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

Answer:

The correct answer is option B.

Explanation:

Michaelis–Menten 's equation:

$v=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}$

$V_{max}=k_{cat}[E_o]$

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 :

$v=\frac{V_{max}}{4}$

[S] =?

$K_m=0.0050 M$

$v=V_{max}\times \frac{[S]}{(K_m+[S])}$

$\frac{V_{max}}{4}=V_{max}\times \frac{[S]}{(0.0050 M+[S])}$

$[S]=\frac{0.005 M}{3}=1.7\times 10^{-3} M$

So, the correct answer is option B.