Enzyme Kinetics
Enzymes are large protein molecules which act as biological catalysts – they speed up
reactions in the body by reducing the activation energy of the reaction. The molecule on
which an enzyme acts is called a substrate, and the place on the enzyme where the
substrate binds is called the active site. Enzymes are specific for a particular substrate
and are not consumed in a reaction.
Catalase
Catalase is an enzyme that catalyzes the breakdown of hydrogen peroxide
into water and oxygen gas. It is composed of 4 polypeptide chains, each
over 500 amino acids long.
__H2O2 ⇌__H2O +__O2
1. Balance the equation above by placing coefficients in front of the molecules.
2. If 4 molecules of O2 are released, how many molecules of H2O2 were broken
down?
3. If catalase can break down 9.3 x 107 molecules of H2O2 per second, how many
water molecules will be produced in 2 seconds?
The energy required (activation energy) to break down hydrogen peroxide into water and
oxygen gas is 18kJ/mol. Catalase reduces the energy cost of this reaction to just 2kJ/mol.
1. a) If 195 kJ of energy was available to do work, calculate the number of H2O2
molecules that can be broken down without catalase.
b) If 195 kJ of energy was available to do work, how many H2O2 molecules
could be broken down with catalase?
2. The First Law of Thermodynamics states that energy cannot be created nor
destroyed. Yet, the products are lower in energy than the reactants in this
reaction.
a. Where did the energy that was stored in the chemical bonds of the
hydrogen peroxide go?
b. Would you expect the surroundings to increase or decrease in temperature
as a result of this energy change?
c. Is this an endothermic (absorb heat energy) or exothermic (release heat
energy) reaction? How do you know?
Increasing substrate concentration will
increase the rate of a reaction up to the
point of saturation. When all enzymes
are being used and are saturated with
substrate, we call the rate of the reaction
the maximum velocity, or Vmax.
What do you think would happen to the
rate of the reaction if the enzyme
concentration was cut in half?
Michaelis-Menton is a model for enzyme
kinetics when very simple kinetics can be
assumed. The equation states that the
velocity of an enzymatic reaction can be
determined if the Vmax, Km and
substrate concentrations are known.
Km, the Michaelis-Menton Constant, is a
variable used to describe the affinity of
an enzyme for its substrate. The larger
the Km value, the less likely the enzyme
will bind the substrate. It is equal to the
substrate concentration at ½ Vmax.
Use the Michaelis-Menten Plot below to answer the following questions.
700
Rate of Reaction (μM/s)
600
500
400
300
200
100
0
0
200
400
600
800
1000
1200
[S] (μM)
a) Label Vmax and Km on the graph.
b) What is the value for Km?
c) What is the Vmax of this enzymatic reaction?
d) Use the Michaelis-Menton equation to calculate the initial rate of the reaction
(vo) when the substrate concentration is 200 μM.
e) If an enzymatic reaction had a maximum velocity of 250 μM/s, and an initial
velocity of 113 μM/s when the concentration of substrate was 76 μM,
calculate the Km for this reaction.
Carbonic Anhydrase
Carbonic Anhydrase is an enzyme that catalyzes the rapid conversion
of CO2 to bicarbonate ion and protons.
__CO2 + __H2O ⇌__HCO3- +__H+
1. Balance the above equation.
2. For this reaction, KM = 2.6 x 10 -2 M and Vmax = 0.4 moles/min.
a) Sketch the expected Michaelis-Menten plot for this reaction. Label the axes
and indicate KM and Vmax on the plot.
b) Calculate the initial velocity of Carbonic Anhydrase when the concentration
of CO2 is 1.5 x 10-2 M.