Practical 2 : Food Test.

Objective 1 :

To test the present of starch, reducing sugar, non-reducing proteins and lipids in a food samples.

Theory :

Plants store glucose as the polysaccharide starch. In order to test for the presence of starch, a solution of Iodine is dropped onto the suspected starch containing moiety, e.g. potato. If a colour changes from deep red to purple/black occurs, this indicates the presence of starch. The reason for this change is that the iodine molecules non-covalently interact with the long starch molecules and this alters the colour obtained.

Some sugars such as glucose are capable of reducing other compounds and are called reducing sugars. When reducing sugars are mixed with Benedicts reagent and heated, a reduction reaction causes the Benedicts reagent to change colour. The colour varies from green to dark red, depending on the amount of and type of sugar.

The copper atoms of biuret solution (CuSO4 and KOH) will react with peptide bonds, producing a colour change. A deep violet or blue colour indicates the presence of proteins and a lighter violet colour indicates the presence of peptides.

Vitamin C is an important anti-oxidant, helps protect against cancers, heart disease, stress, it is part of the cellular chemistry that provides energy it is essential for sperm production, and for making the collagen protein involved in the building and health of cartilage, joints, skin, and blood vessels. Vitamin C helps in maintaining a healthy immune system, it aids in neutralizing pollutants, is needed for antibody production, acts to increase the absorption of nutrients (including iron) in the gut, and thins the blood. Vitamin C is found in green vegetables, fruits and potatoes. It is essential for a healthy diet. The chemical name for vitamin C is ascorbic acid. Methods for the detection of vitamin C involve titrating it against a solution of an oxidizing agent.

Materials :Iodine solution

benedict solution

sodium hydrogen carbonate solution

20% sodium hydroxide solution, hydrochloric acid

1% copper(II)sulphate solution and food samples

(A-raisin/dates, B-pounded groundnut, C-milk, D-honey syrup, E-

mayonaise) the food samples can varies.

Apparatus :

Test-tubestest-tube holdersbeakersBunsen burnerDropperwire gauzetripod standwhite tilefilter paper.

Procedure :

1. Five samples of food labelled A, B, C, D, E are prepared.

2. Five food samples thats be prepared were be used to determine a presence of starch , reducing sugars , non-reducing sugars , proteins and lipids that contained in that food samples.

A) Test for the presence on starch.

Assume food samples were labelled as A-milk, B-honey,C-Peanut butter, D-

blackcurrant/raisin juice and E-Mayonnaise

Procedure :

1. 1cm3 of each food samples are being placed in the respective test tube,

and be labeled as A,B,C,D, and E.

2. Using a syringe, 1cm3 of distilled water were added into each test tube.

3. Then the diluted solution of the food samples were then being stirred by

using a glass rod.

4. Next, 2 drops of each mixture were placed on a white tile, using a

dropper.

5. On each samples, another two drops of iodine were added, to test the

presence of starch.

6. The changes from the food test were observed.

Observations :

1. Peanut butter had changes the colour of iodine solution from brownish to

dark blue.

2. In other food sample such as milk, honey, blackcurrant and mayonnaise

did not cause any colour changes to the iodine solution.

Inferences :

1. The changes colour of iodine solution from brownish to dark blue

indicates the presence of starch in peanut butter.

2. This showed that only peanut butter contained starch.

B) Test for reducing sugars

Procedure :

1. 2ml of each food samples are poured into a test tube respectively. All

the test tube is labeled A, B, C, D and E.

2. 1ml of Benedict’s solution is added to each food samples in the test

tube.

3. The mixture was shaken and then heated by placing all the test tube in

a boiling water bath. 

4. Any change in colour in each mixture is observed and recorded.

Observation :

1.A brick-red precipitate is formed in food samples labelled B, C and E.

2.The other food samples remain in blue colour. 

Inferences :

1.The colour changes indicates the presence of reducing sugars.

2. Food samples such as milk, mayones, and honey has reducing sugars

in it.

C) Test for non-reducing sugars

Assume food samples were labelled as milk=A, honey=B, peanut jam=C ,

blackcurrant/Raisin juice=D, Mayonnaise=E

Procedure :

1. Take 2ml of food sample A and poured into a boiling tube. Three drops of

hydrochloric acid were added.

2. The mixture was heated in boiling water bath for 5 minutes.

3. The boiling tube was taken out from water bath and the mixture was

cooled under running tap.

4. The acid in the mixture was neutralized by sodium hydrogen carbonate

powder until effervescence stopped. 

5. Then, Benedicts’s test was conducted on the mixture.

6. Any colour change was observed in the mixture.

7. Steps 1 to 6 were repeated using food samples B,C,D and E

Observations :

1. Brick red precipitate were found in food samples B, C, D and E

2. Food samples A remained blue in colour. 

Inference :

1. Non reducing sugar were founded in food sample B(peanut jam),

C(blackcurrant/Raisin juice), D(honey) and E(mayonnaise) 

2. Non reducing sugar was absent in food sample A(milk)

D) Test for protein

Assume food samples were labelled as A=milk, B=honey, C=peanut butter,

D=blackcurrant/raisin juice, E=mayonnaise

Procedure :

1) 2 ml of each food sample was poured into a test tube.

2) 1ml of 20% sodium hydroxide solution was added to each of the food

sample and shaken well.

3) A few drops of 1 % copper (II) sulphate solution were added slowly to

each of the mixtures

4) Each mixture was shaken well and allowed to stand.

5) Any changes of colour in the food sample was observed and recorded.

Observation : 

Food sample A & C turns purple. However,the other food samples turns blue

in colour.

Inference :

A food sample A & C contains protein.

E) Test for lipids.

Assume food samples were labelled as A=milk, B=honey, C=peanut butter,

D=blackcurrant/raisin juice, E=mayonnaise

Procedure :

1. A small amount of each of the food samples was rubbed on a piece of

filter paper.

2. Then the filter paper was dried.

3. The filter paper was held against the light.

4. All observations were be recorded.

Observation :

1. The filter paper that was rubbed against food samples B,C and E shows a

translucent mark.

2. The filter paper of the food sample A and D remain opaque.

Inference :

1. Food sample B,C and E contain lipids.

2. Food sample A and D do not contain lipids

Objective 2 : To determine the vitamin C contents in a various fruit juices.

Problem :

Does imported fruits contains more vitamin C than local fruits?

Materials :

1.0 % dichlorophenolindophenol solution (DCPIP)0.1 % ascorbic acid solution freshly prepared guava juice papaya juicemango juice orange juice kiwi juice lime juice

Apparatus :

specimen tubes syringes with needles (1 ml and 5 ml) beakers gauze cloth knife.

Procedure :

1. First ,1 ml of 1.0% DCPIP solution was placed into a specimen tube by using a 1 ml syringe.

2. Then filled a 5 ml syringe with a 0.1 % ascorbic acid solution.3. After that use the needle of the syringe then placed into the DCPIP

solution.4. The ascorbic acid solution was added drop by drop into the DCPIP

solution. 5. The mixture gently stirred with the needle of the syringe. 6. The ascorbic acid solution was continuously added until the DCPIP

solution was decolourised. 7. The volume of ascorbic acid solution used is recorded. 8. Repeated all steps above by using anothers differents fruits juice.9. The volume of fruit juice that required to decolourise the DCPIP solution in

each case was recorded in a table.10.The results are tabulated. The percentage and the concentration of

Vitamin C in each of the fruit juices are calculated using the formulae given.

Results :

Solution/fruit

juice

Volume of solution needed

to decolourise 1ml of

DCPIP solution

Percentage of

Vitamin C in

fruit juice

Vitamin C concentration

in fruit juice

1 2 3 Average

0.1 ascorbic

acid

1.0 1.0 1.0 1.0x0.1=0.1% ×1.0=1.0 mg cm-3

Lime juice 3.2 3.0 3.1 3.1x0.1=0.032% x0.1=0.32 mg cm-3

Pineapple juice 4.9 4.9 4.8 4.9x0.1=0.02% x0.1=0.20 mg cm-3

Orange juice 5.0 5.3 5.2 5.2 x0.1=0.019% x0.1=0.19 mg cm-3

Calculations:

Volume of 0.1 % ascorbic acid used to decolourise DCPIP = x cm³Volume of fruit juice used to decolourise DCPIP = y cm³

x cm³ 0.1 % ascorbic acid ( concentration of 1 mg/cm3) can decolorise 1 cm³ of DCPIP and y cm³ of fruit juice concentration of k mg/cm³ can decolourise 1 cm³ of DCPIP

so,the quantity of ascorbic acid in x cm³ of ascorbic acid 0.1 % = quantity of ascorbic acid in y cm³ of fruit juice

thus,x cm³ X 1 mg/cm³ = y cm³ X k mg/cm³k mg/cm³ = x cm³ X mg/cm³

y cm³

= x mg/cm³ y

Therefore the concentration of ascorbic acid in fruit juice = x mg/cm³ y

Questions :

1. Explain the need for our diet to contain : carbohydrates; sodium

chloride; fresh fruits; and milk

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2. How do vegetarians ensure that they receive a balanced diet?

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3. The body cannot digest fiber. Why is it still important in our diet?

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4. A slightly overweight friend decides to go on a crash diet. She/he

tells you that she/he is going to eat grapefruit and drink black coffee

for two weeks. What advice would you give your friend?

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5. Eskimos who eat mainly fish, seal oil and whale meat rather than

beef and butter have low incidence of heart disease. How do you

explain this?

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Discussion:

There a few type of food samples such as blackcurrant , peanut butter ,

milk , honey syrup and mayonnaise. This food sample was be used to

determine the present of presence of starch, reducing sugars, non-

reducing sugars, proteins ,lipids and vitamin C.

Peanut butter is one of the food sample that contain a starch. This

because when its be test with iodine, the colour of peanut butter from

brownish changes to dark blue. However, the another food samples does

not give any changes in colour. So its shows the starch thus not contains

in its.

For the test to detect a reducing sugars, a food samples is be added with

Benedict’s solution. When a brick red precipitate was formed in the food

sample such as honey, peanut butter and mayonnaise. This will show the

reducing sugar is detected.

This experiment also search the non reducing sugar in a foods sample.

After the test had be done to all foods sample by using hydrochloric acid,

heated, neutralized by sodium hydrogen carbonate and Benedict’s test.

The food sample will give a brick red precipitate to showed the

appearance of non reducing sugar.

Sodium hydroxide solution and copper(II) sulphate solution was be used

to determined a presence of protein. The result that will be showed was

the changes of colour. The food sample that contains a protein will turn to

purple.

To detect a presence of lipids, the food sample will be rubbed on a piece

of filter paper and be dried up. The observation from this steps is a filter

paper will show a translucent mark and the another show a remain

opaque. The food sample that show translucent mark will contain a lipids

and the another is not.

The present of vitamin C that’s contains in fruit is be determined by using

a scientific calculation. The higher the result that be showed by this

calculation, the higher the presentation of vitamin C in that’s food sample.

Conclusion :

1. Peanut butter contained starch because changes of iodine solution from

brownish to dark blue. The food sample that does not showed any

changes do not contain a starch.

2. Milk, Mayonnaise, and Honey has reducing sugars because red

precipitate is formed in food samples

3. Non reducing sugar is present in peanut jam, blackcurrant juice, honey

and mayonnaise because brick red precipitate were found in food

samples

4. Milk and peanut contain protein because a food sample had give a

changes colour to blue.

5. Honey peanut butter and mayonnaise contain lipid because translucent

mark shown at the filter paper.

6. If the blue colour of the DCPIP disappears, then vitamin C is present.

7. The water contained no vitamin C and so did not change the colour of

the DCPIP.