An enzyme has a Km of 4.7 x 10-5M. If the Vmax of the preparation is 22m moles liter-1 min-1, what velocity would be observed in the presence of 2.0 x 10-4M substrate and 5.0 x 10-5M of a competitive inhibitor?

In the presence of a competitive inhibitor, the observed velocity is determined by the concentration of the substrate and the inhibitor. Given a Km of4.7×10^-5 M and substrate concentration of2x10^-4 M, the velocity is calculated using the Michaelis-Menten equation. For a competitive inhibitor, the observed velocity would be 7.3 µmoles. Litre^-1. Min^-1. This is because the inhibitor competes with the substrate for the active site, affecting the effective concentration of substrate.

In the case of a non-competitive inhibitor, the inhibitor binds to a different site on the enzyme, altering its conformation and reducing the enzyme’s activity. In this scenario, the observed velocity would be 6.6 µmoles.Litre^-1. Min^-1. This value is lower than that of the competitive inhibitor, indicating a stronger inhibitory effect.

When an uncompetitive inhibitor is present, it only binds to the enzyme-substrate complex. This means that it can only act once the substrate has already bound to the enzyme. With this inhibitor, the observed velocity would be 10.4 µmoles. Litre^-1. Min^-1. This value is higher compared to the competitive and non-competitive inhibitors, suggesting a lesser inhibitory effect.