enzyme kinetics i 10/15/2009. enzyme kinetics rates of enzyme reactions thermodynamics says i know...
DESCRIPTION
Enzyme kinetics are important 1. Substrate binding constants can be measured as well as inhibitor strengths and maximum catalytic rates. 2. Kinetics alone will not give a chemical mechanism but combined with chemical and structural data mechanisms can be elucidated. 3. Kinetics help understand the enzymes role in metabolic pathways. 4. Under “proper” conditions rates are proportional to enzyme concentrations and these can be determine “ metabolic problems”.TRANSCRIPT
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Enzyme Kinetics I
10/15/2009
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Enzyme Kinetics
Rates of Enzyme Reactions
Thermodynamics says I know the difference between state 1 and state 2 and G = (Gf - Gi)
But Changes in reaction rates in response to differing
conditions is related to path followed by the reaction and
is indicative of the reaction mechanism!!
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Enzyme kinetics are important
1. Substrate binding constants can be measured as well as inhibitor strengths and maximum catalytic rates.
2. Kinetics alone will not give a chemical mechanism but combined with chemical and structural data mechanisms can be elucidated.
3. Kinetics help understand the enzymes role in metabolic pathways.
4. Under “proper” conditions rates are proportional to enzyme concentrations and these can be determine “ metabolic problems”.
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Chemical kinetics and Elementary Reactions
Rate Equations
Consider aA + bB + • • • + zZ. The rate of a reaction is proportional to the frequency with which the reacting molecules simultaneously bump into each other
zba ZBAk Rate
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The order of a reaction = the sum of exponentsGenerally, the order means how many molecules have to bump into each other at one time for a reaction to occur.
A first order reaction one molecule changes to another
A BA second order reaction two molecules react
A + B P + Q
or
2A P
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3rd order rates A + B + C P + Q + R rarely occur
and higher orders are unknown.
Let us look at a first order rate
A B
dtPd
dtAd v
= velocity of the reaction
in Molar per min.
or
moles per min per volume
k = the rate constant of the reaction
AdtAd kv
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Instantaneous rate: the rate of reaction at any specified time point that is the definition of the derivative.
We can predict the shape of the curve if we know the order of the reaction.
A second order reaction: 2A P
2AA kdtdv
Or for A + B P + Q
BABA kdtd
dtdv
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Percent change in A (ratio ) versus time in first and second order reactions
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It is difficult to determine if the reaction is either first or second order by directly plotting changes in
concentration.
AdtAd k
dtd kAA
t
0
A
A
dtk-AA
o
d to kAlnAln
-kto eA A
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However, the natural log of the concentration is directly proportional to the time.
- for a first order reaction-
The rate constant for the first order reaction has units of s-1 or min-1 since velocity = molar/sec
and v = k[A] : k = v/[A]
Gather your data and plot ln[A] vs time.
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The half-life of a first order reaction
2
A A oPlugging in
to rate equation
21
o
A2
A
ln kt
kk693.02lnt
21
The half-life of a first order reaction is the time for half of the reactant which is initially present to decompose or react.32P, a common radioactive isotope, emits an energetic particle and has a half-life of 14 days. 14C has a half life of 5715 years.
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A second order reaction such like 2A P
t
dtk0
A
oA2o
AAd-
ktoA
1A1
When the reciprocal of the concentration is plotted verses time a second order reaction is characteristic of a straight line.
The half-life of a second order reaction is
and shows a dependents on the initial concentration o21 A
1tk
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Kinetics of EnzymesEnzymes follow zero order kinetics when substrate
concentrations are high. Zero order means there is no increase in the rate of the reaction when more substrate is added.
Given the following breakdown of sucrose to glucose and fructose
Sucrose + H20 Glucose + Fructose
O
H
HO
H
HO
H
OH
OHHH
OH
OH
HOH
H OH
O
HH
HO
H
H
H
OH
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P EES S E 21
1-
kk
k
E = Enzyme S = Substrate P = Product
ES = Enzyme-Substrate complex
k1 rate constant for the forward reaction
k-1 = rate constant for the breakdown of the ES to substrate
k2 = rate constant for the formation of the products
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When the substrate concentration becomes large enough to force the equilibrium to form completely all ES the second step in the reaction becomes rate limiting because no more ES can be made and the enzyme-substrate complex is at its maximum value.
ESP2kdt
dv [ES] is the difference between the rates of ES formation minus the rates of its disappearance.
ESESSEES211 kkk
dtd
1
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Assumption of equilibriumk-1>>k2 the formation of product is so much
slower than the formation of the ES complex. That we can assume:
ES
SE
1
1
kkK s
Ks is the dissociation constant for the ES complex.
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Assumption of steady stateTransient phase where in the course of a reaction the concentration of ES does not change
0ES
dtd
2
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ES E E T 3
Combining 1 + 2 + 3
ESk k SES-Ek 21-T1
SEk Sk k kES T1121-
S KSE ES T
M1
21-
kk k K
M
rearranging
Divide by k1 and solve for [ES] Where
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SKSEESP T2
20
Mto
kkdtdv
vo is the initial velocity when the reaction is just starting out.
And is the maximum velocity T2max Ek V
SKSVmax
Mov
The Michaelis - Menten equation
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The Km is the substrate concentration where vo equals one-half Vmax
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The KM widely varies among different enzymes
The KM
can be expressed as:1
2
1
2
1
1 KKkk
kk
kk
sM
As Ks decreases, the affinity for the substrate increases. The KM can be a measure for substrate affinity if k2<k-1
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There are a wide range of KM, Vmax , and efficiency seen in enzymes
But how do we analyze kinetic data?
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The double reciprocal plot
maxmax
M
V1
S1
VK1
ov
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Lineweaver-Burk plot: slope = KM/Vmax,
1/vo intercept is equal to 1/Vmax
the extrapolated x intercept is equal to -1/KM
For small errors in at low [S] leads to large errors in 1/vo
Tmax
EV
catkkcat is how many reactions an enzyme can catalyze per second
The turnover number
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For Michaelis -Menton kinetics k2= kcat
When [S] << KM very little ES is formed and [E] = [E]T
and SEKkSE
Kk
M
catT
M
2 ov
Kcat/KM is a measure of catalytic efficiency
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What is catalytic perfection?
When k2>>k-1 or the ratio 21
21
kkkk
is maximum
Then1
MKkkcat
Or when every substrate that hits the enzyme causes a reaction totake place. This is catalytic perfection.
Diffusion-controlled limit- diffusion rate of a substrate is in the range of 108 to 109 M-1s-1. An enzyme lowers the transition state so there is no activation energy and the catalyzed rate is as fast as molecules collide.
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Lecture 16 – Dr. LeggeThursday 10/15/09Enzyme Kinetics II