LIPIDS PRACTICAL INTRODUCTION 6.2 LIPIDS - BASIC CONCEPTS

Transcription

1 PRACTICAL 6 LIPIDS 6.1 Introduction 6.2 Lipids-Basic Concepts Neutral Fats- Characterization and Assessment Cholesterol- The Derived Lipid Experiment 1 : Determination of the Iodine Number of Lipids using Hanus Method Experiment 2: Determination of the Saponification Number of Fats Experiment 3: Estimation of Cholesterol using Zlatkis Method INTRODUCTION You have studied that lipids together with carbohydrates and proteins are important as food for many animals. Lipids include neutral fats, phospholipids, cholesterol etc. The lipids are also of great biochemical importance because of their role as chief storage form of energy and in cellular structure. In addition, they are commercially important in their role in detergents, soaps and in the chemical and pharmaceutical industry. Fats, because of high degree of concern over health implications and also its extensive use in preparation of food products, have become the area of interest in Various parameters are used for the assessment of the quality of fats and oils. Some of the parameters including determination of moisture content, colour, impurities, acid value, peroxide value and tests for the presence of adulterants in fats and oils, we have already covered in the Principles of Food Science Practical Course - MFNL In this practical, we shall learn about other tests which will help us in characterization of fats and help us assess the purity of fats and oils. Bureau of Indian Standards (BIS) has standardized procedures available for these tests with recommended values and limits. We will be doing some of these tests to assess the purity of fats and oils in this practical. Using the combination of these tests, you will realize, you can determine the identity, the structure, the functional groups, physical properties etc. of the lipid components. Further, the practical shall also focus on methods used for estimation of cholesterol in a given solution. After studying this practical and undertaking the experiments given herewith you will perform tests for partial characterization of fats, differentiate between adulterated and non-adulterated fats and oils, determine the purity of fats and oils based on physical and chemical tests, interpret the results obtained, and estimate the amount ofcholesterol in a given solution. 6.2 LIPIDS - BASIC CONCEPTS You have studied the classification of lipids in your theory lesson. Lipids, you learnt, are classified as simple, compound or derived lipids on the basis of their chemical structure. Simple lipids are esters of fatty acids with various alcohols. They include neutral fats (which are esters of fatty acid with glycerol. Compound lipids are esters of fatty acids containing groups in addition to an alcohol and a fatty acid. Examples include phospholipids, glycolipids etc. Derived liplds are substances derived from above groups by hydrolysis. This group includes fatty acids, glycerol, steroids, sterols, fatty aldehydes and ketone bodies, vitamin A, D, E and K etc. Cholesterol belongs to the family of steroids. Let us learn about neutral fats and the tests involved in the assessment oftheir qual~ty in the next sub-section 157

2 utritional iochemistry Neutral Fats -.Characterization and Assessment and fatty acids. H triglyceride glycerol fatty acid Figure 6.1: Triglyceride When fats are in liquid form they are called oils. Fatty acids are grouped into two classes - saturated and unsaturhted fatty acids. When a fatty acid contains one or more double bonds (4 = C-), it is said to be unsaturated, otherwise it is a saturated fatty acid. They are esterified to glycerol in several permutations and combinations resulting in varying composition of fats and oils. If all the three fatty acids present in a triglyceride are the same, then the fat is referred to as simple glyceride and if they vary then it is referred to as mixed glyceride. A list of fatty acids present in natural fats in given in Table 6.1. You will find the food sources ofthese fatty acids also included in this table for your information. Table 6.1: Fatty acids in natural fats No. of Structure Name Source carbon atoms 4 CH,CH$H,COOH * Butyric acid Butter 6 CH,(CH,),COOH Caproic acid Butter 8 CH,(CHJ,COOH Caprylic acid Coconut oil 10 CH,(CY)FOOH Capric acid Coconut oil 12 CH,(CH,),,COOH Lauric acid Palm kernel oil 14 CH, (CH, ),, COOH Myristicscid Oil of nutmeg 16 CH, (CH2),4COOH Palmitic acid Palm oil 18 CH,(CH,),,COOH Stearic acid Beef tallow 18 CH,(CH,),CH= CH(CH,),COOH Oleic acid Olive oil 18 CH,(CH,),CH= CHCH,CH= CH(CH2),COOH Linolenic acid Soyabean oil 18 CH,CH,(CH= CHCH2)3(CH2)6COOH Linolenic acid Fish oil 20 CH,(CH,),(CH= CHCH,),CH,CH,COOH Arachidonic Liver acid The properties of fats areto a large extent dependent on the properties of the constituent fatty acids. The solubility, melting and boiling points vary with the varying chain length and the degree of unsaturation. You may recall studying about the chemical properties of fatty acids in the Nutritional Biochemistry Theory Course (MFN-002) in Unit 2, section 2.4, which include esterification, hydrogenation, halogenation etc. Recognition of these properties, you will realize, is ~mpottant for a better understanding of fatty acids. Along with the chemical properties of fatty acid, we have also looked at saponification, hydrogenation, hydrolysis which are chemical properties specific to neutral

3 To determine the purity of fats and oils, several physical, chemical and puri9 tests are performed. Some of them we have already covered in the MFNL-008 course. Bureau of Indian Standards (BIS) has standardized procedures available for these tests with recommended values and limits. Two important tests are: Iodine Number Estimation and determining the saponification number of fats. We will be doing these tests to assess the purity of fats and oils. These estimations will help in characterization of fats. Let us get to know about these tests. We begin with the Iodine Number Estimation, Lipids A. Iodine Number Estimation As you may recall studying in the theory course, in Unit 2, fatty acids accept chlorine, bromine and iodine at the double bond when treated with reagents such as iodine monochloride and a fatty halide is formed. This is the halogenation property of unsaturated fats and is used in iodine number estimation. Let us get to know the principle behind this estimation. Iodine number is defined as the number of grams of iodine absorbed by 100 grams of fat..,l gives an idea of the relative degree of unsaturation of fats or the amount ofdoume bonds present in the fats. Iodine number is directly proportional to the degree of In actual practice, Iodine monobromide (IBr) (Hanus method) or Iodine monochloride (ICI) (Wij's method) is used in preference to iodine solution, as these are more reactive and give more reliable and reproducible results. The method is based on the absorption of halogen by the unsaturated fatty acids present in fats. The method employs dissolution of fat in chloroform (CHCI,) and treatment with a known excess of halogen solution. After allowing for absorption of the halogen, the excess halogen is liberated as I, and is back titrated against standard sodium thiosulphate solution using starch as indicator. By using an excess of sodium thiosulphite to react with each double bond in the fatty acid chains and measuring the amount used by back titrating, we can determine the amount of sodium thiosulphate used and thus the relative nuniber of double bonds. We will conduct this experiment later in this practical. Next, let us also learn about the B. h~nification Number of Fats Fkt let us begin by understanding what is meant by saponification. Fats when boiled with alcoholic solution of NaOH or KOH undergo hydrolysis into glycerol and fatty, acids and the latter form soaps with Na or K. The reaction is known as saponification. The sapdnification value is the number of milligram ofpotassium hydroxide required tosaponifi 1 goffat under the conditions specifled. In other terms, the saponification number of fat or oil is the number of milligrams of KOH required to neutralize free or combined fatty acids in one gram of oil or fat. ~a~inification number indicates the molecular weight of the fatty acids present in the oil or fat and is inversely proportional to the molecular weight of fat. Since natural fats and oils are mixtures of fatty acids, it becomes an academic exercise to calculate the saponification number for the component fatty acids. The saponification number 6f a fat is determined by hydrolysis of the fat in an excess standard alcoholic alkali and determination of the excess alkali titrimetrically. Let us get to know the principle behind saponification. 159

4 Yutritional Biochemistry the number of fatty acid chains in the solution. In this experiment, the hydrolyzing agent is KOH. By using an excess of KOH and back titrating the unused excess with hydrochloric acid, we will know the number of moles of KOH used. Like iodine number, this method is also used for estimation of purity of fat. A given amount of fat or oil is refluxed with a known excess of alcoholic KOH and then the excess is back titrated against standard HCI using phenolphthalein as an indicator. This concept will become more clearer as we carry out this technique in Experiment 2 included in this practical. Next, we shall learn about cholestrol and study the methods we can use in the lsooratory for its estimation Cholesterol Cholesterol is an acyclic compound whose structure includes a cyclopentano perhydrophenanthrene nucleus with four fused rings, a single hydroxyl group at carbon 3, double bonds between carbons 5 and six. a hydrocarbon chain with 8 carbon atoms attached to carbon 17 and 2 methyl groups at carbon 10 and 13. Can you draw the structure of cholesterol with this information? Try completing the structure given herewith where the cyclopentano perhydrophenanthrene nucleus with four fused rings is already illustrated. Now refer to the cholesterol structure given in Figure 6.2, and judge your understanding of the structure. Figure 6.2: Cholesterol Cholesterol, we already know, belongs to the family of steroids. It is poorly soluble in water but soluble in fat solvents. It has an alcoholic group and thus forms esters with fatty'acids. In the body cholesterol is esterified with oleic, linoleic and palmitoleic acid. Thus, 70% of the cholesterol present in the blood is esterified whereas only 30% of cholesterol occurs as free cholesterol. The two sources of cholesterol in the body are: 1. Dietary cholesterol - Dietary cholesterol is mainly obtained from foods of animal origin. 2. De novo synthesis -Cholesterol is synthesized in every living cell. The major sites of synthesis are liver, adrenal cortex, aorta, intestine, brain etc.the biosynthesis involves a complex pathway consisting of 24 steps, about which you may recall studying in the theory Course, Unit 7. If not, we suggest you get back to the theory course. Look up Unit 7, sub-section 7.3.3, page 213 for studying the biosynthesis process of cholesterol. The starting material for cholesterol synthesis, you would 160 have noticed. is acetyl CoA.

5 Cholesterol is vital for the body to function properly i.e. for hormones and the bile acid production. Elevated cholesterol levels, on the other hand, have been associated with an increased risk of atherosclerosis (narrowing of the arteries because of fatty deposits). Hence, the importance of estimating cholesterol. Let us then move on to the methods we can use for estimation of cholesterol. Lipids Estimation of cholesterol Many methods for estimation of cholesterol are available, and have been used, which are both chemical and enzymatic. Most of the chemical methods used until recently were based on the ability of cholesterol to be converted to highly coloured substances in strong acid solvents that possess dehydrating, oxidizing and sulphonating properties. For many years, the Liberman Burchard Reaction was used for estimation of cholesterol which gave a green colour resulting from a reaction of cholesterol with acetic anhydride and sulphuric acid. At present, in chemical methods, the method of Zlatkis et al, is most popular. This is a colorimetric method. Colorimetric methods are based on the ability of cholesterol to be converted to highly coloured substances in solvents which have dehydrating, oxidizing and sulphonating properties. Let us study this method of estimation of cholesterol. We begin with the principle behind this estimation. Zlatkis and associates have evolved a method of quantitative determination of cholesterol based on a reagent containing ferric chloride, glacial acetic acid and concentrated sulphuric acid. Cholesterol gives a purple colour with this reagent. The colour development is due to a dehydration reaction in the cholesterol molecule to form 3,5 cholestadiene which is oxidized by H,S04 and polymerises to form a diamer or trimer. The cholastadiene and its polymers react with H,S04 to form mono and disulfonic acids which are highly coloured. In the presence of added metal ions like ferric, the disulphonic acids are preferentially formed. The detailed step-by-step procedure for cholesterol estimation is included in Experiment 3 later in this practical. Study the procedure carefully before conducting the experiment. With this basic understanding about iodine number, saponification technique and methods for cholesterol estimation, we are well equipped to carry out the experiments given in this practical. There are three experiments in this practical. Let us get started with 161

6 DETERMINATION OF THE IODINE NUMBER OF Date: Aim: To determine the Iodine Number of mustard, sesame or groundnut oil by Hanus methqd. Reagents The following reagents are required to conduct this experiment. 1. Hanus solution (Iodine monobromide solution). 2. Chloroform 3. 15% potassium iodide solution 4. Standard sodium thiosulfate solution N 5. 1 % starch solution Principle (We have already described the principle in section 6.2 above. Read and understand the principle and write the same in the space provided herewith). Procedure The technique is conducted in two steps. Follow the procedure given herewith and carry out the experiment. 1. Sample titration With a clean dry pipette, accurately measure 0.3 ml of oil into a clew and dry idination flask (DuplicateJask required). AlternativeIy, weigh and transfer 0.3 g of the oil into the flask for better accuracy. Add 10,ml CHCI, with a measuring cylinder and mix. Using an automatic pipette, add exactly 25 ml Hanus iodine solution. Mix well. Stopper the flask. Add distilled water in the cup around the stopper to prevent loss of iodine. Keep the flask in the dark for minutes. Remove the stopper and let the distilled water in the cup fall into the flask. Add 10 ml of 15% KI solution and 100 ml of freshly boiled and cooled distilled water. Shake. Titrate the iodine liberated in the process with standard sodium thiosulfate solution (0.1 N) stoppering the flask and shaking vigorously from time to time while titrating until a pale yellow colour is obtained. Now add 2 ml of 0.5% starch solution. Continue the titration until the blue colour disappears. After the blue colour disappears from the aqueous phase, note the chloroform layer in the bottom of the flask which may contain untitrated iodine as indicated by a pink or violet colour. By continuous shaking bring this iodine into the aqueous layer. The end pojnt of the titration is reached when the aqueous layer, as well as, the lower chloroform layer are completely colourless. 162 Titrate the duplicate iodine flask in a similar manner.

7 Using an autopipette, measure 25 ml of Hanus iodine reagent accurately and transfer into a 250 ml iodine flask. Add 10 ml CHCI, with a measuring cylinder. Mix. Stopper the flask and add distilled water in the cup around the stopper to prevent loss of iodine. Keep the flask in the dark for minutes. Titrate the contents of the flask according to the method described for sample titration. Since there is no'oil in the blank flask and no reaction will take place between the oil and Hanus solution, the blank titration could be performed without keeping the flask in dark for minutes to reduce the time for analysis. It could also be done with reduced volumes of the reagents as follows: measure accurately 5 ml of Hanus iodine reagent into a 250 ml conical flask, add 2 ml of CHCI, and immediately titrate using 115th volume of all the reagents used in the sample titration except the oil. Precautisnr 1. If using the pipette for transferring the oil, allow the pipette to drain slowly and compietely, as oil takes time to get transferred. 2. Do not pipette Hanus iodine reagent, since the glacial acetic acid vapours are corrosive and can be dangerous. 3. Allow iodination to take place in the dark., 4. Keep the flask stoppered to avoid loss of iodine by sublimation. 5. Always use freshly boiled and cooled distilled water. Method of Calculation 1. Blank Titration: i) Volume of Hanus Iodine = 5 ml ui Volume of 0.1 N Na,S,O, required S.No. Pilot Buret reading (ml) Initial Final Difference Titer value = (a).... ml iii) 5 ml of Hanus Iodine = (a)... ml of 0.1 N Na2S20, :. 25 ml of Hanus Iodine = a x = (b)... ml =... ml of 0.1 N Na2S20,, 2. Sample Titration: \ Volume of oil = 0.3 ml Specific gravity of oil = (c)... (You may use the value 0.91 since this 1s approximately the specific gravity of mustard and peanul 3il). iv) Weight of oil = specific gravity x volume :. Weight of our oil sample = 0.3 x c = 0.3 x c = (a)... gm =... v) Volume of Hanus Iodine = 25 ml 163

8 vi) Volume of 0. I N Na,S,O, required S.No. Pilot Buret reading (ml) Initial Final Difference Titer value = (e) ml =... ml vii) Volume of Hanus Iodine = Volume of 0. IN Na$i,O, added = (b)... ml viii) Volume of Hanus Iodine = Volume of 0.1 N Na,S jo, in excess = (e)....ml ix) Volume of Hanus Iodine = Volume of 0.1 N N%S,O, absorbed =. b - e ml = (f) ml - ml... x) 1000 ml of IN Na,S,O, = 1000 ml of IN lodine (I,) :. (f)... ml of 0.1N N~,s,O, = (f)... ml of 0.1 N 1, xi) 1000 ml of IN I, = 127 g I, (based on molecular weight) : ml of O.lN I, = 12.7 g 1,.a*... :. (f)... ml of 0.1 N I, = 12.7 x (f) = 12.7 x gm = (g) xii) (d)... gm of oil absorbs (g)... gm of 1, :. 100 g of oil absorbs E x 100 =... x 100 = hgm=... d... Result The iodine number of the given oil is... J To help you consolidate your understanding about Iodine Number, we have included some review questions herewith. Answer these questions and see where you stand. Review Questions 1. Define iodine number What is the significance of determination of iodine number What is the indicator used in iodine number titration?

9 Now submit the experiment and the review question answers for evaluation....,,...,..., Counsellor Signature

10 DETERMINATION OFTHE SAPONIFICATION Date: Aim: To determine the saponification number of a given sample of oil. Apparatus Burettes Pipettes Conical flasks 250 ml, I00 ml Volumetric flask I00 ml Reflux condenser Tripod stand Water bath (portable) Reagents 0.5 N alcoholic KOH 0.5 N HCI Solid sodium carbonate 1 % alcoholic solution of methyl orange 1 % alcoholic solution of phenolphthalein Principle (We have already described the principle in section 6.2 above. Read and understand the principle and write the same in the space provided herewith)..- Procedure Follow the steps enumerated herewith and conduct the experiment. 1. Standardization of HCl - Precisely weigh 2.65 g of anhydrous sodium carbonate and transfer it to 100 ml volumetric flask. Make the volume to the mark with distilled water. This is a solution of 0.5 N sodium carbonate. Pipette 10 ml of this solution into a 100 ml conical flask and add a drop of methyl orange as an indicator. Titrate with standard HCI till the colour changes to red. Calculate the normality of the HCI. 2. Sample titration - Measure 2 ml of oil with the help of a pipette and transfer this oil to a 250 ml conical flask. Add 40 ml of alcoholic KOH to the flask. Fit a reflux water condenser on the flask. Put a small water bath on a tripod stand and place the flask fitted with the condenser in the water bath. Heat the water bath for minutes until the fat globulesin the flask are not visible. Cool. Add a few drops of phenolphtalein as an indicator and titrate against the standardized HCI. 3. Blank titration - Pipette 5 ml of alcoholic KOH in a 250 ml conical flask. Titrate against standardized HCI using phenolphtalein as an indicator. The values can later be multiplied by 8. Calculations Now carry out the calculations following the lead given herewith:

11 1. ~tand&dization of HCL: Weight of Na,CO, = 2.65 g 2.65 g of Na,CO, diluted to 100 ml iii) Normality of Na,CO, solution = 0.5 N iiii Volume of Na,CO, solution = 10 ml iv) Volume of HCI required + S;No. Buret reading (ml) Initial Final Difference Pilot Titre value = (a)... v) Now using the formula given below calculate the normality of HCI. N,V, = N,V, where N, = Normality of NqCO, solution 'V, = Volume of Na,CO, solution N, = Normality of HCl V, = Volume of HCl = (a) :. N, = N,V,= 0.5 x 10 = 0.5 x 10 = b N =... N 'ml :. Normality of HCI =... N 2. Blank titration: vi) Volume of alcoholic KOH = 5 ml vii) Volume of (b)... N HCI required viii) 5 ml of alcoholic KOH = (c)... ml of (b)... N HCI 3. Sample tit~ation:... x) Weight of oil = volume x specific gravity Volume of oil = 2 ml Specific gravity of oil = (e)... (You may use the value 0.91 since this is. xi) approximately the specific gravity of mustard and peanut oil). :. Weight ofoil = 2 x e = 2 x... = (f)... &m Volume of alcoholic KOH = 40ml xiii Volume of b... N HCI required

12 Titer value =... mi = g ml xiii) Volume of alcoholic KOH =Volume of N HCI added = (d)... ml xiv)volume of alooholic KOH =Volume of (b)... N HCl in excess = (g)... ml xv) Volume of alcoholic KOH =Volume of(b)... N HCI used = d - g =...ml= h mi' xvi)1000mlof 1 N HCl= l000mlof 1 NKOH :. (h)... ml of (b)... N HCI = (h)... ml of (b)... N KOH xvii) 1000 ml of 1 N KOH = 28 g of KOH (based on molecular weight) : ml of (b)... N KOH = 28 x b = 28 x... = (i)gofkohz... :(h).. ml of (b)... N KOH= g x h x mg of KOH i ooo = ( ) mg = (we have multiplied the above value with 1000 to convert g into mg) xviii) (0... g of oil is saponified by 6).... mg of KOH :. 1 g of oil is saponified by j& =... mg f Result The saponification number of the oil is... To help you consolidate your understanding about saponification number, we have included some review questions herewith. Answer these questions and see where you stand. Review Questions 1. What is saponification? 2. Define saponification number. 3. In what way do saponification number help us characterize fats?. 4. How will you assess the purity of a fat using saponification number? Now submit the experiment and the review question answers for evaluation.... Counsellor signature

13 ESTIMATION OF CHOLESTEROL USING ZLATKIS ~im: To draw a standard curve for cholesterol and estimate the amount of cholesterol Date: in the given solution. Apparatus & Instruments Test tubes Borosil(6x314 inch) Test tube stand Pipettes 1 ml, 5 ml, 10 ml Beakers Measuring cylinders Volumetric flasks I00 ml, 50 ml Glass marker & labels Spectrophotorneterlcolorimeter Single pan balance Cuvettes Tissue roll Reagents

14 2. Colour development for standard solution: a) Take different concentrations ofthis standard solution (1.O, 2.0,3.0,4.0,5.0 and 6.0 ml) in 6 test tubes (S 146) and make the volume in each test tube upto 6 ml with glacial acetic acid. * b) Add 4 ml of colour reagent to each tube carefully from the side of the tube. A ring will be formed. c) Shake well and let it stand for 30 minutes for colour development. 3. Preparation of the unknown solution: Dilute the given stock solution to 50 ml with glacial acetic acid. Mix well. 4. Colour development for unknown solution: Pipette 3.0 rnl of the diluted unknpwn solution in duplicate test tubes Sal and Sa2. Similarly pipette 4.0 ml of the diluted unknown solution in duplicate in tubes Sa 3 and Sa 4. Follow the same procedure as you did for standard solution for colour development. 5. Preparation of blank: Pipette 2.0 ml of glacial acetic acid into a clean dry test tube. The blank will contain no standard cholesterol. Follow the same procedure as you did for standard solution for colour development. 6. Measurement of 0.D: Read the absorbency of all solutions in a colorimeter or spectrophotometer at 540 nm after setting the colorimeter at 100% transmittance using the blank solution. 7. Standard Curve: Plot a standard curve of concentration of cholesterol in standard vs. OD. 8. Determination of cholesterol content: Calculate the amount of cholesterol present in the given solution from standard curve as well as from 0.D for standard solution. Precautions 1. Use aldehyde free glacial acetic acid. 2. Use only glacial acetic for making up and adjusting the volume. Do not use distilled water. 3. Allow 30 minutes for colour development. 4. Before taking the readings, if a coloured layer is formed in the test tube, make sure that the contents in the,test tube are shaked and mixed well once again. 5. Make sure there are no air bubbles when taking the readings. 6. Be very careful while working with glacial acetic acid, since it is very toxic. Observations and Calculations Now record your observations and do the calculations as suggested in the section below. i) Preparation of Standard Cholesterol Weight of cholesterol taken = I00 mg 100 mg cholesterol was dissolved in glacial acetic acid and solution was made up to 100 ml Out of this we took 10 ml & diluted it to 100 ml with glacial acetic acid. :. Concentration of cholesterol solution = 0.1 mg/ml. Now we took different concentrations of cholesterol solution, so calculate for the different concentrations taken (i.e. 2.0 ml, 3.0,4.0, 5.0,6.0) in the space provided herewith. We have calculated for one, now you proceed with the others: Since 1.O ml of cholesterol solution = 0.1 mg cholesterol :. 2.0 ml of standardvcholesterol solution = 0.1 x 2.0 = 0.2 mg cholesterol :. 3.0 ml :. 4.0 ml :. 5.0ml :. 6.0ml

15 I ll m IV V M Volume of Conc. of Glacial Acetic Colour Optical Optical Density Std. Sol. Cholesterol Acid (ml) Reagent Density at for 0.4 mg (in ml) (mg) (mi) 540 nm Cholesterol 1.o O o I 'n m IV Volume of ~lacial Acetic Coloufieagent Optical Std. Sol. Acid (ml) (ml) Density at (in ml) 540 nm 3.0 ml Sal Sa2 4.0 ml w Sal Sa2

16 C1.... mg of cholesterol is present in 3.0 ml of unknown :. I00 ml of unknown contains = C1 x loo=... mg of unknown solution = Dl 3.0 Observed Value = (D, Dl).... * Expected Value = (F)... Now calculate the % error in the space provided using the formula given,, herewith: :. % error (based on optical density) = - F-Dx 100 F v) Determination of Cholestrol Content from Standard Curve On the standard curve prepared earlier, plot the OD of the unknown solution on the y- axis. Now check the corresponding concentration of cholesterol (in the two unknown samples taken) for this OD on the x-axis. Say the value obtained is E and El. So 3 ml of unknown solution conthins E ml of cholesterol :. I00 ml of unknown'solution will contain = E x 100 = F =... 3 Observed Value = F =... Expected value = D =...(Take it from the counsellor) Similarly, using this calculation now you calculate the cholesterol content in the second unknown samples (4.0 ml of which we had taken). Write your calculations here in the space provided. Observed Value = F1 =... Expected value = D =...(Take it from the counsellor) Now calculate the % error in the space provided using the formula given herewith: :. % error (based on o~tical density) = -- D-Fx 100 D Result The concentration of cholesterol in the given solution is ml 172

17 To consolidate your understanding on the topic do attempt the questions included in the review questions included herewith. Review Questions I. Why is it important to estimate cholesterol in blood? What does a high blood cholesterol level predict? What is the role of sulphuric acid in cholesterol estimation? 3. In what form is cholesterol present in blood? Now submit the experiment along with the answers to the review questions for evaluation.... Counsellor signature