Urysohn universal space

A kinase Y and a phosphatase X that act on a protein Z; one possible application for the Goldbeter–Koshland kinetics

The Goldbeter–Koshland kinetics describe a steady-state solution for a 2-state biological system. In this system, the interconversion between these two states is performed by two enzymes with opposing effect. One example would be a protein Z that exists in a phosphorylated form Z_P and in an unphosphorylated form Z; the corresponding kinase Y and phosphatase X interconvert the two forms. In this case we would be interested in the equilibrium concentration of the protein Z (Goldbeter–Koshland kinetics only describe equilibrium properties, thus no dynamics can be modeled). It has many applications in the description of biological systems.

The Goldbeter–Koshland kinetics is described by the Goldbeter–Koshland function:

${\begin{aligned}z={\frac {[Z]}{[Z]_{0}}}=G(v_{1},v_{2},J_{1},J_{2})&={\frac {2v_{1}J_{2}}{B+{\sqrt {B^{2}-4(v_{2}-v_{1})v_{1}J_{2}}}}}\\\end{aligned}}$

with the constants

${\begin{aligned}v_{1}=k_{1}[X];\ v_{2}&=k_{2}[Y];\ J_{1}={\frac {K_{M1}}{[Z]_{0}}};\ J_{2}={\frac {K_{M2}}{[Z]_{0}}};\ B=v_{2}-v_{1}+J_{1}v_{2}+J_{2}v_{1}\end{aligned}}$

Graphically the function takes values between 0 and 1 and has a sigmoid behavior. The smaller the parameters J₁ and J₂ the steeper the function gets and the more of a switch-like behavior is observed. Goldbeter–Koshland kinetics is an example of ultrasensitivity.

Derivation

Since we are looking at equilibrium properties we can write

${\begin{aligned}{\frac {d[Z]}{dt}}\ {\stackrel {!}{=}}\ 0\end{aligned}}$

From Michaelis–Menten kinetics we know that the rate at which Z_P is dephosphorylated is $r_{1}={\frac {k_{1}[X][Z_{P}]}{K_{M1}+[Z_{P}]}}$ and the rate at which Z is phosphorylated is $r_{2}={\frac {k_{2}[Y][Z]}{K_{M2}+[Z]}}$. Here the K_M stand for the Michaelis–Menten constant which describes how well the enzymes X and Y bind and catalyze the conversion whereas the kinetic parameters k₁ and k₂ denote the rate constants for the catalyzed reactions. Assuming that the total concentration of Z is constant we can additionally write that [Z]₀ = [Z_P] + [Z] and we thus get:

${\begin{aligned}{\frac {d[Z]}{dt}}=r_{1}-r_{2}={\frac {k_{1}[X]([Z]_{0}-[Z])}{K_{M1}+([Z]_{0}-[Z])}}&-{\frac {k_{2}[Y][Z]}{K_{M2}+[Z]}}=0\\{\frac {k_{1}[X]([Z]_{0}-[Z])}{K_{M1}+([Z]_{0}-[Z])}}&={\frac {k_{2}[Y][Z]}{K_{M2}+[Z]}}\\{\frac {k_{1}[X](1-{\frac {[Z]}{[Z]_{0}}})}{{\frac {K_{M1}}{[Z]_{0}}}+(1-{\frac {[Z]}{[Z]_{0}}})}}&={\frac {k_{2}[Y]{\frac {[Z]}{[Z]_{0}}}}{{\frac {K_{M2}}{[Z]_{0}}}+{\frac {[Z]}{[Z]_{0}}}}}\\{\frac {v_{1}(1-z)}{J_{1}+(1-z)}}&={\frac {v_{2}z}{J_{2}+z}}\qquad \qquad (1)\end{aligned}}$

with the constants

${\begin{aligned}z={\frac {[Z]}{[Z]_{0}}};\ v_{1}=k_{1}[X];\ v_{2}&=k_{2}[Y];\ J_{1}={\frac {K_{M1}}{[Z]_{0}}};\ J_{2}={\frac {K_{M2}}{[Z]_{0}}};\ \qquad \qquad (2)\end{aligned}}$

If we thus solve the quadratic equation (1) for z we get:

${\begin{aligned}{\frac {v_{1}(1-z)}{J_{1}+(1-z)}}&={\frac {v_{2}z}{J_{2}+z}}\\J_{2}v_{1}+zv_{1}-J_{2}v_{1}z-z^{2}v_{1}&=zv_{2}J_{1}+v_{2}z-z^{2}v_{2}\\z^{2}(v_{2}-v_{1})-z\underbrace {(v_{2}-v_{1}+J_{1}v_{2}+J_{2}v_{1})} _{B}+v_{1}J_{2}&=0\\z={\frac {B-{\sqrt {B^{2}-4(v_{2}-v_{1})v_{1}J_{2}}}}{2(v_{2}-v_{1})}}&={\frac {B-{\sqrt {B^{2}-4(v_{2}-v_{1})v_{1}J_{2}}}}{2(v_{2}-v_{1})}}\cdot {\frac {B+{\sqrt {B^{2}-4(v_{2}-v_{1})v_{1}J_{2}}}}{B+{\sqrt {B^{2}-4(v_{2}-v_{1})v_{1}J_{2}}}}}\\z&={\frac {4(v_{2}-v_{1})v_{1}J_{2}}{2(v_{2}-v_{1})}}\cdot {\frac {1}{B+{\sqrt {B^{2}-4(v_{2}-v_{1})v_{1}J_{2}}}}}\\z&={\frac {2v_{1}J_{2}}{B+{\sqrt {B^{2}-4(v_{2}-v_{1})v_{1}J_{2}}}}}.\qquad \qquad (3)\end{aligned}}$

Thus (3) is a solution to the initial equilibrium problem and describes the equilibrium concentration of [Z] and [Z_P] as a function of the kinetic parameters of the phoshorylation and dephoshorylation reaction and the concentrations of the kinase and phosphotase. The solution is the Goldbeter–Koshland function with the constants from (2):

${\begin{aligned}z={\frac {[Z]}{[Z]_{0}}}=G(v_{1},v_{2},J_{1},J_{2})&={\frac {2v_{1}J_{2}}{B+{\sqrt {B^{2}-4(v_{2}-v_{1})v_{1}J_{2}}}}}.\\\end{aligned}}$

Literature

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