Question

$K_m$ value of enzyme is substrate concentration at

• A. 10 V${}_{max}$
• B. 0.5/1 V${}_{max}$
• C. 1/2 V${}_{max}$
• D. 4/5 V${}_{max}$

Answer 1

Answer: (C)

1/2 V${}_{max}$

The catalytic efficiency of an enzyme is described by the K${}_m$ value or Michaelis Menten constant. The Michaelis constant is the substrate concentration at which the reaction rate is one half the maximum. The K${}_m$ describes the affinity of enzyme for a substrate molecule. Greater the affinity lower is the K${}_m$ value and sooner the V${}_{max}$ can be attained and vice versa.

According to Michaelis-Menten equation $K_m$ is equal to substrate concentration at which the velocity is half the maximum. Michaelis and Menten proposed a hypothesis for enzyme for action according to which the enzyme molecule combines with a substrate complex which further dissociates to form product and enzyme back.

$E+S\underset{K_1}{\overset{K_2}{\to }}E-S\underset{K_3}{\overset{E}{\to }}+P:K_m(Michaclis-MentenConsant)=\frac{K_3+K_2}{K_1}$

They gave the following equation: $V_0=\frac{V_{max}\left(S\right)}{\left(S\right)+K_m}$

It is the statement of the quantitative relationship between the initial velocity $V_0$, the maximum velocity $V_{max}$ the initial substrate concentrations; all related through Michaelis Menten constant $K_m$.

E = free enzyme

S = Substrate

ES = enzyme substrate complex

P = product

K${}_1$ = the rate constant for ES formation

K${}_2$ = the rate constant for dissociation of ES into E and S

K${}_3$ = the rate constant for dissociation of EP complex into E and P.