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Eddy Egan

October 31, 2008

Contreras Pd. AP Biology

Relative Amount of H2O2 Decomposed per second by Catalase measured by use of a

Titrant

Enzymes are globular proteins that catalyze metabolic reactions by reducing the

activation energy needed. This occurs when the substrate binds to the enzyme at its activesite. Once the substrates, or reactants, have absorbed the activation energy needed from

their surroundings, bonds will be broken because the reactants would have become

unstable. This would therefore make them more reactive as the reach their transition state.

(CITE: 1)The reason for this occurrence is because the thermal energy absorbed by thereactants increases their speed, resulting in many forceful collisions, which will break the

reactants bonds. The amount of energy absorbed, the activation energy, needed is lower

by the enzyme, therefore, catalyzing the reaction. Enzymes are very selective in the

reactions which they catalyze, determining the reactions occurring with the cell at anygiven time. (CITE: 2) In order to begin catalysis, the substrate must bind to the enzyme,

forming and enzyme-substrate complex. While joined, the enzymes catalytic actionconverts the substrate into the product. Additionally, enzymes can recognize their own

substrate from a group of many others, further increasing their selectivity (CITE: 2).

These enzymes can even differentiate between isomers. This is a result of their unique

shape, since they are proteins. The specific active site on the enzyme, the region wherethe substrate binds, is the cause of the enzymes specificity. The substrate does not need

to perfectly fit, however, the chemical groups on their surface and the amino acids on the

active site can slightly change there shape to fit better (CITE:2). This is an example of theinduced fit model.

The induced fit model illustrates how the substrates bind to its enzyme to form theenzyme-substrate complex by showing a slight change in the shape of the active site. Noother shape will be able to fit into this active site including isomers (CITE: 2). The only

exceptions are inhibitors, which prevent the normal substrate from binding. This model

differs from the lock and key model, which says that the substrate must fit exactly with itscorresponding enzyme. This model is no longer in use. While substrates are held in the

active site, though, hydrogen bonds and van der Waals forces keep the substrate in place

(CITE: 2) The R groups of several of the amino acids that compose the active site

catalyze the reaction, converting the substrate into the product. The product will then bereleased after the reaction has finished. This process is completed so quickly that one

enzyme can catalyze approximately one thousand reactions per second (CITE: 2). After

the reaction has completed and the product has been released, the enzyme will return toits original shape, allowing it to act upon other molecules. In some cases, the R groups

may be acidic, basic, or alkaline, in an otherwise neutral cellular environment. This

would affect whether the substrate would have been facilitated by H+ or OH- to form theproduct. (CITE: 3)

The initial reaction rates of enzyme-catalyzed reactions are affected by changes in

temperature, pH, enzyme concentration, and substrate concentration. The rate will also be

affected when the enzyme is denatured. This would occur as a result in the drop pH or

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increase in temperature. Denaturing would break the bonds of the enzymes quaternary

structure, including hydrogen and ionic bonds, and van der Waals interactions. This

would cause the enzyme to unravel and lose its shape, which would prevent it fromcatalyzing reactions when it is renatured. (CITE: 4)

As stated before, a catalyst speeds up the process of a reaction, in our lab we used

an catalytic enzyme known as catalase. In the experiment, catalase will catalyze thedecompositions of hydrogen peroxide, producing water and oxygen gas (H2O2 H2O +

O2)

I believe that the enzyme catalase will increase the reactions speed (speed ofH2O2 decomposition into H2O + O2). This is because enzymes catalyze reactions by

lowering their activation energy, therefore increasing the rate of reaction. The rate will

then decrease, though, once the concentration of H2O2begins to substantially decrease.

Procedure:

Exercise 2B: The Base Line Assay

1, 2. 3.

4. Remove a 5 mL sample and place in another beaker to assay for the amount of

H2O2by doing the following5. Put the beaker with the sample over a white piece of paper and use a syringe

to add KMnO4 ONE DROP at a time. Gently swirl the beaker after each drop

is put in. Do this until the solution turns a persistent pink or brown color.6. Record data in Table BL1

Exercise 2C: The uncatalyzed Rate of H2O2 Decomposition1. Pour 10 mL of 1.5% H2O2 in a beaker and store it uncovered at roomtemperature for about 24 hours.

2. Repeat steps 2-5 from exercise 2B to determine the amount of H2O2 still

present.3. Record in Table UD

Exercise 2D: An Enzyme-Catalyzed Rate of H2O2 Decomposition1. 10 seconds:

a. Pour 10 mL of 1.5% H2O2 in a 50 mL beaker and then add 1 mL

catalase extract

i. Catalase will be on ice and may settle to the bottom of thetest tube. To overcome this follow these steps:

1. Tightly close lid on test tube.

2. Invert the test tube and place in a vortex untilcloudy

b. Swirl gently for 10 seconds

c. When time is up, add 10 mL 1.0 M H2SO4 (Added in order to stopany more reactions from occurring)

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2. 30, 60, 90, 120, 180, 360 seconds

a. Repeat steps a-c in step 1, but swirl the length of time until thetimer has reaches 30, 60, 90, 120, 180, and 360 seconds.

3. Each time, obtain a 5 mL sample of the solution and assay for the

amount of H2O2. To find this, repeat step 5 of Exercise 2B and recorddata in Table KT

4. Graph the data from the enzyme-catalyzed reaction.

Analysis:

The rate is highest in the first interval with a rate of 0.20 mL H2O2/sec.

This is because the largest amounts of H2O2 able to be catalyzed were in the first

interval since the reactions were first starting. They decreased over time as moreand more H2O2 was decomposed and less became available. The reaction rate can

be the highest on any point in a graph as long as it contains the most amount of

substrate that is available to react with enzymes.

The rate is lowest at the last interval with a rate of .0005 mL H2O2/sec.This is a result of the very little H2O2 that has not been decomposed in the

solution. All reactions similar to this will slow down when concentration of thesubstrate decreases.

The initial rate of reaction of enzymes is determined by the characteristics

it possesses. This initial reaction rate remains constant for any enzyme and its

substrate at a specific pH and temperature, as long as the substrate is in excess. Itis represented on a graph by the linear portion of the curve. This rate is

determined by finding the slope after choosing two points.

Free energy is the surrounding energy that can be readily absorbed bymolecules and other particles. Part of this energy is absorbed as activation energy

for a reaction. In order to prevent the denaturing of proteins, killing the cell, and

speeding up all reactions, an enzyme is used to lower the activation energy (CITE:2) It does this by stretching out the substrate so that it is more unstable, making it

easier to reach the transition state (CITE: 3) To reach it, free energy is absorbed,

which breaks bonds and forms new ones. By the enzyme lowering the activationenergy barrier, less free energy is needed to be absorbed as activation energy. This

will not affect the free-energy change of a reaction whether an is used or not.

(CITE: 2)

Sulfuric acid was used to stop decomposition in the experiment because ofits ability to denature catalase. Catalase and other enzymes will denature as a

result of a drop in pH of their surroundings. Since catalase is not in its optimal

pH, hydrogen bonds, ionic bonds, and other weak forces are broken, changing theenzymes three-dimensional structure. This prevents further catalyzation of the

decomposition of the H2O2 with this enzyme, inhibiting the function of catalase.

(TALK MORE ABOUT THIS!!) (Tertiary structure too!)By lowering the temperature of an enzymes environment, the rate of

reaction will also decrease. This is because enzymes have an optimum

temperature to function and if it is significantly dropped or raised, the rate of

reaction will slow down. However, icrease in temperature generally almost

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always increases reaction rate until a certain point. Since the particles would be

moving so slowly because of the cold, few collisions would occur. (TALK MORE

ABOUT THIS!!)Hypothesis proved or disproved

The graph shows the amount of H2O2 that has reacted as a function of

time. After the third measured point, the graph shows a decrease in the reactionrate representing the decrease of available H2O2for decomposition. The dotted

line extending from the third point shows how the amount of H2O2 decomposedwould continue to increase at a constant rate if H2O2supplies did not diminish.

Sources of Error:

Misreading values of reactants used

Wrong amounts of solutions added

Further Study: