Question

Enalaprilat is a competitive inhibitor of the angiotensin-converting enzyme (ACEACE), which cleaves the blood-pressure regulating peptide angiotensin I. ACEACE has a KM=52KM=52 μMμM for angiotensin I, which is present in plasma at a concentration of 78 μMμM. When enalaprilat is present at 2.8 nMnM, the activity of ACE in plasma is 13 %% of its uninhibited activity. What is the value of KIKI for enalaprilat?

Answer 1

We know, the Michaelis-Menten equation for an ubinhibited enzyme catalzed reaction is where V_o=initial rate velocity; V_max=maximumvelocity; [S]=substrate concentration; K_m=Michaelis -menten constant. But in presence of a competitive inhibitor, the Michaelis-Menten equation becomes where is the factor increase of the apparent value of K_m in presence of an inhibitor. Here, the enzyme ACE2 has K_m =52 M and angiotensin I has concentration in plasma, [S]=78 M; Enalaprilat is a competitive inhibitor of the ACE2 enzyme and let the equilibrium constant for inhibitor binding be K_I. When the concentration of enalaprilat is 2.8 nM= 2.8 10^-3M, the activity of ACE in plasma is 13% of its uninhibited activity.

Therefore, or,   ( V_max remain unchanged in competitive inhibition)

or,   or,   ( [S]=78 M, K_m =52 M)    

Now, we know the relation between fraction of apparent K_m increase, , inhibitor concentration, [I] and K_I is given by as follows:   

Therefore, K_I value for enalaprilat is 1.68 10^-4M.