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BIOC 460 Summer 2011
Enzymes: Introduction
Firefly bioluminescence is produced by an
oxidation reaction catalyzed by the enzyme firefly
luciferase. The oxidized substrate (product of the
reaction) is in an electronically excited state that
emits light as it returns to the ground state.
Reading: Berg, Tymoczko & Stryer, 6th ed.,
Chapter 8, pp. 205-217 (These pages in textbook
are very important -- concepts of thermodynamics
are fundamental to all of biochemistry.)
•Review thermodynamics practice problems (same as for Lecture 2)
•Enzymes introduction sample problems: EnzIntrodSampleProblems.pdf
(linked in lecture notes directory)
Key Concepts
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Enzymes are powerful and specific biological catalysts.
– Increase rates of (bio)chemical reactions but have no effect on Keq
(and no effect on overall ∆G) 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: the study of reaction rates. Rates depend on rate constants.
– Rate constants depend inversely and exponentially on Arrhenius
activation energy, ∆G‡, the difference in free energy between free
energy of transition state and free energy of reactant(s).
– Rate constants are increased by catalysts (enzymes), because
enzymes decrease ∆G‡.
– Enzymes lower ∆G‡ by affecting either ∆H‡ or ∆S‡ (or both).
– One way enzymes reduce ∆G‡ 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 ∆∆G‡, the decrease in ∆G‡ brought about
by enzyme compared with uncatalyzed reaction's ∆G‡.
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Enzymes: Introduction
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Enzymes are proteins.
– (one exception are ribozymes: catalytic RNA molecules)
biological catalysts
– not chemically altered in reaction
– do not change equilibrium constant (Keq) 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 temps., around
neutral pH, low conc. 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
Enzymes Do Two Things:
• Recognize and bind the substrate (S)
• Catalyze the chemical reaction
Both of these functions contribute to the
SPECIFICITY and RATE ENHANCEMENT
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CATALYTIC POWER OF ENZYMES
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RATE ENHANCEMENT = catalyzed rate constant/uncatalyzed rate
constant = factor by which catalyst increases rate of reaction
Berg et al., Table 8-1
SPECIFICITY OF ENZYMES
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Enzymes very specific
– for substrate acted upon
– for reaction catalyzed
Example: Proteases are a whole class of enzymes that all catalyze
hydrolysis of peptide bonds:
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Substrate Specificity -- proteases as an example
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Substrate specificity (e.g., of proteases) due to precise interaction of
enzyme with substrate
– result of 3-D structure of enzyme active site where substrate has
to bind and be properly oriented for catalysis to occur
(A) Trypsin catalyzes hydrolysis
of peptide bonds on carboxyl
side of Lys or Arg residues
(digestive function in small
intestine, cleaves just about
any protein it encounters after
(eventually) every Lys or Arg)
(B) Thrombin (involved in blood
clotting cascade) catalyzes
hydrolysis of peptide bonds
between Arg and Gly residues
ONLY in specific sequences in
specific protein substrates
(activated only where blood
needs to clot, works only on
very specific target protein)
Berg et al., Fig. 8-1
Enzyme Specificity, continued
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substrate specificity of proteases : Chymotrypsin:
– Recognizes and binds aromatic or hydrophobic amino acid
residues
– The overall catalytic mechanism is EXACTLY the same as trypsin
– evolutionarily related to trypsin
– Genes for trypsin and chymotrypsin are homologous.
• Ancestral gene duplicated and sequences diverged through
evolution.
• Substrate specificities for site of cleavage diverged, but catalytic
mechanism and overall tertiary structure was conserved.
There is also diversity in reactions catalyzed:
Many proteases can ALSO catalyze hydrolysis of carboxylic ester bonds:
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Cofactors
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Some enzymes need cofactors for their activity.
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.)
• Weakly bound coenzymes can associate and dissociate from enzymes
between reaction cycles, behaving like substrates
– sometimes referred to as "cosubstrates“
• Prosthetic groups (e.g. heme in hemoglobin): very tightly bound cofactors
(either coenzymes or metals) remain associated with their enzymes even
between reaction cycles.
Common Coenzymes and Reactions They Mediate
See also Berg et al., 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 (RŠC=O group)
Flavin coenzymes (B2)
Oxidation-reduction (hydrogen atoms) (1 or 2 eŠ transfer)
Lipoic acid
Acyl group transfer
Nicotinamide coenzymes (niacin)
Oxidation reduction (hydride ions H:Š, 2 eŠ 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)
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CHEMICAL KINETICS (review from gen chem)
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For the reaction
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Velocity (rate) of forward reaction = vF = kF [S]eq
Velocity (rate) of reverse reaction = vR = kR [P]eq
At equilibrium, vF = vR, so kF [S]eq = kR [P]eq .
The equilibrium constant is
• k = rate constant
• NOTE:
Rate constants are lower case k's.
Equilibrium constants are upper case K's.
• NOTE: Enzymes do NOT alter Keq.
• As catalysts, enzymes DO increase rate constants and thus increase rates
of reactions.
• COROLLARY: An enzyme that increases kF by a factor of 1010 must also
increase kR by a factor of 1010.
TRANSITION STATE THEORY
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Enzymes: Introduction
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 half-made, half-broken.
Chemical example: an SN2 reaction, attack of a thiolate anion on
iodoacetate: transition state (in brackets)(‡): a trigonal
bipyramid, with 3 covalent bonds + 2 more "half" bonds:
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FREE ENERGY DIAGRAM FOR THE REACTION S → P
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free energy G vs. progress of reaction (i.e., the "reaction coordinate")
• Enzymes decrease
activation energy (∆
∆G‡)
for reactions they
catalyze.
∆G = 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 magnitude of ∆G‡, the
Arrhenius activation energy
(i.e., free energy of
activation for the reaction).
Berg et al., Fig. 8.3
Dependence of rate constant on ∆G‡, the activation energy
Rate constant (k) depends on ∆G‡, the Arrhenius activation energy
(i.e., the free energy of activation for the reaction)
∆G‡ = 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.
∆G‡ has POSITIVE values (∆
∆G‡ > 0) -- it's a free energy BARRIER.
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k is rate constant for the reaction.
κ is Boltzmann’s constant and
h is Planck’s constant.
NOTE: Rate constant k is inversely
and exponentially dependent
on the activation energy, ∆G‡.
Velocity of the reaction:
(rate constant k is what’s inside large brackets.)
How could you increase the
reaction rate of S → P?
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• Rate of S → P = velocity = k [S] = . To increase rate:
1)
2)
increase concentration of a reactant [S], or
increase the rate constant – HOW?
a) increase temperature, or
b) decrease ∆G‡ (catalyst)
• Enzymes increase reaction rates by decreasing ∆G‡ and thus increasing k.
How do enzymes
increase k (i.e.,
decrease ∆G‡)?
step1 step2
step3
Nelson & Cox, Lehninger
Principles of Biochemistry,
4th ed., Fig. 6-3
How do enzymes increase k (decrease ∆G‡)?
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by changing the 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 (e.g., ES and
EP).
•Each step has a
transition state (peak),
so each step has its
own ∆G‡.
step1 step2
step3
•Slowest step in pathway
(the "rate-limiting step")
= step with highest ∆G‡.
•NOTE: Even the highest ∆G‡ (step #2 in figure above, ES < == > EP) for a
catalyzed reaction is less than the ∆G‡ for an uncatalyzed reaction.
•Keq is not affected by the catalyst, and ∆G' is not affected by the catalyst
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Learning Objectives
(See also posted Thermo and Enzymes-Introduction sample problems.)
• 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.
• Define "standard free energy change" and give the symbol for that
parameter.
• Write the mathematical expression relating ∆G ' to Keq', and be able to
interconvert ∆G ' and Keq'.
• Calculate the actual free energy change (∆G'), given the starting
concentrations of appropriate chemical species and either ∆G ' or Keq'.
• Describe the relation between ∆G' and the rate of a reaction; using ∆G',
predict reaction direction.
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Learning Objectives, continued
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Express the velocity of a simple reaction in terms of the rate constant and
the concentration of the reactant.
Express the equilibrium constant of a reaction in terms of the equilibrium
mass action ratio.
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 ∆G for the overall reaction, ∆G‡uncat, and ∆G‡cat.
Indicate (and name) the quantity on a free energy diagram (HINT: it's a
specific kind of ∆G) 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?
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