Investigating the Effect of the Enzyme Catalyse On Hydrogen Peroxide Introduction The aim of this experiment is to determine the effects of varying enzyme (catalyse) on Hydrogen Peroxide. Hydrogen Peroxide + Catalyse à Water + Oxygen 2H2O2 à H2O + O2 + Heat Apparatus & Diagram [IMAGE][IMAGE][IMAGE][IMAGE][IMAGE][IMAGE][IMAGE] Bung Potato Hydrogen Peroxide Water Collected Oxygen Delivery Tube Measuring Cylinder [IMAGE] Using the Equipment Safely It is important that we use the apparatus carefully, as safety will be an issue throughout the whole experiment. We will wear goggles and an apron or lab coat to protect our eyes and clothes. As we are using enzymes and Hydrogen Peroxide we need to be extra careful, ensuring they don't come into contact with our eyes, skin or clothes. Catalyse is an enzyme found in all living cells. It makes Hydrogen Peroxide decompose into water and Oxygen. We will be measuring the amount of Oxygen released from the Hydrogen Peroxide. In order to do this we will use a measuring cylinder. This piece of apparatus measures the amount of Oxygen given off, measuring in cm3. This will help us measure the amount of Oxygen more accurately. To make the test fair, the following parameters must remain constant during the experiment. These parameters must remain constant during the experiment. These parameters are water, Hydrogen Peroxide, Catalyse and the duration of the reactions. By insuring the test is fair, we will gain accurate results. Variables Dependant Variable: Time, Size of 5cm piece of Potato Independent Variable: Amount of Oxygen released Control Variable: Volume of Hydrogen peroxide, size of Potato, concentration of Hydrogen Peroxide Hypothesis I predict that the breakdown of Hydrogen Peroxide will be quicker when the surface area is increased. If you cut the same size piece of potato into smaller pieces, I believe, the breakdown will be faster. I predict that an increase in surface area will result in an increase in
· I predict that the enzyme will work at its best at 37c because that
and this is a result of over production of H O . Although Hydrogen Peroxide left alone will eventually break down, Catalase will speed this reaction up a lot faster therefore in those circumstances it is inserted into the body. Hydrogen Peroxide broken down will produce Water and Oxygen. In our experiment the independent variable is the concentration of the Catalase Enzyme.
Investigating the Effect of Substrate Concentration on Catalase Reaction. Planning -Aim : The aim of the experiment is to examine how the concentration of the substrate (Hydrogen Peroxide, H2O2) affects the rate of reaction. the enzyme (catalase).
To record the results of Oxygen created from decomposing Hydrogen peroxide with catalase that are either heated, cooled of left in room temperature.
How does the temperature (-2°C, 20°C, 30°C, 40°C, 60°C) affect the production of oxygen (cm3) from cow hepatic (the enzyme catalase) when placed in boiling tube with 10 ml of 3% hydrogen peroxide for 1 minute?
== == == = This is what I'm going to be changing in the experiment and this will be the temperature and the concentration of the yeast. There are several variables in this experiment, they are: · Amount Used - Too much or too little of the hydrogen peroxide causes the reaction to speed up/slow down producing different amounts of oxygen.
To investigate the amount of oxygen foam (cm) produced by the enzyme catalase when it breaks down hydrogen peroxide in the animal (liver, milk, honey) and plant cells(potato, purple cabbage) into oxygen and water
Hydrogen peroxide is a toxic by-product of many crucial metabolic reactions for aerobic organisms. Lastly, a colorless dye called guaiacol is added to the reaction so that it can bind to peroxidase, get oxidized while hydrogen peroxide is reduced to water, and form a brown tetraguaiacol. That is an example of an oxidation-reduction reaction, which can be monitored using a spectrophotometer due to the brown nature of tetraguaiacol. With a LabQuest attached to a spectrophotometer, absorbance of the enzyme/substrate solution is monitored versus time when exposed to different conditions.
105-106). It was hypothesized that if the hydrogen peroxide was exposed to the liver, the hydrogen peroxide would decompose faster than the hydrogen peroxide exposed to the potato because complex organisms (like mammals and birds) require more ATP energy and therefore have an increased rate of cellular respiration, producing more hydrogen peroxide as a byproduct (Marziali, 2009). So, the enzymes in a complex organism are more likely to be able to decompose the hydrogen peroxide. This proved true in this lab, however, the enzyme in both potato and liver is the same, so it must be a different factor such as enzyme concentration that caused a different rate of reaction to be observed between the liver and the potato (Nuffield Foundation, 2011; Eed,
oxygen. Before I do this I must do a preliminary plan to see what my
Abstract: Enzymes are catalysts therefore we can state that they work to start a reaction or speed it up. The chemical transformed due to the enzyme (catalase) is known as the substrate. In this lab the chemical used was hydrogen peroxide because it can be broken down by catalase. The substrate in this lab would be hydrogen peroxide and the enzymes used will be catalase which is found in both potatoes and liver. This substrate will fill the active sites on the enzyme and the reaction will vary based on the concentration of both and the different factors in the experiment. Students placed either liver or potatoes in test tubes with the substrate and observed them at different temperatures as well as with different concentrations of the substrate. Upon reviewing observations, it can be concluded that liver contains the greater amount of catalase as its rates of reaction were greater than that of the potato.
The first experiments investigate the order of reaction with respect to the reactants; hydrogen peroxide, potassium iodide and sulphuric acid by varying the concentrations and plotting them against 1/time. An initial rate technique is used in this experiment so ‘the rate of reaction is inversely proportional to time.’ To find the order of reaction in respect to the reactants, 1/time is plotted against the concentration of Hydrogen Peroxide using the equation:
Investigate the Effect of pH on Immobilised Yeast Cells on the Breakdown of Hydrogen Peroxide
Chemical Kinetics is the branch of chemistry that studies the speed at which a chemical reaction occur and the factor that influence this speed. What is meant by the speed of a reaction is the rate at which the concentrations of reactants and products change within a time period. Some reactions occur almost instantaneously, while others take days or years. Chemical kinetics understanding I used in the process of designing drugs, controlling pollution and the processing of food. Most of the time chemical kinetics is used to speed or to increase the rate of a reaction rather than to maximize the amount of product. The rate of a reaction is often expressed in terms of change in concentration (Δ [ ]) per unit of time (Δ t). We can measure the rate of a reaction by monitoring either the decrease in concentration (molarity) of the reactant or the increase in the product concentration.
= == In my investigation to find out how salt solution concentration affect the mass of potatoes, I will investigate how much the mass of a potato changes if I leave it in a beaker of water with a specified salt concentration for half an hour. I will change the salt concentration after each experiment. Background Knowledge --------------------