Download Lecture 10: Enzymes: Introduction

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Lecture 10: Enzymes: Introduction [PDF]
Reading: Berg, Tymoczko & Stryer, Chapter 8, pp. 205-217 (These pages in textbook are very important -- concepts of
thermodynamics are fundamental to all of biochemistry.)
Thermodynamics practice problems (same as for Lecture 2) [PDF]
Enzymes introduction practice problems [PDF]
Updated on: 1/30/07 at 4:45 pm
Key Concepts
Enzymes are biological catalysts, very powerful and very specific.
Enzymes increase RATES of (bio)chemical reactions but have NO EFFECT on Keq (and no effect on
overall ΔG) of the reaction.
Some enzymes need cofactors (inorganic ions or organic/metalloorganic coenzymes, derived from vitamins) for
their catalytic activities.
Different cofactors are useful for different kinds of chemical reactions, including transfers of specific
kinds of groups or transfers of electrons.
Kinetics is the study of reaction rates.
Rates depend on rate constants.
Rate constants depend inversely and exponentially on the Arrhenius ACTIVATION ENERGY, ΔG ‡, the
difference in free energy between the free energy of the transition state and the free energy of the
reactant(s).
Rate constants are increased by catalysts (enzymes), because enzymes decrease ΔG‡.
Enzymes lower ΔG ‡ by affecting either ΔH ‡ or ΔS ‡ (or both).
One way enzymes reduce ΔG ‡ is by tight binding (noncovalent) of the transition state.
Enzymes generally change the pathways by which reactions occur.
Rate enhancement (factor by which enzyme increases the rate of a reaction) is determined by ΔΔG‡, the
magnitude of the decrease in ΔG ‡ brought about by the enzyme compared with the uncatalyzed reaction's
ΔG‡.
Learning Objectives
Terminology: rate enhancement, cofactor, coenzyme, apoenzyme, holoenzyme, prosthetic group,
catalyst, activation energy, transition state. (Review: equilibrium constant, mass action ratio for a
reaction, biochemical standard conditions, standard free energy change, actual free energy change).
Describe the general properties of enzymes as catalysts that are especially important for their roles
as biological catalysts.
Explain the effect of a catalyst on the rate of a reaction, and on the equilibrium constant of a
reaction.
(Thermodynamics Review):
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Explain what thermodynamic parameter determines the direction in which a reaction will go,
and the "rules" about the sign of that parameter as it relates to reaction direction.
Define "standard free energy change" and give the symbol for that parameter.
Write the mathematical expression relating ΔG°' to K eq', and be able to interconvert ΔG°'
and Keq'.
Calculate the actual free energy change (ΔG'), given the starting concentrations of
appropriate chemical species and either ΔG°' or Keq'.
Using ΔG', predict reaction direction.
Describe the relation between ΔG' and the rate of a reaction.
Express the velocity of a simple reaction in terms of the rate constant and the concentration of the
reactant.
(review) Express the equilibrium constant of a reaction in terms of the equilibrium mass action
ratio. Now also express the equilibrium constant of a reaction in terms of the rate constants for the
forward and reverse directions. (Note that equilibrium constants are symbolized with upper case K
and rate constants with lower case k.)
If an enzyme increases the rate constant for the forward reaction by a factor of 108, by what factor
does it increase the rate constant for the back reaction? What is the rate enhancement brought
about by the catalyst for that reaction?
Draw the free energy diagram of a hypothetical reaction and show how a catalyst may increase the
rate of the reaction, pointing out on the diagram ΔG for the overall reaction, ΔG ‡uncat, and ΔG ‡cat .
Indicate (and name) the quantity on a free energy diagram (HINT: it's a specific kind of ΔG) that
determines the magnitude of the rate constant for the reaction at a given temperature. You don't
have to memorize the equation relating this quantity to k.
What reaction parameter (kinetic parameter) do enzymes affect in order to increase the rate?
ENZYMES
Enzymes are proteins, though there are catalytic RNA molecules called ribozymes.
Enzymes as BIOLOGICAL CATALYSTS:
are NOT themselves chemically ALTERED by the reaction
DO NOT CHANGE EQUILIBRIUM CONSTANT for reaction
INCREASE RATE of reaction by providing a pathway of LOWER ACTIVATION ENERGY to get from
reactants to products
operate under PHYSIOLOGICAL CONDITIONS (moderate temperatures, around neutral pH, low
concentrations in aqueous environment…)
work by FORMING COMPLEXES WITH THEIR SUBSTRATES (BINDING), thus providing unique
MICROENVIRONMENT for reaction to proceed (the ACTIVE SITE):
VERY HIGH SPECIFICITY for both reaction catalyzed and substrate used
VERY HIGH CATALYTIC EFFICIENCY
ACTIVITIES of some enzymes REGULATED
CATALYTIC POWER AND SPECIFICITY OF ENZYMES
RATE ENHANCEMENT = catalyzed rate constant/uncatalyzed rate constant = factor by which catalyst
increases rate of reaction
Examples - Berg, Tymoczko & Stryer, 6th ed. Table 8-1:
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SPECIFICITY
for substrate acted upon AND
for reaction catalyzed
Example: Proteases
catalyze hydrolysis of peptide bonds
Substrate specificity:
Berg, Tymoczko & Stryer, 6th ed. Fig. 8-1: Substrate specificity of enzymes (e.g., proteases) due to
precise interaction of enzyme with substrate (result of 3-D structure of enzyme).
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(A) Trypsin catalyzes
hydrolysis of peptide bonds on
carboxyl side of Lys & Arg
residues.
(B) Thrombin (involved in
blood clotting cascade) catalyzes
hydrolysis of peptide bonds
between Arg and Gly residues
in specific sequences in specific
protein substrates.
substrate specificity of chymotrypsin
cleaves on carboxyl side of aromatic & hydrophobic amino acid residues
evolutionarily related to trypsin
Genes for trypsin and chymotrypsin are homologous.
Ancestral gene duplicated and sequences diverged, but catalytic mechanism and overall
tertiary structure was conserved.
Specificity of reaction catalyzed:
Many proteases also catalyze hydrolysis of carboxylic ESTER bonds:
Use of COFACTORS by some enzymes
COFACTORS = small organic or metalloorganic molecules (coenzymes) or metal ions
Cofactors can bind tightly or weakly to enzymes. (Equilibrium below can lie far to left, weak
binding, or far to right, tight binding).
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Prosthetic groups (e.g. heme in hemoglobin): Tightly bound cofactors (either coenzymes or
metals)
-- remain associated with their enzymes even between reaction cycles.
Weakly bound coenzymes (which are NOT prosthetic groups)
can associate and dissociate from enzymes between reaction cycles, behaving like substrates
sometimes referred to as "cosubstrates"
Common Coenzymes and Reactions They Mediate (See also Berg, Tymoczko & Stryer Table 8.2)
(for reference, not for memorization)
Coenzyme (precursor/vitamin)
Reaction Mediated (Group Transferred)
Biotin
Carboxylation (CO2)
Cobalamin (B12)
Alkylation (methyl group), intramolecular rearrangements, and
ribonucleotide reduction
Coenzyme A (pantothenate)
Acyl transfer (R–C=O group)
Flavin coenzymes (B2)
Oxidation-reduction (hydrogen atoms) (1 or 2 e- transfer)
Lipoic acid
Acyl group transfer
Nicotinamide coenzymes (niacin)
Oxidation reduction (hydride ions H:-, 2 e- transfers)
Pyridoxal phosphate (B6)
Amino group transfer (and many other reactions)
Tetrahydrofolate (folic acid)
One-carbon transfer
Thiamine pyrophosphate (B1)
Aldehyde transfer
Uridine diphosphate [UDP]
Sugar transfer (hexose units)
Review of Biological Thermodynamics (will not be covered again in class; go back over notes for
Lecture 2, and Berg, Tymoczko & Stryer 6th ed. Chapter 1, pp. 11-12 and Chapter 8, pp. 208-211)
Many biological reactions involve energy transformations -- examples:
conversion of light energy to chemical bond energy (photosynthesis, with
"mediation" of ion gradient)
conversion of chemical energy in nutrients to another form of chemical bond
energy in ATP (with mediation of metabolic conversions including electron transfers
that drive formation and maintenance of a proton concentration gradient that "powers"
ATP synthesis)
conversion of bond energy in ATP to do cellular work
osmotic work (pumping ions and small molecules)
mechanical work (muscle contraction)
metabolic work (driving energy-requiring biosynthetic reactions)
etc.
change in free energy of any process = ΔG
ΔG = ΔH – TΔS
where ΔH = the change in enthalpy (heat content) of the system
ΔS = the change in entropy of the system that occurs during the process
T ( = absolute temperature, in K) and pressure are assumed not to change during a
biological process.
All reactions/processes proceed toward equilibrium.
The reaction direction required to go toward equilibrium is indicated by the sign of
the change in free energy for that reaction.
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RULES:
ΔG = Gproducts (final state) - Greactants (initial state)
If ΔG < 0, reaction proceeds left to right (as written) -- reaction is exergonic,
free energy decreases in going from reactants to products)
If ΔG = 0, reaction is at equilibrium (no net reaction occurs in either direction)
If ΔG > 0, reaction proceeds right to left (reverse direction from what's written)
(Forward reaction would be endergonic, requiring INPUT of free energy to go
left to right.)
ΔG forward reaction = - ΔG reverse reaction (Just change sign of ΔG for reversing
direction.)
To get a reaction for which ΔG > 0 (a reaction with a positive ΔG) to go
forward, couple it with a reaction for which ΔG < 0 (negative ΔG) to
"drive" the process. (Exergonic reaction provides free energy "input" to
drive endergonic reaction.)
NOTE: ΔG provides no information about the rate of a reaction, only about the
direction in which the reaction must go to achieve equilibrium. (Reaction rate depends on
free energy of activation -- see enzyme kinetics notes.)
Consider the reaction
Equilibrium constant Keq for direction written ( = the equilibrium mass action ratio) =
Direction for reaction to go toward equilibrium depends on 2 things:
1 . equilibrium constant Keq (mass action ratio at equilibrium)
and
2 . actual starting concentrations of A, B, C and D.
ΔG° = standard free energy change = the free energy change for going from standard
conditions to equilibrium.
lets us express equilibrium constant in free energy terms
STANDARD CONDITIONS (the starting conditions for standard free energy change):
1M in every reagent and every product (or 1 atm for gaseous reagents or products) (so
standard conditions mass action ratio = 1),
T = 298 K (that's 25°C), and
(biochemical standard conditions) pH 7.0 and 55.5 M H2O
pH ([H+] concentration) and [H2O] do not change in the reaction, so they're ignored as
reactants or products -- those conditions are accounted for in the biochemical standard free
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energy change (and thus in the biochemical K'eq).
Biochemical free energy changes are indicated with a prime symbol ('), and all ΔG values
discussed in this course will be ΔG' values, whether the "prime" is actually indicated or not.
Relation between standard free energy change (ΔG°') and equilibrium constant K'eq:
ΔG°' = – RTlnK'eq
so ΔG°' and K'eq can be interconverted -- if you know one, you know the other.
Note: minus sign in this equation -standard free energy change is inversely and exponentially related to equilibrium constant:
lnK eq = – ΔG°'/RT, so
K' eq = e– ΔG°'/RT
ACTUAL free energy change ΔG' for any reaction or process depends on 3 things:
1) standard free energy change for that reaction (ΔG°', a "reference" telling where
equilibrium lies)
2) actual starting concentrations of reactants and products (mass action ratio)
3) temperature.
ANY FREE ENERGY CHANGE, FROM ANY STARTING CONDITIONS, CAN BE
DESCRIBED BY THE IMPORTANT EQUATION:
where
is the actual mass action ratio; it is NOT Keq (unless
actual mass action ratio happens to be equilibrium ratio,
in which case reaction is at equilibrium to start with and
ΔG' = 0)
Under ALL starting conditions, the "parent" equation holds:
2 "special cases" of starting conditions:
IF reaction is under standard conditions to start with (m.a. ratio = 1), then RTln(1) = 0,
so
ΔG' = ΔG°'
ΔG°' = the standard free energy change, the free energy change for going from
standard conditions to equilibrium)
IF reaction is at equilibrium to start with:
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ΔG' = 0 and actual mass action ratio = Keq, so
0 = ΔG°' + RTlnK eq
That gives relation between ΔG°' and Keq (described above):
ΔG°' = -RTlnK eq.
BE SURE TO DO THE POSTED THERMODYNAMICS SAMPLE PROBLEMS
[PDF]
CHEMICAL KINETICS:
For the reaction
k = RATE CONSTANT
NOTE: Rate constants are lower case k's.
Equilibrium constants are upper case K's.
Velocity (rate) of forward reaction = v F = k F [S]
Velocity (rate) of reverse reaction = v R = k R [P]
The equilibrium constant is
NOTE: Enzymes do NOT alter Keq.
As catalysts, enzymes DO increase rate constants and thus increase rates of reactions.
COROLLARY: An enzyme that increases k F by a factor of 1010 must also increase k R by a
factor of 1010.
Sample problems [PDF]
TRANSITION STATE THEORY
transition state: an activated complex at the highest free energy point on the reaction coordinate
a PEAK on the free energy diagram
not isolatable as structures (lifetimes ~10 -13 sec) -- they’re "in transition", sort of with bonds halfmade, half-broken.
Chemical example: an S N2 reaction, attack of a thiolate anion on iodoacetate:
transition state (in brackets): a trigonal bipyramid, with 3 covalent bonds + 2 more "half"
bonds:
.
FREE ENERGY DIAGRAM FOR THE REACTION S → P:
free energy G vs. progress of reaction (i.e., the "reaction coordinate")
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Berg, Tymoczko & Stryer, 6th
ed. Fig. 8.3: Enzymes decrease
activation energy (ΔG ‡) for
reactions they catalyze.
ΔG = overall difference in free
energy between final (P) and
starting (S), not affected by
enzyme.
RATE of reaction IS affected by enzyme.
RATE depends on ΔG‡, the Arrhenius activation energy (i.e., the free energy of activation for the
reaction).
ΔG ‡ = G‡ – GS = difference in free energy between transition state and starting state (S in this
case), the "barrier" over which the reaction must go in order to proceed.
ΔG‡, has POSITIVE values (> 0) -- it's a free energy BARRIER.
where k is Boltzmann’s constant and h is Planck’s constant.
NOTE: Rate constant k is inversely and exponentially
dependent on the activation energy, ΔG ‡.
Velocity of reaction:
rate constant k = what's inside
large brackets.
How could you increase the reaction rate of S → P?
Rate of S → P = velocity = k [S]
1) increase concentration of a reactant [S], or
2) increase the rate constant – HOW?
a) increase temperature, or
b) decrease ΔG ‡ (catalyst)
Enzymes increase reaction rates by decreasing ΔG ‡ and thus increasing k, by CHANGING THE
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PATHWAY of reaction and by TIGHTLY BINDING TRANSITION STATE(S).
New pathways: often multiple steps in an enzyme-catalyzed reaction.
INTERMEDIATES are TROUGHS between steps on free energy diagram.
Each step has a transition state (peak), so each step has its own ΔG ‡.
Slowest step in pathway (the "rate-limiting step") = step with highest ΔG ‡.
Nelson & Cox, Lehninger Principles of Biochemistry, 4th ed. (2004) Fig. 6-3 (modified): Reduction
of ΔG ‡ by enzyme, which a) binds transition state very tightly, and b) changes reaction
pathway.
New pathway:
NOTE:
Even the HIGHEST ΔG ‡ (step #2 in figure above, ES < == > EP) for a CATALYZED
reaction is LESS THAN THE ΔG ‡ for an UNcatalyzed reaction.
Keq is NOT AFFECTED BY THE CATALYST and ΔG' is NOT AFFECTED BY THE
CATALYST.
[email protected] .edu
Department of Biochemistry & Molecular Biophysics
The University of Arizona
Copyright (©) 2007
All rights reserved.
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