Question

Renin is an aspartyl protease that has a role in the regulation of blood pressure. The active site has two aspartyl residues, one acting as an acid (pK1 5.0) and the other acting as a base (pK2 of 4.0). If the enzyme activity is measured at pH 6.0, what fraction of the enzyme is expected to be ACTIVE

A. 1% B. 9.0% C. 50% D. 90% E. 95% F. 99%

Answer 1

Henderson-Hasselbach equation

pH = pKa + log ([A-]/[HA]) ----->1

R = [A-]/[HA]

Rewriting equation 1 in terms of pK1 & pK2 and substituting R value

pH = pK1 + log R ----->2

pH = pK1 + log R ----->3

Adding equations 2 & 3,

pH + pH = pK1 + log R + pK2 + log R

2pH = pK1 + pK2 + 2log R

2log R = 2pH - pK1 - pK2

log R² = 2pH - pK1 - pK2

R² = 10^{(2pH - pK1 - pK2)} = 10^{(2 x 6.0 - 5.0 - 4.0)} = 10^{(2 x 6.0 - 5.0 - 4.0)} = 10^3.0 = 1000

R = 31.6

f = 1 / (1+R) = 1 / (1+31.6) = 0.03 = 0.03 x 100 = 3%

f =

Henderson-Hasselbach equation

pH = pKa + log ([A-]/[HA]) ----->1

R = [A-]/[HA]

Rewriting equation 1 in terms of pK1 & pK2 and substituting R value

pH = pK1 + pK2 + log R

log R = pH - pK1 - pK2

R= 10^{(pH - pK1 - pK2)} = 10^{(6.0 - 5.0 - 4.0)} = 10^-3.0 = 0.001

R = 0.001

f = 1 / (1+R) = 1 / (1+0.001) = 0.99 = 0.99 x 100 =99%

Fraction of the enzyme is expected to be ACTIVE =99%