Biochemistry Lab procedures followed in 1st yr MBBS, Kathmandu University

1. DIPESH YADAV
BIOCHEMISTRY
1st year laboratory test and their mechanism

2. CARBOHYDRATE
1. 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.

3. IODINE TEST
1)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

4. 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%>>>>>>Green
1-1.5%>>>>>>>>yellow
1.5-2%>>>>>>>>>orange
2-2.5%>>>>>>>>>>>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.

5. Barfoed’s Test
Procedure:- 1ml of carbohydrate soln + 1ml of Barfoed’s soln
Composition 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.

6. 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-+VE
Glucose—VE
Procedure:-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.

7. OSAZONE TEST
Procedure:- 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

8. Sucrose hydrolysis Test
Procedure:-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.

9. PROTEIN
BIOCHEMISTRY LAB TEST
-Dipesh yadav (NEPALGUNG MEDICAL COLLEGE,MBBS-1)

10. 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

11. Reaction Mechanism
Biuret In alkaline medium, peptide bond
+cupric ion to form violet colored
complex
Ninhydrin rxn 1)AA + ninhydrinketo-acid+
NH3+Co2+Hydrinadantin
2) Hydrinadantin+NH3+Ninhydrin
Ruhemann’s purple
Xanthoproteic rxn (Nitratation rxn)
Phenyl gr. + HNO3nitrophenyl gr.(yellow)
While adding alkali to the nitrophenyl soln
gives orange colour
Million’s rxn Phenolic gr.
+Hgso4mercurictyrosine 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

12. 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

13. TTTHANK YOU!!!!!!!!!!!!!!!!!!