2nd Set of notes
Catalysis and Enzymes:
An enzyme - a catalyst- can increase the rate of rxn by several orders of magnitude. An extreme
case of this is the enzyme _____________
Catalase can increase the rate of decomposition of _____________ by 15 powers of 10! The
reactant, called the _______________________ often binds to the catalyst in an active site, but
the active site alone may not give total catalytic activity. For instance the active site in catalase
contains hematin, a porphyrin molecule with iron and it can be isolated without the rest of
catalase. The activity is 10-8 of the rate of catalase but still 107 times faster than the reaction
without an enzyme Sometimes other molecules called ______________________ must also
bind for the catalytic action to occur.
Others are not quite as good, but, at room temp., all increase the rate primarily by :
From 71 kJ to 8 kJ/mol in catalase.
Some also affect the preexponential factor as the _____________ barrier is also decreased.
Catalase for instance
In Biological Systems most enzymes are proteins.
The reactant
Michaelis-Menten Mechanism
Well known example of reaction in which an intermediate is formed.
This type of mechanism is characteristic of many enzyme-catalyzed reactions where a substrate
is converted to product.
...........E + S k1-><-k1’ ES k2-> P + E
In this mechanism, ES is an intermediate state where the enzyme and the substrate are bound.
P is the product.
Assuming steady state on the enzyme one can write:
d[ES]/dt = 0 = k1[E][S] - k-1’[ES] - k2 [ES]
or
total enzyme is
[E]0 = [ES] + [E]
Typically only a small amount of enzyme is added to the free substrate so that [S] ~ [S]t.
Plugging this in
dividing both sides by
Since d[P]/dt = k2[ES] then:
.......................................
d[P]/dt =
.................................................. = k [E]0 ........ k = k2[S]/(km + [S]) ............ km = (k2 + k1’)/ k1
Rate varies linearly with [E]0 and in a complicated way with [S]
Limiting Cases:
1. [S]>>km ..... [S] cancels in the numerator and the denominator leaving:
.......................... d[P]/dt = k2[E]t ........ (Zero order in [S])
... Rate is constant. So much S is present that the enzyme is saturated. Rate is a maximum.
....k2 is called the maximum turnover number.
vmax = k2[E0]
2. [S] << km[S] ........ then [S] in the denominator is insignificant
.... d[P]/dt = k2[E0] [S] / km
Rate is proportional to both the _____________ and the __________ ______________
concentrations.
Considering the entire equation again:
.......... d[P]/dt = k[E]t .................... k = k2[S]/(km+[S]) ............ km = (k2 + k1’)/ k1
.........................= k2[E]t [S]/(km+[S]) = vmax 1/[S] [S] / ((1/[S](km+[S])
.........................= vmax 1 /( 1+(km/[S])
the rate can also be written in terms of [S]:
......rate = d[P]/dt = -d[S]/dt = vmax 1 /( 1+(km/[S]) .......... so separating variables:
this can be integrated to give:
........................km ln([S]/[S0]) + [S] - [S]0 = -vmax t
without km there is no simple way to plot data to get a straight-line relation between a function of
[S] and time.
Often enzymes act by a simple mechanism with a single enzyme-substrate complex, but even
more complex situations often have the form above for -d[S]/dt.
For purposes of calculations a linearized equation is more useful. There are three used with the
1st below being the most often used.
Lineweaver-Burk
...........1/rate =
.
............Note 1/k also gives
So a plot of ________ vs ___________ gives the value of k2 and km but k1 and k1’ cannot be
determined.
Dixon
......................... [S]/rate = [S]/ =
Eadie-Hofstee
............................  = -km /[S] + vmax
There is statistical error associated with each plot so using all possible combinations the
following are touted to be the best eqns for vmax and km.
..................................... vmax =
....................................... km =
Most enzymatic reactions are reversible, so, there is evidence for an enzyme-product complex as
well.
................................. E + F k1-><-k-1 EF k2-><-k-2 EM k3-><-k-3 E + M
Six kinetic constants here, can’t determine each, can get some expressions to help deal with
kinetics though.
Competition and Inhibition
May have more than one type of enzyme substrate complex that exists between substrates and
products:
.............. E+S k1-><-k-1 ES ->k2 ES’ ->k3 ES” ->k4 E + P
 = kcat[Eo][S]/ (km + [S])
kcat / km is called the ____________________ constant
Allows us to write the relative rates of reaction for two competing substrates, A and B
........................... a/b =
Reversible Inhibition
1. Competitive Inhibition
Enzymes typically are very specific for a particular substrate and reaction, but a molecule that
resembles the substrate which is nearly or completely unreactive can occupy the site and block
the reaction progression. It is a competitive inhibitor:
An operational definition is competitive inhibitors are those that bind reversibly to the active site.
Mechanistically we write:
The new apparent Michaelis constant is given by:
Classic Example is:
2. Noncompetitive Inhibition
Inhibitor molecule binds to the enzyme but not at the active site. It may bind to the enzyme itself
or to the enzyme substrate complex. Products cannot form once it has bound.
This inhibition cannot be reversed by increasing the substrate concentration.
3. Uncompetitive Inhibition
Inhibitor molecule binds to the enzyme substrate complex only. No product results from this.
This Inhibition cannot be reversed by increasing the substrate concentration. It is rarely occurs
in one-substrate systems.
Simplest form of this inhibition occurs when only the value of vmax is affected, but the affinity of
the enzyme for the substrate is not.
Other forms occur in which both vmax and km are affected by the inhibitor.
Irreversible Inhibition
The inhibitor forms a covalent linkage with the enzyme molecule and cannot be removed.
Michaelis-Menten kinetics cannot be applied to irreversible inhibition. Effectiveness of this
inhibitor is determined by the rate at which the binding takes place, not by the equilibrium
constant.
Allosteric Interactions
Some enzymes have kinetics that do not obey the Michaelis-Menten equations. Instead of a
hyperbolic curve they may have a sigmoidal or "S" shaped curve. This is typical of enzymes that
have multiple binding sites where their activity is regulated by the binding of inhibitors or
activators. S shaped curves usually indicate a positive cooperativity where the binding of the
ligand at one site increases the enzyme's affinity for binding aonther ligand at a different site.
Enzymes that show cooperativity are called Allosteric.
An effector is a ligand that can affect the
at a different site on the
Metabolic enzymes are highly regulated, there are inhibitors that decrease their rates and
effectors that increase their rates.
Concerted Model
The Monod-Wyman-Changeux (MWC) theory can describe S-shaped kinetics. It says that the
enzymes consist of two distinct allosteric forms
Allosteric effectors - favor the form that binds the substrate strongly.
Allosteric inhibitors - favor the form that binds the substrate weakly.
Mechaelis-Menten mechanism shows a hyperbolic dependence on substrate concentration with
or without competitive or noncompetitive inhibitors present.
However many enzymes show a sigmoid (S-shaped) dependence on substrate concentration. The
MWC can produce either type of behavior.
An MWC protein has two or more subunits each which can bind a ligand, the sites are assumed
to be identical and independent. Then the protein can exist to two forms,
Two conformations:
One conformation is relaxed - can bind the substrate more tightly
One confromation is tight - binds substrate weakly.
Sequential Model
Proposed by Koshland, Nemethy and Filmer assumes that the affinity of vacant sites for a
particular ligand changes progressively as sites are taken up.