2. • Chemical kinetics is the branch of chemistry
which addresses "how fast do reactions go?"
• Enzymes are the catalysts of biological systems
and are extremely efficient and specific as
catalysts.
• An enzyme accelerates the rate of a reaction
• Enzymes are highly specific.
CHEMICAL KINETICS
3. • A chemical mechanism is the chemical pathway
of conversion of S → P, including the structures
of any intermediates
4. • What we want to be able to determine
– Maximum velocity
– Substrate affinity
– Inhibitor affinity
• What it can tell us
– Flow through metabolic pathways
– Utilization of substrates
• What can we do with the information
– Control and manipulate metabolic events
5. STUDY OF ENZYME KINETICS IS USEFUL FOR MEASURING
– Concentration of an enzyme in a mixture
– Its purity
– Its specificity for different substrates
– Comparison of different forms of the same enzyme
– Effects of inhibitors
- Mechanism, structure of active site
6. ENZYME KINETICS
• The study of rate of the reaction and how it
changes in response to change in experimental
parameters
• These studies include measuring rates of the
enzyme-catalyzed reactions
[S]
Vmax/2
Km
v = Vmax = kcat [E]o
7. • In 1913 Michaelis-Menten derive an equation
which gives the relationship between the [S] &
velocity of the reaction.
• E is the enzyme, S is the "substrate" ES is an
enzyme-substrate complex.
Leonor Michaelis
(1875-1940)
Maud L. Menten
(1879-1960)
MICHAELIS-MENTEN
8. • E+S ES
K1
K-1
ES E+P
K2
K-2
Equation 1
Vo = K2 [ES] Equation 2
• ES is not easily measured thus the term [Et] is
introduced which is the sum of Free enzyme +
substrate bound enzyme
[Et] = [E] + [ES]
[E] = [Et] – [ES]
9. Rate of ES formation = K1([Et]-[ES]) [S] Equation 3
Rate of ES breakdown = K-1([ES]+K2[ES]) Equation 4
K1 ([Et]) – [ES])[S] = K-1 [ES] + K2 [ES] Equation 5
K1 [Et][S] - K1 [ES] [S] = (K-1 + K2) [ES] Equation 6
Step 1 : The rate of formation and break down of ES are
determined by the steps governed by the rate constant
Step 2 : An assumption is made that the initial rate of the
reaction reflects a steady state
10. K1 [Et] [S] = (K1 [S] + K-1 + K2) [ES] Equation 7
[ES] = K1 [Et] [S] Equation 8
K1 [S] + K-1 + K2
[ES] = [Et] [S]
Equation 9
[S] + (K2 + K-1) / K1
Adding the term K1 [ES][S] to both sides of the equation
Solving for ES
Dividing by K1
11. The term (K2 + K-1) / K1 is defined as Km
i.e., [ES] = [Et] [S]
Km + [S]
Equation 10
Vo can be express in terms of [ES]
Vo = K2 [ES]
Vo = K2 [Et] [S] Equation 11
Km + [S]
Vo = Vmax [S]
Km + [S]
Maximum velocity occur when the enzyme is saturated
[ES] = [Et]. Vmax = K2[Et]
12. RECIPROCAL OF MICHEAL MENTEN’S EQUATION
Vo is exactly half of the Vmax
Vmax/2 = Vmax [S] / Km + [S]
On dividing by Vmax
½ = [S] / Km + [S]
Solving for Km, we get Km + [S] = 2 [S]
Km = [S]
13. LINEWEAVER BURK RECIPROCAL PLOT
• It is difficult to determine the limitting value of V
(i.e., Vmax) directly from a plot of V against [S]
• Reciprocal of the MM equation give a straight line
plot
1/V0 = Km + [S]
Vmax [S]
1/Vo = Km + [S]
Vmax [S] Vmax [S]
1/Vo = Km + 1
Vmax [s] Vmax
14. •1/Vo is ploted against 1/[S] a straight line is obtained.
• Slope - Km/Vmax
•An intercept 1/ Vmax on the 1/ Vo axis
•-1/Km on the 1/[S] axis
15. HALDANE AND BRIGGS EQUATION
• In 1925 Briggs and Haldane, came up with a
more generally applicable assumption.
• They assumed that the concentration of ES
does not change with time. This is the Steady
State Assumption (Briggs and Haldane).
[E] + [S] [ES] [P] + [E]
k2
k-1
k1
16. EFFECT OF ENZYME CONCENTRATION
• As the enzyme concentration increases the rate
of enzyme activity increases up to a level where
it becomes constant.
• More the enzymes are available, the more
substrates are broken in less time.
17. •Then becomes constant as the substrate acts as a
limiting factor, which means that there are not
enough substrates to be broken down compared to
the number of enzymes.
18. EFFECT OF PH
• If the pH is too acidic or too basic for an enzyme,
its hydrogen bonds begin to break, causing its
active site to change its shape.
• An altered active site can’t bind with its
substrate so enzyme activity decreases.
19. • If the pH is too unfavorable then covalent bonds
can break, causing the enzyme to denature.
20. EFFECT OF TEMPERATURE
• The temperature increases there is more
movement of molecules and therefore more
collisions between enzymes and substrates – so
the enzyme activity increases.
• There is a limit to which enzyme activity can
increase because at a certain temperature an
enzyme will denature
21. EFFECT OF SUBSTRATE CONCENTARTION
• [S] changes due to the conversion of substrate
to product.
• Measure the initial rate of the reaction
designated V0 ( Initial velocity), when [S] is
much greater than the concentration of enzyme
22. SIGNIFICANCE OF Km AND Vmax
• The Km values of enzymes range widely. Km
values lies between 10-1 & 10-7 M
• The Km value of an enzyme depends on the
particular substrate & also environmental
conditions such as pH, temperature & ionic
strength
• Km is the concentration of substrate at which
half the active sites are occupied
• At lower Km affinity of enzyme is more and at
higher Km affinity of enzyme is low
23. • Km is related to the rate constants of
individual steps in the catalytic scheme
• If K-1 is greater than K2 i.e., dissociation of ES
complex is more rapid than the formation of E
& P then
Km = K-1 / K1
• The dissociation constant of ES complex is
given by
KES = [E] [S] = K-1
[ES] K1
E+S K1 ES K2 E+P
K-1
24. • In some cases the ES complex doesn't breakdown
directly to form free enzyme and product but
they for EP complex. Hence these is an rate
limiting step
Vmax = K3 [Et]
• Turn over number (Kcat) is defined as the number
of substrate converted to product by single
enzyme in a unit of time when enzyme is fully
saturated with substrate.
Vmax = Kcat [Et]
Vo = Vmax [S] / Km + [S]
Vo = Kcat [Et] / Km = [S]
E+S K1 ES K2 EP E+P
K-1 K-2
