How life prepares for an uncertain world

Miraia S. Chiou

Survey · Biology

How life prepares for an uncertain world

Let a receptor learn by waiting, then grow a population across two possible futures.

2026-09-08 · 15 min read · 2 interactive demos

Imagine packing seeds for a planet whose next season might be wet or dry. You could wait for a better forecast. You could pack varieties for both possibilities. Waiting changes what you know; diversifying changes what can survive.

Living systems face versions of both choices. A molecular receptor can delay signaling while a binding encounter unfolds. A population can generate different phenotypes, forms of behavior or physiology, among genetically similar descendants. These strategies make uncertainty something living matter can act on.

Epistemic Physics gives us a way to follow the journey from a physical history to a probability and then to a decision. Our first experiment asks how a receptor’s waiting time can become evidence. Our second asks how a probability can become a population. Together they lead to a surprisingly precise question: how much future growth should a lineage spend to obtain a better answer?

A receptor makes a molecule stay for the interview

A receptor is a molecular structure that responds to binding; its binding partner is called a ligand. Binding can begin a sequence of internal modifications before a signal is transmitted. McKeithan’s 1995 model proposed that this delay could help T-cell receptors distinguish a desired antigen from alternatives through kinetic proofreading: intermediate molecular changes can be interrupted when a ligand detaches. Read McKeithan’s original paper.

Picture an interview with several checkpoints. A visitor must remain present to finish. Visitors who tend to leave quickly rarely reach the final checkpoint, even if they can enter the building just as easily.

Let’s isolate the information in simply remaining present. Suppose the ligand is either “good,” the signal we want to recognize, or “bad,” a confusing alternative. Let its detachment time be the random variable $T$ . Choose exponential detachment, so the probability of remaining bound for at least duration $t$ is

\begin{aligned} Pr (T > t ∣ j) & = e^{- k_{j} t}, \\ j & \in {good, bad} . \end{aligned}

The detachment rate $k_{j}$ has units of inverse time. Exponential waiting is memoryless: conditional on still being bound, the chance of leaving during the next short interval depends on the interval’s duration rather than the binding’s age. Set $k_{good} < k_{bad}$ .

Now observe that a molecule is still bound at time $t$ . The likelihood ratio, the probability of this observation under good divided by its probability under bad, is

\begin{aligned} L (t) & = \frac{Pr (T > t ∣ good)}{Pr (T > t ∣ bad)} \\ = e^{(k_{bad} - k_{good}) t} . \end{aligned}

This starts conditional on an initial binding. Any information in how often the two types bind belongs to an earlier part of the observation model.

Take rates $k_{good} = 0.25$ and $k_{bad} = 1.25$ per time unit. Remaining bound for two units is $e^{2}$ , about $7.39$ , times as likely for the good type. Waiting has changed the relative plausibility of the two explanations.

One encounter can move a belief without becoming many trials

Let $p$ be an observer’s probability that the bound molecule is good. Its odds are $O = p / (1 - p)$ . Bayes’ rule multiplies those odds by $L$ :

O_{t} = O_{0} L (t) .

In Epistemic Physics, the corresponding rapidity coordinate is half the natural logarithm of the odds, $ρ = \frac{1}{2} \ln O$ . Multiplication becomes a shift:

\begin{aligned} ρ_{t} & = ρ_{0} + η (t), \\ η (t) & = \frac{1}{2} \ln L (t) \\ = \frac{k_{bad} - k_{good}}{2} t . \end{aligned}

Here a molecular duration has an exact translation into a movement in evidence coordinates. The natural logarithm measures evidence in nats; one nat corresponds to a likelihood factor of $e$ .

Keep the encounter’s lineage attached to its record. “Still bound at one second” and “still bound at two seconds” are nested observations of the same binding. Once the first second has been counted, the next update uses survival through the additional second. Multiplying the two full survival likelihoods would count the first second twice.

A tracelet can therefore describe one encounter’s time and state history. Several intermediate states within it do not automatically create several independent binding trials. Elapsed duration, checkpoint count, and fresh encounter count answer three different questions.

Build a checkpoint course

Choose a small physical model. A bound receptor needs $n$ forward steps to signal. At every unfinished step, advancement occurs at rate $k > 0$ , while detachment at rate $k_{off}$ ends the encounter and resets the receptor. The rates stay constant throughout this particular course.

For two competing exponential clocks, advancement wins with probability $k / (k + k_{off})$ . It must win $n$ times, giving the completion probability

Q_{n} = {(\frac{k}{k + k_{off}})}^{n} .

For good and bad ligands sharing the same forward rate,

\frac{Q_{n, good}}{Q_{n, bad}} = {(\frac{k + k_{bad}}{k + k_{good}})}^{n} .

Each extra checkpoint amplifies the completion ratio. It also lowers the probability that a good encounter completes. Recognition has acquired a design tradeoff.

Make a prediction: double the forward rate without changing detachment. More molecules should finish. Will good and bad molecules become easier or harder to distinguish by completion alone? Then make their detachment rates equal. Can any number of these checkpoints distinguish them?

A molecule runs the checkpoint course

Build a receptor checkpoint course

Build a course that lets at least 50% of desired molecules signal while admitting at most 10% of decoys. These are design targets for this game. Tap the last checkpoint you want to keep.

Follow one molecule through your course

The design test predicts probabilities across many encounters. Here you follow one sampled encounter. Bind a molecule, then reveal its next molecular event. The button samples both a waiting time and the race outcome; it cannot force an advance. One binding keeps the same encounter number through all intermediate states.

The receptor is unbound. Choose a molecule to begin a new encounter.

This pipeline is one reproducible random trajectory, with a fresh seed for each binding. Release receptor is your intervention; a sampled detachment comes from Next molecular event. Equal step rates and memoryless detachment are the chosen physical rules. Counting chemical fuel would require an additional fuel-use rule.

Change rates, compare curves, and inspect the model

Choose the physical rule

Advancement rate k, in events per time unit

Detachment rates: good ligand 0.25, bad ligand 1.25 per time unit. “Good” is the desired signal; “bad” is a confusing alternative. Changing the rule clears the current encounter.

Exact probabilities across many encounters

Good ligandBad ligand

Good completion probability: 80.0%
Bad completion probability: 44.4%
Good / bad completion ratio: ×1.80
Time among good successes: 0.80 units
Time to a good success, with retries: 1.00 units
Time to a bad success, with retries: 1.00 units

Completing all 1 checkpoints is 1.80 times as likely for the good ligand. More checkpoints reject additional good encounters too. Retrying includes time spent on those failed encounters.

Completion Q = [k / (k + k_off)]ⁿ. Successful-time mean = n / (k + k_off). Retry times assume immediate fresh binding to the same ligand type after every reset, with no search or rebinding delay. All displayed times use one arbitrary time unit.

Open the receptor experiment in the playground.

The upper curves average over encounters analytically. The lower course samples one sequence of waiting times and events. A click advances to the next event, which can be either a forward step or a detachment. Its encounter number stays fixed until a new molecule binds.

Successful encounters are a selected group

How long does a successful signal take? At one checkpoint, the time until either event has mean $1 / (k + k_{off})$ . For exponential races, that duration is independent of which event wins. Conditional on completing all $n$ steps, the mean time is therefore

E [τ ∣ complete] = \frac{n}{k + k_{off}},

where $τ$ is the duration until completion or detachment.

This produces a delightful trap: among successful encounters, a faster-detaching ligand has a shorter mean completion time. Its rare successes disproportionately follow quick paths. Looking only at successful records hides the failed attempts.

For $k_{off} > 0$ , the mean duration of any attempt, including a failed one, is

E [τ] = \frac{1 - Q_{n}}{k_{off}} .

Imagine immediately binding a fresh molecule of the same type after each failure, with independent encounter histories and no time spent searching or rebinding. The mean time until a signal becomes

E [time to signal] = \frac{1 - Q_{n}}{k_{off} Q_{n}} .

With three checkpoints, $k = 1$ and $k_{good} = 0.25$ , the good completion probability is $0.512$ . Good successes take $2.4$ time units on average, while the retry process needs $3.8125$ units on average. Those two clocks answer different experimental questions.

Molecular work adds another layer. Our course has mean forward-step count $k E [τ]$ per encounter. A declared rule assigning energy $ε$ to each forward step, with zero additional reset cost, would give mean modeled debit $ε k E [τ]$ . A chemical interpretation needs the corresponding fuel coupling and reset mechanism. Murugan, Huse and Leibler studied such connections among speed, error and dissipation in richer proofreading networks, relating their exploration of chemical states to microtubule growth. Read their 2012 study.

Put different descendants into different futures

Now leave the receptor and visit a microbial population. Suppose a cell can adopt a wet-ready or dry-ready phenotype. Different descendants can prepare for different conditions. This is bet-hedging: spreading reproductive prospects across environmental possibilities.

Kussell and Leibler compared sensing followed by response with diversity produced by stochastic phenotype switching. Their model linked long-term growth to environmental information and found conditions in which switching could be favored over sensing. Read their 2005 paper.

Our weather jars use a simpler reset rule. Before every generation, divide the surviving population into fraction $q$ wet-ready and fraction $1 - q$ dry-ready. Restore those same fractions at the next generation, whatever happened to the previous mix. The matching phenotype makes two offspring; the other makes none. Fractions can be restored without cost, and population mass is continuous, so we can follow its multiplication without rounding cell counts.

Let $Z_{m}$ denote population size after $m$ generations. One generation gives

\frac{Z_{m + 1}}{Z_{m}} = {\begin{cases} 2 q, & wet, \\ 2 (1 - q), & dry . \end{cases}

Independent seasons are wet with probability $π$ . For $0 < q < 1$ , the long-run average natural-log growth per generation is

\begin{aligned} G (q) & = \ln 2 + π \ln q \\ + (1 - π) \ln (1 - q) . \end{aligned}

Why logarithms? Population factors multiply across generations, while their logarithms add. A large early family cannot compensate for later extinction. If both weather types remain possible, allocating everything to one phenotype eventually encounters the season that removes the entire lineage.

Try an audacious strategy: choose All wet and Wet run first. Step through eight good seasons, then the ninth. Next put one share into the dry jar. Does that small reserve change the fate of the lineage? Finally, match the allocation to the weather frequency.

A population enacts a probability

Put descendants in two weather jars

Wet-ready cells reproduce in wet weather; dry-ready cells reproduce in dry weather. The matching phenotype makes two offspring, the other none. How should you divide the next generation when wet seasons occur with probability π = 2/3?

Before every season: restore the same fractions q and 1 − q among all remaining population mass, then reproduce. This resets the phenotype mix without restoring lost population. The allocation stays fixed through your run.

Choose your allocation

Click a jar to move one of 24 equal shares into it. You can also drag the rose divider, focus it and use arrow keys, or use the allocation buttons. Dots represent shares of a continuous population.

A fixed deck: eight wet, four dry

This is a visible 12-season challenge, with wet frequency 2/3. It illustrates a finite sequence; the expected-growth formula averages over independent seasons with wet probability 2/3. Changing the order restarts the run and keeps your allocation.

1W
2W
3D
4W
5D
6W
7W
8W
9D
10W
11D
12W

Ready to begin. Choose an allocation, then play a season.

Your allocationq = π referenceHalf in each

Your population / starting size: ×1.000
q = π population / starting size: ×1.000
Your realized log growth / season: No seasons yet
Your expected log growth / season: 0.000
Expected loss to q = π: 0.057 nats
Half-in-each population multiple: ×1.000

Each season multiplies population size, so log growth adds along the route. The expected loss from an allocation mismatch is the relative entropy D_KL(π || q), measured in nats per season.

Log means natural logarithm: zero on the graph means the starting population size, and one means a factor e ≈ 2.718. This model uses continuous population mass, cost-free restoration of phenotype fractions, and no sensing, inheritance of phenotype, or integer rounding. All three allocations see the same weather. Their probabilities describe physical allocation; an observer's uncertainty about the weather probability would be a separate distribution.

Open the phenotype experiment in the playground.

The visible deck contains eight wet and four dry seasons. It is a finite challenge whose frequency matches $π = 2 / 3$ . Reordering those same cards changes the population’s route, but every fixed allocation reaches the same final size: its multiplication factors are unchanged. Models with environmental memory or inherited phenotype can make order matter, as we explore in Evolution has an order.

A growth penalty reveals information geometry

To find the best fixed allocation, differentiate:

G^{'} (q) = \frac{π}{q} - \frac{1 - π}{1 - q} .

For $0 < π < 1$ , setting this to zero gives $q_{*} = π$ . The second derivative is negative, so this is the unique maximum. A population with two-thirds wet seasons should assign two-thirds of its mass to the wet-ready phenotype under these reproduction and reset rules.

Subtract the growth of any other allocation from this optimum:

\begin{aligned} G (π) - G (q) \\ = π \ln \frac{π}{q} \\ + (1 - π) \ln \frac{1 - π}{1 - q} \\ = D_{KL} (Bern (π) ‖ Bern (q)) . \end{aligned}

Here $Bern (r)$ means a two-outcome distribution with wet probability $r$ . The Kullback-Leibler divergence, or relative entropy, measures the mismatch between the environmental distribution and the allocation. Its direction matters: the environment supplies the weights $π$ and $1 - π$ .

An information quantity has become a reproductive price, in nats of lost growth per generation. At $π = 0.7$ , the half-and-half allocation loses about $0.0823$ nats per generation to the optimum. Over $m$ generations with matching weather proportions, the optimum’s population advantage is the exponential factor $e^{0.0823 m}$ .

We can also put the allocation in rapidity coordinates. Write $ρ_{q} = \frac{1}{2} \ln [q / (1 - q)]$ . Substitution gives

\begin{aligned} - G (q) & = \ln \cosh ρ_{q} \\ - (2 π - 1) ρ_{q}, \end{aligned}

where the hyperbolic cosine is

\cosh x = \frac{e^{x} + e^{- x}}{2} .

The negative growth is a curved log-cosh term plus a linear tilt, the same functional shape that appears in an EP Beta state’s rapidity potential. Here the surface’s height has a concrete interpretation: lost reproductive growth. Moving its minimum follows the environment’s probabilities.

Put the two experiments beside each other:

A receptor record	A phenotype decision
Survival produces likelihood ratio $L$	Allocation assigns fractions $q, 1 - q$
Odds move by $η = \frac{1}{2} \ln L$	The decision sits at $ρ_{q} = \frac{1}{2} \ln [q / (1 - q)]$
A history changes relative plausibility	A mismatch costs $D_{KL} (π ‖ q)$ in growth

This connection follows the role of information from observation into action. The receptor’s duration supplies evidence; the population’s distribution expresses a choice.

Who knows the weather probability?

The weather jars physically contain different phenotypes. An observer’s uncertainty about $π$ is a separate distribution, whose possible values are weather probabilities.

For example, start an observer with Beta prior weights $1 / 2, 1 / 2$ , totaling one. After $W$ wet and $D$ dry independent season trials, its posterior is

\begin{aligned} π ∣ weather record \\ \sim Beta (W + \frac{1}{2}, D + \frac{1}{2}) . \end{aligned}

The posterior mean is $(W + 1 / 2) / (W + D + 1)$ . If the observer chooses $q$ to maximize expected next-generation log growth, integrating over this posterior makes the optimal allocation equal that mean. Many posterior distributions can recommend the same allocation while retaining different spreads and malleability for future learning.

Each new season contributes one weather trial under the independent-season model. The number of descendant cells exposed to that same season does not multiply the number of independently observed weather events. Biological lineage and environmental history both help explain which observations share a cause.

When is a better forecast worth its cost?

Give our population a perfect forecast of the coming season. It can now assign every descendant to the matching phenotype. Suppose sensing consumes a fraction $c$ of the offspring it would otherwise produce, with the same cost in both weather states. Its growth becomes

G_{sense} = \ln 2 + \ln (1 - c) .

The best unsensed allocation has growth $G (π) = \ln 2 - H (π)$ , where

\begin{aligned} H (π) & = - π \ln π \\ - (1 - π) \ln (1 - π) \end{aligned}

is the weather’s entropy, its average surprise in nats. Perfect sensing pays when

c < 1 - e^{- H (π)} .

For equally likely seasons, the threshold is $c = 1 / 2$ . For a nearly certain season, the affordable cost approaches zero: there is little uncertainty left for sensing to resolve.

This calculation offers a bridge back to the receptor. A longer molecular interview can improve discrimination, but it also delays action. To connect the experiments physically, let receptor signals forecast an environmental state, measure the forecast’s error probabilities and timing, and specify how delay changes reproduction. Then ask which checkpoint count maximizes growth after the population uses the imperfect forecast to allocate its descendants.

An even richer expedition gives weather memory, makes switching costly, and lets cell states persist, as in A cell chooses what to remember. Descendants could inherit a phenotype, an observation history, or a rule for asking the next question. We could watch evolution tune the whole sequence: wait, interpret, choose, reproduce. The organism would be shaping both its answers and the physical questions it can afford to ask.

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