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Delay and Stochastic Extensions · Tutorial 105 of 799

Modeling Delayed Reactions in Stochastic Cellular Systems

Learn how to define delayed-reaction propensities consistently with whether reactants are consumed at initiation or only at completion.

Advanced 10 min read

What You'll Learn

  • Distinguish delayed reactions that consume reactants at initiation from those that change molecular counts at completion.
  • Define mass-action initiation propensities using the eligible reactant molecules.
  • Track reserved reactants so completion-only events cannot use the same molecules twice.
  • Compare how the two conventions affect competing reaction propensities.
  • Identify assumptions and failure modes in delayed stochastic reaction models.

Why the Timing Convention Matters

A delayed reaction has at least two relevant times: when it is initiated and when its effects are completed. The interval between them is not merely a timing detail. If reactants are removed at initiation, other reactions cannot use them during the delay. If molecular counts change only at completion, the model must still say whether those reactants are available to other reactions while the event is pending. Different choices define different stochastic systems and can produce different reaction histories.

The earlier material on cellular continuous-time Markov chains establishes the role of a propensity: conditional on the current state, a reaction channel with propensity \(a_j\) has initiation probability \(a_j\,dt+o(dt)\) over a sufficiently short interval \(dt\). For a delayed channel, this is an initiation propensity. It determines when a reaction begins, not when its products appear.

Definition: A delayed reaction channel has an initiation event, a delay, and a completion event. Its model must specify (1) the state used to calculate the initiation propensity, (2) any molecular-count changes at initiation, and (3) the molecular-count changes at completion. A propensity alone does not specify all three.

Propensities for Eligible Reactants

Let \(x_i\) be the number of molecules of species \(i\), and let \(\nu^-_{ij}\) be the number of molecules of species \(i\) required by reaction channel \(j\). In a stochastic mass-action model, a common propensity is proportional to the number of eligible reactant combinations. For a channel whose rate per eligible combination is \(c_j\), define its total channel initiation hazard directly by

$$ a_j(x)=c_j\prod_i {x_i\choose \nu^-_{ij}}, $$

where \({x_i\choose \nu^-_{ij}}=0\) if \(x_i<\nu^-_{ij}\). The factor counts unordered selections of the required molecules of each species. For example, \(A\to\cdots\) has propensity \(c_jx_A\), while \(2A\to\cdots\) has propensity \(c_j{x_A\choose 2}\). The units of \(c_j\) are inverse time per eligible combination, so \(a_j(x)\) has units of inverse time.

This formula specifies the channel's total hazard. It does not follow merely from assigning each possible molecular combination a marginal small-interval initiation probability: those events could be dependent. If a model instead constructs the channel from separate clocks for combinations, it must specify a construction such as independent clocks, for which simultaneous initiations have probability \(o(dt)\). Otherwise, define the total channel hazard directly, as above.

The standard hazard interpretation assumes that, conditional on the current state and pending-event information, the chance of one channel initiation in \(dt\) is \(a_j\,dt+o(dt)\), and the chance of two or more initiations is \(o(dt)\). Propensities must be nonnegative and must be zero when the required reactants are unavailable. The delay may be fixed, \(\tau_j\), or sampled from a specified distribution. Unless the model says otherwise, the delay is determined at initiation and does not change in response to later state changes.

Two State-Change Conventions

Write \(\nu^-_j\) for the vector of reactants consumed and \(\nu^+_j\) for the vector of products formed. The conventional net stoichiometric change is \(\nu^+_j-\nu^-_j\), but a delayed channel divides that net effect across time differently under the two interpretations below.

Initiation-time consumption: At initiation, remove the reactants, \(x\mapsto x-\nu^-_j\). At completion, add the products, \(x\mapsto x+\nu^+_j\). During the delay, the reactants are no longer available and the products have not yet appeared.
Completion-only molecular-count change: At initiation, leave the molecular counts unchanged. At completion, apply \(x\mapsto x-\nu^-_j+\nu^+_j\). To prevent the same reactants from being committed to multiple pending events, the model must reserve them at initiation or impose another explicit feasibility rule. A reservation is bookkeeping about availability, not a change in the total molecular count.

For completion-only reactions with reservations, let \(r_i\) be the number of molecules of species \(i\) reserved by pending events. The available, unreserved count is \(x_i-r_i\), so the propensity uses those counts:

$$ a_j(x,r)=c_j\prod_i {x_i-r_i\choose \nu^-_{ij}}. $$

At initiation, a new pending event reserves its required reactants, increasing \(r\) by \(\nu^-_j\); total molecular counts \(x\) do not change. At completion, total counts change by \(-\nu^-_j+\nu^+_j\), and the reservation is released. Thus, the molecules committed to that event are removed exactly once. Channels not requiring a reserved molecule may still use other available molecules.

Worked Example: A Delayed First-Order Reaction

Suppose \(A\to B\) has rate \(c=0.40\) per minute per \(A\) molecule and a fixed delay of 3 minutes. Initially, \(x_A=5\), and no event is pending. The initiation propensity is

$$ a(x)=0.40(5)=2.00\ \text{min}^{-1}. $$

Under initiation-time consumption, if one event initiates, the molecular counts immediately become \(x_A=4\), \(x_B=x_B^{\text{initial}}\), and the event is scheduled to add one \(B\) molecule after 3 minutes. Until then, the next initiation propensity for this channel is \(0.40(4)=1.60\ \text{min}^{-1}\).

Under completion-only molecular-count change, initiation leaves the total count at \(x_A=5\), but reserves one \(A\), so \(r_A=1\) and the available count is \(5-1=4\). The subsequent propensity is also \(0.40(4)=1.60\ \text{min}^{-1}\). At completion, the total counts change to \(x_A=4\) and \(x_B=x_B^{\text{initial}}+1\), and the reservation is released. The same later availability in this simple case does not make the two conventions identical: during the delay, the total \(A\) count differs, and other model components may use total counts rather than the available pool.

Deriving and Applying the Channel Hazard

For a mass-action channel, the binomial factors count eligible combinations. With \(x_A=4\) and a reaction requiring two \(A\) molecules, there are \({4\choose 2}=6\) eligible pairs. If the model defines the channel hazard as \(c_j\) per eligible pair, its total propensity is \(6c_j\). More generally, the propensity formula is a direct definition of the total rate at which that channel initiates from its current eligible pool. This avoids relying on an unstated assumption about dependence between possible molecular combinations.

Once a delayed event has initiated, its completion is part of the evolving system. For fixed delay \(\tau_j\), an event initiated at time \(t\) completes at \(t+\tau_j\). For a random delay \(L_j\), the model needs a stated distribution for \(L_j\). Pending events and their scheduled completion times, or an equivalent representation, are needed to determine future state changes. The molecular counts alone generally do not determine the future when pending events exist.

Worked Example: A Second-Order Reaction With Reservations

Consider \(2A\to B\), with a delay of 2 minutes and rate \(c=0.08\) per minute per eligible pair. Initially there are four \(A\) molecules and no reservations. The initiation propensity is

$$ a=0.08{4\choose 2}=0.08(6)=0.48\ \text{min}^{-1}. $$

Under completion-only molecular-count change, one initiation reserves two \(A\) molecules but does not yet change the total count \(x_A=4\) or the \(B\) count. The available count is now \(4-2=2\), so the propensity for another \(2A\to B\) initiation is

$$ a_{\text{after reservation}}=0.08{2\choose 2}=0.08(1)=0.0800\ \text{min}^{-1}. $$

At the first event's completion, the total counts change to \(x_A=2\) and \(x_B=x_B^{\text{initial}}+1\), and its reservation is released; the available count is \(x_A-r_A\), accounting for any other pending reservations. Without reservations, a second event could be initiated from the same molecules already committed to the first. For instance, the original four molecules cannot support three simultaneous \(2A\to B\) events, because those events would require six \(A\) molecules.

Competing Reactions Reveal the Difference

The convention matters most when other channels can use the reactants during the delay. Consider two channels: delayed \(A\to B\) with initiation rate \(0.30\) per minute per available \(A\), and immediate \(A\to C\) with rate \(0.20\) per minute per available \(A\). There are initially three \(A\) molecules. Both channels are assumed to draw from the same available pool; the delayed channel has a fixed 4-minute delay.

Worked Example: A Delayed Event Competing With an Immediate Reaction

Initially, each channel has three eligible \(A\) molecules, so their propensities are \(0.30(3)=0.90\ \text{min}^{-1}\) for delayed \(A\to B\) and \(0.20(3)=0.60\ \text{min}^{-1}\) for immediate \(A\to C\).

Suppose the delayed channel initiates first. With initiation-time consumption, one \(A\) is removed immediately. The immediate channel's propensity becomes \(0.20(2)=0.40\ \text{min}^{-1}\), while the delayed channel's next-initiation propensity is \(0.30(2)=0.60\ \text{min}^{-1}\). The pending event will add \(B\) after 4 minutes.

With completion-only molecular-count change and reservations, total \(A\) remains three, but one is reserved. Both channels use the two available \(A\) molecules, giving propensities \(0.30(2)=0.60\ \text{min}^{-1}\) and \(0.20(2)=0.40\ \text{min}^{-1}\). If the immediate channel then initiates, it consumes one available \(A\); the delayed event still has its reserved \(A\) to consume at completion. In contrast, using total \(A=3\) for subsequent propensities would wrongly allow another reaction to use the committed molecule.

Assumptions, Diagnostics, and Alternatives

A well-defined model states which counts each propensity uses, how reactants are committed, and what happens at completion. If a completion-only model intentionally allows reactants to be used again before completion, then it needs another rule to prevent negative counts or to handle a completion that cannot be fulfilled. Silently allowing an impossible completion is not a harmless approximation; it changes or invalidates the process.

Useful checks include the following:

  • Availability: Verify that every propensity is zero when too few uncommitted reactants are available.
  • Stoichiometric accounting: Confirm that an initiation-time event removes each reactant once and a completion-only event removes each reserved reactant once, at completion.
  • Pending-event consistency: At any time, reservations should equal the reactants committed by all pending completion-only events.
  • Units: Each total propensity should have units of inverse time; delay times should have units of time.
  • Boundary cases: If the delay is zero, initiation and completion effects occur at the same time, so the convention should reduce to the intended instantaneous reaction. If no reactants are available, no new event may initiate.

The main alternative is not a different propensity formula but a different biological interpretation. Initiation-time consumption is appropriate when reactants are committed or depleted as soon as the process begins. Completion-only count changes are appropriate when the molecules remain present but are committed to an ongoing process; the reservation bookkeeping represents that commitment. An Erlang transit chain or another distributed-delay representation instead introduces intermediate states and reaction steps. Such models can represent delay dispersion through explicit stages, but they are not automatically equivalent to a fixed-delay channel with either convention.

Neither convention is universally superior. The choice should reflect when molecules become unavailable and when products become observable in the biological process being represented. A model may also divide effects differently—for example, modifying one species at initiation and producing another at completion—but it must state those changes explicitly and calculate eligibility from the resulting available state.

Key takeaway: A delayed channel needs a total initiation hazard and an explicit state-change schedule. If reactants are not removed from molecular counts until completion, reserve them at initiation or specify another rule that guarantees feasible completions; do not count committed molecules as freely available.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. For \(3A\to C\) with \(x_A=5\) and rate \(c=0.10\) per eligible triple per minute, calculate the mass-action initiation propensity.
  2. A completion-only event requires two \(A\) molecules. There are five total \(A\) molecules, two already reserved, and \(c=0.06\) per pair per minute. What is the initiation propensity for another event?
  3. Why is it not enough to say that each eligible molecular combination has a marginal initiation probability \(c\,dt+o(dt)\) in order to conclude that the total channel hazard is the number of combinations times \(c\)?
  4. Under initiation-time consumption, what changes at initiation and completion for \(A\to B\)?

Solutions. (1) There are \({5\choose3}=10\) eligible triples, so \(a=0.10(10)=1.00\ \text{min}^{-1}\). (2) Three \(A\) molecules are available, so \(a=0.06{3\choose2}=0.06(3)=0.180\ \text{min}^{-1}\). (3) The initiation events for different combinations could be dependent; for example, perfectly correlated events do not produce a total hazard equal to the sum of their marginal hazards. Specify an independent-clock construction with negligible simultaneous events or define the total channel hazard directly. (4) Remove one \(A\) at initiation; at completion, add one \(B\). The delay separates those two changes.

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