# Question b7458

Apr 24, 2016

${k}_{\text{cat" = "250 s"^"-1}}$

#### Explanation:

The Michaelis-Menten model of enzyme catalysis is

E + S stackrelcolor(blue)(k_f)(⇌) ES stackrelcolor(blue)(k_"cat"color(white)(m))(→) E + P
$\textcolor{w h i t e}{m m m m} \stackrel{\textcolor{b l u e}{{k}_{r}}}{\textcolor{w h i t e}{l}}$

If the enzyme concentration is much less than the substrate concentration, the rate of product formation is

(d[P])/dt = V_"max" ([S])/(K_"M" + [S]) = k_"cat" [E]_0 ([S])/(K_"M" + [S])

The maximum rate ${V}_{\text{max}}$ is attained when all of the enzyme molecules are bound to the substrate.

It follows from the Michaelis-Menten equation that

color(blue)(|bar(ul(color(white)(a/a) V_"max" = k_"cat"[E]_0color(white)(a/a)|)))" "#

${\left[E\right]}_{0}$ is the initial enzyme concentration and ${k}_{\text{cat}}$ is the turnover number — the maximum number of substrate molecules converted to product per enzyme molecule per second.

${k}_{\text{cat" = V_"max"/[E]_0 = (30 × 10^"-6" color(red)(cancel(color(black)("mol·L"^"-1")))"min"^"-1")/(2 × 10^"-9" color(red)(cancel(color(black)("mol·L"^"-1")))) = "15 000 min"^"-1" = "250 s"^"-1}}$