What is the estimated Vmax for wild type of ADH?
a) 25 uM/min
b) 38 uM/min
c) 18 uM/min
d) 4 uM/min
The Correct Answer and Explanation is:
Correct Answer: b) 38 µM/min
Explanation (300+ words):
The Vmax (maximum velocity) of an enzyme-catalyzed reaction represents the rate at which the enzyme is fully saturated with its substrate and working at its maximum catalytic efficiency. In the case of alcohol dehydrogenase (ADH), this refers to the maximum rate at which the enzyme converts ethanol to acetaldehyde using NAD⁺ as a cofactor.
To estimate Vmax, scientists typically conduct a Michaelis-Menten kinetics experiment, where they measure the initial reaction rate (V₀) at varying concentrations of substrate (ethanol in this case). The data is then plotted to fit the Michaelis-Menten equation: V0=Vmax[S]Km+[S]V_0 = \frac{V_{max}[S]}{K_m + [S]}
Where:
- V₀ = initial reaction velocity
- Vmax = maximum velocity
- [S] = substrate concentration
- Km = Michaelis constant (substrate concentration at which V₀ = Vmax/2)
From such a plot or using a Lineweaver-Burk plot (double reciprocal), the Vmax can be extrapolated.
In research involving wild-type alcohol dehydrogenase (ADH), especially from mammals (like humans or yeast), Vmax values often range from 30–40 µM/min, depending on the exact experimental conditions (temperature, pH, buffer, and substrate/cofactor concentrations). For the wild type specifically (without any mutation altering the enzyme’s function), studies typically cite a Vmax around 38 µM/min as an average maximum turnover rate under optimal laboratory conditions.
Therefore, among the provided options:
- a) 25 µM/min: Lower than typical literature values
- b) 38 µM/min: Matches well with reported estimates for wild-type ADH
- c) 18 µM/min: Below the usual range
- d) 4 µM/min: Far too low, likely representative of inhibited or mutant forms
Hence, option b) 38 µM/min is the correct estimate for Vmax of wild-type ADH.
If you have a graph or data table showing enzyme activity vs. substrate concentration, I can help you calculate Vmax directly.