Practical 2

UNIVERSITI TEKNOLOGI MARA KAMPUS KUALA TERENGGANU

BIO150 METABOLISM AND CELL DIVISION LABORATORY REPORT TITLE

: THE RATE OF CATALASE REACTION

EXP. NO.

: PRACTICAL 2

NAME

:AHMAD MIRZA RUSYAIDI BIN MOHD RAFI (2011287166)

NAME

:NAZIRUL FAHMI BIN ZULKIFLI (2011408446)

GROUP

: AS120 2 A

DATE

: 12 DECEMBER 2011

LECTURER

: PUAN SARINA

OBJECTIVES

1. To investigate the rate at which the enzyme catalase converts substrate into product. 2. To determine how much substrate has been broken down by catalase at varying time.

INTRODUCTION

Enzyme catalase the reactions by lowering the activation energy necessary for a reaction to occur.The molecule than a enzyme acts on is called the substrate.In an enzyme mediated reaction,substrate molecules are change and the product is formed.

In this experiment you will investigate the rate at which the enzyme catalase converts substrate to product. You will allow catalase to react with hydrogen peroxide for varying amounts of time and then stop the reactions by adding H 2S04.To determine the amount of hydrogen peroxide that remains after the reation, you will do a titration with KMn04.

MATERIALS

Catalase Hydrogen peroxide (H 202) Hydrogen sulphide (H 2S04) KMn04 Pipette 10 ml Beaker 50 ml Burette Conical flask 250 ml Measuring cylinder 10 ml

PROCEDURE

1. 10ml of H2S04 was added to each of 7 beakers. 2. 1ml of catalase was added to the first beaker at 0 second. 3. The reaction was allowed to occur for the time shown on the label. 4. The reaction was stopped by adding 10ml of H2S04 after the time period. 5. The procedure was repeated for each beaker at 10 sec, 30 sec, 60 sec, 120 sec, 180 sec, 360 sec.

Amount of hydrogen peroxide (substrate) that has been broken down by catalase at varying times was determined by measuring the amount of peroxide remaining in each flask.

1. 5ml of sample was removed and transferred to a clean conical flask. 2. Initial burette reading of KMn04 was recorded. 3. KMn04 was added until a faint brown colour persists. That was the end of the titration.(The more KMn04 used, the more peroxide was in the flask) All of the data was recorded in Table 2.1

Beaker

Time

Titration (The amount of KMn04 used)

The rate of enzyme reaction (moles/second)

1

0

14.8

0

2

10

14.5

-0.225

3

30

14.0

-0.067

4

60

13.8

-0.032

5

120

13.0

-0.013

6

180

12.2

-0.11

7

360

7.8

0.003

Table 2.1: Titration of hydrogen peroxide

CALCULATIONS

2H202

2H20 + 02

2 moles H202 react with 1 mole 02 1 mole H202 react with ½ mole 02 Moles of Product

Beaker 1   

 





 

Beaker 2   

 





 

Beaker 3   

 





 

Beaker 4   

 





 

Beaker 5   

 





 

Beaker 6    

 



 

Beaker 7    

 



 

ANALYSIS OF RESULT

Graph of rate of enzyme reaction vs time 0.05

Rate of Enzyme Reaction

r

0 0

10

30

60

120

180

360

-0.05

-0.1 Rate Of Enzyme Reaction

-0.15

-0.2

time -0.25

DISCUSSION

Enzymes is a protein molecules that act as a biological catalyst. It can speed up the chemical reaction by lowering the activation energy. In this experiment, liver was used as the catalase because liver is the place that contains the most enzymes than other parts of body. In this experiment, hydrogen peroxide (H 202) was used because pH value of this solution is very optimum for the enzymes to work efficiently, which is pH 6. The manipulated variable in this experiment was time and the responding variable was the rate of enzyme reaction. Time was manipulated by adding hydrogen sulphide into the solution. Hydrogen sulphide used because its pH value is not optimum for the enzymes to work, so the reaction between the enzymes with hydrogen peroxide can be stopped immediately.

When we put the liver into the solution, enzymes that contained in the liver reacted with hydrogen peroxide. We could see the air bubbles released from the liver. The longer time period, the more the bubbles released. This showed that the longer the time period, the more the reaction between the enzymes and the hydrogen peroxide. The product that produced also higher. We took 5ml of the solution to be titrated with KMn0 4. The solution that contained more products will need less KMn0 4 solution to change its colour into faint brown colour. The rate of enzyme reaction was determined by the amount of KMn0 4 used. The error that happened during the titration process that affected our result of the experiment was the technique of the titration. We should standardize the titration in order to get the accurate result.

After we got all the result for the amount of KMn04 for the titration, we calculated the moles of product and then the rate of enzyme reaction by using the formulae given. The amount of KMn04 that been used suppose to be lower than 10ml, but we got higher than 10 ml. This affected the calculation to find the moles of product as we got negative value except for beaker 7. The negative value of the moles of product then affected the calculation to find the rate of enzymes reaction as we also got negative value for the rate of enzyme reaction. The error that might happened that caused the inaccurate result was the technique when taking the reading of the beaker and the burette. We should place our eyes parallel to the scale in order to the get accurate value.

Based on the graph that we plotted,we can conclude that the result that we achieved from this experiment was not accurate. This is because the graph of rate of enzymes reaction suppost to be increasing linearly to time. There were some errors that happened during the experiment. First, the size of the liver that we used to be put in the beaker was not exactly the same. So, this affected the result of the experiment because small size of liver could react faster than the liver that had a larger size. We should use the exact same size of the liver to make sure the size of the liver did not affect our result.

QUESTION

1. During what time interval is the enzyme working at its maximum velocity? At the first time interval. This is because at during this time interval, the solution has the highest concentration of substrate. The enzyme can react with the substrate at its greatest quantity. The rate of reaction are continuing to decrease until the last time interval because the concentration of substrate left decrease.

2. Suggest what should be done in order to keep the rate constant over the entire time course. By increase the concentration of the substrate. The rate of reaction will increase as the concentration of substrate increase until the enzyme reach its maximum point. When it reach its maximum point, the rate of reaction will stay constant eventhough we increase the concentration of substrate.

3. What is the role sulfuric acid (H 2S04) in this experiment? As we know, in order to enzyme works efficiently, the pH value of the solution needs to be very optimum. During the experiment, the pH value of the hydrogen peroxide is very optimum to the enzyme. Sulphuric acid (H 2S04) used to stop the reaction of the enzyme with hydrogen peroxide because the pH value of the sulphuric acid is not optimum to the enzyme to work which is below than pH 7. It can stop the reaction immediately.

CONCLUSION

Based on the experiment that we had conducted, we can conclude that the rate of enzymes reaction depends on the period of time that given to the enzymes to react with their substrates. The longer the time given to the enzymes to react with the substrates, the faster the rate of enzymes reaction. Although the result that we got from our experiment were not accurate, we had identified the errors that might happened during the experiment and hopefully this can improve our technique for the next experiment.

REFERENCES



http://en.wikipedia.org/wiki/Hydrogen_peroxide



Laboratory Manual Biology 150



Solomon Berg Martin Biology 9 th Edition