PART 1 Introduction
3.3. ENZYME KINETICS 1. Introduction
3.3.4. Models for More Complex Enzyme Kinetics
Example 3.1.
To measure the amount of glucoamylase in a crude enzyme preparation, 1 ml of the crude en- zyme preparation containing 8 mg protein is added to 9 ml of a 4.44% starch solution. One unit of activity of glucoamylase is defined as the amount of enzyme which produces 1 mmol of glucose per min in a 4% solution of Lintner starch at pH 4.5 and at 60∞C. Initial rate exper- iments show that the reaction produces 0.6 mmol of glucose/ml-min. What is the specific ac- tivity of the crude enzyme preparation?
Solution The total amount of glucose made is 10 ml ¥0.6 mmol glucose/ml-min or 6 mmol glucose per min. The specific activity is then:
Vmmust have units such as mmol product/ml-min. Since Vm=k2E0, the dimensions of k2must reflect the definition of units in E0. In the above example we had a concentra- tion of enzyme of 8 mg protein/10 ml solution ◊ 0.75 units/mg protein or 0.6 units/ml. If, for example, Vm=1 mmol/ml-min, then k2=1 mmol/ml-min ∏0.6 units/ml or k2=1.67 mmol/unit-min.
68 Enzymes Chap. 3 versible enzyme inhibitions are competitive, noncompetitive, and uncompetitive inhibi- tions. The substrate may act as an inhibitor in some cases.
Competitive inhibitors are usually substrate analogs and compete with substrate for the active site of the enzyme. The competitive enzyme inhibition scheme can be described as
(3.20)
E S ES E P
I
EI
+ +
+
1
k
k k
K
-
æ Ææ
1
1 2
Figure 3.8. Comparison of Michaelis–
Menten and allosteric enzyme kinetics.
Figure 3.9. Determination of cooperativity coefficient.
a
a
Assuming rapid equilibrium and with the definition of
(3.21)
we can develop the following equation for the rate of enzymatic conversion:
(3.22)
or
(3.23)
where
.
The net effect of competitive inhibition is an increased value of Km¢,appand, there- fore, reduced reaction rate. Competitive inhibition can be overcome by high concentra- tions of substrate. Figure 3.10 describes competitive enzyme inhibition in the form of a double-reciprocal plot.
Noncompetitive inhibitors are not substrate analogs. Inhibitors bind on sites other than the active site and reduce enzyme affinity to the substrate. Noncompetitive enzyme inhibition can be described as follows:
(3.24)
With the definition of
(3.25)
we can develop the following rate equation:
(3.26)
v V
K
K
m m
=
+ + ¢
1 1
1
[ ]
[ ] I
S Ê
ËÁ ˆ
¯˜Ê
ËÁ ˆ
¯˜
K K
v k
m = E S
ES = EI S
ESI = E I
EI = ES I ESI E ] = [E] [ES] [EI] [ESI and = ES
I
0 2
[ ][ ] [ ]
[ ][ ]
[ ] , [ ][ ]
[ ]
[ ][ ]
[ ]
[ + + + ] [ ]
E S ES E P
I I
EII S ESI
+ +
+ +
+
¢
¢
K k
K
K
m
m 2
æ Ææ
K K
m, m K [ ]
app
= Ê + I
ËÁ ˆ
¯˜ 1
1
v V
K
m m
= ¢ +
[ ]
, [ ] S
app S
v V
K K
m
m I
=
¢ + +
[ ] [ ] [ ]
S
I S
È1 ÎÍ
˘
˚˙
K K
v k
m¢ = I =
= + + =
[ ][ ]
[ ] , [ ][ ]
[ ]
[ ] [ ] [ ] [ ] [ ]
E S ES
E I EI
E0 E ES EI and ES2
a
a
a a
70 Enzymes Chap. 3 or
(3.27)
where
The net effect of noncompetitive inhibition is a reduction in Vm. High substrate con- centrations would not overcome noncompetitive inhibition. Other reagents need to be added to block binding of the inhibitor to the enzyme. In some forms of noncompetitive inhibition Vm is reduced and Km¢ is increased. This occurs if the complex ESI can form product.
Uncompetitive inhibitors bind to the ES complex only and have no affinity for the enzyme itself. The scheme for uncompetitive inhibition is
app I
V V
K
m
m
, = [ ]
1
1
Ê + ËÁ
ˆ
¯˜
v V
K
m m
= + ¢
,
[ ]
app
1 S Ê
ËÁ ˆ
¯˜
Figure 3.10. Different forms of inhibited enzyme kinetics.
(3.28)
With the definition of
(3.29)
we can develop the following equation for the rate of reaction:
(3.30)
or
(3.31) The net effect of uncompetitive inhibition is a reduction in both Vmand Km¢values.
Reduction in Vmhas a more pronounced effect than the reduction in Km¢,and the net result is a reduction in reaction rate. Uncompetitive inhibition is described in Fig. 3.10 in the form of a double-reciprocal plot.
High substrate concentrations may cause inhibition in some enzymatic reactions, known as substrate inhibition. Substrate inhibition is graphically described in Fig. 3.11.
The reaction scheme for uncompetitive substrate inhibition is
v V
K
m m
= ¢
, ,
app app
[S]
+[S]
v V
K K
K
m
m
= +
¢ 1
1
[I] [S
[I] [S
l
l
Ê ËÁ ˆ
¯˜
Ê + ËÁ ˆ
¯˜ +
]
]
K K
v k
m¢ = =
= + + =
[ ][ ]
[ ] , [ ][ ]
[ ]
[ ] [ ] [ ] [ ] [ ]
E S ES
ES I ESI
E E ES ESI and 2 ES
1
0
E S ES E P I
ESI
+ +
+
¢
1
K k
K
m
æ Ææ2
Figure 3.11. Comparison of substrate- inhibited and uninhibited enzymatic reactions.
a
a
72 Enzymes Chap. 3 (3.32)
With the definitions of
(3.33) the assumption of rapid equilibrium yields
(3.34)
A double-reciprocal plot describing substrate inhibition is given in Fig. 3.10.
At low substrate concentrations, [S]2/KS1<<1, and inhibition effect is not observed.
The rate is
(3.35)
or
(3.36) A plot of 1/vversus 1/[S] results in a line of slope Km¢/Vmand intercept of 1/Vm.
At high substrate concentrations, K¢m/[S] <<1, and inhibition is dominant. The rate in this case is
(3.37)
or
(3.38) A plot of 1/vversus [S] results in a line of slope 1/Ks1◊Vmand intercept of 1/Vm.
1 1
v= + 1
Vm K VS m
[ ]
S v=+ V
K
m
S
1
1
[ ]S Ê ËÁ
ˆ
¯˜
1 1 1
v= + ¢ V
K
m V
m m
[ ]
S v=+ ¢ V
K
m
1 m
[ ]
S ÈÎÍ ˘
˚˙ v=
¢ V
K K
m
m
[S]
S S
S
+[ ]+ [ ]2
1
KS Km
1 = S ES ¢ =
ES
S E
2 ES
[ ][ ]
[ ]
[ ][ ]
,
[ ]
E S ES E P
S
ES2
+ +
+
¢
K k
K
m
S
2
1
æ Ææ
a
a
The substrate concentration resulting in the maximum reaction rate can be determined by setting dv/d[S] =0. The [S]maxis given by
(3.39) Example 3.2
The following data have been obtained for two different initial enzyme concentrations for an enzyme-catalyzed reaction.
[S]max = K K¢m S
1
v([E0] =0.015 g/l) [S] v([E0] =0.00875 g/l)
(g/l-min) (g/l) (g/l-min)
1.14 20.0 0.67
0.87 10.0 0.51
0.70 6.7 0.41
0.59 5.0 0.34
0.50 4.0 0.29
0.44 3.3
0.39 2.9
0.35 2.5
a. Find Km.
b. Find Vmfor [E0] =0.015 g/l.
c. Find Vmfor [E0] =0.00875 g/l.
d. Find k2.
Solution A Hanes–Woolf plot (Fig. 3.12) can be used to determine Vmand Km.
[S] [S]
v = K +
V V
m
m m
1
[S]/v(E0=0.015) [S]/v(E0=0.00875) [S]
(min) (min) (g/l)
17.5 30 20.0
11.5 20 10.0
9.6 16 6.7
8.5 15 5.0
8.0 14 4.0
7.6 3.3
7.3 2.9
7.1 2.5
From a plot of [S]/v versus [S] for E0 = 0.015 g/l, the slope is found to be 0.6 min/g/l and Vm=1/0.6 =1.7 g/l min. The y-axis intercept is Km/Vm=5.5 min and Km=9.2 g [S]/l.
Also, Vm=k2E0and k2=1.7/0.015 =110 g/g enzyme-min. The Hanes–Woolf plot for E0=0.00875 g/l gives a slope of 1.0 min/g/l and Vm=1.0 g/l-min; k2=Vm/E0=1.0/0.00875 = 114 g/g enzyme-min.
74 Enzymes Chap. 3 Example 3.3
The hydrolysis of urea by urease is an only partially understood reaction and shows inhibi- tion. Data for the hydrolysis of the reaction are given next.
Figure 3.12. Hanes–Woolf plots for E0=0.015 g/l and E0=0.00875 g/l (Example 3.1).
Substrate concentration: 0.2 M 0.02 M
l/v I l/v I
0.22 0 0.68 0
0.33 0.0012 1.02 0.0012
0.51 0.0027 1.50 0.0022
0.76 0.0044 1.83 0.0032
0.88 0.0061 2.04 0.0037
1.10 0.0080 2.72 0.0044
1.15 0.0093 3.46 0.0059
where v=moles/l-min and I is inhibitor molar concentration.
a. Determine the Michaelis–Menten constant (K¢m) for this reaction.
b. What type of inhibition reaction is this? Substantiate the answer.
c. Based on the answer to part b, what is the value of Ki?
Solution A double-reciprocal plot of l/vversus l/[S] for inhibitor concentrations I = 0, 0.0012, 0.0044, and 0.006 indicates that the inhibition is noncompetitive (Fig. 3.13). From the x-axis intercept of the plot, -1/K¢m= -13 andKm¢ ª7.77 ¥10-2M. For [S] =0.2 Mand I =0 from the intercept of 1/Vversus 1/S, l/vm=0.2 and Vm@5 moles/l-min. For I =0.0012 Mand [S] =0.2 M, v=3 moles/l-min. Substituting these values in
gives KI=6 ¥10-3M.
v V
K K
m
I
m
=
+ + ¢
1 [I] 1 [S]
Ê ËÁ ˆ
¯˜Ê ËÁ ˆ
¯˜