substrate concentration and rate of enzyme raction

Published:

Research Question

How does different concentration of hydrogen peroxide (H202 ) may influence the rate of liver (catalase ) reaction ?

Objective

To investigate how substrate concentration may influence the rate of above reaction

Introduction

Chemical reaction in general can be speed up in a number of ways. The important ways are increase the temperature until it reaches the optimum temperature and also by using a catalyst. A catalyst is a substance that accelerates the rate of reaction without being used up itself at it produced another pathway for the reaction to occur. Hydrogen peroxide, H202, is a very reactive chemical which is formed as a by product in cellular respiration. This substance will break down into two harmless substances, water and oxygen.

catalassesese

2H2O2 2H2O + O2

An enzyme, catalase which is found in most tissue from living organism can be use to accelerate the formation of water and oxygen from hydrogen peroxide, H2O2 molecule

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Hyphotesis

When the concentration of hydrogen peroxide (H2O2 ) increases, the rate of liver ( catalase) reaction also increases

Variables

Independent : Different concentration of hydrogen peroxide

Dependent : Rate of catalase reaction

Controlled :

Controlled variables

How to control the variable

1. Mass of liver used

Mass of liver used is constant, 1.00g is measured using electronic balance

2. Initial volume

Initial volume mixture of detergent, hydrogen peroxide and liver is constant, 5 ml

3.Amount of detergent used

2 drops of detergent is used for each experiment

4. Time taken to observe the reaction

30 seconds is constant time to observe the reaction throughout the experiment

Materials

Solution of H2O2 % ( 0.5, 1.0, 2.0, 3.0, 4.0, 6.0) , detergent, distilled water, liver

Apparatus

Stopwatch, forceps, knife, 10 ml measuring cylinder, 100 ml measuring cylinder

Method

1. 7 cubes of liver are cut approximately 1cm Ã- 1cm Ã- 1cm

2. Small measuring cylinder is used to measure 4 ml of 0.5 % of H2O2 with 2 drops of detergent 100 ml measuring cylinder and swirled to mix

3. Using the forceps, one cube liver is taken and placed it in measuring cylinder and immediately stopwatch is started

4. Initial volume of mixture is recorded

5. After 30 second, final volume of mixture is recorded in a table

6. The procedure is repeated with other solutions of H2O2 and also for distilled water as control. Make sure the glassware is cleaned carefully between the procedures

7. This procedure is repeated twice

Data Collection

Qualitative Data

1. Foam was formed and felt a bit of warm when touching the measuring cylinder

2. Reaction is more vigorous when the concentration of H2O2 increases

3. The foam produced is white in colour.

4. The production of foam is increased as the concentration of H2O2 increased.

Quantitative Data

Concentration

Of H2O2 solution,

( % )

Initial volume,ml

±0.5ml

Final Volume of foam produced, ml ( ± 0.5 ml)

1stGroup

2ndGroup

3rdGroup

4thGroup

5thGroup

6thGroup

1st

2nd

1st

2nd

1st

2nd

1st

2nd

1st

2nd

1st

2nd

Distilled water

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

0.5

5.0

15.0

14.0

15.0

17.0

18.0

19.0

15.0

16.0

15.0

15.0

18.0

22.0

1.0

5.0

20.0

18.0

20.0

20.0

19.0

21.0

20.0

20.0

23.0

22.0

23.0

24.0

2.0

5.0

21.0

23.0

25.0

25.0

24.0

24.0

25.0

25.0

24.0

26.0

28.0

27.0

3.0

5.0

30.0

29.0

37.0

40.0

42.0

40.0

29.0

31.0

37.0

35.0

50.0

45.0

4.0

5.0

36.0

33.0

48.0

41.0

46.0

56.0

35.0

38.0

41.0

41.0

52.0

58.0

6.0

5.0

55.0

47.0

55.0

58.0

59.0

56.0

46.0

39.0

58.0

50.0

62.0

59.0

Table 1: The concentration of H2O2 solution and the initial & final volume of foam formed

In order to determine the volume of foam produced, the calculation is as follows:

Volume of foam produced = ( Final volume ) - ( Initial volume )

Example for 0.5 % of H2O2 ;

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= ( 15.0 ) - ( 5.0 )

= 10.0 cm³

Data Processing

Concentration

Of H2O2 solution,

( % )

Volume of foam produced, cm³

1stGroup

2ndGroup

3rdGroup

4thGroup

5thGroup

6thgroup

1st

2nd

1st

2nd

1st

2nd

1st

2nd

1st

2nd

1st

2nd

Distilled water

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

5.0

0.5

10.0

9.0

10.0

12.0

13.0

14.0

10.0

11.0

10.0

10.0

13.0

17.0

1.0

15.0

13.0

15.0

15.0

14.0

16.0

15.0

15.0

18.0

17.0

18.0

19.0

2.0

16.0

18.0

20.0

20.0

19.0

19.0

20.0

20.0

19.0

21.0

23.0

22.0

3.0

25.0

24.0

32.0

35.0

37.0

35.0

24.0

26.0

32.0

30.0

45.0

40.0

4.0

33.0

28.0

43.0

36.0

41.0

51.0

30.0

33.0

36.0

36.0

47.0

53.0

6.0

50.0

42.0

50.0

53.0

54.0

51.0

41.0

34.0

53.0

45.0

57.0

54.0

Table 2: the volume of foam produced

To calculate the average volume of foam produced, the following formula is used:

Average volume of foam produced

= ( 1st Group ) + ( 2nd Group ) + ( 3rd Group ) + ( 4th Group ) + ( 5 Group ) + ( 6th Group )

Total number of readings

Example for 0.5 % is as follows:

Average volume of foam produced

= 10.0 + 9.0 + 10.0 + 12.0 + 13.0 + 14.0 + 10.0 + 11.0 + 10.0 + 10.0 + 13.0 + 17.0

12

= 11.6

The average volume of foam produced in other concentrations is as follows:

Concentration of H2O2,

%

Average volume of foam produced, ml

Distilled Water

0.0

0.5

11.6

1.0

15.8

2.0

19.8

3.0

32.1

4.0

38.9

6.0

48.7

Table 3:The table of average of total volume of the foam

Standard Deviation

∆ V:

Formula :

∑x²= volume of foam produced

n = no of data

x2 = mean volume foam produced

Sample calculation:

0.5 % H2O2

√ 10² + 9² + 10² + 12² + 13² + 14² + 10² + 11² + 10² + 10² + 13² + 17² - ( 11.6)²

12

= 2.13

Concentration of H2O2 solution

Standard deviation of average volume of foam produced

0.0%, distilled water

0.00

0.5 %

2.13

1.0%

2.01

2.0%

1.02

3.0%

6.33

4.0%

7.86

6.0%

6.23

Table 3: Concentration of H2O2 solution and standard deviation of average volume of foam produced

Rate of reaction = ( Average volume of foam - 5 ) ml

30 Seconds

Example for rate of reaction for 0.5% of H2O2:

Rate of reaction

= ( 11.6 -5 ) ml

30 seconds

= 0.220

The rate of reaction of other concentrations is as follows:

Concentration of H2O2 solution

Rate of catalase reaction (ml s-1 )

0.0 %, distilled water

0.0000

0.5 %

0.220

1.0%

0.360

2.0%

0.493

3.0%

0.900

4.0 %

1.130

6.0 %

1.457

Table 4 : Concentration of H2O2 solution and rate of catalase reaction

Standard error

Formula :

∆ ROR = ∆ V + ∆ t

ROR V t

∆ ROR = [ ∆ V + ∆ t ] ROR

V t

∆ ROR = [ ∆ V + 0.1 ] ROR

V 30

Sample of calculation:

2.13 + 0.1 Ã- 0.220

11.6 30

= 0.041

Concentration of H2O2 solution

Standard error for rate of catalase reaction

0.0%, distilled water

0.000

0.5 %

0.041

1.0%

0.047

2.0%

0.027

3.0%

0.180

4.0%

0.232

6.0%

0.191

Table 5 : Concentration of H2O2 solution and standard error for rate of catalase reaction

Discussion

Based on the result of experiment, when distilled water used as substrate, no reaction occurs. It is proved when the initial volume is equal to the initial volume, 5 ml. This is due to the absence of hydrogen peroxide, H2O2 so water and oxygen form thus, result no volume of foam produced. This is act as control of this experiment. In 0.5 % of hydrogen peroxide, H2O2 the rate of catalase reaction is very slow, 0.39 ml s-1 . This condition is happened because the substrate presence in small quantity. From 1% to 4 % of hydrogen peroxide, H2O2 the rate of catalase reaction increases. This is due to more substrate that act as catalyst. During this time, more collision happened between hydrogen peroxide molecule and active site of catalase enzyme in liver. In 6% of hydrogen peroxide H2O2 , the reaction rate is the highest rate as the collision between hydrogen peroxide, H2O2, molecule and active site of the enzyme happened with the greatest frequency. Catalyst provides an alternative path by lowering the activation energy required for the reaction.

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Limitation

1. The cube sizes of the livers were not 1cm3 so the surface areas of the livers are different and do not have same mass. This condition can affect the accuracy of the result by getting inaccurate rate of catalase reaction .

2. The livers used were not fresh and cold, thus, will slow down the reaction rate. The reaction is slow is due to the less effective collision happened between the hydrogen peroxide molecule and active site of catalase enzyme in liver.

3. The measuring cylinder used is too big. Distribution of foam in the measuring cylinder is too big which result inaccuracy.

4. The fan is opened during the measuring of liver using electronic balance. Presence of air will affect the accuracy mass of liver that being measured.

Suggestion

1. Use sharp knife and forcep to slice the livers so that the the liver cube sizes are fixed. Then, measure the mass of liver accurately

2. Use the fresh livers which are fresh and do not been frozen

3. Use a smaller measuring cylinder or use measuring cylinder with narrower scale.

4. Switch off the fan throughout the measuring mass of liver process

Conclusion

The rate of catalase reaction increases as the substrate increases from 0.0 % to 6 %. Hence, the rate of catalase reaction increases from 0.00 mls-1 to 1.62 mls-1. The hypothesis is accepted.