Biochemistry for Medical Laboratory Science Lab - MLSBCHML PDF

BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] LESSON 1: LABORATORY GLASSWARES The control of pH is important in organisms and their cells...

BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] LESSON 1: LABORATORY GLASSWARES The control of pH is important in organisms and their cells because chemical reactions and processes are affected by the Performing biochemical analysis in the laboratory hydrogen ion concentration. requires the use of particular glasswares and Nemo (Clownfish) requires a pH between 7.8 and 8.4 laboratory equipment. Each apparatus performs a specific purpose which → ACID may be similar but most often unique. An acid is a compound that can donate a hydrogen ion. Students at the beginning of the class must be properly oriented with these apparatuses and must be able to → BASE utilize them correctly. A base is a substance that accepts hydrogen ions. There are different types and kinds of glassware in the laboratory. They are commonly utilized to carry out varying functions such as: ○ measurement of solutions and liquids (e.g. graduated cylinder, serological pipettes) 1. carrier or container of reagents (e.g. beaker, Erlenmeyer flask) 2. reaction vessel (e.g. test tube) 3. grounding vessel (e.g. mortar and pestle) 4. vessel for heating to extremely high temperatures → BACKGROUND (e.g. crucible with cover, evaporating dish) Chemists have tried to define acids and bases in 5. for transferring and mixing chemicals and liquids relation to their compositions and (e.g. stirring rod, glass funnel) and many more molecular structures. 6. Other laboratory apparatus are also available to facilitate easy handling of laboratory glasswares (e.g. 1. Svante Arrhenius test tube rack, crucible tong, test tube holder, wire defines acids as substances that gauze, tripod) produce H+ ions in aqueous 7. Specialized automated and semi-automated solution while bases are equipment are also utilized in the Biochemistry substances that produce OH- ions laboratory to help carry out the experiments such as in aqueous solution. Spectrophotometer, analytical balance and micropipettes. 2. Gilbert N. (G.N.) Lewis Mentioned that acids are electron-pair acceptors and bases are electron-pair donors. LESSON 2: PH MEASUREMENT AND BUFFER PREPARATION However, the two mentioned definitions have limitations. Thus, → PH the most useful and accepted pH is a measure of how acidic/basic a solution is. definition of acids and bases The range goes from 0 to 14. nowadays are those proposed by Johannes Bronsted ○ If pH < 7, then the solution is acidic. and Thomas Lowry, and it is known as the ○ If pH = 7, then the solution is neutral. Bronsted-Lowry theory. ○ If pH > 7, then the solution is basic. The letters pH stand for potential of hydrogen; Reaction between HCl with water: pH is a measure of the hydrogen ion (H+) concentration in an aqueous solution; pH is also expressed as the negative logarithm of the hydrogen-ion concentration. pH = - log [H+] → POH In contrast, pOH stands for potential of Hydroxide It is a measure of hydroxide ion (OH–) concentration, and; It is expressed as the negative logarithm of the hydroxide-ion concentration. pOH = - log [OH-] HCl (Hydrochloric acid) is an acid because it donates a proton making Cl- (Chloride) while water is a base because it accepts a proton making H3O+. Furthermore, the theory explains that for every acid-base reaction, there is a creation of a conjugate acid-base pair. In the above example Cl- is the conjugate base of HCl and H3O+ is the conjugate acid of water as shown below. 3. Søren Peter Lauritz Sørensen (S.P.L.) Sørensen introduced the pH scale that measures the strength of an aqueous acidic or basic solution. NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] It converts the H+ concentrations to pH using the For Example: formula: pH = - log [H+] A solution has a pH of 4. Find the pOH and [OH-]. The precise definition of pH is "the negative common pH + pOH = 14 logarithm of the activity of hydrogen ions in solution.” pOH = 14 - pH pOH = 14 - 4 → COMPUTATIONS pOH = 10 [OH-] = 10-pOH = 10-10 [OH-] = 1 x 10-10 M Find the pOH and [OH-] of a solution with a pH of 8.4. pH + pOH = 14 pOH = 14 - pH pOH = 14 – 8.4 pOH = 5.6 [OH-] = 10-pOH = 10-5.6 [OH-] = 2.5 x 10-6 M Interpretation of Results Computation of pH & [H+] The pH of a solution is the negative logarithm of the hydrogen-ion concentration. pH = -log[H+] For example: Compute the pH of pure water with [H+] = 1.0 × 10 -7 M (Molarity). Substituting into the pH expression: → BUFFERS Buffers prevent changes in pH. pH = -log[H+] Buffers resist changes in the pH even when acids or = -log[1.0 × 10-7] bases are added. = -(-7.00) Buffers are a mixture of a weak acid or alkali and one pH = 7.00 of its salts. Ex: acetic acid + sodium acetate The ability of buffers to resist large changes in pH is Compute for the pH of solution with [H+] = 2.3 × 10 -5 M governed by Le Chatellier's principle. pH = -log[H+] La Chatellier's Principle - A principle of equilibrium = -log[2.3 × 10-5] shift due to changes in buffer conditions. pH = 4.64 (Acidic) Buffers in Human Blood You can also compute for the [H+] if pH is known using In our blood, carbonic acid is the most important this formula: buffer. [H+] = 10-pH This solution maintains our blood pH to facilitate *Use the 10^x key on your calculator transport of oxygen from the lungs to the cells. ○ pH of blood is normally slightly basic (7.35 to For Example: 7.45) What is the [H+] of a solution with a pH of 9.14? [H+] = 10-pH → MEASURING PH QUALITATIVELY AND QUANTITATIVELY = 10-9.14 1. Litmus Test [H+] = 7.24 × 10-10 M Litmus Test is a simple test to check if a substance is acidic or basic using a litmus paper. What is the [H+] of a solution with a pH of 5.4? There are two types of litmus paper available that can be used to identify acids and bases – red litmus paper [H+] = 10-pH and blue litmus paper. = 10-5.4 Blue litmus paper turns red for acidic pH [H+] = 3.98 x 10-6 M Red litmus paper turns blue for basic pH No color change for neutral pH Computation of pOH The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. pOH = -log[OH–] [OH-] = 10-pOH *Use the 10^x key on your calculator Constant: pH + pOH = 14 NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] 2. Indicator Paper Indicator paper is impregnated with organic compounds that change their color at different pH values. The color shown by the paper is then compared with a color standard usually provided by the manufacturer. 3. pH Meter The pH meter should be calibrated first before operating the device. The standard procedure for calibrating a pH meter is to calibrate it at three different pHs (pH 7, pH 4, and pH 10). → MONOSACCHARIDE After calibration, all that needs to be done is to insert Simplest sugar and cannot be hydrolyzed further the electrodes of the pH meter into the solution to be - CHO or the aldehyde groups are widely tested and read the pH flashed on the screen. distributed in plants and even found in certain animal tissues such as liver and muscles.. Classification is by the number of carbon atoms they contain: ○ Pentoses: five-carbon atoms Aldopentose = ribose & xylose ○ Hexoses: six-carbon atoms Aldohexose = glucose, galactose Ketohexose = fructose On hydrolysis, disaccharides yield two monosaccharide units. LESSON 3: QUALITATIVE TESTS FOR CARBOHYDRATES → WHAT IS CARBOHYDRATES One of the three main macronutrients alongside proteins and fats. Primary source of energy for the body. Made up of three main types of components such as sugars, starches and fibers. Polyhydroxyaldehydes (aldoses) & polyhyroxyketones (ketoses) ○ Aldehydes - has a carbon, hydrogen, and oxygen at the end (carbonyl group) Glucose ○ Ketone - carbonyl group within the chain Fructose General Formula: (CH2O)n=>3 Classification is based on the number of monosaccharide units they contain: ○ Monosaccharides - one sugar ○ Disaccharides - two sugar ○ Oligosaccharides - few or little (3-10 mono) Two general classes of carbohydrate test reagents based on ○ Polysaccharides - many (11 or more mono) the type of reaction involved. 1. Dehydrating acids (2-step analysis) Converts pentoses into furfural and hexoses into 5-hydroxymethylfurfural (5-HMF) which then reacts with phenolic compounds. ○ Pentoses → Furfural Undergoes a dehydration reaction where water is being removed = C5H4O2 ○ Hexoses → 5-HMF Dehydration reaction = C6H6O3’ Production of highly colored products ○ Involves the conversion of carbohydrates into colored compounds through dehydration and subsequent reactions NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] ○ Furfural & Phenol = furfural phenol Interpretation of Result condensates which are used in the Ketose reacts to produce a deep cherry red color productions of colored reacids. Aldoses may react slightly to produce a faint pink to ○ HMF & Phenolic Acids or compounds such as cherry red color if the test is prolonged. in the presence of ammonia = formation of The product and reaction time of the oxidation colored compounds, melanoidin which is a reaction helps to distinguish between carbohydrates. deep brown color. Sucrose and insulin also give a positive result for this Molisch’s, Anthrone, Bial’s & Seliwanoff’s tests test as these are hydrolyzed by acid to give fructose. 2. Copper (II) ions containing solutions → MOLISCH’S TEST CHO reduces copper (II) ions into copper (I) oxide A chemical test that checks for the presence of ○ Commonly observed in the Benedict test or carbohydrates in an analyte. It was discovered by the Fehling’s test, which are used to detect Czech-Austrian botanist Hans Molisch. Analytes are the presence of reducing sugars. treated with Molisch's reagent (a solution of a-naphthol ○ In the presence of an aldehyde group such as and ethanol), which is then mixed with concentrated in reducing sugars like glucose or fructose, H2SO4 (sulphuric acid) before the test is carried out. the sugar acts as in reducing agents. A sensitive qualitative test used to detect the presence ○ Copper (II) ions are present in the solution in of carbohydrates by adding α-naphthol and the form of copper (II) sulfate or could be concentrated sulfuric acid, which forms a violet ring if copper (II) tartrate in fehling solution. carbohydrates are present. This test can detect all Reducing sugars include aldoses containing either a types of carbohydrates, including monosaccharides, free aldehyde group or cyclic hemiacetal disaccharides, and polysaccharides (Zubrick, 2020). ○ This can reduce copper (II) ions or copper (I) The presence of carbohydrates can be identified by the oxide. presence of purple or purplish-red colored layers ○ Presence of a reducing sugar is indicated by a created by the test solution. Therefore, the Molisch test color change in the solution due to the results are positive. In the absence of purple or formation of a red or orange precipitate. purplish-red tinted layers in the test solution, the solution does not contain carbohydrate molecules. A. GENERAL TESTS FOR CARBOHYDRATES Therefore, the Molisch test results are negative. → SELIWANOFF’S TEST Test Objective To detect the presence of ketohexoses in a given sample To distinguish ketosis from aldoses Aldose Ketose Contains 1 aldehyde group Contains 1 ketone group per per molecule molecule In Seliwanoff’s test, react In Seliwanoff’s test, reacts slowly and produce a light with resorcinol to give a → BIAL’S TEST pink color deep cherry-red color Test Objective To detect the presence of carbohydrates To distinguish the pentoses and pentosans from other Principle of the Test derivatives of carbohydrates like the hexoses. The reagent of this test consists of resorcinol in 6M HCl The acid hydrolysis of polysaccharides and In chemistry, a pentose is a monosaccharide (simple sugar) oligosaccharides yields simpler sugars with five carbon atoms (C5H10O) Ketoses are more rapidly dehydrated than aldoses Ketoses undergo dehydration in the presence of concentrated acid to yield hydroxymethyl furfural that condense with resorcinol Principle of the Test This test is based on the principle that under hydrolysis Figure: Seliwanoff’s test with fructose as an example. pentosans are hydrolyzed into pentoses. Further, pentoses are dehydrated to yield furfural, which in turn condense with orcinol to form a blue-green precipitate. The intensity of the color developed depends on the concentration of HCl, ferric chloride, orcinol, and the duration of boiling. NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] → BARFOED’S TEST Test Objective A chemical test used to detect the presence of reducing monosaccharides To distinguish reducing monosaccharide from disaccharides Principle of the Test Made up of copper acetate (cupric ions) in a dilute Interpretation of Result solution of acetic acid (acidic medium) The presence of a blue-green complex indicates the Reducing Monosaccharides are strong reducing agent: presence of pentoses in the sample. React within 3 minutes Reducing Disaccharides have to first to get hydrolyzed B. REDUCING PROPERTY TEST OF SUGAR in the acidic solution and then react with the reagent: React in about >3 minutes → BENEDICT’S TEST Test Objective Determine the presence of absence of reducing sugar in the solution All monosaccharides are reducing sugars A reducing sugar has free aldehyde group or free ketone group Principle of the Test Benedict’s reagent contains copper (II) ions in an alkaline solution with sodium citrate to keep the cupric ions in solution. Test is performed by heating the reducing sugar solution with Benedict’s reagent. Alkaline condition of this test causes isomeric transformation of ketoses to aldoses resulting in the reduction of blue cupric ion to cuprous oxide Interpretation of Result Positive Result: Brick red precipitate at the bottom of the tube ○ Monosaccharide: ≤ 3 minutes ○ Disaccharide: ≥ 3 minutes The difference in the time of appearance of precipitate thus helps Interpretation of Result distinguish reducing monosaccharides from reducing disaccharides → FEHLING’S TEST Fehling’s test is conducted in laboratory experiments to analyze the components of a solution and detect the presence of reducing sugars in the sample (Sapkota, 2022b). More so, it differentiates aldehydes and ketones. Fehling’s test involves a redox reaction where copper(II) ions (Cu²⁺) in Fehling’s solution are reduced to copper(I) oxide (Cu₂O) by the aldehyde group of reducing sugars, resulting in a color change from blue to red or orange (Miller, 2012). A positive result is indicated by the formation of a red or orange precipitate, which confirms the presence of reducing sugars. If no precipitate forms and the solution remains blue, the test is negative, indicating the absence of reducing sugars (Miller, 2012). NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] Principle of the Test Lactose also yields a music acid, due to the hydrolysis of the glycosidic bond between glucose and galactose subunits of the carbohydrate. Other monosaccharides like glucose also have a similar structure; however, the resultant precipitate formed in glucose is water-soluble under room temperature Interpretation of Result The formation of crystals at the bottom of the tube indicates a positive result which means that the sample solution has galactose or its derivatives. → TOLLEN’S TEST The absence of such crystals indicates a negative This is a qualitative test used to distinguish between result and represents that the sample doesn't have aldehyde and ketones. It is based on the galactose or its derivative. The solution may still have oxidation-reduction reaction between aldehydes and other carbohydrates. reagent (Prasad, 2024). It undergoes an oxidation reaction. The Tollen’s → FERMENTATION reagent, which is composed of Silver nitrate and Occurs in yeast cells and bacteria, as well as in the Ammonia, is a mild oxidizing agent where it can oxidize muscles of animals. A glucose breakdown pathway aldehyde but not ketones (AESL, n.d.). occurs anaerobically. In a broader sense, fermentation The result of Tollen’s test for benzaldehyde is the is the development of carbon dioxide gas as the formation of a silver mirror, indicating the presence of changes caused by yeasts and other microorganisms an aldehyde in the solution. This produces a shiny growing anaerobically, without air. surface. A negative result will show no change in appearance or color. C. SPECIAL TEST FOR SELECTED SUGARS → INVERSION OF SUCROSE → MUCIC ACID TEST Sucrose inversion is the process by which sucrose, A test that is highly specific and is used for the often known as table sugar, is chemically broken down detection of the presence of galactose and lactose. It into a solution of glucose and fructose. Alternatively is also termed galactaric acid, named after the product referred to as hydrolysis, this procedure involves the of the reaction addition of water. A chemical reaction called hydrolysis is what turns Test Purpose sucrose into glucose. It involves adding water to break To detect the presence of galactose and lactose in a down complicated molecules into simpler ones. In given sample particular, hydrolysis breaks down sucrose, a To distinguish between the galactose containing disaccharide made up of glucose and fructose units, saccharides and other sugars into its monosaccharides. Water is very important to this process because it makes it easier for the bond Principle of the Test between the glucose and fructose to break. Sugar can Aldohexoses (monosaccharide) upon treatment with be broken down by acids like hydrochloric acid or potent oxidizing agents like nitric acid yield saccharic enzymes like invertase, which speed up the process. acids (dicarboxylic acids) Nitric acid has capacity to oxidize both aldehyde and → PHENYLHYDRAZONE TEST / OSAZONE TEST primary alcoholic groups present at C1 and C6 Phenylhydrazone or Osazone test is a chemical test respectively of galactose to yield an insoluble used to detect and differentiate reducing sugars. precipitate (rod-shaped crystals) of mucic acid under Differentiation of reducing sugars is based on the time high temperature. of appearance of the complex and relative solubility in hot water. Reaction Examples of reducing sugars are: glucose, fructose, glyceraldehyde and galactose. Test Objective To detect reducing sugars To differentiate reducing sugars from non-reducing sugars. To distinguish different reducing sugars between each other. NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] Principle of the Test Prostaglandins are a group of bio The reagent for this test consists of phenylhydrazine in active lipids derived from acetate buffer arachidonic acid which is a type of The test is based on the fact that carbohydrates with fatty acid. free or potentially free carbonyl groups react with They act as signaling molecules and phenylhydrazine to form osazone have a wide range of physiological The condensation-oxidation-condensation reaction effects. between three molecules of phenylhydrazine and In terms of function, prostaglandins carbon 1 and 2 of aldoses and ketoses yields 1, play a key role in the inflammatory 2-diphenylhydrazone, which is known as osazone response. They mediate various aspects of inflammation, including vasodilation, increased blood flow, and the recruitment of immune cells to sites of injury or infections. Lipid Profile Interpretation of Result Osazone appears as yellow-colored crystals of characteristic shape, solubility, melting point, and time of formation. Osazone is different for different sugars. → EXTRACTION OF LIPIDS FROM EGG YOLK Principle Many Phospholipids are insoluble in acetone and they precipitate out whereas triglycerides, sterol, pigments are soluble in acetone. Egg and egg yolk is very important sources of Lecithin which is a generic term to designate any group of yellow-brownish fatty substances occurring in animal and plant tissues, which are amphiphilic – they attract both water and fatty substances (and so are both hydrophilic and lipophilic) LESSON 4: LIPIDS ○ Used for smoothing food textures, dissolving powders (emulsifying), homogenizing liquid → LIPIDS mixtures, and repelling sticking materials. Amphipathic molecules that encompass fatty acids and Lecithin can easily be extracted chemically using derivatives such as diglycerides, triglycerides, solvents such as hexane, ethanol, acetone, petroleum phospholipids, and cholesterol ether, benzene, etc., or extraction can be done Roles in human body: mechanically. ○ Provide energy for the body ○ It is usually available from sources such as Triglycerides stored in adipose tissue soybeans, eggs, milk, marine sources, serve as a long term energy reserve. rapeseed, cottonseed, and sunflower. It has ○ Act as chemical messenger low solubility in water, but is an excellent Certain lipids access signaling emulsifier. molecules that regulate various physiological functions, including inflammation, blood clotting, and immune responses. ○ Involved in the maintenance of body temperature Adipose tissue insulates the body, helping to maintain body temperature ○ Involved in membrane layer formation Fat fads around vital organs such as the kidneys and heart, access cushioning to protect them from physical damage. ○ Involved in the formation of prostaglandins and mediation of inflammation NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] → ACROLEIN TEST Positive result: Translucent spots will appear on the Principle filter paper. Acrolein test is used to detect the presence of glycerol Negative result: Translucent spot will not appear on or fat. When fat is treated strongly in the presence of a the filter paper. dehydrating agent like potassium bisulphate (KHSO4), the glycerol portion of the molecule is dehydrated to form an unsaturated aldehyde, acrolein that has a pungent irritating odor. Positive result: If glycerol is present in the sample, it will give a pungent smell. Negative result: If glycerol is absent in a sample, it will not produce a pungent smell. → LIEBERMANN-BURCHARD TEST → TEST FOR UNSATURATION Test Objective Principle To detect the presence of cholesterol. Unsaturation test is used to detect the unsaturated fatty acids or double bonds in a lipid sample. All the Principle neutral fat contains glycerides of fatty acids. Double A chemical estimation of cholesterol, the cholesterol bonds are found in the structure of unsaturated fatty reacts as a typical alcohol with a strong concentrated acids, which becomes saturated by taking up either acid and the product is bromine or iodine. If the lipid contains more colored substances. unsaturated fatty acids or Acetic anhydride are used more double bonds, it will as solvent and dehydrating take more iodine. agents. Positive result: Pink color Positive result: It indicates will disappear by the cholesterol in a sample by addition of unsaturated giving bluish-green color to fatty acids. the solution. Negative result: Pink Negative result: The color color will not disappear. of the solution will not change. → COPPER ACETATE TEST Principle This test used to distinguish between oil or neutral fat and fatty acid saturated and unsaturated. The copper acetate solution does not react with the oils, while saturated and unsaturated fatty acids react with copper acetate to form copper salt. unsaturated fatty acids can only be extracted by petroleum ether. Unsaturated fatty acid: 2 layer appearance - upper layer color green and the lower layer color blue. Saturated fatty acid: 2 layer appearance – appearance of white blue precipitate at the bottom of the upper layer. LESSON 5: HYDROLYSIS OF DISACCHARIDES AND POLYSACCHARIDES → HYDROLYSIS A fundamental chemical process that plays a crucial role in the breakdown of complex carbohydrates such as disaccharides and polysaccharides into simpler sugars. → SPOT/TRANSLUCENT TEST Principle A translucent spot test is a preliminary test for the lipids, which is characterized by a translucent and greasy spot. The lipid will not wet the filter paper, unlike water. The lipids will form a greasy or translucent spot due to their greasy texture, and penetrate the filter paper. Unlike lipids, the spot of water will disappear from the paper. NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] Disaccharides and polysaccharides belong to → DIGESTIVE ENZYMES PRODUCED BY SMALL INTESTINE carbohydrates. carbohydrates are a diverse group of 1. Maltase organic compounds that play critical roles in biological Breaks maltose (a type of sugar) into glucose systems serving as energy sources and signaling 2. Lactase molecules. Responsible for breaking down lactose, the sugar They can be categorized into simple sugar such as present in milk, into glucose and galactose. monosaccharides and complex carbohydrates 3. Sucrose including disaccharides and polysaccharides. Breaks sucrose, a common sugar, into glucose and Disaccharides are composed of two monosaccharides fructose. units linked by glycosidic bonds. While polysaccharides are long chains of monosaccharide units. → APPLICATION OF HYDROLYSIS In the hydrolysis of disaccharides, the glycosidic bond Beyond biological systems, hydrolysis reactions have connecting the two sugar units is cleaved by the significant industrial applications. Enzymatic hydrolysis addition of water, resulting in the formation of two is used in the production of high fructose corn syrup monosaccharides molecules. cornstarch, which is widely used as a sweetener in For example, sucrose, a common disaccharides found processed foods. as table sugar hydrolyzes into glucose and fructose. Similarly, the hydrolysis of cellulose, a polysaccharide This reaction is catalyzed by the enzyme sucrase in found in the plants cells walls, into glucose is essential biological systems. for biofuel production, turning agricultural base into Polysaccharides such as starch and cellulose as you sustainable energy resources. can see here in this diagram we have a branch of Some of the applications in the industry of hydrolysis carbohydrates simple or complex can be found in soap production, pharmaceuticals, and Starch, fibers, glycogen, which undergo hydrolysis to textiles. yield their constituent monosaccharides. Starch, storage polysaccharide in plants, is broken → POST LAB 5 DISCUSSION down into glucose units through a series of enzymatic I. Using Litmus Paper reactions involving amylase. Cellulose, the structural polysaccharide in plant cell 1. Sucrose + HCl walls, is hydrolyzed by cellulase enzymes into glucose Hydrolysis of sucrose by HCl should yield glucose and through this process. fructose, but initially, you will have an acidic solution. Upon adding NaOH, the solution should turn neutral, → DISACCHARIDES indicated by a color change in litmus paper from red to Are composed of two monosaccharide units linked blue. together are hydrolyzed into their constituent sugar For example, sucrose, a common disaccharide found in 2. Starch + HCl table sugar is hydrolyzed into glucose and fructose. Hydrolysis of starch by HCl produces simpler sugars Similarly, lactose, the sugar found in milk, is broken like maltose and glucose. The solution should is initially down into glucose and galactose. be acidic With NaOH addition, you should observe a color → POLYSACCHARIDES change from red to blue as the solution becomes Our long chains of monosaccharide units include neutral. The number of drops required to neutralize the starch, glycogen, and cellulose. These complex solution gives an indication of the extent of acid carbohydrates are hydrolyzed into simple sugars present. through a series of enzymatic reactions. Starch, for instance, is broken down into maltose and II. Using Iodine’s Test eventually into glucose, while cellulose is hydrolyzed 1. Starch into glucose units, through this process is more Iodine turns blue-black in the presence of starch. If challenging due to structural complexity of cellulose. starch was present before hydrolysis, you should see a blue-black color. Hydrolysis of disaccharides and polysaccharides is essential for various biological and industrial processes. 2. Sucrose Iodine will not change color if sucrose is present, as sucrose does not react with iodine. 3. Post-Hydrolysis Solutions After hydrolysis, starch should be broken down into simpler sugars, which do not react with iodine. Thus, a blue-black solar should fade or disappear if starch has been hydrolyzed. III. Using Benedict’s Test 1. Reducing Sugars (e.g., Glucose, Fructose) Benedict’s reagent changes color based on the concentration of reducing sugars: ○ Blue = no reducing sugars ○ Green/Yellow = low concentration of reducing sugars ○ Orange/Red = moderate to high concentration of reducing sugars ○ Brick Red = high concentration of reducing sugars NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025 BIOCHEMISTRY FOR MEDICAL LABORATORY SCIENCE [LAB] - MLSBCHML MED232 [2ndYR-T1] 2. Sucrose Before hydrolysis, sucrose will not react with Benedict’s reagent (solution remains blue). After hydrolysis, the presence of glucose and fructose should result in a color change, indicating reducing sugars. 3. Starch Starch itself will not change color with Benedict’s reagent. However, after hydrolysis, the resulting glucose and maltose will cause a color change if present. NU MOA (National University - Mall of Asia) John Christhoper B. Dela Cruz Dr. Floyd Rey P. Plando A.Y. 2024-2025