PART II

Approach to Medical Sciences Biochemistry

PAGE Approach to Medical Sciences Biochemistry

Table of contents:

page

1- Qualitative Analysis of Lipids... 02

2- Qualitative Identification of Carbohydrates (I).... . 07

3- Qualitative Identification of Carbohydrates (II) . 11

4- Tutorial to assess the student understanding of lessons 1,2,3. .

5- Ionic Properties of Amino Acids. 14

6- Protein size and Separation. 18

7- Enzyme Activity. ..... 22

8- Tutorial to assess the student understanding of lessons 5,6,7..

QUALITATIVE ANALYSIS OF LIPIDS

Introduction :

Lipids are naturally occurring compounds that are esters of long chain fatty acids. They are insoluble in water but soluble in fat solvents such as acetone, alcohol, chloroform or benzene.

Alkaline hydrolysis (known as saponification) gives rise to the alcohol and the Na or K salts of the constituent fatty acids.

Chemically, lipids can be divided into two main groups:

1- Simple lipids.

2- Compound lipids.

Steroids and the fat soluble vitamins are also considered as lipids because of their similar solubility characteristics and are known as derived lipids.

Major Roles of Biological Lipids:Lipids of physiological importance for humans have four major functions:

1. They serve as structural components of biological membranes.

2. They provide energy reserves, predominantly in the form of triacylglycerols.

3. Both lipids and lipid derivatives serve as vitamins and hormones.

4. Lipophilic bile acids aid in lipid solubilization.

1- Simple lipids:

Esters of glycerol and fatty acids are known as acylglycerols or glycerides.

Triglyceride

The acylglycerols are known as neutral lipids, they are called fats or oils, depending on whether they are solid or liquid at room temperature.

Fatty acids:Fatty acids are long-chain hydrocarbon molecules containing a carboxylic acid moiety at one end.

Fatty acids that contain no carbon-carbon double bonds are termed saturated fatty acids; those that contain double bonds are unsaturated fatty acids.

Saturated fatty acids of less than eight carbon atoms are liquid at physiological temperature, whereas those containing more than ten are solid. The presence of double bonds in fatty acids significantly lowers the melting point relative to a saturated fatty acid.

2- Compound lipids:

Complete hydrolysis of a compound lipid yields at least one other component as well as the usual alcohol and fatty acids. These compounds are essential structural components of cell membranes. 3- Derived lipids:

Include Steroids and fat soluble vitamins.

Cholesterol:

Cholesterol is an extremely important biological molecule that has roles in membrane structure as well as being a precursor for the synthesis of the steroid hormones and bile acids.

The synthesis and utilization of cholesterol must be tightly regulated in order to prevent over-accumulation and abnormal deposition within the body. Of particular importance clinically is the abnormal deposition of cholesterol and cholesterol-rich lipoproteins in the coronary arteries. Such deposition, eventually leading to atherosclerosis, is the leading contributory factor in diseases of the coronary arteries.

Cholesterol

PRINCIPLES:

The main groups of lipids have different solubility characteristics. When fats and oils are heated with alkali, free fatty acids and glycerol are liberated and this process is known as saponification. The excess alkali present reacts with the liberated fatty acids to form the Na or K salts which give the solution a characteristic soap appearance. Soaps are soluble in water but precipitated on the addition of excess NaCl. The Mg and Cl salts on the other hand are insoluble and give rise to the cum formed when soap is lathered in hard water.

The fatty acids in animal fats are usually fully saturated whereas those found in vegetable oils contain one or more double bonds. Hydrogenation of the double bonds converts the oils into solid fats and this is carried out commercially for the production of margarine.

Halogens also readily add across the double bonds and the decolorization of a solution of bromine or iodine by a lipid indicates the presence of double bonds.

MATERIALS:

Reagents

Ethanol (absolute).

Diethyl ether.

Petroleum ether.

Chloroform.

Benzene.

Copper acetate (1% w/v).

Alco. KOH (100 g/l in ethanol).

HCl. Conc.

NaOH (1N).

NaCl (solid).

Bromine solution (1 ml Br2/20 ml Chloroform).

H2SO4 Conc.

KI (10% w/v).

Standard liquids

Fatty acids: palmitic acid, Stearic acid.

Fats: Butter.

Oils: Olive oil.

Phospholipids: Lecithin.

Sterols: cholesterol (solid and 0.5% in ethanol).

1- SOLUBILITY:

Note the physical state of:

a) Palmetic acid.

b) Stearic acid.

c) Oleic acid.

d) Olive oil.

e) Butter.

f) Water.

g) Ethanol.

h) Diethyl ether.

i) Petroleum ether.

j) Chloroform.

Carefully observe the differences between the above groups of lipids, write your remarks.

2- GREASE STAIN TEST:

Place one drop of the above lipids on a filter and leave to dry, observe the formation of a clear grease spot and give your remarks.

3- COPPER ACETATE TEST:

Dissolve a few drops of the oil in 3 ml. of petroleum ether and add equal volumes of copper acetate (1%), mix once by inversion (Dont shake). Leave the tube until the emulsion will separate into two layers.

A. Saturated fatty acid: upper layer is clear and precipitate in the lower layer.

B. Unsaturated FA: upper layer is greenish-Blue color and lower layer is colorless.

C. If the two layers are clear then the test.4- TEST FOR UNSATURATION:

Add one drop of Olive oil, one spatula-point of Butter and one spatula-point of Lecithin to separate dry test tubes, and dissolve the lipids in about 1 ml. of chloroform.

Add 1 ml. of chloroform to other test tube to act as Blank.

By means of Pasteur pipette add drop wise a solution of bromine in chloroform until define yellow color is produced, Note the number of drops in each case, and comment on the results.

CH3 (CH2)7 CH=CH (CH2)7 COOH + Br2Oleic acid Bromine

H H

CH3 (CH2)7 C C (CH2)7 COOH

Br Br

Dibromo-Stearic acid

5- TEST FOR CHOLESTEROL (Liebermans test):

Add 10 drops of acetic anhydride and 2 drops of conc. H2SO4 to 2 ml of each of the following:

a) Cholesterol solution in chloroform (0.5 %).

b) Egg yolk solution in chloroform (0.5%).

c) Butter solution in chloroform (0.5%).

d) Chloroform.

Give possible interpretation of the reaction in each case.

CARBOHYDRATES

Introduction:

Sugars can be defined as polyhydroxy aldehydes or ketones. Hence the simplest sugars contain at least three carbons. The most common are the aldo- and keto-trioses, tetroses, pentoses, and hexoses. The simplest 3C sugars are glyceraldehye and dihydroxyacetone

Carbohydrates can be classified as either monosaccharides, oligosaccharides or polysaccharides. Anywhere from two to ten monosaccharide units, linked by glycosidic bonds, make up an oligosaccharide. Polysaccharides are much larger, containing hundreds of monosaccharide units. The presence of the hydroxyl groups allows carbohydrates to interact with the aqueous environment and to participate in hydrogen bonding, both within and between chains. Derivatives of the carbohydrates may contain nitrogens, phosphates and sulfur compounds. Carbohydrates also can combine with lipid to form glycolipids or with protein to form glycoproteins.

Monosaccharides:

The monosaccharides commonly found in humans are classified according to the number of carbons they contain in their backbone structures. The major monosaccharides contain four to six carbon atoms.

Cyclic Fischer Projection of

-D-Glucose Haworth Projection of

-D-Glucose

Disaccharides:Covalent bonds between the anomeric hydroxyl of a cyclic sugar and the hydroxyl of a second sugar are termed glycosidic bonds, and the resultant molecules are glycosides. The linkage of two monosaccharides to form disaccharides involves a glycosidic bond. Physiogically important disaccharides are sucrose, lactose and maltose.

Sucrose: prevalent in sugar cane and sugar beets, is composed of glucose and fructose through an -(1,2)-glycosidic bond.

Sucrose

Lactose: is found exclusively in the milk of mammals and consists of galactose and glucose in a -(1,4) glycosidic bond.

Lactose

Maltose: the major degradation product of starch, is composed of 2 glucose monomers in an -(1,4) glycosidic bond.

Maltose

Polysaccharides:

Most of the carbohydrates found in nature occur in the form of high molecular weight polymers called polysaccharides. The monomeric building blocks used to generate polysaccharides can be varied; in all cases, however, the predominant monosaccharide found in polysaccharides is D-glucose. When polysaccharides are composed of a single monosaccharide building block, they are termed homopolysaccharides. While the Polysaccharides composed of more than one type of monosaccharide are termed heteropolysaccharides.

PREPARATION OF REAGENTS

1. Molischs reagent

5% ( naphthal in alcohol, i.e., 5g of (naphthal dissolved in 100ml of ethanol.2. Anthrone test:

Anthrone (2g/L in conc H2SO4).3. Iodine solution

0.005% in 3% KI, i.e., 3g of KI dissolved in 100ml water and then 5mg of iodine is dissolved.

4. Benedicts solution

17.3g of sodium citrate and 10g of sodium carbonate are dissolved in 75ml of water. 1.73g of CuSO4.5H2O is dissolved in 20ml of water. Mix the CuSO4 solution with alkaline citrate with constant stirring, finally the whole volume is made up to 100ml with water.

5. Barfoeds reagent

13.3g of copper acetate in 200ml of water and add 2ml of glacial acetic acid.6. Fearons test:

(methylamine Hydrochloride (MH) 5% in H2O + NaOH (20%).7. Bials reagent

Dissolve 300mg of orcinol in 100ml of concentrated HCl.

8. Seliwanoffs reagent

Dissolve 50g of resorcinol in 100ml of con.HCl in the ratio of 1:2.

9. Concentrated HCl

10. Concentrated H2SO411. Osazone Reagent

Phenyl hydrazine hydrochloride

Sodium acetate

Acetic acid

Mechanistic principles of qualitative identification of Carbohydrates

PRINCIPLE OF REACTIONS:1. Molischs test:

Con. H2SO4 dehydrates carbohydrates to form furfural and its derivatives. This product combines with sulphonated ( naphthal to give purple colour.

2. Anthrone test:

The same principle outlined above, except that the furfural reacts with anthrone (10-keto-9,10-dihydroanthracene) to give a blue-green complex.

3. Iodine test:

Iodine forms a coloured absorption complex with polysaccharides due to the formation of micellae aggregate. Iodine will form a polysaccharide inclusion complex.

4. Benedicts test

Carbohydrates with a potential aldehyde or ketone group have reducing property when placed in an alkaline solution. Cupric ions present in the solution will be reduced to cuprous ion. This will give a red coloured precipitate. Moreover, this test is more specific for reducing sugars.5. Barfoed test:

Barfoeds reagent is weakly acidic and it is only reduced by monosaccharides. Prolonged boiling may hydrolyze the disaccharide to give false positive test.

6. Bials test:

When pentose is heated with con.HCl, furfural, which condenses with orcinol in the prescence of ferric ion to give a blue green colour.7. Seliwanoffs test:

Ketoses are dehydrated more rapidly than aldose to give a furfural derivatives, which then condenses with resorcinol to form a red colour complex.

8. Osazone test:

Compounds containing aldehyde and keto groups form crystalline osazone with phenyl hydrazine hydrochloride. Osazone crystals have characteristic shape and melting point which helps in the identification of reducing sugar.

GENERAL PROCEDURE FOR QUALITATIVE ANALYSIS OF CARBOHYDRATES

No. EXPERIMENTOBSERVATION ONCLUSION

1.Molischs test

To 1ml of test solution, add 2 drops of Molischs reagent. Then add con. H2SO4 carefully along the sides of the test tube.Violet coloured ring is formed at the junction of the 2 layers.Presence of carbohydrate.

2. Anthrone test

To 5 drops of sugar solution add 2 ml. Anthrone reagent.Blue green color complex is formedPresence of carbohydrate.

3. Iodine test

To 1ml of the test solution, 2 drops of iodine is added and observe the colour change.(i) Deep blue colour is obtained.

(ii)Dark brown colour is obtained.

(ii)No characteristic

colour change.Presence of polysaccharide.

(Starch)

Presence of polysaccharide. (Glycogen)

Absence of polysaccharide.

4.Benedicts test

5ml of Benedicts reagent is mixed with 1ml of test solution and the contents are boiled for a few minutes.(i)Orange red precipitate is obtained.


colour change.Presence of reducing sugar.

Absence of reducing sugar.

5.Barfoeds test

To 2ml of test solution, 2ml of Barfoeds reagent is added and boiled for 3 minutes and the colour change is noted.(i)Brick red precipitate is obtained at the bottom of test tube.


colour change.Presence of reducing monosaccharide

Absence of reducing monosaccharide.

6.FEARON'S TEST

To 1ml of the test solution, 2ml of Fearons reagent is added and the content is heated. Then NaOH was added to the cold mixture.i)Red coloration appear.

ii)no color change.Presence of reducing disaccharide.

Absence of reducing disaccharide.

7.Bials test

To 1ml of the test solution, 5ml of Bials reagent is added. The contents are boiled and cooled.

(i)Blue green colour

is obtained.


colour change.Presence of pentose sugar.

Absence of pentose sugar.

8.Seliwanoffs test

To 1ml of the test solution, 3ml of Seliwanoffs reagent is added and the contents are boiled(i)Cherry red colour

is obtained.


colour change.

Presence of fructose.

Absence of fructose.

9.Osazone test

To 1ml of the test solution, , add 2-3 drops of glacial acetic acid, followed by the addition of a pinch of phenyl hydrazine hydrochloride and twice the amount of sodium acetate and the contents are boiled(i)Yellow colour crystals are formed in 5 minutes, as observed through a microscope.

Haystack structure form of fructosazone.

(ii) Yellow colour crystals are formed in 5-10 minutes, as observed through a microscope.

Haystack structure form of glucosazone

(iii) Yellow colour crystals are formed in 10 minutes.

Broad fan structure crystals of galactosazone are observed through microscope.

(iv) Yellow colour crystals are formed in 20-25 minutes,

Sunflower shaped crystals of maltosazone are observed through the microscope.

(v)Yellow colour crystals are formed in 5-10 minutes,

scattered needles shaped crystals of Xylosazone are observed through the microscope.

(vi) Yellow colour crystals are formed in 20-30 minutes,

Powderpuff shaped crystals of Lactosazone are observed through the microscope.Presence of fructose is confirmed

Presence of glucose is confirmed.

Presence of galactose is confirmed.

Presence of Maltose is confirmed.

Presence of Xylose is confirmed.

Presence of Lactose is confirmed.

IONIC PROPERTIES OF AMINO ACIDS

OBJECTIVES:

1- To investigate the ionization of compounds with two or more dissociable groups.

2- To demonstrate the particular importance of histidine as a potential buffer at physiologic pH.

Introduction:

Amino acids serve as basic building blocks for proteins. There are a large number of chemically possible amino acids, but only a few of these occur naturally.

In case of the 22 or so amino acids found in proteins, nearly all of them are -amino acids, where the amino group is present on the -carbon atom.

The strong positive charge on the -NH3+ group induces a tendency for the COOH group to lose a proton, so that amino acids are strong acids. The pKa for (the COOH of) Glycine, for example (2.3), is much lower than acetic acid (4.75), and other amino acids containing an aliphatic side chain have similar pK values to Glycine.Ionic properties of amino acids:

Amino acids are polyprotic, i.e. they all contain at least two dissociable protons, when dissolved in water, The proton from the -COOH group transfers to the -NH2 end of the molecule, because the NH2 group is a stronger base than -COO- forming the so called (zwitter ion) which contain both positive and negative charges ,while the net charge on the ion will be zero.

If the zwitter ion is treated with acid, H+ will be added to the COO- to form COOH, the resulting shape is the cation bearing a net positive charge. Similarly, treating the zwitter ion with base will result in the loss of the removable proton attached to the NH3+ group to form NH2, and the result anion bears a net negative charge. The following pH dependent equilibrium can be drawn.

K1 K2H3N+-CHR-COOH H3N+-CHR-COO- H2N-CHR-COO-

(K1 , K2: are the ionization constants of the COOH and NH3 respectively.)

e.g. : Glycine :

pKa= 2.34 9.60H3N+---CH2---COOH H++H3N+---CH2---COO- H++H2N---CH2---COO-

Gly+ Glyo Gly-Buffering:

According to Henderson-Hasselbalch equation:

pH = pKa + log[A-]/[HA]At the point of the dissociation where the concentration of the conjugate base [A-] is equal to that of the acid [HA]: pH = pKa + log[1]The log of 1 = 0. Thus, at the mid-point of titration of a weak acid: pH = pKaAt this point, when the pH = pKa, the slope of the curve (i.e. the change in pH with addition of base or acid) is at a minimum, so the buffer solution best resists addition of either acid or base, and hence has its greatest buffering ability.As a general rule, buffer solution can be made for a weak acid/base in the range of 1 pH unit from the pKa of the weak acids. addition of strong base produces weak conjugate base: CH3CO2H + OH- ------------> CH3CO2- + H2O

addition of strong acid produces weak acid: H3O+ + CH3CO2------------> CH3CO2H + H2O

Blood Buffering:

The pH of blood is maintained in a narrow range around 7.4. Even relatively small changes in this value of blood pH can lead to severe metabolic consequences. Therefore, blood buffering is extremely important in order to maintain homeostasis. The primary buffers in blood are hemoglobin in erythrocytes and bicarbonate ion (HCO3-) in the plasma. Buffering by hemoglobin is accomplished by ionization of the imidazole ring of histidines in the protein.

EXPERIMENT-1

Preparation of Normal Titration Curve for Glutamic Acid

The dissociation of glutamic acid can be represented as:

We are going to use the pH meter to explore the acid-base behavior of Glutamic acid .

Materials:

0.05M HCl. 0.05M Glu. 0.05M Hi. Ph meter.

Buret

Beaker.

pH standards.

0.05M NaOH.

Procedure:

1- Titrate 10 ml. of 0.05M Glutamic acid against 0.05M NaOH. Repeat the titration with 0.05M HCl.

2- Record your data in the given table.

3- Sketch a curve from your data on a graph paper.

(Plot pH (Yaxis) versus volume of NaOH expended (Xaxis).Volume (ml.) NaOH 0.05MpHVolume (ml.) HCl. 0.05MpH

0.00.0

0.50.5

1.01.0

1.51.5

2.02.0

2.52.5

3.03.0

3.53.5

4.04.0

4.54.5

5.05.0

EXPERIMENT-2

Preparation of Normal Titration Curve for Histidine

Repeat the above procedure using 0.05M Histidine.

The dissociation of Histidine can be represented as:

Here , is the ionizable groups of Histidine:

PROTEIN SIZE AND SEPARATIONPROTEINS:

Proteins are polymers of the bifunctional monomers, amino acids. The twenty common naturally-occurring amino acids each contain an -carbon, an -amino group, an -carboxylic acid group, and an -side chain or side group. These side chains (or R groups) may be either nonpolar, polar and uncharged, or charged, depending on the pH and pKa of the ionizable group.

Amino acids form polymers, the amino group from one amino acid attached to the carboxylic group of the adjacent amino acid, the resulting link between them is an amide link which biochemists call a peptide bond. In this reaction, water is released. In a reverse reaction, the peptide bond can be cleaved by water (hydrolysis).

O O

H2N C ------ HN C OH C C R1 R2 Peptide bond

When two amino acids link together to form an amide link, the resulting structure is called a dipeptide. Likewise, we can have tripeptides, tetrapeptides, and other polypeptides. At some point, when the structure is long enough, it is called a protein. There are many different ways to represent the structure of a polypeptide or protein.

Separation of proteins

The molecular weights of proteins are very high. Due to their wide variety of amino acid configurations, proteins behave very differently. These differences constitute the bases of the biochemical function of proteins. And these basic differences are also the parameters which are used to separate proteins:

1- Physical Size:

Which reflects molecular weight of the protein, Gel filtration uses this parameter, also referred to as Exclusion chromatography.

2- Electric Charge:

Some amino acid residues are positively charged while others are negatively charged. Variations in the pH of an amino acid system cause variations in the charge of amino acid residues. Consequently, the net surface charge of a protein (comprised of amino acid residues) also varies with its environment. It is these variations in the net charge of proteins which allow them to be separated by such techniques as ion exchange chromatography or by electrophoretic techniques.

3- Hydrophobic character:

Hydrophobic regions available to interact with a hydrophobic stationary phase provide the important characteristic of proteins which is used in adsorption chromatography.

4- 3-dimentional substructure:

Provide the basis for very specific separation methods. Bio-specific affinity chromatography is used to separate proteins according to their biological functions. Proteins, such as enzymes, antibodies and glycoproteins are particularly appropriate for separation by affinity chromatography.

Experiment: EXCLUSION CHROMATOGRAPHYObjectives:

a- To demonstrate the principle of molecular exclusion (gel permeation) chromatography using a bead of G-100 Sephadex gel to separate a mixture of:

Fluorescin : Yellow M.wt. = 332 (size of a dipeptide).

Hemoglobin : Red M.wt. = 68,000 (protein).

Blue Dextran : Blue M.wt. = 200,000 (large protein).

b- To show that the technique can be used to determine the Molecular Weight of a newly discovered protein.

Principle:

In gel filtration the gel acts as a molecular sieve separating molecules with differences in molecular size and weight. The gel matrix contains numerous porous beads (stationary phase) with (mobile phase) in between. If the ample of mixture is applied at the top of the column, the large molecules in the sample will not be able to enter the pores in the bead but will pass between them and so be eluted first, smaller molecules that have access to the pores are retarded in the gel to a certain extent and will therefore be eluted after the large molecules in order of decreasing M.wt. and size.

Materials:

Sephadex G-100: 0.5 g of (S.G-100) /100 ml. of 0.3% (w/v) NaCl; leave to swell for 1 hr. prior to experiment.

Eluant: 0.3% (w/v) NaCl.

Mixture of : Flurescin, hemoglobin and blue dextran (each 0.1 g/10 ml. H2O). To prepare Hemoglobin dilute 1 ml. of blood to 10 ml. with H2O.

Chromatography column (0.1 to 1.5 cm. diameters).

Microperpox peristaltic pump: Flow rate = 0.5 ml/min.

Disposable syringes (5ml, needle size 20 G).

Method:

1. Set up the column replacing the waist flask by a fraction collector.

2. Pour slurry of Sephadex G-100, into the column and allow the gel beads to settle. Once about 10 cm. have settled, allow the solvent to run out of the button of the column. Do not allow the liquid level to fall below the top of the Sephadex. Continue to add Sephadex until you have a column whose settled height is at least 15 cm.

3. Carefully place a small filter paper disc on top of the gel bed. This will prevent disturbance of the surface when sample or solvent are added.

4. Let the liquid level fall to the filter paper disc. Carefully add 0.2 ml. of the colored mixture with a pipette to the top of the column and let this run into the gel. Then add 0.5 ml. of the eluant and run this into the gel.

5. Fill the column with the eluant by connecting it to the reservoir and elute the sample, collecting fractions of ml. in numbered test tubes (Flow rate = 0.5 ml/min.)

6. Stop collecting fractions when the last visible band has been eluted and record the appearance of your fractions.

7. Repeat the procedure after adding 1% of protein to the mixture.

Treatment of data:

1. Measure the absorbance of the known fractions:

Fluroscin : 490 nm.

Hemoglobin : 420 nm.

Blue Dextran : 630 nm.

2. Plot a graph of absorbance versus fraction numbers.

3. Measure the absorbance of the fractions containing the protein of unknown M.wt.

Plot M.wt. versus fraction numbers and read the M.wt. of the unknown protein.

Fig.: Exclusion Chromatography.

Separation of the sampleTheory of separation

ENZYME ACTIVITY

Enzymes are biological catalysts that increase the rate of chemical reactions in the living organisms. Unlike most inorganic catalysts, enzymes have a very narrow specificity, i.e., they will only catalyze a comparatively small range of reactions or, in some cases, only one reaction.The overall reaction can be represented as follows:

The activity of enzymes can be measured by measuring the concentration of their products and this approach is widely used in clinical biochemistry and medical laboratories.

Enzymes function only under certain conditions of pH, temperature, substrate concentration, cofactors, etc. and some of these properties are illustrated in the following experiments.

ALP Reaction:

The enzyme catalyze the following reaction:

p-Nitrophenyl phosphate p-Nitrophenol

(Substrate) (Yellow colour)

Materials:

1. Carbonate buffer with different pH.

2. p-Nitrophenyl phosphate (0.01 M)

3. NaOH 1M.

4. Alkaline phosphatase Enzyme (0.1 mg/ml.)

5. p-Nitrophenol (50 mol/L.)

Experiment -1- Effect of Temperature

This experiment demonstrates the apparent existence of an optimum, above and below which the overall rate of the reaction decreases. The effect is the resultant of tow opposing processes:

An increase in temperature increases the velocity of the a chemical reaction.

An increase in temperature causes an increased rate of thermal inactivation of the enzyme, with the result that the velocity of the enzymatic reaction is diminished.

Set the following tubes and incubate them at the stated temperature:

Contents (ml.)Tube No.

12345678910

Buffer (pH9.2)2222222222

Substrate0.01M2222222222

Temp. ()1837455580

Absorbance

Conc. of the product

1. When the tubes have been given time to reach the desired temperature, add 2 ml. of enzyme to tubes: 1,3,5,7 and 9. mix thoroughly and incubate for your chosen time (Experiment Time course).

2. At the end of the incubation period stop the reaction by adding 2 ml. of 0.1 M NaOH to all tubes, and then 2 ml. of enzyme to tubes: 2,4,6,8, and 10. Mix well.

3. Read the absorbance starting with the faint tube.

4. After substracting the values of the Blank (i.e. the even numbered tubes); use the calibration curve to calculate the amount of p-Nitrophenol liberated, and plot these values (which give reaction velocity v) against temperature.

5. Calibration curve:

6. Prepare a standard curve. Stock of p-Nitrophenol (50 umol/L.). dissolve 278.4 gm/dl. In the buffer and dilute 1 ml. of the solution to 100 ml. with the buffer. Working solutions: 5,10,20,30,40 and 50 umol. (final concentrations).

7. Add 3 ml. of NaOH (1M) to 1 ml.of each of the above solutions into a test tube, mix well and read the absorbance at 405 nm. Plot a graph of absorbance against concentration.

8. Use the standard curve to calculate the amount of product (p-Nitrophenol) liberated. Plot this against reaction temperature. Measure the initial rate and choose a suitable incubation temperature for experiment 2.

Experiment -2- Optimum pHSet the following tubes in a batch at 37 using buffers of different pH.


123456789(BL.)

Buffer pH8.59.09.29.61010.51111.59.2

Buffer ml.222222222

Substrate 0.01M222222222

Enzyme

222222222

Absorbance


Incubate the tubes at 37 for your chosen period (time course), then add 2 ml. of 1 M NaOH and mix. After adding 2 ml. of enzyme to the blank tube 9, read the absorbance of all tubes.

Experiment -3- The time course of the reactionSet the tubes at 37 as follows:


123456789

Buffer pH =(9.2)2222222222

Substrate 0.01M2222222222

Enzyme

2222222222

Time (min.)

Absorbance


1. Pre-incubate the tubes for 5 mts.

2. Stop the reaction by adding 2 ml. of 1 M NaOH at the following times:

3. Tubes: (1,2:4 mts.);(3,4:8 mts.);(5,6:12 mts.);(7,8:16 mts.);(9,10:20 mts.).

4. Blank: add 2 ml. of enzyme to tubes: 2,4,6,8 and 10. mix well.

5. Read the absorbance of the solutions (all tubes) in order of increasing color density at 405 nm.

6. prepare a standard curve (calibration curve). Stock of p-Nitrophenol (50umol/L.). dissolve 278.4 mg/dl. In the buffer and dilute 1 ml. of the solution to 100 ml. with the buffer. Working solutions: 5,10,20,30,40 and 50 umol. (Final concentration.).

7. add 3 ml. of NaOH (1M) to 1 ml. of each of the abovesolutions into a test tube, mix well and read the absorbance at 405 nm. Plot a graph of absorbance against concentration.

8. use the standard curve to calculate the amount of product (p-Nitrophenol) liberated, and plot this against reaction time (point-c). measure the initial rate and choose a suitable incubation period for experiment 2.

Experiment -4- Effect of enzyme concentration:Set the following tubes and add the enzyme last:

Contents (ml)Tube No.

0123456

Buffer pH 9.22222222

Subst. 0.01M2222222

Water21.81.510.70.30.0

Enzyme0.00.20.511.31.72

Absorbance


1. Incubate the tubes for a suitable period.

2. Stop the reaction by adding 2 ml. of 1M NaOH.

3. Read the absorbance of the solution.(405 nm)

4. Use the standard curve to calculate the amount of p-Nitrophenol liberated in each tube and plot this against the volume of enzyme added. Choose the enzyme concentration at which the activity is linear ( data based on linear activity or rate gives precise and accurate estimates of the enzyme parameters such as Michael's constant (Km) of the enzyme.

Further readings:

1. Skoog, D. A.; Principles of Instrumental Analysis, 6th ed.; Thompson Brooks/Cole: Belmont, CA, 2006, Chapter 28.

2. Smith AL (Ed) et al. (1997). Oxford dictionary of biochemistry and molecular biology. Oxford [Oxfordshire]: Oxford University Press. ISBN 0-19-854768-4.

3. Grisham, Charles M.; Reginald H. Garrett (1999). Biochemistry. Philadelphia: Saunders College Pub. pp.4267. ISBN 0-03-022318-0.

4. Plummer D. T.: An Introduction to Principle Biochemistry, 2nd. ed. 1978.

UMM AL-QURA UNIVERSITY

Faculty of Medicine

Department of BIOCHEMISTRY

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Approach to Medical Sciences Biochemistry

Laboratory Manual

(Part II)

P

Documents

PART II