Download Testing for Biologically Important Molecules

Testing for Biologically Important Molecules
General Principles
H 1 O
C
There are four major classes of organic compounds found in living organisms - Carbohydrates, Lipids, Proteins and Nucleic
Acids. The chemical properties of the different classes depend on
the presence of their specific functional groups.
In general, the larger molecules in each class are formed by joining one or more building block molecules (monomers) together
in a Dehydration Synthesis reaction to create a polymer. This is
an energy-requiring process in which a hydroxyl of one unit is
removed, and hydrogen is removed from the other creating a
molecule of water. The two subunits are bonded covalently. It is
also referred to as Condensation.
Hydrolysis is an energy releasing process which breaks the bond
between the subunits and requires the addition of a water molecule. Large polymers are broken down into the smaller monomers by a reaction which is the reverse of dehydration synthesis.
The larger molecule is split apart (“Lysed”) by water (“Hydro”),
with a hydroxyl of water going to one group, and hydrogen to the
other. This is also referred to as a Catabolic process.
HO
OH
HO
OH
HO
OH
2 H2O
Dehydration
Synthesis
2
H
H OH
4 6
OH
HO
3
H
H
4
OH
H
5
OH
6
HO
HO
1
CH2OH
HO
3
O
H
O
HO 5
4 CH2OH
OH H 6
HOH2C
2 H
HO 3
H
H
4
H
5
OH
6
OH
-Fructose
(Ketone)
CH2OH
Maltose, sucrose and lactose have the same molecular formula,
C12H22O11. However, they are isomers since they are have different structures.
C6H12O6 + C6H12O6  C12H22O11 + H2O
Glucose + Glucose  Maltose (malt sugar)
Fructose + Glucose  Sucrose (cane sugar)
Galactose + Glucose  Lactose (milk sugar)
OH
HO
HO
O
1
O
2
4
HO
-Glucose
(Aldehyde)
CH2OH
Hydrolysis
2 H2O
HO
5
1 OH
3 H 2HO
H
H
6
5
2
3
O
HO 1
1
HO
OH
OH
O
HO 5
2
HO 3
OH
-Glucose
4
6 OH
-Fructose
In this exercise you will learn about the structure, properties, and
how to test for the presence of these organic molecules.
Dehydration
Synthesis
I. Carbohydrates
H2O
OH
Carbohydrates are classified according to the number of sugar
molecules they contain. Monosaccharides, such as glucose,
fructose, ribose, and galactose, contain only one sugar monomer. Disaccharides, such as sucrose, maltose and lactose, contain two sugar monomers linked together. Polysaccharides, such
as starch, glycogen, cellulose and chitin, contain many sugar
monomers linked together.
HO
HO
O
HO
HO
1 2
O OH
O
OH
OH
Sucrose
-(1,2) Linkage
Monosaccharides have the formula (CH2O)n, where n is generally
an integer from 3 to 8. Monosaccharides contain many hydroxyl
groups and either a ketone or an aldehyde functional group.
These polar groups make sugar very soluble in water.
Polysaccharides are formed by linking many monosaccharides
together. They are used as energy storage by both plants and
animals. Plants store their glucose as Starch while animals produce Glycogen, which is stored in liver and muscle cells.
Glucose (C6H12O6) and contains an aldehyde group. Fructose,
has the same formula as glucose, but it is a ketone. This difference in structure (Isomers) gives the two monosaccharides
slightly different chemical properties.
Polysaccharides are also important as structural compounds.
Plant cell walls contain the rigid polymer Cellulose. Fungi cell
walls and the exoskeletons of arthropods contain a polysaccharide called Chitin.
Biologically Important Molecules
Reducing Sugar
Tests for Carbohydrates
Nonreducing Sugar
H OH
Oxidation/Reduction (Redox) Reactions: Oxidation is
generally referred to as the loss of electrons, while Reduction is
the gain of electrons.
H OH
HO
HO
HO
H HO
H
HO
HO
HO
OH
H
O
H HO
H
R
H
Benedict’s reaction for Reducing Sugars is shown as:
Cu2+ + e–  Cu+
Copper (II) ions (Cu2+) are reduced to copper (I) ions (Cu+) by
gaining electrons from the reducing sugars. The reducing sugar is
oxidized as a result of giving up its electron.
It is often difficult to see where electrons are flowing when you
can’t see an ion charge. However, since the movement of hydrogen often follows the electrons, we could describe oxidation as
the loss of hydrogen and reduction as the gain of hydrogen.
Remember that all monosaccharides are reducing sugars and that
some disaccharides are reducing sugars, while others are not. If
the aldehyde or ketone groups are tied to another compound (this
is represented by the R in the diagram above), they are not free to
open, and cannot reduce the Cu2+. Therefore, chains of glucose,
such as starch, are not reducing sugars.
Again, disaccharides may or not be reducing. It depends on what
carbons are being linked together.
Sucrose
-(1,2)-Linkage
OH
The reduction of pyruvate to lactate at the end of glycolysis
(fermentation) demonstrates this:
O
HO
HO
HO
HO
+ H2 = Reduction
OH
H3C
- H2 = Oxidation
Pyruvic Acid
Acetal
Nonreducing
Lactic Acid
-Glucose
OH
1) Benedict’s Test for Reducing Sugars
4
HO
HO
Benedict’s Reagent contains copper (II) ions which cause the
reagent to be blue colored. It can detect the presence of reducing
sugars and is used to monitor glucose levels in urine.
Carbohydrates which can be oxidized by Cu2+ and are classified
as Reducing Sugars. When a solution containing Benedict’s
reagent and a reducing sugar are heated, the copper (II) ions in
are reduced to copper (I) ions and the solution changes from the
initial blue color to green, orange, red-orange, or brick-red (depending upon the quantity of reducing sugar).
H
Cu2+
+
(complex)
C
O-
O
(HOCH)n
CH2OH
OH
-
Cu2O
+
Brick-Red
C
O
(HOCH)n
-Glucose
O
1H
HO
Acetal
Nonreducing
Hemi-acetal
Reducing
OH
4
O
HO
O
HO
Polysaccharides do not test positive for reducing sugars unless
they are broken down to form smaller units.
OH
HO
HO
O
OH
HO
O
HO
O
OH
HO
O
HO
CH2OH
A brick-red precipitate of copper (I) oxide (Cu2O), may appear in
the bottom of the tube. The higher the concentration of reducing
sugar present, the more precipitate will form in the tube.
H
1
OH
Maltose
(Disaccharide)
-(1,4) Linkage
Increasing amounts of reducing sugar
OH
OH
O C
HC OH
H3C
C O
Blue
O
OH
O C
O OH
1 2
O
OH
HO
O
HO
Amylose (Starch)
O
HO
OH
You will not be asked details of which sugars are reducing sugars and which are not. You should know that some
are and some aren’t and that polysaccarides are not.
Green  Orange  Red  Brown
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Biologically Important Molecules
2) Iodine Test for Starch
Lugol's iodine reagent (I2KI) is useful to distinguish starch from
other polysaccharides. The -(1,4) linkage between monomers in
starch cause the helical structure of the polysaccharide chain.
Saturated Fatty Acids: Saturated fatty acids (lard, bacon
fat, butter, etc.) contain no double bonds, a maximum of hydrogen atoms, and pack very closely together.
Unsaturated Fatty Acids: Double bonds between carbon
atoms cause fatty acids chains to bend or kink. The fatty acids of
unsaturated fats contain at least one double bond, fewer hydrogen
atoms, and the fatty acid chains cannot pack as closely together
because at least one of the chains has a kink or bend.
Since it takes more energy to break tightly packed saturated fatty
acids apart, they are solid at room temperature. The less "orderly" structure of unsaturated fats is responsible for their lower
melting point (i.e. liquid at room temperature).
O
Iodine (I2) can fit into the helices and will cause a dark bluepurple-black color. Monosaccharides and other polysaccharides
cause no color change; the solution remains brownish yellow.
II. Lipids
OH HO
O
Oleic Acid
OH
Glycerol
Dehydration
Synthesis
2 H 2O
Lipids include fats and oils (important as energy storage compounds), phospholipids and glycolipids (part of the structure of
cell membranes), waxes (protective surface coatings on many
plants and animals), and steroids (found in some cell membranes
and many hormones).
Sudan IV (Red) is a lipid soluble dye. When Sudan red is added
to a mixture of lipids and water, the dye will move into the lipid
layer, coloring it red.
Carbon 9
OH HO
Lipids are organic molecules that are insoluble in water and other
polar solvents. They are, however, very soluble in nonpolar solvents, such as gasoline, chloroform, benzene, and ether.
3) Sudan IV Test for Lipids
Stearic Acid
Ester Bonds
O
O
O
O
Cis
OH
When testing a solution for lipids there are two results you
should be looking for; 1) Do you get a separation of layers (water
& lipid)?, and 2) Does the dye migrate toward one of the layers?
If the mixtures are all water soluble, then the Sudan IV will form
small micelles and disperse throughout the solution.
A very simple test for lipids is based on their ability to produce
translucent grease-marks on unglazed paper. This is a lipid test
most of us has “performed” when eating French Fries out of their
paper wrappings or when wiping our hands on paper napkins.
Fatty Acids: Long carbon chains containing hydrogen with a
carboxyl group (–COOH) on one end, which makes the molecule
an acid. The carboxyl group is involved in bonding each fatty
acid to the glycerol molecule through dehydration synthesis.
When these fatty acids combine with the three hydroxyl groups
on glycerol, they will either form a triglyceride (3 fatty acids) or
a phospholipid (2 fatty acids and a phosphate).
Making Margarine from Plant Oils
Many clinical studies have shown a correlation between the melting points of dietary fatty acids and hardening of the arteries.
Saturated fatty acids appear to be the major problem; therefore
unsaturated vegetable oils tend to be healthier.
However, most people don’t want to “Pour” oil on their morning
toast, so vegetable oils can be made solid by adding hydrogen in
a process called “Hydrogenation”.
Unfortunately, hydrogenation creates its own set of problems.
The catalytic process can also cause cis-trans isomerization, and
unfortunately trans fatty acids appear to be as bad or possibly
worse than saturated fats.
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Biologically Important Molecules
Several Biologically Important Fatty Acids
III. Proteins
The building blocks of proteins are the twenty different amino
acids all of which similar structure. At the center of the molecule
is the alpha carbon which is bonded to four different groups: an
amino group (–NH2), a carboxyl group (–COOH), a hydrogen
atom, and the variable R group (also called the side chain). The
amino acids have identical structures except for their R groups.
O
H2N

O
H
C
R
C OH

N
O H
H
R
Amino Acid 2
Amino Acid 1
Dehydration
Synthesis
H 2O
Peptide
Bond
Stearic, Oleic, Linoleic, & Linolenic Acids
O
H2N
HO
H
R
OH
N H C
H
O
Dipeptide
O
Lauric Acid
MP: 44 °C
Polypeptides are formed by joining amino acids together in a
long, unbranched chain. The amino acids are linked together by
peptide bonds formed when the carboxyl group of one amino
acid reacts with the amino group of the next amino acid in a dehydration synthesis reaction.
O
HO
C
R
Stearic Acid
MP: 70 °C
HO
O
Side
Chains
Oleic Acid
MP: 13 °C
OH
O
HO
Linoleic Acid
(Omega-6)
MP: -5 °C
Cis
Isomers
TRANS vs. CIS
Linolenic Acid
(Omega-3)
MP: -11 °C
O
O
H
N
H2N
N
H
O
HO
CH3
O
CH3
H
N
N
H
O
O
HO
R
OH
CH3
R
HN
NH
H2N
Five Amino Acide Residues
4
Biologically Important Molecules
4) Biuret Test for Proteins
Experiments to Identify Compounds
Biuret Reagent is a light blue solution (Cu2+) which turns purple
when mixed with polypeptides containing at least four peptide
bonds. The purple color is formed when copper (II) ions in the
Biuret reagent react with the lone pair of electrons on the N in
the peptide bonds to form a complex. Cu2+ forms a “tetradentate”
coordination complex through the four nitrogen donor atoms.
In this experiment we will be conducting tests to identify four
types of compounds:
O
R
O
C
R
C
O
C
NH

Iodine Test for Starch

Sudan IV Test for Lipids

Biuret Test for proteins
Each of the tests involves a control and an unknown solution.
Controls are a known solution. We use controls to confirm that
our procedure is detecting what we expect it to detect. We compare the unknown’s response with the control. There is one positive and one negative control for each of the following tests. Just
because a substance reacts to the test does not mean it is a control
– we are simply testing it to obtain the result.
HN
Cu2+
C
Benedict’s Test for Reducing Sugars
R
HN
O

NH
R
Biuret - Cu2+ Complex
Cu2+ Coordinated w/4 Peptide Amides
R
R
R
Cu
R
R
R
A Positive Control contains the variable for which you are testing; it reacts positively and demonstrates the test's ability to detect what you expect. For example, if you are testing for protein
in unknown solutions, then an appropriate positive control is a
solution known to contain protein. A positive reaction shows that
your test reacts correctly; it also shows you what a positive test
looks like.
A Negative Control does not contain the variable for which you
are searching. It contains only the solvent (often distilled water
with no solute) and does not react in the test. A negative control
shows you what a negative result looks like.
Controls are important because they reveal the specificity of a
particular test. For example, if water and a reducing sugar solution react similarly in a particular test, the test cannot distinguish
water from the sugar. But if the sugar solution reacts differently
from distilled water, the test can distinguish water from reducing
sugars. In this instance, the distilled water is a negative control
for the test, and a known reducing sugar solution is a positive
control.
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Biologically Important Molecules
Procedures
Sudan IV Test for Lipids:
NEVER Mix Pipettes from one solution to another!!!
Benedict’s Test for Reducing Sugars:
1.
Set up 8 tubes; label each as shown in the table below.
2.
Add 1 mL of each sample to be tested. Make sure you stir
the solution before pipeting it into your tube.
3.
Add 2 mL of Benedict’s reagent to each tube.
4.
Place the tubes in a beaker of boiling water for 3 minutes.
5.
Remove the tubes and allow them to cool.
6.
Examine each tube and record the solution color in the table.
7.
Rinse out the tubes; use the same labels for the iodine test.
Iodine Test for Starch:
1.
2.
3.
4.
5.
6.
7.
Set up 6 tubes; label each as indicated in the table.
Add 3 mL of distilled water to each tube.
Add about 1 mL of each sample to be tested.
DO NOT add Sudan IV to the first tube (Salad Oil #1)
Add 5 drops of Sudan IV to all tubes, EXCEPT the first.
Mix each tube very well then let settle for 2 minutes.
Examine each tube and record the Colors and Separation.
Results of Sudan IV Tests
Tube Layer
Contents (Y/N)
Salad
Oil #1
1.
Set up 8 tubes as shown in the table below. (You will set up
the same 8 tubes twice.)
Salad
Oil #2
2.
Add 1 mL of each sample to be tested. Make sure you stir
the solution before pipeting it into your tube.
Water
3.
Add 7-9 drops of iodine (IKI) to each tube. DO NOT heat.
4.
Examine each tube and record the color in the table.
Known
Lipid
Results of Benedict’s & Iodine/Starch Tests
Tube
Contents
Water
Glucose
Sucrose
Reducing
Sugar
Starch
Potato
Juice
Onion
Juice
Unknown
Benedict’s Test
Color
Result
(+/-)
Description
Honey
Iodine/Starch Test
Color
Result
(+/-)
Unknown
Biuret Test for Proteins
1.
2.
3.
4.
5.
Set up 6 tubes; label each as indicated in the table.
Add 2 mL of 2.5% sodium hydroxide (NaOH) to each tube.
Add 2 mL of each sample to be tested.
Add 3-5 drops of Biuret reagent to each tube, mix well and
let settle for 2 minutes.
Examine each tube and record the colors.
Results of Biuret Tests
Tube
Contents
Color
Result
(+/-)
Water
Amino Acids
Egg Albumen
Honey
Protein Solution
Unknown
6
Biologically Important Molecules
Due Date:
Name:
Please hand in this and the last page.
1) Look at the table for each experiment and list the positive control, the negative control and the positive results for each biochemical
test.
Reducing Sugars :
+ Control:
– Control:
Positive Results:
Starch:
+ Control:
– Control:
Positive Results:
Fats:
+ Control:
– Control:
Positive Results:
Proteins:
+ Control:
– Control:
Positive Results:
2.Define in your own words “Positive Control” and “Negative Control”. Explain why they are used in experiments.
Positive Control:
Negative Control:
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Biologically Important Molecules
3) Give the expected results (i.e., state if the result will be positive or negative) for a Iodine test of (A) starch and (B) starch that has
been hydrolyzed. Hydrolyzed means that the polymer has undergone hydrolysis.
A) Starch:
B) Hydrolyzed Starch:
4) Give the expected results (i.e., state if the result will be positive or negative) for a Biuret test for a protein (A) before and (B)
after hydrolysis.
A) Protein:
B) Hydrolyzed Protein:
5) You did two different tests for different forms of carbohydrates. For what two forms of carbohydrates did you test?
6) Based upon your tests, molecules from which of the four categories was in the unknown solution (there can be more than one).
7) Since the onion rings you are eating pass the “translucent bag” test you know they have lipids. (Not much of a surprise since they
are deep-fat fried in oil.) You ask the chef for the recipe and are told that the batter is made of flour, eggs and water and of course the
onions. If you ground up the onion rings, and performed all the tests correctly please state what color you would expect for each of
these chemical tests:
Benedict’s (reducing sugars):
Iodine (starch):
Biruet (protein):
8