Lecture 13

Chemical Reaction Engineering (CRE) is the

field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
 Lecture 13
Enzymatic Reactions
 Michealis-Menten Kinetics
 Lineweaver-Burk Plot
 Enzyme Inhibition
 Competitive
 Uncompetitive
 Non-Competitive

2
 Robert S. Langer
Chemical Engineer, MIT

• Robert S. Langer is one of 12 Institute Professors at MIT;

being an Institute Professor is the highest honor that can be
awarded to a faculty member. Dr. Langer has written more
than 1,480 articles.
• He also has over 1,360 issued and pending patents
worldwide. Dr. Langer’s patents have been licensed or
sublicensed to over 400 pharmaceutical, chemical,
biotechnology and medical device companies.
A catalyst lowers activation energy.

 Catalysts are substances that speed up chemical

reactions.
 decrease activation energy
 increase reaction rate
 Enzymes allow chemical
reactions to occur under tightly
controlled conditions.
 Enzymes are catalysts in living things.
 Enzymes are needed for almost all processes.

–Most enzymes are proteins.

 Disruptions in homeostasis can prevent
enzymes from functioning.

– Enzymes function best in a small range of conditions.

– Changes in temperature and pH can break hydrogen bonds.
– An enzyme’s function depends on its structure.
 An enzyme’s structure allows only certain reactants to
bind to the enzyme.

– substrates
– active site
substrates
(reactants)

enzyme

Substrates bind to an
enzyme at certain places called active
sites.
  The lock-and-key model helps
illustrate how enzymes function.

– substrates brought together

– bonds in substrates weakened

Substrates bind to an The enzyme brings The catalyzed reaction forms

enzyme at certain places substrates together and a product that is released
called active sites. weakens their bonds. from the enzyme.
 Enzymes
Michaelis-Menten Kinetics
Enzymes are protein-like substances with catalytic properties.

Enzyme Unease
[From Biochemistry, 3/E by Stryer, copywrited 1988 by Lubert Stryer. Used with
permission of W.H. Freeman and Company.]
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 Enzymes
Enzymes provide a pathway for the substrate to
proceed at a faster rate. The substrate, S, reacts
to form a product P.

S Slow P

E•S
Fast

 can only catalyze only one reaction.

A given enzyme
Example, Urea is decomposed by the enzyme urease.
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 Enzymes - Urease
A given enzyme can only catalyze only one reaction. Urea is
decomposed by the enzyme urease, as shown below.

NH2CONH 2 + UREASE ⎯H⎯→

⎯
2O
2NH3 + CO2 + UREASE
S + E ⎯H⎯
⎯
2O
→P + E
The corresponding mechanism is:

E + S ⎯⎯→ E • S
k1

E • S ⎯⎯→ E + S
k2

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E • S + W ⎯⎯→ P + E
k3
 Enzymes - Michaelis-Menten Kinetics
rP = k3 (E • S )(W )
rE •S = 0 = k1 (E )(S ) − k 2 (E • S ) − k3W (E • S )

k1 (E )(S )
(E • S ) =
k2 + k3W

Et = (E ) + (E • S )

(E ) = Et
 k1S 
1 +  
14  k 2 + k3W 
 Enzymes - Michaelis-Menten Kinetics
Vmax
k cat
 
rP = k3 (E • S )(W ) =
k3W Et S kcat Et S
=
k 2 + k3W K + S
+S M
k1
 
KM

rP = k3 (E • S )(W ) =
Vmax S
Km + S
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 Enzymes - Michaelis-Menten Kinetics
Vmax=kcatEt

Turnover Number: kcat

Number of substrate molecules (moles) converted to
product in a given time (s) on a single enzyme molecule
(molecules/molecule/time)
kcat
For the reaction: H2O2 + E →H2O + O + E

40,000,000 molecules of H2O2 converted to product per

second on a single enzyme molecule.

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 Enzymes - Michaelis-Menten Kinetics
Michaelis-Menten Equation
VmaxS
rP = −rS =
KM + S
(Michaelis-Menten plot)
Vmax VmaxS1/ 2
Vmax Solving: =
2 K M + S1/ 2
-rs KM=S1/2
therefore KM is the
concentration at which the rate
17 S1/2 CS is half the maximum rate.
 Enzymes - Michaelis-Menten Kinetics
1 1 KM  1 
Inverting yields: = +  
− rS Vmax Vmax  S 
Lineweaver-Burk Plot

1/-rS slope = KM/Vmax

1/Vmax

18 1/S
 Types of Enzyme Inhibition
Competitive
E + I  I • E (inactive)

Uncompetitive
E • S + I  I • E • S (inactive )

Non-competitive
E • S + I  I • E • S (inactive )
I • E + S  I • E • S (inactive )
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 Competitive Inhibition

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 Competitive Inhibition
E + S
⎯⎯→
k1
⎯⎯ E • S ⎯⎯→
k3
E+P
k2

E + I
⎯⎯→
k4
⎯⎯ E • I (inactive )
k5

1) Mechanisms:
E + S → E S E S → E + S
E S → P + E E + I → EI
EI → E + I
rP = k 3C ES
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 Competitive Inhibition
2) Rate Laws:
rES = 0 = k1CSC E − k 2 C ES − k 3C ES

k1CSCE CSCE
CES = =
k 2 + k3 Km
k 3 CS C E
rP =
Km
rIE = 0 = k 4 C I C E − k 5C IE
CI CE k5
CIE = KI =
22 KI k4
 Competitive Inhibition
CEtot
C Etot = CE + C ES + CIE CE =
CS C I
1+ +
k 3C Etot CS Km KI
rP =
CI K m
K m + CS +
KI

Vmax CS
− rS =
 CI 
CS + K m 1 + 
 KI 

1 1 km  CI  1
= + 1 + 
23 − rS Vmax Vmax  K I  CS
 Competitive Inhibition
From before (no competition): 1 = 1 + K M 1
Increasing C
− rS Vmax Vmax CS
I

Competitive

1 No Inhibition Competitive
rS KM
slope = 1 1 K M  CI  1
Vmax = + 1 + 
− rS Vmax Vmax  K I  CS
1
Intercept = 1
Vmax
CS

Intercept does not change, slope increases as

24 inhibitor concentration increases
 Uncompetitive Inhibition

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 Uncompetitive Inhibition
Inhibition only has affinity for enzyme-substrate complex
⎯⎯→
k1

E +S E • S ⎯⎯→
k3
P
⎯⎯
k2

⎯⎯→
k4

I + E •S I • E • S (inactive )
⎯⎯
k5

Developing the rate law:

rP = −rS = kcat (E • S )

rE •S = 0 = k1 (E )(S ) − k 2 (E • S ) − kcat (E • S ) − k 4 (I )(E • S ) + k5 (I • E • S ) (1)

rI•E•S = 0 = k 4 (I )(E • S ) − k5 (I • E • S ) (2)
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 Uncompetitive Inhibition
Adding (1) and (2)
k1 (E )(S ) − k 2 (E • S ) − kcat (E • S ) = 0

(E • S ) = k1 (E )(S ) = (E )(S )
k 2 + kcat KM
From (2)
(I • E • S ) = k4 (I )(E • S ) = (I )(E • S ) = (I )(E )(S )
k5 KI KI KM
k5
KI =
k4
kcat (E )(S )
rp = kcat (E • S ) =
27 KM
 Uncompetitive Inhibition
Total enzyme
Et = (E ) + (E • S ) + (I • E • S )

= (E )1 +
( S ) (I )(S ) 
+ 
 KM KI KM 
kcat Et (S )
rp =

K M 1 +
( S ) (I )(S ) 
+ 
 KM KI KM 
Vmax (S )
− rS = rP =
 (I ) 
K M + (S )1 + 
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 KI 
 Uncompetitive Inhibition
1   (I )  
K M + (S )1 +
1
=  
− rS Vmax (S ) 

 KI 


1 K  1  1  (I ) 
= M   + 1 + 
− rS Vmax  (S )  Vmax  KI 

Slope remains the

same but intercept
changes as inhibitor
concentration is
increased

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Lineweaver-Burk Plot for uncompetitive inhibition
 Non-competitive Inhibition

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 Non-competitive Inhibition
E+S E·S P+E

+I -I -I +I
(inactive)I.E + S I.E.S (inactive)
Increasing I

1
−
rS Vmax CS
No Inhibition − rS =
 
(k M + CS )1 + CI 
Both slope and intercept  kI 
changes
1 1  C I  k M  1  C I 
= 1 +  +  1 + 
1 − rS Vmax  k I  Vmax  CS  k I 
CS
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 Summary: Types of Enzyme Inhibition

Lineweaver–Burk plots for three types of enzyme inhibition.

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