Modelling attempt 1: exponential decay A reasonable assumption is that an enzyme is more busy with the elimination of alcohol the more it is likely to interact with an alcohol molecule, and that the frequency of interaction is larger when the alcohol concentration is greater in the tissues where the enzyme is. When there is a rapid exchange between the bloodstream and the liver, then we may suppose that the concentration in the tissues is equal or proportional to the blood alcohol concentration. So \[\frac{\dd C}{\dd t}=-r\cdot C\] for some positive parameter \(r\). The solution is \[C(t)=C_0\cdot e^{-r\cdot t}\] Unfortunately, this model does not describe properly measured blood alcohol levels after alcohol consumption: At a high alcohol permillage the elimination seems to reach a maximum speed.

Modelling attempt 2: the Wagner Model What goes wrong in the first attempt at modelling of alcohol breakdown? Essentially, the point is that we have not considered yet in the model that the interaction between the enzyme and the alcohol molecule takes time to reach proper conditions to form an enzyme-alcohol complex.

We assume that the average waiting time \(t_w\) to realize an interaction between the enzyme and the alcohol molecule is inversely proportional to the blood alcohol concentration: \(C\), say \[t_w=\frac{\alpha}{C}\] for some positive parameter \(\alpha\). Once the enzyme molecule and the alcohol molecule "found each other" there is still an average processing time \(t_v\) needed. We assume that this processing time does not depend on the concentration \(C\) but is constant. The average time required for the enzyme molecule to convert one alcohol molecule is set equal to the sum of the waiting and processing time. For conversion of \(N\) alcohol molecules we expect the time \(t\) needed for one enzyme molecule to achieve this equals \[t=N\cdot (t_w+t_v)\tiny.\] Then we get the following expression for the reaction rate \(r\) per enzyme molecule, that is, the number of alcohol molecules converted per unit of time: \[r=\frac{N}{t}=\frac{1}{t_w+t_v}\] Substitution of the assumed expression for the average waiting time then gives \[\begin{aligned}r&=\frac{1}{\frac{\alpha}{C}+t_v}\\ \\ &=\frac{C}{\alpha + t_v\cdot C} \\ \\ &=\frac{1}{t_v}\cdot \frac{C}{\frac{\alpha}{t_v}+C}\end{aligned}\] But in the liver is more than one enzyme molecule present and therefore we must for the conversion rate of alcohol through the liver multiply the previous formula by a constant \(\beta\) and then we have: \[V_d\cdot \frac{\dd C}{\dd t} = - \frac{\beta}{t_v}\cdot \frac{C}{\frac{\alpha}{t_v}+C}\] where \(V_d\) is the volume of distribution of alcohol in the central compartment. We rewrite this model as \[V_d\cdot \frac{\dd C}{\dd t} = - \frac{v_{\text{max}}\cdot C}{K_m+C}\] where \(K_m\) is called the Michaelis-Menten constant (value is about 30 mg/L), and \(v_{\text{max}}\) is the maximum elimination rate of alcohol (equivalent to approximately 200 mg/min).