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Breakdown of H202 by catalase

Asked by payne6 | Jan 19, 2006 | A Level > Chemistry > Coursework
payne6

Investigating the effect of enzyme concentration on the rate of breakdown of hydrogen peroxide by catalyse.

Any ideas will be welcome on how do go about this. :D

SZPoints: 6 | Difficulty: Hard | Comments: 13 | Answered (goto answer)

12 Comments, 1 Answer

suzie_1, 19 Jan, 2006 @ 23:04
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suzie_1
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Enzymes such as Catalase are protein molecules, which are found in living cells. They are used to speed up specific reaction within the cell. They are all very specific as each enzyme just performs one particular reaction.

Catalase is an enzyme found in food such as potato and liver. It is used for removing Hydrogen Peroxide from cells. Hydrogen Peroxide is the poisonous by-product of metabolism; this compound can kill cells. Catalase speeds up the decomposition of Hydrogen Peroxide into water and oxygen because the shape of its active site matches the shape of hydrogen peroxide molecule. This type of reaction where a molecule is broken down into smaller pieces is called a catabolic reaction.coeg egr seegegw oreg egk ineg foeg eg.

This is the reaction equation for the breakdown of hydrogen peroxide into water and oxygen.

Catalase

2H2O2 2H2O + O2

The higher the temperature, the higher the rate of reaction up to a certain point. This is because as particles gain heat they move about much more with more vigour, this increases the chances of a successful collision. The reacting particles will collide more frequently and with more energy, this will increase the rate of reaction. However, enzymes have a specific shape that is important for their reaction. If the amino acid molecules gain too much energy and move about too much, the hydrogen and ionic bonds holding the chain of the enzyme in its 3D shape will break. This means that the enzyme can no longer act as a catalyst, i.e. the catalyst would be denatured.

When the concentration of Hydrogen Peroxide is increased, the rate of reaction increases at a directionally proportional rate until the concentration of Hydrogen Peroxide reaches about 10vol. If the concentration were doubled, I would expect the amount of Oxygen released to be a figure twice as much.
From 25vol to 10vol shows a directionally proportional decrease in reactivity rate, after 10vol the rate of reaction slows down. This is shown by the gradient on the graph going down. At this point virtually all active sites are occupied making the active sites saturated with Hydrogen Peroxide. With an increase the concentration of Hydrogen Peroxide, the number of active sites increases, hence, makes a more violent reaction (Quicker).


hope it helps

XLNC, 19 Jan, 2006 @ 23:26
Accepted Answer
XLNC
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Why did man invent biology, its a pain! mwahaha

Okay, I just happen to have done the very same coursework a year ago.But the unfortunate thing is, I only  have ''bits'' of it. I'm going to copy paste the work I have for your use. Oh, and if you do get arrest by the police for plagarism, i'll bail you out :)

Introduction:

In this experiment, I am going to investigate how different enzyme concentrations affect the rate of reaction. I am going to use different liver masses as a source of Catalase (enzyme) and Hydrogen Peroxide as the substrate.
Hydrogen Peroxide is a chemical compound of Hydrogen and Oxygen with the formula H2O2. Pure, anhydrous Hydrogen Peroxide is a colourless, syrupy liquid that is a strong oxidizing agent (which is harmful) and, in water solution, a weak acid. It has an exact gravity of 1.44. It blisters the skin. When reacting with Catalase it gives off Oxygen and water.
Catalase behaves as a catalyst for the conversion of hydrogen peroxide into water and Oxygen. Catalase is an example of a particularly efficient enzyme. Catalase is a biological enzyme, which speeds up a chemical reaction. Catalase is a biological enzyme that brings about (catalyses) the reaction by which hydrogen peroxide is decomposed to water and oxygen. Catalase is found in the liver. Catalase prevents the build up of and protects the body tissues from damage by Peroxide, which is continuously produced by numerous metabolic reactions (the chemical reactions that take place within each cell of a living organism). The Catalase converts the Hydrogen Peroxide (which is harmful) to Water and Oxygen (which are harmless).
In order to do this experiment I am going to use different liver masses as a source of the enzyme Catalase (these are 0g- control, 1g, 2g, 3g, 4g and 5g) on the substrate, Hydrogen peroxide (20 Vol). Hence the rate of reaction can be measured in terms of how much Oxygen is produced in the given time. An enzyme molecule contains many active sites, the area to which the substrate molecules bind to. The active site is precisely the right shape for the substrate molecules to slot into. All enzymes have a substrate molecule, which has a complimentary shape to the enzyme. As molecules of Hydrogen Peroxide collide into the active site of Catalase, temporary bonds form between the substrate and enzyme. As these bonds form, the Hydrogen Peroxide molecules are pulled into a slightly different shape, which makes it easier for the bonds to be broken. These bonds break and result in a product being formed. The products formed are Oxygen and Water. Due to the fact that Oxygen is a gas, it is easy to see the Oxygen being formed, in contrast to Water, so I will measure the volume of Oxygen being produced.

Aim:

To investigate the impact of different enzyme concentrations on the rate of reaction

Hypothesis:
Activation energy is needed by every reaction to occur. A high enzyme concentration inevitably increases the rate of reaction. There is a directly proportional ratio (positive correlation) between the two, until a peak rate of reaction occurs where the rate cannot increase any further. This will produce a straight-line graph through the origin, which will eventually flatten off at peak rate. 
 

Alternative hypothesis:
Having acknowledged the background information, I predict that as the concentration of the enzyme Catalase increases, the rate of reaction will increases proportionally up to a point where the reaction cannot proceed.
As 0.5g being the highest mass (enzyme concentration), I predict it will hydrolyse the Hydrogen Peroxide the fastest in contrast to all the other masses, so more product is produced. This is because the higher the mass of the liver, the more enzyme molecules in the solution, and therefore a large number of active sites available. This means that there is a greater chance of a substrate molecules colliding with an active site, resulting in more successful collisions as there are more active sites available.
As 0.1g being the lowest mass (enzyme concentration), I predict it will hydrolyse the Hydrogen Peroxide the slowest in contrast to all the other masses, so less product is produced. This is because the lower the mass of the liver, the less enzyme molecules in the solution, and therefore a small number of active sites available. This means that there is a little chance of a substrate molecules colliding with an active site, resulting in less successful collisions as there are less active sites available.

I predict that as the liver masses increase (numerically), the volume of Oxygen will increase by a certain amount, rougly by 10 ml each time. I also predict that the reaction will slow down and eventually stop at a point as the substrate concentration will be the limting factor. This will cause the graph curve off

Equipement and Usage ( it was acutally in a table format but it didnt come out!)


Liver (masses of 0g- control. 1g, 2g, 3g, 4g, and 5g)
Required for investigating how different enzyme concentrations affect the rate of reaction.
1 Knife Used to cut the liver into tiny cubes of different masses.
1 Plastic tile To aid in cutting the liver.
Hydrogen Peroxide – (20 Vol) (Volume = 3cm3) Used as the substrate to produce the product, which in turn determines the rate of reaction.
1 Syringe 5ml Used for accurately measuring the volumes of the enzyme and substrate.
1 Stop watch Used to note the time of reaction between enzyme and substrate. Used to obtain a fair test.
Buffer of the pH of 7
(Volume = 1cm3) Used to keep the pH constant to obtain a fair test.
1 Plastic measuring cylinder (100ml) Used to measure the amount of Oxygen produced
1 Trough (bucket) Filled with water to assure no water escapes from the plastic measuring cylinder. This means that the Oxygen can be read of accurately as water is displaced.
1 Weighing scale Used to accurately weight different masses of liver.
1 Delivery/Rubber tube with stopper. Used to transport the Oxygen to the plastic measuring cylinder
Goggles Used to protect the eyes from Hydrogen Peroxide.
3 beakers Used as a ‘container’ to store the buffer, Hydrogen Peroxide and liver, once they have been measured.

Method: 
„Ï Set up the conical flask with a stopper and delivery/rubber tube connected to it.
„Ï Get hold of a knife and cut out 1g of liver (Catalase) on the plastic tile, weigh it to confrim its size.
„Ï Measure 1cm3 of the buffer solution by using a 5ml syringe for accuracy and place it in a sutiable container.
„Ï Measure 3cm3 of Hydrogen Peroxide (20 Vol– standard state) using a 5ml syringe for accuracy. Now place the Hydrogen Peroxide in a suitable container (take care when handling the Hydrogen Peroxide).
„Ï Fill the trough and the plastic measuring cylinder with water. Care should be taken to avoid any 'gaps' between the trough and the plastic measuring cylinder. Simultaneously make sure the delivery/rubber tube is in contact with the plastic measuring cylinder, as shown in the diagram.
„Ï Now place the buffer, the 1g of liver (Catalase) and Hydrogen Peroxide in the conical flask. Make sure the stopper is thoroughly tightened.
„Ï Get hold of a stop watch and time for 1 minute. At the end of the 1 minute period, note down the volume of Oxygen produced.
„Ï Now remove the stopper and delivery/rubber tube and thoroughly wash the conical flask before investigating the next experiment
„Ï Repeat the test three times for each liver mass. 18 readings must be taken in total (liver masses : 0g – control, 1g, 2g, 3g, 4g, 5g)

Note: For the control – (0g of liver), instead of liver (Catalase) add water.

Getting useful and reliable results:

As mentioned in the method, 18 readings in total must be taken. This is because an average can be worked out for the volume of Oxygen collected, giving me reliable results. A control will enable me to see the facts. It will help me to prove that it is the enzyme concentration that affects the rate of reaction.

The Independent Variables
This is a factor which is changed in the experiment. In this case, the independent variables are the liver masses (enzyme concentration).

The controlled Variable
The controlled variable is the factor that I will keep constant throughout the experiment. In this case the controlled variables are: the volume of Hydrogen Peroxide, timing, pH and substrate concentration.
In order to obtain a fair test, the following factors need to be controlled:
• The pH – The pH can slow down the enzyme hydrolysing the substrate. Extremes of pH can denature the enzyme so the reaction would not proceed. This will lead to an unfair test if not controlled.
• The substrate concentration – If the substrate concentration is increased each time, reactions will occur very fast. However if the substrate concentration is decreased each time, reactions will occur very slow. This will lead to unreliable results. So controlling the substrate concentration is very important in order to gain a fair test and good results.
• The volume of substrate (Hydrogen Peroxide) added – If the volume of the substrate is increased, this means more of it has been added so again, reactions will occur fast. However if the volume of the substrate is decreased, this means less of it has been added so reactions occur slowly. So the volume of the substrate need to be controlled.
• Time – It is important that the volume of Oxygen produced is always noted after the 1 minute interval. This will assure a fair test.
All of these variables will need to be controlled by me, and to be kept the same throughout my experiment, in order to keep it fair and for the results to be accurate. I am only testing the effect of enzyme concentration on the rate of reaction so the other factors will have to remain the same, so that they do not affect my results.
I have chosen a buffer with of the pH of 7 as this very close to the optimum pH of the enzyme Catalase ( pH of 7.6). The buffer will keep the pH constant throughout the experiment. As 7 is close to the optimum pH of Catalase, this means that the enzyme will work the efficiently in hydrolysing the Hydrogen peroxide, providing me with accurate results.
I choose the time interval to be 1 minute as this is the best time in providing accurate results. At this time interval the appropriate amount of Oxygen is produced i.e. not a lot, nor to little. This means that the amount of water displaced in the plastic measuring cylinder can be read accurately.
I did not use a water bath as the reaction between Catalase and Hydrogen Peroxide is exothermic. This means if attended to put in the water bath, additional heat will being given out changing the temperature of the water bath itself. This will lead to an unfair test.

Accuracy:
To get accurate measurements of the buffer and substrate I will use a 5ml syringe. To control the substrate and buffer I will use the same volume and concentration each time, so that it will not affect my results. I have decided to control the pH and keep it the same so that this will not affect the rate of reaction by using a buffer. The time will also be controlled by using a stop watch to assure a fair test.

From here your on your OWN pal. I dont have rest! But it follows on as ''results, conclusion and evaluation''

Hope it helps. Best regards.

chinaman5, 20 Jan, 2006 @ 08:51
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chinaman5
send txt message to chinaman5 deepskybluestar (26) chinaman5's blog
..................................omg
payne6, 20 Jan, 2006 @ 17:02
Author Comment
Hey
Thank you for your replies.
You should not have posted your coursework man, it makes me feel as though I have a massive unfair advantage.
Both of your replies were exactly what I needed to know.
100 points would be given to both of you, but that not possible to do here.
suzie_1, 20 Jan, 2006 @ 18:18
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suzie_1
send txt message to suzie_1 bronzestar (78) suzie_1's blog
hey he copied and paste it from coursework.info dats not fair
XLNC, 20 Jan, 2006 @ 18:21
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XLNC
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I posted my coursework because its of no use to me. May I remind you nearly all of the people doing a certain coursework will seek information from a variety of reources -perhaps better than what i posted. So dont feel ''advantaged'' :).

Best regards.

suzie_1, 20 Jan, 2006 @ 18:31
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suzie_1
send txt message to suzie_1 bronzestar (78) suzie_1's blog
lol u got urself a gold medal
XLNC, 20 Jan, 2006 @ 20:47
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XLNC
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I've got two now! One from ''someone'' and one now.
suzie_1, 20 Jan, 2006 @ 21:39
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suzie_1
send txt message to suzie_1 bronzestar (78) suzie_1's blog
im happy for you
payne6, 21 Jan, 2006 @ 00:07
Author Comment
Is sahika right?
suzie_1, 21 Jan, 2006 @ 01:55
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suzie_1
send txt message to suzie_1 bronzestar (78) suzie_1's blog
yh
k.facia, 21 Jan, 2006 @ 02:07
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k.facia
send txt message to k.facia k.facia's blog

im here

suzie_1, 21 Jan, 2006 @ 02:08
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suzie_1
send txt message to suzie_1 bronzestar (78) suzie_1's blog
hello

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