Vmax is the product of catalytic constant on enzyme concentration. It not depends from enzyme concentration, because it use the concentration of active enzyme molecules ("active centers") in solution.

At the other hand, activity of enzyme preparation in units depends from its purity - you can have (for example) the samples with the 20 or 80% of active enzyme molecules, and other is the inert (not active) material.

For true description of kinetic experiments it is necessary to determine (from the independent experiment) the percentage of active enzyme molecules in your sample (titration of active centers) and then use this parameter for the conversion of weight concentration of emzyme to molar concentration of active enzyme molecules in reaction mixture.

I agree with Andrey. Nowadays, we use kcat instead of Vmax since the wierd unit "umol/mg/min" makes the comparison of two reactions inconvenient. For example, even if we know Vmax1=10umol/mg/min and Vmax2 = 100umol/mg/min we still have no idea which reaction is faster....So we have to search all the paper to find concentrations of enzymes, as well as the extinction coefficients , and don't forget to do the math........

Decades ago, people could not determine the exact amount of protein in solution so they have no other choice but Vmax. But It's out of date. So always do a titration of active enzyme before you go any further

As the answers so far have explained, yes, Vmax will depend on your active enzyme concentration, and in a rigorous study of Michaelis-Menten kinetics will be reported rather as kcat (=Vmax/[E]) in units of s-1, where [E] is your concentration of active enzyme that has been determined by an active site titration. Such a study would also report pH, T, solvents or buffers and any other parameter that might be expected to impact on enzyme activity.

The second unit of "enzyme activity" to which you refer is more commonly called "specific activity" and is specific to a particular assay. It does not define the saturation kinetics of the enzyme but is carried out at a single fixed substrate concentration. One can define the conditions, concentrations, and units of a specific activity assay to be whatever one wants; the point would be that if the same assay is used in another lab, then my 10 units of enzyme X should represent the same amount of activity as your 10 units of enzyme X. Obviously this approach does not represent the same rigorous determination of kinetics behavior as a Michaelis-Menten study; however it is practical and often used in biological experiments to specify, for example, how much of an (often commercially purchased) enzyme was used to treat a dish of cells or digest a tube of DNA. Units of specific activity can also be used to ensure that different batches or lots of enzyme that may differ in purity can be applied in units of equivalent activity.

I hope this is helpful.

Confirming the previous answers, yes, Vmax depends on enzyme concentration. Vmax = kcat*Eo

You can take a look to the Fig attached where the effect of enzyme concentrations is represented in the Michaelis-Menten plot (Fig. 1) and conversion curve plot (Fig. 2).

Pedro is correct. Concisely, Vmax is a maximal RATE (e.g., mol L-1 s-1) and is defined as kcat [E], where the RATE CONSTANT kcat is defined in reciprocal time units (e.g., s-1), that is, a first order rate constant.. So Vmax is certainly dependent on enzyme concentration. Some investigators report bastardized versions of maximal rate constant, e.g. mmol product/min/mg protein. These are neither Vmax nor kcat but can be converted to kcat if you work out all the molar concentrations. The convention in enzymology is to report first and second order rate constants, e.g., kcat and kcat/Km, both of which are fundamental intrinsic rates constants for enzyme-catalyzed reactions.

No. Vmax does not depend upon enzyme concentration. The better way to show enzymatic reactions is to show Kcat.

No , vmax depend on substrate concentration

https://www.researchgate.net/post/Does_Km_and_Kcat_differ_with_enzyme_concentration

 No, because it depends on multiple factors as mentioned above. It only relates to the enzyme concentration when such enzyme is pure, actually, Vmax =Kcat [E], which is a linear relationship. For that reason, it is used the Vmax for the calculation of the catalytic efficiency (Vmax/Km) in complex systems instead of the turnover rate (kcat/Km).