This document discusses enzyme kinetics and catalytic mechanisms. It provides details on: 1. The catalytic triad of serine proteases (Ser195, His57, Asp102) and how inhibitors like DIPF can label the active site serine, disabling the enzyme. 2. The kinetic steps of enzyme-catalyzed reactions, including substrate binding, formation of a tetrahedral intermediate, and release of products. 3. How transition state theory is used to explain the kinetic barrier for reactions, and how enzymes lower this activation energy to greatly increase reaction rates. 4. How reaction rates depend on parameters like temperature, activation energy, and catalysts through equations derived from Arrhenius kinetics

1. M.Prasad Naidu
MSc Medical Biochemistry,
Ph.D.Research Scholar

2. •Diverse and widespread proteolytic enzymes
•Involved in digestion, development, clotting,
inflammation…
•Common catalytic mechanism

3. P-Nitrophenolate is very
yellow while the acetate is
colorless. This is an example
of an artificial substrate!

4. The kinetics show
1. A “burst phase” where the product is rapidly
formed with amounts stoichiometric with the
enzyme.
2. Slower steady state that is independent of
substrate concentration.

5. A covalent bond between a Serine and the substrate
suggests an “active Serine”. These Serines can be
labeled with inhibitors such as diidopropyl
phosphofluoridate specifically killing the enzyme.
Ser 195 is specifically labeled

6. DIPF is extremely toxic because other active Serines
can be labeled. Such as acetylcholine esterase.
Nerve gases, serin gas,
are very toxic!! Many
insecticides also work
this way.

7. His 57 is a second important catalytic residue. A
substrate containing a reactive group binds at the
active site of the enzyme and reacts with a nearby
reactive amino acid group. A Trojan horse effect.
Tosyl-L-phenylalanine chloromethyl ketone (TPCK)

8. The reaction of TPCK with His 57 of chymotrypsin

9. Bovine
Trypsin

11. Bovine trypsin catalytic triad

13. Catalytic mechanism
1. After the substrate binds Ser 195 nucleophilically
attacks the scissile peptide bond to form a transition state
complex called the tetrahedral intermediate (covalent
catalysis) the imidazole His 52 takes up the proton Asp 102
is hydrogen bonded to His 57. Without Asp 102 the rate of
catalysis is only 0.05% of wild-type.
2. Tetrahedral intermediate decomposes to the acyl-
enzyme intermediate. His 57 acts as an acid donating a
proton.
3. The enzyme is deacylated by the reverse of step 1 with
water the attacking nucleophile and Ser 195 as the leaving
group.

14. 1. Conformational distortion forms the tetrahedral
intermediate and causes the carboxyl to move close to the
oxyanion hole
2. Now it forms two hydrogen bonds with the enzyme that
cannot form when the carbonyl is in its normal conformation.
3. Distortion caused by the enzyme binding allows the
hydrogen bonds to be maximal.

15. Triad charge transfer complex stabilization

16. Rates of Enzyme Reactions
How fast do reactions take place
•Reaction rates
Thermodynamics says I know the difference between
state 1 and state 2 and DG = (Gf - Gi)
But
Changes in reaction rates in response to differing
conditions is related to path followed by the reaction
and
is indicative of the reaction mechanism!!

17. 1. Substrate binding constants can be measured as
well as inhibitor strengths and maximum catalytic
rates.
2. Kinetics alone will not give a chemical mechanism
but combined with chemical and structural data
mechanisms can be elucidated.
3. Kinetics help understand the enzymes role in
metabolic pathways.
4. Under “proper” conditions rates are proportional to
enzyme concentrations and these can be determine “
metabolic problems”.

18. Chemical kinetics and Elementary
Reactions
A simple reaction like A  B may proceed through several
elementary reactions like A  I1  I2  B Where I1 and I2 are
intermediates.
The characterization of elementary reactions comprising an
overall reaction process constitutes its mechanistic
description.
Rate Equations
Consider aA + bB + • • • + zZ. The rate of a reaction is
proportional to the frequency with which the reacting
molecules simultaneously bump into each other
     zba
ZBAkRate 

19. The order of a reaction = the sum of exponents
Generally, the order means how many molecules have to bump into
each other at one time for a reaction to occur.
A first order reaction one molecule changes to
another
A  B
A second order reaction two molecules react
A + B  P + Q
or
2A  P

20. 3rd order rates A + B + C  P + Q + R rarely occur
and higher orders are unknown.
Let us look at a first order rate
A  B
   
dt
Pd
dt
Ad
v
n = velocity of the reaction
in Molar per min.
or
moles per min per volume
k = the rate constant of the
reaction
   A
dt
Ad
kv 

21. Instantaneous rate: the rate of reaction at any specified
time point that is the definition of the derivative.
We can predict the shape of the curve if we know the
order of the reaction.
A second order reaction: 2A  P
   2
A
A
k
dt
d
v 
Or for A + B  P + Q
      BA
BA
k
dt
d
dt
d
v 

22. Percent change in A (ratio ) versus time in first and
second order reactions

23. It is difficult to determine if the reaction is either first or
second order by directly plotting changes in
concentration.
   A
dt
Ad
k
 
 
dt
d
k
A
A

 
t
0
A
A
dtk-
A
A
o
d
    to kAlnAln 
    -kt
o eAA 

24. However, the natural log of the concentration is
directly proportional to the time.
- for a first order reaction-
The rate constant for the
first order reaction has
units of s-1 or min-1 since
velocity = molar/sec
and v = k[A] : k = v/[A]
Gather your data and plot
ln[A] vs time.

25.    
2
A
A o
 Plugging in
to rate equation
 
  2
1
o
A
2
A
ln kt












kk
693.02ln
t
2
1 

26. The half-life of a first order reaction can be used to
determine the amount of material left after a length
of time.
The time for half of the reactant which is initially
present to decompose or change.
32P, a common radioactive isotope, emits an
energetic b particle and has a half-life of 14 days.
14C has a half life of 5715 years.

27. A second order reaction such like 2A  P
 
  
 
 
t
dtk
0
A
oA
2
o
A
Ad
-
   
kt
oA
1
A
1
When the reciprocal of the concentration is plotted verses time a
second order reaction is characteristic of a straight line.
The half-life of a second order reaction is
and shows a dependents on the initial concentration
 o2
1
A
1
t
k


28. A bimolecular reaction A + B C A B + C at some point
in the reaction coordinate an intermediate ternary complex will exist
A B C
This forms in the process of bond formation and bond breakage
and is called a transition state.
Ha + Hb Hc Ha Hb + Hc
This is a molecule of H2 gas reforming by a collision

29. An energy contour of
the hydrogen reaction
as the three molecules
approach the transition
state at location c.
This is called a saddle
point and has a higher
energy than the
starting or ending
point.

30. Energy diagrams for the
transition state using the
hydrogen molecule
Transition state diagram
for a spontaneous reaction.
X‡ is the symbol for the
species in the transition
state

31.       Xk'BAk
dt
Pd

QPBA 
For the reaction
‡ Where [X] is the
concentration of the
transition state species
 
  BA
X
K 
‡
‡
GRTlnK- D
‡ ‡
DG‡ is the Gibbs free energy of the activated
complex.
k' = rate constant for the decom-
position of the activated complex

32.     BAek'
t
P RT
G
-
D

d
d
‡
The greater the DG‡, the more unstable the transition
state and the slower the reaction proceeds.
This hump is the activation barrier or kinetic barrier for a reaction.
The activated complex is held together by a weak bond that would
fly apart during the first vibration of the bond and can be
expressed by k' = kn where n is the vibrational frequency of the
bond that breaks the activated complex and k is the probability
that it goes towards the formation of products.

33. Now we have to define n. E = hn and n = E/h where h
is Planks constant relating frequency to Energy. Also
through a statistical treatment of a classical
oscillator E= KbT where Kb is Boltzmann constant.
By putting the two together
h
TK
k b

RT
G
b
h
TK
k
D

 e
And
The rate of reaction decreases as its free energy of
activation, DG‡ increases
or
the reaction speeds up when thermal energy is added
‡

34. Consider
PA 21
 kk
I
If one reaction step is much slower than all the rest this step
acts as a “bottleneck” and is said to be the rate-limiting step

36. Catalysts act to lower the activation barrier of the reaction being
catalyzed by the enzyme.
Where DDG‡
cat = DG‡
uncat- DG‡
cat
The rate of a reaction is increased by RT
Gcat
e
DD
DDG‡
cat = 5.71 kJ/mol is a ten fold increase in rate.
This is half of a hydrogen bond!!
DDG‡
cat = 34.25 kJ/mol produces a million fold
increase in rate!!
Rate enhancement is a sensitive function of DDG‡
cat