Question

Determine the value of the turnover number ( k_cat ) for the enzyme carbonic anhydrase, given that

V _max =249μmol⋅L ^−1 ⋅s ^−1
and [E] _t=2.73nmol⋅L ^−1
.

a) 91.21 s⁻¹
b) 108.39 s⁻¹
c) 121.65 s⁻¹
d) 137.23 s⁻¹

Answer 1

The turnover number (k_cat) for the enzyme carbonic anhydrase is determined by dividing the given V_max by the total enzyme concentration [E]_t, resulting in a value of 91.21 s⁻¹.

Step-by-step explanation:

The student has asked to determine the value of the turnover number (kcat) for the enzyme carbonic anhydrase. Given that Vmax = 249 μmol·L− 1 ·s− 1 and [E]t = 2.73 nmol·L− 1, we can calculate kcat using the formula kcat = Vmax / [E]t.

First, we need to convert the values to compatible units. To convert the enzyme concentration from nmol/L to μmol/L, we multiply by 1,000, which gives us 2.73 μmol/L.

Vmax = 249 μmol·L− 1 ·s− 1
[E]t = 2.73 μmol·L− 1
kcat = 249 μmol·L− 1 ·s− 1 / 2.73 μmol·L− 1
kcat = 91.21 s− 1

Therefore, the correct answer is a) 91.21 s− 1.