Transcript Document

Biochemistry 412
Enzyme Kinetics I
March 29th, 2005
Reading:
Mathews & van Holde, Biochemistry, Benjamin/Cummings
Publishing Co., Redwood City, CA, pp. 341-364 in 1990
edition (or equivalent pages in a later edition)
Other (optional) resources:
http://web.mit.edu/esgbio/www/eb/ebdir.html
http://web.indstate.edu/thcme/mwking/enzyme-kinetics.html
>>> And special thanks for this lecture goes to Dr. Gabriel
Fenteany, Department of Chemistry, University of Illinois at
Chicago (www.chem.uic.edu/fenteany/teaching/452), whose slides I
liberally borrowed!
Enzymes Are Uniquely Powerful Catalysts
• Enzymes are proteins that can accelerate
biochemical reactions often by factors of
106 to 1012! This is much higher than
chemical catalysts.
• Enzymes can be extremely specific in terms
of reaction substrates and products.
• Enzymes catalyze reactions under mild
conditions (e.g. pH 7.4, 37ºC).
• The catalytic activities of many enzymes
can be regulated by allosteric effectors.
For example:
QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture.
Triose Phosphate Isomerase
And…
Chemical Kinetics
Irreversible First-Order Reactions
k
AB
v = d[B]/dt = -d[A]/dt = k[A]
(k = first-order rate constant (s-1))
Rearrange:
d[A]/[A] = dln[A] = -kdt
Integrate and express [A] as a
function of time (t):
[A]/[A]o = e –kt
or
[A]= [A]o e –kt
([A]o = initial concentration)
Reversible First-Order Reactions
k1
A 
 B
k-1
v = -d[A]/dt = k1[A] - k-1[B]
At equilibrium: k1[A]eq - k-1[B]eq = 0
[B]eq/[A]eq = k1/k-1 = Keq
Second-Order Reactions
k
2A  P
v = -d[A]/dt = k[A]2
kP
A+B
v = -d[A]/dt = -d[B]/dt = k[A][B]
(k = second-order rate constant (M-1s-1))
Change in [A] with time:
1/[A] = 1/[A]o + kt
Note: third-order
reactions rare, fourthand higher-order
reactions unknown.
Free Energy Diagrams
K = e –∆Gº/RT
For A  A‡
[A]‡/[A]o = e –∆Gº‡/RT
[A]‡ = [A]o e –∆Gº‡/RT
K = equilibrium constant
‡ = transition state
[A]‡ = concentration of molecules having the activation energy
[A]o = total concentration
–∆Gº‡ = standard free energy change of activation (activation energy)
Relationship of Reaction Rate Constant to
Activation Energy and Temperature:
The Arrhenius Equation
Reaction rate constant (k) determined by
activation energy (∆Gº‡) and temperature (T)
and proportional to frequency of forming
product (Q = kBT/h, where kB = Boltzmann’s
constant, h = Planck’s constant):
k = Q e -G°‡/RT =
(kBT/h) e -G°‡/RT
(G = H - T S)
k = Q e S°‡/R  e - H°‡/RT
k = Q´e -H°‡/RT
(Q´ = Q e -S°‡/R)
ln k = ln Q´ - H°‡/RT
ln k
L-malate 
fumarate + H20
The Transition State Energy Barrier Opposes the
Reaction in Both Directions
K = k1/k-1
K = (Q e
K=e
-G1°‡/RT)/(Q
e -G °‡/RT)
-1
-(G1°‡ - G-1°‡)/RT
∆Gº = G1º‡ - G-1º‡
K = e –∆Gº/RT
Equilibrium constant K says nothing about rate
of reaction, only free energy difference between
final and initial states.
Effect of a Catalyst on Activation Energy
•Catalysts do not affect GA (initial) or GB (final) and so do not affect overall
free energy change (∆Gº = GB - GA) or equilibrium constant K.
•Equilibrium concentrations of A and B still determined solely by overall free
energy change.
•Catalysts only affect ∆Gº‡, lowering the activation energy.
•They accelerate both the forward and reverse reaction (increase kinetic rate
constants k1 and k-1).
Intermediate States in Multi-step Reactions
Q: How do enzymes work?
A: A number of ways: “propinquity”, catalytic
groups at active site, catalytic metals at
active site, etc. (see assigned reading).
>>> however, primitive enzymes may have
behaved like catalytic antibodies, which can
accelerate reactions merely by binding to and
increasing the relative concentration of the
transition state ([A]‡ = [A]o e –∆Gº‡/RT effect).
Enzyme Kinetics
Types of Enzymes
1.
2.
3.
4.
5.
6.
Oxidoreductases catalyze oxidation-reduction reactions.
Transferases catalyze transfer of functional groups from one
molecule to another.
Hydrolases catalyze hydrolytic cleavage.
Lyases catalyze removal of a group from or addition of a group to a
double bond, or other cleavages involving electron rearrangement.
Isomerases catalyze intramolecular rearrangement.
Ligases catalyze reactions in which two molecules are joined.
Two Models for
Enzyme-Substrate Interaction
Induced Conformational Change in
Hexokinase
Triose Phosphate Isomerase
QuickTime™ and a TIFF ( Uncompr essed) decompressor are needed to see this pictur e.
Mutational analysis to study
enzyme mechanism:
•Site-directed mutagenesis substitution mutation at specific
position in sequence
•Deletion mutation
Free Energy Barrier to the Glyceraldehyde-3-Phosphate
<-> Dihydroxyacetone Phosphate Reaction
Mutant in flexible
loop that closes
over active site.
OK, but now let’s talk about kinetics….
The Effect of Substrate Concentration on
Reaction Velocity
Q: for a fixed amount of enzyme, what happens if
you keep adding more and more substrate?
The Steady State in Enzyme Kinetics
Michaelis-Menten Kinetics (1)
v = k2[ES], if this is the rate-limiting step*
E = free enzyme, S = substrate
ES = enzyme-substrate complex
P = product
[Enzyme]total = [E]t = [E] + [ES]
How to solve for [ES]?
1. Assume equilibrium, if k-1 >> k2:
KS = k-1/k1 = [E][S]/[ES]
or
2. Assume steady state:
(Briggs and
d[ES]/dt = 0
(Michaelis and
Menten, 1913)
Haldane, 1925)
[*Note: v is always measured as an initial rate!]
Michaelis-Menten Kinetics Continued (2)
Because of steady state assumption:
d[ES]/dt = k-1[ES] + k2[ES] - k1[E][S] = 0
So: k1[E][S] = k-1[ES] + k2[ES]
Rearranging: [ES] = (k1/(k-1 + k2))[E][S]
Substituting (the “M” constant* = KM = (k-1 + k2)/k1):
[ES] = ([E][S])/KM
So: KM[ES] = [E][S]
*Note: Briggs & Haldane came up with this,
but they lost out when it came time to name things!
Michaelis-Menten Kinetics Continued (3)
Substituting ([E] = [E]t - [ES]):
KM[ES] = [E]t[S] - [ES][S]
Rearranging: [ES](KM + [S]) = [E]t[S]
So: [ES] = [E]t[S]/(KM + [S])
Now we can substitute for [ES] in the rate equation
v = k2[ES], so…
The Michaelis-Menten Equation (4)
v = k2[E]t[S]/(KM + [S])
or
v = Vmax[S]/(KM + [S])
(since Vmax = k2[E]t)
At [S] « Km, Vo is proportional to [S]
At [S] » Km, Vo = Vmax
A Lineweaver-Burk Plot
An Eadie-Hofstee Plot
Multi-step Reactions
k1
k2
k3
E+S 
 ES  EP  E + P
k-1
v = kcat[E]t[S]/(KM + [S])
(kcat = general rate constant that
incorporates k2 and k3)
kcat, KM, and kcat/Km: Catalytic Efficiency
=> “Perfect enzyme”
Diffusion-controlled
limit: 108-109 M-1s-1
Substrate
preferences for
chymotrypsin