BIOCHEMISTRY JOURNALGROUP1
Subcellular Fractionation
I. Background of the experiment
Organelles are membrane-enclosed vesicles inside all eukaryotic cells that function in a specific way and a variety of important cellular processes.
The basic principle for all microscopes is that the cell is composed of smaller physical units, the organelles. Definition of the organelles is possible with microscopy, but the function of individual organelles is often beyond the ability of observations through a microscope. We are able to increase our chemical knowledge of organelle function by isolating organelles into reasonably pure fractions.
A host of fractionation procedures are employed by cell biologists. Each organelle has characteristics (size, shape and density for example) which make it different from other organelles within the same cell. If the cell is broken open in a gentle manner, each of its organelles can be subsequently isolated. The process of breaking open cells is homogenization and the subsequent isolation of organelles is fractionation. Isolating the organelles requires the use of physical chemistry techniques, and those techniques can range from the use of simple sieves, gravity sedimentation or differential precipitation, to ultracentrifugation of fluorescent labeled organelles in computer generated density gradients.
Subcellular Fractionation involves the separation of different cell components from homogenous sets, usually organelles done by subjecting the cells to differential centrifugation
The arrangement of macromolecules within a cell is as important to cellular function as their catalytic activities. Cellular compartmentalization provides efficiency by bringing together related compounds that can interfere with each other (i.e. lysosomal hydrolytic enzymes). Cellular compartmentalization is accomplished in part by various subcellular organelles. In this lab module, we will isolate several subcellularorganelle fractions from liver cells, and will examine their various properties. The method we will use to separate the various organelles utilizes differential centrifugation to isolate components of different densities. With this technique, the heaviest or most dense organelles, nuclei pellet in less time and with less force than is required to pellet lighter organelles such as mitochondria. First, a cell homogenate is made by rupturing the cell membranes in the tissue. The homogenate is then centrifuged for a short period of time to remove cell debris and nuclei.
The supernatant is then transferred to another tube and centrifuged longer to pellet the lighter mitochondria.
For this type of fractionation, which tissue we use, and the method of homogenization are dictated by the biological system. Homogeneous cell populations from cell culture are well suited for cell fractionation. Some tissues, like those in the liver also have one cell type that predominates, so are also well suited. Most chlorophyll-free plant tissues are acceptable for preparing mitochondria, but recently-harvested plant tissues are usually required, making their use uncertain during winter months. Once a cell type is chose (we will work with liver), it is important to obtain the organelles in a biochemically active, morphologically whole state. Homogenizers are used to break open the cells without damaging the organelles. Homogenizers have a precise clearance between the glass tube and pestle, which breaks the cell membrane leaving the smaller organelle membranes intact. The homogenization buffer is a solution which often includes sucrose to partially dehydrate the organelles, keeping them intact.
No technique used to isolate organelles is perfect. It is very difficult to get pure unbroken preparations of any organelle. Techniques providing optimal isolation of one organelle may completely rupture another organelle. Thus methods are often used to measure the contamination of one organelle fraction by another. This can be done by analyzing each organelle fraction for organelle-specific marker enzymes.
PROCEDURE FOR SUBCELLULAR FRACTIONATION:
Cell lysis or extraction of the cell componentsCell lysis or cellular disruption is a cell biology method for the release of biological
molecules including organelles, proteins, DNA, RNA and lipids from inside a cell
For most cells, mild osmosis is usually enough to lyze the cells. This is done by lowering the ionic strength of the surrounding solution, causing the cells to swell and burst releasing their contents.
Mild surfactants are often used in addition to mechanical lysis methods.
PROCESS:
Cut 10 g of fresh chicken liver and lyse by soaking in distilled water or tap water for at least one hour.
Cell Homogenization
Homogenization involves in breaking open the cells and releases the cellular components into the resultant homogenate. It is also a process in which a mixture is made uniform throughout. Generally this procedure involves reducing the size of the particles of one component of the mixture and dispersing them evenly throughout the other component.
PROCESS:
Homogenize the cells using normal
saline solution in an approximately 1:3 ratio
Purification or Differential centrifugation
Differential centrifugation is a common procedure in microbiology and cytology used to separate certain organelles from whole cells for further analysis of specific parts of cells. In the process, a tissue sample is first homogenized to break the cell membranes and mix up the cell contents. The homogenate is then subjected to repeated centrifugations, each time removing the pellet and increasing the centrifugal force. Finally, purification may be done through equilibrium sedimentation, and the desired layer is extracted for further analysis. It also separates the individual components within the homogenate according to density, size and shape.
PROCESS:
Supernatant 1: nuclear fraction contains the nuclei and any unbroken cellsSupernatant 2: mitochondrial fraction contains the mitochondria, lysosomes and
microbodiesSupernatant 3: ribosomal fraction contains the ribosomes and microsomes consisting of endoplasmic reticulum,Golgi and plasma membrane fragments
II. CHEMICAL PRICIPLE OF THE DIFFERENT QUALITATIVE TEST
TEST FOR CARBOHYDRATES
Molisch Test
PRINCIPLE:
Molisch test is a sensitive chemical test for the presence of carbohydrates, based on the dehydration of the carbohydrate by sulfuric acid to produce an aldehyde, which condenses with two molecules of phenol (usually α-naphthol, though other phenols (e.g. resorcinol, thymol) also give colored products) resulting in a red- or purple-colored compound.
TEST FOR LIPIDS
Sudan Test
PRINCIPLE:
Sudan test is a test for lipids such as fats and oils based on their ability to selectively absorb pigments in fat-dyes such as the Sudan IV. The test will make red globs appear if it is indeed a lipid.
To test for the presence of lipids in a solution you will use a Sudan IV Test. In this test dark red Sudan IV is added to a solution along with ethanol to dissolve any possible lipids. If lipids are present the Sudan IV will stain them reddish-orange, giving a positive test.
TEST FOR NUCLEIC ACIDS
Bial’s testPRINCIPLE:Bial’s test is a chemical test for the presence of pentoses. The components
include orcinol, hydrochloric acid, and ferric chloride. In the presence of a pentose, the pentose will be dehydrated to form furfural. The solution will turn bluish and a precipitate may form.
The test reagent dehydrates pentoses to form furfural. Furfural further reacts with orcinol and the iron ion present in the test reagent to produce a bluish product.
III. POSITIVE RESULTSMOLISCH TEST
Adding 3 ml of Molisch reagent and 3ml
of freshly prepared H2S04
POSITIVE RESULT: appearance of violet ring
BIURET TEST
0.5 ml cell suspension add 0.5 ml of
10% NaOH and 0.5 ml of 0.5% Cu S04.
POSITIVE RESULT: violet in color
SUDAN TEST
0.5 ml of cell suspension add 0.5 ml of
chloroform and a pinch of Sudan IV
crystals
POSITIVE RESULT: red in color
BIAL’S TEST
Add 0.5 ml of Bial’s reagent to 0.5 ml of cell suspension then heat in water bath for 10 mins.
POSITIVE RESULT: blue green in color
IV. SIGNIFICANCE AND APLLICATION OF THE EXPERIMENT
Scientists use this tool to increase their knowledge of organelle functions. To be able to do so they isolate organelles into pure groups, such as isolating the mitochondria or the nucleus.
Cell fractionation also allows researchers to prepare specific cell components in bulk for research in the functions and structures of cell organelles. This method has already resulted in the knowledge several cell organelles' functions.
For example, by centrifugation a specific cell fraction was determined to have enzymes that function in cellular respiration. This unknown cell fraction was rich in mitochondrias. Therefore there researchers obtained evidence that helped determine mitochondrias was the site of cellular respiration.
Cell Membrane Permeability and Osmosis
I.Background of the Experiment
The cell membrane is a biological membrane that separates the interior of all cells from the outside environment. The cell membrane is selectively-permeable to ions and organic molecules and controls the movement of substances in and out of cells. It consists of the phospholipid bilayer with embedded proteins. Cell membranes are involved in a variety of cellular processes such as cell adhesion, ion conductivity and cell signaling and serve as the attachment surface for the extracellular glycocalyx and cell wall and intracellular cytoskeleton.
Fluid mosaic modelAccording to the fluid mosaic model of S. J. Singer and Garth Nicolson,
the biological membranes can be considered as a two-dimensional liquid where all lipid and protein molecules diffuse more or less easily. This picture may be valid in the space scale of 10 nm. However, the plasma membranes contain different structures or domains that can be classified as: (a) protein-protein complexes; (b) lipid rafts, and (c) pickets and fences formed by the actin-based cytoskeleton.
Lipid bilayer
Lipid bilayers go through a self assembly process in the formation of membranes. The cell membrane consists primarily of a thin layer of amphipathic phospholipids which spontaneously arrange so that the hydrophobic "tail" regions are shielded from the surrounding polar fluid, causing the more hydrophilic "head" regions to associate with the cytosolic and extracellular faces of the resulting bilayer. This forms a continuous, spherical lipid bilayer. Forces such as Van der Waal, electrostatic, hyrdogen bonds, and noncovalent interactions, are all forces that contribute to the formation of the lipid bilayer. Overall, hydrophobic interactions are the major driving force in the formation of lipid bilayers
Membrane Transport
Passive transport involves carriers, channels, or direct diffusion through a membrane.This type of transport always operates from regions of greater concentration to regions of lesser concentration. No external source of energy is required.
Types: · Simple diffusion - a substance passes through a membrane without the aid of an intermediary such as a integral membrane protein.· Channel diffusion -· Facilitated diffusion - it involves proteins known as carriers, which, as in the case of ion channels, are specific for a certain type of solute and can transport substances in either direction across the membrane. However, unlike channels, they facilitate the movements of solutes across the membrane by physically binding to them on one side of the membrane and releasing them on the other side.
Active transport involves the use of proteins that don't just passively facilitate the transport of substances across the cell membrane, but require the use of cellular energy (usually ATP) to actively pump substances into or out of the cell. This type of transport operates from region of lesser concentration to regions of greater concentration.
There are two types of active transport:· Primary active transport – directly uses ATP ex. Sodium pump· Secondary active transport– does not directly use ATP. It takes advantage of a previously existing concentration gradient (via carriers).
Osmosis
Osmosis is the movement of water molecules through a selectively-permeable membrane down a water potential gradient. More specifically, it is the movement of water across a selectively permeable membrane from an area of high water potential (low solute concentration) to an area of low water potential (high solute concentration). It may also be used to describe a physical process in which any solvent moves, without input of energy, across a semi-permeable membrane (permeable to the solvent, but not the solute) separating two solutions of different concentrations. Osmosis releases energy, and can be made to do work, but is a passive process, like diffusion.
II. Chemical Principle of Each Test
OsmosisA. Hemolysis
Hemolysis, breakdown or destruction of red blood cells so that the contained hemoglobin is freed into the surrounding medium. Antibody (lysin) attaches to the red cell but cannot cause bursting in the absence of a normal blood component called complement. Apart from normal breakdown of aged red blood cells, hemolysis is abnormal in the living but may be caused by inherited defects in the blood cells (e.g.,hereditary spherocytosis, thalassemia), by chemicals, venoms, the toxic products of microorganisms, transfusion of the wrong blood type, or Rh incompatibility of fetal and maternal blood, a condition called erythroblastosis fetalis. It is a major finding in hemolytic anemia.
Hemolysis may be produced in the laboratory by various physical agents (heat, freezing, flooding with water, sound); in certain situations it is used as a specific laboratory test for antigen–antibody reactions. Hemolysis caused by physical agents is rare in the living because of body buffering systems; but in the disease paroxysmal cold hemoglobinuria, exposure to cold causes self-produced hemolyzing agents to destroy the individual’s own red cells
Osmotic Pressure
Osmotic pressure is the pressure which needs to be applied to a solution to prevent the inward flow of water across a semipermeable membrane.
Tonicity: Hypertonicity - a solution having a greater concentration of impermeable
solutes on the external side of the membrane. When a cell’s cytoplasm is bathed in a hypertonic solution the water will be drawn into the solution and out of the cell by osmosis. If water molecules continue to diffuse out of the cell, it will cause the cell to shrink, or crenate.
Hypotonicity - is a solution having a lesser concentration of impermeable solutes on the external side of the membrane. When a cell’s cytoplasm is bathed in a hypotonic solution the water will be drawn out of the solution and into the cell by osmosis. If water molecules continue to diffuse into the cell, it will cause the cell to swell, up to the point that cytolysis (rupture) may occur
Isotonicity - A condition or property of a solution in which its solute concentration is the same as the solute concentration of another solution with which it is compared.
Permeability
The rate of passage depends on the pressure, concentration, and temperature of the molecules or solutes on either side, as well as the permeability of the membrane to each solute. Depending on the membrane and the solute, permeability may depend on solute size, solubility, properties, or chemistry. How the membrane is constructed to be selective in its permeability will determine the rate and the permeability.
Gibbs Donnan Equilibrium
Donnan equilibrium (which ca also be referred to as the Gibbs-Donnanequilibrium) describes the equilibrium that exists between two solutions that are separated by a membrane. The membrane is constructed such that it allows the passage of certain charged components(ions) of the solutions. The membrane, however, does not allow the passage of all the ions present in the solutions and is thus a selectively permeable membrane. Donnanequilibrium is named after Frederick George Donnan ,who proved its existence in biological cells. J. Willard Gibbshad predicted the effect some 30years before. The impermeability of the membrane is typically related to the size of the particular ion. Anion can be too large to pass through the pores of the membrane to the other side. The concentration of those ions that can pass freely though the membrane is the same on both sides of the membrane. As well, the total number of charged molecules on either side of the membrane is equal. A consequence of the selective permeability of the membrane barrier is the development of an electrical potential between the two sides of the membrane. The two solutions vary in osmotic pressure, with one solutionhaving more of a certain type(species) or types of ion that does the other solution. As a result, the passage of some ions across the membrane willbe promoted
III. Positive Results
Osmosis
Hemolysis
- Hemolysis occured after 5 more drops of distilled water.
- The supertanant becomes clear red.
Osmotic Pressure
TEST TUBE NUMBER #1
whole blood + 0.25% NaCl
Hypotonic
TEST TUBE NUMBER #2
whole blood + 0.9% NaCl
Isotonic
TEST TUBE NUMBER #3
whole blood + 10% NaCl
Hypertonic
Permeability
Test tube 1: 5 ml NSS + 2ml whole blood Physiological concentration
Test tube 2: 5 ml NSS +0.2g urea/glucose+ 2ml whole blood Glucose enters cell
Test tube 3: : 5 ml NSS +0.2g sucrose+ 2ml whole blood sucrose cant enter the cell
Gibbs Donnan Equilibrium Gelatin – impermeable substance
Colloidal bag – semipermeable membrane
Phenolphthalein – indicator
NaOH – permeable substance
Cl
K
concentration gradient
electrochemical gradient
concentration gradient
electrochemical gradient
IV. Significance and Application of the Experiment
Cell membraneCell membrane is what keeps the shape and size of the cell, it doesn't
control what goes in and out (the nucleus does that) but it can see everything which goes in and out. Without the cell membrane the cell would distort and probably mix in with other cells.
Cells need to maintain a very specific environment to function properly. This means pH and other ion concentrations need to be tightly regulated, so a semi permeable membrane allows for regulation of these concentrations via protein channels. Remember that even a small change in pH can cause proteins to lose their secondary and tertiary structure.
Osmolsis
In Plants:1.Absorbption of water from the soil.2. Opening and closing of stomata.
In Animals: 1.Fresh water animals have to maintain their osmotic pressure with the external environment. Thus they are either having a contractile vacuole (as in amoeba)
Gibbs donnan equilibrium
The presence of negatively charged, impermeant proteins in the plasma space alters the distribution of diffusible ions in the plasma and interstitial fluid (ISF) compartments to preserve electroneutrality. We have derived a new mathematical model to define the quantitative interrelationship between the Gibbs-Donnan equilibrium, the osmolality of body fluid compartments, and the plasma water Na+ concentration ([Na+]pw) and validated the model using empirical data from the literature. The new model can account for the alterations in all ionic concentrations (Na+ and non-Na+ ions) between the plasma and ISF due to Gibbs-Donnan equilibrium. In addition to the effect of Gibbs-Donnan equilibrium on Na+ distribution between plasma and ISF, our model predicts that the altered distribution of osmotically active non-Na+ ions will also have a modulating effect on the [Na+]pw by affecting the distribution of H2O between the plasma and ISF. The new physiological insights provided by this model can for the first time provide a basis for understanding quantitatively how changes in the plasma protein concentration modulate the [Na+]pw. Moreover, this model defines all known physiological factors that may modulate the [Na+]pw and is especially helpful in conceptually understanding the pathophysiological basis of the dysnatremias.
DNA Extraction From Kiwi Fruit
I. Background of the Experiment:
DNA is in the cells of all living organisms. DNA is essentially the blueprint for an organism. It encodes genetic information in the nucleus of cells. It determines the structure, function and behavior of the cell/organism. Deoxyribonucleic acid is the genetic material present in all organisms, from bacteria to humans. A single subunit of DNA is called a nucleotide and consists of a nitrogen-containing base, a sugar, and a phosphate group. Hundreds of thousands of nucleotides are hooked together to form a chain, and two chains are paired together and twisted into a double helix to form the DNA finished molecule. This procedure is designed to extract DNA from kiwi fruit in sufficient quantity to be seen and spooled.
II. Chemical Principle of Each Test:
Test for Ribose (Molisch Test)The Molisch test uses concentrated sulfuric acid as the dehydrating acid.
This acid dehydrates all carbohydrates, so the test is used to distinguish between carbohydrates and non-carbohydrates. The dehydration products of carbohydrates, furfural or 5-hydroxymethylfurfural, result from the reaction of the sulfuric acid with pentoses and/or hexoses. These products condense with α-naphthol to yield a purple condensation product.
Carbohydrate dehydration product purple product
Test for Phosphates Phosphate Test gave out a yellow result for the precipitate which simply
means that a phosphodiester bond exists between DNA and RNA between the 3' Carbon atom and the 5' Carbon of the ribose sugar.
Test for Purines (Murexide Test)
A reaction giving rise to murexide when uric acid or a related compound is heated with nitric acid and the product is treated with ammonia—called also murexide reaction
Test for Pyrimidines (Wheeler-Johnson Test)
Wheeler-Johnson is a qualitative test for the pyrimidine bases cytosine and uracil, which produces a green coloration when the sample is treated with bromine water. The addition of barium hydroxide will turn the liquid purple.
III. Positive Results:
Test for Ribose (Molisch Test) The positive reaction of Molisch
test is the formation of violet ring onthe solution which indicates thepresence of carbohydrates.
Test for Phosphates The positive reaction of this test is theformation of yellow precipitate on the solution which indicates the
presence of phosphate on the solution.
Murexide Test
The positive reaction of Murexide test is
the formation of red to brown residue
which indicates the presence of purines
on the solution.
Wheeler-Johnson Test
The positive reaction of Wheeler-Johnson
test is the formation of violet precipitate
which indicates the presence of
pyrimidine on the solution.
IV. Significance and Application
The DNA is the biological molecule that stores all the genetic information of the cell (in some viruses RNA may function as the molecule that stores the genetic information). Everything that the cells has to do, at what time in its life cycle, and how it has to do it is determined by the information contained in the DNA molecule. In addition, DNA functions as the molecule that carries on the genetic information from parent to offspring.
RNA is made when the complex biochemical decodification machinery of the cell acts on the DNA to extract the information needed for a particular function. RNA is a key factor for protein synthesis. RNA is responsible for transferring the information contained in the DNA to make a particular protein needed in a specific process for a specific function. Messenger RNA (mRNA) is the nucleic acid that brings information (from the nucleus to the cytoplasm) about which protein to make, and transfer RNA (tRNA) is responsible for transporting aminoacids to the ribosomes to make the required proteins. There is also regulatory RNA, that is RNA molecules capable of regulating gene expression by different mechanisms such as interference or blocking.
Role of Nucleic Acids in DiseasesWhen an error occurs in any of the steps involved in expressing the genetic information
contained in DNA a genetic disease may occur. Understanding how nucleic acids store and deliver genetic information within the cells is necessary to understand diseases and to devise strategies for disease treatment. Many genetic diseases cannot be cured at the moment, but recognizing the importance of nucleic acids in these diseases may be the key that eventually unlocks a cure.
Isolation of Casein From Milk
Chemical Characterization and Denaturation of Proteins
I. Background of Experiment
I.I. Introduction
Milk is probably the most nutritionally-complete food that can be found in nature. All kinds of milk, human or animal, contain vitamins (principally thiamine, riboflavin, pantothenic acid, and vitamins A, B12, and D), minerals (calcium, potassium, sodium, phosphorus, and trace metals), proteins (mostly casein), carbohydrates (principally lactose), and lipids (fats).
The average composition of whole cow’s milk is 87.1% water, 3.4% protein, 3.9% fats, 4.9% carbohydrates, and 0.7% minerals. The only important nutrients lacking in milk are iron
and vitamin C. There are three kinds of proteins in milk: casein, lactalbumins, and lactoglobulins. All three are globular proteins which tend to fold back on themselves into compact, nearly spheroidal units, and are more easily solubilizedin water as colloidal suspensions than fibrous proteins.
Casein, the main protein in milk, is a phosphoprotein which has phosphate groups that are attached to some of the amino acid side chains. These are attached mainly to the hydroxyl groups of the serine and threonine moieties.
Casein exists in milk as the calcium salt, calcium caseinate. Calcium caseinate has an isoelectric (neutrally) point at pH 4.6. Therefore, it is insoluble in solutions of pH less than 4.6. The pH of milk is about 6.6; therefore, casein has a negative charge at this pH and is solubilized as a salt. If acid is added to milk, the negative charges on the outer surface of the casein micelles are neutralized (the phosphate groups are protonated), and the neutral protein precipitates with the calcium ions remaining in the solution.
Ca2+Caseinate + 2HCl → Casein ↓ + CaCl2
When milk sours, lactic acid is produced by bacterial action, and the consequent lowering of the pH causes the same clotting reaction. The isolation of casein from milk will be carried out in this experiment.
By the isolation of casein from milk, the protein is denatured. Denaturation of proteins involves the disruption and possible destruction of both the secondary and tertiary structures. Since denaturation reactions are not strong enough to break the peptide bonds, the primary structure (sequence of amino acids) remains the same after a denaturationprocess. Denaturation disrupts the normal alpha-helix and beta sheets in a protein and uncoils it into a random shape.
I.II. Procedure
Weigh out 5g of powdered non-fat dry milk and dissolve it in 20ml of warm water.
Bring the temperature of the solution to 55°C (do not exceed 60°C).
Then add dropwise a solution of 10% HOAc while stirring with a stirring rod. Do not add the acid too rapidly. Continue the acid addition (lightly less than 2ml will be required), keeping the beaker on the water bath, until the liquid changes from milky to almost clear and the casein no longer separates.
Do not add too much acid.
Stir the precipitated casein until it forms a large amorphous mass, then remove it with a stirring rod and place it in another beaker.
Collect the casein by filtration to remove as much water as possible. Press the solid with a spatula. Place the casein in a 100ml beaker and add 5ml of a mixture of 1:1 ethyl ether and ethanol (CAUTION: HIGHLY FLAMMABLE – NO FLAMES). Stir the casein in the ether for a few minutes, decant the ether, and filter the product.
Place the casein between several layers of paper towels to help dry the product, and let it stand in the air for 10 minutes.
Divide the wet product in half, and weigh the two portions. Place one portion in a 125ml Erlenmeyer flask with 35ml of water and 0.5ml of 1M NaOH, stopper the mixture, and shake it to ensure solution of as much of the casein as possible, and save it for use in the chemical tests to be performed next.
II. Chemical Principle of each Test
II.I. Color Reaction TestsUsing 0.2g of casein for each test tube, then perform the following tests:
Biuret Test Add 0.5ml each of 10% NaOH and 0.5% CuSO4 to the test sample.
Principle: It is a chemical test used for detecting the presence of peptide bonds. Copper (II) ions, in basic solution, will
complex with the nitrogen atoms of four peptide (amide) bonds to form violet colored complex products. The general illustration of the reaction is:
These peptide bonds do not have to be in any order or any sequence in the protein. Any four peptide bonds can coordinate with the copper ion to form the color.
Molisch Test
To 1ml of cell suspension, add 2 drops of Molisch reagent. Mix thoroughly and tilt the tube, and then slowly add, allowing to run on the side, 40 drops of concentrated H2SO4.
Principle:
It is a test for the presence of carbohydrates. Therefore, glycoproteins give positive results to Molisch’s test because they are proteins in complex with carbohydrates. Glycoproteins are very common; therefore many proteins that are not glycoproteins give a negative result to Molisch’s test.
Millon’s Test
Add 0.5ml of Millon’s reagent to the sample and note the result.
Principle:
It is a test specific for tyrosine, the only amino acid containing a phenol group, hydroxyl group attached to a benzene ring. In Millon’s test, the phenol group of tyrosine is first nitrated by nitric acid in the test solution. Then the nitrated tyrosine complexes mercury (I) and mercury (II) ions in the solution to form red precipitate or a red solution, both positive results.
Ninhydrin Test
Add 0.5ml of 1% Ninhydrin solution to the sample and heat in water bath for 5 minutes.
Principle:
It is test for detecting free amino groups. Therefore, molecules with a free –NH2 group (a free amino group) will form a purple-blue color with the Ninhydrin reagent. If all amino groups are part of a peptide bond, no color will result.Ninhydrin is most commonly used as a forensic chemical to detect “fingerprints,” as amines left over from proteins sloughed off in fingerprints react with Ninhydrin giving a characteristic purple color.
Sulfur Test
Add 0.5ml each of 20% NaOH and 10% Lead acetate to the sample. Heat the test sample in water bath for 10 minutes.
Principle:
It is a test for proteins containing sulfur (in cysteine and cystine) which give a positive result when heated with lead acetate in alkaline medium.
Sulfur-containing protein NaOH → S2 Pb2+ → PbS
II.II. Protein Denaturation
Using 1ml of albumin (egg white) for each test tube, perform the following:
By Heat
Heat the sample in water bath for 5 minutes and observe for changes.
Principle:
Heat can be used to disrupt hydrogen bonds and non-polar hydrophobic interactions. This occurs because heat increases the kinetic energy and causes the molecules to vibrate so rapidly and violently that the bonds are disrupted. The proteins in eggs denature and coagulate during cooking. Other foods are cooked to denature the proteins to make it easier for enzymes to digest them.
By AlcoholAdd 1ml of ethyl alcohol to egg albumin and observe for precipitation.
Principle: Hydrogen bonding occurs between amide groups in the secondary protein structure.
Hydrogen bonding between “side chains” occurs in tertiary protein structure in a variety of amino acid combinations. All of these are disruptedby the addition of another alcohol.
Alcohol denatures proteins by disrupting the side chain intramolecular hydrogen bonding. New hydrogen bonds are formed instead between the new alcohol molecule and the protein side chains.
By Inorganic AcidsTo the sample, add 3 drops of concentrated H2SO4.
Principle:If the pH is lowered far below the isoelectric point, the protein will lose its negative
and contain only positive charges. The like charges will repel each other and prevent the protein from aggregating as readily. In areas of large charge density, the intramolecularrepulsion may be great enough to cause unfolding of the protein. This will have an effect similar to that of mild heat treatment on the protein structure. In some cases the unfolding may be extensive enough to expose hydrophobic groups and cause irreversible aggregation. Until this occurs such unfolding will be largely reversible.
By Heavy Metal SaltsAdd 2 drops of 5% FeCl3 to the egg albumin. You may do the same procedure
using 10% Lead acetate.
Principle:Heavy metal salts act to denature proteins in much the same manner as
acids and bases. Heavy metal salts usually contain Hg+2, Pb+2, Ag+1 Tl+1, Cd+2 and other metals with high atomic weights. Since salts are ionic they disrupt salt bridges in proteins. The reaction of a heavy metal salt with a protein usually leads to an insoluble metal protein salt.
By Alkaloidal ReagentsAdd 2 drops of 5% Tannic acid to the egg albumin. You may perform the same
procedure using 1ml of 5% Picric acid.
Principle:Alkaloidal reagents (e.g. tannate and trichloroacetate) are high
molecular weight anions. The negative charge of these anions counteracts the positive charge of the amino group in proteins giving a precipitate.
III. Positive Results
III.I. Color Reaction Tests
Biuret Test
Positive result:The positive reaction for the Biuret test is the
formation of the violet color that indicates the presence of peptide bonds. Solutions without peptide bonds (such as free amino acids) will not form the violet color and will test negative in the biuret test for the presence of peptide bonds.
Molisch TestPositive result:The positive reaction for the Molisch test is the formation of violet ring that indicates the presence of carbohydrates.
Millon’sTestPositive result:
Proteins that contain tyrosine will therefore yield apositive result of pink to dark-red color. The Millon’s reagentis a solution of mercuric and mercurous ions in nitric and nitrous acids. The red color is probably due to a mercury salt of nitrated tyrosine.
Ninhydrin TestPositive result:
A positive Ninhydrin reaction is the purple-blue color (lavender solution) which indicates the presence of free amino groups. Only FREE amino groups (like those in free amino acids and protein side chains) will give a positive reaction with the Ninhydrinreagent. An exception is the amino acid proline, which contains a secondary amino group (instead of the usual primary amino group found in most amino acids and amino acid side chains) and will produce an orange color instead of the purple-blue.
Solutions with no (or very few) free amino groups will not produce the purpleblue color, and hence, the Ninhydrin test is interpreted as being negative for the presence of free amino groups. Protein chains have a free amino group at the beginning of the protein, but this may not be enough to produce a positive reaction.
Reduced Sulfur TestPositive result:The positive reaction for the Reduced Sulfurtest is the black solution/ deposit of lead sulfide(PbS) when heated with lead acetate in alkaline medium which indicates the presence of proteins containing sulfur.
III.II. Protein Denaturation
By HeatResult:
When the egg albumin is heated in thewater bath for 5 minutes, coagulationoccured turned out like a hardboiled egg.
By AlcoholResult:When 1ml of ethyl alcohol was added to egg albumin, coagulation occurred but not all proteins, some liquid portion of the egg albumin did not coagulate.
By Inorganic AcidsResult:When 3 drops of concentrated H2SO4was added to the egg albumin sample, the H2SO4 settled at the bottom part of the test tube and coagulation occurred.
By Heavy Metal SaltsResult:
When 2 drops of 5% FeCl3 or 2 drops of 10% Lead Acetate was added to 1ml eggalbumin, coagulation occurred and the egg albumin became brown solution; therefore the proteins were denatured.
By Alkaloidal ReagentsResult:When 2 drops of 5% Tannic acid or 1ml of 5% Picric acid was added to 1ml of egg albumin, coagulation occurred and the egg albumin became yellow solution with a formation of darker yellow precipitate.
IV. Significance and Application of the Experiments
Casein is the principal protein found in milk from which it has been extracted commercially for most of the 20th century. It is responsible for the white, opaque appearance of milk in which it is combined with calcium and phosphorus as clusters of casein molecules, called micelles.
The major uses of casein until the 1960s were in technical, non-food applications such as adhesives for wood, in paper coating, leather finishing and in synthetic fibers, as well as plastics for buttons, buckles etc. During the past 30 years, however, the principal use of casein products has been as an ingredient in foods to enhance their physical (so-called functional) properties, such as whipping and foaming, water binding and thickening, emulsification and texture, and to improve their nutrition.
In New Zealand, casein is precipitated from the skim milk that is produced after centrifugal separation of whole milk. The skim milk may be acidified to produce acid casein or treated with an enzyme, resulting in the so-called rennet casein. The precipitated casein in curd is separated from the whey, washed and dried. Water-soluble derivatives of acid caseins, produced by reaction with alkalis, are called caseinates.
Casein is generally not consumed as a food on its own. Casein products are used mainly as ingredients in foods for the purpose of either modifying the physical properties of that food product or providing nutritional supplementation to it. As a consequence, they usually form a relatively minor proportion of the food.
Casein has been used commercially in non-food technical applications since the mid-19th century, initially in adhesives and water-based paints. These applications have multiplied during the 20th century.
Carbohydrate Chemistry
I. Background of the Experiment
Carbohydrates are the most abundant biomolecules on earth. They have either an aldehyde or a keto group in addition to many hydroxyl groups. Dihydroxyacetone is the simplest carbohydrate with a keto group. Carbohydrates are defined as the polyhydroxyaldehydes or polyhydroxy ketones or substances that yield such compounds on hydrolysis. The empirical formula of carbohydrates is (CH2O) n. Carbohydrates serve as energy stores, structural elements and they are precursors for many organic compounds like fats and amino acids. Carbohydrates are classified into four categories. Monosaccharides (simple sugars) containing a single polyhydroxy aldehyde or ketone unit. Ex: Glucose. Disaccharides consist of two monosaccharides linked by glycosidic linkage. Ex: Maltose. Oligosaccharides consist of three to twelve monosaccharides. Ex: Maltotriose which is a trisaccharide, made up of three glucose units. Polysaccharides consist of more than twelve monosaccharides. Ex: Cellulose.
Classification
Monosaccharides: They are the simplest Carbohydrates; if they are further broken, doesn’t
possess the characteristics carbohydrates. Monosaccharides are colorless, crystalline substances which have a sweet taste and are soluble in water. They have a single carbon chain having a free aldehyde or ketone group. Examples: Glucose (aldo-hexose), Fructose (keto- hexose).
Disaccharides: They are formed by the condensation of two monosaccharide
molecules. The monosaccharides are connected by a glycosidic linkage. Some common disaccharides are:
Glucose + Glucose = MaltoseGlucose + Galactose = LactoseGlucose + Fructose = Sucrose
Polysaccharides:
They are formed by the condensation of a large number of monosaccharide molecules and have a high molecular weight, hence are slightly soluble in cold water. They form a colloidal solution when heated with water. They are not sweet and don’t exhibit any of the properties of the aldehyde or ketone groups as in case of monosaccharides and disaccharides. They are further divided into:
I. Homopolysaccharides - They are made up of the same kind of monosaccharide units. E.g.: starch, glycogen, and cellulose. They are polymers of glucose and inulin (polymer of fructose).
II. Heteroploysaccharides - They are made up of different kinds of monosaccharide units, and are further divide into two main groups:
a) Animal origin: E.g. mucoplysaccharide group which includes hyaluronic acid, heparin, chondriotin sulphate, blood group, and serum mucoids.
b) Plant origin: The most common example is the mucilage group which includes agar, vegetable gums and Dectins. In plant cell walls, cellulose occurs in densely packed fibrils called hemicelluloses.
Starch is stored in the form of glucose in plants.Glycogen is stored in the form of glucose in Animals.Cellulose is a structural polysaccharide of plant cells. Although it is present in the human diet, but it cannot be digested because humans lack the enzyme cellulase.
Chemical types of Carbohydrates
I. Reducing Sugars: Reducing sugars are those which have potentially an aldehyde and
ketone group in their structure. In other words, these sugars have a free anumeric carbon atom; by virtue of this, they have the ability to give a H atom to other substances .i.e. hey have the ability to reduce other substances and themselves get oxidized. All monosaccharides i.e. glucose, fructose, and galactose are reducing sugars because they have a free carbon atom. Disaccharides as maltose and lactose are also reducing sugars.
II. Non-reducing sugars: Their anumeric carbon atoms are engaged in making bonds with each
other; therefore, they don’t have a free aldehyde or ketone group in their structure, and thereby cannot reduce other substances. Some disaccharides like sucrose and all polysaccharides are non-reducing sugars. However, on hydrolysis with an acid, base, and an enzyme, they yield smaller units which have a reducing capacity.
Chemical Principle of Test
Color Reaction Tests for Carbohydrates
Molisch Test The Molisch test uses concentrated sulfuric acid as the dehydrating
acid. This acid dehydrates all carbohydrates, so the test is used to distinguish between carbohydrates and non-carbohydrates. The dehydration products of carbohydrates, furfural or 5-hydroxymethylfurfural, result from the reaction of the sulfuric acid with pentoses and/or hexoses. These products condense with α-naphthol to yield a purple condensation product.
Carbohydrate dehydration product -> purple productFurfural
Bial’s TestA test uses concentrated hydrochloric acid as the dehydrating
acid and orcinol with a iron (III) chloride as the condensation reagent. Bial's test is used to distinguished between pentose and hexoses. Pentoses subjected to the test yield a blue or green condensation product, while hexoses yield a muddy brown-to-gray condensation product.
pentose dehydrating product blue or green condensation product
furfural
hexose dehydrating product muddy brown-gray condensation product
5-hydroxymethylfurfural
Moore’s TestMoore’s test is used to identify the presence of reducing sugars.
Monosaccharide such as glucose, fructose, and galactose are reducing sugar having free carbon atom. Disaccharides as maltose and lactose are also reducing sugars. This test gives the positive result of brown solution with caramel scent.
Seliwanoff’s TestSeliwanoff’s test is used to identify ketohexoses like fructose.
Seliwanoff’s test uses 6M hydrochloric acid as the dehydrating agent and resorcinol as the condensation reagent. Ketopentoses and ketohexoses react within 2 minutes to form a cherry red solution when mixed with Seliwanoff’s reagent. Aldopentoses react after 2 minutes to form a blue-green condensation product, which may further change to a peach color.
Ketose dehydration product cherry-red product
Iodine Test
Starches form deeply colored blue-black complexes with iodine. Starches contain α -amylose, a helical saccharide polymer, and amylopectin. Iodine forms a large complex polysaccharide with the α-amylose helix, producing the blue-black color. Simpler oligosaccharides and monosaccharides do not form this complex with iodine. Thus, the I2/KI test can be used to distinguish starches from other carbohydrates.
Test for Reducing Sugar
Benedict’s TestBenedict's test uses a mixture of copper (II) sulfate, sodium
citrate, and sodium carbonate in a mildly basic solution. This reagent is used as a general test for detecting reducing sugars. If the saccharide is reducing sugar, it will reduce the copper (II) ions to copper (I) oxide, a red precipitate.
R-CHO ( reducing carbohydrates) + 2Cu + 5OH- R-CO2 (carbohydrate ion) + Cu2O (s,red) + 3H2O
Fehling’s TestFehling's tests for aldehydes are used extensively in
carbohydrate chemistry. A positive result is indicated by the formation of a brick red precipitate. Like other aldehydes, aldoses are easily oxidized to yield carboxylic acids. Cupric ion complexed with tartrate ion is reduced to cuprous oxide.
Hydrolysis of Oligosaccharides and polysaccharides
Sucrose Hydrolysis
Hydrolysis breaks the glycosidic bond, converting sucrose into glucose and fructose. Hydrolysis is, however, so slow that solutions of sucrose can sit for years with negligible change. Hydrolysis can be accelerated with acids, such as hydrochloric acid.
Sucrose has been hydrolyzed in the presence of concentrated HCL into its reducing monosaccharide constituents, glucose and fructose. Hence the test for reducing sugars, are positive.
Sodium hydroxide is added for the purpose of neutralizing excessive HCL, and to create an alkaline medium which provides the optimum pH for the reducing ability of reducing sugars.
Starch Hydrolysis
Hydrolysis of Starch into Glucose:
Polymers are broken down by hydrolysis which is essentially the reverse of condensation. An –OH group from water attaches to one monometer and a H attaches to the other. This is a hydrolysis reaction because water (hydro) is used to break (lyse) a bond. When a bond is broken, energy is released.
Polysaccharides such as starch, dextrin and glycogen, give positive iodine test. Starch is a non reducing polysaccharide therefore it does not give positive result with Benedict’s, and Fehling’s reagents. However after hydrolysis into monosaccharide by the actions of strong acid, its components (glucose molecules) give all the test positive.
PRINCIPLE:
Heating of starch in the presence of conc. HCl causes its hydrolysis into glucose. Because glucose have free Aldehyde group, therefore it is a strongly reducing monosaccharide, and hence Benedict’s, Selivanoff’s and Osazone tests become positive.
Sodium hydroxide is added to neutralize excessive HCl, because the reducing ability of reducing sugars is high in alkaline medium, and hence gives good results of Benedicts ,Selivanoff’sand Osazone tests.
Erythrodextrin give red colour. Its further hydrolysis produces achrodextrins, which gives negative iodine, test. When the iodine test becomes negative, we heat test tubes for two minutes more. The reason being is to provide time to complete hydrolysis of achrodextrin into maltose and maltose into glucose.
III. Positive Result
Color Reaction Tests for Carbohydrates
To 10% solution of the following sugars: glucose, sucrose, fructose, lactose, galactose, perform the following tests:
Molisch Test
To 1 ml of cell suspension, add 3ml of Molisch reagent. Mix thouroughly. Tilt the tube, then slowly add, allowing to run on the side, 3 mL of freshly prepared concentrated H2SO4.
RESULT:
The positive reaction for Molisch test
is the formation of violet ring which indicates
the presence of carbohydrates on the sample
solution.
Bial’s Test Add 0.5mL of Bial’s reagent to 0.5mL sugar solution then heat in water bath for 10 minutes.RESULT:
The positive reaction for Bial’s test is theformation of blue-bluegreen color in the solutionwhich indicates the presence of peptose sugar onthe sample solution.
Moores’s TestTo 1mL of sugar solution, add 1 mL of 30% NaOH then heat in water bath for 10 minutes.
RESULT: The positive reaction for Moore’s test is the formation of brown solution with caramel scent which indicates the presence of
reducing sugar in the sample solution.
Seliwanoff’s Test
Add 0.5 mL of Seliwanoff’s reagent to 0.5 mL of sugar
solution then heat in water bath for 10 minutes.
RESULT:
The positive reaction for Seliwanoff’s test is the
formation of red solution with brown precipitate which
indicates the presence of ketose sugar on the sample solution.
Iodine Test
To 1 mL of sugar solution add 3 drops of iodine solution.
RESULT:
The positive reaction for Iodine test is the formation of dark blue to black solution which indicates the presence of polysaccharide,
particularly starch on the sample solution.
Test for Reducing Sugars
Benedict’s Test
Add 0.5 mL of Benedict’s reagent and heat in water
bath for 5 minutes.
RESULT:
The positive reaction for Benedicts’s test is the formation
of brick red precipitate on the solutionwhich indicates the
presence of a reducing sugar.
Fehling’s Test
Add 0.5 mL each of Fehling’s A and Fehling’s B then heat in water bath for 5 minutes.
RESULT:
The positive reaction for Fehling’s test is the formation of brick red precipitate on the solution which indicates the presence of a reducing sugar.
Hydrolysis of Oligosaccharides and Polysaccharides
Sucrose HydrolysisTo 3mL of sucrose solution, add 2 drops of concentrated HCl then
heat in water bath for 20 minutes. To 0.5 mL each of the hydrolyzed solution, perform Fehling’s Test and Benedict’s Test.
RESULT:
The positive result for Benedicts
test and Fehlings test is the formation
of brick red precipitate on the solution
which indicates the presence of reducing
sugar, glucose and fructose which was the
hydrolyzed form of sucrose.
Starch Hydrolysis
RESULT:
The positive reaction of starch in Iodine test is the formation of dark blue to black solution. Upon treating with concentrated HCl the solution gives positive result to Fehlings and Benedict’s test with a positive result of brick red precipitate on the solution which indicates the presence of glucose that is the product of hydrolyzed starch.
IV. Significance and Application of the Experiment
Each group of carbohydrates can be observed through different qualitative test for carbohydrates that exhibits changes in color base on certain reactions after reacting on specific functional groups in them. Molisch test resulted in purple color involves dehydration by H2SO4 and condensation with α-naphthol, Iodine test exhibits blue-black complex that involves complexation where insertion of iodine into the helical structures of starch and glycogen occurred. Benedict’s test resulted in red precipitate is a test for reducing sugars, Barfoed’stest that forms rusty color distinguished reducing monosaccharides from reducing disaccharides, Seliwanoff’s test produced cherry red color due to condensation in its mechanism, and the Bial’s test that gave out blue precipitate due to condensation with Orcinol. Two unknown carbohydrate samples were identified as fructose and lactose base on the color change observed in their reactons on certain reagents.
Many carbohydrates can be identified using condensation reagents, which react with the carbohydrates to produce highly colored products. Specific nature of condensation reaction varies with the carbohydrate under investigation (3). Six different test were conducted in this experiment to identify the given unknown carbohydrate samples and at the same time, to understand the mechanisms that undergone on certain reactions occurred. Six qualitative tests for carbohydrates are the Molisch’s test, Benedict’s test, Iodine test, Barfoed’s test, Seliwanoff’s test, and Bial’s test.
Benedict’s Test
CLINICAL APPLICATIONS:
•This is widely employed test for detection of glucose in urine.
•It is commonly used for preliminary screening for hyperglycemia and for monitoring the effect of Treatment
Sucrose Hydrolysis
CLINICAL CORRELATION
•Digestion of sucrose in the body.
•Dissacharidase deficiency.
•Role of Sucrose when given as intra- venous.
•Sources of sucrose.
•Invert Sugar.
•Structure and component linkage.
Lipid Chemistry
I. Background of the Experiment:
Lipids are generally characterized as being insoluble in water but soluble in organic non-polar solvents such as ether, chloroform, and carbonic tetrachloride to name a few. The simplest unit of lipid is fatty acids which are classified as either saturated or unsaturated based on the presence or absence of double bonds. Most lipids are amphipathic, that is, they contain both polar and non-polar parts; except for waxes which is completely hydrophobic due to its long hydrocarbon tail.
Aside from waxes, lipids can also be classified as triacylglycerol, sphingolipids, glycolipids, and steroids. Classification is basically based on the back bone unit present.
II. Chemical Principle of each test:
Lipid Solubility TestThis indicates that the coconut oil is soluble only in organic
compounds such as ether, chloroform, and carbonic tetrachloride. Note: High Carbon = High Unsaturation, it means more drops are needed.
Test for UnsaturationThere are two way to test the unsaturation of lipids, these are:
Iodine Absorption Test and Bromine Absorption Test. Note: the more double bonds, the more drops will be needed for color reaction.
Lipid EmulsificationEmulsify means it allows the substance to mix. It maybe defined
as Miscible or Immiscible. Note: in order to be miscible, you need solutions such as: dishwashing liquid, Na2CO3 or also known as washing soda.
Saponification
* This is simply means soap making. It is the production of glycerol/sodium soap or sodium salt. The term saponification is the name given to the chemical reaction that occurs when a vegetable oil or animal fat is mixed with a strong alkali. The products of the reaction are two: soap and glycerin. Water is also present, but it does not enter into the chemical reaction. The water is only a vehicle for the alkali, which is otherwise a dry powder.
Acrolein Test
It is also known as Fat Denaturation (Transidity of Lipids).
Glycerol - > Potassium Sulfate = Unsaturated Aldehyde >> Acrolein Test (test for the presence of fatty acids/glycerin)
III. Positive Results:
Lipid Solubility Test
TEST TUBE NUMBER #1
Coconut oil + H2O
Result: Insoluble
TEST TUBE NUMBER #2
Coconut oil + CHCl3
Result: Soluble
TEST TUBE NUMBER #3
Coconut oil + ethanol at
room temperature)
Result: Insoluble
TEST TUBE NUMBER #4
Coconut oil + hot ethanol
Result: Insoluble
Test for Unsaturation
TEST TUBE NUMBER #1Coconut oil
Result: Saturated
TEST TUBE NUMBER #2Olive oil
Result: Most unsaturated
TEST TUBE NUMBER #3Palmitic acid
Result: More saturated
Lipid Emulsification
TEST TUBE NUMBER #1H2O + coconut oilResult: Immiscible
TEST TUBE NUMBER #2H2O + coconut oil + dishwashing liquid
Result: Miscible
TEST TUBE NUMBER #3H2O + coconut oil + Na2CO3
Result: Miscible
Saponification
This results from the formation of soap.
Acrolein Test
The positive reaction of Acrolein test is the formation of colorless or yellow liquid with a disagreeable solution which indicates the presence of glycerol and fatty acid.
IV. Significance and Application of the Experiment:
The lipid solubility test is used to determine the presence of saturated and unsaturated fatty acids. In the experiment if theres a formation of layer on the solution, it means that the mixture is insoluble with one another while if the solution doesn’t form any layer, meaning they are soluble mixture. We can use it to test the saturation. Saturated food are more harmful for people while unsaturated are less harmful for people.
For the test for unsaturation, it indicates that the more unsaturated or the more the double bond the sample has, the more drops of iodine needed for absorption.
Emulsifier such as dishwashing liquid can help mix two solution that don’t normally mix together. Like in washing the dishes, the dishwashing liquid that is normally made from fats, uses to clean our dishes only because the fats on the liquid and the stains on our dishes mix together making them easy to remove and wash.
Saponification is the process used to make our soaps which we used on our daily activities such as cleaning dishes, washing clothes, cleaning house or even during bathing
Catalase Activity
I. Background of the Experiment
Enzymes are globular proteins, responsible for most of the chemical activities of living organisms. They act as catalysts, substances that speed up chemical reactions without being destroyed or altered during the process. One enzyme may catalyze thousands of reactions every second.
Enzymes catalyze chemical reactions involving the substrates. Substrates are molecules upon which enzymes act. During a reaction, the substrate binds with the enzyme’s active site, and an enzyme-substrate complex is formed. The substrate is transformed into one or more products, which are then released from the active site. Almost all processes in a biological cell need enzymes to occur at significant rates. Since enzymes are selective for their substrates and speed up only a few reactions from among many possibilities, the set of enzymes made in a cell determines which metabolic pathways occur in that cell. In this experiment, we have studied the activity of the enzyme catalase, an enzyme that functions to catalyze the decomposition of hydrogen peroxide to oxygen and water. Hydrogen peroxide is naturally produced in organisms as a by-product of oxidative metabolism. Without the presence of catalases, decomposition of the concentrations of the harmful hydrogen peroxide in our body to water and oxygen would not be possible.
Almost all enzymes are composed of proteins and are found in all tissues and fluids of the body. Intracellular enzymes catalyze the reactions of metabolic pathway. Plasma membrane enzymes regulate catalysis within cells in response to extracellular signals, and enzymes of the circulatory system responsible for regulating the clotting of blood. Almost every significant life process is dependent on enzyme activity. As catalysts, enzymes excel all other chemicals in their power and specificity.
II.Chemical principle of each test
A. NORMAL CATALASE ACTIVITY
PRINCIPLE:
Catalase is a common enzyme found in nearly all living organisms that are exposed to oxygen, where it functions to catalyze the decomposition of hydrogen peroxide to water and oxygen.
The reaction of catalase in the decomposition of hydrogen peroxide is:
2 H2O2 → 2 H2O + O2
This can be tested by taking a solution of the enzyme and dipping a piece of paper into it, then seeing how long it takes to rise in a tube of hydrogen peroxide
IS CATALASE REUSABLE?
PRINCIPLE:Catalase is reusable because during the chemical reaction, they
are not permanently changed or destroyed which makes them able to still have an end product of hydrogen peroxide or bubble formation.
OCCURRENCE OF CATALASE
PRINCIPLE:In living systems the optimal ranges of temperature, pH and salt
concentration for a given enzyme are the ranges found. Meaning, those who exhibit the characteristic with the most optimum value of the factors that affect catalase activity will have the fastest rate of reaction.
EFFECTS OF TEMPERATURE ON CATALASE ACTIVITY
PRINCIPLE:
In general, chemical reactions speed up as the temperature is raised. When the temperature increases, more of the reacting molecules have the kinetic energy required to undergo the reaction. Enzyme catalyzed reactions also tend to go faster with increasing temperature until a temperature optimum is reached. Above this value the conformation of the enzyme molecule is disrupted. Changing the conformation of the enzyme results in less efficient binding of the substrate. Temperatures above 40-50°C denature many enzymes.
EFFECTS OF PH ON CATALASE ACTIVITY
PRINCIPLE:
pH is a measure of the acidity or hydrogen ion concentration of a solution. It is measured on a scale of 0-14 with pH values below 7 being acidic, values above 7 being basic and a value around 7 is neutral. As the pH drops into the acidic range an enzyme tends to gain hydrogen ions from the solution. As the pH moves into the basic range the enzyme tends to lose hydrogen ions to the solution. In both cases the changes produced in the chemical bonds of the enzyme molecule result in a change in conformation that decreases enzyme activity.
III. RESULTS WITH ILLUSTRATIONS
NORMAL CATALASE ACTIVITY
PROCESS: Place 2mL of 3% hydrogen peroxide solution into a clean test tube. Using forceps and scissors, cut a small piece of liver and add it to the test tube. Push it
into the hydrogen peroxide. Through this investigation, estimate the time passed (how rapidly the solution bubbles)
before the bubbles appear, in seconds. Record the time in DATA TABLE 1, and DATA TABLE 2 as the rate room temperature. Feel the temperature of the test tube with your hand. (Note if it is warm or not)
RESULTS:
The estimated time before bubbles formed when the liver came into contact with hydrogen peroxide at room temperature, was 2 seconds.
IS CATALASE REUSABLE?PROCESS: Place 2mL of 3% H2O2 solution into a clean test tube and add a small
piece of liver. Pour off the liquid into a second test tube. Add another 2mL of H2O2 to the liver remaining in the first test tube. Observe.
RESULTS:
The reusability of the enzyme Catalasewas tested by pouring unused H2O2 over the chicken liver used for previous part. Results agreed with the established biological fact that enzymes don't change, therefore enzymes, particularly Catalase in this experiment, are/is reusable
OCCURRENCE OF CATALASE
PROCESS: Place 2mL of H2O2 in each of 3 clean test tubes.
Tube 1: add a small piece of potato Tube 2: add a loopwhole of Staphylococcus aureus Tube 3: add a small piece of apple
As you add each test substance, record the reaction rate (in seconds) for each tube in TABLE 1.
RESULTS:After performing the experiments, we can infer that catalase is more concentrated in animal cells and bacteria because their reaction time to form bubbles or froth is faster (1 second for the bacteria and 2 seconds for the liver) than the potato (65 seconds) and apple (112 seconds) samples which contain plant cells.
EFFECTS OF TEMPERATURE ON CATALASE ACTIVITY
PROCESS: Put a piece of clean liver in a clean test tube and soak it in 3mL water. Place this tube in a boiling
water bath for 5 minutes. Remove the test tube from the water bath, allow it to cool, then pour out the water. Add 3mL of
H2O2. Note the time it takes before the bubbles appear. Record the time in DATA TABLE 2. Put equal quantities of liver into 2 clean test tubs and 2ML H2O2 into 2 other test tubes. Put one test tube of liver and one of H2O2 into each of the following water baths: Ice bath: 0 degree
Celsius
Warm bath: 37 degrees Celsius After 3 minutes, pour each tube of H2O2 into the corresponding tube of liver and observe the
reaction. Record the reaction time in DATA TABLE 2. You recorded the time for ROOM TEMPERATURE earlier.
The reaction rate of liver stored at room temperature compared to iced
liver is faster. This was so because as long as the enzymes don't get
denatured, the enzymes react faster as it gets near the optimum temperature.
RESULTS:SAMPLE
SPEED OF ENZYME ACTIVITY
0 degrees Celsius
6 seconds
Room tempertaure
2 seconds
37 degrees Celcius
1second
100 degrees Celcius
60 seconds
EFFECTS OF PH ON CATALASE ACTIVITY
PROCESS: Add 2mL H2O2 to each of 3 clean test tubes. Treat each tubes as follows:
Tube 1: 5 drops 1M HCl at a time until acid Ph
Tube 2: 5 drops 1M NaOH at a time until basic pH
Tube 3: 1 drop H2O Add a small piece of liver to each test tube. Estimate the reaction (in seconds) and record in DATA
TABLE 3.
RESULT:
Reaction rate was fastest at Neutral pH. This was
so because the optimum pH for the enzyme Catalase is usually,
around 7. Preceded by the reaction on basic solution, that had
occurred after 2 seconds and the reaction on acid pH took the
slowest enzyme activity rate, which formed froth after 10 seconds.
RESULTS: SAMPLE SPEED OF ENZYME ACTIVITY
acid pH 10 seconds
basic pH 2 seconds
neutral pH 1 seconds
IV. Significance and Application of the Experiment
Hydrogen peroxide is a harmful by-product of many normal metabolic processes: to prevent damage, it must be quickly converted into other, less dangerous substances. To this end, catalase is frequently used by cells to rapidly catalyze the decomposition of hydrogen peroxide into less reactive gaseous oxygen and water molecules.
The true biological significance of catalase is not always straightforward to assess: Mice genetically engineered to lack catalase are phenotypicallynormal, indicating that this enzyme is dispensable in animals under some conditions. A Catalase deficiency may increase the likelihood of developing Type II Diabetes. Some human beings have very low levels of catalase (acatalasia), yet show few ill effects. It is likely that the predominant scavengers of H2O2 in normal mammalian cells areperoxiredoxins rather than catalase.
Human catalase works at an optimum temperature of 37°C, which is approximately the temperature of the human body. In contrast, catalaseisolated from thehyperthermophile archaea Pyrobaculum calidifontis has a temperature optimum of 90°C.
Catalase is usually located in a cellular organelle called the peroxisome. Peroxisomes in plant cells are involved in photorespiration (the use of oxygen and production of carbon dioxide) and symbiotic nitrogen fixation (the breaking apart of diatomic nitrogen (N2) to reactive nitrogen atoms). Hydrogen peroxide is used as a potent antimicrobial agent when cells are infected with a pathogen. Pathogens that are catalase-positive, such as Mycobacterium tuberculosis, Legionella pneumophila, and Campylobacter jejuni, make catalase in order to deactivate the peroxide radicals, thus allowing them to survive unharmed within the host.
Catalase is used in the food industry for removing hydrogen peroxide from milk prior to cheeseproduction. Another use is in food wrappers where it prevents food from oxidizing. Catalase is also used in the textile industry, removing hydrogen peroxide from fabrics to make sure the material is peroxide-free.
A minor use is in contact lens hygiene - a few lens-cleaning products disinfect the lens using a hydrogen peroxide solution; a solution containing catalase is then used to decompose the hydrogen peroxide before the lens is used again. Recently, catalase has also begun to be used in the aesthetics industry. Several mask treatments combine the enzyme with hydrogen peroxide on the face with the intent of increasing cellular oxygenation in the upper layers of the epidermis.
Digestion of Proteins and Carbohydrates
I. Background of the Experiment
Digestion occurs in all mammals, including human beings. There are multiple steps involved in the breaking down of food, each of which is related to specific digestive enzymes. Some of these are found in the mouth, while others are found in the intestines. Pepsin is a material that has its own unique function in the digestion of food.
Pepsin is a digestive enzyme that is released in the stomach as pepsinogen. The release of hydrochloric acid stimulates the release of this basic form of pepsin. When pepsinogen is exposed to the hydrochloric acid in the stomach, the pepsinogen unfolds and breaks into pepsin. Pepsin is made out of hydro chlorine acid and pepsi.
The main function of pepsin is to break down proteins that are found in foods such as meat and eggs into smaller pieces (polypeptides). It breaks down proteins only at certain points, so the protein is not digested completely to the amino acid level. In order for that to occur, the food has to pass into the intestines, where other enzymes complete the digestion process.
Amylase is an enzyme that catalyses the breakdown of starch into sugar. Amylase is present in human saliva, where it begins the chemical process of digestion. The pancreas also makes amylase (alpha amylase) to hydrolyse dietary starch into disaccharides and trisaccharides which are converted by other enzymes to glucose to supply the body with energy. Plants and some bacteria also produce amylase. Amylase is a digestive enzyme classified as a saccharidase (an enzyme that cleaves polysaccharides). It is mainly a constituent of pancreatic juice and saliva, needed for the breakdown of long-chain carbohydrates (such as starch) into smaller units.
The primary function of the enzyme amylase is to break down starches in food so that they can be used by the body. Amylase is also synthesized in the fruit of plants during ripening, causing them to become sweeter. Amylase helps you digest carbohydrates into simple sugar so that our cells can use them for energy pepsin is used to digest protein and they are absorbed into the bloodstream from the small intestine.
II. Chemical Principle of Each Test
A. Protein DigestionThe Action of Pepsin
Pepsin is an enzyme that catalyses the breakdown of protein into smaller pieces (polypeptides). Protein has an optimum pH of 1.5 – 1.6.
Biuret TestThe Biuret test is a chemical test used for detecting the presence
of peptide bonds. In the presence of peptides, a copper(II) ion forms a violet-colored complex in an alkaline solution.
B. Carbohydrates Digestion (Digestion in the Mouth)
Amylase is an enzyme that catalyses the breakdown of starch into sugar. Amylase is present in human saliva, where it begins the chemical process of digestion. It has an optimum pH of 6.8 - 7.0.
Iodine TestStarches form deeply colored blue-black complexes with iodine. Starches contain α -amylose, a helical saccharide polymer, and amylopectin. Iodine forms a large complex polysaccharide with the α-amylose helix, producing the blue-black color. Simpler oligosaccharides and monosaccharides do not form this complex with iodine. Thus, the I2/KI test can be used to distinguish starches from other carbohydrates.
Benedict’s TestBenedict's test uses a mixture of copper (II) sulfate, sodium citrate, and sodium carbonate in a mildly basic solution. This reagent is used as a general test for detecting reducing sugars. If the saccharide is reducing sugar, it will reduce the copper (II) ions to copper (I) oxide, a red precipitate.
R-CHO (reducing carbohydrates) + 2Cu + 5OH- --- R-CO2 (carbohydrate ion) + Cu2O (s,red) + 3H2O
III. Positive Result:
Protein Digestion
The Action of Pepsin
TEST TUBE NUMBER # 1
Pepsin + Egg Albumin solution
Positive: violet solution
TEST TUBE NUMBER # 2
Pepsin + Glacial Acetic Acid
+ Egg Albumin Solution
Positive: light blue solution
TEST TUBE NUMBER # 3
Egg Albumin
Positive: blue-violet solution
TEST TUBE NUMBER # 4
Pepsin + milk solution
Positive: violet solution
TEST TUBE NUMBER #5
Pepsin + Glacial Acetic Acid + Milk Solution
Positive: whitish solution w/ precipitate
TEST TUBE NUMBER # 6
Milk Solution
Positive: blue-violet solution
Carbohydrates Digestion (Digestion in the Mouth)
Iodine Test- The positive reaction for Iodine test is the formation of dark blue to black solution whichindicates the presence of polysaccharide on the sample solution.- On the first test, Iodine test is positive- After 2 test the second part becomes negative.
Benedict’s TestThe positive reaction for Benedicts’s test is the formation of brick red precipitate on the solutionwhich indicates the presence of a reducing sugar.
IV. Significance and Application of the Experiment
Protein Digestion:
Digestion of proteins takes place in different stages. The initial phase of protein digestion and absorption occurs in the stomach and the latter takes place in the small intestine. Proteins provide the body with amino acids, which are required by the body, but unfortunately, the body is incapable of producing them by its own. Well, the protein digestion process basically consists of the breakdown of proteins into amino acids, which is the absorbable form of protein molecules.
Carbohydrate Digestion:
Digestion of carbohydrate involves conversion of the large molecules of carbohydrates like disaccharides and polysaccharides into simple monosaccharide molecules which can be easily absorbed by the body. The first step of carbohydrate digestion will start the moment we put the food in mouth. As we chew the food, the saliva released by the salivary glands of the mouth starts its work of breaking up of the carbohydrates. This is possible because of the presence of a special enzyme named amylase in the saliva. Then we swallow the food and it goes to the stomach. Here the digestive acids secreted by the glands of the stomach play a major role in processing the carbohydrate molecules further. The digestive enzymes of the stomach also helps in this work but they do not have much significant role in carbohydrate digestion in the stomach.
Group 1 sec:2MLS-K
Members:Ilarde, QueenieJose, Emelyn ThereseLagat, Janine JoyReyes,QueenieTan, Hiyas Mina
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