For an enzyme in the absence of inhibitor Vmax is 10 μM.s-1 and its Km is 2 mM. In the presence of a 3 mM inhibitor the apparent Vmax is 6.67 μM.s-1 and the apparent Km is 4 mM.

The question seems to be incomplete. However, based on the information provided, it seems like you might be asking about the type of inhibition this enzyme is experiencing.

Enzyme inhibitors can work in different ways, but they generally function by preventing the enzyme from catalyzing a reaction. There are two main types of inhibitors: competitive and non-competitive.

Competitive inhibitors bind to the active site of the enzyme, preventing the substrate from binding. This increases the apparent Km (the concentration of substrate needed to reach half the maximum reaction rate), but does not change the Vmax (the maximum reaction rate).

Non-competitive inhibitors bind to a different site on the enzyme, changing its shape and making it less effective. This decreases the apparent Vmax, but does not change the Km.

In this case, both the apparent Vmax and the apparent Km have changed in the presence of the inhibitor. This suggests that the inhibitor is a mixed inhibitor, which can bind to the enzyme whether or not the substrate has already bound, and can bind to the same or different site as the substrate.

The Vmax has decreased, which is consistent with non-competitive or mixed inhibition. The Km has also increased, which is consistent with competitive or mixed inhibition. Therefore, the inhibitor is likely a mixed inhibitor.

However, without more information or a specific question, it's hard to provide a more detailed answer.