Question

For enzymes in which the slowest (rate-limiting) step is the reaction:

K2

ES → E+P

K_m becomes equivalent to:

A) k_cat.

B) the [S] where V₀ = V_max.

C) the dissociation constant, K_d, for the ES complex.

D) the maximal velocity.

E) the turnover number.

The answer is C), could you please explain?

Answer 1

  thank you very much,hope this will helpful for you,good luck from my side

P= Product E = Enzyme s = substrate Date: Page The Enzyme Reaction occur in two Stage (1) E ts__k? → ES K -1 (2) ES Krew E + P { kd=dissociation Because Es dissociate de The french mathematicians Leonor michalis and maud menten derived a mathematical equation that describes The kinetic Behavior of most Enzyme He assumed (1 (2) is • sets The Enzyme Reaction occur in two stage The Rate of the second stage of the Reaction CESZE+P) slower than the first for this reason the second stage of the Reaction the rate of the overall Reaction See. Second Stage is Rate liniting. Now velocity (Rate) of the Reaction at Vo = kg CES] or Vo= kd [ES] Problem is How kn = kd Vmax Vmax [s] max Vo = (S)+(5] 2 (3) 2 if the breakdown of ES E TP I truly is rate limiting then K2 can be I neglected from the km Constent and kon= kaiki under these Km =

Page: equivalent o the dissociation Condition km is Constant - - kort Ets k-1 is dissociation Constant ka Eski - K [E] [1] - (K-1) - (km = KES = TES] Ki Extra fact km lower – higher affinity of the enzyme for its substrate km higher - lower affinity of the enzyme for its substrate