Enzyme

Enzymes: Introduction
Enzymes are proteins.
– (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 very specific
– for substrate acted upon
– for reaction catalyzed
• Example: Proteases are a whole class of enzymes that all catalyze
hydrolysis of peptide bonds:
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
Berg et al., Fig. 8-1
(A) Trypsin catalyzes hydrolysis
of peptide bonds on carboxyl
side of Lys and Arg residues
(digestive function in small
intestine, cleaves just about
any protein it encounters after
(eventually) every Lys and Arg)
(B) Thrombin (involved in blood
clotting cascade) catalyzes
hydrolysis of peptide bonds
between Arg and Gly residues
in specific sequences in
specific protein substrates
(activated only where blood
needs to clot, works only on
very specific target protein)
• substrate specificity of proteases --
another example, chymotrypsin:
– cleaves on carboxyl side of aromatic and hydrophobic amino acid
residues
– 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.
Specificity of reaction catalyzed:
Many proteases also catalyze hydrolysis of carboxylic ester bonds:
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.)
• 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"
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 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:
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.
• k is rate constant for the reaction.
• ? 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
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.
• k is rate constant for the reaction.
• ? 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