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CP504 – ppt_Set 02
Enzyme kinetics and associated reactor design:
Introduction to enzymes,
enzyme catalyzed reactions and
simple enzyme kinetics
- learn about enzymes
- learn about enzyme catalyzed reactions
- study the kinetics of simple enzyme catalyzed reactions
Prof. R. Shanthini
09 Nov 2012
What is an Enzyme?
Enzymes are mostly proteins, and hence they consists of
amino acids.
Enzymes are present in all living cells, where they help
converting nutrients into energy and fresh cell material.
Enzymes breakdown of food materials into simpler
compounds.
Examples:
- pepsin, trypsin and peptidases break down proteins
into amino acids
- lipases split fats into glycerol and fatty acids
- amylases break down starch into simple sugars
Prof. R. Shanthini
09 Nov 2012
What is an Enzyme?
Enzymes are very efficient (biological) catalysts.
Enzyme catalytic function is very specific and effective.
Enzymes bind temporarily to one or more of the reactants of
the reaction they catalyze.
Enzymes does not get consumed in the reaction that it
catalyses.
Prof. R. Shanthini
09 Nov 2012
How does an Enzyme help?
Enzymes speed up reactions enormously. To understand
how they do this, examine the concepts of activation
energy & the transition state.
In order to react, the molecules involved are distorted,
strained or forced to have an unlikely electronic
arrangement. That is the molecules must pass through a
high energy state.
This high energy state is called the transition state.
The energy required to achieve it is called the activation
energy for the reaction.
Prof. R. Shanthini
09 Nov 2012
How does an Enzyme help?
The higher the free energy change for the transition barrier,
the slower the reaction rate.
Prof. R. Shanthini
09 Nov 2012
How does an Enzyme help?
Enzymes lower energy barrier by forcing the reacting molecules
through a different transition state. This transition state involves
interactions with the enzyme.
Enzyme
Prof. R. Shanthini
09 Nov 2012
Enzyme classification
Oxidoreductase: transfer oxygen atoms or electron
Transferase: transfer a group (amine, phosphate, aldehyde,
oxo, sulphur, etc)
Hydrolase: hydrolysis
Lyase: transfer non-hydrolytic group from substrate
Isomerase: isomerazion reactions
Ligase: bonds synthesis, using energy from ATPs
Prof. R. Shanthini
09 Nov 2012
Examples
Examples of
of enzyme
Enzymecatalyzed
Catalysedreactions
Reactions
Example 1:
CO2+ H2O
Carbonic anhydrase
H2CO3
Carbonic anhydrase is found in red blood cells.
It catalyzes the above reaction enabling red blood cells
to transport carbon dioxide from the tissues (high CO2)
to the lungs (low CO2).
One molecule of carbonic anhydrase can process
millions of molecules of CO2 per second.
Prof. R. Shanthini
09 Nov 2012
Examples of enzyme catalyzed reactions
Example 2:
2H2O2
Catalase
2H2O + O2
Catalase is found abundantly in the liver and in the red
blood cells.
One molecule of catalase can breakdown millions of
molecules of hydrogen peroxide per second.
Hydrogen peroxide is a by-product of many normal
metabolic processes.
It is a powerful oxidizing agent and is potentially
damaging to cells which must be quickly converted into
less dangerous substances.
Prof. R. Shanthini
09 Nov 2012
Industrial use of catalase
- in the food industry for removing hydrogen peroxide from milk
prior to cheese production
- in food-wrappers to prevent food from oxidizing
- in the textile industry to remove hydrogen peroxide from
fabrics to make sure the material is peroxide-free
- to decompose the hydrogen peroxide which is used (in some
cases) to disinfect the contact lens
Prof. R. Shanthini
09 Nov 2012
Examples of Industrial Enzymes
See the hand out on the same topic
Prof. R. Shanthini
09 Nov 2012
More on enzymes
Enzymes are very specific.
Absolute specificity - the enzyme will catalyze only one
reaction
Group specificity - the enzyme will act only on molecules that
have specific functional groups, such as amino, phosphate or
methyl groups
Linkage specificity - the enzyme will act on a particular type of
chemical bond regardless of the rest of the molecular structure
Stereochemical specificity - the enzyme will act on a particular
steric or optical isomer
Prof. R. Shanthini
09 Nov 2012
Prof. R. Shanthini
09 Nov 2012
Source: http://waynesword.palomar.edu/molecu1.htm
E+S
Prof. R. Shanthini
09 Nov 2012
ES
Source: http://waynesword.palomar.edu/molecu1.htm
Lock & Key Theory Of Enzyme Specificity
(postulated in 1894 by Emil Fischer)
E+S
Prof. R. Shanthini
09 Nov 2012
ES
E+P
Source: http://waynesword.palomar.edu/molecu1.htm
Prof. R. Shanthini
09 Nov 2012
Active Site Of Enzyme Blocked By Poison Molecule
Prof. R. Shanthini
09 Nov 2012
Source: http://waynesword.palomar.edu/molecu1.htm
Induced Fit Model
(postulated in 1958 by Daniel Koshland )
E+S
ES
E+P
Binding of the first substrate induces a conformational shift that
helps binding of the second substrate with far lower energy than
otherwise required. When catalysis is complete, the product is
released, and the enzyme returns to its uninduced state.
Prof. R. Shanthini
09 Nov 2012
Source: http://www.mun.ca/biology/scarr/Induced-Fit_Model.html
Simple Enzyme Kinetics
E+S
k1
k3
ES
E+P
k2
which is equivalent to
[E]
S
Prof. R. Shanthini
09 Nov 2012
P
S
for substrate (reactant)
E
for enzyme
ES
for enzyme-substrate complex
P
for product
Michaelis-Menten approach to the rate equation:
E+S
k1
ES
k3
E+P
k2
Assumptions:
1. Product releasing step is slower and it determines the
reaction rate
2. ES forming reaction is at equilibrium
3. Conservation of mass (CE0 = CE + CES)
Initial concentration of E
Concentration of E at time t
Prof. R. Shanthini
09 Nov 2012
Concentration of ES at time t
Michaelis-Menten approach to the rate equation:
E+S
k1
ES
k3
E+P
k2
Product formation (= substrate utilization) rate:
rP = - rS = k3 CES
(1)
Since ES forming reaction is at equilibrium, we get
k1 CE CS = k2 CES
Prof. R. Shanthini
09 Nov 2012
(2)
Michaelis-Menten approach to the rate equation:
E+S
k1
ES
k3
E+P
k2
Using CE0 = CE + CES in (2) to eliminate CE, we get
k1 (CE0 – CES) CS = k2 CES
which is rearranged to give
CES =
Prof. R. Shanthini
09 Nov 2012
CE0CS
k2/k1 + CS
(3)
Michaelis-Menten approach to the rate equation:
E+S
k1
k3
ES
E+P
k2
Using (3) in (1), we get
rP = - rS =
k3CE0CS
k2/k1 + CS
=
rmaxCS
KM + CS
where rmax = k3CE0
(5)
and KM = k2 / k1
(6)
Prof. R. Shanthini
09 Nov 2012
(4)
Other terminology used
Catalytic step
E+S
k1
k3
ES
E+P
k2
Substrate binding step
k3 = kcat
rmax = k3CE0 = kcatCE0
KM = k2 / k1
Prof. R. Shanthini
09 Nov 2012
(6)
(5a)
Briggs-Haldane approach to the rate equation:
E+S
k1
ES
k3
E+P
k2
Assumptions:
1. Steady-state of the intermediate complex ES
2. Conservation of mass (CE0 = CE + CES)
Initial concentration of E
Concentration of E at time t
Concentration of ES at time t
Prof. R. Shanthini
09 Nov 2012
Briggs-Haldane approach to the rate equation:
E+S
k1
ES
k3
E+P
k2
Product formation rate:
rP = k3 CES
(7)
Substrate utilization rate:
rs = - k1 CECS + k2 CES
(8)
Since steady-state of the intermediate complex ES is assumed,
we get
k1 CECS = k2 CES + k3 CES
Prof. R. Shanthini
09 Nov 2012
(9)
Briggs-Haldane approach to the rate equation:
E+S
k1
ES
k3
E+P
k2
Combining (7), (8) and (9), we get
rP = - rS = k3 CES
(10)
Using CE0 = CE + CES in (9) to eliminate CE, we get
k1 (CE0 - CES)CS = (k2 + k3)CES
which is rearranged to give
CES =
Prof. R. Shanthini
09 Nov 2012
CE0CS
(k2+k3)/k1 + CS
(11)
Briggs-Haldane approach to the rate equation:
E+S
k1
k3
ES
E+P
k2
Combining (10) and (11), we get
rP = - rS =
k3CE0CS
(k2+k3)/k1 +CS
where rmax = k3CE0
and KM = (k2 + k3) / k1
=
rmaxCS
KM + CS
(5)
(13)
When k3 << k2 (i.e. product forming step is slow),
Prof. R. Shanthini
09 Nov 2012
KM = k2 / k1
(6)
(12)
Simple Enzyme Kinetics (in summary)
[E]
S
rP = - r S =
P
rmaxCS
KM + CS
where rmax = k3CE0 = kcatCE0
and KM = f(rate constants)
rmax is proportional to the initial concentration of the enzyme
KM is a constant
Prof. R. Shanthini
09 Nov 2012