Question

1. According to the Michaelis-Menten equation, when is the reaction velocity at half the maximal value (v_o = 0.5 V_max)

A. when [S] = 0.5 K_m

B. when [S] = 1.0 K_m

C. when v_o = k_cat [E]

D. when v_o = k_cat / K_m

2. The value of V_max depends on ...

A. k₁ and k_-1

B. k₁ and k₂

C. k₁ only

D. k₂ only

3. When is K_m approximately equal to K_d?

A. when k₁ is much less than k_-1

B. when k₂ is much less than k_-1

C. when k_-1 is much less than k₁

D. when k_-1 is much less than k₂

Answer 1

Question 1

Ans : Option B

$\odot$ When [S] = 1.0 [Km]

Reason :

The Michaelis Menten Kinetic Equation:

V_o = (V_max [S]) / (K_m + [S]) ;

When [S] << K_m,

V_o = V_max [S] / K_m

When [S] >> K_m,

V_o = V_max [S] / [S]

V_o = V_max

When [S] = K_m

V_o = V_max K_m / (K_m + K_m)

V_o = V_max K_m / 2 K_m

V_o = V_max / 2

V_o = 0.5 V_max

Thus [S] = 1.0 Km at which reaction is half its maximal value.

Question 2

Ans : Option D

$\odot$ k₂ only

Reason : Vmax is the product of catalytic constant on enzyme concentration.

That is,

V_max = k_cat [E_T]

k_cat = catalytic constant = turnover number

Here k₂ = k_cat

The higher the k₂ is, the more substrates get turned over in one second. Therefore V_max depends on k₂ only.

Question 3

Ans : Option B

$\odot$ When k₂ is much less than k_-1

Reason : The QSSA (quasi-steady-state approximation) defines the Km value as:

K_m = k_-1 + k₂ / k₁

where,

k₁ is the rate constant

k_-1 is the unproductive dissociation rate constant of ES – complex

k₂ = kcat is the catalytic rate constant or turnover number

When k₂ << k_-1,

K_m = k_-1 + k₂ / k₁

K_m ≈ k_-1 / k₁

k_-1 / k₁ = K_d

Therefore ,

Km ≈ Kd