Question

Enalapril is an anti-hypertension "prodrug" (1.6., a drug precursor) that is inactive until the ethyl ester (arrow in figure) is hydrolyzed by esterases present in blood plasma. The active drug is the dicarboxylic acid ("enalaprilat") that results from this hydrolysis reaction Enalaprilat is a competitive inhibitor of the angiotensin-converting enzyme (ACE), which Cleaves the blood-pressure regulating peptide angiotensin I. ACE has a K_M = 52 mu M for angiotensin I, which is present in plasma at a concentration d 75 mu M When enalaprilat is protent at 2 4 nM, the activity of ACE in plasma is 10 % of its uninhibited activity What is the value of K_1 for enalaprilat? Express your answer to two significant figures and in units of M.

Answer 1

The K1 value can be calculated by the following equation,

$\alpha = \sqrt{K1/C}$

Given that, alpha ie.degree of ionization = 10% = 10/100

C= concentration of enalpril = 2.4 nM = 2.4*10^-8 M

$\alpha^2 = {K1/C}$

$K1 = \alpha ^2 * C$

= $(10/100)^2*2.4*10^-8$