Reaction Rate Between The Catalase And Hydrogen example essay topic

Enzyme Investigation Enzymes such as Catalase are protein molecules, which are found in living cells. They are used to speed up specific reactions 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 to remove Hydrogen Peroxide from cells.

Catalase speeds up the decomposition of Hydrogen Peroxide into water and oxygen because the shape of its active site matches the shape of the Hydrogen peroxide molecule. This type of reaction where a molecule is broken down into smaller pieces is called a catabolic reaction. This is the reaction equation for the breakdown of hydrogen peroxide into water and oxygen: Hydrogen Peroxide + Catalase = Water + Oxygen. 2 H 202 2 H 20 + 02 Aim: The aim of this experiment was to examine the effect of catalase on the breakdown of the substrate hydrogen peroxide. The catalase used was obtained from a piece of liver.

An enzyme is a biological catalyst. Catalysts speed up reactions. Most enzymes optimum working temperature is 37 (body temperature), however the optimum working temperature of catalase is 30. Enzymes have their optimum temperature which is when the molecules of the enzyme are moving at their fastest. This makes it more likely that the enzyme molecules will collide with substrate and react more successfully. At 40, catalase becomes denatured, which means the molecular bonds of the enzyme are broken, the active site (the area on the enzyme where the reaction takes place) is deformed and the enzyme becomes useless.

At 60 catalase is completely destroyed or is said to have denatured. The molecular bonds of the enzyme are weak and are broken by slight changes in its surroundings, like temperature or pH. The concentration of the catalase, compared to the concentration of substrate (in this case, hydrogen peroxide) is important because one is directly proportional to the other. This is because the number of collisions between the catalase molecules and substrate molecules is increased. Eventually, with no more substrate molecules available, the reaction will have reached its maximum level. As the concentration of catalase is increased, the surface area will increase.

The larger surface area will provide more active sites for the hydrogen peroxide to react with, increasing the speed in which the reaction takes place, otherwise known as the reaction rate. If I had carried out the experiment, I would have predicted that the more catalase used will speed up the reaction rate between the catalase and hydrogen peroxide. My predicted graph will resemble the graph on the next page: As the concentration of catalase is increased the rate of reaction will increase forming a series of positively correlated lines, increasing in gradient as the catalase concentration increases. The particular enzymes in question are breaking enzymes and they will engulf the substrate and separate into smaller pieces and release it.

The enzymes in question will not die or be used up and will continue working until all the substrate is reacted with. My predicted rate of reaction against temperature should resemble this graph: To work out the rate of reaction you would divide the vertical by the horizontal or otherwise: Y X Here the rate of reaction gradually rises until it reaches its optimum temperature that is 37-40 and a steep decrease in the rate of reaction suggests that the enzymes are breaking up and denaturing (or being destroyed by the temperature). Analysis: The Graph to show volume of gas Against Time shows that in general, as the concentration of catalase increases, the volume of oxygen also increases. 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. With more successful collisions with more energy, the amount of oxygen should be produced at a faster rate, which is how the graph has turned out. The graph more or less resembles my predicted graph and my prediction. Looking at the graph I will explain why the set temperatures on the graph are correct or not. 18 C -The line of best fit produced fits in with my prediction and the theories stated on the previous pages.

Whilst the optimum temperature of a catalase is higher than that of room temperature (app 20 C), this shows that oxygen has been given off at an acceptable rate at room temperature. There was enough successful collisions and energy to produce oxygen from the reaction, so is not regarded as an anomalous result. 33 C - In my opinion I would regard this scatter as anomalous. This is because it does not tie in with the rate of reaction theory.

33 C is near an enzymes optimum temperature and so should produce more oxygen in a shorter space of time, but this is not the case. This could have been due to a recording error or a disruption to the independent variables. It could also be the result of a method with several problems. The liver was always used at room temperature. Not is this an inconsistent value due to doing the experiments over a series of days, adding cold liver to warm H 202 cools the enzyme. As you increase the temperature percentage, you also increase the error percentage.

The fact that the liver was not warm proves the theory of kinetics is true because the fact that it may have been cooled affects the results but not the conclusion. The enzymes would have taken more energy from the temperature to start reacting, therefore taking longer for the reaction to start producing oxygen. Other flaws could have been made if the liver was not weighed accurately or the amount of hydrogen peroxide was not measured accurately. 50 C: - This goes back to the normal trend or pattern of the best fit line.

This is because there are more successful collisions with more energy, especially with the heat, the amount of oxygen is being produced at its fastest rate and follows the normal rules in the previous set of rate of reaction statements. 70 C: - Here the enzymes have been destroyed or denatured. This is because the temperature as killed off the enzymes, therefore causing them to stop working or reacting at all with the substrate. Hence, the small amount of oxygen produced at a very slow rate.

Judging from the Graph to show rate of reaction Against Temperature, this concludes and proves my predicted graph. Here the optimum temperature seems to be 50 C, where the enzymes are working at their quickest and tails off at 70 C where we clearly know that the enzymes have denatured. Evaluation: The method seems to have been carried out well and with some accuracy. The majority of results agreed with my original prediction and predicted graphs, which support how accurate the method, must have been carried out.

The results seem to follow the same pattern. Some anomalous results occurred, especially when plotting the volume of gas produced at 33 C. This could have been due to reading errors or slight changes in the independent variables. Another possibility is that one sample could have contained more liver than the others. More reliable results could have been obtained by using more accurate equipment like a burette when measuring the hydrogen peroxide or scales to weight the liver correct to one decimal place for each sample. The experiment could not be repeated with the same catalase and hydrogen peroxide samples, as they would have been contaminated with each other.

A test using different samples would yield different results. Further testing, using more catalase or allowing the reaction to continue for longer would show if the reaction would level off or not as drawn in on the predicted graph. Obviously, the experiment could have been improved by repeating the experiment more times to calculate a better average. I think there was sufficient evidence to draw firm conclusions from the results as it agreed with my prediction and proves many of the reaction theories.

Many variations of this experiment could back up the evidence I already have. The independent variables could be changed. For instance: the hydrogen peroxide concentration, differently shaped sizes of the liver and surface area. How much enzyme is in 1 gram of liver is questionable. The same weight doesn't truly determine the same amount of enzyme contained in the liver. So if a piece of the liver was collected from the centre of the liver it may contain more or less enzyme than the edge or vice verse.

It may even contain the same amount of enzyme all over. The liver was probably kept over a series of days and this would of affected the amount of water in the liver. If the water evaporated the catalase would be in a dehydrated state. This would mean that the weight may stay the same or decrease and the amount of enzyme would also increase or decrease. To overcome this problem, it would be wise to dehydrate all the liver or keep the whole of the liver moist.