Question

An enzyme catalyzes a reaction with a K m of 6.50 mM and a V max of 4.25 mM ⋅ s − 1 . Calculate the reaction velocity, v 0 , for each substrate concentration.

Answer 1

The given question is incomplete. The complete question is as follows.

An enzyme catalyzes a reaction with a K m of 6.50 mM and a V max of 4.25
`mM⋅s^(-1)`. Calculate the reaction velocity, v 0 , for each substrate concentration.

(a) 4.00 mM, (b) 6.50 mM

Step-by-step explanation:

It is given that the enzyme catalyzes a reaction with a
`K_(m)` of 6.50 mM and a
`V_(max)` of 4.25
`mMs^(-1)`.

So, we will calculate the reaction velocity (
`v_(0)`), for the following substrate concentrations as follows.

`(1)/(V) = ((K_(m) + [S]))/((V_(max) * [S]))`

V =
`((V_(max) * [S]))/((K_(m) + [S]))`

Putting the given values into the above formula as follows.

V =
`((4.25 * [S]))/((6.5 + [S]))`

a) It is given that [S] = 4 mM

So, V =
`((4.25 * [S]))/((6.5 + [S]))`

V =
`((4.25 * 4))/((6.5 + 4))`

V = 1.619 mM/s

Hence, the reaction velocity (), for substrate concentration 4 mM is 1.619 mM/s.

b) It is given that [S] = 6.5 mM

So, V =
`((4.25 * [S]))/((6.5 + [S]))`

V =
`((4.25 * 6.5))/((6.5 + 6.5))`

V = 2.125 mM/s

Hence, the reaction velocity (), for substrate concentration 6.5 mM is 1.619 mM/s.