Dimensional Dependence of Binding Kinetics

Abstract

In the context of protein–protein binding, the dissociation constant is used to describe the affinity between two proteins. For protein–protein interactions, most experimentally-measured dissociation constants are measured in solution and reported in units of volume concentration. However, many protein interactions take place on membranes. These interactions have dissociation constants with units of areal concentration, rather than volume concentration. Here, we present a novel, stochastic approach to understanding the dimensional dependence of binding kinetics. Using stochastic exit time calculations, in discrete and continuous space, we derive general reaction rates for protein–protein binding in one, two, and three dimensions and demonstrate that dimensionality greatly affects binding kinetics. Further, we present a formula to transform three-dimensional experimentally-measured dissociation constants to two-dimensional dissociation constants. This conversion can be used to mathematically model binding events that occur on membranes.

Appendix

Theorem 1

For a matrix, ${\textbf {W}}$, of the form ${\textbf {W}}=\delta {\textbf {L}} -k {\textbf {E}}$ where ${\textbf {L}}$ is a matrix with zero column sum, $\{\delta ,k\} \in \mathbb {R}$, and ${\textbf {E}}={\textbf {ee}}^T$ with

$$\begin{aligned} {\textbf {e}}= \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \end{aligned}$$

${\textbf {W}}^{-1}=\frac{{\textbf {A}}}{\delta }-\frac{\pmb {\Pi }}{k}$ where ${\textbf {A}}$ is independent of both $\delta$ and k, and $\pmb {\Pi }= \pmb {\eta } {\textbf {1}}^T$ where $\pmb {\eta }$ is an element of the null(L) normalized such that $\pmb {\eta }(1)=1$.

Proof of Theorem 1.

For a matrix ${\textbf {W}}$ which can be written as ${\textbf {W}}=\delta {\textbf {L}} -k {\textbf {E}}$ where ${\textbf {L}}$ is a matrix with zero column sum and ${\textbf {E}}={\textbf {ee}}^T$ with

$$\begin{aligned} {\textbf {e}}= \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \end{aligned}$$

(112)

it follows that

$$\begin{aligned}{} & {} \frac{d{\textbf {W}}}{dk}=-{\textbf {E}} \end{aligned}$$

(113)

$$\begin{aligned}{} & {} \frac{d{\textbf {W}}}{d\delta }={\textbf {L}}. \end{aligned}$$

(114)

Observe that

$$\begin{aligned} 0= & {} \frac{d}{dk}({\textbf {W}}{} {\textbf {W}}^{-1}) \end{aligned}$$

(115)

$$\begin{aligned}= & {} \frac{d{\textbf {W}}}{dk}{} {\textbf {W}}^{-1}+{\textbf {W}}\frac{d{\textbf {W}}^{-1}}{dk} \end{aligned}$$

(116)

$$\begin{aligned}= & {} -{\textbf {E}}{} {\textbf {W}}^{-1}+{\textbf {W}}\frac{d{\textbf {W}}^{-1}}{dk}. \end{aligned}$$

(117)

Therefore,

$$\begin{aligned} {\textbf {W}}\frac{d{\textbf {W}}^{-1}}{dk}={\textbf {E}}{} {\textbf {W}}^{-1}. \end{aligned}$$

(118)

Now consider the matrix ${\textbf {W}}(\frac{d{\textbf {W}}^{-1}}{dk}-\frac{\pmb {\Pi }}{k^2}){\textbf {W}}$.

$$\begin{aligned} {\textbf {W}}\left( \frac{d{\textbf {W}}^{-1}}{dk}-\frac{\pmb {\Pi }}{k^2}\right) {\textbf {W}}= & {} {\textbf {W}}\frac{d{\textbf {W}}^{-1}}{dk}{} {\textbf {W}}-\frac{{\textbf {W}}\pmb {\Pi } {\textbf {W}}}{k^2} \end{aligned}$$

(119)

$$\begin{aligned}= & {} {\textbf {E}}{} {\textbf {W}}^{-1}{} {\textbf {W}}-\frac{{\textbf {W}}\pmb {\Pi } {\textbf {W}}}{k^2} \end{aligned}$$

(120)

$$\begin{aligned}= & {} {\textbf {E}}-\frac{(\delta {\textbf {L}}-k{\textbf {E}})\pmb {\eta } {\textbf {1}}^T(\delta {\textbf {L}}-k{\textbf {E}})}{k^2}. \end{aligned}$$

(121)

Recall that ${\textbf {L}}\pmb {\eta } = 0$ and that ${\textbf {1}}^T{\textbf {L}}=0$. Therefore,

$$\begin{aligned} {\textbf {E}}-\frac{(\delta {\textbf {L}}-k{\textbf {E}})\pmb {\eta } {\textbf {1}}^T(\delta {\textbf {L}}-k{\textbf {E}})}{k^2}= & {} {\textbf {E}}-\frac{k^2{\textbf {E}}\pmb {\eta }{} {\textbf {1}}^T{\textbf {E}}}{k^2} \end{aligned}$$

(122)

$$\begin{aligned}= & {} {\textbf {E}}-{\textbf {ee}}^T\pmb {\eta }{} {\textbf {1}}^T{\textbf {ee}}^T \end{aligned}$$

(123)

$$\begin{aligned}= & {} {\textbf {E}}-{\textbf {E}} \end{aligned}$$

(124)

$$\begin{aligned}= & {} 0. \end{aligned}$$

(125)

Therefore,

$$\begin{aligned} {\textbf {W}}\left( \frac{d{\textbf {W}}^{-1}}{dk}-\frac{\pmb {\Pi }}{k^2}\right) {\textbf {W}} = 0 \end{aligned}$$

(126)

and since ${\textbf {W}}$ is nonsingular

$$\begin{aligned} \frac{d{\textbf {W}}^{-1}}{dk}-\frac{\pmb {\Pi }}{k^2} = 0 \end{aligned}$$

(127)

This can be rewritten as

$$\begin{aligned} \frac{d}{dk}\left( {\textbf {W}}^{-1}+\frac{\pmb {\Pi }}{k}\right) = 0. \end{aligned}$$

(128)

which leads to the conclusion that ${\textbf {W}}^{-1}+\frac{\pmb {\Pi }}{k}$ is independent of k. Therefore,

$$\begin{aligned} {\textbf {W}}^{-1}+\frac{\pmb {\Pi }}{k} = f(\delta ). \end{aligned}$$

(129)

It now remains to prove that ${\textbf {A}}$ is independent of $\delta$. To do this, we first show that ${\textbf {E}}{} {\textbf {A}}={\textbf {A}}{} {\textbf {E}}=0$.

Assume that ${\textbf {W}}^{-1} = \frac{{\textbf {A}}}{\delta }-\frac{\pmb {\Pi }}{k}$. Since ${\textbf {W}}{} {\textbf {W}}^{-1}={\textbf {I}}$, it follows that

$$\begin{aligned} {\textbf {I}}= & {} (\delta {\textbf {L}}-k{\textbf {E}})\left( \frac{{\textbf {A}}}{\delta }-\frac{\pmb {\Pi }}{k}\right) \end{aligned}$$

(130)

$$\begin{aligned}= & {} \frac{\delta {\textbf {L}}{} {\textbf {A}}}{\delta }-\frac{\delta {\textbf {L}}\pmb {\Pi }}{k}-\frac{k{\textbf {E}}{} {\textbf {A}}}{\delta }+\frac{k{\textbf {E}}\pmb {\Pi }}{k} \end{aligned}$$

(131)

$$\begin{aligned}= & {} \frac{{\textbf {W}}{} {\textbf {A}}}{\delta }+{\textbf {E}}\pmb {\Pi }. \end{aligned}$$

(132)

Rearranging,

$$\begin{aligned} \frac{{\textbf {W}}{} {\textbf {A}}}{\delta }= & {} {\textbf {I}}-{\textbf {E}}\pmb {\Pi }. \end{aligned}$$

(133)

Notice that ${\textbf {I}}-{\textbf {E}}\pmb {\Pi }$ has the structure

$$\begin{aligned} \begin{bmatrix} 0 &{} -1 &{} -1 &{} \cdots &{} -1 \\ &{} 1 &{} &{} &{} \\ &{} &{} 1 &{} &{} \\ &{} &{} &{} \ddots &{} \\ &{} &{} &{} &{} 1 \end{bmatrix}. \end{aligned}$$

(134)

Observe that $({\textbf {I}}-{\textbf {E}}\pmb {\Pi }){\textbf {E}}$ is a matrix of all zeros.

$$\begin{aligned} \begin{bmatrix} 0 &{} -1 &{} -1 &{} \cdots &{} -1 \\ &{} 1 &{} &{} &{} \\ &{} &{} 1 &{} &{} \\ &{} &{} &{} \ddots &{} \\ &{} &{} &{} &{} 1 \end{bmatrix} \begin{bmatrix} 1 &{} 0 &{} 0 &{} \cdots &{} 0 \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \end{bmatrix} = \begin{bmatrix} 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 \end{bmatrix}. \end{aligned}$$

(135)

Therefore, $\frac{{\textbf {W}}{} {\textbf {A}}{} {\textbf {E}}}{\delta } = ({\textbf {I}}-{\textbf {E}}\pmb {\Pi }){\textbf {E}} = 0$. Since ${\textbf {W}}$ is nonsingular, ${\textbf {A}}{} {\textbf {E}}=0$. We can similarly show that ${\textbf {E}}{} {\textbf {A}}=0$. We now show that ${\textbf {A}}$ is independent of $\delta$.

$$\begin{aligned} 0= & {} \frac{d}{d\delta }({\textbf {W}}{} {\textbf {W}}^{-1}) \end{aligned}$$

(136)

$$\begin{aligned}= & {} \frac{d{\textbf {W}}}{d\delta }{} {\textbf {W}}^{-1}+{\textbf {W}}\frac{d{\textbf {W}}^{-1}}{d\delta } \end{aligned}$$

(137)

$$\begin{aligned}= & {} {\textbf {L}}\left( \frac{{\textbf {A}}}{\delta }-\frac{\pmb {\Pi }}{k}\right) +(\delta {\textbf {L}}-k{\textbf {E}})\left( \frac{d{\textbf {A}}}{\delta }\frac{1}{\delta }-\frac{{\textbf {A}}}{\delta ^2}\right) \end{aligned}$$

(138)

$$\begin{aligned}= & {} \frac{{\textbf {L}}{} {\textbf {A}}}{\delta }-\frac{{\textbf {L}}\pmb {\Pi }}{k}+\frac{{\textbf {W}}}{\delta }\frac{d{\textbf {A}}}{d\delta }-\frac{\delta {\textbf {L}}{} {\textbf {A}}}{\delta ^2}+\frac{k{\textbf {E}}{} {\textbf {A}}}{\delta ^2} \end{aligned}$$

(139)

$$\begin{aligned}= & {} \frac{{\textbf {W}}}{\delta }\frac{d{\textbf {A}}}{d\delta }. \end{aligned}$$

(140)

Therefore, $\frac{d{\textbf {A}}}{d\delta }=0$ and ${\textbf {A}}$ is indeed independent of $\delta$ which proves that ${\textbf {W}}^{-1}=\frac{{\textbf {A}}}{\delta }-\frac{\pmb {\Pi }}{k}$ with ${\textbf {A}}$ independent of $\delta$ and k.