slow rise and a sharp step with the same final concentration ratio. Which one crosses the boundary between fates? How would the answer change near the toggle’s production threshold? We now have a concrete route from sensing a changing world to choosing what part of that encounter will endure.

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