substrate concentration and rate of enzyme raction Essay

Research Question

How does different concentration of H peroxide ( H202 ) may act upon the rate of liver ( catalase ) reaction?

Aim

To look into how substrate concentration may act upon the rate of above reaction

Introduction

Chemical reaction in general can be speed up in a figure of ways. The of import ways are increase the temperature until it reaches the optimal temperature and besides by utilizing a accelerator.

A accelerator is a substance that accelerates the rate of reaction without being used up itself at it produced another tract for the reaction to happen. Hydrogen peroxide, H202, is a really reactive chemical which is formed as a by merchandise in cellular respiration. This substance will interrupt down into two harmless substances, H2O and O.catalassesese2H2O2 2H2O + O2An enzyme, catalase which is found in most tissue from populating being can be use to speed up the formation of H2O and O from H peroxide, H2O2 moleculeHyphotesisWhen the concentration of H peroxide ( H2O2 ) increases, the rate of liver ( catalase ) reaction besides increasesVariablesMugwump: Different concentration of H peroxideDependent: Rate of catalase reactionControlled:Controlled variablesHow to command the variable1.

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Mass of liver usedMass of liver used is changeless, 1.00g is measured utilizing electronic balance2. Initial volumeInitial volume mixture of detergent, H peroxide and liver is changeless, 5 milliliter3.Amount of detergent used2 beads of detergent is used for each experiment4.

Time taken to detect the reaction30 seconds is changeless clip to detect the reaction throughout the experimentMaterialsSolution of H2O2 % ( 0.5, 1.0, 2.0, 3.0, 4.0, 6.

0 ) , detergent, distilled H2O, liverApparatusStopwatch, forceps, knife, 10 milliliter mensurating cylinder, 100 milliliter mensurating cylinderMethod1. 7 regular hexahedrons of liver are cut about 1cm A- 1cm A- 1cm2. Small mensurating cylinder is used to mensurate 4 milliliter of 0.

5 % of H2O2 with 2 beads of detergent 100 milliliter mensurating cylinder and swirled to blend3. Using the forceps, one regular hexahedron liver is taken and placed it in mensurating cylinder and instantly stop watch is started4. Initial volume of mixture is recorded5.

After 30 2nd, concluding volume of mixture is recorded in a tabular array6. The process is repeated with other solutions of H2O2 and besides for distilled H2O as control. Make certain the glasswork is cleaned carefully between the processs7.

This process is repeated twiceData CollectionQualitative Data1. Foam was formed and felt a spot of warm when touching the measurement cylinder2. Chemical reaction is more vigorous when the concentration of H2O2 additions3. The froth produced is white in coloring material.4.

The production of froth is increased as the concentration of H2O2 increased.Quantitative Data

Concentration

Of H2O2 solution,

( % )

Initial volume, milliliter

A±0.5ml

Concluding Volume of froth produced, ml ( A± 0.

5 milliliter )

1stGroup

2ndGroup

3rdGroup

4thGroup

5thGroup

6thGroup

1st

2nd

1st

2nd

1st

2nd

1st

2nd

1st

2nd

1st

2nd

Distilled H2O5.05.05.05.05.05.

05.05.05.05.05.

05.05.00.55.

015.014.015.017.018.

019.015.016.015.015.018.022.01.

05.020.018.020.020.019.

021.020.020.023.

022.023.024.

02.05.021.023.025.

025.024.024.

025.025.024.026.028.027.03.05.

030.029.037.040.042.040.

029.031.037.

035.050.045.04.05.036.033.048.

041.046.056.035.038.041.041.

052.058.06.05.055.047.

055.058.059.056.046.039.

058.050.062.059.0

Table 1: The concentration of H2O2 solution and the initial & A ; concluding volume of froth formed

In order to find the volume of froth produced, the computation is as follows:Volume of froth produced = ( Final volume ) – ( Initial volume )Example for 0.

5 % of H2O2 ;= ( 15.0 ) – ( 5.0 )= 10.0 cmA?Datas Processing

Concentration

Of H2O2 solution,

( % )

Volume of froth produced, cmA?

1stGroup

2ndGroup

3rdGroup

4thGroup

5thGroup

6thgroup

1st

2nd

1st

2nd

1st

2nd

1st

2nd

1st

2nd

1st

2nd

Distilled H2O

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 froth produced

To cipher the mean volume of froth produced, the undermentioned expression is used:Average volume of froth produced= ( 1st Group ) + ( 2nd Group ) + ( 3rd Group ) + ( 4th Group ) + ( 5 Group ) + ( 6th Group )Entire figure of readingsExample for 0.5 % is as follows:Average volume of froth produced= 10.0 + 9.0 + 10.0 + 12.0 + 13.0 + 14.

0 + 10.0 + 11.0 + 10.0 + 10.0 + 13.

0 + 17.012= 11.6The mean volume of froth produced in other concentrations is as follows:

Concentration of H2O2,

%

Average volume of froth produced, milliliter

Distilled Water0.00.511.61.

015.82.019.

83.032.14.

038.96.048.7

Table 3: The tabular array of norm of entire volume of the froth

Standard Deviationa?† V:Formula:a?‘xA?= volume of froth producedn = no of informationsx2 = mean volume froth producedSample computation:0.

5 % H2O2a?s 10A? + 9A? + 10A? + 12A? + 13A? + 14A? + 10A? + 11A? + 10A? + 10A? + 13A? + 17A? – ( 11.6 ) A?12= 2.13

Concentration of H2O2 solution

Standard divergence of mean volume of froth produced

0.0 % , distilled H2O0.

000.5 %2.131.0 %2.012.0 %1.023.

0 %6.334.0 %7.866.0 %6.23

Table 3: Concentration of H2O2 solution and standard divergence of mean volume of froth produced

Rate of reaction = ( Average volume of foam – 5 ) milliliter30 SecondsExamples for rate of reaction for 0.

5 % of H2O2:Rate of reaction= ( 11.6 -5 ) milliliter30 seconds= 0.220The rate of reaction of other concentrations is as follows:

Concentration of H2O2 solution

Rate of catalase reaction ( ml s-1 )

0.0 % , distilled H2O0.00000.5 %0.2201.

0 %0.3602.0 %0.4933.

0 %0.9004.0 %1.1306.0 %1.

457

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

Standard mistakeFormula:a?† ROR = a?† V + a?† TROR V Ta?† ROR = [ a?† V + a?† T ] RORV Ta?† ROR = [ a?† V + 0.1 ] RORV 30Sample of computation:2.13 + 0.1 A- 0.22011.6 30= 0.041

Concentration of H2O2 solution

Standard mistake for rate of catalase reaction

0.0 % , distilled H2O0.

0000.5 %0.0411.0 %0.0472.0 %0.

0273.0 %0.1804.

0 %0.2326.0 %0.191

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

DiscussionBased on the consequence of experiment, when distilled H2O used as substrate, no reaction occurs. It is proved when the initial volume is equal to the initial volume, 5 milliliter. This is due to the absence of H peroxide, H2O2 so H2O and O signifier therefore, result no volume of froth produced.

This is act as control of this experiment. In 0.5 % of H peroxide, H2O2 the rate of catalase reaction is really slow, 0.39 ml s-1. This status is happened because the substrate presence in little measure. From 1 % to 4 % of H peroxide, H2O2 the rate of catalase reaction additions.

This is due to more substrate that act as accelerator. During this clip, more hit happened between H peroxide molecule and active site of catalase enzyme in liver. In 6 % of H peroxide H2O2, the reaction rate is the highest rate as the hit between H peroxide, H2O2, molecule and active site of the enzyme happened with the greatest frequence. Catalyst provides an alternate way by take downing the activation energy required for the reaction.Restriction1. The regular hexahedron sizes of the livers were non 1cm3 so the surface countries of the livers are different and do non hold same mass.

This status can impact the truth of the consequence by acquiring inaccurate rate of catalase reaction.2. The livers used were non fresh and cold, therefore, will decelerate down the reaction rate.

The reaction is slow is due to the less effectual hit happened between the H peroxide molecule and active site of catalase enzyme in liver.3. The mensurating cylinder used is excessively large. Distribution of froth in the measurement cylinder is excessively large which consequence inaccuracy.4. The fan is opened during the measurement of liver utilizing electronic balance.

Presence of air will impact the truth mass of liver that being measured.Suggestion1. Use crisp knife and forcep to slit the livers so that the the liver regular hexahedron sizes are fixed. Then, mensurate the mass of liver accurately2. Use the fresh livers which are fresh and do non been frozen3. Use a smaller measurement cylinder or usage mensurating cylinder with narrower graduated table.

4. Switch over off the fan throughout the mensurating mass of liver procedureDecisionThe rate of catalase reaction additions as the substrate additions from 0.0 % to 6 % .

Hence, the rate of catalase reaction additions from 0.00 mls-1 to 1.62 mls-1. The hypothesis is accepted.

x

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