Download - Lecture 12 - Enzyme Kinetics I
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8/18/2019 Lecture 12 - Enzyme Kinetics I
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Enzyme Kinetics I
Maud Menten Leonor Michaelis
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Enzyme Kinetics
What is it?
Enzymes are molecular machines – as with any machine or mechanistic
process in nature, it is worthwhile to understand:
the individual steps taken by the machine in its process
the rate at which the machine can operate
the environment in which the machine best operates
+ +
Enzyme(E)
Enzyme(E)
Enzyme-Substrate
Complex(ES)
Substrate(S)
Product(P)
k1
k-1
k2
k-2
+ +
Enzyme(E)
Enzyme(E)
Enzyme-Substrate
Complex(ES)
Substrate(S)
Product(P)
k1
k-1
k2
k-2
Enzyme kinetics is a quantitative approach to understanding the functions of and
mechanisms employed by enzymes.
It is based upon determining the rates at which enzymes catalyze reactions
and the environmental factors that influences these rates.
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Why analyze enzyme kinetics?
1. The kinetic analysis of enzyme behavior lends insight into the molecular
mechanisms used by enzymes to convert substrate to product.
2. Kinetic analyses provide both the experimental means and strategy
towards understanding readily measurable quantities that can be used to
describe enzyme behaviour. Quantities derived through enzyme kinetics are
the universal language by which enzymes are compared by biochemists.
3. Enzymes are used in many industrial processes or constitute a componentof useful products. Useful enzymes are identified by their kinetic properties
e.g. their affinity for substrate and the efficiency with which they carry out
reactions.
4. In medicine: many chemicals living organisms may ingest either astherapeutic drugs or as toxins mediate their effects by binding to enzymes
and altering the rate at which they catalyze reactions.
5. The affinity of an enzyme for substrate and the rate at which it converts
substrate to product is a consequence of its evolved role in the metabolicpathways that define life.
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The kinetic analysis of enzymes preceded knowledge of their molecular
nature
pre 1900
Pasteur posited that the special activities of living cells that could mediatechemical changes were a special property of life, inseparable from life.
The concept of Vitalism.
around 1900
Eduard Buchner discovered that these chemical activities were the
properties of molecules that could function separate from living
organisms.
The term “Enzymes” was coined and over the next few decades it was
established that these molecules were proteins.
post – 1900
the emergence of theories that described the action of enzymes and their
behavior
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The rate of an enzyme catalyzed reaction (v) is dependent on substrate
concentration [S]
r e a c
t i o n r
a t e , v
substrate concentration, [S]
r e a c
t i o n r
a t e , v
substrate concentration, [S]
Led Victor Henri in 1903 to propose that reaction proceed via an
interaction between enzyme (E) and substrate (S).
ES complex
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r e a c
t i o n r
a t e , v
substrate concentration, [S]
r e a c
t i o n r
a t e , v
substrate concentration, [S]
in 1913, Leonor Michaelis and Maud Menten proposed a general theory of
enzyme activity:
ax
b + xy =
E + S ES E + Pk
-1
k 1
k -2
k 2
fast slow
E + S ES E + Pk
-1
k 1
k -1
k 1
k -2
k 2
k -2
k 2
fast slow
v =Vmax[S]
Km + [S]v =
Vmax
[S]
Km + [S]
Vmax is the maximum velocity
of the rate of reaction that occurs
when enzyme is saturated.
KM is a measure of the apparent
affinity of the enzyme for substrate
Vmax
KM
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Derivation of the M-M equation
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk -1
k 1
k -2
k 2
fast slow
E + S ES E + Pk -1
k 1
k -1
k 1
k -2
k 2
k -2
k 2
fast slow
reaction rates are always measured
as initial velocities, before there is
appreciable depletion of substrate
or formation of product
approximation
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk -1
k 1
k -2
k 2
fast slow
E + S ES E + Pk -1
k 1
k -1
k 1
k -2
k 2
k -2
k 2
fast slow
reaction rates are always measured
as initial velocities, before there is
appreciable depletion of substrate
or formation of product
approximation
-the overall rate of rx (v) is limited by the conversion of ES → E + P
-2 factors influence this conversion: rate constant (k2) and concentration of ES, [ES]
therefore, v = k2[ES] Equation 1
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E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
Two important assumptions are made during M-M analysis:1) substrate is in vast excess i.e. [S] >>[E]
2) steady-state assumption, that ES is forming and breaking down
at same rate.
k1[E][S] = (k-1 + k2)[ES]
collect terms on right side equation
solve for [ES]
k1[E][S] = k-1[ES] + k2[ES]
formation breakdown
k1[E][S] = k-1[ES] + k2[ES]
formation breakdown
[ES] =k1[E][S]
(k-1 + k2)
next page
error in text pg 231 line 15 shouldread breakdown not formation
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E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
[ES] =k1[E][S]
(k-1 + k2)
since all 3 rate constants are on same side of equation,
we can combine these into one term
k1
(k-1 + k2)=
1
KM
(k-1 + k2)
k1or KM =
k1
(k-1 + k2)=
1
KM
(k-1 + k2)
k1or KM = KM is called the Michaelis constant
[ES] =
[E][S]
KM
[E][S]
KMEquation 2[ES] =
[E][S]
KM
[E][S]
KMEquation 2
substitute KM into above equation
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E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
[ES] =[E][S]
KM
[E][S]
KMEquation 2[ES] =
[E][S]
KM
[E][S]
KMEquation 2
the total amount of enzyme, ET, is constant throughout experiment
(i.e. the catalyst is not consumed).
But it may be either free (not bound to substrate), E or bound
to substrate, ES.
therefore,
[ET] = [E] + [ES]
[E] = [ET
] - [ES]
we can solve for [E]
and substitute this definition of [E] into Equation 2, so
[ES] =([ET] – [ES])[S]
KM
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[ES] = ([ET]
–[ES])[S]
KM
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
[ES] =[ET][S] – [ES][S]
KM[ES] =
[ET][S] – [ES][S]
KM
[ET][S] = [ES]KM + [ES][S]
[ET][S] = (KM + [S])[ES]
[ES] =[ET][S]
KM + [S][ES] =
[ET][S]
KM + [S]
multiply through by [S]
solve for [ET][S]
collect terms
solve for [ES]
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[ES] =[ET][S]
KM + [S][ES] =
[ET][S]
KM + [S]
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
remember Equation 1: v = k2[ES]then
[ES] =v
k2[ES] =
v
k2
Now, the maximum rate (Vmax) occurs when all enzyme bound to substrate.That is, when [ET] = [ES]
Under this condition, Equation 1 becomes: Vmax = k2[ET]
v
k2
[ET][S]
KM + [S]=
vk2[ET][S]
KM + [S]=v
k2[ET][S]
KM + [S]=
substitute term for [ES]
solve for v
v =
Vmax[S]
KM + [S]
substitute in Vmax
the Michaelis-Menten equation
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r e a c t i o n r
a t e
, v
substrate concentration, [S]
r e a c t i o n r
a t e
, v
substrate concentration, [S]
ax
b + xy =
v =Vmax[S]
Km + [S]v =
Vmax[S]
Km + [S]
Vmax
KM
The Significance of Vmax, Km and other kinetic quantities
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Vmax
Vmax is the maximal velocity of the enzyme
In general the rate of a reaction is dependent on the rate constant k2 and the
substrate concentration. Therefore v = k2[ES]
but, when substrate concentration is so high that all of the enzyme is saturated
then the enzyme is operating as fast as it can and Vmax = k2[ET] where ET is
the total Enzyme in the reaction ([ES]=[ET] when the enzyme is saturated).
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk -1
k 1
k -2
k 2
fast slow
E + S ES E + Pk -1
k 1
k -1
k 1
k -2
k 2
k -2
k 2
fast slow
reaction rates are always measured
as initial velocities, before there is
appreciable depletion of substrate
or formation of product
approximation
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk -1
k 1
k -2
k 2
fast slow
E + S ES E + Pk -1
k 1
k -1
k 1
k -2
k 2
k -2
k 2
fast slow
reaction rates are always measured
as initial velocities, before there is
appreciable depletion of substrate
or formation of product
approximation
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r e a c t i o n r
a t e , v
substrate concentration, [S]KM
Vmax
½ Vmax
r e a c t i o n r
a t e , v
substrate concentration, [S]KM
Vmax
½ Vmax
v =Vmax[S]Km + [S]
v =Vmax[S]Km + [S]
KMKM is the substrate concentration at which the enzyme is rate is half its maximal
velocity (Vmax) or put another way:
KM is the substrate concentration where the enzyme is half-saturated.
On a Michaelis-Menten plot, KM equals the [S] at which the reaction rate
is half-maximal (1/2 Vmax)
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KM
E + S ES E + Pk
-1
k 1
k 2
E + S ES E + Pk
-1
k 1
k 2
E + S ES E + Pk
-1
k 1
k 2
E + S ES E + Pk
-1
k 1
k 2
Km is actually a combined rate constant that describes the formation and breakdown
of the ES complex
in most reactions, k2 is usually much lower than k1
therefore Km approximately reduces to:
v =Vmax[S]
KM + [S]
(k-1 + k2)
k1 KM =
(k-1 + k2)
k1 KM =
k1
(k-1 + k2)=
1
KM
k1
(k-1 + k2)=
1
KM
k-1
k1KM =
Note that when k1 > k-1, the formation of ES is favored and KM is small
when k-1 > k1, the dissociation if ES is favored and KM is large
that is, small KM indicates high affinity of enzyme for substrate
high KM indicates low affinity of enzyme for substrate
KM is a measure of the affinity of an enzyme for its substrate
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KcatE + S ES E + P
k -1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
kcat is equivalent to the rate limiting step, e.g. k2
therefore, k2 = kcat
and when all enzyme is bound to substrate,
Vmax = k2[ET]
therefore, k2 = = kcat Vmax
[ET]
kcat is also referred to as the turnover number (units of s-1
)
that is, the maximum number of substrate molecules per active site
converted to product per second
kcat is the turnover number
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also known as the Specificity constant
the parameters of KM or kcat alone are insufficient for comparing
the overall efficiency of an enzyme
a better measure is to calculate the ratio of kcat to KM
it is a measure of the overall rate of reaction where ultimately every
reaction relies on an encounter of Enzyme with Substrate
the upper limit of kcat/KM is defined by the rate at which E and S
can diffuse together in an aqueous solution
(about 108 to 109 M-1s-1)
some enzymes have specificity constants approaching this value
they are said to be “catalytically perfect” enzymes
Enzyme Substrate Kcat
(s-1)
KM
(M)
kcat /KM
(M-1s-1)
Catalase H2O2 4 x 107 1 x 100 4 x 107
-lactamase penicillin 2 x 103 2 x 10-5 1 x108
kcat
KM
kcat
KM
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Practical Determination of Kinetic Parameters in Laboratory
Hypothetical case:
You work for an industrial company that uses the enzyme Cellulase in the
formulation of laundry detergents.
It is your job to determine the kinetic attributes of 5 different Cellulase
enzymes to assess their suitability for use as an additive in laundry detergent.
Background
Cellulase is an enzyme that breaks down Cellulose into glucose molecules
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CO2 + H2O
glucosecellulose
cotton
O2
Energy
Cellulases are enzymes that bind
cellulose and chemically cleaves
it back to free glucose molecules
http://www.google.ca/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=LWsRCD4h98T0hM&tbnid=OF3weCUF0rk-mM:&ved=0CAUQjRw&url=http://science.howstuffworks.com/environmental/green-tech/energy-production/artificial-photosynthesis.htm&ei=txZKUsCoMofP2QW41oGIDQ&bvm=bv.53371865,d.b2I&psig=AFQjCNFIC1tONuAv9fhps7d45NE01NHU9A&ust=1380673566474132
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light
new shirt
surface
wear and tear after cellulase treatment
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Your job
Carry out kinetic analysis of 5 cellulase enzymes
Cellulase 1Cellulase 2
Cellulase 3
Cellulase 4
Cellulase 5
-determine the most active enzyme of the five.
M ll l ti it b i i f l t ti
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time
[ g l u c o s e ]
Measure cellulase enzyme activity by increase in free glucose concentration
The Progress curve of reaction at defined concentration of cellulose, [S].
y = m x + b
m=d[glucose]/dt
Enzyme + cellulose → Enzyme + glucose
The rate (v) is the increase
in glucose liberatedfrom cellulose over time
slope of the line = initial rate
or velocity (v) of the reaction
don’t confuse this with the
M-M plot, which has axes
of rate vs [S]
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Determination of reaction rates at varying substrate concentrations
etc.
[S]=4[S]=3
[S]=2
[S]=1
time
[ g l u c o s e ]
multiple progress curves
etc.
[S]=4[S]=3
[S]=2
[S]=1
time
[ g l u c o s e ]
multiple progress curves [S] v
1 0.001
2 0.002
3 0.004
4 0.007
5 0.010
6 0.013
Next, plot Rates vs [S]
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Plotting substrate concentration vs rate – The Michaelis-Menten curve
[substrate], [S]
rate, v
o
o
o
o
o
oo
o
o
o
o
oo o o
oo Vmax
½ Vmax
KM
v =Vmax[S]
KM + [S]
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So,
Having generated a Michaelis-Menten Plot specific for the Cellulase #1:
You now know the Vmax for the enzyme.
You can now determine KM for the enzyme since KM is the substrate concentration
at 1/2Vmax
Also, you can calculate Kcat since you know how much enzyme you used in the assay
and
Kcat =
Therefore you can also calculate the specificity constant since
Specificity Constant = Kcat/KM
Vmax
[ET]
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Enzyme kcat
(s-1)
KM
(M)
kcat /KM
(M-1s-1)
Cellulase 1 4 x 104 1 x 10-2 4 x 106
Cellulase 2 1 x 102 4 x 10-4 3 x 105
Cellulase 3 2 x 105 4 x 10-2 5 x 106
Cellulase 4 4 x 101 4 x 10-3 1 x 104
Cellulase 5 2 x 103
2 x 10-5
1 x108
Therefore, you would report that Cellulase #5 is the most active enzyme
Note that Cellulase #1 has a higher turnover number (kcat) than Cellulase #5
But Cellulase #5 has a much higher affinity for substrate (KM).
Overall, based on the specificity constant (kcat/KM) Cellulase #5 is the most active
enzyme.
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Understand the meanings of
Vmax
the relationship between Vmax and KM
the meaning of Km in terms of substrate affinity
the turnover number, Kcat
Specificity constant Kcat/KM
the general process for determining Vmax and KM in the lab
E + S ES E + Pk
-1
k 1
k 2
fast slow
E + S ES E + Pk
-1
k 1
k 2
fast slow
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Recommended Reading
Stryer 8th Edition
Ch 8 Enzymes: Basic Concepts and Kinetics Pgs 216-245.
Stryer 7th Edition
Ch 8 Enzymes: Basic Concepts and Kinetics Pgs 219-248.
If you would like to look at and discuss your exam:
Email me ([email protected]) with BIOL 2023 in subject line.
List 3 days/times you can meet and I will pick one that fits my schedule.
I cannot respond to emails that do not conform to above specs.