ABSTRACT

Amino acids is a monomer for protein and a building blocks for our body, it contains many types of amino acids. HPLC is a high-performance liquid chromatography that can be used to determine amino acids. It can detect types of amino acids by allow the component flow through it column and detect by the detectors that lead signals of peak for every compound. HPLC is well known for its advantages include rapid analysis, specificity, high sensitivity and simplified ample preparation. It is widely used in food industry. Nearly anything found in food can be detected and quantified by HPLC provided it can be solubelized in a suitable solvent. ^[2][3]

There are 16 types of amino acids that detected in this experiment and each of them results in different peak frequency and value. Those are aspartic acid, glutamic acid , serine, glycine, histidine, arginine,theonine, alanine,proline, tyrosine, valine, methionine,cystine, isoleucine, leucine and phenylalanine. This experiment done with the sample must first be hydrolysis. It can be kept for about 4 weeks in -20⁰C. After adding with the redyring solution, sample is then can be inject to the HPLC, but need to be mixed with 100µl of ample diluents and vortex. In this experiment, glutamic acid has the highest peak while methionine has the lowest peak. This shows that amino acid contain a lot of glutamic acid and have low amount of methionine acid compound.

INTRODUCTION

High-performance liquid chromatography is a form of column chromatography used frequently in biochemistry and analytical chemistry to separate, identify, and quantify compounds. HPLC utilizes a column that holds chromatographic packing material (stationary phase), a pump that moves the mobile phase through the column, and a detector that shows the retention times of the molecules. Retention time varies depending on the interactions between the stationary phase, the molecules being analyzed, and the solvent used. The sample to be analyzed is introduced in small volume to the stream of mobile phase. The analyte’s motion through the column is slowed by specific chemical or physical interactions with the stationary phase as it traverses (passes through) the length of the column. The amount of retardation depends on the nature of the analyte, stationary phase and mobile phase composition. The time at which a specific analyte elutes (comes out of the end of the column) is called the retention time; the retention time under particular conditions is considered a reasonably unique identifying characteristic of a given analyte. The use of smaller particle size column packing (which creates higher backpressure) increases the linear velocity (speed) giving the components less time to diffuse within the column, leading to improved resolution in the resulting chromatogram. Common solvents used include any miscible combination of water or various organic liquids (the most common are methanol and acetonitrile). Water may contain buffers or salts to assist in the separation of the analyte components, or compounds such as trifluoroacetic acid which acts as an ion pairing agent.^[1][2]

                        The choice of solvents, additives and gradient depend on the nature of the stationary phase and the analyte. Often a series of tests are performed on the analyte and a number of trial runs may be processed in order to find the HPLC method which gives the best separation of peaks.

OBJECTIVES

1.      To determine amino acids by using HPLC.

2.      To study the HPLC and the functions.

PROCEDURE

Sample hydrolysis

0.25g of sample was weight in a close test tube

↓

15ml of 6N HCl was added and vortex

↓

Nitrogen gas was flushed for 1 minute

↓

Put in oven of 110˚C for 24 hour and was cooled

↓

10ml of AABA (internal standard) was added

↓

The sample was poured in the 50ml volumetric flask and was made up with ionized water

↓

The sample was filtered with the filter paper (throw 5-11 ml of upper layer)

↓

The hydrolysis sample can be kept for 4 weeks at -20˚C.

Prepare Redrying Solution

Solution prepared by the following ration

Methanol + Water + Triethylamine (2:2:1)

Derivitization Reagent

A solution is prepared by the following ration

Methanol + Water + Triethylamine + Phenylisocyanate (7:1:1:1)

Preparing Sample and Standard Derivation

Sample filtered with 0.2µm, cellulose nitrate membrane filter/nylon type

↓

10µL sample added to a vial and 10µL standard in another vial

↓

Both vials dried under vacuum for 30 min

↓

20µL of redrying solution added into sample and standard vial and is vortexed

↓

Vial is dried under vacuum for 30 min

↓

20µL of derivation reagent added, vortexed and left at room temp for 20 min

↓

Vials vacuumed dry for 30 min

↓

Prepared sample can be kept up to 4 weeks

Sample and Standard Injection

Sample and standard must mix with 100µL sample diluent (mobile phase A) and vortexed for 15 min before injected into the HPLC

↓

Inject Volume:

Standard- 8µL

Blank- 8µL

                                                                Sample- 20µL

↓

Internal Standard: 0.2578g AABA make up to 1 litre 0.1N HCL

APPENDICES

RFF (in standard) = area of AABA (in standard)

              area of AA (in standard)

RFF (in sample) = RFF (in standard) x area of AA (in sample) x constant (µM/mL)

                                            area of AABA (in sample) x weight of sample

Original Concentration = RRF (in sample) x molecular weight of AA

where: AABA – α-amino butyric acid

            AA – amino acid

            Weight of sample – 0.25g

            Constant (µM/mL) – 0.025 for all AA accept Cystine, the value is 0.0125

asp, Aspartic Acid

RFF (in standard) = 1892240.028

                                   1688979.199

                                = 1.1203

RFF ( in sample) = 1.1203 x 1318482.020 x 0.025

                                         676994.410 x 0.25

                              = 0.2182

Original concentration = 0.2182 x 133.1

                                        = 29.0415 mg/g of sample

glu, glutamic acid

RFF (in standard) = 1892240.028

                                   1643835.424

                                = 1.1511

RFF ( in sample) = 1.1511 x 4977412.074 x 0.025

                                         676994.410 x 0.25

                              = 0.8463

Original concentration = 0.8463 x 147.1

                                        = 124.4928 mg/g of sample

ser, serine

RFF (in standard) = 1892240.028

                                   1914917.706

                                = 0.9982

RFF ( in sample) = 0.9982 x 1083984.016 x 0.025

                                         676994.410 x 0.25

                              = 0.1582

Original concentration = 0.1582 x 105.1

                                        = 16.6290 mg/g of sample

gly, glycine

RFF (in standard) = 1892240.028

                                   1995224.030

                                = 0.9484

RFF ( in sample) = 0.9484 x 2671557.640 x 0.025

                                         676994.410 x 0.25

                              = 0.3743

Original concentration = 0.3743 x 75.07

                                        = 28.0951 mg/g of sample

his, histidine

RFF (in standard) = 1892240.028

                                   1859657.337

                                = 1.0175

RFF ( in sample) = 1.0175 x 512243.208 x 0.025

                                         676994.410 x 0.25

                              = 0.0770

Original concentration = 0.0770 x 155.2

                                        = 11.9489 mg/g of sample

arg, arginine

RFF (in standard) = 1892240.028

                                   1959651.814

                                = 0.9656

RFF ( in sample) = 0.9656 x 3244574.805 x 0.025

                                         676994.410 x 0.25

                              = 0.4628

Original concentration = 0.4628 x 174.2

                                        = 80.6154 mg/g of sample

thr, threonine

RFF (in standard) = 1892240.028

                                   2221909.248

                                = 0.8516

RFF ( in sample) = 0.8516 x 1010463.956 x 0.025

                                         676994.410 x 0.25

                              = 0.1271

Original concentration = 0.1271 x 119.1

                                        = 15.1390 mg/g of sample

ala, alanine

RFF (in standard) = 1892240.028

                                   1937958.429

                                = 0.9764

RFF ( in sample) = 0.9764 x 1828142.530 x 0.025

                                         676994.410 x 0.25

                              = 0.2637

Original concentration = 0.2637 x 89.09

                                        = 23.4901 mg/g of sample

pro, praline

RFF (in standard) = 1892240.028

                                   1936384.521

                                = 0.9772

RFF ( in sample) = 0.9772 x 1752266.426 x 0.025

                                         676994.410 x 0.25

                              = 0.2529

Original concentration = 0.2529 x 115.1

                                        = 29.1122 mg/g of sample

tyr, tyrosine

RFF (in standard) = 1892240.028

                                   2273277.640

                                =0.8324

RFF ( in sample) = 0.8324 x 1213144.253 x 0.025

                                         676994.410 x 0.25

                              = 0.1492

Original concentration = 0.1492 x 181.2

                                        = 27.0277 mg/g of sample

val, valine

RFF (in standard) = 1892240.028

                                   1736244.160

                                = 1.0898

RFF ( in sample) = 1.0898 x 507815.342 x 0.025

                                         676994.410 x 0.25

                              = 0.0817

Original concentration = 0.0817x 117.2

                                        = 9.5811 mg/g of sample

met, methionine

RFF (in standard) = 1892240.028

                                   2231294.299

                                = 0.8480

RFF ( in sample) = 0.8480 x 190627.203 x 0.025

                                         676994.410 x 0.25

                              = 0.0239

Original concentration = 0.0239 x 149.2

                                        = 3.5628 mg/g of sample

cys, cystine

RFF (in standard) = 1892240.028

                                   874905.571

                                = 2.1628

RFF ( in sample) = 2.1628 x 477754.407 x 0.0125

                                         676994.410 x 0.25

                              = 0.0763

Original concentration = 0.0763 x 240.3

                                        = 18.3383 mg/g of sample

ile, isoleucine

RFF (in standard) = 1892240.028

                                   2207383.863

                                = 0.8572

RFF ( in sample) = 0.8572 x 900820.875 x 0.025

                                         676994.410 x 0.25

                              = 0.1141

Original concentration = 0.1141 x 131.2

                                        = 14.9653 mg/g of sample

leu, leucine

RFF (in standard) = 1892240.028

                                   1954516.281

                                = 0.9681

RFF ( in sample) = 0.9681 x 2776271.180 x 0.025

                                         676994.410 x 0.25

                              = 0.3970

Original concentration = 0.3970 x 131.2

                                        = 52.0892 mg/g of sample

phe, phenylalanine

RFF (in standard) = 1892240.028

                                   1906228.028

                                = 0.9927

RFF ( in sample) = 0.9927 x 1080960.372 x 0.025

                                         676994.410 x 0.25

                              =0.1585

Original concentration = 0.1585 x 165.2

                                        = 26.1840 mg/g of sample

NH₃, ammonia

RFF (in standard) = 1892240.028

                                   335200.805

                                = 3.6451

RFF ( in sample) = 3.6451 x 379073.606 x 0.025

                                         676994.410 x 0.25

                              = 0.3161

Original concentration = 0.3161 x 53.49

                                        = 16.9076 mg/g of sample

CONCLUSION

REFERENCE

1.      Displacement Chromatography 101.Sachem, Inc. Austin, TX 78737

2.      A recent book provides a comprehensive treatment of the theory of high-performance gradient chromatography: Lloyd R. Snyder and John W. Dolan (2006). High-Performance Gradient Elution: The Practical Application of the Linear-Solvent-Strength Model. Wiley Interscience. ISBN 0471706469. 

3.      Food chemistry a laboratory manual,page 142-143, Dennis D.Miller; Department of Food Science Cornell University Ithaca, NY. Published by JOHN WILEY & SONS,INC 1998.

3. MATERIALS AND METHODS

3.1. Moisture content
The oven was set up at temperature 105 oC before crucible can be put in it. Clean and dried crucible was put in the oven for 30 minute to remove all moisture. The crucible were took out from the oven and been cooled to room temperature before it can be weight. The crucible been weight and the data was recorded. 5 grams of sample yogurt and dadih were put in the each crucible. The crucible then had been put in the oven back for at least 8 hours. After that, the crucible was cooled in the dessicator until it becomes room temperature. The crucible was weight and the data was recorded. The amount of moisture in yogurt and dadih was calculated.

3.2. Ash Content
Crucible was put in the oven at 105 oC for at least 60 minutes. Then the crucible was cooled to room temperature in the dessicator and been weight. The weight of crucible was recorded. 5 gram of food products which is yogurt and dadih were been put in the crucible each. Then the crucibles were put in the muffle furnace at temperature 425 oC. After the entire particle had been burned off, where there was no black particle in the food sample, the crucibles were weight and the data were recorded. The amount of ash was calculated.

3.3. Crude Protein
To make the experiment of crude protein and crude fat, the sample had been dried for one day in oven to remove all the moisture is the food sample. It is because the experiment is used in dried sample. 0.15 gram of sample yogurt and dadih were weight and put in the Micro Kjedhal test tube. 0.8g of mixed catalyst was being added to the test tube, followed by 2.5mL concentrated sulphuric acid and the solution was swirled to mix them well. The tube was heated slowly on a heating coil under fume hood. The solution was boiled it until clear or blue green colour solution was obtained. The test tube was cooled to 40 oC before 5 mL distilled water were added. The digested product was transferred into distillation tube. 10mL 45% NaOH solution was added slowly to separate the two layers of solution. Then distillation tube was fixed to the condenser neatly. 10mL 0.5N boric acid was added into a conical flask (for distillation product), followed by 2 drops of indicator. The conical flask was placed at the distillate platform and the tip of distillation tube immersed into acid solution. The content of distillation flask was mixed by swirl it gently. Steam was purged into the flask. The ammonia solution was being let to distil into the conical flask. After distillation product was mixed, titration was carried out to titrate unreacted boric acid with 0.05N H2SO4 till neutral. The same procedures were repeated for another 2 tubes and a blank.

3.4. Crude Fat
Round bottom flask were put in oven before it been weight. 3 grams of dried sample yogurt and dadih was been weight and put into the thimble. The thimble was inserted into the Soxhlet apparatus. 200ml of petroleum ether had been pouring into round bottom flask after it had been weight. The Soxhlet was connected to the reflux and round bottom flask. Sample was refluxed continuously for about 8 hours. After 8 hours, petroleum ether was evaporated from the round bottom flask using rotary evaporator. The round bottom flask was placed in desiccators to remove all moisture in it. Round bottom flask together with the extracted fat was weighted. The data were recorded.

3.5. Crude Fiber
Crucible and ashless filter paper was dried in an oven for 1hour at 105°C. Both crucible and ashless paper was cooled in dessicator. The weight of filter paper was calculated. 2g of defatted sample was weighted into 500ml conical flask. 200ml of sulphuric acid was added and dissolved using spatula. The conical flask was attached to the reflux condenser. The sample was boiled for 30 min. The conical flask was turned every 5 min to make sure that the sample was always in contact with the acid. The hydrolyzed mixture was filtered through filter paper Whatman No. 541. The residue was rinsed with distilled water to remove the acid from the filtrate. The residue was placed back to its conical flask and 200ml boiled sodium hydroxide (NaOH) was added. The condenser was attached and was boiled for another 30 min. The hydrolyzed mixtures that had been boiled were filtered through dried and weighed filter paper. The residue was rinsed with boiled distilled water until the filtrate is free from alkaline. Rinse again with small amount of alcohol and then drain it. The residue and filter paper was placed into the crucible and dried in oven at 105°C until the constant weight is achieved. The crucible was placed in the muffle furnace at 550 oC and burn completely. The crucible was placed into the dessicator. The constant weight was calculated.

3.6. Somogyi-Nelson Method
The sample that been used in this test was yogurt and dadih which is both is semi solid product. Because of that, 5 grams of each product was taken and was first blend with 10ml distilled water in order to homogenize it. The sample was later transferred to 250 ml volumetric flask and was dilute till reached the mark.. This sample can be called as sample A for each product. For each of it, 10 ml of sample were taken from sample A and been pipette into 100 ml volumetric flask. The samples were diluted to 100 ml with distilled water. To calculate the reducing sugar, the blank sample had to be constructed to get the graft. Blank and standard solution is prepared as shown as below:
Tube 1 Tube 2 Tube 3 Tube 4 Tube 5 Tube 6 Tube 7
Glucose solution (100g/ml) (ml) - 0.5 1.0 1.5 2.0 - -
Glucose solution (300g/ml) (ml) - - - - 1.0 1.5
Distilled water (ml) 2.0 1.5 1.0 0.5 - 1.0 0.5

A sample solution containing not more than 2 ml of solution B was made. A series of sample with 0.5, 1.0 and 1.5 ml of sample was made and all of it was make it up to volume of 2ml with distilled water. All samples, blank, and standard sugar solution was added with 2ml copper reagent and mix the contents well. Samples, blank and standard sugar solution were heated 10 min in a vigorously boiling water bath and then cooled under running water for 5 min. 1 ml of arsenomolybdate reagents were added to all the tubes and mix well. The solution was diluted to 10 ml using volumetric flask when all the cuprous oxide is dissolved after mixing. The solution was allowed to stand at least 15min but not more than 40 min. Read the absorbance at 520 nm using a spectrophotometer.

4. RESULT
4.1. Determination of Moisture Content

Yogurt:
sample Weight of crucible Weight of sample weight of crucible + sample
(after drying process in oven)

1 54.741 4.0561 51.800

2 54.688 4.0913 57.577
3 55.155 4.0441 58.116

Percentage of moisture:
sample % moisture % total solids (xi-x)2 Standard deviation
1 10.97 89.03 2.1609

1.739

2 14.39 85.61 3.6481
3 11.95
88.05 0.2401

Dadih:
sample Weight of crucible Weight of sample weight of crucible + sample
(after drying process in oven)

1 48.881 4.0113 51.800

2 54.748 4.0224 57.577
3 55.195 4.0135 58.116

Percentage of moisture:
sample % moisture % total solids (xi-x)2 Standard deviation
1 27.22 72.78 0.6561

1.4029
2 29.65 70.35 2.6244
3 27.22 72.78 0.6561

4.2. Determination of Ash Content
Yogurt:
sample Weight of crucible Weight of sample weight of crucible + sample
(after drying processin muffle furnace)

1 48.932 4.0618
48.994

2
48.832 4.0119 48.891
3 48.831 4.0327 48.883

Percentage of ash:
sample % ash (xi-x)2 Standard deviation
1 1.53 0.0004

0.038
2 1.47 0.0016
3 1.54 0.0009

Dadih:

sample Weight of crucible Weight of sample weight of crucible + sample
(after drying processin muffle furnace)

1 55.0685 4.0971
55.0797

2
56.0112 4.0351 56.0304
3 55.0449 4.0551 55.0587

Percentage of ash:
sample % ash (xi-x)2 Standard deviation
1 0.273 0.0081

0.1033
2 0.476 0.0127
3 0.340 0.00053

4.3. Determination of Crude Protein

Yogurt:

Weight of sample (g) Volume of H2SO4 (mL) Percentage of Nitrogen (%) Percentage of Protein (%)
0.1487 7.0 3.248 20.67
0.1490 6.8 3.147
20.07
0.1488 6.8 3.152
20.10
0.0 0.10 - -
Average percentage = 20.28
Normality of Nitrogen = 0.05N

Weight of sample (g) Volume of H2SO4 (mL) Percentage of Nitrogen (%) Percentage of Protein (%)
0.1501 0.5 0.1865 1.1898
0.1491 0.6 0.2347 1.4974

0.1483 0.5 0.1888 1.2045
0.0 0.10 - -
Average percentage = 1.297
Dadih:

Normality of Nitrogen = 0.05N

4.4. Determination of Crude Fat

Yogurt:
Weight of thimble (g) Weight of thimble + sample (g) Weight of sample (g) Weight of round bottom flask (g) Weight of round bottom flask + oil (g) Weight of oil (g) Percentage of oil in sample (%)
2.3363 5.0317 118.401 118.485 0.0894 1.667%
2.3455 5.0228 118.644 118.707 0.0624 1.242%
Average percentage = 1.455%

Dadih:
Weight of thimble (g) Weight of thimble + sample (g) Weight of sample (g) Weight of round bottom flask (g) Weight of round bottom flask + oil (g) Weight of oil (g) Percentage of oil in sample (%)
2.3213 5.0161 120.048 120.096 0.0482 0.961%
2.3325 5.0213 119.024 119.039 0.0153 0.305%
Average percentage = 0.633%

4.5. Determination of Crude Fiber
Yogurt:
Weight of sample (g) Weight of filter paper (g) Weight of crucible + filter paper + dried residue (g) Weight of crucible + ash (g) Percentage crude fibre (%)
2.0026 0.2950 48.090 47.394 20.04
2.0092 0.3076 56.225 55.433 24.14
2.0183 0.2991 51.956 51.268 19.32
Average percentage = 21.17

Dadih:
Weight of sample (g) Weight of filter paper (g) Weight of crucible + filter paper + dried residue (g) Weight of crucible + ash (g) Percentage crude fibre (%)
2.0508 0.2932 48.386 47.892 3.07
2.0672 0.2887 50.411 50.089 1.611
2.0574 0.2901 42.421 42.079 2.522
Average percentage = 3.602

4.6. Determination of reducing sugar
Standard Solution:
Tube 1 Tube 2 Tube 3 Tube 4 Tube 5 Tube 6 Tube 7
Glucose solution
(100 µg/ml)
(mL) - 0.5 1.0 1.5 2.0 - -
Glucose solution
(300 µg/ml)
(mL) - - - - - 1.0 1.5
Distilled water
(mL) 2.0 1.5 1.0 0.5 - 1.0 0.5
Glucose concentration
(µg/ml) 0 25 50 75 100 150 225
Absorbance - 0.421 0.522 0.751 0.924 1.455 2.237
- 0.397 0.515 0.736 0.943 1.476 2.229
Average - 0.409 0.518 0.744 0.934 1.466 2.233

Yogurt (solution B)
Tube 1 Tube 2 Tube 3
Solution B
(mL) 0.5 1.0 1.5
Distilled water
(mL) 1.5 1.0 0.5
Absorbance
0.262 0.442 0.771
Glucose concentration (µg/ml) 23.89 43.89 59.11
Reducing sugar content g/ml 0.0478 0.0439 0.0536

Dadih (solution B)
Solution B
(mL) 0.5 1.0 1.5
Distilled water
(mL) 1.5 1.0 0.5
Absorbance
0.667 1.254 1.75
Glucose concentration (µg/ml)

68.889
134.111
189.22
Reducing sugar content g/ml 0.138 0.134 0.126

5. DISCUSSION

In this experiment, we have done the two analysis to determine the major component of both yogurt and dadih sample. Those were moisture, ash, protein, fat, fiber and also the reducing sugar content for both samples. Those two analysis are proximate analysis and Somogyi-nelson. These two AOAC standard analysis are carried out, and the result is later been compare each other to see the differences. This could be referring as the compared nutritional value in both yogurt and dadih. To determine the moisture content of yogurt and dadih, we used Oven Drying Method. The moisture content is determined by measuring the mass of food before and after the water is removed by evaporation. Furthermore, in determination of ash, we use Dry Ashing Method. In this method, we used muffle furnace. Water and other volatile materials are vaporized and organic substances are burned in the presence of oxygen and carbon dioxide. Most mineral are converted to oxides, sulfates, phosphates and silicates. Because of our samples are in liquid and semi-solid condition, it need to be dried in the oven for one day this is because we need our sample in dried condition in order to follow the proximate analysis method which require sample in dried except for moisture and ash. After dried, samples can now be used. To determine protein, we use Kjeldahl Method. In this method, protein and other components in a sample are digested with H2SO4 in the presence of catalyst. The total organic N is converted to (NH4)2SO4. The digest is neutralized with alkali and distilled into a boric acid. The borate anions formed are titrated with standardized acid, which is converted to nitrogen in the sample. In addition, to determine fat, we use Semi continuous solvent extraction by using Soxhlet apparatus. Solvent Extraction Method is based on the solubility of the lipid in the organic solvent used. In this method, the sample was placed in the Soxhlet apparatus and extract with petroleum ether. Residue substance in flask is fat. And for the fiber, we use Gravimetric Method, using Crude Fiber. In this method, crude fiber was determined by extraction with 0.25N H2SO4 and 0.313N NaOH. Insoluble residue is collected and ash content is subtracted. The last one is to determine the reducing sugar content, in this method, samples are not dried and it is in origin form. Cu2+(cupric ion) is undergoing reduction process to Cu+(cuprous ion) by reducing sugars. Cu+ ions reduce an arseomolybdate complex that is measured spectrophotometrically. The reaction must be used with a standard curve of the sugar being determined or glucose.This experiment is to compare the differences nutritional value between yogurt and dadih, to begin with moisture, dadih contain high amount of sugar which is 28.03% compare to yogurt 12.44%. This can be referring to dadih ingredients that contain 58 ml in order to produce one cup. Compare to yogurt that contain low amount of water even it is viscous. Same goes to ash content. Ash is an inorganic residue from the incineration or oxidation of organic matter in foodstuff which mean of total mineral of a food. Yogurt is also containing high ash content. From the tabulated result, yogurt consists of 1.51% compare to dadih which only contain 0.363%. This might be because, dadih is a simple combination of ingredient such as sugar, water, milk, salt and flavors. In yogurt however, contain a more complex formulation in order to give this fermented milk product healthy to be consume. Thus, referring it package, it contain calcium which is a mineral that is unanalyzed in this study. After that, samples are dried in muffle furnace, there is no black particles appeared which means that both samples are burned completely. Yogurt is known for it high protein content for as long as ever since it was consume by people around the globe. Compare to dadih that is usually known as an Asian fermented milk product and normally made based from home kitchen. This might be the answer for the high content of protein in yogurt 20.28% compare to dadih which only consist of 1.297%. As well as protein yogurt also contain high amount of fat 1.455% compare to 0.633% of dadih. Even our yogurt sample is a low fat product, but compare to dadih, it’s still higher. Fiber is a polysaccharide and lignin that are not digested by human enzymes. Also includes resistant starch, which is a plant polysaccharide. For both sample, cellulose is the component of fiber that determined. To be equal, both of our samples contain of strawberry flavors. However, yogurt contain raw strawberry that been sliced to small pieces compare to dadih which only added strawberry flavors (without raw strawberry fruit). This explained, why yogurt contain high amount of fiber 21.17% compare to 3.602% of dadih. Carbohydrate by difference is calculated in by subtracting from 100, the average quantity expressed as a percentage of water, protein, fat, dietary fiber, ash and if quantified, any other unavailable carbohydrate and the substances. From this experiment, the result shows that dadih contain high amount of carbohydrate 72.93% compare to yogurt which is 44.29%. In determination of reducing sugar using Somogyi-Nelson method, dadih contain high amount of reducing sugar compare to yogurt for all three solution 1, 2 and 3. This lead the dadih to have high amount of energy which is 1285.26kJ compare to yogurt 1116.33 kJ. Carbohydrate is a group of nutrient important in the diet as a source of energy. Carbohydrates contain of three major groups, monosaccharide, disaccharides and polysaccharides. Dadih contain 72.93% of carbohydrates, this lead to it to have high amount of energy compare to yogurt. Even thou yogurt also contains high fiber content that is because raw strawberry slice fruit id there inside the yogurt. Fruit as we know contain fiber which is also polysaccharides. In standard dadih producing formulation, about one spoon of sugar is required. Now this one spoon of sugar is high enough for dadih compare to yogurt. During the experiment of reducing sugar, both samples are not in aqueous solution, Somogyi-Nelson method require sample that is in aqueous such as carbonated drink. This is contrasting with dadih and yogurt which are in semi-solid and liquid condition. In standard method, 5ml of aqueous sample is used, so we weight about 5gram of both samples and then mix with 10ml water. We blend it until it is homogenous and then transferred to volumetric flask and later we dilute it till 250ml. When we measured our samples in spectrometric, we are having problem with the result as it give a negative value. But after keep on repeating the experiment, lastly we successfully obtained the result that is relevant enough to show that dadih contain high amount of sugar compare to yogurt. Both product are likely to be consume by people, the difference is that dadih is a non commercial product and usually only people around Asia such as Indonesia and Malaysia as well as other Asian country really consume it. In contrast with yogurt, that is well known around the globe is produced commercially. Dadih is a home made product, where by the method of making this fermented milk is under no specific control condition. No improvement in ingredients. Yogurt is industrial dairy producing product, so every significant aspect is in control condition. This also comes up with many improvement such as adding calcium and vitamin. The samples that we used contain high amount of protein, this is due to the amount of lactobacillus bacteria that is used. Even thou both samples contain the probiotic bacteria that is benefited for health; past research had shown that yogurt can prevent antibiotic-associated diarrhea. One study suggests that eating yoghurt containing L. acidophilus helps prevent vulvovaginal candidiasis, though the evidence is not conclusive .Dadih however, only using these bacteria without control condition such as the temperature, the amount of bacteria and the species of bacteria. Industrial product however, is controlling every aspect of it. This might explained why yogurt is high in protein. So based on the compared nutritional value between yogurt and dadih, we assumed that yogurt is a healthy food to be consumed. Dadih however is still a good choice but due to its high sugar content, it is not recommended to be consuming often. But still, it does provide more energy and it is tastier than yogurt, a good reason for it as to be a potential marketed fermented milk product.

6. APPENDICES

6.1. Determination of moisture content:

Moisture content is calculated based on percentage of wet weight

% of wet – weight = a – b x 100
a
Where; a = weight of sample used (g)
b = dry weight of sample (g)

Moisture content in yogurt

1 2
3
Weight of yogurt, a (g) 4.0561 4.0913
4.0441
Weight of dry dried yogurt, b (g) 3.6112 3.5023
3.5610

1) % of wet – weight = (4.0561 – 3.6112) x 100 = 10.97%
4.0561

2) % of wet – weight = (4.0913 – 3,5023) x 100 = 14.39%
4.0913

3) % of wet – weight = (4.0441 – 3,5610) x 100 = 11.95%
4.0441

Mean,x = 10.97+ 14.39+11.95 = 12.44% 3

1) (xi – x)2 = (10.97-12.44)2 = 2.1609

2) (xi – x)2 = (14.35 – 12.44)2 = 3.6481

3) (xi – x)2 = (11.95 – 12.44)2 = 0.2401

Total, ∑ (xi – x)2 = 2.1609+3.6481+0.241 = 6.05

Standard deviation (σ),

= 6.05 = 1.739
2

Coefficient of variation( CV) : 1.739 x 100 = 13.98
12.44

Moisture content of dadih

1 2 3
Weight of yogurt, a (g) 4.0113 4.0224
4.0135
Weight of dry dried yogurt, b (g) 2.9194 2.8297
2.9212

1) % of wet – weight = (4.0113 – 2.9194) x 100 = 27.22%
4.0113

2) % of wet – weight = (4.0224 – 2,8297) x 100 = 29.65%
4.0224

3) % of wet – weight = (4.0135 – 2,9212) x 100 = 27.22%
4.0135

Mean,x = 27.22+ 29.65+27.22 = 28.03% 3

1) (xi – x)2 = (27.22-28.03)2 = 0.6561

2) (xi – x)2 = (29.65 – 28.03)2 = 2.6244

3) (xi – x)2 = (27.22 – 28.03)2 = 0.6561

Total, ∑ (xi – x)2 = 0.6561+2.6244+0.6561 = 3.9366

Standard deviation (σ),

= 3.9366 = 1.4029
2

Coefficient of variation( CV) : 1.4029 x 100 = 5.005
28.03

6.2. Determination of ash content:

% ash = (weight of ash + weight of crucible) – (weight of crucible) x 100
Weight of sample

Ash content of yogurt

1 2 3
Weight of sample (g) 4.0618

4.0119 4.0327
Weight of crucible (g) 48.932 48.832 48.831
Weight of crucible + dried sample (g) 48.994 48.891 48.883

1) % ash = (48.994-48.932) x 100 = 1.53%
4.0618

2) % ash = (48.891-48.832) x 100 = 1.47%
4.0119

3) % ash = (48.893-48.831) x 100 = 1.54%
4.0327

Mean = 1.53 + 1.47+1.54 = 1.51%
3

1) (xi – x)2 = (1.53– 1.51)2 = 0.0004

2) (xi – x)2 = (1.47 – 1.51)2 = 0.0016

2) (xi – x)2 = (1.54 – 1.51)2 = 0.0009

Total, ∑ (xi – x)2 = 0.03 +0.0009 = 0.0029

Standard deviation (σ),

= 0.0029 = 0.038
2

Coefficient of variation( CV) : 0.038 x 100 = 2.521
1.51

Ash content of Dadih

1 2 3
Weight of sample (g) 4.0971

4.0351 4.0551
Weight of crucible (g) 55.0685 56.0112 55.0449
Weight of crucible + dried sample (g) 55.0797 56.0304 55.0587

1) % ash = (55.0797-55.0685) x 100 = 0.273%
4.0971

2) % ash = (56.0304-56.0112) x 100 = 0.476%
4.0351

3) % ash = (55.0587-55.0449) x 100 = 0.340%
4.0551

Mean = 0.273+0.476+0.34 = 0.363%
3

1) (xi – x)2 = (0.273– 0.363)2 = 0.0081

2) (xi – x)2 = (0.476 – 0.363)2 = 0.0127

3) (xi – x)2 = (0.340 – 0.363)2 = 0.00053

Total, ∑ (xi – x)2 = 0.0081 + 0.0127 + 0.00053= 0.0213

Standard deviation (σ),

= 0.0213 = 0.1033
2

Coefficient of variation( CV) : 0.1033x 100 = 28.44
0.363

6.3. Determination of Crude Protein:

Crude protein in yogurt

Sample Weight of sample (g) Volume of H2SO4 (mL)
1
0.1487 7.0
2 0.1490 6.8
3 0.1488 6.8
Blank 0.0 0.1

% of Nitrogen = (Is – Ib) x N x 1.4
W

Where,

Is = Volume of H2SO4 used to titrate boric acid

Ib = Volume of H2SO4 used to titrate blank

W = Weight of sample (g)

N = Normality of H2SO4 (0.05N)

1) % of Nitrogen = (7.0 – 0.1) x 0.05 x1.4
0.1487

= 3.248%

2) % of Nitrogen = (6.8 – 0.1) x 0.05 x1.4
0.1490

= 3.147%

3) % of Nitrogen = (6.8 – 0.1) x 0.05 x1.4
0.1488

= 3.152%

Therefore,
% Protein = % nitrogen x conversion factor (CF)

With CF of milk based product = 6.38

1) % Protein = 3.24 x 6.38
= 20.67%
2) % Protein = 3.147 x 6.38
= 20.07%
2) % Protein = 3.152 x 6.38
= 20.10%

Mean,x = (20.67+20.07+20.10) = 20.28
3

1) (xi – x)2 = (20.67-20.28)2 = 0.152
2) (xi – x)2 = (20.07-20.28)2 = 0.0441
3) (xi – x)2 = (20.10-20.28)2 = 0.0324

Total, ∑ (xi – x)2 = 1.831 + 0.441 +0.0324 = 0.2286

Standard deviation (σ),

= 0.2286 = 0.2887
2

Coefficient of variation( CV) : 0.2887 x 100 = 1.424
20.28

Crude protein in Dadih

Sample Weight of sample (g) Volume of H2SO4 (mL)
1
0.1501 0.5
2 0.1491 0.6
3 0.1483 0.5
Blank 0.0 0.1

% of Nitrogen = (Is – Ib) x N x 1.4
W

Where,

Is = Volume of H2SO4 used to titrate boric acid

Ib = Volume of H2SO4 used to titrate blank

W = Weight of sample (g)

N = Normality of H2SO4 (0.05N)

1) % of Nitrogen = (0.5– 0.1) x 0.05 x1.4
0.1501

= 0.1865%

2) % of Nitrogen = (0.6 – 0.1) x 0.05 x1.4
0.1491

= 0.2347%

3) % of Nitrogen = (0.5 – 0.1) x 0.05 x1.4
0.1483

= 0.1888%

Therefore,
% Protein = % nitrogen x conversion factor (CF)

With CF of milk based product = 6.38

1) % Protein = 0.1865 x 6.38
= 1.1898%

2) % Protein = 0.2347 x 6.38
= 1.4974%

3) % Protein = 0.1888 x 6.38
= 1.2045%

Mean,x = (1.1898+1.4974+1.2045) = 1.297
3

1) (xi – x)2 = (1.1899 – 1.297)2 = 0.012

2) (xi – x)2 = (1.497 – 1.297)2 = 0.040

2) (xi – x)2 = (1.2045 – 1.297)2 = 0.009

Total, ∑ (xi – x)2 = 0.012 + 0.040+0.009 = 0.1312

Standard deviation (σ),

= 0.1312 = 0.256
2

Coefficient of variation( CV) : 0.256 x 100 = 19.75
1.297

6.4. Determination of Crude Fat:

Crude fat in Yogurt

sample Weight of sample (g) Weight of oil (g)
1 5.0317 0.0839
2 5.0228 0.0624

% of oil in sample = M x 100
W

Where,
M = weight of oil (g)
W = weight of sample(g)

1) % of oil in sample = 0.0839 x 100
5.0317
= 1.667%

2) % of oil in sample = 0.0624 x 100
5.0228
= 1.242%

Mean, = 1.667+1.242 = 1.455%
2

Crude fat in Dadih

sample Weight of sample (g) Weight of oil (g)
1 5.0161 0.0482
2 5.0213 0.0153

% of oil in sample = M x 100
W

Where,
M = weight of oil (g)
W = weight of sample(g)

1) % of oil in sample = 0.0482 x 100
5.0161
= 0.961%

2) % of oil in sample = 0.0153 x 100
5.0213

= 0.305%

Mean, = 0.961+0.305 = 0.633%
2

6.5. Determination of Crude Fiber:

Crude fiber in Yogurt

Weight of sample (g) Weight of filter paper (g) Weight of crucible + filter paper + dried residue (g) Weight of crucible + ash (g) Percentage crude fibre (%)
2.0026 0.2950 48.090 47.394 20.04
2.0092 0.3076 56.225 55.433 24.14
2.0183 0.2991 51.956 51.268 19.32
Average percentage = 21.17

% crude fibre = (S – K) – A x 100
W

Where,
S = Weight of crucible + filter paper + dried residue (g)
K = Weight of filter paper (g)
A = Weight of crucible + ash (g)
W = Weight of sample (g)

1) % crude fibre = (48.090– 0.2950) – 47.394 x 100
2.0026

= 20.04%

2) % crude fibre = (56.225– 0.3076) – 55.433 x 100
2.0092

= 24.14%

3. % crude fibre = (51.956– 0.2991) – 51.267 x 100
2.0183

= 19.32%

Average percentage = 20.04+24.14+19.32
3
= 21.17

Crude fiber in Dadih

Weight of sample (g) Weight of filter paper (g) Weight of crucible + filter paper + dried residue (g) Weight of crucible + ash (g) Percentage crude fibre (%)
2.0508 0.2932 48.386 47.892 3.07
2.0672 0.2887 50.411 50.089 1.611
2.0574 0.2901 42.421 42.079 2.522
Average percentage = 3.602

% crude fibre = (S – K) – A x 100
W

Where,
S = Weight of crucible + filter paper + dried residue (g)
K = Weight of filter paper (g)
A = Weight of crucible + ash (g)
W = Weight of sample (g)

1) % crude fibre = (48.386– 0.2932) – 48.155 x 100
2.0508

= 3.07%

2) % crude fibre = (50.411– 0.2887) – 50.089 x 100
2.0672

= 1.611%

3. % crude fibre = (42.421– 0.2901) – 42.079 x 100
2.0574

= 2.522%
Average percentage = 3.07+1.611+2.522
3
= 3.602%

6.6. Determination of Carbohydrates (By Difference) and Energy:

Yogurt:

% of carbohydrate = 100 – (%moisture +% ash + % protein + % fat + %fibre)

= 100 – (12.44+ 1.51 + 20.28 + 1.455 + 21.17)

= 43.15%

Energy Calculation:

Protein(g) x 4kcal/g = 20.28(4kcal)
= 81.12kcal
CHO (g) x 4kcal/g = 43.15(4kcal)
= 172.6kcal
Fat (g) x 9kcal/g = 1.455(9kcal)
= 13.09kcal

Total, 81.12kcal + 172.58kcal + 13.09kcal = 266.81kcal

Then convert to kJ = 266.81kcal x 4.184 kJ

= 1116.33 kJ

(1 kcal = 4.184kJ)

Dadih:

% of carbohydrate = 100 – (%moisture +% ash + % protein + % fat + %fibre)

= 100 – (20.03+ 0.363 + 1.297 + 0.633 + 3.602 )

= 74.08%

Energy Calculation:

Protein(g) x 4kcal/g = 1.297(4kcal)
= 5.188kcal
CHO (g) x 4kcal/g = 74.08(4kcal)
= 296.3kcal
Fat (g) x 9kcal/g = 0.633(9kcal)
= 5.697kcal

Total, 5.188+296.3+5.697 = 307.19kcal

Then convert to kJ = 307.19kcal x 4.184 kJ

= 1285.26kJ

(1 kcal = 4.184kJ)

6.7. Determination of reducing sugar:

Sample Yogurt (solution B)

Tube 1 Tube 2 Tube 3
Solution B
(mL) 0.5 1.0 1.5
Distilled water
(mL) 1.5 1.0 0.5
Absorbance
0.262 0.442 0.771

Sample Dadih (solution B)

Solution B
(mL) 0.5 1.0 1.5
Distilled water
(mL) 1.5 1.0 0.5
Absorbance
0.667 1.254 1.75

Yogurt:

Dilution factor:

Solution 1 = 0.5ml of solution B + 1.5 ml of DW

- Total DF = 250 x 100 x 2
5 10 0.5
= 2000 (solution B was diluted 2000x)

Reducing sugar content of yogurt in g/ml:

Sample : 0.5 ml absorbance : 0.262

From the equation that we obtain from graph:
y = 0.009x + 0.047
0.262 = 0.009x + 0.047
x = 23.89

RS content = 2000 x X value (µg/ml) x 1g/106 µg
= 2000 x 23.89 value (µg/ml) x 1g/106 µg
= 0.0478g/ml

Dilution factor:

Solution 2 = 1 ml of solution B + 1 ml of DW

- Total DF = 250 x 100 x 2
5 10 1
= 1000 ( solution B was diluted 1000x)

Reducing sugar content of sample in g/ml:

Sample : 1.0 ml absorbance : 0.442

From the equation that we obtain from graph:
y = 0.009x + 0.047
0.442 = 0.009x + 0.047
x = 43.89

RS content = 1000 x X value (µg/ml) x 1g/106 µg
= 1000 x 43.89 (µg/ml) x 1g/106 µg
= 0.0439g/ml

Dilution factor:

Solution 3 = 1.5 ml of solution B + 0.5 ml of DW

- Total DF = 250 x 100 x 2
5 10 1.5
= 666.667 ( solution B was diluted 666.667x)

Reducing sugar content of sample in g/ml:

Sample : 1.5 ml absorbance : 0.771

From the equation that we obtain from graph:
y = 0.009x + 0.047
0.771 = 0.009x + 0.047
= 59.11

RS content = 666.667 x X value (µg/ml) x 1g/106 µg
= 666.667 x 59.11 (µg/ml) x 1g/106 µg
= 0.0536g/ml

Dadih:

Dilution factor:

Solution 1 = 0.5ml of solution B + 1.5 ml of DW

- Total DF = 250 x 100 x 2
5 10 0.5
= 2000 (solution B was diluted 2000x)

Reducing sugar content of yogurt in g/ml:

Sample : 0.5 ml absorbance : 0.667

From the equation that we obtain from graph:
y = 0.009x + 0.047
0.667= 0.009x + 0.047
x = 68.889

RS content = 2000 x X value (µg/ml) x 1g/106 µg
= 2000 x 68.889value (µg/ml) x 1g/106 µg
= 0.138g/ml

Dilution factor:

Solution 2 = 1 ml of solution B + 1 ml of DW

- Total DF = 250 x 100 x 2
5 10 1
= 1000 ( solution B was diluted 1000x)

Reducing sugar content of sample in g/ml:

Sample : 1.0 ml absorbance : 1.254

From the equation that we obtain from graph:
y = 0.009x + 0.047
1.254 = 0.009x + 0.047
x = 134.111

RS content = 1000 x X value (µg/ml) x 1g/106 µg
= 1000 x 134.111 (µg/ml) x 1g/106 µg
= 0.134g/ml

Dilution factor:

Solution 3 = 1.5 ml of solution B + 0.5 ml of DW

- Total DF = 250 x 100 x 2
5 10 1.5
= 666.667 ( solution B was diluted 666.667x)

Reducing sugar content of sample in g/ml:

Sample : 1.5 ml absorbance : 1.75

From the equation that we obtain from graph:
y = 0.009x + 0.047
1.75= 0.009x + 0.047
= 189.22

RS content = 666.667 x X value (µg/ml) x 1g/106 µg
= 666.667 x 189.22 (µg/ml) x 1g/106 µg
= 0.126g/ml

7. CONCLUSION

Yogurt contain higher amount of ash, protein, fat and crude fiber compare to dadih while dadih contain high amount of moisture, carbohydrate and amount of energy than yogurt.
Nutritional value Yogurt Dadih
Moisture 12.44% 28.03%
Ash 1.51% 0.363%
Protein 20.28% 1.297%
Fat 1.455% 0.633%
Crude fibre 21.17% 3.602%
Carbohydrate 44.29% 72.93%
Energy 1116.33 kJ 1285.26kJ
Dadih content high amount of reducing sugar compare to yogurt.
solution Yogurt Dadih
Solution 1 0.0478g/ml 0.138g/ml
Solution 2 0.0439g/ml 0.134g/ml
Solution 3 0.0536g/ml 0.126g/ml

8. ACKNOWLEDGEMENT

Deeply thankful to our lecture Dr. Faridah Abas for guiding us until the end of this study.and also thank you to miss siti hafizah Ibrahim, miss norhafiza abdul razak, miss rosmawati Othman, mr azman abu yamin and mr zolkifli johari for helping with the sample procedures and organizing/monitoring us during this experiment was conduct until it finish.

9. REFERENCES

1. S. Suzanne Nielsen, Food analysis 2nd edition, KLUWERACADEMIC/PLENUM PUBLISHERS, page 355-356.
2. Dennis D. Miller ,Food Chemistry a Laboratory Manual,,Department of Food Science
Cornell University. A Wiley-Interscience Publication, 2000, page 54-56.
3. Ceirwyn S.James, Analytical chemistry of foods, An Aspen publication,Aspen publishers,inc. Gaithersburg, Maryland 1999, page 76-79
4. Surono and A. Hosono, FERMENTED MILKS | Types and Standards of Identity, journal of food science,I. S. Center for the Assessment and Application of Biotechnology, Tangerang, Indonesia, Shinshu University, Nagano-Minamiminowa, Japan, elsevier 2002.
5. Takano and N. Yamamoto, FERMENTED MILKS | Health Effects of Fermented Milks, journal of food science, T. Calpis Co. Ltd, Kanagawa, Japan,Elsevier 2002.