# Understanding association kinetics

I would like to understand the classic kinetic model of association / dissociation that tries to describe the concentration of a compound $$[\ce{AB}]$$. Let's say we have a model:

$$\ce{A + B} \xrightleftharpoons[k_{\text{off}}]{k_{\text{on}}} \ce{AB}$$

So $$k_{\text{on}}$$ is the rate of binding and $$k_{\text{off}}$$ is the rate of unbinding and the rate equation is:

$$\frac{d[\ce{AB}]}{dt}=k_{\text{on}}[\ce{A}][\ce{B}]-k_{\text{off}}[\ce{AB}]$$

I don't understand why texts say that if $$k_{\text{on}}$$ and $$k_{\text{off}}$$ are both constant then we reach a steady state, i.e., $$k_{\text{on}}\ce{AB}=k_{\text{off}}[\ce{AB}]$$ I mean: if, for example, the rate of binding, is greater then the rate of unbinding, shouldn't the concentration of the final compound increase up to infinity? What I have in mind is: we produce 4 proteins per second and degrade 2 proteins per second, then the net production is 2 proteins per second so, as time goes by, the concentration should increase and doesn't reach a plateau.

Why do we instead reach a plateau in the final concentration? I can understand the plateau in the plot of the reaction velocity but still, if the velocity is constant, still the concentration of the final compound should increase.

• As @David said, this reaction has nothing to do with Michaelin-Menten model of enzyme catalysis. It looks like one reversible reaction (which can be physical or chemical). Commented May 29 at 8:34
• I edited to remove the references to Michaelis-Menten, as we seem to all be in agreement that this is not that. Commented May 29 at 16:01
• @MaximilianPress — That’s a lot better. However, unless either A or B are a protein — not stated — we are left with a question for SE Chemistry. Commented May 29 at 17:31

There is a misunderstanding here that is very common when first learning about chemical kinetics. First you say:

if $$k_{\rm on}$$ and $$k_{\rm off}$$ are both constant

but then:

if the velocity is constant

This is not one and the same!

$$k$$ is the rate constant of reaction.* The actual reaction rate (i.e., the change in concentration over time) is given in your example by:

$$v_{\rm on} = k_{\rm on}[A][B]$$ for the forward (binding) reaction, and

$$v_{\rm off} = k_{\rm off}[AB]$$ for the reverse (unbinding) reaction.

Suppose that $$k_{\rm on} > k_{\rm off}$$. Then at first $$v_{\rm on} > v_{\rm off}$$ and the product AB is produced faster than it degrades. Therefore the concentration of AB increases. On the other hand, A and B are consumed in the reaction, so their concentrations decrease.

Now look at the equations for reaction rates above! The rate of the forward reaction, $$v_{\rm on}$$, depends on concentrations of starting materials, $$[A][B]$$; and the rate of the reverse reaction, $$v_{\rm off}$$, depends on the concentration of product, $$[AB]$$. Therefore, as the reaction proceeds, $$v_{\rm on}$$ must decrease and $$v_{\rm off}$$ must increase until they reach equilibrium where both rates are equal.

* Note: As a convention, reaction rate constants are usually denoted by lowercase $$k$$. Uppercase $$K$$ is reserved for equilibrium constants, or the Michaelis constant $$K_M$$, or a number of other constants.

• "Suppose that 𝑘on>𝑘off." But kon is a second-order rate constant whereas koff is a first-order rate constant, and have very different units (M(-1).s(-1) versus s(-1), for example) and as such is it surely meaningless to compare them? Commented May 28 at 14:09
• @user338907 Not necessarily. For example, the equilibrium constant is equal to the ratio of forward and reverse reaction rate constants, so here K = k_on/k_off. Then k_on > k_off means that K > 1, i.e., equilibrium favours the products. Commented May 29 at 9:52
• You cannot directly compare the magnitude of a first order rate-constant with a second-order rate constant. This IS a fundamental mistake in dealing with enzyme kinetics. How can something with dimensions of M(-1).s(-1) be greater or less than (or equal to) something with dimensions of s(-1)? To take a more obvious example, how can pH 7 be considered greater than or less than 100 degrees celsius? (Going back to MM, the steady-state dfn of Km is the sum of two first-order rate constants, divided by a second-order rate content, a good eg of the value of dimen. analysis) Commented May 29 at 10:29
• This is probably a more relevant answer than mine. Commented May 29 at 16:01

Update: As noted by commenter, the example stated in the question is not actually Michaelis-Menten but is rather an example of binding kinetics. See also the other answer, which is likely more relevant.

I would suggest, if you would like to find the steady-state concentration of $$[AB]$$, that you can solve this yourself mathematically. Try differentiating and setting the $$\Delta [AB]$$ side of the equation to zero, and solving. You will find:

$$\Delta [AB] = (k_{on}[A][B] - k_{off}[AB])\Delta t$$

$$0 = (k_{on}[A][B] - k_{off}[AB])\Delta t$$

Rearranging:

$$k_{on}[A][B] = k_{off}[AB]$$

And finally:

$$\frac{[AB]}{[A][B]} = \frac{k_{on}}{k_{off}}$$

Note that if the $$k$$ are constant, this amounts to the law of mass action.

You can also do a forward simulation (see R code):

# association/dissociation kinetics
kinetic_sim = function(k_on, k_off, a, b) {
ab = 0
ab_ts = c()
a_ts = c()
b_ts = c()
for (i in 1:1000) {
ab_old = ab
ab = ab * (1 - k_off) + k_on * (a * b)
a = a - a*b*k_on + ab_old * k_off
b = b - a*b*k_on + ab_old * k_off
#cat(i, a, b, ab, k_on*a*b, k_off*ab, "\n", sep=" ")
ab_ts = append(ab_ts, ab)
a_ts = append(a_ts, a)
b_ts = append(b_ts, b)

}
return(list(ab_ts, a_ts, b_ts))
}
out = kinetic_sim(.1, .2, 1, 1)
plot(out[[1]] / (out[[2]] * out[[3]]), ylab="[AB] / ([A] * [B])", xlab="time step")
plot(out[[1]], ylab="[AB]", xlab="time step")


You can see that both the ratio and the final concentration equilibrate.

• A few suggestions. 1. The reaction example shown by the OP does not depict an enzyme catalysis. So using the term "Michaelis-Menten" would be misleading. 2. The system has 3 variables but there are only two independent differential equations. The third equation should be the conservation of mass (a + b + ab = constant). Alternatively (deviating from OP's example), you can model the system as an open system with rates of synthesis and degradation of the different components. Commented May 29 at 8:43
• @WYSIWYG thanks for the thoughtful comment. 1) You are absolutely correct that this is not michaelis-menten, my brain went immediately to the kinetic example shown without thinking about its name. 2) you are correct, the constant mass is implicitly modeled as shown in the simulation. Commented May 29 at 15:51