LAB REPORT 1-4

DEPARTMENT OF BIOLOGY

FACULTY OF SCIENCE & MATHEMATICS

SULTAN IDRIS EDUCATION UNIVERSITY

FOOD SCIENCE

SBK 3033

LABORATORY REPORT

NAME	MATRIC NUMBER
SHAFIQAH BINTI HANAFI	D20141067057
NOR ASHLIA BINTI SAIDI	D20141067059
NAZIYAHTUL BINTI ANIFAH	D20141067075

LECTURER’S NAME : PROFESSOR DR HASIMAH ALIMON

GROUP                         : GROUP A

LABORATORY 1 : STARCH ANALYSIS

EXPERIMENT 1

RESULT

Substance	Colour change after adding iodine
10% glucose solution	Yellow colour and bubble form
1% starch solution	Dark blue colour
1% albumen solution	Light yellow colour
water	Yellow colour

           

                             

DISCUSSION

1) The substance selected for testing are examples of three of the principle chemical substances in cells, sugar (glucose), starch, protein (albumen). With which of these substances did iodine react to give a colour change?

The substance that react with the iodine solution is 1% starch solution because the colour are changes from colourless to dark blue solution. This is because the starch iodide complex are formed as charge that can transfer between the starch and iodide ion that caused changes the spacing between the energy levels. Then, the starch iodide complex absorb light at different wavelength caused the formation of dark blue colour.

In starch, there is amylose that function to form a deep blue colour when iodine is adding. The molecule in iodine slips inside of the amylose coil. Iodine is not very soluble in water, so the iodine solution is making by dissolving it in water with the presence of potassium iodide. Thus, a linear triiodide ion complex with is soluble that slips into the coil of the starch cause blue-black colour become intense.

2) Does your result indicate that there is, for example, no sugar and no protein that will give a colour change with iodine?

Yes, the tests solution that have no sugar and protein will give reaction and changes colour with iodine solution. For example, the colour of 1% starch solution that have no sugar and protein changes to dark blue due to the presence of starch in the solution.

3) What experiments would you have to carry out in order to give a confident answer to question 2?

There are several experiment that can conduct for the presence of starch, protein and glucose. The iodine test are used for the presence of starch. The solution colour that have been tested will change to dark blue colour when iodine solution are added due to presence of starch in the solution. Besides, the Biuret test are used for the presence of protein and amino acid. The basis for the formation of protein is peptide bonds.The solution colour that have been tested with the Biuret test will change to purple colour due to the presence of protein in the solution. Then, the Benedict’s test are used for the presence of reducing sugar. The solution colour that have been tested with Benedict’s solution will change to orange or yellow due to presence of reducing sugar in the solution. It depends on the amount of sugar inside the solution when glucose is presence.

4) Does the result indicate that starch will always react with iodine solution to give a colour change?

Yes but the iodine test cannot be performed on the very dark solids or liquids which does not permit the observation of colour change in the solution. This is because the iodine test will change the colour of solution to dark blue due to presence of the starch.

5) What was the point of the water in tube 4?

The water is act as constant variable. This is because, water has a neutral pH and tasteless. We can make a comparison with the others test tube that contains starch, sugar and protein. The water will change the colour only when the iodine is drop into it without any changes of composition.

CONCLUSION

Only 1% starch solution give a positive result in this experiment when added 3 drops of iodine solution while others not.

EXPERIMENT 2

RESULT

Food sample	Colour change with iodine	Interpretation of result
1    Potato	Dark blue colour	Starch present
2    Onion	Yellow precipitate form	Starch absent
3    Bread	Dark blue colour	Starch present
4    Banana	Yellowish brown precipitate form	Starch absent
5    Apple	Light green colour	Starch absent
6    Dried milk	No Colour change with blue precipitate form	Starch absent

DISCUSSION

            Starch are carbohydrate that exists in two types of molecules which are amylose and amylopectin. Based on the theory, the reaction of iodine with starch will form intensely coloured starch-iodine complex form or dark blue colour (Taryn Biggs, n.d.). The amylose molecules that consist of single and unbranched chains of glucose molecules caused the iodine solution stuck in the coils of beta amylose molecules. This cause the transfer of charge between the starch and iodine and form the blue black colour to the mixture.

Based on the experiment, we used 6 different of food sample in order to identify the present of the starch on it. We use the iodine solution in order to test the present of the starch. The solution of the food sample with the 5 drops of the iodine solution determined the present of the starch. The solution will remain theoretically change to blue if there is the present of the starch.

The first sample is the sample of potato. As what we know potatoes are vegetables but they contain a lot of starch that make them more like rice, pasta and bread in terms of nutrition. In this experiment, the theory was proven to be true. It is because the colour change when the iodine was dropped in the potato sample. The solution turns to blue that can be seen in the picture from the previous pages. The solution colour changes showed that the potato sample contained starch.

Next, the second food sample that had been tested was onion. Based on what had been observed by us, the solution of onion sample formed a yellow precipitate in the test tube after tested with the iodine. It showed that the starch do not present in the onion. It is true that onion contain carbohydrates but it do not have starch in it. This is because the onion provided a different amount of carbohydrates compared to potato sample.

The third food sample solution was bread. Same as before, the solution was tested by 5 drops of iodine solution in the test tube. After dropped the iodine in the test tube, we could observe that the solution turns to dark blue colour. This changes showed that the bread contain starch in it. Bread is a carbohydrates food and it also known as starchy food that the food contain many starch that was the reason behind the positive result of the test in this experiment when being tested by the bread sample.

            Based on the experiment, the result obtain shown that there are absence of starch on banana, apple and dried milk. The colour presence on the banana solution are yellowish brown precipitate, light green colour on apple solution and no changes with blue precipitate on dried milk solution.

The colour presence on the banana are yellowish brown due to the presence of starch are decreases on the ripen fruits. This is because, the starch in the banana have transformed to sugars when the banana are ripen. Besides, there are presence of simple sugars such as fructose, glucose and lactose in the ripen banana caused the blue black colour does not present when added iodine solution.

            Besides, the colour presence on the apple are light green because the starch on the ripen apple have been changes to the sugar by natural enzymes in the fruit and caused undetectable when the iodide solution are added on the ripen apple. Besides, there are presence of simple sugars such as fructose, glucose and lactose in the ripen apple caused the blue black colour does not present when added iodine solution.

            Lastly, the solutioncolour of dried milk have no changes but have little blue precipitate at the bottom of test tube. This is because the dried milk contain the natural carbohydrates which are lactose (Natalie stain, 2015). In addition, the dried milk are higher with protein caused there are no changes when the dried milk are tested with the iodine solution.

            There are some precaution during conduct the experiment which are carefully while handling the iodine solution because the solution can stain clothing, equipment and skin. Besides, the iodine solution must far away from mouth and not try on any tested foods, as iodine can be poisonous. In addition, the test tube, mortar and pestle must be clean for food test to ensure the result are accurate due to the sensitivity of the starch test.

CONCLUSION

Based on the experiment, the food test that contain starch are potato and bread, whereas the onion, banana, apple and dried milk does not contain the starch. This result are obtain from the colour changes of the food test solution when iodine solution are added. The food test colour that contain starch will change from clear solution to dark blue solution when iodine solution are added.

LABORATORY 2 : PROTEIN ANALYSIS

EXPERIMENT 1 : THE TEST FOR PROTEIN

RESULT

	Substance	Reaction with copper sulphate and sodium hydroxide
1	1% starch solution into tube 1	Colour turn to pale blue
2	10% glucose solution into tube 2	Colour turn to pale blue
3	1% albumen solution into tube 3	Colour turn to purple
4	Water into tube 4	Colour turn to pale turquoise

Table 1 : Reaction with copper sulphate and sodium hydroxide

DISCUSSION

1. The substances selected for testing are examples of three of the principles types of chemical substances in cells: starch, sugar (glucose), protein (albumen). With which of these samples did the reaction give a purple colour?

Albumen solution gives a positive result while others are not. A protein composed of amino acid. The sodium hydroxide has raised the pH of the solution to alkaline level, the important component is the copper (II) ions from the CuSO_4.  If there are present of peptide bond in this alkaline solution, the copper (II) ions will form a coordination complex with four nitrogen atoms involved in peptide bonds.  Copper sulphate solution is blue colour, but when the copper (II) ions are coordinated with the nitrogen atoms of these peptide bonds, the colour of the solution changes from blue to purple. This colour change is depending on the number of peptide bonds in the solution. Thus, the more protein contains in the solution, the more intense the change. If the peptides are very short, the solution will turn pink, rather than purple.

2. The substance that gave a purple colour was a single example of its class of substances. Would you expect all other samples of this class to give the same reaction?

Other sample gives the same reaction because there are not contain protein. In addition, the starch contains many carbohydrates whereas the glucose is a sugar that include in monosaccharide group. Both starch and glucose does not have peptide bonds where can found in the protein. Therefore, both starch and glucose does not react with the copper sulphate and sodium hydroxide.

3. What was the point of using test tube 4 with the water?

The water is act as constant variable. This is because, water has a neutral pH and tasteless. We can make a comparison with the others test tube that contains starch, glucose and albumen. The water will change the colour to blue when the CuSO₄ is drop into it without any changes of composition.

EXPERIMENT 2: PROTEIN CHARACTERISTIC

RESULT

·         Part 1 : Precipitation of casein from milk with an acid (vinegar)

Substance	Weight (g)
1.    Empty beaker	067.32
2.    Beaker + milk	182.48
3.    Milk	114.65
4.    Curds	028.24

·         Part 2 : Enzymatic coagulation of the casein from milk with rennet

substance	Weight (g)
1.    Empty beaker	71.04
2.    Beaker + milk	185.69
3.    Milk	114.65
4.    Curds	35.89

·         Part 3 : Coagulation of protein from soymilk using a salt (magnesium sulfate)

Substance	Weight (g)
1.    Empty beaker	100.62
2.    Beaker + soymilk	225.95
3.    Soymilk	125.33
4.    Curds	18.93

Result (the comparison of the three milk )

Substance	Weight of milk/soymilk (g)	Weight of curd (g)	Describe the curd (color, texture)
Milk + acid	115.16	28.24	White, fine granules
Milk + rennet	114.65	35.89	White, fluffy, spongy, thick
Soymilk + Epsom salt	125.33	18.93	Light brown, fine granules

  

DISCUSSION

1.  1.  Compare the weights of the curds from the milk (acid and rennet) with that from the soymilk.

The milk + rennet curd weighed the most with 35.89 compared to the milk + acid curd and also the soymilk + epsom salt curd. The soymilk + epsom curd weighed the less with only 18.93 g compared to the other two. The milk + acid curd weighed with 28.24 g.

2.    2.  Why did the casein that was coagulated with the rennet weigh more than the casein that was precipitated with the acid?

The casein that was coagulated with the rennet weigh more than the casein that was precipitated with the acid because the rennet have the enzyme called rennase or rennin that reacts with milk to produces the coagulation of milk contain milk protein and fat meanwhile the acid coagulation only contains casein.

3.    3. Compare the amount of acid casein precipitated from the whole milk with the amount of soy protein coagulated from the soymilk. How do your results compare with the Nutrition Facts label for each product?

The whole milk give out the results of more casein precipitate rather than the soymilk. The nutrition facts label on the box of the whole milk (3.84 g) of protein per 120 mm , while the soymilk only (2.76 g) of protein per 120 mm. The result that gained from the experiment was accepted that the whole milk give out more curds than the soymilk. 

4.    4. How did the biuret test indicate the presence of proteins?

The biuret test can be used to indicate the presence of the proteins because the peptide bonds occur with the same frequency per amino in the peptide. The test is done to show the peptide bonds, which are the basis for the formation of the proteins. These bond will change the blue colour of biuret regent to purple colour.

CONCLUSION

As a conclusion, based on the experiment 1, only the albumin solution gives a positive result while others are not because the albumin contain peptide bond that react with the copper sulphate and sodium hydroxide caused the colour changes to purple colour. In addition, based on the experiment 2, the protein that contain on the soymilk and also whole milk are different because of the amount of casein in both of the milk. Besides that it can be conclude that the amount of protein will also affected by the enzymes that reacts with the milk.

LABORATORY 3 : LIPID ANALYSIS

EXPERIMENT 1 : THE TEST FOR LIPID

RESULT

		Result when added to water
1	Oil dissolved in alcohol	3	The solution turn oily and cloudy
2	Alcohol alone	4	No changes in the solution

DISCUSSION

1.    What was the only difference between the contents of tubes 1 and 2?

The contents that found in test tube 1 and 2 are the presence of one drop of vegetable oil. The one drop of the vegetable oil are added in the test tube 1 that contain alcohol whereas there are only alcohol solution in the test tube 2.

2.   What was the visible difference between tubes 3 and 4 after adding the contents of tubes1 and tube 2 to the water in them?

The difference that can be observed from the test tubes 3 are the solution turn oily and cloudy whereas the solution in the test tube 4 are no changes with colourless liquid due to absence of lipid substance in the solution.

3.    How would you attempt to explain the appearance of the liquid in tube 3?

The appearance of the solution in the test tube 3 are oily and cloudy in colour. This is because the presence of vegetables oil in the test tube 1 that added into test tube 3 caused the mixture of water and alcohol are turn cloudy. The appearance of the solution due to the fats that coming out from the alcohol to form an emulsion in the water. The emulsion is the layer of cloudy white suspension forms at the top of the solution that found in the food with higher lipid content such as vegetables oil.

4.  What difficulties can you foresee in using this test with samples which contain both lipids and water?

Generally, lipids are insoluble in water. When mix the alcohol to water, it not dissolved. There is presence of bubbles and was hot too when we mixed the water. The alcohol molecules fall to the spaces between the water molecules that there is oxygen between the spaces. Therefore, the alcohol molecules push the oxygen out of the spaces and produces bubbles.

5.  Water and alcohol mix in all proportions. What precautions must be taken in selecting the alcohol for use in this experiment and why is this precaution necessary?

The most common type of alcohol used in schools and colleges are ethanol and methanol. Firstly, review the Safety Data Sheets (SDS) before using alcohol and make students aware of all hazards. Next, handle alcohols in a fume hood or captured to prevent any flammable from reaching any source of ignition.

6.  How would you design an experiment to find out the sensitivity of this test for fats?

Test for Fats (lipids) by using Sudan IV. The chemical Sudan IV is not soluble in water but, it soluble in lipids. In this test, the chemical Sudan IV need add to a solution along with ethanol to dissolve any possible lipids. The positive result shows the formation stain reddish-orange colour in the test tube. In chemistry, a dilution is a process to reduce the concentration of a substance in a solution. In this experiment, we should use a serial dilution, which means the dilution is repeated to amplify the dilution factor quickly. It’s commonly performed in experiments requiring highly dilute solutions with great accuracy.

EXPERIMENT 2 : CHARACTERISTIC OF LIPID IN FOOD

RESULT

A) Visual evidence of invisible fats from food

Table 1 – Visual observations of fat

Food	Describe what you see on the paper towel
Chocolate chips	Grease spot spreads beyond the chocolate
Potato chips	The grease spot are spreads more on the paper towel
Sunflower seeds	The grease spot are shown on the seeds that touch on the paper towel

B) Quantitative measurement of invisible fats from foods

Table 2 – Extraction of lipids

Food	Weight of beaker (g)	Weight of beaker with raw food (g)	Weight of raw food (g)	Weight of beaker with dried food (g)	Weight lost from food (g)	% Lipid extraction
Chocolate chips	32.81	37.79	4.98	36.90	0.89	17.87
Potato chips	30.52	35.48	4.96	34.19	1.29	26.00
Sunflower seeds	40.20	45.08	4.88	43.76	1.32	27.04

(weight of beaker with raw food) – (weight of beaker) = weight of raw food

(weight of beaker with raw food) – (weight of beaker with dried food) = weight lost from food

(weight lost from food / weight of raw food) x 100 = % lipid extracted

Table 3– Description of fats

Food	Color	Texture	Odor	Viscosity
Chocolate chips	Light brown	Waxy	Chocolate smells	Dry and hard
Potato chips	Light yellow	Oily	corn smells	Thick oil
Sunflower seeds	Yellow	Oily	Sunflower seeds smells	Thick oil

DISCUSSION

1.  How can you tell that the dark wet spot on the paper towel is fat and not water?

The paper towel feel greasy when we touch it and it is also can be seen transparent.

 2. Rank from most to least the percentage of lipid extracted from all three foods. Look at the Nutrition Facts label on the packages of all three foods and rank them. Did your ranking agree with the ranking of the product labels?

Sunflower seed, potato chips and chocolate chips. The arrangement of the lipid extracted from all three food is same with the nutrition facts label on the packages with the least is the chocolate chips with only 0.0055 gram per 5 gram. The potato chips with 0.7654 gram per 5 gram . The sunflower seed with the most 0.8468 gram per 5 gram. The result gained parallel with the nutrition facts behind the products.

3. Determine which lipids contained saturated and unsaturated fatty acids in this experiment, based on your descriptions of the fats in the Petri dishes.

The lipid contained in both potato chips and sunflower seed were the unsaturated fat which there are at least one double bond within the fatty acid chain. The chocolate chips contained the saturated which include animal fat products such as cream, cheese, butter, other whole milk dairy products and fatty meats which also contain dietary cholestrol. Certain vegetable products have high saturated fat content, such as coconut oil and palm kernel oil. The difference between the saturated and unsaturated fats are the saturated fats are solid at room temperature, while unsaturated fats are liquid at room temperature. This is because saturated and unsaturated fats are differ in their chemical structures. Saturated fats have no double bond between molecules, which means there are no gaps and the fat is saturated with hydrogen molecules.

CONLUSION

As a conclusion, the oil vegetable contain lipids that caused the solution changes cloudy and oily whereas the test tube that does not contain the vegetable oil remain unchanged. Then, all the food sample contained lipid but the amount was not the same with each other as the composition of the food is not the same. The fatter contained in the food, the lipid extracted from the food sample will also become higher. The type of the fat also will affect the amount of the lipid extracted. 

LABORATORY 4 : VITAMIN C ANALYSIS

EXPERIMENT 1 : DETERMINATION OF VITAMIN C CONTENT IN FRUIT JUICES

RESULT

Table 1

Tube	Liquid	Volume needed to Decolorize DCPIP (ml)
1	0.1% ascorbic acid	0.05
2	Fresh lemon juice	0.10
3	Fresh orange juice	0.05
4	Canned or bottled orange juice	0.10
5	Papaya	0.05
6	Pumpkin	1.20
7	Pineapple	0.05
8	Orange juice that has stood in an open beaker for three days	0.10

DISCUSSION

Vitamin C are water soluble vitamin that naturally present in some food and dietary supplement. Vitamin C is the body's primary water-soluble antioxidant, defending all aqueous areas of the body against free radicals that attack and damage normal cells. In addition, vitamin C is vital for the proper function of the immune system that cause the nutrient to turn to for the prevention of recurrent ear infections, colds, and flu (St. Patrick, 2017).

Ascorbic acid need 0.05 ml to decolorized the DCPIP solution because it also known as vitamin C. Ascorbic acid is a potent water soluble natural anti-oxidant. Vitamin C helps prevent cell damage caused by free radicals in our body. This vitamin is helpful in preventing scurvy.

Fresh orange juice only needs 0.05 ml to give a colorless to the DCPIP solution in a test tube. Based on the self-nutrition data website, raw orange contain 136mg per 100g serving of vitamin C. This fruits is very low in saturated fat, cholesterol and sodium. It also a good source of vitamin C, thiamin, folate and potassium. Based on the Authority nutrition website by Dr. Atil Arnarson, potassium can lower the blood pressure in people with hypertension and has beneficial effects on cardiovascular health. 

Fresh lemon juices need to added 0.10 ml in DCPIP solution because it has 129 mg per 100 g serving of vitamin C less than fresh orange juices. Based on the ‘nutrition and you’ website, lemon has zero saturated fat, low glycemic fruits that called blood sugar level and excellent source of fiber. Citric acid in lemon is a natural preservative, aids in smooth digestion and helps dissolve kidney stones.

As for the bottled orange juice, the result that we gained that the volume of vitamin c contained in it was a bit lower than the fresh orange juice as it need 0.10 ml volume of DCPIP to decolourised the solution meanwhile the fresh orange only needed 0.05 ml of DCPIP.  The result maybe because of the loss of the vitamin c due to the oxidation by a residual air layer trapped within the container at higher storage temperatures.

Then the volume of DCPIP needed by a papaya solution becomes decolourised was 0.05 ml. this showed that papaya also contained high number of vitamin c in it. According to a research, papaya providing a whopping 144% of the daily recommended value of vitamin c per serving which is great as an infection fighter as well as a free radical-scavenging.

Lastly was the pumpkin. The pumpkin showed the highest volume of DCPIP to become decolourised with 1.20 ml. the pumpkin do contained the vitamin c but low in a lower amount compared to the other. It is because the pumpkin is loaded with important antioxidant, beta-carotene that converted into vitamin A in the body.

            Pineapple are one of the fruits that content higher number of vitamin C. Based on the experiment, there only 0.05 ml volume of pineapple extract that needed to decolourize the 1 ml DCPIP solution in a test tube. This show that higher contain of vitamin C in the pure extract pineapple. In addition, the pineapple are also excellent source of manganese that act as essential cofactor in a number of enzymes important energy production and antioxidant defenses. Besides, the losses of the vitamin C are less than 5% in mango, strawberry and watermelon pieces, 10% in pineapple pieces, and 12% in kiwifruit slices based on researchers study. This show that the value of pineapple vitamin C loss after 6 day are in the middle of other fruit tested.

            Besides, the content of vitamin c in the orange juice that has stood in an open beaker for 3 days are reduced compare to the number of the fresh orange juice. This is because the vitamin c that easily oxidized cause the oxygen in the air drastically reduce the vitamin c content in the orange juice. Then, according to Nagy and Smoot, temperature and storage time affects the percent of vitamin C content of orange fruits and orange juice. Based on the experiment, the amount of the 3 days extract juice for decolourize DCPIP are higher which is 0.10 ml than the amount of fresh orange juice. This show that, there are factor that cause the vitamin c content in the extract juice reduce which are climate condition, types of citrus fruit, type of container and handling and storage.

            There are some precaution during conduct the experiment which are students need to measure the volume of the extract juice and ascorbic acid accurately. The eyes of the observer should perpendicular to the scale of the syringe to get an accurate volume needed. Besides, the extract of juices should drop by drop in the test tube that contain DCPIP solution to get accurate volume for the DCPIP solution decolourize. Then, the test tube that have been used for other extract should be rinsed because the sensitivity of vitamin c. Then, student should carefully handle and observe of the volume of extract needed for the DCPIP solution to decolourize because the coloured fruit juices will not give the colourless solution and the end point is when the blue colour disappear and original colour return. This is important to ensure that result obtain are accurate.

CONCLUSION

As a conclusion the amount of vitamin c in the food sample depended by the amount of the volume of DCPIP that needed to decolourise the solution in the test tubes. As for the citrus fruits, the orange have the most of the vitamin c meanwhile the papaya also give out the same result eventhough it was higher of vitamin c rather than the citrus per gram. The fruit that had the lowest number of vitamin c was the pumpkin.