Question

Th e enzyme-catalyzed conversion of a substrate at 25°C has a Michaelis constant of 0.045 mol dm−3. Th e rate of the reaction is 1.15 mmol dm−3 s−1 when the substrate concentration is 0.110 mol dm−3. What is the maximum velocity of this reaction?

Answer 1

Answer:

$1.620\times 10^{-3} mol/dm^3 s$is the maximum velocity of this reaction.

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 :

$K_m=0.045 mol/dm^3$

$v=1.15 mmol/dm^3 s=1.15\times 10^{-3} mol/dm^3 s$

$[S]=0.110 mol/dm^3$

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

$1.15\times 10^{-3} mol/dm^3 s=V_{max}\times \frac{0.110 mol/dm^3}{[(0.045 mol/dm^3)+(0.110 mol/dm^3)]}$

$V_{max}=\frac{1.15\times 10^{-3} mol/dm^3 s\times [(0.045 mol/dm^3)+(0.110 mol/dm^3)]}{0.110 mol/dm^3}=1.620\times 10^{-3} mol/dm^3 s$

$1.620\times 10^{-3} mol/dm^3 s$is the maximum velocity of this reaction.