Question

Alcohol dehydrogenase (ADH) is and enzyme that can catalyze the oxidation of a number of alcohols to the corresponding aldehyde. Two such reactions are:

ethanol CH₃-CH₂OH ————> CH₃-CHO

butanol CH₃-CH₂-CH₂-CH₂OH ———> CH₃-CH₂-CH₂-CHO

The Km for ethanol is 10^-5M and the Km for butanol is 10^-3M. Both substrates bind at the same active site, and the turnover number for each substrate is the same.

Two experiments are carried out to determine the Vmax of this amount of enzyme for ethanol oxidation. In one experiments 10^-2 M butanol is included and in the other experiment no butanol is present. The measured Vmax for ethanol will ______.

Higher with butanol.

Lower with butanol.

The same in both tubes.

Answer 1

Answer: Vmax is reduced in the presence of butanol

Explanation:
Vmax = Rate of reaction when the enzyme is saturated with the substrate.
In the given case, both ethanol and butanol are substrates for alcohol dehydrogenase.
If only ethanol is present, the competition for the active site is absent.
When butanol molecules are also present, both ethanol and butanol molecules compete for the same active site.
So, Vmax for ethanol will be reduced in the ethanol+butanol compared to the unmixed reaction.