Laboratory work. Determination of catalase activity
(According to A.N. Bahu and A.I. Oparina)
All diverse transformations that make up the basis of the life of the body occur with the participation of biological catalysts - enzymes that are specific proteins.
Thanks to enzymes, one of the wonderful features of living cells is manifested - the ability to implement the most complex reactions in a very short time and at a relatively low body temperature. The biological significance of this phenomenon is very large.
To study the action of enzymes, it is necessary to allocate them from alive tissues. Usually, for this purpose, an easily accessible source rich in this enzyme is selected. In order to translate the enzyme into the solution, it is necessary to destroy the cell shells, which is achieved using a homogenizer, rubbing into a mortar with pestle, freezing and thawing, autolysis, etc. Usually, the destruction of cellular shells is produced at low temperatures, due to which all enzymatic processes are suspended in tissues.
The group of redox enzymes includes a hem containing enzyme catalasewhich catalyzes the reaction of the decomposition of hydrogen peroxide with the release of molecular oxygen:
2N 2 O 2 - > 2 H 2 O + O 2
Catalase is located in most fabrics of a living organism.
Principle of the method.
Hydrogen peroxide is decomposed by catalase. The excess of hydrogen peroxide is titrated by potassium permanganate in an acidic environment. The reaction is under the equation:
2 KMNO 4 + 5 H 2 O 2 + 3 H 2 SO 4 - > 2 MNSO 4 + K 2 SO 4 + 8 H 2 O + 5 O 2
In the experiment, the number of the remaining non-destructive hydrogen peroxide is determined, in the control - the total number of hydrogen peroxide (catalase in the control sample is inactivated by boiling).
Responted the results of the experience from the results of control, learn the number of hydrogen peroxide destroyed at a certain period of time, which makes it possible to judge the activity of the catalase.
Equipment: 1. Burents direct by 50 and 100 ml (1 pcs.)
2. Cylinder on 10 ml
3. mortar with pestle
4. Conical flasks for 200-250 ml (2 pcs.)
5. Technical scales with variations
6. Pipette on 10 ml "Enzyme exhaust"
7. Pipette "H 2 SO 4"
8. Boiling water bath
9. Spatula
10. Flask Mernery per 100 ml
11. Voronek
12. Filter paper
Materials: 1. Fresh vegetable material (carrots or potatoes)
2. 0.1 H 2 O 2 solution
3. 0.1N KMNO4 solution
4. 10% solution H 2 SO 4
5. Saco 3 (Cryst.)
6. Sand quartz
Progress:
2 g of raw potatoes (or carrots) are triturated with quartz sand in a mortar, gradually adding 2-3 ml of water. To reduce acid reaction, add calcium carbonate spike to stop excretion of carbon dioxide bubbles. The mass is quantitatively transferred to the measuring flask and bring to 100 ml with water. The mixture is left to stand for 30-60 minutes, after which it is filtered.
In a conical flask on 200 ml, 25 ml of 0.1H hydrogen peroxide solution is taken from a burette and a pipette of 20 ml of an enzyme exhaust is added there. After 30 minutes, the action of the enzyme is stopped by adding 5 ml of a 10% solution of sulfuric acid and titrate a mixture of 0.1H potassium permanganate solution (before the formation of a resistant for about 1 min of pink staining). The number of milliliters of the potassium permanganate solution, which went to the titration of the remaining hydrogen peroxide was noted.
At the same time, they are controlled with inactivated heating in a boiling water bath for 5 minutes by an enzyme solution (20 ml) to this solution after cooling, 25 ml of 0.1H hydrogen peroxide solution is added. The mixture is left to stand for 30 minutes, after which 5 ml of a 10% solution of sulfuric acid is added and titated with 0.1 h potassium permanganate solution. The number of milliliters permanganate potassium, which went to title the total amount of hydrogen peroxide.
In terms of the difference between the experimental and control titration, the amount of permanganate is found equivalent to the amount of decomposed hydrogen peroxide. According to the reaction equation between KMNO 4 and H 2 O 2, 1 ml of 0.1 H permanent solution of potassium permanganate corresponds to 1.7 mg of hydrogen peroxide.
Example of calculation.
Of the 1.25 g of carrots prepared a catalase exhaust with a volume of 100 ml. 15.5 ml testing was expended on the titration of the experimental sample, the control-30.2 ml of 0.1 H permanent solution of potassium permanganate. The amount of decomposed hydrogen peroxide in the sample is equivalent to 30.2-15.5 \u003d 14.7 ml of 0.1 H permanent solution of potassium permanganate and, therefore, 14.7 * 1.7 \u003d 24.99 mg.
In 1 g of crude carrots, it contains the amount of catalase, capable of decomposing in 30 minutes (24.99 * 100) / (20 * 1.25) \u003d 99.96 mg of hydrogen peroxide, and for 1 min - 99.96 / 30 \u003d 3.33 mg. As
1 μmol of hydrogen peroxide is 0.034 mg, then 33.3 / 0.034 \u003d 100 e of catalases are present in 1 g of carrots.
Laboratory work number 4.
Preparation of sacrament from yeast cells. The specificity of the action of enzymes.
All diverse chemical transformations, which form the basis of the life of the body, proceed with the participation of biological catalysts - enzymes, which are with n e c and f and h c and m and b e l k and m and.
Thanks to enzymes, one of the remarkable features of living cells is manifested - the ability to implement the most complex reactions in a very short time and relatively at low body temperature.
The study of the properties of enzymes, the conditions of their action, the determination of the content of enzymes in various organs and tissues is of great importance for the correct understanding of the most complex processes of the body's life.
One of the most important properties of enzymes is the specificity of their action with respect to a specific substrate. The specificity of the catalytic properties of enzymes is manifested in the fact that the enzyme, as a rule, acts only on a certain substance. On the strict specificity of enzymes indicates the fact that in cases of stereoisomeria, a certain enzyme catalyzes only one stereoisomer splitting. The specificity of enzymes is their essential biological property, without which ordered metabolism is impossible.
In this paper, the processes of splitting by the enzyme saharase substrate - sucrose and the splitting of starch by the enzyme - amylase, which is contained in the human saliva.
Extraction of sacrament from yeast cells
Materials and reagents:
· Pressed yeast - 10 g
· Homogenizer (mortar with pestle)
· Quartz sand
· Distilled water
Progress
10 grams of dry yeast put into a mortar, add 10 ml of distilled water and homogenize in a porcelain mortar with a small amount of quartz sand for the destruction of cell walls. Then a porcelain mortar with a homogenizat is placed in a drying cabinet with a temperature of 60 0 sleep 30-40 minutes.
After the specified time expires, remove the mortar, cool, add 30 ml of distilled water and rub the contents of the mortar to a homogeneous mass.
Then, homogenizate, to precipitate the cell mass, centrifuged at 3000 rpm for 15 minutes. The resulting supernatant is the extract of sugar.
LABORATORY WORK
Determination of catalase activity (1.11.1.6)
1) with potassium permanganate and catalase calculation(Bach and Zubkov method)
Principle of the method. The catalase enzyme is contained in large quantities in red blood cells, as well as in all tissues and body fluids. The biological role of the catalase is to neutralize hydrogen peroxide (n2 O 2. ) by decomposition on molecular oxygen and water:
2N 2 O 2 → O 2 + 2N 2 O
The activity of the enzyme is expressed using a catalase and catalase indicator.Catalase number They call the amount of hydrogen peroxide, which splits 1.0 μl of blood over a certain period of time. The amount of precipitated hydrogen peroxide is judged by the difference in the number of potassium permanganate, which went to the titration of control and prototypes.
Catalase indicator Calculate the ratio of the catalase number to the amount of millions of erythrocytes in 1.0 μl of the blood under study.
Reagents: 1) 1% solution H 2 O 2; 2) 10% sulfuric acid solution; 3) 0.1 M Potassium permanganate solution (for titration).
Object of study: Hemolyzate obtained by breeding blood distilled water in a 1: 1000 ratio.
Cooking : A 10 ml of distilled water is poured into a dimensional flask per 100 ml and 0.1 ml of blood is added to micropipette. Pipette is washed several times the same solution. Water is added to the flask to the label and obtain the base solution of blood (1: 1000), which is used to determine the catalase number.
Progress.
In two tetting flasks (experience and control) poured 7 ml of water and in each add 1 ml of the main blood solution. Each flask is made exactly 2 ml. 3 ml is added to the control flask to split the catalase. Both flasks are incubated at room temperature for 30 minutes. Then, 3 ml of 10% sulfate acid solution is poured into an experimental flask. The contents of both flasks are titrated by 0.1 M potassium permanganate solution until pink staining appears. Multiple the difference between the results of experience and control by 1.7 and receive a catalase number of blood.
Example of calculation : Mol Equivalent N2 O 2. equals 17 g. So, in 1 ml of 0.1 M solution contains 1.7 mg2 O 2. ; Multiplying 1.7 to the difference between the amount of ml of 0.1 M of the potassium permanganate solution, which went to title the contents of the control and experimental flask, receive the number of mg2 O 2. which splits 1 μl of the blood under study, that is, they immediately calculate the catalase.
Norms: catalase ranges from 10 to 15 units,
Catalase indicator is 2-3x10 -: 6 , in the clinic it is used more often
Clinical and diagnostic value.High catalase activity is observed in pernicious and macrocytic anemia, as well as when entering alcohol, caffeine, acetone tel to the body. The activity of catalase in the blood is reduced with cancer, anemia, tuberculosis and other diseases.
2) using ammonium molybdate
Catalase is one of the most phylogenetically ancient enzymes of the antioxidant system of the organism, refers to the class of oxidoreductases, catalyzing the redox reactions, is included in the hydroperoxidase group using the substrate2 O 2 (2 H 2 O 2 → 2N 2 O + O 2 ) or organic hydropercycles, therefore, along with catalase, it has peroxidase activity. According to the structure - hemoprotein containing 4 hem groups. Catalase is an intracellular enzyme. In the circulating blood, the larger share of the enzyme is localized in the cytoplasm of erythrocytes.
Catalase functions:
Participates in the protection of the body from endogenous hydrogen peroxide, which is generated as a result of the functioning of aerobic dehydrogenases;
Suppresses the formation of hydroxyl radicals;
Protects against oxidation of hemoglobin and contributes to the transfer of oxygen inside the cellular structures;
Participates in the oxidative metabolism of some amino acids;
Protects from the oxidation of the SH-Group, including those included in the active center of many enzymes and functionally active proteins.
Principle of the method. The activity of the catalase is determined by the conversion of the substrate enzyme (hydrogen peroxide), which is capable of the formation of a painted complex with ammonium molybdate salts.
Reagents. 1) 0.03% solution H 2 O 2 ; 2) 4% ammonium molybdate solution.
Biological material:blood serum.
Progress. Experienced test: To 2.0 ml of a 0.03% hydrogen peroxide solution add 0.1 ml of serum. INhaving a test Instead of serum - 0.1 ml of distilled water.IN control sample Instead of peroxide, 2.0 ml of distilled water is added. Samples incubate 10 min at 37° C, then stop the reaction by adding 1.0 ml of a 4% solution of ammonium molybdate.Centrifuge 10 min at 4000 rpm. Measure the optical density of idle and prototypes against control at a wavelength of 410 nm.
Payment Catalase activity in serum:
A (MKAT / L) \u003d
where e op and e - extinction of experienced and idle samples,
3.1 - general breeding,
T - time incubation, C,
V. - the volume of the injured sample, l,
22,210 3 - Extinction coefficient,mmol -1 cm -1.
Catalase activity in serum 2,6+ 0.5 μUT / l
Registration of work.
Record the principle of the method. Fix the results, draw out.
The principle of the method:molybdenum-oxid ammonium with hydrogen peroxide solution forms a comprehensive yellow compound. Catalase destroys hydrogen peroxide according to the following formula:
2N 2 O 2 \u003d 2N 2 O + O 2
The degree of decrease in the intensity of the color of the solution is proportional to the activity of the catalase.
Progress:a 4 ml of 0.03% hydrogen peroxide solution is added to the experimental and control tube, 0.2 ml of hemolyzed blood is added to the experimental test tube (1: 1000 dilution). Inhibit the sample 20 minutes at 37 0 C. Then in both test tubes are made in 2 ml of ammonium molybdate solution, and in the control - an additional 0.2 ml of hemolyzate. Stir. Measure the optical density of the experimental and control sample against the water at a blue filter. Determine the activity of catalase by the formula:
(E K - E 0). 5600.
A \u003d ------------------- where
And - the activeness of the catalase (ICMOL N 2 O 2 / min per ml of blood);
E to-extinction of control; E 0 - extinction of experience;
In the time of incubation, min; 5600 - coefficient
Normally, the activity of the catalase is 135 μmol H 2 O 2 / min per ml
blood (in saliva -12-16 μmol). The activity of catalase in the blood may decrease with anemia, tumor growth, tuberculosis, some other diseases, and increase in acute inflammatory processes (in saliva with inflammatory processes of the oral mucosa).
Agranulated and briefly answer the following:
1. Can the oxidation of CH 3 - CH 2 - CH 2 - it ®
CH 3 - CH 2 - CH \u003d O in oxygen-free medium? What conditions are it necessary for this? Write a reaction scheme.
2. Can it and with what condition in an oxygen-free medium occur the oxidation by type CH 3 - CH 2 - CH \u003d O ® CH 3 - CH 2 - Soam? Specify the necessary components, make a reaction scheme. How to clarify the appearance of two oxygen atoms in the reaction product.
3. Is the oxygen exceptional (only) a finite hydrogen acceptor in a tissue respiratory chain and in general in biological oxidation?
4. Why was the oxidation in wildlife identified with burning outside the body? Name the external signs of similarity between the burning and the oxidation process in the body. Name the differences between these processes.
5. Write the dehydrogenation reactions of the connections:
R-CH 2 -CH 2 - R; R-CH \u003d O; R - Sony -R; R-CH \u003d CH-R
6. Set compliance:
Redox potential: A. + 0.82; B. +0.10; V. + 0.25; G.- 0.22.
Component CPE: 1.Ubikinon; 2.Corodorod; 3.FMN; 4. Cytrochrom
7. Which of the listed compounds are substrates of the FAD-dependent dehydrogenase: glucose, sucrose; succinic acid; Glycerin Aldehyde, Nadn +.
8. Write an electron and protons scheme from isocatrate to oxygen (isocitraterathydhydrogenase - over-dependent enzyme) and specify
the name of all enzyme complexes.
9. Medicinal preparations - Bantician derivatives:
A. Remote a hypnotic action.
B. Activate fabric breathing.
Q. Inhibit Nadn-dehydrogennase.
D. cause a hypoenergetic state.
Laboratory work number 1
The role of enzymes in accelerating the reaction in the cell
(Identification of catalase activity)
Purpose: Detect the action of the catalase enzyme in vegetable and animal cells, compare the enzymatic activity of natural and damaged cell boiling cells.
Equipment: 3% solution of hydrogen peroxide, pieces of raw and boiled potatoes and meat (liver, lungs), test tubes.
Detection of catalase activity
Catalase is an enzyme that catalyzing hydrogen peroxide decomposition with the formation of molecular oxygen released in the form of gas bubbles:
catalase
2 H 2 O 2 _ → 2H 2 O + O 2
The hydrogen peroxide is formed in some plant and animal cells as a by-product of oxidative reaction reactions. This compound is toxic for cells, and catalase ensures efficient removal. Catalase is one of the fastest working enzymes: one catalase molecule decomposes in one second to 200,000 hydrogen peroxide molecules. Catalase is localized in membrane bubbles of cells - microthels and peroxyms.
Progress
Take 4 clean tubes and place a small amount of fine potatoes in the first one of them, in the second - a bit of boiled potatoes, in the third - finely chopped pieces of meat (liver, lightning), in fourth - a little crushed boiled meat. Add 3-4 ml to each test tube. 3% hydrogen peroxide solution. Purchase what is happening in test tubes. Surveillance results include in the table.
Enzymatic activity of natural and damaged cells
| An object | Phenomena observed in the test tube | Explanation of observations |
| Raw potatoes | _____________________________ _____________________________ _____________________________ _____________________________ | _____________________________ _____________________________ _____________________________ _____________________________ _____________________________ |
| Boiled potatoes | _____________________________ _____________________________ _____________________________ _____________________________ _____________________________ | _____________________________ _____________________________ _____________________________ _____________________________ _____________________________ |
| Raw meat | _____________________________ _____________________________ _____________________________ _____________________________ _____________________________ | _____________________________ _____________________________ _____________________________ _____________________________ _____________________________ |
| Boiled meat | _____________________________ _____________________________ _____________________________ _____________________________ _____________________________ | _____________________________ _____________________________ _____________________________ _____________________________ _____________________________ |
Explain the results obtained. Take the conclusion about the catalytic activity of catalase in alive and dead cells.
Output: ________
_______________
_______________
_____________
Control questions:
1. What is enzymes? What structure of proteins is created by their activity? ___________
2. What properties do enzymes have? ______
3. What is called an active enzyme center? How many such centers can be in the enzyme? _
4. Due to what enzymes accelerate chemical reactions? ___________________________
5. * Alcohol, phenol, chlorine and other antiseptics are used in medicine to treat areas of body contaminated by pathogenic flora. Explain why ._____
Gr .___ 3.
Laboratory work number 2
Detection of organic substances
Purpose: Eleterate organic substances in tissues (starch, protein, fats) and explore their properties.
Equipment:a gauze bag, grinding wheat grains (wheat flour), 5% iodine solution, sunflower seeds (or any other oilseed area: cotton, LNA, peanuts, soybean, etc.).
Organic compounds - Carbon-containing substances characteristic of wildlife are an average of 20-30% of the mass of the cells of living organisms. The main properties of cells and organisms are determined by organic polymers: proteins, carbohydrates, nucleic acids, as well as complex compounds - fats and a number of hormone molecules, pigments, separate nucleotides, in particular ATP. In addition to organic substances in cells, mineral substances and water are contained, but the content of organic substances is always higher. The amount of organic substances may be different.
Proteins -irregular, or informational, polymers whose monomers are amino acids.
In its composition, proteins are divided into:
- simple - Consist of alone amino acids. For example, vegetable proteins are prolaminones, they are contained in the gluten seeds of cereals, do not dissolve in water;
- sophisticated - In addition to amino acids, other organic compounds (nucleic acids, lipids, carbohydrates), phosphorus compounds, metals are in its composition. Accordingly, they wear names: nucleoproteis, lipoproteins, glycoproteins, phospho- and metalloproteins.
Carbohydrates - Compounds containing carbon, hydrogen and oxygen. Divided into mono-, di- and polysaccharides. Polysaccharides are high-molecular carbohydrates consisting of a large number of monosaccharides, their molecular weight is large, molecules have a linear or branched structure. In functionality, the polysaccharides of backup and structural purposes are distinguished. Not soluble in water starch - main reserve polysaccharide of vegetable cells (polymer ά - glucose); Under action on it, iodine shines; Contained in large numbers in potato tubers, fruits, seeds. Glycogen- polysaccharide contained in human body tissues and animals, as well as in mushrooms and yeast, plays an important role in the transformation of carbohydrates in cells. Fiber (cellulose) - the main structural polysaccharide of plants cell shells.
Lipids and lipoids- Fats and leafy substances - organic compounds with different structure. They do not dissolve in water, but they are well dissolved in organic compounds: ether, gasoline, chloroform, etc.
According to the chemical structure of the lipid - the compounds of glycerol - trotham alcohol - with high molecular organic acids (fat), do not have a polymer structure.
Composition of seeds
Laboratory work number 1Topic: Catalytic activity of enzymes in living cells
Purpose: Identify the catalytic function of proteins in living cells, form knowledge about the role of enzymes in cells, consolidate the ability to work with a microscope, conduct experiments and explain the results of the work. Equipment: Raw and boiled potatoes, elephant leaf (other plants), fresh 3% hydrogen peroxide solution, test tubes, tweezers, sand, mortar and pestle, notebook, pen, simple pencil, line.
Progress:
Prepare two, and place in the first little sand, in the second - a piece of raw potatoes, to the third - a slice of boiled potatoes, add a bit of hydrogen peroxide into each of the tubes. Jump out what will happen in each of the tubes.
Grind a piece of raw potatoes into a mortar with a small amount of sand.
Transfer chopped potatoes along with sand into the test tube and drop a little hydrogen peroxide.
Compare the activity of crushed and whole plant tissue.
Make a table showing the activity of each tissue with different processing. Explain the results obtained. Answer the questions:
Observations
Hydrogen peroxide and raw potatoes
Oxygen is released, the protein decomposes to the primary structure and turns into a foam
Hydrogen peroxide and boiled potatoes
No reactions
Answer questions: What test tubes showed the activity of the catalase enzyme? Explain why.
Catalase enzyme catalyzing the decomposition reaction of hydrogen peroxide into water and molecular oxygen: H2O2 + H2O2 \u003d O2 + 2N2O. The biological role of K. is to degradation of hydrogen peroxide formed in cells as a result of a series of flavoprotein oxidases (xanthine oxidase, glucose oxidase, monoaminoxidase, etc.), and ensuring the effective protection of cellular structures from destruction under the action of hydrogen peroxide. The genetically determined deficiency of K. is one of the causes of the so-called acatalance - a hereditary disease, clinically manifested by the ulceration of the nasal mucosa and the oral cavity, sometimes sharply pronounced atrophic changes in the alveolar partitions and the dental loss. Activity manifested in 1.3 test tubes, because They had raw foods containing proteins. And in the rest of the test tubes there were products with a protected protected in the cooking process and the reaction was not manifested. Therefore, the organism is better absorbed by the products containing protein.
How does the enzyme activity and dead fabrics manifest?Explain the observed phenomenon. In the dead fabrics there is no activity of enzymes, because The protein in them was destroyed during cooking. And in alive tissues, oxygen was released when interacting with hydrogen peroxide, and the protein split to the primary structure was transformed into a foam.
How does the chopping of the fabric affect the activity of the enzyme in the living tissues of plants and animals? When grinding live tissue, the reaction passes faster, because The area of \u200b\u200bcontact of protein and hydrogen peroxide increases how would you suggest measuring the decomposition rate of hydrogen peroxide? V \u003d kc (a) C (b) where V is the speed of the chemical reaction K - the rate constant C - a change in concentration as you think, whether all the living organisms contain a catalase enzyme that provides a decomposition of hydrogen peroxide?
Justify the answer. Since this is an enzyme class oxidoreductaz, it catalyzes the decomposition of toxic for living cells of hydrogen peroxide into water and oxygen. Contained in lysosomes. It can be concluded that is contained in all cells of living organisms. Explain your observations. Word output.
Conclusion: protein is contained only in living products, and in the boiled products protein destroyed, so no reaction is happening to them. If the products are crushed, the reaction will pass faster.