I am trying to understand the following equations from a paper by Wang et al. [1, p. 5 at the SI]:

Michaelis–Menten Representation of the Kinesin Cycle

We assumed that there is a strong coupling between ATP turnover rate of kinesin and kinesin stepping on an MT. Each chemical step in the chemomechanical cycle of kinesin was assumed to be irreversible, except ATP binding and releasing (Fig. 1, step i). Consider the following scheme:

$$\large\ce{0 <->[$k_\mathrm{ATP}\lbrack\ce{S}\rbrack$][$k_\mathrm{-ATP}$] 1 ->[$k_\mathrm{s}$] 2 ->[$k_\mathrm{attach}$] 3 ->[$k_\mathrm{ADP}$] 4 ->[$k_\mathrm{h}$] 5 ->[$k_\mathrm{-P}$] 6}$$

Assume that $F_i(t)$ is the probability for the system to reach state $6$ if, at time $t = 0$, the system is in state $i.$ This function is governed by the backward master equations:

$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)\label{eqn:1}\tag{S1}$$

Why does equation \eqref{eqn:1} read as

$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)$$

and not

$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{-ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)?$$

I would like to change/modulate $S(t)=\cos(t)$ in time and see how it changes the states $F_i(t)$


1 Wang, Q.; Diehl, M. R.; Jana, B.; Cheung, M. S.; Kolomeisky, A. B.; Onuchic, J. N. Molecular Origin of the Weak Susceptibility of Kinesin Velocity to Loads and Its Relation to the Collective Behavior of Kinesins. PNAS 2017, 114 (41), E8611–E8617. https://doi.org/10/gdj4h5.

  • 2
    $\begingroup$ I'm still not satisfied that my answer gives a clear explanation, so I changed the title of the question in the hopes of attracting attention from someone more expert. This really isn't about Michaelis-Menten kinetics in the traditional sense. $\endgroup$
    – Andrew
    Commented Jun 20, 2019 at 11:57
  • $\begingroup$ @Andrew I started a bounty. I hope it will help us to get a canonical answer. $\endgroup$
    – 0x90
    Commented Jun 21, 2019 at 4:16
  • $\begingroup$ @Andrew Here is a paper that explains forward Master equations (very similar to chemical kinetics) and backward Master equations (used for first passage problems, and looking unfamiliar: arxiv.org/pdf/1503.00291.pdf $\endgroup$
    – Karsten
    Commented Jun 23, 2019 at 15:38

2 Answers 2


UPDATED: I wrote a first answer assuming that $t>0$ which got close to what's in the paper, but not quite the same. Thanks go to Karsten Theis for pointing out that $t<0$ in these "backward equations". Here's a corrected explanation:

The scenario is that we have a system that can be in any of seven consecutive states (numbered 0 through 6). At some time $t<0$, we want to know something about the probability of a system in each state reaching state 6 by time $t=0$, which is expressed as $F_i(t)$ where $i$ indicates the state number.

We start with a system in state 6 at time $t$. Since reversal back to state 5 is not permitted and there is no state beyond state 6, any system that is in state 6 will remain in state 6 up to (and past) time $t=0$. Thus the probability $F_6(t)=1$. In the paper, they use a Dirac $\delta$ function to indicate this, because that is necessary when $F_6(t)$ is used in the other equations.

For the other five states, the equations do not give explicit expressions for the probabilities, but instead express the change in probability over time as a function of the probability in a differential equation.

Let's consider a system that at time $t$ is in state $5$. During a time interval $\Delta t$, two things can happen: Either (1) the system remains in state $5$ or (2) the system switches to state $6$.

One assumption they seem to make is that the only way the probability can change is if the system changes state. So if the system remains in state 5, its probability $F_5(t+\Delta t)$ remains equal to $F_5(t)$. If it changes to state 6, then its probability increases to $F_6(t)$. The change is thus a weighted combination of these two.

In more detail, if we consider the probability of switching from state 5 to state 6 during the interval $\Delta t$ to be $P_{56}$, we can say that our new probability is the weighted sum

$F_5(t+\Delta t)=P_{56}F_6(t)+(1-P_{56})F_5(t)$,

so our change in probability is

$\Delta F_5=P_{56}F_6(t)+(1-P_{56})F_5(t)-F_5(t)=P_{56}[F_6(t)-F_5(t)]$.

Our last step is to figure out what $P_{56}$ is, and it is given by the rate constant for the change from state $5$ to state $6$. The rate constant is the average frequency of the reaction, so the probability of the reaction occurring in a given time interval is the rate constant times the length of time. So we replace $P_{56}$ with $k_{-P}\Delta t$ and get

$\Delta F_5(t)=k_{-P}\Delta t[F_6(t)-F_5(t)]$.

Dividing by $\Delta t$, we have

$\dfrac{\Delta F_5(t)}{\Delta t}=k_{-P}[F_6(t)-F_5(t)]$.

As $\Delta t$ goes to $0$, this becomes the derivative:


You can repeat this for the other five states to generate the other equations in the paper. The key point is that even when back reactions are permitted (such as from state 1 to state 0), the probability function is only for the systems that began in the indicated state, so the rate constant for a system coming back to that state does not enter into the equation. An implicit assumption is that the system does not make two moves during the time interval $dt$. That is, it cannot move from state 0 to state 1 and back to state 0.

  • $\begingroup$ They used δ in two contexts: δ(t) is the delta function, so it is a probability of one at time zero, just like you thought it should be. δ is also the step size, the multiplier in the $V$ equation. $\endgroup$
    – Karsten
    Commented Jun 13, 2019 at 13:56
  • $\begingroup$ How would it not reach state 6? The first step never reaches equilibrium because product is continuously removed. Therefor, I would think the reaction would go to completion. $\endgroup$
    – Karsten
    Commented Jun 13, 2019 at 14:18
  • $\begingroup$ @KarstenTheis - I agree that the scheme as written suggests that everything eventually gets to state 6, but in that case all of the probabilities would be 1, so there must be some leakage that isn't shown. Perhaps the kinesin irreversibly detaches before reaching state 6? I didn't read through the paper. There could be an explanation in there somewhere. $\endgroup$
    – Andrew
    Commented Jun 13, 2019 at 15:58
  • $\begingroup$ @KarstenTheis - also thanks for the clarification on the delta function. But wouldn't that be equal to 0 at $t \neq 0$? I would expect $F_6(t)$ to be 1 for all t, not just t=0. $\endgroup$
    – Andrew
    Commented Jun 13, 2019 at 15:59
  • $\begingroup$ These are not cumulative probabilities to be in a state, but probabilities of state transitions. t = 0 is the reference point where all species transition from state 5 to state 6. The other equations describe transition probabilities for t < 0, hence reverse master equation, see en.wikipedia.org/wiki/Kolmogorov_backward_equations_(diffusion) $\endgroup$
    – Karsten
    Commented Jun 13, 2019 at 16:21

First passage problems

This is a first passage problem, asking when a system reaches a certain final state for the first time . If you were to use a simulation to explore this, you would erase parts of trajectories right after reaching the final state before summing up results. The Backwards Master Equation is especially suited for first passage problems (one way to think about it is that there is complete knowledge at the end of the trajectory, and less and less knowledge as you go backward in time). For folks familiar with chemical kinetics, this is very confusing because the typical case is maximal knowledge at the initial state, with stuff diverging as time passes.

Sketch of Derivation

This follows https://arxiv.org/pdf/1503.00291.pdf, equations (15) - (17). I changed the name of the conditional probability to stay closer to the paper the OP asked about (the conditional probability is called $S$ and the transition probabilities are called $\alpha$ in the derivation I am paraphrasing, but here I am using $F$ and $k$ instead.)

$F_i(0,t)$ is the conditional probability that given a system that is a state $i$ at time 0 will transition into the final state 6 at time t. To derive the Backward Master Equation, we consider two exclusive events shortly after $t = 0$.

  1. The system transitioned into a different state $j$. The probability that this happened is governed by the reaction rates $k_{ij}$ with state $i$ as reactant and state $j$ as product. This is where the asymmetry in the final equations comes from. The conditional probability to reach the final state at time = $t$ for a system that transitioned into state $j$ is $F_j(\Delta t,t)$.
  2. The system is still in state $i$. The probability that this happened is one minus that the system transitioned into a different state $j$ during that short time. The conditional probability to reach the final state is now $F_i(\Delta t,t)$.

From this, we get:

$$F_i(0,t) = \big(\sum_j{k_{ij} \cdot F_j(0 + \Delta t,t)}\big) + (1 - \sum_j{k_{ij})} \cdot F_i(0 + \Delta t,t)$$

Again, we only consider reaction out of state $i$, not into state $i$. The conditional probabilities are stationary (not dependent on time, the system has no memory), so

$$F_i(t_1, t_2) = F_i(0, t_2 - t_1)$$

Using that, we can rewrite as:

$$F_i(0,t) = \big(\sum_j{k_{ij} \cdot F_j(0,t - \Delta t)}\big) + (1 - \sum_j{k_{ij})} \cdot F_i(0,t - \Delta t)$$

Now we can make $\Delta t$ infinitesimally small and write the differential:

$$\frac{dF_i(0,t)}{dt} = \big(\sum_j{k_{ij} \cdot F_j(0,t)}\big) - \sum_j{k_{ij}} \cdot F_i(0,t)$$

In the paper, they used $F_i(t)$ instead of $F_i(t_1, t_2)$, but we needed to allow the first time point to be different from zero in the derivation, so I included it as a argument in the function $F$.

Comparison to (Forward) Master Equation

The Forward Master Equation matches what we usually do in chemical kinetics. The variable $t$ is conventional time, and we figure out the concentration of a species at time $t + dt$ by asking about the rates to make that species and to use up that species. So both $k_{ATP}$ and $k_{-ATP}$ would appear in the same differential equation. The way the Backward Master Equation is derived (we are asking about state $i$ at a certain time $t$, and getting information information about it by going backwards from time $t + \Delta t$), we only have to consider paths from that state to other states.

Modulating [ATP]

This is just a guess. If the ATP concentration changes rapidly compared to the timescale of the reaction steps, you can just use an average rate. If it is changed very slowly, the overall speed of the motor would change proportionally (because the system has sufficient time to "settle in"). If the modulation matches the timescale of the individual steps, it will be complicated. An analogy is NMR and the timescale of conformational change (if the conformational change is fast, you get one averaged signal; if it is slow, you get two separate signals; if it is intermediate, it gets complicated and you lose it).


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