Question

An enzyme, Pittpantherase, has a KM of 8mM. At a substrate concentration of 4mM, the velocity of the reaction was 2mM/min. What is Vmax for this enzyme

Answer 1

Answer: [Vmax]= 6mM/min.

Step-by-step explanation:

Enzymes are proteins that produce a specific chemical change in a substance, called substrate, resulting in a product. By following the rate of product appearance (or disappearance of the substrate) as a function of time, the kinetics of the reaction is obtained. As the reaction proceeds, the rate of product formation decreases and the substrate is consumed.

The rate [V] indicates the amount of substrate that is converted into product per second. With increasing substrate concentrations [S], the enzyme approaches its maximum velocity [Vmax], but does not reach it. For this reason, there is no specific value of [S] for Vmax. However, a characteristic parameter of the enzyme can be defined using the substrate concentration at which half of the maximum velocity (Vmax/2) is reached. The initial velocity of the reaction is equal to the slope of the zero-time forward curve and is measured before 10% of the total substrate is consumed, so that the substrate concentration can be considered as essentially constant throughout the experiment. So, although it is impossible to measure exactly the substrate concentration that gives Vmax, enzymes can be characterized by the substrate concentration at which the reaction rate is half the maximum rate. This substrate concentration is known as the Michaelis-Menten constant (KM). For enzymes exhibiting simple Michaelis-Menten kinetics this constant represents the dissociation constant of the enzyme-substrate complex (or the inverse of the enzyme-substrate affinity). Thus, Km indicates the affinity of the enzyme for its substrate, since the higher the Km the lower the affinity. And the lower the Km the higher the affinity.

The formula for the calculation is as follows:

`V_(0)= ([Vmax] x [S])/([S]+[KM])`

So if we replace the numbers, we get:

`2mM/min= ([Vmax] x [4mM])/([4mM]+[8mM])`

If we clear the equation, we obtain that [Vmax]= 6mM/min.