Download Chapter 3: Enzymes: Structure and Function

Chapter 3: Enzymes: Structure and Function
Enzymes actt as the
E
th body’s
b d ’ catalysts
t l t by
b complexing
l i the
th reaction's
ti ' participants
ti i
t in
i the
th correctt
arrangement to react, lowering the activation energy, Ea, to react, but G stays the same.
H
HO
B
+
O
TS1
+
C
H
enzyme
active site
O
O
C
H3C
enzyme
active site
B
complex
forms
=
O
C
C
H3C
O
lactic acid = Product
C
H3C
+
+
O
TS3
disassociate
O
C
C
H3C
+
O
TS2
O
C
enzyme
active site
H
HO
NH2
O
NAD+
N
O
Pyruvic acid = Substrate
Lactate dehydrogenase
LDH
(enzyme)
H
R
O
O
S + E
Enzymes
lower the Ea
of a reaction
Substrate
G =
PE
potential
energy
TS1
Substrate
exergonic
TS3
Ea2
G =
Product
Product
POR = progress of reaction
with the enzyme
POR = progress of reaction
without the enzyme
-Ea2
10 2.3RT
-Ea1
10 2.3RT
-Ea2
2 3RT
= 10 2.3RT
-(4 - 20)
14
1.4
= 10
H
HO
TS2
11 43
= 1011.43
= 3 x 1011
Enzyme catalyzed reactions
are always
l
f
faster.
A
Assumed,
d
Ea1 = 20 kcal/mole and
Ea2 = 4 kcal/mole.
N
attack site
(2 faces)
O
NH2
H H
P
EP
OH
HO
O
O
ES
N
H H H H
Ea is much lower so reaction is much faster.
Ea1
kenz
=
kno enz
P
N
O
O
Ea is too high so
reaction is too slow.
TS
exergonic
O
NAD+
Nicotinamide
adenosine dinucleotide
(reducing agent )
cofactor
site
P
H
O
NADH
(cofactor)
H
PE
potential
energy
N
R
N
N
H
OH
1. Provides a reaction surface
(the active site)
2. Provides a suitable environment
(hydrophobic)
3. Brings reactants together
4. Positions reactants correctly for
reaction
5. Weakens bonds in the reactants
6. Provides acid / base catalysis
7. Provides nucleophilic groups
8 Stabilises
8.
St bili the
th transition
t
iti state
t t with
ith
intermolecular bonds
© Oxford University Press, 2013
1
The active site is often a hydrophobic hollow or cleft with key polar (or nonpolar) amino acids in key
locations on the enzyme surface that can accept substrates and cofactors. The enzyme contains
amino acids that interact with the substrate and cofactor in the usual way (ionic interactions, H bonds,
dipole-dipole, dispersion forces and covalent bonds) which all help repeatedly catalyze the reaction
(catch and release). It is usually proposed that the transition state complex is stabilized, lowering the
activation energy which accelerates the reaction rate. Rather than the old 'lock and key' model, it is
proposed that the enzyme and substrate influence one another to form a stronger interaction. This is
called the 'induced fit' model.
Identification of active sites is crucial in the process of drug discovery. The 3-D structure of the
enzyme is analysed to identify active sites and design drugs which can fit into them. The most
common ways to do this are x-ray crystollography, NMR analysis and computer modeling.
Inhibitors bind to an enzyme's active site and block interaction with natural substrates. Knowing the
strength of binding between the active site and an enzyme inhibitor is an important strategy in drug
design.
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2
active site
Interactions between the substrate,
substrate
cofactors and the enzyme can be
very complicated.
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3
Substrate binding uses the usual forces of interaction.
1. Ionic
2. H-bonding
3. Dispersion forces
4 Dipole-dipole
4.
Dipole dipole
vdw
5. Covalent bonds
interaction
6. Pi stacking
S
H-bond
Active site
H
Ser
ionic
bond
O
Phe
CO2
Asp
Enzyme
4
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Induced fit - active site of the enzyme and the substrate alter shapes to maximise
intermolecular bonding.
bonding
S
Ser O
S
Ser
Phe
H
CO2
Asp
Intermolecular bonds not optimum
length for maximum bonding
Induced
fit
O H
Phe
CO2
Asp
Intermolecular bond lengths
optimised Susceptible bonds in
optimised.
substrate strained. Susceptible
bonds in substrate more easily
broken
5
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Binding of pyruvic acid in LDH (lactic dehydrogenase enzyme)
1.
2.
3.
Ionic bonding
H-bonding
Dispersion
i
i forces
f
O
H-Bond
H
O
O
C
H3C
dispersion
vdw-interactions
vdw
interactions
C
O
+
H3N
Ionic bond
6
© Oxford University Press, 2013
Catalysis mechanisms – necessary functions
Acid/base catalysis
Histidine
H
O
N
C
CH
Protein
Protein
O
N
C
pKa  6.0
p
H
B
H
N
B
H
H
Non-ionised
Acts as a basic catalyst
(proton 'sink')
N
N
Cysteine H
O
H
b
C
CH
Protein
CH2
a
H
O
C
CH
Protein
H
Where is the best nucleophile?
a. serine oxygen
b. cysteine sulfur
c. tyrosine
i oxygen
d. all are similar
CH2
b
H
a. mainly on a
b. mainly on b
c. on both a and b
Tyrosine H
O
N
C
O
N
Protein
N
Where is the + charge?
H
Protein
N
H
Ionised
Acts as an acid catalyst
(proton source)
Nucleophilic residues
Serine
N
CH2
a
N
Protein
CH
Protein
N
CH2
H
Protein
CH
CH2
N
Protein
N
H
H
S
H
c
O
7
© Oxford University Press, 2013
Serine acting as a nucleophile
S i
Serine
O
H
C
N
CH
Protein
B
H
H
S
N
C
H
B
acyl CoA
H
H
N
C
CH
R
B
S
H
thiol
H
O
B
H
B
O
C
Protein
N
CH2
O
H
O
H
H
O
O
H
Protein
O
C
R
R
Protein
N
CH2
R
C
Serine
CH
Protein
H
S
Threonine is
also possible.
O
R
N
CH2
O
Protein
H
R
B
H
O
O
As carboxylate
at body pH.
C
O
8
© Oxford University Press, 2013
R
Mechanism for chymotrypsin uses 3 amino acids at the active site
Catalytic triad of serine, histidine and aspartate
H
N
N
H
O
O
O
Serine
Histidine
Aspartic acid
Chymotrypsin
9
© Oxford University Press, 2013
Mechanism for chymotrypsin
Chymotrypsin
C
N
O
H
O
H
Protein
Chymotrypsin
N
R
H
C
N
Protein
CH
Protein
CH
Protein
N
H
R
N
H
N
H
O
H
O
Serine
Aspartic acid
Histidine
Chymotrypsin
Chymotrypsin
Chymotrypsin
O
C
O
H
Protein
O
N
C
CH
R
N
N
H
H
O
H
CH
R
H
O
Protein
H2 N
C
N
Protein
R
H
O
H
CH
O
H
N
O
O
Serine
Aspartic acid
Histidine
Chymotrypsin
N
O
O
Serine
Protein
H
N
O
O
H
N
N
H
O
O
O
Histidine
Chymotrypsin
Aspartic acid
Serine
Histidine
Aspartic acid
Chymotrypsin
© Oxford University Press, 2013
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Mechanism for chymotrypsin
:O:
C
NH protein
Ser
C
C
H
Ser
His
Asp
Ser
C
:O :
O
..
His
O
O
N H
Ser
Asp
His
Asp
O
:O :
Ser
N H
His
protein
C
O
O
Asp
OH
O
H
O
:N
:O :
C
:N
H
O
C
..
: O : ..
OH
protein
..
H
H
O
N H
N
:O :
O
.. H
:O:
protein
protein
H
O
Asp
:
:O
O
NH protein
protein
N H
:N
O
:
..
:O :
NH protein
Ser
His
..
:O :
protein
N H
:N
H
:
..
:O
:
protein
N
N H
His
O
Asp
O
..
: OH
Ser
:N
N H
His
O
O
Asp
© Oxford University Press, 2013
11
Overall Process of Enzyme Catalysis
S
P
S
EE
E
catch
E+S
P
E
react
ES
E
release
EP
E+P
1 Binding interactions must be strong enough to hold the substrate
1.
sufficiently long for the reaction to occur
2. Interactions must be weak enough to allow the product to depart
3. Interactions stabilize the transition state,, loweringg Ea
4. Designing molecules with stronger binding interactions results in
enzyme inhibitors which block the active site
12
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Regulation of Enzymes
1.. Many
a y eenzymes
y es aaree regulated
egu a ed by age
agentss w
within thee cell
ce
2. Regulation may enhance or inhibit the enzyme
3. The products of some enzyme-catalysed reactions may act as inhibitors
4. Often they bind to a binding site called an allosteric binding site
NH2
Example
N
N
O
O
P
O
O
O
H
H
OH
H
OH
H
O
AMP
= negative feedback reduces
enzyme activity
HO
HO
H
H
H
= stimulates enzyme activity
H O
H O
HO
N
H OH
H OH
Glycogen
N
OH
OH
Phosphorylase a
H
H
H
OH
O
Glucose-1-phosphate
O
P
HO
OH
© Oxford University Press, 2013
n
13
Regulation of Enzymes
Active
A
ti site
it
unrecognisable
Active site
ACTIVE SITE
(open)
Enzyme
ENZYME
Allosteric
bindingg site
Induced
fit
(open)
Enzyme
ENZYME
Allosteric
inhibitor
1. Inhibitor binds reversibly to an allosteric binding site (molecule near end of
pathway
2 Intermolecular bonds are formed (the usual kinds)
2.
3. Induced fit of allosteric inhibitor alters the shape of the enzyme
4. Active site is distorted and is not recognised by the substrate (catalysis slows
or stops)
5. Increasing substrate concentration does not reverse inhibition
6. Inhibitor differs in structure to the substrate (different enzyme location)
© Oxford University Press, 2013
14
Regulation of Enzymes
Biosynthetic pathway
S
P
P’
P’’
P’’’
(open)
Enzyme
ENZYME
Inhibition
Feedback control
Enzymes with
E
i h allosteric
ll
i sites
i are often
f at the
h start off a biosynthetic
bi
h i
pathway. The enzyme is controlled by the final product of the pathway.
The final product binds to the allosteric site and switches off the enzyme
as it builds up in concentration
concentration.
15
© Oxford University Press, 2013
Regulation of Enzymes
1 External
1.
E t
l signals
i l can regulate
l t the
th activity
ti it off enzymes
(e.g. neurotransmitters or hormones)
2. Chemical messenger initiates a signal cascade which activates
enzymes called protein kinases
3. Protein kinases phosphorylate target enzymes to affect activity
Example
Phosphorylase b
(inactive)
Protein
kinase
Signal
cascade
p y
a
Phosphorylase
(active)
Gl
Glycogen
Gl
Glucose-1-phosphate
1 h h t
inside
cell
Adrenaline,
outside cell,
reacts with
different types of
cells in different
ways.
16
© Oxford University Press, 2013
Enzyme kinetics can be used to study factors important to enzyme behavior.
Michaelis-Menton derivation
k1
E
+
k2
S
ES
E
k-1
dP
dt
= o = k2 [ES]
+
P
rate of product formation
assume: k-1 >> k2
assume: [S] >> [E] so [ES]  constant = steady state assumption
d[ES] = 0 = k [E][S] - k [ES] - k [ES]
1
-1
2
dt
k1[E][S] = k-1[ES] + k2 [ES] = (k-1 + k2)[ES]
Etotal = [ET] = [E] + [ES]
(rearranged)
[E] = [ET] - [ES]
[ES] = [ET] - [E]
k1([ET] - [ES]) ([S]) = (k-1 + k2) [ES]
substitution: [E] = [ET] - [ES]
k1 [ET] [S] - k1 [ES] [S] = (k-1 + k2) [ES]
k1 [ET] [S] = k1 [ES] [S] + (k-1 + k2) [ES]
1
k1
x k1 [ET] [S] = k1 [ES] [S] + (k-1 + k2) [ES]
( k1 [ET] [S] = k1 [ES] [S] + (k-1 + k2) [ES])
(k1)
(k1)
(continued on next slide)
17
© Oxford University Press, 2013
[ET] [S] =
[ES] =
(k1[S] + k-1 + k2)
k1
k1
(k1[S] + k-1 + k2)
1
[ES] =
dP
dt
[S] + KM
= o = k2 [ES]
defined KM =
(k-1 + k2)
k1
[ES]
[ET] [S]
(rearranged)
(1/k1)
k1
(1/k1)
(k1[S] + k-1 + k2)
algebra (substitution)
[ET] [S]
= k2
[ET] [S]
[S] + KM
=
Vmax[S]
[S] + KM
1
=
([S]) +
=
(k-1 + k2)
k1
1
[S] + KM
Michaelis-Menton Eq. Vmax = k2 [ET]
18
© Oxford University Press, 2013
dP
dt
k1 [E ][S]
T
= k2
[S] + KM
=
Vmax[S]
Michaelis-Menton Eq.
[S] + KM
Vmax = k2[ET]
Vmax
The smaller the KM,
the tighter the binding.
when KM = [S]
(1/2) Vmax
KM
Vmax[S]
dP =
dt
Vmax[S]
=
[S] + KM
[S] + [S]
=
defined
(k-1 + k2)
=
k1
KM 
1
2 Vmax
unbound
bound
KM
1
Another way to plot the data (inverse).
initial
rate
y=
=
Vmax[S]
dP
dt
[S] + KM
Vmax[S]
=
KM
+
Vmax[S]
[S]
Vmax[S]
[S] + KM
1
Lineweaver-Burk Plots
1
=
1

=

y
=
KM
Vmax
1
[S]
m (x)
+
1
Vmax
+
b
y intercept
x intercept
_
1
Vmax
1
KM
m = slope =
1
x=
[S]
KM
Vmax
x
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