Question

At what substrate concentration would an enzyme with a K_cat of 30.0 s⁻¹ and a Km of 0.0050 M operate at one-quarter of its maximum rate?

A. 0.33 x 10⁻³ M
B. 1.7 x 10⁻³ M
C. 2.33 x 10⁻³M
D. 8.73 x 10⁻³ M

Answer 1

The correct answer is option B.

Step-by-step explanation:

Michaelis–Menten 's equation:

`v=V_(max)* ([S])/((K_m+[S]))=k_(cat)[E_o]* ([S])/((K_m+[S]))`

`V_(max)=k_(cat)[E_o]`

v = rate of formation of products

[S] = Concatenation of substrate = ?

`[K_m]` = Michaelis constant

`V_(max)`= Maximum rate achieved

`k_(cat)` = Catalytic rate of the system

`E_o` = initial concentration of enzyme

We have :

`v=(V_(max))/(4)`

[S] =?

`K_m=0.0050 M`

`v=V_(max)* ([S])/((K_m+[S]))`

`(V_(max))/(4)=V_(max)* ([S])/((0.0050 M+[S]))`

`[S]=(0.005 M)/(3)=1.7* 10^(-3) M`

So, the correct answer is option B.