1
$\begingroup$

It has been a little while since I have taken ordinary differential equation so I am struggling with what is a fairly simple problem. I thought providing some context would be helpful/interesting and also allow google to find this question.

I did an experiment, and got some data that shows how the amount of analyte bound to ligand changes with time. The x axis is time in seconds. The y axis is simply "Response" so it doesn't have concentration units, but I can still use the data to determine the binding kinetics. The amount of analyte in solution that is bound to the ligand over the course of the experiment is much less than the total amount in solution so the amount of analyte in solution, A0, is assumed to be a constant. The rate constants $k_d$ and $k_a$ are also constant throughout the reaction.

The one-to-one binding kinetics follows the following model:
$L+A \underset{k_a}{\stackrel{k_d}{\rightleftharpoons}} LA$

The differential equations for this model are:

$\begin{equation} \frac{d[L]}{dt} = - (k_a[L]A_0-k_d[LA]) \end{equation}$

$\begin{equation} \frac{d[LA]}{dt}=k_a[L]A_0-k_d[LA] \end{equation}$

These are the differential equations that I need to solve. What would be the best technique to use to solve them?

I tried doing an eigenvalue method but I ended up with a really ugly quadratic equation. I also tried Laplace transforms, but I need to review some things before I can pull that off.

Additional Information: I found a website, that provides the following set of solutions, where R is the measured response and Rmax (from what I can tell) is the max response. These are shown below. I am not sure if these formulas are correct. I believe C in them refers to A0.

$R = R_{eq}(1-e^{-(k_aC+k_d)(t-t_0)})$

$R=R_0e^{-k_d(t-t_0)}$

$R_{eq} = \frac{k_aC}{k_aC+k_d}R_{max}$

$R_{max} = max response$

I am including these in the answer for completeness, but I would really like to see a solution method to the system of ODEs above.

$\endgroup$
1
$\begingroup$

Let $[L]=y$ and $[LA]=x$, then clearly (by adding both of your ode’s together):

$$x’+y’=0$$

Therefore,

$$x+y=C$$

$$x=C-y$$

Where $C=x(0)+y(0)=[LA]_0+[L]_0$.


Because,

$$y’=k_d x-A_0 k_a y$$

We get:

$$y’=k_d (C-y)-A_0k_a y$$

$$y’+(k_d+k_aA_0) y=k_d C$$

A first order linear ode, that can be solved by integrating factors.

$\endgroup$
  • $\begingroup$ Thank you. For completeness the integrating factor is e^(kd+kaA0)t $\endgroup$ – dantonio Sep 16 '18 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.