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
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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
Investigation of How the Concentration of Catalase Enzyme Affects the Rate of Reaction Aim: To find out how the concentration of Catalase Enzyme will affect the enzyme activity and the rate of reaction towards Hydrogen Peroxide. (H O ) Prediction: I predict that with the higher concentration of enzyme, the likelihood of it breaking down molecules will be greater because there will be more enzymes to work at the substrate and the chances of it colliding will be higher making the activity time quicker. Equipment: · Syringe · Measuring Cylinder (×2) · Knife · Blender · Beakers (×2) · Balance · Hydrochloric Acid in a beaker · Stop clock · Potato · Water in a beaker Preliminary Experiment: In this experiment we will be using an enzyme called Catalase. By using different amounts of this enzyme we will be diluting it with water to test how the concentration of Catalase affects the rate of reaction with Hydrogen Peroxide.
This experiment was conducted to determine the effects of pH and temperature on peroxidase from a potato. The optimum temperature for peroxidase was determined to be 23°C, because it had a rate of absorbance of 0.3493, higher than the other temperatures evaluated. A temperature of 48°C is inefficient of speeding up peroxidase activity because its rate of absorbance was 0.001.
· I predict that the enzyme will work at its best at 37c because that
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?
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.
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).
Investigate the Effect of pH on Immobilised Yeast Cells on the Breakdown of Hydrogen Peroxide
To record the results of Oxygen created from decomposing Hydrogen peroxide with catalase that are either heated, cooled of left in room temperature.
* Size of potatoes * Diameter of each potato tube * Time in sugar solution We need to make sure in both experiments the fair test lists are used and the procedures are carried out. This needs to be done otherwise my results will not be accurate and will look odd. Method: Firstly we got out all our equipment.
== == == = 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.
How the Concentration of the Substrate Affects the Reaction in the Catalase Inside Potato Cells Introduction Enzymes are made of proteins and they speed up reactions, this means that they act as catalysts. Hydrogen peroxide is a byproduct of our cell's activities and is very toxic. The enzymes in our bodies break down the hydrogen peroxide at certain temperatures they work best at body temperature, which is approximately 37 degrees. At high temperatures, the cells begin to denature. This means that the hydrogen peroxide is prevented from being broken down because they will not 'fit' into the enzyme.[IMAGE] Objective I am going to find out how the concentration of the substrate, hydrogen peroxide affects the reaction in the catalase inside the potato cells.
= == 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 --------------------
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,
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.
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: