Question

Bioprocces , enzyme kinetics

2. Angiotensin ll is a chemical that induces the increase of blood pressure. By inhibiting the activity of Angiotensin Converting Enzyme (ACE), which produces angiotensin I, hypertension (high blood pressure) can be treated. A potential inhibitor has been developed, and the kinetic data together with inhibition kinetic data, obtained by ACE reaction with and without an inhibitor, are shown in the following table. Initial reaction rate (mM/h) Substrate concentration No Inhibitor 1.95 mM Inhibitor (mM) 0.25 1.02 0.56 0.33 1.39 0.75 0.40 1.67 0.85 1.89 1.00 0.50 0.60 2.08 1.28 0.75 2.44 1.39 1.00 2.50 1.82 a) What type of inhibition kinetics is this newly-developed inhibitor based on? b) Determine Vmau KM, and K

Answer 1

To study an enzyme different parameters must be taken into account, such as at which temperature it functions, how much affinity it has for it substrate, the efficiency rate and possible inhibitors. Enzyme kinetics tell us how fast an enzyme is and how much affinity it has for a substrate in a certain state. The change of states let us prove to which condition the enzyme is resistant and to which it is not. A usual condition that is proved is inhibitors.

Enzyme kinetics look at how much substrate it takes to reach a maximum velocity, this is called Vmax and in this point the enzyme is working at full speed. Since this is difficult to establish precisely it usually used the ½ Vmax. Where ½ Vmax is reached, the concentration of substrate is called Km. Km is a measurement of affinity, if Km is high the affinity is low and vice versa.

There are two types of inhibitors, competitive and non-competitive. Competitive inhibitors have a similar structure to the substrate, thus they can interact with the enzyme blocking the substrate. The competitive inhibitors affect the Km but not the Vmax. On the other hand, non-competitive inhibitors do not interfere with the active site but they impede the structural changes done by the enzyme to transform the substrate, this inhibitors affect Vmax, which can never be reached, but not Km.

In this case we must star by graphing the concentration (X) to the initial reaction rate( Y), this must be done for both kinetics.

Substrate Concentration (X)	Inhibitor- (Y)	Inhibitor+ (Y)
0,25	1,02	0,56
0,33	1,39	0,75
0,4	1,67	0,85
0,5	1,89	1
0,6	2,08	1,28
0,75	2,44	1,39
1	2,5	1,82

Inhibition 3 2,5 2 1,5 1 0,5 0,33 0,4 0,6 0,75 0,25 0,5 1 mM In hibitor- (Y) In hibitor+ (Y) 4/ww

1. In this case by the graph we can conclude it is a competitive inhibition as Vmax will be reached but in a higher concentration which means that Km as increased.
2. Vmax can be calculated with the line formula mx+b=Y. Where 1/Vmax= b

In this case the formula can be obtained by

1.39 1.02 Y2 Y1 0,37 4.625 т 3 1 0.33 0.25 X2 X1 0.08

Then for b you can use X=0.25

тx + b 3D у У

4.625 0.25 +b = 1.02

b 3D у/mx b

1.02 1.02 b 4.625 0.25 0.882 1.15625

1 = b Vтах

1 Vmax b

1 Vтах b

1 1 Vтах b = 1.13mM/h 0.882

кm т Vтах

Vтах — km т ж

4.625 1.13= 5.24 km

With inhibitor

0.75 0.56 Y2 Y1 0,19 2.375 т — X2 X1 0.33 0.25 0.08

0.56 b 2.375 0.25 0.94

1 1 1.06mM/h Vтах b 0.94

1.06 2.375 = 5.55 km

So without inhibitor it has a

Km= 5.24 mM and Vmax= 1.13mM/h

And with inhibitor

Km= 5.55mM and Vmax =1.06mM/h

And Ki is calculated by using the Kma

This is where Y=0 in the inhibited line

тx + b 3D у У

тx + b — 0

-b тх —

0.94 b 0.3957 2.375 т l

0.94 b 0.3957 2.375 т l

1 0.3957kma

1 kтa 2.52 0.3957

And the non inhibited

1 1 kma 0.21 0.882/4.625 4.6289

П, [I] kтa — km[1 + ki

km[] kта — km + ki

km[I] кта — kт ki

kта — km 2.52 0.21 2.31 5.64 ki km[I] 0.21 1.95 0.4095