Upload
dipesh-yadav
View
220
Download
3
Embed Size (px)
Citation preview
DIPESH YADAVBIOCHEMISTRY
1st year laboratory test and their mechanism
CARBOHYDRATE1. Molisch’s Test A)procedure:-1 ml of carbohydrate soln + 2-3 drops of α-napthol + 2 ml
conc. H2so4
B)Observation:-Violet ring at the junction of the two liquids C)Mechanism:- It is a general test for the detection of carbohydrates. The strong H2so4 Hydrolyses carbohydrate (poly- and disaccharides) to liberate monosac-Charides. The monosaccharide get dehydrated to form furfural (from pe-Toses) or hydroxy methylfurfural (from hexoses) which condense with α-naphthol to form a violet coloured complex.
IODINE TEST1)Procedure:- 2ml of carbohydrate soln + few drops of iodine soln Deep blue color:- with starch(non-reducing polysaccharides)
Purple color:-Dextrin soln
2)Mechanism:-Starch>soluble starch>Amylodextrin(purple)>Erythrodextrin(red)> Arcrodextrin(No colour)> maltose( end product of enzymatic hydrolysis of strach)
Polysaccharides combine with iodine to form a coloured complex. -ve proceeds for benedict’s test+ve proceeds for acid hydrolysis
Benedict’s test >5 ml Bendict’s reagent+ 1ml of carbohydrate soln>heat for 1-2 minutes>semiquantative quantative test because cuprous oxide gives different
colour with different concn of carbohydrate soln. 0.5-1%>>>>>>Green1-1.5%>>>>>>>>yellow1.5-2%>>>>>>>>>orange2-2.5%>>>>>>>>>>>Red composition of Benedict’s reagent:- copper sulphate (BLUE SOLN), Sodium citrate(prevents cuprous ion) Sodium carbonate (mild alkali)
Mechanism :- This is a test for the identification of reducing sugars, which form enediols (predominantly under alkaline conditions). The enediol forrms of sugars reduce cupric ions (cu++) of copper sulfate to coprous ions(cu+) which form a yellow precipiate of cuprous hydroxide or a brick red precipatate of cuprous oxide.
Barfoed’s TestProcedure:- 1ml of carbohydrate soln + 1ml of Barfoed’s solnComposition of Barfoed’s reagent <<<Cupric acetate <<< Acetic Acid
MECHANISM:-The principle of this test is the same as that of a Benedict’s test except that the reduction is carried in mild acidic medium. Since acidic medium is not favorable for reduction, only strong reducing sugars (monosaccharide) give this test positive. Thus, Barfoed’s test serves as a key reaction to distingush monosaccharide form disaccharides.
Observation :- Scanty re d ppt at the bottom of the test tube.
Seliwanoff’s test CLINICAL SIGNIFICANCE :- Thrombosis ,Degeneration, Necrosis Composition of Seliwanoff’s reagent:- Resor cinol(ALDEHYDE COMPOUND) HCL(Dehydrates Ketohexose more rapidally to form furfural derivatives)Fructose-+VEGlucose—VEProcedure:-1ml of carbohydrate soln+ 3ml of Seliwanoff reagent heat for 30 sec.
MECHANISM:- This is a specific test for ketohexoses. Concentrated HCL dehydrates Ketohexoses to form furfural derivatives which condense with resorcinol to give a cherry red complex.
OSAZONE TESTProcedure:- 2ml of sugar soln+2ml of osazone reagent mix and heat over boiling water bath.
Osazone reagent :- 3 molecules of phenyl hyadrazine Osazone –yellow crystals, crystalline derivatives of sugar and performed only with reducing sugar
Mechanism:-Phenylhydrazine in acetic acid, when boiled with reducing sugars forms osazones. The first two carbons(c1 and c2) are involved in this rxn. The sugars that differ in their configuration on these two carbons give the same type of osazones, since the difference is marked by binding with phenyl hydrazine. Thus, glucose, fructose, and mannose give the same type (needle shaped), maltose- sunflower shaped while lacose gives powder puff shaped
Sucrose hydrolysis TestProcedure:-2ml of sucrose solution in test tube+ 5 drops of conc. Hcl boil for 1 min over small flame and cool the contents and add 40% NaOH soln.
Mechanism:- Sucrose is a non- reducing sugar. Hence, it does not give Benedict’s and Barfoed’s tests. Sucrose can be hydrolyzed by conc. HCL, to be con verted to converted to glucose and fructose which answer the reducing rxns. However, after sucrose hydrolysis, the medium has to be made alkaline (by adding 40% NaOH) for effective reduction process.
PROTEINBIOCHEMISTRY LAB TEST-Dipesh yadav (NEPALGUNG MEDICAL COLLEGE,MBBS-1)
Color rxn of protein
Reaction Specific group or amino acid
Biuret Two peptide linkages
Ninhydrin rxn Alfa-amino acids
Xanthoproteic rxn Aromatic amino acid(phe, Tyr, Trp)
Million’s rxn Phenolic group(tyr)
Hopkins’s rxn(Aldehyde test) Indole ring (Trp)
Sakaguchi’s test Guanidino group(Arg)
Sulphur test Sulfhydryl group(Cystine & Cystiene)
Pauly’s test Imidazole ring(His)
Molisch’s test Carbohydrate moiety
Test for organic phosphate Casein
Reaction Mechanism
Biuret In alkaline medium, peptide bond +cupric ion to form violet colored complex
Ninhydrin rxn 1)AA + ninhydrinketo-acid+NH3+Co2+Hydrinadantin
2) Hydrinadantin+NH3+Ninhydrin
Ruhemann’s purple
Xanthoproteic rxn (Nitratation rxn) Phenyl gr. + HNO3nitrophenyl gr.(yellow)While adding alkali to the nitrophenyl soln
gives orange colour
Million’s rxn Phenolic gr.+Hgso4mercurictyrosine complex which on nitration give red color with sod. nitrate
Hopkins’s rxn(Aldehyde test) INDOLE gr.+Formaldehyde+(in presence of oxidizing agent H2SO4 with mercuric sulphate)violet colored complex
Sakaguchi’s test In alkaline medm(40% NaOH), α-napathol+Guanidino Gr.complex which is oxidized by sod. Hypobromide to give red colour complex
Sulphur test When cysteine and cystine are boiled with NaOH, organic sulfur is converted to inorganic sodium sulfide. This reacts with lead aceatate to form black ppt. of lead sulfide. Methionine doesnot give this test, since sulfur of methionine isnot split by alkali
Pauly’s test Diazotised sulfanilic acid reacts with imidazole ring om alkaline medium to form a red colored complex.
Molish’s test The proteins containing carbohydrate(Glycoprotein ) give +ve to this test.(ALBUMIN)
Test for organic phosphate
TTTHANK YOU!!!!!!!!!!!!!!!!!!