Click for printer friendely version of this HowTo Macroscopic/Deterministic Behavior Drug-Receptor Model Ino_title Example 2.8.1.1   Example 2.8.1.2 (Drug-Receptor Model I)   Let's now look at a few examples of systems that lead to first order differential equations. Consider the process of a neurotransmitter binding to a receptor. Let $N$ be the concentration of the neurotransmitter, and $R_o$ be the number of occupied receptors where $R_{\textrm{max}}$ is the total number of receptors. The number of unoccupied receptors, $R_u$, is thus $R_{\textrm{max}} - R_o$. Thus, the reaction between neurotransmitter, unbound receptors and bound receptors can be encapsulated with the formula, $$N + R_u \underset{l}{\overset{k}{\rightleftharpoons}} R_o,$$ where $k$ is the proportionality constant for binding and $l$ is the proportionality constant for unbinding. The rate of change of occupied receptors is thus, $$\frac{\textrm{d}R_o}{\textrm{d}t} = kNR_u - lR_o = kN(R_{\textrm{max}} - R_o) - lR_o.$$ (2.8.1) If we convert Equation 2.8.1 to represent the change in the fraction of bound receptors by dividing by $R_{\textrm{max}}$, and rearrange the terms a little bit, then we can solve it by using what is called an integration factor.2.3That is, if we let $b = \frac{R_o}{R_{\textrm{max}}}$, we can rewrite Equation 2.8.1 as $$\frac{\textrm{d}b}{\textrm{d}t} = kN(1-b)-lb.$$ (2.8.2) Rearranging the terms gives us: $$\frac{\textrm{d}b}{\textrm{d}t} = -(kN+l)b + kN,$$ (2.8.3) or $$\frac{\textrm{d}b}{\textrm{d}t} + (kN+l)b = kN.$$ (2.8.4) With Equation 2.8.4 in the exact same form as Equation 2.10.15, the form required for our general solution, we can easily solve for $b(t)$. First, we will determine the integrating factor. That is, from Equations 2.8.4 and 2.10.17 $$\mu(t) = \exp \left( \int (kN + l) \textrm{d}t \right)$$ $$= e^{(kN + l)t}.$$ (2.8.5) We can now plug the integrating factor, along with bits from Equation 2.8.4 into Equation 2.10.16 and solve for $b(t)$. Thus, $$b(t) = e^{-(kN + l)t} \left( kN \int e^{(kN + l)t}\textrm{d}t + C \right)$$ $$= e^{-(kN + l)t} \frac{kN}{kN + l}e^{(kN + l)t} + Ce^{-(kN + l)t}$$ $$=\frac{kN}{kN + l} + Ce^{-(kN + l)t}.$$ (2.8.6) Since we know that at time $t=0$ that some fraction of receptors are occupied, $b(0)$, we can solve for $C$. That is, $$b(0) = \frac{kN}{kN + l} + C$$ and simple rearrangement gives us, $$C = b(0) - \frac{kN}{kN + l}.$$ Thus, our general solution for $b(t)$ is $$b(t) =\frac{kN}{kN + l} + \left( b(0) - \frac{kN}{kN + l} \right) e^{-(kN + l)t}.$$ (2.8.7) $\vert\boldsymbol{\vert}$ Phase Plane Analysisno_title Example 2.8.1.3   Example 2.8.1.4 (Phase Plane Analysis)   If we return to Equation 2.8.3, we can determine the asymptotic behavior of the solution without having to solve for it.2.4First, we will determine any points where the system is at equilibrium. That is, determine where the derivative is zero. $$\frac{\textrm{d}b}{\textrm{d}t} \stackrel{\mathrm{set}}{=} 0$$ $$kN - (kN+l)b = 0$$ $$b = \frac{kN}{(kN + l)}$$ Thus, when $b(0) = kN/(kN + 1)$, for all $t$, we are at an equilibrium and will not move from it. When $b(0) < kN/(kN + 1)$, then the slope for all $t$ is positive (to see this, try plugging in $b(0) = kN/2(kN +1)$), and thus, as $t \rightarrow \infty$, $b(t)$ approaches $kN/(kN + 1)$ from below. If $b(0) > kN/(kN + 1)$, then the slope for all $t$ will be negative and as $t \rightarrow \infty$, $b(t)$ approaches the equilibrium from above. Note that $b = 1/2$ when $N = l/k$. This is called the equilibrium dissociation constant and is the concentration of drug where the fraction of bound receptors is 1/2. Thus, from the kinetics of binding and unbinding, you can directly get a feel for the concentration of drug required to have half the receptors occupied. $\vert\boldsymbol{\vert}$ Click for printer friendely version of this HowTo

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