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

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.  

     $\frac{1}{V} = \frac{(K_{m} + [S])}{(V_{max} \times [S])}$

                 V = $\frac{(V_{max} \times [S])}{(K_{m} + [S])}$

Putting the given values into the above formula as follows.

          V = $\frac{(4.25 \times [S])}{(6.5 + [S])}$

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

So,     V = $\frac{(4.25 \times [S])}{(6.5 + [S])}$

          V = $\frac{(4.25 \times 4)}{(6.5 + 4)}$

          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 = $\frac{(4.25 \times [S])}{(6.5 + [S])}$

            V = $\frac{(4.25 \times 6.5)}{(6.5 + 6.5)}$

            V = 2.125 mM/s

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