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Biology (AS-LEVEL)
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Biology Practicals (9700/33) [All syllabus content with requisite details] Compiled by: Aarjit Adhikary, A level, NP704
A. Cell Structure
(1) [PA] use an eyepiece graticule and stage micrometer scale to measure cells and be familiar with units (millimeter, micrometer, nanometer) used in cell studies;
Calibration with a standard stage micrometer with pitch [0.1 mm]
Magnification (X)
Number of ocular
divisions [od]
number of stage
division [sd]
Value of one ocular division
[sd*0.1/od] / (mm)
10 40 5 0.01240 65 2 0.003
(2) [PA] compare and contrast the structure of typical animal and plant cells;
Additional differences include: Plasmodesmata, tonoplast, cellulose barrier/lipid & protein barrier, membranes adapted for lysis/plasmolysis.
(3) [PA] draw and label low power plan diagrams of tissues and organs (including transverse sections of stems, roots and leaves) and calculate the linear magnification of drawings;
Characteristics and roles;
Root and Stem:
▪ Epidermis: surface of the stem made of a number of layers often with a waxy cuticle to reduce water loss.
▪ Cortex Tissue: Forming a cylinder of tissue around the outer edge of the stem. Often contains cells with secondary thickening in the cell walls which provides additional support.
▪ Vascular bundle: contains xylem, phloem and cambium tissue.
▪ Xylem: a longitudinal set of tubes that conduct water and mineral salts from the roots upward
through the stem to the leaves.
▪ Phloem (sieve elements) transports sap through the plant tissue in a number of possible directions.
▪ Vascular cambium is a type of lateral meristem that forms a vertical cylinder in the stem. The cambium produces the secondary xylem and phloem through cell division in the vertical plane.
▪ In the centre of the stem can be found the pith tissue composed of thin walled cells called parenchyma. In some plants this section can degenerate to leave a hollow stem.
Leaves:
▪ Cuticle is a waxy layer which reduces water loss through the upper epidermis.
▪ Upper epidermis is a flattened layer of cell that forms the surface of the leaf and makes the cuticle.
▪ Palisade Layer: This is the main photosynthetic region of the leaf.
▪ Vascular bundle: contains the transport system and vascular meristem tissue (x-xylem, p-phloem).
▪ Spongy mesophyll: contains spaces that allows the movement of gases and water through the leaf tissue..
▪ Lower epidermis: bottom surface layer of tissues which contains the guard cells that form each stoma.
(4) [PA] calculate linear magnification of drawings and photographs;
m = (I/O) [I measured with a 30 cm rule and O with a calibrated eyepiece graticule]
(5) [PA] calculate the actual sizes of specimens from drawings and photographs.
A scale bar is usually given. In that case, you measure the scale bar, not the cells!
B. Biological Molecules
(6) [PA] carry out tests for educing and non-reducing sugars (including using color standards as a semi-quantitative use of the Benedict's test), the iodine in potassium iodide solution test for starch, the emulsion test for lipids and the Biuret test for proteins.; Test Observation Basis of testBenedict's test (Reducing sugars)*
Add 2 cm^3 of a solution to a test-tube. Add equal volume of Benedict's reagent. Shake and bring gently to boil (80-90C)
The initial blue coloration of the mixture turns green, then yellow and may finally form a brick red ppt.
Benedict's solution contains Copper Sulphate. Reducing sugars reduce the blue cupric Cu(II) ion to cuprous Cu(I) ion resulting in a red ppt. of CuO.
Benedict's test (Non-reducing sugars)**
Add 2 cm^3 of a solution to a test-tube. Add 1 cm^3 of dilute hydrochloric acid. Boil for one minute. Carefully neutralize the solution with sodium bicarbonate (base). Check with pH paper. Avoid vigorous effervescence. Carry our Benedict's test.
As Benedict's test
A disaccharide can be hydrolyzed to its monosaccharide constituents by boiling with dilute hydrochloric acid. Eg. Sucrose in hydrolyzed to glucose and fructose, both of which are reducing sugars and give a positive result with the Benedict's test.
Iodine/Potassium iodide test (Starch)
Add 2 cm^3 of a solution to a test-tube. Add a few drops of I2/KI solution. Alternatively, add the latter to the solid form of starch.
A blue-black coloration
A polyiodide complex is formed with starch.
Sudan III (Lipids)
Add 2 cm^3 of sample to an equal volume of water in a test-tube. Add a few drops of Sudan III and shake.
A red-stained layer separates on the surface of the water, which remains uncolored.
Fat globules are stained red and are less dense than water.
Emulsion test (Lipids)
Add 2 cm^3 of a solution with 2 cm^3 of absolute ethanol (>95% Vol) to a test-tube. Dissolve the lipid by shaking vigorously. Add an equal volume of cold water.
A cloudy white suspension
Lipids are immiscible with water. Adding water to a solution of lipid and alcohol forms an emulsion reflecting light and giving an opalescent appearance.
Grease spot test (Lipids)
Rub a drop of a the sample into a piece of paper/fabric. Allow time for any water to evaporate. Gentle warming will speed up the process.
A permanent transparent spot/stain on the paper/fabric.
Unsaturated fats have double-bonded carbon kinks capable of disrupting natural polymers.
Biuret's test (Proteins)
Add 2 cm^3 of a solution in a test-tube. Add an equal volume of 5% Potassium hydroxide followed by two drops of 1% Copper sulphate and mix. No heating is required.
A mauve/purple coloration
Nitrogen atoms in the peptide chain react with Copper(II) ions to form a purple complex.
DCPIP test (ascorbic acids)
Add 1 cm^3 of the blue DCPIP solution to a test-tube. With a syringe, add 1 cm^3 of solution gradually. Avoid stirring as this result in the oxidation of the sample. Add until the blue color of the dye just disappears.
Blue color of the dye disappears leaving a colorless solution
DCPIP(dichlorophenolindophenol) is a blue colored dye which gets reduced to a colorless compound by ascorbic acid, a strong reducing agent.
Test Observation Basis of testBenedict's test (Reducing sugars)*
Add 2 cm^3 of a solution to a test-tube. Add equal volume of Benedict's reagent. Shake and bring gently to boil (80-90C)
The initial blue coloration of the mixture turns green, then yellow and may finally form a brick red ppt.
Benedict's solution contains Copper Sulphate. Reducing sugars reduce the blue cupric Cu(II) ion to cuprous Cu(I) ion resulting in a red ppt. of CuO.
Benedict's test (Non-reducing sugars)**
Add 2 cm^3 of a solution to a test-tube. Add 1 cm^3 of dilute hydrochloric acid. Boil for one minute. Carefully neutralize the solution with sodium bicarbonate (base). Check with pH paper. Avoid vigorous effervescence. Carry our Benedict's test.
As Benedict's test
A disaccharide can be hydrolyzed to its monosaccharide constituents by boiling with dilute hydrochloric acid. Eg. Sucrose in hydrolyzed to glucose and fructose, both of which are reducing sugars and give a positive result with the Benedict's test.
Iodine/Potassium iodide test (Starch)
Add 2 cm^3 of a solution to a test-tube. Add a few drops of I2/KI solution. Alternatively, add the latter to the solid form of starch.
A blue-black coloration
A polyiodide complex is formed with starch.
Sudan III (Lipids)
Add 2 cm^3 of sample to an equal volume of water in a test-tube. Add a few drops of Sudan III and shake.
A red-stained layer separates on the surface of the water, which remains uncolored.
Fat globules are stained red and are less dense than water.
Emulsion test (Lipids)
Add 2 cm^3 of a solution with 2 cm^3 of absolute ethanol (>95% Vol) to a test-tube. Dissolve the lipid by shaking vigorously. Add an equal volume of cold water.
A cloudy white suspension
Lipids are immiscible with water. Adding water to a solution of lipid and alcohol forms an emulsion reflecting light and giving an opalescent appearance.
Grease spot test (Lipids)
Rub a drop of a the sample into a piece of paper/fabric. Allow time for any water to evaporate. Gentle warming will speed up the process.
A permanent transparent spot/stain on the paper/fabric.
Unsaturated fats have double-bonded carbon kinks capable of disrupting natural polymers.
Biuret's test (Proteins)
Add 2 cm^3 of a solution in a test-tube. Add an equal volume of 5% Potassium hydroxide followed by two drops of 1% Copper sulphate and mix. No heating is required.
A mauve/purple coloration
Nitrogen atoms in the peptide chain react with Copper(II) ions to form a purple complex.
DCPIP test (ascorbic acids)
Add 1 cm^3 of the blue DCPIP solution to a test-tube. With a syringe, add 1 cm^3 of solution gradually. Avoid stirring as this result in the oxidation of the sample. Add until the blue color of the dye just disappears.
Blue color of the dye disappears leaving a colorless solution
DCPIP(dichlorophenolindophenol) is a blue colored dye which gets reduced to a colorless compound by ascorbic acid, a strong reducing agent.
Test Observation Basis of testBenedict's test (Reducing sugars)*
Add 2 cm^3 of a solution to a test-tube. Add equal volume of Benedict's reagent. Shake and bring gently to boil (80-90C)
The initial blue coloration of the mixture turns green, then yellow and may finally form a brick red ppt.
Benedict's solution contains Copper Sulphate. Reducing sugars reduce the blue cupric Cu(II) ion to cuprous Cu(I) ion resulting in a red ppt. of CuO.
Benedict's test (Non-reducing sugars)**
Add 2 cm^3 of a solution to a test-tube. Add 1 cm^3 of dilute hydrochloric acid. Boil for one minute. Carefully neutralize the solution with sodium bicarbonate (base). Check with pH paper. Avoid vigorous effervescence. Carry our Benedict's test.
As Benedict's test
A disaccharide can be hydrolyzed to its monosaccharide constituents by boiling with dilute hydrochloric acid. Eg. Sucrose in hydrolyzed to glucose and fructose, both of which are reducing sugars and give a positive result with the Benedict's test.
Iodine/Potassium iodide test (Starch)
Add 2 cm^3 of a solution to a test-tube. Add a few drops of I2/KI solution. Alternatively, add the latter to the solid form of starch.
A blue-black coloration
A polyiodide complex is formed with starch.
Sudan III (Lipids)
Add 2 cm^3 of sample to an equal volume of water in a test-tube. Add a few drops of Sudan III and shake.
A red-stained layer separates on the surface of the water, which remains uncolored.
Fat globules are stained red and are less dense than water.
Emulsion test (Lipids)
Add 2 cm^3 of a solution with 2 cm^3 of absolute ethanol (>95% Vol) to a test-tube. Dissolve the lipid by shaking vigorously. Add an equal volume of cold water.
A cloudy white suspension
Lipids are immiscible with water. Adding water to a solution of lipid and alcohol forms an emulsion reflecting light and giving an opalescent appearance.
Grease spot test (Lipids)
Rub a drop of a the sample into a piece of paper/fabric. Allow time for any water to evaporate. Gentle warming will speed up the process.
A permanent transparent spot/stain on the paper/fabric.
Unsaturated fats have double-bonded carbon kinks capable of disrupting natural polymers.
Biuret's test (Proteins)
Add 2 cm^3 of a solution in a test-tube. Add an equal volume of 5% Potassium hydroxide followed by two drops of 1% Copper sulphate and mix. No heating is required.
A mauve/purple coloration
Nitrogen atoms in the peptide chain react with Copper(II) ions to form a purple complex.
DCPIP test (ascorbic acids)
Add 1 cm^3 of the blue DCPIP solution to a test-tube. With a syringe, add 1 cm^3 of solution gradually. Avoid stirring as this result in the oxidation of the sample. Add until the blue color of the dye just disappears.
Blue color of the dye disappears leaving a colorless solution
DCPIP(dichlorophenolindophenol) is a blue colored dye which gets reduced to a colorless compound by ascorbic acid, a strong reducing agent.
Test Observation Basis of testBenedict's test (Reducing sugars)*
Add 2 cm^3 of a solution to a test-tube. Add equal volume of Benedict's reagent. Shake and bring gently to boil (80-90C)
The initial blue coloration of the mixture turns green, then yellow and may finally form a brick red ppt.
Benedict's solution contains Copper Sulphate. Reducing sugars reduce the blue cupric Cu(II) ion to cuprous Cu(I) ion resulting in a red ppt. of CuO.
Benedict's test (Non-reducing sugars)**
Add 2 cm^3 of a solution to a test-tube. Add 1 cm^3 of dilute hydrochloric acid. Boil for one minute. Carefully neutralize the solution with sodium bicarbonate (base). Check with pH paper. Avoid vigorous effervescence. Carry our Benedict's test.
As Benedict's test
A disaccharide can be hydrolyzed to its monosaccharide constituents by boiling with dilute hydrochloric acid. Eg. Sucrose in hydrolyzed to glucose and fructose, both of which are reducing sugars and give a positive result with the Benedict's test.
Iodine/Potassium iodide test (Starch)
Add 2 cm^3 of a solution to a test-tube. Add a few drops of I2/KI solution. Alternatively, add the latter to the solid form of starch.
A blue-black coloration
A polyiodide complex is formed with starch.
Sudan III (Lipids)
Add 2 cm^3 of sample to an equal volume of water in a test-tube. Add a few drops of Sudan III and shake.
A red-stained layer separates on the surface of the water, which remains uncolored.
Fat globules are stained red and are less dense than water.
Emulsion test (Lipids)
Add 2 cm^3 of a solution with 2 cm^3 of absolute ethanol (>95% Vol) to a test-tube. Dissolve the lipid by shaking vigorously. Add an equal volume of cold water.
A cloudy white suspension
Lipids are immiscible with water. Adding water to a solution of lipid and alcohol forms an emulsion reflecting light and giving an opalescent appearance.
Grease spot test (Lipids)
Rub a drop of a the sample into a piece of paper/fabric. Allow time for any water to evaporate. Gentle warming will speed up the process.
A permanent transparent spot/stain on the paper/fabric.
Unsaturated fats have double-bonded carbon kinks capable of disrupting natural polymers.
Biuret's test (Proteins)
Add 2 cm^3 of a solution in a test-tube. Add an equal volume of 5% Potassium hydroxide followed by two drops of 1% Copper sulphate and mix. No heating is required.
A mauve/purple coloration
Nitrogen atoms in the peptide chain react with Copper(II) ions to form a purple complex.
DCPIP test (ascorbic acids)
Add 1 cm^3 of the blue DCPIP solution to a test-tube. With a syringe, add 1 cm^3 of solution gradually. Avoid stirring as this result in the oxidation of the sample. Add until the blue color of the dye just disappears.
Blue color of the dye disappears leaving a colorless solution
DCPIP(dichlorophenolindophenol) is a blue colored dye which gets reduced to a colorless compound by ascorbic acid, a strong reducing agent.
Test Observation Basis of testBenedict's test (Reducing sugars)*
Add 2 cm^3 of a solution to a test-tube. Add equal volume of Benedict's reagent. Shake and bring gently to boil (80-90C)
The initial blue coloration of the mixture turns green, then yellow and may finally form a brick red ppt.
Benedict's solution contains Copper Sulphate. Reducing sugars reduce the blue cupric Cu(II) ion to cuprous Cu(I) ion resulting in a red ppt. of CuO.
Benedict's test (Non-reducing sugars)**
Add 2 cm^3 of a solution to a test-tube. Add 1 cm^3 of dilute hydrochloric acid. Boil for one minute. Carefully neutralize the solution with sodium bicarbonate (base). Check with pH paper. Avoid vigorous effervescence. Carry our Benedict's test.
As Benedict's test
A disaccharide can be hydrolyzed to its monosaccharide constituents by boiling with dilute hydrochloric acid. Eg. Sucrose in hydrolyzed to glucose and fructose, both of which are reducing sugars and give a positive result with the Benedict's test.
Iodine/Potassium iodide test (Starch)
Add 2 cm^3 of a solution to a test-tube. Add a few drops of I2/KI solution. Alternatively, add the latter to the solid form of starch.
A blue-black coloration
A polyiodide complex is formed with starch.
Sudan III (Lipids)
Add 2 cm^3 of sample to an equal volume of water in a test-tube. Add a few drops of Sudan III and shake.
A red-stained layer separates on the surface of the water, which remains uncolored.
Fat globules are stained red and are less dense than water.
Emulsion test (Lipids)
Add 2 cm^3 of a solution with 2 cm^3 of absolute ethanol (>95% Vol) to a test-tube. Dissolve the lipid by shaking vigorously. Add an equal volume of cold water.
A cloudy white suspension
Lipids are immiscible with water. Adding water to a solution of lipid and alcohol forms an emulsion reflecting light and giving an opalescent appearance.
Grease spot test (Lipids)
Rub a drop of a the sample into a piece of paper/fabric. Allow time for any water to evaporate. Gentle warming will speed up the process.
A permanent transparent spot/stain on the paper/fabric.
Unsaturated fats have double-bonded carbon kinks capable of disrupting natural polymers.
Biuret's test (Proteins)
Add 2 cm^3 of a solution in a test-tube. Add an equal volume of 5% Potassium hydroxide followed by two drops of 1% Copper sulphate and mix. No heating is required.
A mauve/purple coloration
Nitrogen atoms in the peptide chain react with Copper(II) ions to form a purple complex.
DCPIP test (ascorbic acids)
Add 1 cm^3 of the blue DCPIP solution to a test-tube. With a syringe, add 1 cm^3 of solution gradually. Avoid stirring as this result in the oxidation of the sample. Add until the blue color of the dye just disappears.
Blue color of the dye disappears leaving a colorless solution
DCPIP(dichlorophenolindophenol) is a blue colored dye which gets reduced to a colorless compound by ascorbic acid, a strong reducing agent.
Test Observation Basis of testBenedict's test (Reducing sugars)*
Add 2 cm^3 of a solution to a test-tube. Add equal volume of Benedict's reagent. Shake and bring gently to boil (80-90C)
The initial blue coloration of the mixture turns green, then yellow and may finally form a brick red ppt.
Benedict's solution contains Copper Sulphate. Reducing sugars reduce the blue cupric Cu(II) ion to cuprous Cu(I) ion resulting in a red ppt. of CuO.
Benedict's test (Non-reducing sugars)**
Add 2 cm^3 of a solution to a test-tube. Add 1 cm^3 of dilute hydrochloric acid. Boil for one minute. Carefully neutralize the solution with sodium bicarbonate (base). Check with pH paper. Avoid vigorous effervescence. Carry our Benedict's test.
As Benedict's test
A disaccharide can be hydrolyzed to its monosaccharide constituents by boiling with dilute hydrochloric acid. Eg. Sucrose in hydrolyzed to glucose and fructose, both of which are reducing sugars and give a positive result with the Benedict's test.
Iodine/Potassium iodide test (Starch)
Add 2 cm^3 of a solution to a test-tube. Add a few drops of I2/KI solution. Alternatively, add the latter to the solid form of starch.
A blue-black coloration
A polyiodide complex is formed with starch.
Sudan III (Lipids)
Add 2 cm^3 of sample to an equal volume of water in a test-tube. Add a few drops of Sudan III and shake.
A red-stained layer separates on the surface of the water, which remains uncolored.
Fat globules are stained red and are less dense than water.
Emulsion test (Lipids)
Add 2 cm^3 of a solution with 2 cm^3 of absolute ethanol (>95% Vol) to a test-tube. Dissolve the lipid by shaking vigorously. Add an equal volume of cold water.
A cloudy white suspension
Lipids are immiscible with water. Adding water to a solution of lipid and alcohol forms an emulsion reflecting light and giving an opalescent appearance.
Grease spot test (Lipids)
Rub a drop of a the sample into a piece of paper/fabric. Allow time for any water to evaporate. Gentle warming will speed up the process.
A permanent transparent spot/stain on the paper/fabric.
Unsaturated fats have double-bonded carbon kinks capable of disrupting natural polymers.
Biuret's test (Proteins)
Add 2 cm^3 of a solution in a test-tube. Add an equal volume of 5% Potassium hydroxide followed by two drops of 1% Copper sulphate and mix. No heating is required.
A mauve/purple coloration
Nitrogen atoms in the peptide chain react with Copper(II) ions to form a purple complex.
DCPIP test (ascorbic acids)
Add 1 cm^3 of the blue DCPIP solution to a test-tube. With a syringe, add 1 cm^3 of solution gradually. Avoid stirring as this result in the oxidation of the sample. Add until the blue color of the dye just disappears.
Blue color of the dye disappears leaving a colorless solution
DCPIP(dichlorophenolindophenol) is a blue colored dye which gets reduced to a colorless compound by ascorbic acid, a strong reducing agent.
*The test is semi-quantitative, i.e. a rough estimation of the amount of reducing sugar present can be made. GYOR color change and the 'heaviness' of the ppt correlates with the the amount/concentration of the sugar being tested. **A relatively 'heavier' ppt. is seen if both reducing and non-reducing sugars are present.
C. Enzymes
(7) [PA] follow the progress of an enzyme catalyzed reaction by measuring rates of formation of products (e.g. measuring V o2 in the dissociation of H2O2 via
catalase) or rates of disappearance of substrate (e.g. using amylase);
(8) [PA] investigate and explain the effects of temperature, pH, enzyme concentration and substrate concentration on the rate of enzyme-catalyzed reactions;
Sample Experiment
Introduction
One of the by-products of many cellular reactions is hydrogen peroxide (H2O2). It is extremely toxic to living cells. All aerobic organisms use oxygen for respiration or oxidation of nutrients. During reduction of molecular oxygen to water, hydrogen peroxide is generated. Two examples of reactions that produce H2O2 are conversions of amino acids into "fuel" molecules and conversion of lipids to carbohydrates. It can damage DNA, protein and lipid membranes and may even be a causative factor in cancer. There are some human immune system cells that actually use H2O2 to kill foreign invaders. The catalase enzyme is specific for the hydrolysis of H2O2:
Catalase is found in animal and plant tissues, and is especially abundant in plant storage organs such as potato tubers, corms, and in the fleshy parts of fruits. You will use catalase isolated from potato tubers and measure its rate of activity under different conditions.
Like other enzymes, catalase is a protein. Enzymes speed up chemical reactions by reducing the activation energy required to convert substrate(s) into product(s). Enzymes have specialized binding sites to do this.
Because enzymes are proteins, they are somewhat fragile. They can be denatured by heat, and can easily be broken down by proteases when cells are homogenized. To preserve activity of proteins in solution, it is important to keep the solutions on ice until you are ready to use them. Denaturing conditions, such as boiling, can also be used as evidence to show that an enzyme-based reaction is protein-dependent.
In the experimental protocol described here a filter paper disk will be immersed in a solution of the enzyme, then placed in the hydrogen peroxide. The oxygen produced from the subsequent reaction becomes trapped in the disc and will give it buoyancy. The time measured from the moment the disc touches the bottom of the container substrate to the time it reaches the surface of the solution is an indirect, but easily quantifiable measure of the rate of the enzyme activity.
Experimental Procedure
Extraction of Catalase
1. Peel a fresh potato tuber and cut the tissue into small cubes. Weigh out 50 g of potato cubes.
2. Place the potato cubes, 50 mL of cold distilled water, and a small amount of crushed ice in a blender.
3. Homogenize for 30 seconds at high speed. From this point on, the enzyme preparation must be carried out in an ice bath.
4. Filter the potato extract, then pour the filtrate into a 100 mL graduated cylinder.
5. Add ice-cold distilled water to bring up the final volume to 100 mL. Mix well. This extract is the 100% enzyme solution.
6. Note: This rough 100% enzyme solution should work OK although it is worth testing it before proceeding with the experiment. At room temperature (approx. 20°C) in a 1% H2O2 solution it would be sensible if the disk took about 20 seconds to rise in the beaker you are using. If it is faster than this, dilute the enzyme and use that as the 100% solution. If it is slower, prepare the extract again, starting with an increased amount of potato cubes. If the filter rises too quickly at room temperature, then the reaction at higher temperatures will be too quick to measure. If the filter rises too slowly, then the lower temperatures will take forever.
7. Keep your catalase preparation on ice.
The Effect of Temperature on Enzyme Activity
1. Label five 50 mL beakers with the temperature for testing (0, 10, 20, 30, and 40°C).
2. Add 40 mL of 1% hydrogen peroxide solution taken from the appropriate temperature water bath to each beaker.
3. Put the beakers in the appropriate water bath.4. Using forceps, immerse a filter paper disk into the
catalase solution you have prepared.5. Allow the disc to absorb the enzyme solution for 5
seconds, then remove it and drain off the excess enzyme solution by touching the filter paper to the edge of the beaker.
6. Drop the disc into the first substrate solution.7. The oxygen produced from the breakdown of the
hydrogen peroxide by catalase becomes trapped in the fibers of the disc causing the disc to float to the surface of the solution.
8. The time (t) in seconds, from the second the disc touches the solution to the time it again reaches the surface is an indirect measure of enzyme activity.
9. Remove the disk from the beaker once it reaches the surface and dispose of it.
10. Record the time taken in a table in your notebook.
11.Clean the beaker and repeat the procedure until you have 5 replicates at the first temperature.
12. Repeat for each temperature so you have data for 0, 10, 20, 30, and 40°C.
13. Calculate mean and standard deviation for each temperature.
14. Construct and label a graph of your results.
The Effect of Enzyme Concentration on Reaction Rate
It is important to demonstrate that the enzyme assay shows that the enzyme actually follows accepted chemical principles. One way to demonstrate this is by determining the effect of enzyme concentration on the rate of activity while using a substrate concentration, in this case H2O2, that is in excess. This can be easily demonstrated with the experimental system used in the previous section. This set of experiments can be done most conveniently at room temperature.
1. Label 5 tubes 100%, 75%, 50%, 25%, and 0%.2. Add 1, 0.75, 0.5, and 0.25 mL of 100% enzyme
extract to the first 4 tubes .3. Using a separate pipette, skip the first tube (100%
enzyme) add 0.25, 0.5, 0.75, and 1 mL ice cold distilled water to the last four tubes and mix. Each tube will have now contain 1 mL of solution at the indicated enzyme concentration.
4. Keep all tubes on ice throughout.5. Set up 5 beakers, each with 40 ml 1% H2O2 on the
bench.6. Check that the temperature in each one has
equilibrated to room temperature.7. Using the 100% enzyme solution, measure the
reaction time using the disk enzyme assay for 5 replicate disks.
8. Repeat the procedure for each of the other four enzyme concentrations.
9. Plot the rate of reaction about enzyme concentration.
(9) [PA] investigate the effects on plant cells of immersion in solutions of different water potentials;
Water flows to regions from a higher potential to a lower potential down the water potential gradient, especially via the selectively-permeable plasma membrane by osmosis. kPa can affect the movement of water down the water potential gradient. Sufficient kPa can 'push' water up the gradient as well. Usually, a calibration table of kPa to water potential is provided if you need to calculate the kPa outside the cell or the water potential inside/outside the cell body.
(10) [PA] describe, with the aid of diagrams, the behavior of chromosomes during the mitotic cell cycle and the associated behavior of the nuclear envelope, cell membrane, centrioles and spindle fibers (names of the main stages are expected);
Introduction
One of the by-products of many cellular reactions is hydrogen peroxide (H2O2). It is extremely toxic to living cells. All aerobic organisms use oxygen for respiration or oxidation of nutrients. During reduction of molecular oxygen to water, hydrogen peroxide is generated. Two examples of reactions that produce H2O2 are conversions of amino acids into "fuel" molecules and conversion of lipids to carbohydrates. It can damage DNA, protein and lipid membranes and may even be a causative factor in cancer. There are some human immune system cells that actually use H2O2 to kill foreign invaders. The catalase enzyme is specific for the hydrolysis of H2O2:
Catalase is found in animal and plant tissues, and is especially abundant in plant storage organs such as potato tubers, corms, and in the fleshy parts of fruits. You will use catalase isolated from potato tubers and measure its rate of activity under different conditions.
Like other enzymes, catalase is a protein. Enzymes speed up chemical reactions by reducing the activation energy required to convert substrate(s) into product(s). Enzymes have specialized binding sites to do this.
Because enzymes are proteins, they are somewhat fragile. They can be denatured by heat, and can easily be broken down by proteases when cells are homogenized. To preserve activity of proteins in solution, it is important to keep the solutions on ice until you are ready to use them. Denaturing conditions, such as boiling, can also be used as evidence to show that an enzyme-based reaction is protein-dependent.
In the experimental protocol described here a filter paper disk will be immersed in a solution of the enzyme, then placed in the hydrogen peroxide. The oxygen produced from the subsequent reaction becomes trapped in the disc and will give it buoyancy. The time measured from the moment the disc touches the bottom of the container substrate to the time it reaches the surface of the solution is an indirect, but easily quantifiable measure of the rate of the enzyme activity.
Experimental Procedure
Extraction of Catalase
1. Peel a fresh potato tuber and cut the tissue into small cubes. Weigh out 50 g of potato cubes.
2. Place the potato cubes, 50 mL of cold distilled water, and a small amount of crushed ice in a blender.
3. Homogenize for 30 seconds at high speed. From this point on, the enzyme preparation must be carried out in an ice bath.
4. Filter the potato extract, then pour the filtrate into a 100 mL graduated cylinder.
5. Add ice-cold distilled water to bring up the final volume to 100 mL. Mix well. This extract is the 100% enzyme solution.
6. Note: This rough 100% enzyme solution should work OK although it is worth testing it before proceeding with the experiment. At room temperature (approx. 20°C) in a 1% H2O2 solution it would be sensible if the disk took about 20 seconds to rise in the beaker you are using. If it is faster than this, dilute the enzyme and use that as the 100% solution. If it is slower, prepare the extract again, starting with an increased amount of potato cubes. If the filter rises too quickly at room temperature, then the reaction at higher temperatures will be too quick to measure. If the filter rises too slowly, then the lower temperatures will take forever.
7. Keep your catalase preparation on ice.
The Effect of Temperature on Enzyme Activity
1. Label five 50 mL beakers with the temperature for testing (0, 10, 20, 30, and 40°C).
2. Add 40 mL of 1% hydrogen peroxide solution taken from the appropriate temperature water bath to each beaker.
3. Put the beakers in the appropriate water bath.4. Using forceps, immerse a filter paper disk into the
catalase solution you have prepared.5. Allow the disc to absorb the enzyme solution for 5
seconds, then remove it and drain off the excess enzyme solution by touching the filter paper to the edge of the beaker.
6. Drop the disc into the first substrate solution.7. The oxygen produced from the breakdown of the
hydrogen peroxide by catalase becomes trapped in the fibers of the disc causing the disc to float to the surface of the solution.
8. The time (t) in seconds, from the second the disc touches the solution to the time it again reaches the surface is an indirect measure of enzyme activity.
9. Remove the disk from the beaker once it reaches the surface and dispose of it.
10. Record the time taken in a table in your notebook.
11.Clean the beaker and repeat the procedure until you have 5 replicates at the first temperature.
12. Repeat for each temperature so you have data for 0, 10, 20, 30, and 40°C.
13. Calculate mean and standard deviation for each temperature.
14. Construct and label a graph of your results.
The Effect of Enzyme Concentration on Reaction Rate
It is important to demonstrate that the enzyme assay shows that the enzyme actually follows accepted chemical principles. One way to demonstrate this is by determining the effect of enzyme concentration on the rate of activity while using a substrate concentration, in this case H2O2, that is in excess. This can be easily demonstrated with the experimental system used in the previous section. This set of experiments can be done most conveniently at room temperature.
1. Label 5 tubes 100%, 75%, 50%, 25%, and 0%.2. Add 1, 0.75, 0.5, and 0.25 mL of 100% enzyme
extract to the first 4 tubes .3. Using a separate pipette, skip the first tube (100%
enzyme) add 0.25, 0.5, 0.75, and 1 mL ice cold distilled water to the last four tubes and mix. Each tube will have now contain 1 mL of solution at the indicated enzyme concentration.
4. Keep all tubes on ice throughout.5. Set up 5 beakers, each with 40 ml 1% H2O2 on the
bench.6. Check that the temperature in each one has
equilibrated to room temperature.7. Using the 100% enzyme solution, measure the
reaction time using the disk enzyme assay for 5 replicate disks.
8. Repeat the procedure for each of the other four enzyme concentrations.
9. Plot the rate of reaction about enzyme concentration.
(9) [PA] investigate the effects on plant cells of immersion in solutions of different water potentials;
Water flows to regions from a higher potential to a lower potential down the water potential gradient, especially via the selectively-permeable plasma membrane by osmosis. kPa can affect the movement of water down the water potential gradient. Sufficient kPa can 'push' water up the gradient as well. Usually, a calibration table of kPa to water potential is provided if you need to calculate the kPa outside the cell or the water potential inside/outside the cell body.
(10) [PA] describe, with the aid of diagrams, the behavior of chromosomes during the mitotic cell cycle and the associated behavior of the nuclear envelope, cell membrane, centrioles and spindle fibers (names of the main stages are expected);
Introduction
One of the by-products of many cellular reactions is hydrogen peroxide (H2O2). It is extremely toxic to living cells. All aerobic organisms use oxygen for respiration or oxidation of nutrients. During reduction of molecular oxygen to water, hydrogen peroxide is generated. Two examples of reactions that produce H2O2 are conversions of amino acids into "fuel" molecules and conversion of lipids to carbohydrates. It can damage DNA, protein and lipid membranes and may even be a causative factor in cancer. There are some human immune system cells that actually use H2O2 to kill foreign invaders. The catalase enzyme is specific for the hydrolysis of H2O2:
Catalase is found in animal and plant tissues, and is especially abundant in plant storage organs such as potato tubers, corms, and in the fleshy parts of fruits. You will use catalase isolated from potato tubers and measure its rate of activity under different conditions.
Like other enzymes, catalase is a protein. Enzymes speed up chemical reactions by reducing the activation energy required to convert substrate(s) into product(s). Enzymes have specialized binding sites to do this.
Because enzymes are proteins, they are somewhat fragile. They can be denatured by heat, and can easily be broken down by proteases when cells are homogenized. To preserve activity of proteins in solution, it is important to keep the solutions on ice until you are ready to use them. Denaturing conditions, such as boiling, can also be used as evidence to show that an enzyme-based reaction is protein-dependent.
In the experimental protocol described here a filter paper disk will be immersed in a solution of the enzyme, then placed in the hydrogen peroxide. The oxygen produced from the subsequent reaction becomes trapped in the disc and will give it buoyancy. The time measured from the moment the disc touches the bottom of the container substrate to the time it reaches the surface of the solution is an indirect, but easily quantifiable measure of the rate of the enzyme activity.
Experimental Procedure
Extraction of Catalase
1. Peel a fresh potato tuber and cut the tissue into small cubes. Weigh out 50 g of potato cubes.
2. Place the potato cubes, 50 mL of cold distilled water, and a small amount of crushed ice in a blender.
3. Homogenize for 30 seconds at high speed. From this point on, the enzyme preparation must be carried out in an ice bath.
4. Filter the potato extract, then pour the filtrate into a 100 mL graduated cylinder.
5. Add ice-cold distilled water to bring up the final volume to 100 mL. Mix well. This extract is the 100% enzyme solution.
6. Note: This rough 100% enzyme solution should work OK although it is worth testing it before proceeding with the experiment. At room temperature (approx. 20°C) in a 1% H2O2 solution it would be sensible if the disk took about 20 seconds to rise in the beaker you are using. If it is faster than this, dilute the enzyme and use that as the 100% solution. If it is slower, prepare the extract again, starting with an increased amount of potato cubes. If the filter rises too quickly at room temperature, then the reaction at higher temperatures will be too quick to measure. If the filter rises too slowly, then the lower temperatures will take forever.
7. Keep your catalase preparation on ice.
The Effect of Temperature on Enzyme Activity
1. Label five 50 mL beakers with the temperature for testing (0, 10, 20, 30, and 40°C).
2. Add 40 mL of 1% hydrogen peroxide solution taken from the appropriate temperature water bath to each beaker.
3. Put the beakers in the appropriate water bath.4. Using forceps, immerse a filter paper disk into the
catalase solution you have prepared.5. Allow the disc to absorb the enzyme solution for 5
seconds, then remove it and drain off the excess enzyme solution by touching the filter paper to the edge of the beaker.
6. Drop the disc into the first substrate solution.7. The oxygen produced from the breakdown of the
hydrogen peroxide by catalase becomes trapped in the fibers of the disc causing the disc to float to the surface of the solution.
8. The time (t) in seconds, from the second the disc touches the solution to the time it again reaches the surface is an indirect measure of enzyme activity.
9. Remove the disk from the beaker once it reaches the surface and dispose of it.
10. Record the time taken in a table in your notebook.
11.Clean the beaker and repeat the procedure until you have 5 replicates at the first temperature.
12. Repeat for each temperature so you have data for 0, 10, 20, 30, and 40°C.
13. Calculate mean and standard deviation for each temperature.
14. Construct and label a graph of your results.
The Effect of Enzyme Concentration on Reaction Rate
It is important to demonstrate that the enzyme assay shows that the enzyme actually follows accepted chemical principles. One way to demonstrate this is by determining the effect of enzyme concentration on the rate of activity while using a substrate concentration, in this case H2O2, that is in excess. This can be easily demonstrated with the experimental system used in the previous section. This set of experiments can be done most conveniently at room temperature.
1. Label 5 tubes 100%, 75%, 50%, 25%, and 0%.2. Add 1, 0.75, 0.5, and 0.25 mL of 100% enzyme
extract to the first 4 tubes .3. Using a separate pipette, skip the first tube (100%
enzyme) add 0.25, 0.5, 0.75, and 1 mL ice cold distilled water to the last four tubes and mix. Each tube will have now contain 1 mL of solution at the indicated enzyme concentration.
4. Keep all tubes on ice throughout.5. Set up 5 beakers, each with 40 ml 1% H2O2 on the
bench.6. Check that the temperature in each one has
equilibrated to room temperature.7. Using the 100% enzyme solution, measure the
reaction time using the disk enzyme assay for 5 replicate disks.
8. Repeat the procedure for each of the other four enzyme concentrations.
9. Plot the rate of reaction about enzyme concentration.
(9) [PA] investigate the effects on plant cells of immersion in solutions of different water potentials;
Water flows to regions from a higher potential to a lower potential down the water potential gradient, especially via the selectively-permeable plasma membrane by osmosis. kPa can affect the movement of water down the water potential gradient. Sufficient kPa can 'push' water up the gradient as well. Usually, a calibration table of kPa to water potential is provided if you need to calculate the kPa outside the cell or the water potential inside/outside the cell body.
(10) [PA] describe, with the aid of diagrams, the behavior of chromosomes during the mitotic cell cycle and the associated behavior of the nuclear envelope, cell membrane, centrioles and spindle fibers (names of the main stages are expected);
Introduction
One of the by-products of many cellular reactions is hydrogen peroxide (H2O2). It is extremely toxic to living cells. All aerobic organisms use oxygen for respiration or oxidation of nutrients. During reduction of molecular oxygen to water, hydrogen peroxide is generated. Two examples of reactions that produce H2O2 are conversions of amino acids into "fuel" molecules and conversion of lipids to carbohydrates. It can damage DNA, protein and lipid membranes and may even be a causative factor in cancer. There are some human immune system cells that actually use H2O2 to kill foreign invaders. The catalase enzyme is specific for the hydrolysis of H2O2:
Catalase is found in animal and plant tissues, and is especially abundant in plant storage organs such as potato tubers, corms, and in the fleshy parts of fruits. You will use catalase isolated from potato tubers and measure its rate of activity under different conditions.
Like other enzymes, catalase is a protein. Enzymes speed up chemical reactions by reducing the activation energy required to convert substrate(s) into product(s). Enzymes have specialized binding sites to do this.
Because enzymes are proteins, they are somewhat fragile. They can be denatured by heat, and can easily be broken down by proteases when cells are homogenized. To preserve activity of proteins in solution, it is important to keep the solutions on ice until you are ready to use them. Denaturing conditions, such as boiling, can also be used as evidence to show that an enzyme-based reaction is protein-dependent.
In the experimental protocol described here a filter paper disk will be immersed in a solution of the enzyme, then placed in the hydrogen peroxide. The oxygen produced from the subsequent reaction becomes trapped in the disc and will give it buoyancy. The time measured from the moment the disc touches the bottom of the container substrate to the time it reaches the surface of the solution is an indirect, but easily quantifiable measure of the rate of the enzyme activity.
Experimental Procedure
Extraction of Catalase
1. Peel a fresh potato tuber and cut the tissue into small cubes. Weigh out 50 g of potato cubes.
2. Place the potato cubes, 50 mL of cold distilled water, and a small amount of crushed ice in a blender.
3. Homogenize for 30 seconds at high speed. From this point on, the enzyme preparation must be carried out in an ice bath.
4. Filter the potato extract, then pour the filtrate into a 100 mL graduated cylinder.
5. Add ice-cold distilled water to bring up the final volume to 100 mL. Mix well. This extract is the 100% enzyme solution.
6. Note: This rough 100% enzyme solution should work OK although it is worth testing it before proceeding with the experiment. At room temperature (approx. 20°C) in a 1% H2O2 solution it would be sensible if the disk took about 20 seconds to rise in the beaker you are using. If it is faster than this, dilute the enzyme and use that as the 100% solution. If it is slower, prepare the extract again, starting with an increased amount of potato cubes. If the filter rises too quickly at room temperature, then the reaction at higher temperatures will be too quick to measure. If the filter rises too slowly, then the lower temperatures will take forever.
7. Keep your catalase preparation on ice.
The Effect of Temperature on Enzyme Activity
1. Label five 50 mL beakers with the temperature for testing (0, 10, 20, 30, and 40°C).
2. Add 40 mL of 1% hydrogen peroxide solution taken from the appropriate temperature water bath to each beaker.
3. Put the beakers in the appropriate water bath.4. Using forceps, immerse a filter paper disk into the
catalase solution you have prepared.5. Allow the disc to absorb the enzyme solution for 5
seconds, then remove it and drain off the excess enzyme solution by touching the filter paper to the edge of the beaker.
6. Drop the disc into the first substrate solution.7. The oxygen produced from the breakdown of the
hydrogen peroxide by catalase becomes trapped in the fibers of the disc causing the disc to float to the surface of the solution.
8. The time (t) in seconds, from the second the disc touches the solution to the time it again reaches the surface is an indirect measure of enzyme activity.
9. Remove the disk from the beaker once it reaches the surface and dispose of it.
10. Record the time taken in a table in your notebook.
11.Clean the beaker and repeat the procedure until you have 5 replicates at the first temperature.
12. Repeat for each temperature so you have data for 0, 10, 20, 30, and 40°C.
13. Calculate mean and standard deviation for each temperature.
14. Construct and label a graph of your results.
The Effect of Enzyme Concentration on Reaction Rate
It is important to demonstrate that the enzyme assay shows that the enzyme actually follows accepted chemical principles. One way to demonstrate this is by determining the effect of enzyme concentration on the rate of activity while using a substrate concentration, in this case H2O2, that is in excess. This can be easily demonstrated with the experimental system used in the previous section. This set of experiments can be done most conveniently at room temperature.
1. Label 5 tubes 100%, 75%, 50%, 25%, and 0%.2. Add 1, 0.75, 0.5, and 0.25 mL of 100% enzyme
extract to the first 4 tubes .3. Using a separate pipette, skip the first tube (100%
enzyme) add 0.25, 0.5, 0.75, and 1 mL ice cold distilled water to the last four tubes and mix. Each tube will have now contain 1 mL of solution at the indicated enzyme concentration.
4. Keep all tubes on ice throughout.5. Set up 5 beakers, each with 40 ml 1% H2O2 on the
bench.6. Check that the temperature in each one has
equilibrated to room temperature.7. Using the 100% enzyme solution, measure the
reaction time using the disk enzyme assay for 5 replicate disks.
8. Repeat the procedure for each of the other four enzyme concentrations.
9. Plot the rate of reaction about enzyme concentration.
(9) [PA] investigate the effects on plant cells of immersion in solutions of different water potentials;
Water flows to regions from a higher potential to a lower potential down the water potential gradient, especially via the selectively-permeable plasma membrane by osmosis. kPa can affect the movement of water down the water potential gradient. Sufficient kPa can 'push' water up the gradient as well. Usually, a calibration table of kPa to water potential is provided if you need to calculate the kPa outside the cell or the water potential inside/outside the cell body.
(10) [PA] describe, with the aid of diagrams, the behavior of chromosomes during the mitotic cell cycle and the associated behavior of the nuclear envelope, cell membrane, centrioles and spindle fibers (names of the main stages are expected);
Introduction
One of the by-products of many cellular reactions is hydrogen peroxide (H2O2). It is extremely toxic to living cells. All aerobic organisms use oxygen for respiration or oxidation of nutrients. During reduction of molecular oxygen to water, hydrogen peroxide is generated. Two examples of reactions that produce H2O2 are conversions of amino acids into "fuel" molecules and conversion of lipids to carbohydrates. It can damage DNA, protein and lipid membranes and may even be a causative factor in cancer. There are some human immune system cells that actually use H2O2 to kill foreign invaders. The catalase enzyme is specific for the hydrolysis of H2O2:
Catalase is found in animal and plant tissues, and is especially abundant in plant storage organs such as potato tubers, corms, and in the fleshy parts of fruits. You will use catalase isolated from potato tubers and measure its rate of activity under different conditions.
Like other enzymes, catalase is a protein. Enzymes speed up chemical reactions by reducing the activation energy required to convert substrate(s) into product(s). Enzymes have specialized binding sites to do this.
Because enzymes are proteins, they are somewhat fragile. They can be denatured by heat, and can easily be broken down by proteases when cells are homogenized. To preserve activity of proteins in solution, it is important to keep the solutions on ice until you are ready to use them. Denaturing conditions, such as boiling, can also be used as evidence to show that an enzyme-based reaction is protein-dependent.
In the experimental protocol described here a filter paper disk will be immersed in a solution of the enzyme, then placed in the hydrogen peroxide. The oxygen produced from the subsequent reaction becomes trapped in the disc and will give it buoyancy. The time measured from the moment the disc touches the bottom of the container substrate to the time it reaches the surface of the solution is an indirect, but easily quantifiable measure of the rate of the enzyme activity.
Experimental Procedure
Extraction of Catalase
1. Peel a fresh potato tuber and cut the tissue into small cubes. Weigh out 50 g of potato cubes.
2. Place the potato cubes, 50 mL of cold distilled water, and a small amount of crushed ice in a blender.
3. Homogenize for 30 seconds at high speed. From this point on, the enzyme preparation must be carried out in an ice bath.
4. Filter the potato extract, then pour the filtrate into a 100 mL graduated cylinder.
5. Add ice-cold distilled water to bring up the final volume to 100 mL. Mix well. This extract is the 100% enzyme solution.
6. Note: This rough 100% enzyme solution should work OK although it is worth testing it before proceeding with the experiment. At room temperature (approx. 20°C) in a 1% H2O2 solution it would be sensible if the disk took about 20 seconds to rise in the beaker you are using. If it is faster than this, dilute the enzyme and use that as the 100% solution. If it is slower, prepare the extract again, starting with an increased amount of potato cubes. If the filter rises too quickly at room temperature, then the reaction at higher temperatures will be too quick to measure. If the filter rises too slowly, then the lower temperatures will take forever.
7. Keep your catalase preparation on ice.
The Effect of Temperature on Enzyme Activity
1. Label five 50 mL beakers with the temperature for testing (0, 10, 20, 30, and 40°C).
2. Add 40 mL of 1% hydrogen peroxide solution taken from the appropriate temperature water bath to each beaker.
3. Put the beakers in the appropriate water bath.4. Using forceps, immerse a filter paper disk into the
catalase solution you have prepared.5. Allow the disc to absorb the enzyme solution for 5
seconds, then remove it and drain off the excess enzyme solution by touching the filter paper to the edge of the beaker.
6. Drop the disc into the first substrate solution.7. The oxygen produced from the breakdown of the
hydrogen peroxide by catalase becomes trapped in the fibers of the disc causing the disc to float to the surface of the solution.
8. The time (t) in seconds, from the second the disc touches the solution to the time it again reaches the surface is an indirect measure of enzyme activity.
9. Remove the disk from the beaker once it reaches the surface and dispose of it.
10. Record the time taken in a table in your notebook.
11.Clean the beaker and repeat the procedure until you have 5 replicates at the first temperature.
12. Repeat for each temperature so you have data for 0, 10, 20, 30, and 40°C.
13. Calculate mean and standard deviation for each temperature.
14. Construct and label a graph of your results.
The Effect of Enzyme Concentration on Reaction Rate
It is important to demonstrate that the enzyme assay shows that the enzyme actually follows accepted chemical principles. One way to demonstrate this is by determining the effect of enzyme concentration on the rate of activity while using a substrate concentration, in this case H2O2, that is in excess. This can be easily demonstrated with the experimental system used in the previous section. This set of experiments can be done most conveniently at room temperature.
1. Label 5 tubes 100%, 75%, 50%, 25%, and 0%.2. Add 1, 0.75, 0.5, and 0.25 mL of 100% enzyme
extract to the first 4 tubes .3. Using a separate pipette, skip the first tube (100%
enzyme) add 0.25, 0.5, 0.75, and 1 mL ice cold distilled water to the last four tubes and mix. Each tube will have now contain 1 mL of solution at the indicated enzyme concentration.
4. Keep all tubes on ice throughout.5. Set up 5 beakers, each with 40 ml 1% H2O2 on the
bench.6. Check that the temperature in each one has
equilibrated to room temperature.7. Using the 100% enzyme solution, measure the
reaction time using the disk enzyme assay for 5 replicate disks.
8. Repeat the procedure for each of the other four enzyme concentrations.
9. Plot the rate of reaction about enzyme concentration.
(9) [PA] investigate the effects on plant cells of immersion in solutions of different water potentials;
Water flows to regions from a higher potential to a lower potential down the water potential gradient, especially via the selectively-permeable plasma membrane by osmosis. kPa can affect the movement of water down the water potential gradient. Sufficient kPa can 'push' water up the gradient as well. Usually, a calibration table of kPa to water potential is provided if you need to calculate the kPa outside the cell or the water potential inside/outside the cell body.
(10) [PA] describe, with the aid of diagrams, the behavior of chromosomes during the mitotic cell cycle and the associated behavior of the nuclear envelope, cell membrane, centrioles and spindle fibers (names of the main stages are expected);
Introduction
One of the by-products of many cellular reactions is hydrogen peroxide (H2O2). It is extremely toxic to living cells. All aerobic organisms use oxygen for respiration or oxidation of nutrients. During reduction of molecular oxygen to water, hydrogen peroxide is generated. Two examples of reactions that produce H2O2 are conversions of amino acids into "fuel" molecules and conversion of lipids to carbohydrates. It can damage DNA, protein and lipid membranes and may even be a causative factor in cancer. There are some human immune system cells that actually use H2O2 to kill foreign invaders. The catalase enzyme is specific for the hydrolysis of H2O2:
Catalase is found in animal and plant tissues, and is especially abundant in plant storage organs such as potato tubers, corms, and in the fleshy parts of fruits. You will use catalase isolated from potato tubers and measure its rate of activity under different conditions.
Like other enzymes, catalase is a protein. Enzymes speed up chemical reactions by reducing the activation energy required to convert substrate(s) into product(s). Enzymes have specialized binding sites to do this.
Because enzymes are proteins, they are somewhat fragile. They can be denatured by heat, and can easily be broken down by proteases when cells are homogenized. To preserve activity of proteins in solution, it is important to keep the solutions on ice until you are ready to use them. Denaturing conditions, such as boiling, can also be used as evidence to show that an enzyme-based reaction is protein-dependent.
In the experimental protocol described here a filter paper disk will be immersed in a solution of the enzyme, then placed in the hydrogen peroxide. The oxygen produced from the subsequent reaction becomes trapped in the disc and will give it buoyancy. The time measured from the moment the disc touches the bottom of the container substrate to the time it reaches the surface of the solution is an indirect, but easily quantifiable measure of the rate of the enzyme activity.
Experimental Procedure
Extraction of Catalase
1. Peel a fresh potato tuber and cut the tissue into small cubes. Weigh out 50 g of potato cubes.
2. Place the potato cubes, 50 mL of cold distilled water, and a small amount of crushed ice in a blender.
3. Homogenize for 30 seconds at high speed. From this point on, the enzyme preparation must be carried out in an ice bath.
4. Filter the potato extract, then pour the filtrate into a 100 mL graduated cylinder.
5. Add ice-cold distilled water to bring up the final volume to 100 mL. Mix well. This extract is the 100% enzyme solution.
6. Note: This rough 100% enzyme solution should work OK although it is worth testing it before proceeding with the experiment. At room temperature (approx. 20°C) in a 1% H2O2 solution it would be sensible if the disk took about 20 seconds to rise in the beaker you are using. If it is faster than this, dilute the enzyme and use that as the 100% solution. If it is slower, prepare the extract again, starting with an increased amount of potato cubes. If the filter rises too quickly at room temperature, then the reaction at higher temperatures will be too quick to measure. If the filter rises too slowly, then the lower temperatures will take forever.
7. Keep your catalase preparation on ice.
The Effect of Temperature on Enzyme Activity
1. Label five 50 mL beakers with the temperature for testing (0, 10, 20, 30, and 40°C).
2. Add 40 mL of 1% hydrogen peroxide solution taken from the appropriate temperature water bath to each beaker.
3. Put the beakers in the appropriate water bath.4. Using forceps, immerse a filter paper disk into the
catalase solution you have prepared.5. Allow the disc to absorb the enzyme solution for 5
seconds, then remove it and drain off the excess enzyme solution by touching the filter paper to the edge of the beaker.
6. Drop the disc into the first substrate solution.7. The oxygen produced from the breakdown of the
hydrogen peroxide by catalase becomes trapped in the fibers of the disc causing the disc to float to the surface of the solution.
8. The time (t) in seconds, from the second the disc touches the solution to the time it again reaches the surface is an indirect measure of enzyme activity.
9. Remove the disk from the beaker once it reaches the surface and dispose of it.
10. Record the time taken in a table in your notebook.
11.Clean the beaker and repeat the procedure until you have 5 replicates at the first temperature.
12. Repeat for each temperature so you have data for 0, 10, 20, 30, and 40°C.
13. Calculate mean and standard deviation for each temperature.
14. Construct and label a graph of your results.
The Effect of Enzyme Concentration on Reaction Rate
It is important to demonstrate that the enzyme assay shows that the enzyme actually follows accepted chemical principles. One way to demonstrate this is by determining the effect of enzyme concentration on the rate of activity while using a substrate concentration, in this case H2O2, that is in excess. This can be easily demonstrated with the experimental system used in the previous section. This set of experiments can be done most conveniently at room temperature.
1. Label 5 tubes 100%, 75%, 50%, 25%, and 0%.2. Add 1, 0.75, 0.5, and 0.25 mL of 100% enzyme
extract to the first 4 tubes .3. Using a separate pipette, skip the first tube (100%
enzyme) add 0.25, 0.5, 0.75, and 1 mL ice cold distilled water to the last four tubes and mix. Each tube will have now contain 1 mL of solution at the indicated enzyme concentration.
4. Keep all tubes on ice throughout.5. Set up 5 beakers, each with 40 ml 1% H2O2 on the
bench.6. Check that the temperature in each one has
equilibrated to room temperature.7. Using the 100% enzyme solution, measure the
reaction time using the disk enzyme assay for 5 replicate disks.
8. Repeat the procedure for each of the other four enzyme concentrations.
9. Plot the rate of reaction about enzyme concentration.
(9) [PA] investigate the effects on plant cells of immersion in solutions of different water potentials;
Water flows to regions from a higher potential to a lower potential down the water potential gradient, especially via the selectively-permeable plasma membrane by osmosis. kPa can affect the movement of water down the water potential gradient. Sufficient kPa can 'push' water up the gradient as well. Usually, a calibration table of kPa to water potential is provided if you need to calculate the kPa outside the cell or the water potential inside/outside the cell body.
(10) [PA] describe, with the aid of diagrams, the behavior of chromosomes during the mitotic cell cycle and the associated behavior of the nuclear envelope, cell membrane, centrioles and spindle fibers (names of the main stages are expected);
Introduction
One of the by-products of many cellular reactions is hydrogen peroxide (H2O2). It is extremely toxic to living cells. All aerobic organisms use oxygen for respiration or oxidation of nutrients. During reduction of molecular oxygen to water, hydrogen peroxide is generated. Two examples of reactions that produce H2O2 are conversions of amino acids into "fuel" molecules and conversion of lipids to carbohydrates. It can damage DNA, protein and lipid membranes and may even be a causative factor in cancer. There are some human immune system cells that actually use H2O2 to kill foreign invaders. The catalase enzyme is specific for the hydrolysis of H2O2:
Catalase is found in animal and plant tissues, and is especially abundant in plant storage organs such as potato tubers, corms, and in the fleshy parts of fruits. You will use catalase isolated from potato tubers and measure its rate of activity under different conditions.
Like other enzymes, catalase is a protein. Enzymes speed up chemical reactions by reducing the activation energy required to convert substrate(s) into product(s). Enzymes have specialized binding sites to do this.
Because enzymes are proteins, they are somewhat fragile. They can be denatured by heat, and can easily be broken down by proteases when cells are homogenized. To preserve activity of proteins in solution, it is important to keep the solutions on ice until you are ready to use them. Denaturing conditions, such as boiling, can also be used as evidence to show that an enzyme-based reaction is protein-dependent.
In the experimental protocol described here a filter paper disk will be immersed in a solution of the enzyme, then placed in the hydrogen peroxide. The oxygen produced from the subsequent reaction becomes trapped in the disc and will give it buoyancy. The time measured from the moment the disc touches the bottom of the container substrate to the time it reaches the surface of the solution is an indirect, but easily quantifiable measure of the rate of the enzyme activity.
Experimental Procedure
Extraction of Catalase
1. Peel a fresh potato tuber and cut the tissue into small cubes. Weigh out 50 g of potato cubes.
2. Place the potato cubes, 50 mL of cold distilled water, and a small amount of crushed ice in a blender.
3. Homogenize for 30 seconds at high speed. From this point on, the enzyme preparation must be carried out in an ice bath.
4. Filter the potato extract, then pour the filtrate into a 100 mL graduated cylinder.
5. Add ice-cold distilled water to bring up the final volume to 100 mL. Mix well. This extract is the 100% enzyme solution.
6. Note: This rough 100% enzyme solution should work OK although it is worth testing it before proceeding with the experiment. At room temperature (approx. 20°C) in a 1% H2O2 solution it would be sensible if the disk took about 20 seconds to rise in the beaker you are using. If it is faster than this, dilute the enzyme and use that as the 100% solution. If it is slower, prepare the extract again, starting with an increased amount of potato cubes. If the filter rises too quickly at room temperature, then the reaction at higher temperatures will be too quick to measure. If the filter rises too slowly, then the lower temperatures will take forever.
7. Keep your catalase preparation on ice.
The Effect of Temperature on Enzyme Activity
1. Label five 50 mL beakers with the temperature for testing (0, 10, 20, 30, and 40°C).
2. Add 40 mL of 1% hydrogen peroxide solution taken from the appropriate temperature water bath to each beaker.
3. Put the beakers in the appropriate water bath.4. Using forceps, immerse a filter paper disk into the
catalase solution you have prepared.5. Allow the disc to absorb the enzyme solution for 5
seconds, then remove it and drain off the excess enzyme solution by touching the filter paper to the edge of the beaker.
6. Drop the disc into the first substrate solution.7. The oxygen produced from the breakdown of the
hydrogen peroxide by catalase becomes trapped in the fibers of the disc causing the disc to float to the surface of the solution.
8. The time (t) in seconds, from the second the disc touches the solution to the time it again reaches the surface is an indirect measure of enzyme activity.
9. Remove the disk from the beaker once it reaches the surface and dispose of it.
10. Record the time taken in a table in your notebook.
11.Clean the beaker and repeat the procedure until you have 5 replicates at the first temperature.
12. Repeat for each temperature so you have data for 0, 10, 20, 30, and 40°C.
13. Calculate mean and standard deviation for each temperature.
14. Construct and label a graph of your results.
The Effect of Enzyme Concentration on Reaction Rate
It is important to demonstrate that the enzyme assay shows that the enzyme actually follows accepted chemical principles. One way to demonstrate this is by determining the effect of enzyme concentration on the rate of activity while using a substrate concentration, in this case H2O2, that is in excess. This can be easily demonstrated with the experimental system used in the previous section. This set of experiments can be done most conveniently at room temperature.
1. Label 5 tubes 100%, 75%, 50%, 25%, and 0%.2. Add 1, 0.75, 0.5, and 0.25 mL of 100% enzyme
extract to the first 4 tubes .3. Using a separate pipette, skip the first tube (100%
enzyme) add 0.25, 0.5, 0.75, and 1 mL ice cold distilled water to the last four tubes and mix. Each tube will have now contain 1 mL of solution at the indicated enzyme concentration.
4. Keep all tubes on ice throughout.5. Set up 5 beakers, each with 40 ml 1% H2O2 on the
bench.6. Check that the temperature in each one has
equilibrated to room temperature.7. Using the 100% enzyme solution, measure the
reaction time using the disk enzyme assay for 5 replicate disks.
8. Repeat the procedure for each of the other four enzyme concentrations.
9. Plot the rate of reaction about enzyme concentration.
(9) [PA] investigate the effects on plant cells of immersion in solutions of different water potentials;
Water flows to regions from a higher potential to a lower potential down the water potential gradient, especially via the selectively-permeable plasma membrane by osmosis. kPa can affect the movement of water down the water potential gradient. Sufficient kPa can 'push' water up the gradient as well. Usually, a calibration table of kPa to water potential is provided if you need to calculate the kPa outside the cell or the water potential inside/outside the cell body.
(10) [PA] describe, with the aid of diagrams, the behavior of chromosomes during the mitotic cell cycle and the associated behavior of the nuclear envelope, cell membrane, centrioles and spindle fibers (names of the main stages are expected);
