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.
- Where did the energy that was stored in the chemical bonds of the hydrogen peroxide go?
- Would you expect the surroundings to increase or decrease in temperature as a result of this energy change?
- 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.
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.