biochemistry post lab finals

BIOCHEMISTRY LABORATORYGeromil J. Lara, RMT, MSMT

QUALITATIVE TESTS FOR CARBOHYDRATES

MOLISCH TEST

– Carbohydrates are dehydrated by concentrated sulfuric acid to form hydroxymethylfurfural will react with alpha-naphthol (Molisch reagent) to yield a purple condensation product

– Not specific for carbohydrates


ANTHRONE TEST– Sugars react with the anthrone reagent

under acidic conditions to yield a blue-green color

– The sample is mixed with sulfuric acid and the anthrone reagent and then boiled until the reaction is completed.

– Determines both reducing and non-reducing sugars because of the presence of the strongly oxidizing sulfuric acid


OSAZONE FORMATION (Phenylhydrazine Test)– Reducing sugars form characteristic

osazone crystals when heated with an excess phenylhydrazine

– This property is attributed to the presence of aldehyde or ketone group in their molecules

– -OH group immediately adjacent to the keto group is oxidized to a keto group and adds to phenylhydrazine to form the yellow to pale orange osazones

GLUCOSAZONE

MALTOSAZONE

GALACTOSAZONE

LACTOSAZONE


MOORE’S TEST– When reducing sugar is heated with an

alkali it turns yellow to orange– Finally dark brown liberating the odor of

caramel, which becomes more marked upon acidification

– Due to the liberation of aldehyde which subsequently polymerizes to form resinous substance, caramel


FEHLING’S TEST– Fehling I consists of 7 g of hydrated

copper(II) sulfate dissolved in 100 mL of distilled water

– Fehling II is made by dissolving 35 g of potassium sodium tartrate and 10 g of sodium hydroxide in 100 mL of distilled water

– Equal volume – deep blue solution


FEHLING’S TEST– Aldolases are easily oxidized to yield

carboxylic acids– Cupric ion complexed with tartrate ion is

reduced to cuprous oxide (brick red ppt.)

Tartrate ions as a complexing agent to keep the copper ion in solution.

Without the tartrate ions, cupric hydroxide would precipitate from the basic solution.

The tartrate ion is unable to complex cuprous ion Cu+, so the reduction of Cu2+ to Cu+ by reducing sugars results in the formation of an orange to red precipitate of Cu2O.


NYLANDER’S TEST– Nylander’s Reagent

• A solution of Rochelle salt (potassium sodium tartrate), potassium or sodium hydroxide, and bismuth subnitrate in water

Bi(OH)2NO3 + KOH Bi(OH)3 + KNO3

reducing sugar

2 Bi(OH)3 2 Bi + 3 O + 3 H2O

heat (black precipitate)


BARFOED’S TEST– Barfoed’s Reagent

• Cupric acetate and acetic acid

• Reducing monosaccharides are oxidized by the copper ion in solution to form a carboxylic acid and a reddish precipitate of copper (I) oxide within three minutes.


PICRIC ACID TEST

– Picric acid (yellow) is reduced to picramic acid (mahogany red) in alkaline solution

– Presence of reducing sugars


SELIWANOFF’S TEST– a chemical test which distinguishes

between aldose and ketose sugars– is based on the fact that, when heated,

ketoses (e.g. fructose) are more rapidly dehydrated than aldoses

– Seliwanoff’s Reagent• Resorcinol and concentrated sulfuric acid


SELIWANOFF’S TEST– Seliwanoff’s Reagent

• Resorcinol and concentrated sulfuric acid

– Hydrolysis of polysaccharides and oligosaccharides yield simple sugars

– dehydrated ketose then reacts with the resorcinol to produce a deep cherry red color

– aldoses may react slightly to produce a faint pink color• Fructose (fast-reacting) and sucrose

http://en.wikipedia.org/wiki/File:Seliwanow.svg


INVERSION OF SUCROSE– Catalyzed by H ions to

form glucose and fructose• Sucrose is dextrorotatory

– Neutralized by barium hydroxide

– Barium sulfate = precipitate

– Glucose and Fructose = filtrate – Seliwanoff’s Test


IODINE TEST– Iodine in potassium iodide

• Reacts with starch to form a deep purple color

– Heating of the colored solution• Lighter in color or diminishing purple color

– Add few drops of sodium thiosulfate• To hydrolyze the starch to split fragments of

gradually diminishing molecular size and complexity, with simultaneous production of maltose (colorless)

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MUCIC ACID TEST– Sugar + Nitric Acid

– Oxidation of galactose and hydrolysis of lactose will yield insoluble mucic acid

– Mucic acid crystals – also known as galactaric acid• Melts at 213 degrees Celsius• Insoluble in alcohol and nearly insoluble in

cold water

CHARACERIZATION OF LIPIDS

SOLUBILITY– Insoluble in water

– Insoluble in ordinary solvents

– Readily dissolve in chloroform, benzene, ether, boiling alcohol and other organic solvents


FORMATION OF TRANSLUCENT SPOT– Lipids have a characteristic greasy feel– When brought in contact with a substance

like paper, penetrate through it producing a translucent spot• Fats are non-volatile• In RT, the spot of water can absorb enough

heat from the air and evaporized• But the spot of grease can never absorb

enough heat to evaporized• When the liquid is inside the sheet of paper, it

diffracts light – TRANSLUCENT PHENOMENON


REACTION OF FATS– Fatty acids are carboxylic acids and are

therefore weak acids– For fatty acids, the value of pKa is around

4.5. Therefore, generally speaking, fatty acids are neutral below pH 4.5 and charged above pH 4.5

– Fats containing high unsaturated fatty acids are neutral in reaction, but when exposed to air become acidic due to hydrolysis which results from the liberation of volatile fatty acids


ACROLEIN FORMATION

- 2H2O

Glycerol Acrolein (acrid odor)

dehydrating agent

(potassium bisulfate)


EMULSIFICATION OF FATS


EMULSION– Is a mixture of two or more materials that

are ordinarily immiscible– Droplets of the dispersed component

rapidly coalesce to form a separate layer– Emulsifying agent must be present to

stabilize the emulsion

– Lecithin in the egg will serve as emulsifier


SAPONIFICATION OF LARD– Alcoholic Potash

• KOH dissolved in ethanol• To neutralized fatty acids in the lard

– A metallic salt of fatty acid is formed

– Hydrophobic tails extend into the greasy droplets whereas the polar heads of the soap molecules face toward the water


SAPONIFICATION OF LARD– Sodium Carbonate – to produce hard soap

Palmitin Palmitate + Glycerol

Soap


SEPARATION OF CHOLESTEROL AND TRIGLYCERIDES– Precipitate = cholesterol digitonide after

digitonin precipitation– Supernatant = triglycerides

– PRECIPITATE tested for Salkowski Test and Liebermann-Burchard Test• Test for cholesterol


SALKOWSKI TEST– The presence of a double bond in one

cholesterol rings is responsible for its ability to form color products in the presence of concentrated inorganic acids

– Sulfuric acid• Results in dehydration of cholesterol molecule

with a formation of a red bicholestadien disulphonate

– Bluish color between the 1st layer (chloroform) and 2nd layer (H2SO4)


LIEBERMAN-BURCHARD TEST– Deep green color (+)

– due to the hydroxyl group (-OH) of cholesterol reacting with the reagents (acetic anhydride and concentrated sulfuric acid) and increasing the conjugation of the un-saturation in the adjacent fused ring

TRANSITIONAL PAGE

biochemistry post lab finals

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