# How to calculate the kinetic order of an enzymatic reaction?

This question is concerning the metabolism of ethanol by an alcohol dehydrogenase enzyme.

Usually people metabolize alcohol equivalent to "one beer per hour". One beer is said to contain 33cL. The ethanol content of a beer is 4.6%(V/V). If this approximation is correct for an average lean person of 75kg after drinking two beers, then what is the kinetic order of reaction of the enzymatic reaction metabolizing the alcohol?

The alcohol content in one beer is: $$(33 \cdot 0.046)cL = 1.518cL$$

The ethanol content in two beers is then: $$(2 \cdot 1.518)cL = 3.036cL$$

How do I link this to the weight of the person (the 75 kg)? And how to I determine of this follows a zero, first, or second-order reaction?

I know that a first order reaction is given by the following rate expression: $$-\frac{dA}{dt} = k \cdot A$$

The second order reaction is given by the following rate expression: $$-\frac{dA}{dt} = k \cdot A^2$$

• This reaction is zero order (except at very, very low concentrations) and this is why it can be used in breathylisers. Jan 7, 2019 at 8:00
• @porphyrin Could you please elaborate on how you determine that it is a zero order reaction? Jan 7, 2019 at 14:41
• People do not metabolize one beer per hour, see e.g. here
– Karsten
Jan 7, 2019 at 21:33
• ... and here.
– Karsten
Jan 7, 2019 at 21:44

The legal limit in some states of 0.08% BAC corresponds to about 17 mM ethanol in the blood. The $$K_M$$ value for human liver alcohol dehydrogenase is 0.5 mM according to this paper. So at this particular legal limit, the enzyme works at 97% of its maximal capacity. ($$v = v_{max} \frac{[S]}{[S] + K_M}$$).
For ethanol concentrations below the $$K_M$$, the reaction is approximately first order in ethanol, so it gets slower and slower as the ethanol concentration approaches very small concentration.
While the $$K_M$$ is the same for different individuals, the amount of enzyme is not (depends on the size and health of the liver, gender, and on genetics). That's where the body mass comes in.