# Derivation of time probability distributions for single molecule kinetics

If we have a protein $$\ce{A}$$ that can bind to protein $$\ce{B}$$:

$$\ce{A + B <=>[k_\mathrm{fwd,1}][k_\mathrm{rev,1}] AB}$$

and the bound complex $$\ce{AB}$$ can be converted into $$\ce{C}$$:

$$\ce{AB <=>[k_\mathrm{fwd,2}][k_\mathrm{rev,2}] C}$$

Assume we have a volume of 1 and some initial amount of $$\ce{A}$$ and $$\ce{B}$$ (which are $$A_0,$$ $$B_0$$) and that $$C_0 = 0.$$ We then measure the amount of time elapsed before the first molecule of $$C$$ is formed. The experiment is repeated many times, and the frequency of occurence of first $$C$$ during each time interval 0 to $$t_1$$, $$t_1$$ to $$t_2$$, $$t_2$$ to $$t_3$$, etc. is plotted, so that the result is a probability distribution of the time elapsed before the first molecule of $$C$$ is formed.

I would like to know what is the mathematical function that describes this distribution and how is it derived?

UPDATE: this is not a homework question and the point is not a specific answer. I would like references to literature/textbooks that derives these distributions analytically for simple examples like the above would be helpful. Have not seen a setup where distribution of duration is derived. Commenters say it's standard. Please give references.

• This is a very standard set-up for a kinetics problem. The typical way involves making the steady-state approximation which says that the change in concentration of AB with time is zero (essentially because it is eliminated immediately). If one can't make an assumption like this, there typically won't be a closed-form solution. One can always solve these types of differential equations numerically though. Basically I'm saying you probably need to give more info for this to be answerable. – jheindel Jun 3 '19 at 3:44
• It would help if you could at least give the equilibrium constants for the two steps to define the distribution of species at equilibrium. "The first C" is a strange thing to ask about - the standard kinetics are about bulk properties. – Karsten Theis Jun 3 '19 at 11:24
• @jheindel happy to assume AB doesn't change with time. do you have a link to an example where durations are computed for a problem of this sort? – user9277 Jun 3 '19 at 17:43
• not a homework question. happy with references instead of answer. closing it as off topic is silly – user9277 Jun 4 '19 at 1:14
• Here's a more basic write-up of the derivation for a unimolecular first-order reaction. Extending it to the more complex system is a matter of a bit more math using the same concepts. makarov.cm.utexas.edu/resources/Lecture-notes,-tutorials-etc./… – Andrew Jun 5 '19 at 16:30

To see how this calculation is done, let us consider the simpler case of just the second step shown above, ie $$\ce{AB ->[k_{f2}] C}$$. Since we are interested in the situation where no $$C$$ has yet formed, the reverse reaction can be ignored. I will assume that this reaction is first order in $$AB$$, so $$k_{f2}$$ has units of $$s^{-1}$$.

In this case, the rate constant $$k_{f2}$$ represents the average frequency of the reaction.

Now let's consider that we establish an infinite number of evenly spaced time intervals of length $$\Delta t$$. When we consider the probability distribution, we need to consider two factors:

1) The probability of reaction occurring in an interval $$t_n + \Delta t$$ and

2) The probability that the reaction has not yet occurred at $$t_n$$

The first value is simply the product of the rate constant $$k_{f2}$$ and the length of the time interval: $$k_{f2}\Delta t$$.

The second probability is slightly more complicated, since it is the chance of the reaction $$not$$ occurring in each of the $$n$$ intervals preceding $$t_n$$. That probability is given by $$1-(k_{f2}\Delta t)^n$$. Conveniently, as our time interval $$\Delta t$$ goes to $$0$$, this probability becomes a continuous function $$e^{-k_{f2}t}$$.

To get our final probability, we simply multiply these two probabilities:

$$[\text{total probability of reaction occurring in interval }t+dt] = [\text{probability of occurring in interval }t+dt] \times [\text{probability that reaction has not occurred by time }t]=k_{f2}dt\times e^{-k_{f2}t}$$

The probability as a continuous function of $$t$$ therefore can be written as

$$p_{\ce{AB->C}}(t)=k_{f2}e^{-k_{f2}t}$$.

As expected, integration from $$t=0$$ to $$t=\infty$$ gives an overall probability of $$1$$.

For the two-step process, if $$\ce{A+B <=> AB}$$ establishes equilibrium very fast relative to the $$\ce{AB->C}$$ reaction, then the above is a reasonable approximation (after adjustment of $$k_{f2}$$ to reflect the equilibrium concentration of $$AB$$). If the equilibrium is not established quickly, then you would use the same reasoning to determine the probability first of $$AB$$ having formed by time $$t$$ and then the probability of it reacting based on how long it has been in existence.

I have not read Makarov's textbook on single molecule kinetics, but I assume it covers these kinetics in much greater detail and might be a good resource:

Single Molecule Science: Physical Principles and Models 1st Edition by Dmitrii E. Makarov; CRC Press 2015; ISBN 9781466559516

• So with the combination of the two reactions as given in the OPs question, would there be a lag time (for AB to build up) and then exponential decay? – Karsten Theis Jun 12 '19 at 12:40
• Yes, instead of a y-intercept of $k_{f2}$, the probability would start at 0 and increase with the concentration of AB. – Andrew Jun 12 '19 at 14:41
• @Andrew: thank for un-closing my q and for very helpful explanation – user9277 Jun 13 '19 at 15:26
• @Andrew: quick clarification: shouldn't you take into account not only the rate $k_{f2}$ but also the amount of $AB$ you have formed thus far in time? the formation of $C$ should depend on the rate and on that abundance – user9277 Jun 13 '19 at 16:30
• Yes, the k should be scaled by a factor [AB]/[1 M] for concentrations other than 1 M. I left that out for simplicity, but I should have noted it. – Andrew Jun 13 '19 at 16:50

This is an alternative derivation to that given by @Andrew in his answer for the same first order reaction.

We suppose that the species has a probability $$p$$ of reacting in time interval $$\Delta t$$ and that this is independent of any past history, i.e. it depends only on the length of $$\Delta t$$ and for a sufficiently short time is proportional to this time, i.e. $$p=k\Delta t$$. The probability of not reacting in the time interval is $$1-p=1-k\Delta t$$.

If there is no reaction in this time interval and also in the next, the chance of surviving both is $$(1-k\Delta t)^2$$; and, by induction, for $$n$$ such intervals is $$(1-k\Delta t)^n$$.

If the total time $$t$$ is defined as $$t=n\Delta t$$ then the survival probability is $$(1-kt/n)^n$$. The chance of surviving unreacted at time $$t$$ is just this value when $$\Delta t$$ is made infinitesimally small, that is in the limit of $$n \to \infty$$ and this produces $$e^{-kt}$$ for the limiting value, when we recall the definition $$e^x= \lim_{n\to \infty} (1+x/n)^n$$.

As we do not consider just one molecule but many (initially $$n_0$$) the fraction remaining unreacted ($$n$$) at time $$t$$ is $$\displaystyle \frac{n}{n_0}=e^{-kt}$$.

We can also calculate the probability $$q(t)$$ of a molecule lasting a time interval that has a value between $$t \to t+dt$$. This is calculated as the product of not reacting between $$0\to t$$ and of reacting from time $$t\to t+dt$$. The first of these is $$e^{-n_0kt}$$ and the second, the probability that any of the $$n_0$$ molecules react in time $$dt$$ is $$n_okdt$$, thus the total probability is $$q(t)dt=n_0ke^{-n_0kt}dt$$. To test this experimentally a large number $$s$$ of time intervals of short length $$\Delta t$$ are needed then the number $$t\to t+\Delta t$$ should be $$sn_0ke^{-n_0kt}\Delta t$$. A plot of the log of the number of such intervals vs time should be linear.

Notes:

The chance of not reacting $$0\to t$$ can be found using the Binomial Distribution. This describes the very general case that given a large set of objects in which the chance of having an object of type $$w$$ is $$p$$, and then $$n$$ objects are withdrawn then the chance that exactly $$r$$ of these are of type $$w$$ is $$\displaystyle W(r)=\frac{n!}{(n-r)!r!}p^r(1-p)^{n-r}$$.

In our example we will have $$m$$ reactions in time $$t$$ from $$n_0$$ initial molecules. We make the connections $$n_0\equiv n$$ the number chosen or withdrawn, and $$m\equiv r$$ where the $$m$$ have the property that these many react in time $$t$$, thus $$\displaystyle W(r)=\frac{n_0!}{(n_0-m)!m!}p^m(1-p)^{n_0-m}$$. Furthermore, the chance of not reacting in time $$t$$ is $$(1-p)$$ and this is equal to the number that survive unreacted to the total number which is $$n/n_0$$, and is derived above, thus $$1-p=e^{-kt}$$.

Thus the chance of $$m$$ reacting is $$\displaystyle W(m)=\frac{n_0!}{n_0!0!} (1-e^{-kt} )^m (e^{-kt})^{n_0-m}$$.

In the case of no reaction $$0\to t$$ the probability is $$W(0)=e^{-n_0kt}$$