A 5 00l reaction co ntaining 05 pmol of an enzyme following

A 5 00l reaction co ntaining 0.5 pmol of an enzyme following Michaelis – Menten kinetics with saturating substrate concentration is studied . For this enzyme k1 = 1 x 10 9 M - 1 s - 1 , k - 1 = 2 x 10 3 s - 1 , and k2 = 1 x 10 5 s - 1 .

a . What is the Vmax for the enzyme considering that it has only one substrate binding site ? b. What are the values of Km and Ks? Is this a rapid equilibrium enzyme or does it follow steady state kinetics? Explain why .

c . What is the substrate concentration needed to achieve half maximal velocity? Explain the basis for your answer.

d . If the substrate concentration in part c is doubled , how many fold will the initial velocity increase?

e . What is the Km i f the concentration of the enzyme is doubled? f . What is the efficiency of this enzyme?

g. Would this enzyme be considered a perfect enzyme? Explain.

h . If 0.25 pmol of an irreversible or classic non - competitive inhibitor is added before starting the reaction. What is the resulting Km for the enzyme?

Solution

a) K_{2 =} V_max E_t ; Given K₂ =1 x 10 ⁵ s ^{- 1} , E_t = 0.5 pmol

V_max = K₂/E_{t =} 1 x 10 ⁵ s ^{- 1}/ 0.5 x 10¹² mol = 2 x 10¹⁷ s^-1mol^-1.

b) K_M = (K_-1 + K₂) / K_{1 .} Put the values in the equation and get the answer.

k1 = 1 x 10 ⁹ M ^{- 1} s ^{- 1} , k - 1 = 2 x 10 ³ s ^{- 1 ,} k2 = 1 x 10 5 s - 1

K_M = (2 x 10 ³ + 1 x 10 ⁵ ) / 1 x 10 ⁹ = 1 x 10^-4 M

Ks = k_-1 / k₁ = 2 x 10 ³ s ^{- 1} /1 x 10 ⁹ M ^{- 1} s - 1 = 2 x 10⁶ M. Since Km is not equal to Ks, hence it is not the rapid equilibrium enzyme.