The Effect of Catalase Concentration on Rate of Reaction

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The following experiment details the effect of different concentrations of catalase on the production of oxygen and water through the breakdown of Hydrogen Peroxide. In this experiment paper disc where coated in varying concentrations of catalase, 0, 25, 50 75 and 100%. The time taken for the disc to float between two markers on the side of a glass was then recorded. This experiment demonstrates that the higher the concentration of enzyme used the greater the production of oxygen on the paper disc. The oxygen then produced on the disc gives it greater buoyancy allowing it to move past the markers faster. Aim

To find the effect of the enzyme concentration on the reaction between Catalase and hydrogen peroxide. Introduction
Enzymes are proteins that act as catalysts (a substance that increases or decreases the rate of a reaction) 2. Enzymes bind to a molecule called a substrate, converting it into a product. Nearly all of the chemical reactions that occur in a biological cell need enzymes to make them occur. Enzymes like all catalysts lower the activation energy needed for a reaction to take place3. They dramatically speed up the rate at which the reactions take place3. They are not consumed by the reaction and remain unchanged by the reaction it self3. Enzymes have been linked to nearly 4000 biochemical reactions. Enzymes are very specific to the reaction in which they catalyse3. In 1894 it was suggested because of the specific nature of enzymes for their substrate the enzyme posses a complementary shape to the substrate4. The enzyme and the substrate fit exactly together without any need for changes in structure. This was called the lock and key model as the substrate fits exactly into the hole on the enzyme4. Although this model explains the specificity of the reaction it does not explain how the enzyme lowers the activation energy for the reaction to take place. In 1958 Daniel Koshland suggested a more flexible model5. The Induced Fit model suggested that the enzyme is more flexible and the substrate does not bind to a rigid active site5. The amino acids and their side chains that make up the active site are thought to remodel themselves around the substrate. This allows for chemical bonds to be formed between the substrate and enzyme; altering the shape and charge of the substrate. This distortion of the bound substrate reduces the energy needed to split the substrate into its products. Vmax is the maximum initial velocity (Vo) that an enzyme can achieve. Initial velocity is defined as the catalytic rate when substrate concentration is high, enough to saturate the enzyme, and the product concentration is low enough to neglect the rate of the reverse reaction. Therefore, the Vmax is the maximum catalytic rate that can be achieved by a particular enzyme. Km is determined as the substrate concentration at which 1/2 Vmax is achieved. This kinetic parameter therefore importantly defines the affinity of the substrate for the enzyme. In order to break down the substrate into a product the enzyme must first form an enzyme substrate complex non as a noncovalent ES complex. The formula to describe this reaction is shown below6:

E + SESE + P
Where E= enzyme, S =Substrate, P = product.
K1is the forward rate constant for substrate binding
K2 is the catalytic rate constant
From the Michaelis-Menten equation the velocity Vo of a reaction is proportional to the Vmax.
Vo = Vmax x [S]/Km + [S]
These two parameters for a specific enzyme define7:
Vmax – as the rate at which a substrate will be converted to product once bound to the enzyme. Km - how effectively the enzyme would bind the substrate, hence affinity. The Vmax is proportional to K2 the rate at which the Enzyme breaks down the substrate to a product. Therefore the more enzyme that is present the greater the rate at which the product is produced. Therefore: Vmax is proportional to the Enzyme total ET

As the rate of the reaction...
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