Click for printer friendely version of this HowTo Microscopic/Probabilistic Behavior Drug-Receptor Model IIno_title Example 2.8.2.1   Example 2.8.2.2 (Drug-Receptor Model II)   Example 2.8.1.1 is probably the most useful derivation in biological models. It is also simple with only a few assumptions. If we look at the system more closely we see that binding comes from molecules colliding with receptors and only every once in a while is the collision sufficiently strong that an ``event'' takes place, producing a complex of ``molecule bound to a receptor.'' The collisions are due to thermal motion and, up to a certain point, the hotter the solution the more vigorous the collisions and the more binding events. However, beyond that point, the complex can also vibrate and fall apart. Thus, at thermal equilibrium, both binding and unbinding events are constantly happening. In Example 2.8.1.1 we were assuming that the probability of a binding event or an unbinding event is constant in time. In this example we will take into account the microscopic view of binding and unbinding and demonstrate that this this is a pretty solid assumption to make. The probability that a receptor is occupied at time $t$ is $P_o(t)$ and the probability that a receptor is unoccupied at time $t$ is $P_u(t)$. We also know that $P_o(t) = 1 - P_u(t)$. The probability that an unoccupied receptor will become occupied depends on the collision rate which in turn depends on the concentration of hormones, neurotransmitters or whatever molecule is involved in the reaction. We will denote the concentration of the molecule as, $N$. The probability that an unoccupied site at time $t$ will become occupied during the next increment of time, $\Delta$ is $$P_{u \rightarrow o} = \lambda N \Delta,$$ (2.8.8) where $\lambda$ is the proportionality constant for the binding rate per molecule of drug. For a receptor is bound, there are two possibilities for its state after $\Delta$. It can either become unoccupied or remain occupied. The probability that an occupied receptor will become unoccupied during the time interval, $\Delta$ is a fixed rate, $$P_{o \rightarrow u} = \mu\Delta.$$ (2.8.9) The probability of an occupied site remaining occupied during $\Delta$ is simply $$P_{o \rightarrow o} = 1 - P_{o \rightarrow u} = 1 - \mu\Delta.$$ (2.8.10) To determine the probability that a site will be occupied at time $t + \Delta$ we need to consider two possibilities. Either the site was empty at time $t$ and became occupied during $\Delta$, or the site was occupied at time $t$ and it did not become unoccupied during the interval $\Delta$. Thus, $$P_o(t+\Delta) = P_u(t)P_{u \rightarrow o} + P_o(t)P_{o \rightarrow o}$$ $$= P_u(t)\lambda N\Delta + P_o(t)(1-\mu\Delta)$$ (2.8.11) We can now rearrange terms in Equation 2.8.11 and make a difference equation, $$\frac{P_o(t+\Delta) - P_o(t)}{\Delta} = \lambda N P_u(t) - \mu P_o(t)$$ $$= \lambda N (1-P_o(t)) - \mu P_o(t).$$ (2.8.12) If we now let $\Delta$ go to zero, we will end up with a differential equation for the probability that a receptor site will become occupied: $$\frac{\mathrm{d}P_o}{\mathrm{d}t}=\lambda N(1-P_o) - \mu P_o.$$ (2.8.13) If we compare Equation 2.8.13 to Equation 2.8.2 we notice a striking similarity. Notice that $$P_o(t) = \frac{R_{o}(t)}{R_{\textrm{max}}} = b(t),$$ where $b(t)$ is the fraction of bound receptors at time $t$. Also, both $\lambda$ and $\mu$ map to the constants used in Equation 2.8.2. Thus, once we take away the assumption that there is a constant event probability over time, we end up with the same general equation. $\vert\boldsymbol{\vert}$ Click for printer friendely version of this HowTo

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