Order reaction
First order reaction
For irreversible reaction A  B the velocity of reaction (Rate) is given by
V = d [B] / dT
or
V = d [A]/dT
These equations are equally valid for this reaction so,
V = d [B] / dT = - d [A] / dT = k [A]
k = the rate constant of this reaction (unit : second-1)
This kind of reaction is known as First order reaction because the rate of the reaction is
depend on the first power of concentration of substrate
If k1 = large  the reaction is fast,
If k1 = small  the reaction is slow
A plot of this equation shows that concentration of A decreases exponentially with time
The amount of time it takes for half of A to be lost  half life (t½)
Reversible reaction
The first order reaction is too simple to explain the reaction occurred in biological
system. As many of reactions is reversible where the concentration of product becomes
very important for the reverse reaction
So for reaction
AB
k1 = the rate constant for the reaction moving rightward
k-1 = the rate constant for the reaction moving leftward
so the molecule A is consumed for the reaction to the right and is formed for the reaction
to the left. Such a reaction approaches the state of equilibrium at which point the rate of
forward and the reverse reaction becomes equal. At the same time the rate of reaction
becomes zero.
-k1 [A] + k-1 [B] = 0
K = equilibrium constant = [B]/[A] = k1 / k-1
For a reversible reaction, the equilibrium constant is always the ratio of forward and
reverse rate constant
Second order reaction
A reaction typically occurs when two molecules comes together to form product.
2 A  A2
with a rate constant given by k2
The rate of such a reaction is proportional to the second power of the concentration of
reactant.
V = - d [A] / dT = - k2 [A]2
Why we learn enzyme kinetics?
Enzyme kinetics is important to understand
1. the mechanism of enzyme action
2. the role of an enzyme in the cell
3. how the activity of enzyme can be controlled
When we talk about enzyme kinetics  it relates to
the velocity of PRODUCT being formed or
the velocity of SUBSTRATE being changed to product
The reaction rate (same as velocity or rate of formation of products) can be written as
V = k2[ES]
[ES] is usually not a measurable concentration. Easily measurable items are the substrate
(or product) and the total concentration of enzyme, which is the sum of the free enzyme
and complexed enzyme.
That is [E]t = [E] + [ES],
where [E]t is total enzyme, [E] is free enzyme, and [ES] is complexed enzyme.
Briggs and Haldane proposed a model that avoided the equilibrium assumption. It
assumes that the more ES that is present, the faster ES will dissociated either to products
or back to reactants. Therefore, when the reaction is started by mixing enzymes and
substrates, the ES concentration builds up at first, but quickly reaches a steady state, in
which it remains almost constant. The steady state will persist until almost all of the
substrate has been consumed
If the steady state, rates of formation and breakdown of ES are equal,
Rearranging 11.17 gives equation 11.18,
Combining the three rate constants of equation 11.18 into one, KM, yields
KM = (k-1 + k2)/k1
Equation 11.18 becomes
KM[ES] = [E][S]
Because [E] = [E]t - [ES],
KM[ES] = [E]t - [S] - [ES][S]
Solving for [ES],
[ES] = [E]t[S]/(KM+[S])
Substituting this into the earlier velocity equation,
V = k2[E]t[S]/(KM+[S])
This last equation is the Michaelis-Menten equation,
KM is called the Michaelis constant. KM has units of concentration and, because it is a
ratio of the rate constants of a reaction, KM is characteristic of the reaction. A given
enzyme acting upon a given substrate has a distinct KM.
KM, is often associated with the affinity of the enzyme for substrate, but this is not
always correct. A more accurate statement is that, for reactions obeying MichaelisMenten kinetics,
KM is a measure of the substrate concentration required for effective catalysis to occur.
 an enzyme with a high KM requires  a higher substrate concentration to achieve
a given reaction velocity than an enzyme with a low KM
Km value = the concentration of substrate that give half the maximum velocity of
enzymic reaction
Velocity of enzyme reaction depend on
- affinity of Enzyme-Substrate Binding and is expressed as Km value
- The lower the Km value, the higher the velocity the higher the activity
Example
Hexokinase  Isoenzyme I – IV
(Isoenzyme IV  glukokinase)
Glukokinase is confined to the liver and has a high Km for glucose (10 mmol dm-3)
Isoenzymes I- III  have a much wider tissue distribution and a low Km (40 μmol dm-3)
At the prevailing levels of blood glucose in a fasting subject  isoenzymes I-III are
working essentially at their maximum velocity , whereas isoenzyme IV is only working at
about 25% of its maximum velocity.
After an intake of carbohydrate, the level of blood glucose rises isoenzyme IV can
work at about half its maximum velocity.
The liver isoenzyme (IV) with the extra glucose  converting it to D-glucose 6-P  first
step in the process of storage as glycogen.
Km value  depend on enzyme conformation
How do we obtain kinetic data?
Aim  measure the rate formation of product (or disappearance of substrate) under
specific condition
Example :
D-glucose + ATP D-glucose 6-phosphate + ADP
Rate of reaction could be monitored by
- removing samples from the reaction mixture at known time after addition of
enzyme  stopping the reaction quickly
- then measuring the product formed
 called stop and sample / discontinuous assay
Continuous assay  continuous measurement of some property which changes during
the course of the reaction
Use absorbance of the product by coupling the production of D-glucose 6-phosphate to
the reduction of NADP+ to NADPH using glucose 6-phosphate dehydrogenase
NADP+  does not absorb at 340 nm but NADPH does
Important  must provide sufficient coupling enzyme and substrates so that the Dglucose 6-phosphate formed in the first step  immediately converted to second product
So the coupling reaction is not the rate limiting factors
Another way  involving radioactively labeled substrate  very sensitive and very
helpful when the substrate and enzyme concentration are very low
Important precaution to obtain reliable data:
1. The substrates, buffer, should be of as high a purity as possible  contaminants affect
the activity of enzymes
2. The enzyme should be stable and does not contain any compound which interferes
with the assays
3. The parameters that affect the activity of enzymes (ex. pH, temperature, etc)  must
be stabilized by use of buffers, thermostatted baths
.
the rate of a chemical reaction
 is described by the number of molecules of reactant(s) that are converted into
product(s) in a specified time period
Reaction rate is always dependent on
1. the concentration of the chemicals involved in the process
2. the rate constants that are characteristic of the reaction
AB
Rate of reaction :
- [A] = k [B]
The rate of enzyme reactions is influenced by many factors including
1. enzyme concentration,
2. substrate concentration,
3. temperature,
4. pH
5. the presence of inhibitors.
Rate constants are simply proportionality constants that provide a quantitative
connection between chemical concentrations and reaction rates
Each chemical reaction has characteristic values for its rate constants; these in
turn directly relate to the equilibrium constant for that reaction.
Thus, reaction can be rewritten as an equilibrium expression in order to show the
relationship between
1. reaction rates,
2. rate constants and
3. the equilibrium constant
The rate constant for the forward reaction is defined as k+1 and the reverse as k-1.
At equilibrium, the rate of the forward reaction is equal to the rate of the reverse
reaction leading to the equilibrium constant of the reaction and is expressed
by:
[B]/[A] = k+1/k-1 = Keq
This equation demonstrates that the equilibrium constant for a chemical reaction
is not only equal to the equilibrium ratio of product and reactant concentrations,
but is also equal to the ratio of the characteristic rate constants of the reaction.
Enzymes increase reaction rates by decreasing the amount of energy required to
form a complex of reactants that is competent to produce reaction products.
This complex is known as the activated state or transition state complex for the
reaction. Enzymes and other catalysts accelerate reactions by lowering the
energy of the transition state. The free energy required to form an activated
complex is much lower in the catalyzed reaction.
The amount of energy required to achieve the transition state is lowered;
consequently, at any instant a greater proportion of the molecules in the
population can achieve the transition state
Michaelis-Menten Kinetics
The catalytic event that converts substrate to product involves the formation of a
transition state, and it occurs most easily at a specific binding site on the
enzyme. This site, called the catalytic site of the enzyme,
Catalytic site has been evolutionarily structured
 to provide specific, high-affinity binding of substrate(s) and
 to provide an environment that favors the catalytic events.
The complex that forms when substrate(s) and enzyme combine is called the
enzyme substrate (ES) complex. Reaction products arise when the ES
complex breaks down releasing free enzyme.
The series of enzyme-substrate reaction can be shown:
E + S <---> ES <---> ES* <---> EP <---> E + P
The kinetics of simple reactions like that above were first characterized by
biochemists Michaelis and Menten.
The Michaelis-Menten equation:
Michaelils –Menten equation is a quantitative description of the relationship
among the rate of an enzyme- catalyzed reaction [v1], the concentration of
substrate [S] and two constants, Vmax and Km
The symbols used in the Michaelis-Menton equation refer to the reaction rate
[v1], maximum reaction rate (Vmax), substrate concentration [S] and the MichaelisMenton constant (Km).
The Michaelis-Menten equation has the same form as the equation for a
rectangular hyperbola; graphical analysis of reaction rate (v) versus substrate
concentration [S] produces a hyperbolic rate plot.
Lineweaver-Burk Plot
To avoid dealing with curvilinear plots of enzyme catalyzed reactions,
biochemists Lineweaver and Burk introduced an analysis of enzyme kinetics
based on the following rearrangement of the Michaelis-Menten equation:
[1/v] = [Km (1)/ Vmax[S] + (1)/Vmax]
Plots of 1/v versus 1/[S] yield straight lines having a slope of Km/Vmax and an
intercept on the ordinate at 1/Vmax.
A Lineweaver-Burk Plot
Km - the Michaelis Constant
...is a constant that is independent of [S] or [E]
...is a mathematical interpretation of an enzyme action
...is a substrate concentration... at which enzyme rate is equal
to ½ Vmax