For the preparation of Part 1, pieces were cut from the potato and were blended. The prepared suspensions were broken cells of the potato used as the extraction. The process of the suspension was the homogenization of the organism and later was centrifuged so that some of the substances reached the bottom (insoluble particles) and used the liquid as the enzyme(Schultz, 2006). The enzyme was brown colored known as the catechol oxidase, and that product can be used to measure the reaction rate using a spectrometer (Schultz, 2006). We started with twice the volume of the stock, so for the case of Part 1 was to begin with 6 ml of full strength enzyme. Two test tubes were used for the dilution. The first contained 5ml of buffered catechol and the second with 5ml of distilled …show more content…
This procedure help decide if an inhibitor acted as a competitive or a noncompetitive inhibitor. Inhibitors used in this experiment are poisons. The one used in this experiment was phenolthiourea which is known to react with copper, if present(Schultz, 2006). Having used the comparison of the percentage inhibition of and inhibitor at different concentrations was what determined the effect of the inhibitor. Part 4 started with the serial dilution as done in Part 3 with catechol but using 10ml instead and was diluted with water. Dilutions were made for full strength all the way halved to 1/16 (lack of time resulted only up to 1/8 dilution). 5 test tubes were prepared for dilution respectively to 5 spec tubes that had the inhibitor and water and ready for the enzyme addition. Recordings were done every 60 seconds for 3 minutes. Reaction rate was then calculated after time ended. After having used the inhibitor, the steps were repeated but replace the inhibitor with water as control and experimented for the rates without the inhibitor. Percentages were graphed by the percentage inhibition versus the substrate concentration for the
The control for both curves was the beaker with 0% concentration of substrate, which produced no enzyme activity, as there were no substrate molecules for...
Catecholase is an enzyme formed by catechol and oxygen used to interlock oxygen at relative settings, and it is present in plants and crustaceans (Sanyal et. al, 2014). For example, in most fruits and vegetables, the bruised or exposed area of the pant becomes brown due to the reaction of catechol becoming oxidized and oxygen becoming reduced by gaining hydrogen to form water, which then creates a chain that is is the structural backbone of dark melanoid pigments (Helms et al., 1998). However, not all fruits and plants darken at the same rate. This leads to question the enzymatic strength of catecholase and how nearby surroundings affect its activity. The catecholase enzyme has an optimal temperature of approximately 40°C (Helms et al., 1998). Anything above that level would denature the tertiary or primary structure of the protein and cause it to be inoperable. At low temperatures, enzymes have a slower catalyzing rate. Enzymes also function under optimal pH level or else they will also denature, so an average quantity of ions, not too high or low, present within a solution could determine the efficiency of an enzyme (Helms et al., 1998). Also, if more enzymes were added to the concentration, the solution would have a more active sites available for substrates and allow the reaction rate to increase if excess substrate is present (Helms et al., 1998). However, if more
The shape of the molecules is changing and so the enzyme molecules can no longer fit into the gaps in the substrate that they need to and therefore the enzymes have de – natured and can no longer function as they are supposed to and cannot do their job correctly. Changing the temperature: Five different temperatures could be investigated. Water baths were used to maintain a constant temperature. Water baths were set up at 40 degrees, 60 degrees and 80 degrees (Celsius). Room temperature investigations were also carried out (20 degrees).
The independent variable for this experiment is the enzyme concentration, and the range chosen is from 1% to 5% with the measurements of 1, 2, 4, and 5%. The dependant variable to be measured is the absorbance of the absorbance of the solution within a colorimeter, Equipments: Iodine solution: used to test for present of starch - Amylase solution - 1% starch solution - 1 pipette - 3 syringes - 8 test tubes – Stop clock - Water bath at 37oc - Distilled water- colorimeter Method: = == ==
With this information we were able to identify any patterns and similarities. Hypothesis: The higher the temperature of water, potato and H²O², the rate at which the Enzyme will work will be faster therefore producing more oxygen. The reaction will be the same without the catalase (potato). Therefore in both experiments the Enzyme will work more rapidly and produce more oxygen. Aim: To test the hypothesis.
In this experiment as a whole, there were three individual experiments conducted, each with an individualized hypothesis. For the effect of temperature on enzyme activity, catalase activity will be decreased when catalase is exposed to temperatures greater than or less approximately 23 degrees Celsius. For the effect of enzyme concentration on enzyme activity, a concentration of greater or less than approximately 50% enzymes, the less active catalase will be. Lastly, the more the pH buffer deviates from a basic pH of 7, the less active catalase will be.
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).
I believe that at the time of the flood the water table of his land
Materials used in the experiment included 5-7 g of the potato tissue, 50ml of 2.0M phosphate buffer coffee filter and guaiacol dye.
The Effect of Surface Area on the Rate of Reaction Between Catalase from a Potato and Hydrogen Peroxide
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.
Osmosis in Potato Tubes Osmosis: Osmosis is the movement of water molecules through a semi-permeable membrane from a high concentration to a low concentration. Diagram: [IMAGE] [IMAGE] Aim: To see the effects of different concentration of sugar solution on Osmosis in potato tubes. Key factor: In the investigation we change the sugar solution from: 0%-10%-20%-30%-40%-50% this is the independent variable; the dependant variable is the change in mass. Prediction: I predict that all the potato tubes in pure water or low concentration sugar solution will swell because water enters their cells by osmosis.
How the Concentration of the Substrate Affects the Reaction in the Catalase Inside Potato Cells
Purpose: This lab gives the idea about the enzyme. We will do two different experiments. Enzyme is a protein that made of strings of amino acids and it is helping to produce chemical reactions in the quickest way. In the first experiment, we are testing water, sucrose solution, salt solution, and hydrogen peroxide to see which can increase the bubbles. So we can understand that enzyme producing chemical reactions in the speed. In the second experiment, we are using temperature of room, boiling water, refrigerator, and freezer to see what will effect the enzyme.
Researchers then hypothesized that the results would indicate the greatest amount of potato enzyme activity level will take place at room temperature. In this experiment, researchers used potato extract and different temperature levels to test the hypothesis. Moreover, researchers wanted to test the color intensity scale and how specific catechol oxidase is for catechol. In this experiment, researchers used dH2O, catechol solution, hydroquinone, and potato extract. Lastly, researchers tested the substrate concentration and how it has an effect on enzyme activity.