Detection Of Biological Molecules Essay Research Paper

Detection Of Biological Molecules Essay, Research Paper

Detection of Biological Molecules

Introduction: Without carbon, nitrogen, hydrogen, sulfur, oxygen and phosphorus,

life wouldn’t exist. These are the most abundant elements in living organisms.

These elements are held together by covalent bonds, ionic bonds, hydrogen bonds,

and disulfide bonds. Covalent bonds are especially strong, thus, are present in

monomers, the building blocks of life. These monomers combine to make polymers,

which is a long chain of monomers strung together. Biological molecules can be

distinguished by their functional groups. For example, an amino group is

present in amino acids, and a carboxyl group can always be found in fatty acids.

The groups can be separated into two more categories, the polar, hydrophilic,

and the nonpolar, hydrophobic. A fatty acid is nonpolar, hence it doesn’t mix

with water. Molecules of a certain class have similar chemical properties

because they have the same functional groups. A chemical test that is sensitive

to these groups can be used to identify molecules that are in that class. This

lab is broken down into four different sections, the Benedict’s test for

reducing sugars, the iodine test for the presence of starch, the Sudan III test

for fatty acids, and the Biuret test for amino groups present in proteins. The

last part of this lab takes an unknown substance and by the four tests,

determine what the substance is.


Introduction: Monosaccharides and disaccharides can be detected because of

their free aldehyde groups, thus, testing positive for the Benedict’s test.

Such sugars act as a reducing agent, and is called a reducing sugar. By mixing

the sugar solution with the Benedict’s solution and adding heat, an oxidation-

reduction reaction will occur. The sugar will oxidize, gaining an oxygen, and

the Benedict’s reagent will reduce, loosing an oxygen. If the resulting solution

is red orange, it tests positive, a change to green indicates a smaller amount

of reducing sugar, and if it remains blue, it tests negative.

Materials: onion juice5 test tubes1 beaker potato juice

rulerhot plate deionized waterpermanent

marker5 tongs glucose solutionlabels starch solution6 barrel

pipettes Benedict’s reagent5 toothpicks

Procedure: 1.Marked 5 test tubes at 1 cm and 3 cm from the bottom. Label test

tubes #1-#5. 2.Used 5 different barrel pipettes, added onion juice up

to the 1 cm mark of the first

test tube, potato juice to the 1 cm mark of the second, deionized water

up to the 1

cm mark of the third, glucose solution to the 1 cm mark of the fourth,

and the

starch solution to the 1 cm mark of the fifth test tube. 3.Used the

last barrel pipette, added Benedict’s Reagent to the 3 cm mark of all 5

test tubes and mix with a toothpick. 4.Heated all 5 tubes for 3

minutes in a boiling water bath, using a beaker, water, and

a hot plate. 5.Removed the tubes using tongs. Recorded colors

on the following table. 6.Cleaned out the 5 test tubes with deionized



Benedict’s Test Results

Discussion: From the results, the Benedict’s test was successful. Onion juice

contains glucose, and of course, glucose would test positive. Starch doesn’t

have a free aldehyde group, and neither does potato juice, which contains starch.

Water doesn’t have glucose monomers in it, and was tested to make sure the end

result would be negative, a blue color.


Introduction:The iodine test is used to distinguish starch from

monosaccharides, disaccharides, and other polysaccharides. Because of it’s

unique coiled geometric configuration, it reacts with iodine to produce a blue-

black color and tests positive. A yellowish brown color indicates that the test

is negative.

Materials: 6 barrel pipettespotato juicestarch solution 5 test

tubeswateriodine solution onion juice

glucose solution5 toothpicks

Procedure: 1.Used 5 barrel pipettes, filled test tube #1 with onion juice,

second with potato

juice, third with water, fourth with glucose solution, and fifth with

starch solution. 2.Added 3 drops of iodine solution with a barrel pipette,

to each test tube. Mixed

with 5 different toothpicks. 3.Observed reactions and recorded

in the table below. Cleaned out the 5 test tubes. Data:

Iodine Test Results

Discussion:The iodine test was successful. Potato juice and starch were

the only two substances containing starch. Again, glucose and onion juice

contains glucose, while water doesn’t contain starch or glucose and was just

tested to make sure the test was done properly.


Introduction: Sudan III test detects the hydrocarbon groups that are remaining

in the molecule. Due to the fact that the hydrocarbon groups are nonpolar, and

stick tightly together with their polar surroundings, it is called a hydrophobic

interaction and this is the basis for the Sudan III test. If the end result is

a visible orange, it tests positive.

Material: scissorsdeionized watermargarineSudan

III solution petri dishstarchethyl alcohol

forceps lead pencilcream5 barrel pipettes filter paper

cooking oilblow dryer

Procedure: 1.Cut a piece of filter paper so it would fit into a petri dish. 2.

Used a lead pencil, and marked W for water, S for starch, K for cream, C


cooking oil and M for margarine. Draw a small circle next to each

letter for the

solution to be placed. 3.Dissolve starch, cream, cooking oil and

margarine in ethyl alcohol. 4.Used a barrel pipette for each solution, added a

small drop from each solution to

the appropriate circled spot on the filter paper. 5.Allowed the

filter paper to dry completely using a blow dryer. 6.Soaked the paper in the

Sudan III solution for 3 minutes. 7.Used forceps to remove the paper from

the stain. 8.Marinated the paper in a water bath in the petri dish, changed

water frequently. 9.Examined the intensity of orange stains of the 5 spots.

Record in the table below. 10. Completely dried the filter paper, and

washed the petri dish.

Data: Sudan III Test Results

Filter paper:

Discussion: The results indicate that the Sudan III test was sucessful. Water

and starch definitely doesn’t contain any fatty substances. Cream and cooking

oil no doubtedly does contain lipids. It was surprising to find that margarine

doesn’t contain any fat.


Introduction: In a peptide bond of a protein, the bond amino group is

sufficiently reactive to change the Biuret reagent from blue to purple. This

test is based on the interaction between the copper ions in the Biuret reagent

and the amino groups in the peptide bonds.

Materials: 6 test tubesegg white solutionstarch

solution6 toothpicks rulerchicken soup

solutiongelatin6 parafilm sheets permanent

markerdeionized watersodium hydroxide labels

glucose solutioncopper sulfate

Procedures: 1.Used 6 test tubes, and labeled them at 3cm and 5cm from the

bottom. Labeled

each #1 to #6. 2.Added egg white solution to the 3cm mark of the

first tube, chicken soup solution

to the 3-cm mark of the second tube, water to the 3 cm mark of the third

test tube,

glucose solution to the fourth, starch to the fifth, and gelatin to the

sixth, all at the

3 cm mark. 3.Added sodium hydroxide to the 5 cm mark of each tube and

mix with 6 different

toothpicks. 4.Added 5 drops of Biuret test reagent, 1% copper sulfate,

to each tube and mix

by placing a parafilm sheet over the test tube opening, and shake

vigorously. 5.Held the test tubes against a white piece of paper, and recorded

the colors and

results. Discarded the chemicals, and washed the test tubes.


Biuret Test Results

Discussion: The Biuret test seemed to have been successful. Glucose and starch

are both carbohydrates, while water has no proteins. Egg white definitely has

proteins, and so does gelatin. Chicken soup had a hint of protein content.

Unknown Chemical # 143

Introduction: By performing the Benedict’s Test, the Iodine Test, the Sudan III

Test, and the Biuret Test, chemical #143 should be identified.

Materials: materials from the Benedict’s Testmaterials from the Sudan

III Test Materials from the Iodine Testmaterials from the

Biuret Test

Procedures: 1.Performed the Benedict’s Test, and recorded results. 2.

Performed the Iodine Test, and recorded results. 3.Performed the Sudan III

Test, and recorded results. 4.Performed the Biuret Test, and recorded results.

Data: Properties of Chemical #143

chemical #143 was a white powderish substance.

Conclusion: After ruling out the obvious wrong substances from the list like

ground coffee, egg white and yolk, table sugar and salt, syrup and honey, the

small amount of proteins was taken into factor. That also eliminated powdered

skim milk, and soy flour. The low, or none fat content ruled out some more

choices like enriched flour. The only choices left was corn starch, glucose,

and potato starch. Because of the low reducing sugar, glucose can be ruled out


The starch content of substance #143 was very high. The protein content was

around the 10% range, so potato starch would be a better guess then corn starch.

But corn starch contained only a trace of fat when potato starch contained 0.8%.

But 0.8% is very insignificant. The most educated guess to what chemical #143

is potato starch.


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