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,
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
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.
SUDAN III TEST
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
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
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
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
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.