Practical Course of Biochemistry-I & the Activity Notebook Clinical pharmacy Level 2 By Staff Members of Biochemistry Department (2015 2016) 1/48
Attendance sheet Student Name:..................................................................................... ID:.......................................................................................................... Serial Title Date Signature 1 Monosaccharides 2 Disaccharides 3 Polysaccharides 4 Carbohydrate s Revision 5 Albumin 6 Neutral proteins 7 Alkaline proteins 8 Protein s Revision 9 NPN 10 General Scheme 11 Final Revision 12 Exam 2/48
Carbohydrates Definition: Carbohydrates are polyhydroxy aldehydic or polyhydroxy ketonic compounds or any compound giving these substances on hydrolysis. Classifications:- 1- According to type of carbonyl group: a-aldoses: Contain aldehydic group. Example: glucose. b-ketoses: Contain ketonic group. Example: Fructose. 2-According to number of sugar units: A-Monosaccharides (simple sugar):- - Composed of one sugar only. - Can t give simpler forms on hydrolysis. - Examples: Glucose and fructose. B-Oligosaccharides:- Composed of 2-10 units linked together through glycosidic linkage (between carbonyl group of one sugar with OH group of another one as lactose or between 2 carbonyl groups as sucrose). - They are sub classified according to number of sugar units into: 1- Disaccharides: composed of 2 units.they are either : a- Reducing: Examples: Lactose and maltose. b- Non-reducing: Example: Sucrose. 2- Trisaccharides. 3- Tetrasaccharides. C- Polysaccharides:- They are composed of more than 10 sugar units linked together through glycosidic linkage. Examples: Starch and dextrin. 3/48
Scheme of carbohydrates A-Physical characters:- 1-Colour:.... (As you see in your sample) 2-Odour:.... (As you see in your sample) 3-Action to litmus paper: neutral. 4-Aspect: clear or turbid (As you see in your sample) Monosaccharides B-Chemical characters:- 1- Molisch s test (general test for all carbohydrates): Procedure:-1 ml sugar solution +few drops of alcoholic alpha naphthol shake well then add 1 ml conc. sulphuric on the wall of the test tube carefully violet ring formed at the junction between 2 layers spreads by shaking. Comment:- 1- Dehydration of sugar to furfural as in fructose or hydroxymethyl furfural (HMF) as in glucose by conc. sulphuric. 2- Condensation of furfural with 2 molecules of alpha naphthol giving violet ring. N.B:- Molisch s test is positive with all carbohydrates even carbonyl group is occupied as in disaccharides and polysaccharides which are first hydrolyzed to monosaccharides by conc. sulphuric. 2- Reduction of Cu ++ ions (for reducing sugars): Sugar solution (reducing agent) + Cu ++ /OH - (oxidizing agent) Sugar acid+ Cu 2 O (cuprous oxide, red ppt) N.B:- Oxidation potential of Cu ++ ions increases in alkaline medium: increasing alkalinity of OH - will increase the oxidation potential of Cu ++ and decrease the selectivity of the test. 4/48
Types of tests:- a- Fehling s test:- This test is for reducing sugar either monosaccharide or disaccharide. The reagents are: a-fehling A: CuSO 4 b-fehling B: 30% NaOH+ Roschell salt (sodium potassium tartarate). Procedure: 1 ml Fehling A+I ml Fehling B Shake well till dark clear blue solution then add 1ml sugar solution Heat for 2-5 mins on direct flame Cu 2 O red ppt N.B: Disadvantages of Fehling s test: a- Difficult handling of 2 separate bottles. b- The medium is highly alkaline so increasing oxidation potential of cupric ion become strong oxidizing agent can be reduced by any reducing agent. b- Benedict s test:- Benedict s reagent: CuSO 4 /sodium citrate/na 2 Co 3 in one bottle. Procedure: 1 ml sample+1 ml reagent then heat for 2-5 min on direct flame Red ppt is obtained. Advantages of benedict s reagent: a- One bottle is easly to be handled. b- Na 2 Co 3 renders the medium less alkaline so reducing the oxidation potential of Cu ++ and increasing selectivity of the test. c- Barfoed s test:- Test for reducing monosaccharides. Barfoed s reagent is cupper acetate/gatial acetic acid. Procedure:- 1 ml sugar solution+1 ml reagen heat on water bath for 5 min little red ppt within 5 min. Comment: Monosaccharides reduce Cu ++ to Cu 2 O (red ppt). 5/48
N.B:- 1- The medium is weakly acidic (the oxidation potential is very low) so only monsaccharides are oxidized after 2-3 min (maximum 5 minutes). 2-Disaccharides give positive results after prolonged heating due to hydrolysis to monosaccharides. 3- Ketose test:- This test is used for differentiation between aldose (glucose) and ketose (fructose). Procedure: 1ml sugar solution + 3ml conc. HCl fructose, Heat for 3 min reddish brown color in case of While in case of glucose no change in color or just pale yellow color appears. Comment: 1- In case of fructose: Dehydration of fructose and formation of furfural which is reddish brown in color. 2- In case of glucose: Dehydration to HMF which is colorless. 4- Moor's test: It is the action of 30% NaOH. Procedure: 2 ml sample+3 ml 30%NaOH boil for 2-3 min yellowish brown solution + burnt sugar odor (caramel odor). Comment: it is a reducing sugar contains free carbonyl groups undergo aldol condensation (caramelization) reaction in alkaline medium. 5-Osazone test:- Procedure: a- For monosaccharides: 2ml sample + 1gm phenyl hydrazine HCL + 5ml acetate buffer Mix well put in W.B. for 15-20 min crystallization occurs on hot and appears as raphides (long needles) on the wall of the test tube without cooling. 6/48
b- For disaccharides: 2ml sample + 1gm phenyl hydrazine HCL + 5ml acetate buffer mix well then put in W.B. for 45 min cool slowly till crystallization transfer crystals carefully on a slide and examine under microscope the shape of these crystals under microscope are: a- Clusters Lactose b- Rosettes Maltose Comment: Formation of osazone crystals due to the reaction of one sugar + 3 phenyl hydrazine HCl in three successive steps: condensation, oxidation then condensation. Advantages of osazone test: 1- Differentiation between mono and disaccharides (so differentiate between glucose and lactose in urine of pregnant or lactating women diagnosis of D.M). (N.B: Benedict s test can t be used for detection of DM in case of pregnant and lactating women). 2- Sure detection of glucose in urine. 3- Differentiation between lactose and maltose. 7/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Glucose II) Chemical characters: Test Observation Comment 8/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Fructose II) Chemical characters: Test Observation Comment 9/48
Disaccharides They contain 2-monosaccharide units. They are classified into: a- Reducing disaccharides: examples: Maltose and Lactose b- Non-reducing disaccharides: example: Sucrose Chemical reactions for disaccharides: A- Reactions of reducing Disaccharides (Lactose & Maltose): as mentioned previously in monosaccharides. B- Non-Reducing disaccharides: (Sucrose or cane-sugar): It is formed by the condensation of a molecule of α-d-glucose with a molecule of β-dfructose through α-1,2-glucosidic linkage. It is non-reducing because as the two carbonyl atoms form the glucosidic linkage; thus no free carbonyl group is present in the molecule can t form osazone crystals and negative in all tests for reducing sugars. Chemical tests for sucrose: Test Observation Comment 1-Molisch test (general for all carbohydrates) Violet ring at the junction of 2 layers spreads by shaking. 1-hydrolysis of sucrose by conc. H 2 So 4 to simple sugars (glucose and fructose). 2-Dehydration to HMF. 3-Condensation between HMF and 2 molecules of alcoholic alpha naphthol. 2-Reduction of Cu ++ : a-fehling s test: b-benedict s test: No red ppt No red ppt Due to the absence of free carbonyl group (non-reducing sugar) c-barfoed s test: 3-Moor`s test: (for reducing sugars) No little red ppt within 5 min No yellowish brown color and Due to the absence of free carbonyl group (non-reducing sugar) 10/48
4-Osazone test: 5-ketose test: 6-Fehling s after hydrolysis: Procedure: 2 ml sucrose solution + 2 ml dil. H 2 SO 4 boil 2-3 min cooling neutralize excess acid with 2ml 30 % NaOH add 2ml of previously mixed Fehling s reagents. no burnt sugar odor. No crystal formation Reddish brown color. Red ppt. Due to the absence of free carbonyl group. 1- Due to hydrolysis into glucose and fructose. 2- Dehydration of fructose into furfural which has reddish brown color. The acid hydrolyzes sucrose to glucose and fructose both of which reduce Cu ++ to Cu 2 O (red ppt). N.B: Neutralization is essential otherwise the red precipitate Cu 2 O dissolves in acidic medium, thus interfering with the test. 11/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Lactose II) Chemical characters: Test Observation Comment 12/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Maltose II) Chemical characters: Test Observation Comment 13/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Sucrose II) Chemical characters: Test Observation Comment 14/48
Polysaccharides 1-Starch Starch: It is non-reducing sugar as there is no free carbonyl group. Chemical characters: a- Molisch s test: As all carbohydrates. b- Tests for the free carbonyl groups: Fehling ' s, Benedict's, Barfoed ' s, Moor's, Ketose and Osazone: all are negative with starch. Special tests of starch: 1-Iodine test: Procedure: Test Observation Comment 1- Conc iodine Dark blue color Due to adsorption of I 2 molecules on amylose chain giving starch- I 2 complex 2- Dil. I 2 Blue color The same as above. 3- Heating Disappearance of the color Due to physical desorption of iodine (destruction of complex) 4- Cooling Reappearance of blue color Due to readsorption of I 2 molecules on amylose chain giving I 2 -starch complex. 2-Stepwise acid hydrolysis: 20 ml starch paste + 5 ml cone. HCI diluted 1 in 5 and mix divide into 5 test tubes Boiling water bath. Remove each test tube at time intervals 2, 5, 10, 15, 20 min divide each tube into 2 portions 1st for I 2 test and 2 nd is neutralized by 30%NaOH apply Benedict s or Fehling s test. 15/48
Time I 2 Reaction Benedict's Reaction Comment 2 Blue color No change Starch 5 Bluish violet color Greenish yellow color Amylodextrin 10 Reddish violet color Brownish color Erythrodextrin 15 Reddish brown color Reddish brown precipitate Achrodextrin 20 Yellowish brown color Red precipitate Maltose + glucose 3- Salting out (Precipitation by (NH 4 ) 2 SO 4 ): Definition of salting out: Decrease in the solubility of solute by addition of large amount of neutral salt depending on the affinity of both to H 2 O. Salting in: It is the increase in the solubility of solute by addition of amount of neutral salt. Test Observation Comment a) Half saturation: using saturated solution of (NH 4 ) 2 SO 4 b) Full saturation: using (NH 4 ) 2 SO 4 powder. Small white ppt. Heavy white creamy ppt. Ammonium Sulphate has higher affinity to water than starch so it removes H 2 O molecules surrounding starch, leading to precipitation of starch. 2- Dextrin Composition: 1- They are the products of partial hydrolysis of starch (mixture of amylodextrin, erythrodextrins and achrodextrins). 2- The commercial dextrin contains traces of glucose & maltose little reducing properties with Fehling's (the strongest due to high alkalinity). Scheme of dextrins: 1- Physical characters: The same as in mono- and disaccharides. 2- Chemical reactions: a- Molisch s test: As with all carbohydrates 16/48
b- Tests for free carbonyl groups: i- Fehling's +ve after prolonged heating little red ppt as it contains traces of maltose and glucose which reduce Cu +2 Cu 2 O because Cu +2 has high oxidation potential in highly alkaline medium. ii- -ve Benedict's, -ve Barfoed, s, -ve Moor's, -ve Osazone 3- Special tests of dextrins: a- 1 2 test the same as starch but violet color obtained in all steps instead of blue color in case of starch due to the formation of dextrin I 2 complex. b- Salting out: only give precipitate with full saturation. Test Observation Comment a- half saturation No ppt It is fairly soluble. b- full saturation White ppt Due to higher affinity of (NH 4 ) 2 SO 4 to H 2 O than dextrins. 17/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Starch II) Chemical characters: Test Observation Comment 18/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Dextrin II) Chemical characters: Test Observation Comment 19/48
Proteins Definition: High molecular weight organic compounds composed of amino acids linked together through peptide (amide) linkage between COOH & NH 2 groups. O + H 3 N CH C OH H + H + O + H 3 N CH C O - H 2 N O CH C O - CH 3 ph 3 ph 7 ph 10 (acidic med.) (neutral) (alkaline) Net charge= +1 net charge= 0 net charge= -1 ZWITTER ION CH 3 H + CH 3 Iso-electric point: ph at which amino acid occur as zwitterion. Denaturation of protein: Change in the physical, chemical and biological properties of protein a- Examples of physical denaturing agents: heat, UV light... b- Examples of chemical denaturing agents: conc. acid, conc. alkali (30% NaOH) and heavy metals Classification of proteins 1- Heat coagulable protein: example: albumin. 2- Neutral proteins: example: peptone, gelatin 3- Alkaline proteins (negatively charged proteins): example: metaprotein and caseinogen. 20/48
Chemical reactions: I- Color reactions: 1) Biuret Test: general test for all proteins Procedure: 1 ml Sample + 1 ml 10% NaOH shake well add few drops of dil. CuS0 4 Violet color. Comment This color is due to co-ordination between Cu +2 and at least 2 peptide linkages in the same molecule of protein N.B.: The intensity of the color depends on the number of peptide linkages present in the protein Peptone has low no. of peptide linkages rose or pink color. 2) Xanthoproteic test: for aromatic amino acid (phenylalanine, tyrosine & tryptophan): Procedure Comment a- 1 ml protein + 1 ml cone. HNO 3 white precipitate. b- Heat for 2 min yellow ppt or color c- Cooling then 1 ml 30% NaOH orange color. a- Due to chemical denaturation. b- Due to nitration of aromatic rings Nitro group acts as a chromophore induces yellow color. c- Due to entrance of OH acts as an auxochrome intensify the color to orange. 3) Millon's Test for phenolic amino acid (tyrosine) Millon's reagent = Mercuric and mercurous nitrate & nitrite in excess nitric acid and little nitrous acid. Procedure Comment a- 1 ml sample + 1 ml Millon's reagent white ppt. b- Heat 2 min yellow color. c- Heat 5 min brick red mass a- Due to chemical denaturation of protein. b- Due to nitration of aromatic ring NO 2 is a chromophore induces yellow color. c- Due to formation of mercuric salt of nitrated tyrosine. 4) Rosenheim's Test (for amino acid containing indol group like tryptophan): Rosenheim's Reagent: FeCl 3 / HCHO. Procedure: 1 ml sample + few drops of Rosenheim's reagent add 1 ml conc. H 2 SO 4 slowly on the wall of test tube Violet ring at the junction of 2 layers Comment Due to condensation between 1 molecule of HCHO and 2 molecules of tryptophan (with the α-h of indol). 21/48
N.B: a- Role of FeCl 3 initiate reaction (catalyst) b- Role of cone. H 2 SO 4 dehydrating agent to remove water (2 H from tryptophan and 1 O from HCHO). 5) Molisch Test for glycoprotein: Example: Egg albumin contains glucose as impurities during its separation from egg. Procedure: as in carbohydrates. 6) Sulfur test for sulphur containing amino acids: Procedure: 1 ml sample + 1 ml 30% NaOH boil on direct flame for 5 minutes + 1 ml Pb(Ac) 2 Observation: yellowish brown color Comment: due to formation of PbS. N.B.: 1- Protein + Pb(AC) 2 (without NaOH) no brown color as S in protein is present in organic combination not free. 2- Boiling with 30% NaOH break the bond C SH in cysteine or cystine amino acids give S as Na 2 S 7) Sakaguchi (test for basic amino acid containing guanidine group as arginine): Procedure: 1 ml sample + few drops alcoholic α-naphthol + 1 ml of Na hypobromite (NaOBr) Observation: Intense red color. Comment: Due to condensation of 2 molecules of alcholic α -naphthol with the guanido group of 1 molecule of arginine N.B.: Role of Sodium hypobromite: proton acceptor. II) Precipitation Reactions: 1- Mineral acids: conc.h 2 SO 4, conc.hno 3, conc. HCl,. 2- Organic acids: Trichloroacetic acid. 3- Heavy metals: FeCl 3, lead acetate, CuSO 4. The previous reagents give white ppt due to chemical denaturation. 4- Bulky or complexing agents: Examples: a- Picric acid yellow ppt. b- Tannic acid white ppt. Precipitation is due to bulky salt formation. 22/48
5- Salting out: for albumin solution a) Half saturation: no ppt as albumin is fairly soluble. b) Full saturation: white ppt as ammonium Sulphate has higher affinity for H 2 O than albumin so remove albumin from its solution. III) Heat coagulation reaction: Physical denaturation: Specific test for albumin: Fill the test tube to the half with albumin + heat top of the solution on direct flame for 10 minutes white opaque mass (coagulum) add few drops of dil. acetic acid: a- If insoluble true coagulum as in case of albumin b- If soluble false coagulum Comment: Albumin is a heat-coagulable protein true coagulation is formed due to irreversible physical denaturation in which all linkages in albumin are broken because albumin is sensitive to heat 23/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Albumin II) Chemical characters: Test Observation Comment 24/48
Neutral proteins a- Type of peptide: b- Solubility: d- Contents : e- Scheme: I-Physical characters: 1-color: 2-odor: 3-aspect: 4-action to L.P. II-Chemical: a- Color tests: 1- Biuret test using very dil.cuso 4 : 2-Xanthoproteic 3- Millon, s 4-Rosenheim, s 5-molisch, s 6-sulfur test: 7- Sakagouchi: Peptone Short derived peptide. Highly water soluble All amino acids as albumin but fewer in number Rose or pink color as it is short peptide containing few no. of peptide linkage. Conc.HNO 3 no white ppt (as it is short peptide). Boil yellow color. 30% NaOH orange color. a- No ppt b- Brick red mass. Violet ring Violet ring Brown color Intense red color Gelatin Long derived peptide. Sparingly soluble. All amino acids except sulpher containing, aromatic and phenolic amino acids. Violet color. White ppt Pale yellow color as it is poor in aromatic amino acids. a- White ppt b- No brick red mass as it is poor in phenolic a.a. No violet ring due to absence of tryptophan. No violet ring due to absence of CHO. No brown color as it is poor in S containing amino acids. Intense red color. 25/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Peptone II) Chemical characters: Test Observation Comment 26/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Gelatin II) Chemical characters: Test Observation Comment 27/48
Alkaline proteins a- Source: b-impurities: c- Color reactions: Metaproteins Alkaline hydrolysis (30% NaOH) of albumin. Glucose Positive with all color tests. Milk Lactose Caseinogen Positive with all color reactions except: : 1- Sulfur test pale yellow color as it is poor in S containing amino acids. 2- Rosenheim s test: as it is poor in tryptophan a.a. 2- Coagulation at isoelectric point (IEP): Add dil acetic acid step wise till the formation of flocculum (IEP) Heating the flocculum for 5 min on direct flame opaque mass (coagulum) is formed then add excess dil. HAc: A- If the opaque mass does not dissolve it is a true coagulum (as in case of metaprotein which is a protein sensitive to coagulation as it is derived from albumin) B- If soluble false coagulation it is caseinogen (not coagulable protein - not sensitive for coagulation). Comment for Meta protein: 1- Flocculation formed by addition of dil. acetic acid due to precipitation because of zwitter ion formation. 2- True coagulation is formed due to irreversible physical denaturation in which all linkages in Meta protein are broken because Meta protein is sensitive to heat as it is derived from albumin. Comment for caseinogens: 1- Flocculation formed by addition of dil. acetic acid due to precipitation because of zwitterion formation. 2- False coagulation is formed because caseinogen is not coagulable protein and not sensitive for coagulation. N.B.: The tests for differentiation between metaprotein and caseinogens are: a- Sulfur test. b- Coagulation at IEP. 28/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for m-protein II) Chemical characters: Test Observation Comment 29/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Caseinogen II) Chemical characters: Test Observation Comment 30/48
Non-protein Nitrogenous Compound (NPN compounds) 1- Sodium Urate 1- Uric acid: It is the metabolic end product of purine bases (Adenine and Guanine) of nucleoproteins and nucleic acids; DNA and RNA. Prosperities of uric acid: 1- Solubility: Sparingly soluble in water but soluble in strong alkali as 30% NaOH. That is why; the sample is prepared as Na urate. 2- Alkaline to LP. 1- Physical characters: Scheme for sodium urate sample identification a- Color:.. (As you see in your sample) b- Odor:.. (As you see in your sample) c- Aspect:.. (As you see in your sample) d- Action to LP:.. (As you see in your sample) 2- Chemical reactions: a- Fehling s test: Procedure: As previous Observation: Little red precipitate is formed Comment: Formation of Cu 2 O as uric acid has little reducing power. N.B.: Avoid excess Fehling s reagent, otherwise, white precipitate of Cu urate is formed. b- Follin s test: Follin's reagent is 10% phsophotungestic acid. Procedure: sample+ Follin's reagent deep blue color Comment: reduction of phosphotungestic acid to tungsten blue. 31/48
c- Muroxide test (specific for xanthine nucleus): Procedure: Evaporate 2 ml sample in a porcelain dish till the formation of white residue add 1 ml conc. HNO 3 evaporate till dryness red residue due to the formation of alloxanthine add 1 ml dil NH 4 OH or 10% NaOH violet color is formed due to the formation of sodium or ammonium purpurate. 2- Urea Urea: It is the metabolic end product of protein formed through urea cycle. 1- Physical characters: Scheme for urea sample identification a- Color:.. (As you see in your sample) b- Odor:.. (As you see in your sample) c- Aspect:.. (As you see in your sample) d- Action to L.P.:.. (As you see in your sample) 2- Chemical reactions: a- Biuret test: i- On cold: Observation: No violet color is obtained. ii-: After fusion (in porcelain dish or in test tube): Procedure: Evaporate 2 ml of urea sample till dryness and formation of white residue (biuret residue) dissolve this residue in the list volume of NaOH 10% add few drops of NaOH 10% violet color is formed 32/48
Comment: Formation of biuret residue contains 2 peptide linkage which coordinates with Cu 2+ forming violet colored complex. b- Hypobromite test: Procedure: 1 ml urea+1 ml NaOH 10%+ few drops of sodium hypobromite weak effervescence Comment: this effervescence is due to the evolution of nitrogen gas. CO(NH 2 ) 2 + NaOBr + 2NaOH N 2 + 3NaBr + Na 2 CO 3 + 3H 2 O c- Urease paper test: Procedure: Urea sample+ 1 drop of phenol red indicator insert filter paper impregnated with urease enzyme color change from yellow to pink color Comment: Urea (neutral medium) + phenol red indicator urease enzyme NH 3 (alkaline medium) By action of urease enzyme on urea, the medium changes from neutral (yellow color of indicator) to alkaline (pink color of indicator). 33/48
Scheme for Urea I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: II) Chemical characters: Test Observation Comment 34/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Sodium Urate II) Chemical characters: Test Observation Comment 35/48
Scheme for Identification of Simple Liquid Unknown 1- Biuret test: A- If violet color obtained, it is a protein sample B- If no violet color obtained, it may be a carbohydrate or non-protein nitrogenous compound A- If violet color obtained: 1- Apply heat coagulation test: a- If opaque mass insoluble in dil acetic acid obtained, it is albumin sample. Apply other 2 confirmatory tests. b- If no opaque mass obtained, apply litmus paper test: i- If neutral protein, it may be peptone or gelatin. ii- If alkaline protein, it may be metaprotein or caseinogens. Tests for differentiation between peptone and gelatin: a- Biuret test using very dilute CuSo 4 : Peptone rose or pink color Gelatin violet color b- Half saturation test: If white ppt is obtained If no ppt it is peptone it is gelatin Tests for differentiation between metaprotein and caseinogen a- Sulfer test: metaprotein yellowish brown color Caseinogen pale yellow color b- Coagulation at IEP: If true coagulum is formed metaprotein If false coagulum is formed caseinogens. 36/48
B- If no violet color is obtained with biuret test: 1- Apply molisch s test: a- If violet ring is obtained it is carbohydrate b- If no violet ring is obtained it is non-protein nitrogenous compound a- If violet ring is obtained 1- Apply iodine test: a- If violet color is obtained It is dextrin sample Confirmatory test for dextrin: apply Fehling s test. b- If blue color is obtained it is starch sample Identify its physical characters. c- If no characteristic color it is not polysaccharide sample If it is not polysaccharide: 2- Apply Fehling s test: a- If no red ppt It may be sucrose Apply Fehling s after hydrolysis and ketose test b- If red ppt It is reducing sugar If it is reducing sugar: 3- Apply both barfoed s and osazone tests: a- If red ppt formed within 5 minutes in barfoed s and yellow crystals formed on hot after 20 minutes in osazone test It may be glucose or fructose Apply ketose test: i- If reddish brown color it is fructose ii- If no reddish brown color It is glucose b- If no red ppt within 5 minutes in barfoed s test and yellow crystals formed after cooling after 45 minutes in osazone test it may be maltose or lactose See osazone crystals under the microscope: i- If clusters It is lactose ii- If rosettes It is maltose 37/48
b- If no violet ring is obtained with molisch s test: Apply litmus paper: a- If neutral It may be urea Apply biuret after fusion and sodium hopbromide tests. b- If alkaline It is sodium urate Apply muroxide. 38/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Unknown 1 II) Chemical characters: Test Observation Comment 39/48
III) Special tests: (any three positive tests): 1-2- 3- IV) Conclusion: unknown no ( ) is................................................. 40/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Unknown 2 II) Chemical characters: Test Observation Comment 41/48
III) Special tests: (any three positive tests): 1-2- 3- IV) Conclusion: unknown no ( ) is................................................. 42/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Unknown 3 II) Chemical characters: Test Observation Comment 43/48
III) Special tests: (any three positive tests): 1-2- 3- IV) Conclusion: unknown no ( ) is................................................. 44/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Unknown 4 II) Chemical characters: Test Observation Comment 45/48
III) Special tests: (any three positive tests): 1-2- 3- IV) Conclusion: unknown no ( ) is................................................. 46/48
I) Physical characters: 1. Color: 2. Odor: 3. Aspect: 4. Reaction to litmus paper: Scheme for Unknown 5 II) Chemical characters: Test Observation Comment 47/48
III) Special tests: (any three positive tests): 1-2- 3- IV) Conclusion: unknown no ( ) is................................................. The End 48/48