Buzzer

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Cartoon of a Buzzer
The Buzzer is a network motif that consists of a species $R$ that is phosphorylated by the Kinase $E_K$ and dephosphorylated by the Phosphatase $E_P$ (see picture on the right). $E_K$ is the signal for the reaction $R \rightarrow RP$ and $E_P$ is the signal for the reaction $RP \rightarrow R$. We assume that the total amount of $R$ stays constant $R_T=R+RP$.

Let us first study the equation for the case of pure Kinase signaling. We call the signal $S=E_K$. Following mass action kinetics we obtain the equation \begin{equation} \frac{dRP}{dt}=k_1 S \cdot R -k_2 RP \end{equation} The steady state concentration $RP_{St}$ for $\frac{dRP}{dt}=0$ can by use of $R_T=R+RP$ be written as \begin{align} 0&=k_1S \cdot (R_T-RP_{St}) -k_2 RP_{St}\\ RP_{St}(k_2+S k_1)&=k_1S \cdot R_T\\ RP_{St}&=\frac{k_1S \cdot R_T}{k_2+S k_1} \end{align} thus we obtain for the steady state concentration $RP_{St}$ \begin{equation} RP_{St}=\frac{S \cdot R_T}{k_2/k_1+S} \end{equation} If we plot $RP_{St}$ as a function of the signal, we find a hyperbolic signal-response curve.

Hyperbolic-response-curve.png

To study both phosphorylation and dephosphorylation events characteristic for a Buzzer (see picture upper right), we consider the Michaelis-Menten kinetics for phosphorylation. \begin{equation} E_K+R \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} E_KR \overset{k_2}{\rightarrow}E_K+RP \end{equation} \begin{equation} v_K=\frac{k_2 E_K \cdot R}{R + K_{m1}} \end{equation}

and dephosphorylation \begin{equation} RP+E_P \underset{k_{-3}}{\overset{k_3}{\rightleftarrows}} RPE_P \overset{k_4}{\rightarrow}E+R \end{equation} \begin{equation} v_P=\frac{k_4 E_P \cdot RP}{RP + K_{m2}} \end{equation}

The change in concentration of $R$ is given by the difference \begin{equation} \frac{d R}{dt}=v_P-v_K=\frac{k_4 E_P \cdot RP}{RP + K_{m2}}-\frac{k_2 E_K \cdot R}{R + K_m1} \end{equation} The steady state concentration of $R$ is given by $\frac{d R}{dt}=0$ and by use of $R_T=R+RP$ we find \begin{align} \frac{d R}{dt}&=0\\ \frac{k_4 E_P \cdot RP}{RP + K_{m2}}&=\frac{k_2 E_K \cdot R}{R + K_{m1}} \\ \frac{k_4 E_P \cdot (R_T-R)}{(R_T-R) + K_{m2}}&=\frac{k_2 E_K \cdot R}{R + K_{m1}} \\ \end{align} By renaming $v_1=k_2E_K$ and $v_2=k_4E_P$ and pulling out $R_T$ we get \begin{align} \frac{v_2 \cdot (1-R/R_T)}{(1-R/R_T) + K_{m2}/R_T}&=\frac{v_1 \cdot R/R_T}{R/R_T + K_{m1}/R_T} \end{align} Now we call $x=\frac{R}{R_T}$, $J_1=\frac{K_{m1}}{R_T}$ and $J_2=\frac{K_{m2}}{R_T}$ \begin{align} \frac{v_2 \cdot (1-x)}{(1-x) + J_2}&=\frac{v_1 \cdot x}{x + J_1}\\ v_2 \cdot (1-x)[x + J_1]&=v_1 \cdot x[(1-x) + J_2]\\ J_2 v_1+ z v_1 - J_2 v_1 x - x^2 v_1 &= x v_2 J_1+ v_2 x - x^2 v_2\\ x^2 (v_2 - v_1) - x \underbrace{(v_2 - v_1 + J_1 v_2 + J_2 v_1)}_{B} + v_1 J_2 &= 0\\ \end{align} This is a quadratic equation where we are interested only in one solution \begin{align} x = \frac{B - \sqrt{B^2 - 4 (v_2 - v_1) v_1 J_2}}{2 (v_2 - v_1)} &= \frac{B - \sqrt{B^2 - 4 (v_2 - v_1) v_1 J_2}}{2 (v_2 - v_1)} \cdot \frac{B + \sqrt{B^2 - 4 (v_2 - v_1) v_1 J_2}}{B + \sqrt{B^2 - 4 (v_2 - v_1) v_1 J_2}}\\ x &= \frac{ 4 (v_2 - v_1) v_1 J_2}{2 (v_2 - v_1)} \cdot \frac{1}{B + \sqrt{B^2 - 4 (v_2 - v_1) v_1 J_2}}\\ x &= \frac{ 2 v_1 J_2}{B + \sqrt{B^2 - 4 (v_2 - v_1) v_1 J_2}}. \qquad \qquad (3) \end{align}

We arrived at the Goldbeter-Koshland function \begin{align} x = \frac{R}{R_T } = G(v_1, v_2, J_1, J_2) &= \frac{ 2 v_1 J_2}{B + \sqrt{B^2 - 4 (v_2 - v_1) v_1 J_2}}.\\ \end{align} which gives the relative steady state concentration $x$ as a function of $v_1, v_2, J_1, J_2$.

How can we interpret this solution? If we plot both $v_K$ and $v_P$ in the same graph, then their intersection gives the steady state solution $\frac{dR}{dt}=0$. Notice, $v_1$ is proportional to $E_K$ and $v_2$ is proportional to $E_P$. If we keep the signal $E_P$ and hence $v_2$ fixed and vary the signal strength of $E_K=v_1/k_2$ how does the response $R_{St}$ change?

Buzzersol.png

The graphic shows, that the steady state concentration rises rapidly near the point $v_1\approx v_2$. If we plot $G(v_1, v_2, J_1, J_2)$ as a function of $v_1$ we find the sigmoidal signal response curve

Goldbeterkoshlandfkt.png
Mathematica code
J1 := 1/10
J2 := 1/50
v2 := 3
B[v1_] := v2 - v1 + J1*v2 + v1*J2
Plot[{2 v1*J2/(B[v1] + (B[v1]^2 - 4 (v2 - v1) v1*J2)^(1/2))}, {v1, 0, 
 10}, PlotRange -> {{0, 10}, {0, 1}}, 
AxesLabel -> {"\!\(\*SubscriptBox[\"E\", \"K\"]\)", R}, 
BaseStyle -> {FontWeight -> "Bold", FontSize -> 20}, Ticks -> None ]



Further Reading:

  • Zoltan Szallasi et al. - System Modeling in Cellular Biology: From Concepts to Nuts and Bolts
  • ALBERT GOLDBETER AND DANIEL E. KOSHLAND JR. - An amplified sensitivity arising from covalent modification in biological systems
  • John J Tyson, Katherine C Chen and Bela Novak - Sniffers, buzzers, toggles and blinkers: dynamics of regulatory and signaling pathways in the cell [[1]]

Video Lectures:

  • Network Dynamics and Cell Physiology by John J. Tyson - Lecture 2: Network motifs: sniffers, buzzers, toggles and blinkers [2]