Lecture 10
prabeshrajjk
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Oct 26, 2014
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Lecture 10 Enzyme Kinetics
Rate constant (k) measures how rapidly a reaction occurs
A B + C
k
1
k
-1
Rate (v, velocity) = (rate constant) (concentration of reactants)
v= k
1
[A]
1
st
order reaction (rate dependent on concentration of 1 reactant)
v= k
-1
[B][C]
2
nd
order reaction (rate dependent on concentration of 2 reactants)
Zero order reaction (rate is independent of reactant concentration)
Rate constants and reaction order
A B + C
k
1
k
-1
Rate (v, velocity) = (rate constant) (concentration of reactants)
v= k
1
[A]
1
st
order reaction (rate dependent on concentration of 1 reactant)
v= k
-1
[B][C]
2
nd
order reaction (rate dependent on concentration of 2 reactants)
Zero order reaction (rate is independent of reactant concentration)
Rate constants and reaction order
Sample questions
The rate of a second order reaction depends on the concentration of
_________.
•(a) one substrate
•(b) two substrates
•(c) three substrates
•(d) none of the above
The rate of a second order reaction depends on the concentration of
_________.
•(a) one substrate
•(b) two substrates
•(c) three substrates
•(d) none of the above
E + S ES E + P
k
1
k
-1
k
2
k
-2
E
S+ E
S E+P
k
1
k
-1
k
2
k
-2
E
S+ E
S E+P
Initial Velocities
[S] = 1 mM
d[P]/dT = Vo
1 mM
[P]
time
Hold [E] constant
[E]<<<<<[S]
[S] = 1 mM
d[P]/dT = Vo
1 mM
[P]
time
Hold [E] constant
[E]<<<<<[S]
Initial Velocities
[S] = 1 mM
[S] = 5 mM
[S] = 10 mMd[P]/dT = Vo
10 mM
d[P]/dT = Vo
5 mM
d[P]/dT = Vo
1 mM
[P]
time
[S] = 1 mM
[S] = 5 mM
[S] = 10 mMd[P]/dT = Vo
10 mM
d[P]/dT = Vo
5 mM
d[P]/dT = Vo
1 mM
[P]
time
Plot Vo vs. [S]
Vo
1 mM
Vo
5 mM
Vo
10 mM
Vo
1 mM
Vo
5 mM
Vo
10 mM
1)Measurements made to measure initial velocity (v
o
). At
v
o
very little product formed. Therefore, the rate at
which E + P react to form ES is negligible and k
-2
is 0.
Therefore
Initial Velocity Assumption
E + S ES E + P
k
1
k
-1
k
2
E
S+ E
S
k
-2
E+P
o
). At
v
o
very little product formed. Therefore, the rate at
which E + P react to form ES is negligible and k
-2
is 0.
Therefore
Initial Velocity Assumption
E + S ES E + P
k
1
k
-1
k
2
E
S+ E
S
k
-2
E+P
Steady State Assumption
E + S ES E + P
k
1
k
-1
k
2
Steady state Assumption = [ES] is constant. The rate of ES
formation equals the rate of ES breakdown
E
S+ E
S E+P
E + S ES E + P
k
1
k
-1
k
2
Steady state Assumption = [ES] is constant. The rate of ES
formation equals the rate of ES breakdown
E
S+ E
S E+P
Data from a single experiment
performed with at a single [S].
(single point on Vo vs. [S] plot)
performed with at a single [S].
(single point on Vo vs. [S] plot)
E + S ES
k
1
E
S+ E
S
Rate of ES formation
Rate = k
1
[E] [S]
k
1
E
S+ E
S
Rate of ES formation
Rate = k
1
[E] [S]
ES E + P
k
2
E
S E+P
ES E + S
k
-1
E
S+E
S
Rate of ES breakdown
Rate = (k
2
[ES]) + (k
-1
[ES])
Rate = [ES](k
2
+ k
-1
)
k
2
E
S E+P
ES E + S
k
-1
E
S+E
S
Rate of ES breakdown
Rate = (k
2
[ES]) + (k
-1
[ES])
Rate = [ES](k
2
+ k
-1
)
If the rate of ES formation equals the rate of ES
breakdown
1) k
1
[E][S] = [ES](k
-1
+ k
2
)
2) (k
-1+ k
2) / k
1 =
[E][S] / [ES]
3) (k
-1
+ k
2
) / k
1
= K
m
(Michaelis constant)
breakdown
1) k
1
[E][S] = [ES](k
-1
+ k
2
)
2) (k
-1+ k
2) / k
1 =
[E][S] / [ES]
3) (k
-1
+ k
2
) / k
1
= K
m
(Michaelis constant)
Michaelis-Menton Derivation
1. The overall rate of product formation: v = k
2
[ES]
2. Rate of formation of [ES]:v
f = k
1[E][S]
3. Rate of decomposition of [ES]:
v
d
= k
-1
[ES] + k
2
[ES]
4. Rate of ES formation = Rate of ES decomposition
(steady state)
5. So: k
1
[E][S] = k
-1
[ES] + k
2
[ES]
E + S ES E + P
k
1 k
2
k
-1
Not required to know
1. The overall rate of product formation: v = k
2
[ES]
2. Rate of formation of [ES]:v
f = k
1[E][S]
3. Rate of decomposition of [ES]:
v
d
= k
-1
[ES] + k
2
[ES]
4. Rate of ES formation = Rate of ES decomposition
(steady state)
5. So: k
1
[E][S] = k
-1
[ES] + k
2
[ES]
E + S ES E + P
k
1 k
2
k
-1
Not required to know
Michaelis-Menton Derivation
6. In solving for [ES], use the enzyme balance to
eliminate [E]. E
T
= [E] + [ES]
7. k
1
(E
T
- [ES])[S] = k
-1
[ES] + k
2
[ES]
k
1
E
T
[S] - k
1
[ES][S] = k
-1
[ES] + k
2
[ES]
8. Rearrange and combine [ES] terms:
k
1 E
T[S] = (k
-1 + k
2 + k
1 [S])[ES]
k
1 E
T[S]
9. Solve for [ES] = -----------------------
(k
-1
+ k
2
+ k
1
[S])
Not required to know
6. In solving for [ES], use the enzyme balance to
eliminate [E]. E
T
= [E] + [ES]
7. k
1
(E
T
- [ES])[S] = k
-1
[ES] + k
2
[ES]
k
1
E
T
[S] - k
1
[ES][S] = k
-1
[ES] + k
2
[ES]
8. Rearrange and combine [ES] terms:
k
1 E
T[S] = (k
-1 + k
2 + k
1 [S])[ES]
k
1 E
T[S]
9. Solve for [ES] = -----------------------
(k
-1
+ k
2
+ k
1
[S])
Not required to know
Michaelis-Menton Derivation
E
T
[S]
10. Divide through by k
1
:
[ES] = -----------------------
(k
-1
+ k
2
)/k
1
+ [S]
11. Defined Michaelis constant: K
M
= (k
-1
+ k
2
) / k
1
12. Substitute K
M
into the equation in step 10.
13. Then substitute [ES] into v = k
2
[ES] from step1
and replace V
max
with k
2
E
T
to give:
V
max
[S]
v
o
= -----------
K
M
+ [S]
Not required to know
E
T
[S]
10. Divide through by k
1
:
[ES] = -----------------------
(k
-1
+ k
2
)/k
1
+ [S]
11. Defined Michaelis constant: K
M
= (k
-1
+ k
2
) / k
1
12. Substitute K
M
into the equation in step 10.
13. Then substitute [ES] into v = k
2
[ES] from step1
and replace V
max
with k
2
E
T
to give:
V
max
[S]
v
o
= -----------
K
M
+ [S]
Not required to know
Km = [S] at ½ V
max
(units moles/L=M)
(1/2 of enzyme bound to S)
V
max
= velocity where all of the
enzyme is bound to substrate
(enzyme is saturated with S)
max
(units moles/L=M)
(1/2 of enzyme bound to S)
V
max
= velocity where all of the
enzyme is bound to substrate
(enzyme is saturated with S)
Understanding V
max
The theoretical maximal velocity
•V
max
is a constant
•V
max
is the theoretical maximal rate of the
reaction - but it is NEVER achieved in reality
•To reach V
max
would require that ALL enzyme
molecules are tightly bound with substrate
•V
max
is asymptotically approached as
substrate is increased
max
The theoretical maximal velocity
•V
max
is a constant
•V
max
is the theoretical maximal rate of the
reaction - but it is NEVER achieved in reality
•To reach V
max
would require that ALL enzyme
molecules are tightly bound with substrate
•V
max
is asymptotically approached as
substrate is increased
What does K
m
mean?
1.K
m
= [S] at ½ V
max
2.K
m
is a constant; K
m
is a combination of rate
constants describing the formation and breakdown
of the ES complex
3.K
m
is usually a little higher than the physiological [S]
m
mean?
1.K
m
= [S] at ½ V
max
2.K
m
is a constant; K
m
is a combination of rate
constants describing the formation and breakdown
of the ES complex
3.K
m
is usually a little higher than the physiological [S]
What does Km mean?
4.K
m
represents the amount of substrate required to
bind ½ of the available enzyme (binding constant of
the enzyme for substrate)
5.K
m
can be used to evaluate the specificity of an
enzyme for a substrate
6.Small K
m
means tight binding; high K
m
means weak
binding
Glucose Km = 8 X 10
-6
Allose Km = 8 X 10
-3
Mannose Km = 5 X 10
-6
Hexose Kinase
Glucose + ATP <-> Glucose-6-P + ADP
4.K
m
represents the amount of substrate required to
bind ½ of the available enzyme (binding constant of
the enzyme for substrate)
5.K
m
can be used to evaluate the specificity of an
enzyme for a substrate
6.Small K
m
means tight binding; high K
m
means weak
binding
Glucose Km = 8 X 10
-6
Allose Km = 8 X 10
-3
Mannose Km = 5 X 10
-6
Hexose Kinase
Glucose + ATP <-> Glucose-6-P + ADP
Sample questions
•How does the Michaelis-Menten equation
explain why the rate of an enzyme-catalyzed
reaction reaches a maximum value at high
substrate?
•At high So, Km <<<< So (numerically), so the term Km +
So in the M-M equation becomes equal to So. Vo =
(Vmax So)/So, and So cancels. Therefore at high So
then, Vo = Vmax.
•How does the Michaelis-Menten equation
explain why the rate of an enzyme-catalyzed
reaction reaches a maximum value at high
substrate?
•At high So, Km <<<< So (numerically), so the term Km +
So in the M-M equation becomes equal to So. Vo =
(Vmax So)/So, and So cancels. Therefore at high So
then, Vo = Vmax.
The turnover number
A measure of catalytic activity
•k
cat
, the turnover number, is the number of
substrate molecules converted to product
per enzyme molecule per unit of time,
when E is saturated with substrate.
•If the Michaelis-Menten model fits, k
2 = k
cat
= V
max
/E
t
A measure of catalytic activity
•k
cat
, the turnover number, is the number of
substrate molecules converted to product
per enzyme molecule per unit of time,
when E is saturated with substrate.
•If the Michaelis-Menten model fits, k
2 = k
cat
= V
max
/E
t
What does k
cat
mean?
1.k
cat
is the 1
st
order rate constant describing
ES E+P
2.Also known as the turnover number because it
describes the number of reactions that a molecule of
enzyme can catalyze per second under optimal
condition.
3.Most enzyme have k
cat
values between 10
2
and 10
3
s
-1
4.For simple reactions k
2
= k
cat
, for multistep reactions
k
cat
= rate limiting step
E + S ES E + P
k
1
k
-1
k
cat
cat
mean?
1.k
cat
is the 1
st
order rate constant describing
ES E+P
2.Also known as the turnover number because it
describes the number of reactions that a molecule of
enzyme can catalyze per second under optimal
condition.
3.Most enzyme have k
cat
values between 10
2
and 10
3
s
-1
4.For simple reactions k
2
= k
cat
, for multistep reactions
k
cat
= rate limiting step
E + S ES E + P
k
1
k
-1
k
cat
What does k
cat
/K
m
mean?
•It measures how the enzyme
performs when S is low
•k
cat
/K
m
describes an enzymes
preference for different substrates =
specificity constant
•The upper limit for k
cat
/K
m
is the
diffusion limit - substrate diffuse into
the active site, or product diffuse out
•Catalytic perfection when k
cat
/K
m
=
diffusion rate
The catalytic efficiency
cat
/K
m
mean?
•It measures how the enzyme
performs when S is low
•k
cat
/K
m
describes an enzymes
preference for different substrates =
specificity constant
•The upper limit for k
cat
/K
m
is the
diffusion limit - substrate diffuse into
the active site, or product diffuse out
•Catalytic perfection when k
cat
/K
m
=
diffusion rate
The catalytic efficiency
k
cat
/K
M
k
cat/K
M is taken to be a measure of the efficiency
of an enzyme.
Rewriting k
cat
/K
M
in terms of the kinetic constants
gives:
k
cat
k
1
k
2
---- = -----------
K
M
k
-1
+ k
2
So, where k
2
is small, the denominator becomes
k
-1
and k
cat
/K
M
is small.
cat
/K
M
k
cat/K
M is taken to be a measure of the efficiency
of an enzyme.
Rewriting k
cat
/K
M
in terms of the kinetic constants
gives:
k
cat
k
1
k
2
---- = -----------
K
M
k
-1
+ k
2
So, where k
2
is small, the denominator becomes
k
-1
and k
cat
/K
M
is small.
Short summary
•Km substrate specificity; substrate binding
•k
cat
, the turnover number
•k
cat
/K
m
the catalytic efficiency
•Km substrate specificity; substrate binding
•k
cat
, the turnover number
•k
cat
/K
m
the catalytic efficiency
Sample questions
Which of the following kinetic parameters best describes how
well suited a specific compound functions as a substrate
for a particular enzyme?
•(a) Km
•(b) Vmax
•(c) kcat
•(d) kcat/Km
Which of the following kinetic parameters best describes how
well suited a specific compound functions as a substrate
for a particular enzyme?
•(a) Km
•(b) Vmax
•(c) kcat
•(d) kcat/Km
Sample questions
The rate-determining step of Michaelis Menten kinetics is
•A.the complex formation step
•B.the complex dissociation step to produce product
•C.the product formation step
•D.Both (a)and(c)
The rate-determining step of Michaelis Menten kinetics is
•A.the complex formation step
•B.the complex dissociation step to produce product
•C.the product formation step
•D.Both (a)and(c)
Limitations of Michaelis-Menten
model
1.Some enzyme catalyzed reactions show more complex behavior
E + S<->ES<->EZ<->EP<-> E + P
Michaelis-Menten can look only at rate limiting step
2.Often more than one substrate
E+S
1
<->ES
1
+S
2
<->ES
1
S
2
<->EP
1
P
2
<-> EP
2
+P
1
<-> E+P
2
Must
optimize one substrate then calculate kinetic parameters for the
other
3.Assumes k
-2
= 0
4.Assume steady state conditions
model
1.Some enzyme catalyzed reactions show more complex behavior
E + S<->ES<->EZ<->EP<-> E + P
Michaelis-Menten can look only at rate limiting step
2.Often more than one substrate
E+S
1
<->ES
1
+S
2
<->ES
1
S
2
<->EP
1
P
2
<-> EP
2
+P
1
<-> E+P
2
Must
optimize one substrate then calculate kinetic parameters for the
other
3.Assumes k
-2
= 0
4.Assume steady state conditions
The dual nature of the
Michaelis-Menten equation
Combination of 0-order and 1st-order kinetics
•When S is low, the equation for rate is 1st
order in S
•When S is high, the equation for rate is 0-
order in S
•The Michaelis-Menten equation describes a
hyperbolic dependence of v on S
Michaelis-Menten equation
Combination of 0-order and 1st-order kinetics
•When S is low, the equation for rate is 1st
order in S
•When S is high, the equation for rate is 0-
order in S
•The Michaelis-Menten equation describes a
hyperbolic dependence of v on S
How do you get values for V
max
, K
m
and k
cat
?
•Can determine K
m
and V
max
experimentally
•Km can be determined without an absolutely pure
enzyme
•K
cat
values can be determined if V
max
is known and
the absolute concentration of enzyme is known (V
max
= k
cat
[E
total
]
max
, K
m
and k
cat
?
•Can determine K
m
and V
max
experimentally
•Km can be determined without an absolutely pure
enzyme
•K
cat
values can be determined if V
max
is known and
the absolute concentration of enzyme is known (V
max
= k
cat
[E
total
]
B
B
B
B
B
B
B
0
0.05
0.1
0.15
0.2
0.25
012345678910
V
o
[S]
[S] Vo
0.5 0.075
0.75 0.09
2 0.152
4 0.196
6 0.21
8 0.214
10 0.23
V max
Km
Km ~ 1.3 mM
Vmax ~ 0.25
B
B
B
B
B
B
0
0.05
0.1
0.15
0.2
0.25
012345678910
V
o
[S]
[S] Vo
0.5 0.075
0.75 0.09
2 0.152
4 0.196
6 0.21
8 0.214
10 0.23
V max
Km
Km ~ 1.3 mM
Vmax ~ 0.25
Lineweaver-Burke Plots
(double reciprocal plots)
•Plot 1/[S] vs 1/V
o
•L-B equation for straight
line
•X-intercept = -1/Km
•Y-intercept = 1/Vmax
•Easier to extrapolate
values w/ straight line vs
hyperbolic curve
(double reciprocal plots)
•Plot 1/[S] vs 1/V
o
•L-B equation for straight
line
•X-intercept = -1/Km
•Y-intercept = 1/Vmax
•Easier to extrapolate
values w/ straight line vs
hyperbolic curve
Sample questions
•For an enzyme (5 μM) , the following
initial velocities have been reported
depending on the substrate
concentration:
•(a) Draw a Michaelis-Menten plot for
this enzyme.
•(b) Draw a Lineweaver-Burke plot for
this enzyme.
•(c) Determine Km and Vmax for this
enzyme
•(d) Indicate in both graphs (a & b)
where Vmax and Km can be
recognized.
•(e) Calculate the turnover number and
the catalytic efficiency for this enzyme.
[Substrate],
mM
v0, mM/s
0.02 10.83
0.04 18.57
0.07 26.76
0.1 32.50
0.15 39.00
0.2 43.33
0.3 48.75
0.5 54.17
0.7 56.88
kcat = Vmax / [E]totalcatalytic efficiency: kcat/Km
•For an enzyme (5 μM) , the following
initial velocities have been reported
depending on the substrate
concentration:
•(a) Draw a Michaelis-Menten plot for
this enzyme.
•(b) Draw a Lineweaver-Burke plot for
this enzyme.
•(c) Determine Km and Vmax for this
enzyme
•(d) Indicate in both graphs (a & b)
where Vmax and Km can be
recognized.
•(e) Calculate the turnover number and
the catalytic efficiency for this enzyme.
[Substrate],
mM
v0, mM/s
0.02 10.83
0.04 18.57
0.07 26.76
0.1 32.50
0.15 39.00
0.2 43.33
0.3 48.75
0.5 54.17
0.7 56.88
kcat = Vmax / [E]totalcatalytic efficiency: kcat/Km
Answer
•(a)
•(b)
•(c) Km and Vmax can be determined from the intercepts in the Lineweaver-
Burke plot:
1/Vmax = 0.015 s/mM Vmax = 66 mM/s
-1/Km = -10 mM Km = 0.1 mM
•(e) kcat = Vmax / [E]total = 65 mM/s ÷ 5 μM
= 65 mM/s ÷ 0.005 mM = 13000/s
catalytic efficiency: kcat/Km = 13000/s ÷ 0.1 mM
= 13000/s ÷ 0.0001 M
= 1.3×10
8
M/s
•(a)
•(b)
•(c) Km and Vmax can be determined from the intercepts in the Lineweaver-
Burke plot:
1/Vmax = 0.015 s/mM Vmax = 66 mM/s
-1/Km = -10 mM Km = 0.1 mM
•(e) kcat = Vmax / [E]total = 65 mM/s ÷ 5 μM
= 65 mM/s ÷ 0.005 mM = 13000/s
catalytic efficiency: kcat/Km = 13000/s ÷ 0.1 mM
= 13000/s ÷ 0.0001 M
= 1.3×10
8
M/s
Enzyme Inhibition
•Inhibitor – substance that binds to an
enzyme and interferes with its activity
•Can prevent formation of ES complex or
prevent ES breakdown to E + P.
Reversible versus Irreversible
•Reversible inhibitors interact with an
enzyme via noncovalent associations
•Irreversible inhibitors interact with an
enzyme via covalent associations
•Inhibitor – substance that binds to an
enzyme and interferes with its activity
•Can prevent formation of ES complex or
prevent ES breakdown to E + P.
Reversible versus Irreversible
•Reversible inhibitors interact with an
enzyme via noncovalent associations
•Irreversible inhibitors interact with an
enzyme via covalent associations
Reversible Inhibitors
E + S <-> ES -> E + P
E + I <-> EI
Ki = [E][I]/[EI]
•Competitive
•Uncompetitive
•Non-competitive
E + S <-> ES -> E + P
E + I <-> EI
Ki = [E][I]/[EI]
•Competitive
•Uncompetitive
•Non-competitive
Classes of Reversible Inhibition
Two real, one hypothetical
•Competitive inhibition - inhibitor (I) competes with the
substrate for the active site of the enzyme
•Non-competitive inhibition - inhibitor (I) binds to an enzyme
somewhere other than the active site. It can binds either ES
or E. A non-competitive inhibitor reacts with the enzyme-
substrate complex, and slows the rate of reaction to form the
enzyme-product complex.
•Uncompetitive inhibition - inhibitor (I) binds only to ES, not to
E. This is a hypothetical case that has never been
documented for a real enzyme, but which makes a useful
contrast to competitive inhibition
Two real, one hypothetical
•Competitive inhibition - inhibitor (I) competes with the
substrate for the active site of the enzyme
•Non-competitive inhibition - inhibitor (I) binds to an enzyme
somewhere other than the active site. It can binds either ES
or E. A non-competitive inhibitor reacts with the enzyme-
substrate complex, and slows the rate of reaction to form the
enzyme-product complex.
•Uncompetitive inhibition - inhibitor (I) binds only to ES, not to
E. This is a hypothetical case that has never been
documented for a real enzyme, but which makes a useful
contrast to competitive inhibition
•Competitive inhibitor: Vmax stays the same, but Km increases
•Non-competitive inhibitor decreases the turnover number of the
enzyme rather than preventing substrate binding- Vmax decreases
but Km stays the same. This cannot be overcome with an increase
in substrate concentration.
•Non-competitive inhibitor decreases the turnover number of the
enzyme rather than preventing substrate binding- Vmax decreases
but Km stays the same. This cannot be overcome with an increase
in substrate concentration.
Enzyme Inhibition
Noncovalent binding:
Competitive (I binds only to E)
Uncompetitive (I binds only to ES)
Noncompetitive (I binds to E or ES)
Covalent binding – irreversible
Group Specific
Substrate Analogs
(bound to the active site and prevent further reactions)
Noncovalent binding:
Competitive (I binds only to E)
Uncompetitive (I binds only to ES)
Noncompetitive (I binds to E or ES)
Covalent binding – irreversible
Group Specific
Substrate Analogs
(bound to the active site and prevent further reactions)
Competitive Inhibitor (CI)
•CI binds free enzyme
•Competes with substrate for enzyme binding.
•Raises Km without effecting Vmax
•Can relieve inhibition with more S
•CI binds free enzyme
•Competes with substrate for enzyme binding.
•Raises Km without effecting Vmax
•Can relieve inhibition with more S
Competitive Inhibitors look like substrate
NH
2
C
O
HO NH
2
S
O
H2N
O
PABA
Sulfanilamide
NH
2
C
O
HO NH
2
S
O
H2N
O
PABA
Sulfanilamide
Non-competitive Inhibitor (NI)
•NI can bind free E or ES complex
•Lowers Vmax, but Km remains the same
•NI’s don’t bind to S binding site therefore don’t effect Km
•Alters conformation of enzyme to effect catalysis but not
substrate binding
•NI can bind free E or ES complex
•Lowers Vmax, but Km remains the same
•NI’s don’t bind to S binding site therefore don’t effect Km
•Alters conformation of enzyme to effect catalysis but not
substrate binding
Uncompetitive Inhibitor (UI)
•UI binds ES complex
•Prevents ES from proceeding to E + P or back to E + S.
•Lowers Km & Vmax, but ratio of Km/Vmax remains the
same
•Occurs with multisubstrate enzymes
•UI binds ES complex
•Prevents ES from proceeding to E + P or back to E + S.
•Lowers Km & Vmax, but ratio of Km/Vmax remains the
same
•Occurs with multisubstrate enzymes
Sample questions
Which of the following binds to an enzyme at its active site?
•A) irreversible inhibitor
•B) reversible competitive inhibitor
•C) reversible noncompetitive inhibitor
•D) more than one correct response
•E) no correct response
An uncompetitive inhibitor binds to _____.
•(a) E
•(b) ES
•(c) P
•(d) a and b
•(e) a and c
Which of the following binds to an enzyme at its active site?
•A) irreversible inhibitor
•B) reversible competitive inhibitor
•C) reversible noncompetitive inhibitor
•D) more than one correct response
•E) no correct response
An uncompetitive inhibitor binds to _____.
•(a) E
•(b) ES
•(c) P
•(d) a and b
•(e) a and c
Sample questions
A reversible inhibitor that can bind to either E alone or the ES complex
is referred to as a _____.
•(a) competitive inhibitor.
•(b) non-competitive inhibitor.
•(c) uncompetitive inhibitor.
•(d) suicide inhibitor.
•(e) irreversible inhibitor.
A reversible inhibitor that can bind to either E alone or the ES complex
is referred to as a _____.
•(a) competitive inhibitor.
•(b) non-competitive inhibitor.
•(c) uncompetitive inhibitor.
•(d) suicide inhibitor.
•(e) irreversible inhibitor.
Sample questions
A competitive inhibitor of an enzyme is usually
•A.a highly reactive compound
•B.a metal ion such as Hg
2+
or Pb
2+
•C.structurally similar to the substrate.
•D.water insoluble
The enzyme inhibition can occur by
•A.reversible inhibitors
•B.irreversible inhibitors
•C.Both (a) and (b)
•D.None of these
A competitive inhibitor of an enzyme is usually
•A.a highly reactive compound
•B.a metal ion such as Hg
2+
or Pb
2+
•C.structurally similar to the substrate.
•D.water insoluble
The enzyme inhibition can occur by
•A.reversible inhibitors
•B.irreversible inhibitors
•C.Both (a) and (b)
•D.None of these
Sample questions
In a Lineweaver-Burk Plot, competitive
inhibitor shows which of the following
effect?
•A.It moves the entire curve to right
•B.It moves the entire curve to left
•C.It changes the x-intercept
•D.It has no effect on the slope
In a Lineweaver-Burk Plot, competitive
inhibitor shows which of the following
effect?
•A.It moves the entire curve to right
•B.It moves the entire curve to left
•C.It changes the x-intercept
•D.It has no effect on the slope
Sample questions
Non-competitive inhibitor of an enzyme catalyzed
reaction
•A.decreases V
max
•B.binds to ES
•C.both (a) and (b)
•D.can actually increase reaction velocity in rare
cases
Non-competitive inhibitor of an enzyme catalyzed
reaction
•A.decreases V
max
•B.binds to ES
•C.both (a) and (b)
•D.can actually increase reaction velocity in rare
cases
Sample questions
A classical uncompetitive inhibitor is a compound that binds
•A.reversibly to the enzyme substrate complex yielding an
inactive ESI complex
•B.irreversibly to the enzyme substrate complex yielding an
inactive ESI complex
•C.reversibly to the enzyme substrate complex yielding an
active ESI complex
•D.irreversibly to the enzyme substrate complex yielding an
active ESI complex
A classical uncompetitive inhibitor is a compound that binds
•A.reversibly to the enzyme substrate complex yielding an
inactive ESI complex
•B.irreversibly to the enzyme substrate complex yielding an
inactive ESI complex
•C.reversibly to the enzyme substrate complex yielding an
active ESI complex
•D.irreversibly to the enzyme substrate complex yielding an
active ESI complex
Kinetics of Multisubstrate
Reactions
E + A + B <-> E + P + Q
•Sequential Reactions
a)ordered
b)random
•Ping-pong Reactions
Reactions
E + A + B <-> E + P + Q
•Sequential Reactions
a)ordered
b)random
•Ping-pong Reactions
Sequential Reactions
E EA (EAB) (EPQ) EQ E
A B P Q
AB
PQ
A B
E
EA
EB
(EAB)(EPQ)
P Q
EQ
EP
E
Ordered
Random
E EA (EAB) (EPQ) EQ E
A B P Q
AB
PQ
A B
E
EA
EB
(EAB)(EPQ)
P Q
EQ
EP
E
Ordered
Random
Ordered Sequential
Ordered Random
Ping-Pong Reactions
E (EA)(FP) (F) (FB)(EQ) E
A BP Q
•In Ping-Pong reactions first product released
before second substrate binds
•When E binds A, E changes to F
•When F binds B, F changes back to E
E (EA)(FP) (F) (FB)(EQ) E
A BP Q
•In Ping-Pong reactions first product released
before second substrate binds
•When E binds A, E changes to F
•When F binds B, F changes back to E
Lineweaver-Burke Plot of
Multisubstrate Reactions
Increasing
[B]
Increasing
[B]
Sequential Ping-Pong
Vmax doesn’t change
Km changes
Both Vmax & Km change
1/Vo
1/[S]
1/Vo
1/[S]
Multisubstrate Reactions
Increasing
[B]
Increasing
[B]
Sequential Ping-Pong
Vmax doesn’t change
Km changes
Both Vmax & Km change
1/Vo
1/[S]
1/Vo
1/[S]
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