Measurement of the rate constants for a simple enzymatic reaction obeying Michae
ID: 1024866 • Letter: M
Question
Measurement of the rate constants for a simple enzymatic reaction obeying Michaelis Menten kinetics gave the following results: k1 2x10 M sec k1 1x10 sec kcat = 50 sec-l What is the apparent dissociation constant for the ES complex (K)? Ignoring kcat, what is the true dissociation constant for the ES complex (K) If the Kea value were similar in magnitude to carbonic anhydrase, (40,000 sec"), would the K, overestimate or underestimate the actual affinity of the ES complex? If the kinetic measurements were made using 2 nanomoles of enzyme in I ml of buffer and saturating amounts of substrate, what would Vimax equal? (for this question, use kat 50 sec.) Again, using 2 nanomoles of enzyme per l mL of buffer, what concentration of substrate would give 0 75Explanation / Answer
Michaelis-Menton equation is represented as-
V (rate) = (Kcat .Cs.CE)/(Km+Cs)
Where Cs = concentration of free substrate at time t
CE = concentration of free enzyme at time t
Km (apparent dissociation constant) = (K-1+Kcat)/K1
So apparent dissociation constant (Km ) = (1000+50)/(2×108) M
= 5.25× 10-6 M
True dissociation constant (ignoring Kcat) Kd = [1000/(2× 108)] M
= 5× 10-6 M
For Kcat = 40000 sec-1 , Km = (1000+40000)/(2×108) M
= 2.05× 10-4 M
We know higher the value of Km lower the affinity of ES complex . Here Km value is highly increased compared to its actual Km . Hence underestimation of affinity of ES complex would happen in this case
We know Vmax = Kcat × E0 (E0 = initial concentration of enzyme)
So Vmax = 50× 2× 10-9 mol.L-1.sec-1 (E0 = 2× 10-9 mol.L-1)
= 10-7 mol.L-1.sec-1