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Lab Handout - MOLECULES OF LIVING SYSTEMS
INTRODUCTION
Most chemicals in living organisms are compounds that contain carbon as the main structural
element. Because of this, life is said to be carbon-based. These biological molecules also contain
other elements such as hydrogen, oxygen, nitrogen, phosphorus, and a few other elements. These
biochemicais are organized into macromolecules that can be classified as carbohydrates, proteins,
lipids, or nucleic acids. In turn, these biological compounds are composed of smaller subunits. or
monomers, that are linked by strong chemical bonds called covalent bonds.
Typically, the monomers of macromolecules are linked together by a special type of chemical
reaction called dehydration synthesis. In this process, water is removed from two monomers,
which become linked together by various types of bonds. The joining of two monomers forms a
dimer. Three or more linked monomers form a polymer.
Carbohydrates, which are complex macromolecules, are composed of monomers called
monosaccharides such as glucose and fructose. These monosaccharides may come together to form
more complex disaccharides and polysaccharides such as sucrose, maltose, starch, cellulose, and
glycogen. Carbohydrates serve a number of important functions in living organisms, such as
providing energy to cells, acting as storage products for energy that may be needed in the future,
and adding structure to a cell.
Many lipids, which include fats and oils, may also play important roles in energy storage. Lipids
also compose the plasma membranes of all living organisms and some act as internal signals
(hormones). The building blocks of lipids are very long chains of carbon and hydrogen called fatty
acids.
Proteins are some of the most important biological chemicals in the cells of all organisms. Some
proteins, the enzymes, act as biological catalysts that speed the rates of millions of chemical
reactions. Other proteins play important structural roles in the cell by forming the cytoskeleton.
Proteins are composed of long sequences of amino acids.
An essential type of biological molecule is the nucleic acid. Chief among these nucleic acids is DNA,
which serves as the reservoir of the genetic material of all living organisms. Other nucleic acids
include RNA, which plays an important role in deciphering the genetic code of DNA, and ATP, a
molecule that stores enormous amounts of energy in its chemical bonds to be used by millions of
cellular processes.
Biological molecules tend to be fairly complex in their structure. This complexity arises from the
fact that carbon can form up to four covalent bonds with other atoms. It is important to also
understand that molecules have three-dimensional structure. They are not flat as usually drawn on
a page. The three-dimensional shape of molecules plays an important role in determining the kinds
of interactions that can occur between molecules. In this exercise, we will make use of molecular
models to illustrate the three-dimensional nature of biological molecules. The kit we will be using
contains pieces that are color-coded for particular atoms as follows:
Color
Black
White
Red
Blue
Green, Orange, or Purple
Element
Carbon
Hydrogen
Oxygen
Nitrogen
Any other element or group
Symbol
C
H
O
N
-
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FUNCTIONAL GROUPS
All biological molecules have distinct chemical and physical properties that determine their
functions. These chemical properties are partly the result of functional groups, which are attached
to the main body of the molecule. These functional groups will determine how the molecules will
bond to one another, or they may determine how the molecules interact with one another as units
in an even larger molecule. Functional groups are formed by a specific arrangement of atoms that
holds a particular chemical property. The same carbon skeleton will behave differently depending
on the functional group attached. An analogy is a Swiss army knife: one handle, various
attachments serve different purposes.
One of our first tasks will be to construct some of these functional groups and discuss their
properties. Below you see some examples of common functional groups attached to carbon
skeletons:
Hydroxyl group
Carbonyl group
Amino group
Carboxyl group
To help in constructing the molecules, remember that hydrogen will always have one bond
(represented in the kit with a plastic peg), oxygen will always have two bonds, nitrogen will have
three bonds, and carbon will have four bonds (with exceptions).
I PROCEDURE FOR FUNCTIONAL GROUP SYNTHESIS
Construct one of the simplest organic molecules, methane (CH4), by inserting four bonds
with hydrogens into a carbon atom. Notice that the model has a 3-dimensional shape.
2. Now remove one of the hydrogens to give rise to a free radical called methyl. Methyl is a
functional group that is commonly seen in organic molecules.
3. At the location on the methyl group where you removed the hydrogen, attach a hydroxyl
group (-OH) by first attaching an oxygen atom, and then attach a hydrogen atom to this
oxygen atom. The resultant molecule is called methyl alcohol or methanol (wood alcoholcan be toxic). Make a diagram of this molecule below.
1.
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4. Another simple organic molecule is called formic acid ( H-C-OH ), produced by ants and
delivered when they sting. To construct this compound, two curve-shaped bonds (the long
slender ones) must be inserted between the oxygen and the carbon atoms, representing a
double bond.
5. Now remove the hydrogen that is attached directly to the carbon atom. This will result in
the formation of a carboxyl functional group. The presence of this group on organic
molecules causes them to behave like organic acids. Examples of this are amino acids, the
building blocks of proteins.
6. Construct an ammonia molecule (NH 3). Remove one of the hydrogen atoms from the
nitrogen to create the non-ionized version of the amino or amine functional group.
Amines impart alkaline or basic characteristics to the organic molecules they are bound to.
Amines are also present on amino acids.
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IDENTIFICATION AND STRUCTURE OF MAJOR BIOLOGICAL MOLECULES
Many biochemical tests are available to identify the major types of organic compounds. Each of
these tests is composed of three or more components: an unknown solution that is to be
identified, a control solution that can be used as a reference for the test, and an indicator
substance, which reacts in a specific visible way with only one type of molecule. The unknown
solutions may or may not contain the substance that you are trying to detect, but the control is
always composed of a. known solution that will react in a predictable way during the test.
Typically, there are two types of controls: a positive control and a negative control. Positive
controls contain the variable for which we are testing. They react positively with the indicator and
show that your test is reacting correctly. A negative control does not contain the variable for
which we are testing. Usually negative controls are composed of the solvent (water) minus the
solute and do not react with the indicator. When the correct indicator reacts with its target
solution (the unknown), a visible change in either color or physical state will occur. This tells us
that a particular biological molecule is present.
CARBOHYDRATE MOLECULES
Carbohydrates include such biological molecules as the sugars and other polysaccharides. They are
composed of building blocks called monosaccharides that occur in repeating patterns. The basic
formula for a monosaccharide is (CH2O)x, where x may indicate almost any number. For example,
in glucose (C6H1206)x, x = 6. Monosaccharides such as glucose can occur as either long linear
chains or as rings that are connected end-to-end.
II PROCEDURE FOR THE STRUCTURE OF CARBOHYDRATFS
1. Construct a molecule of befa-glucose as follows: (note: the white atoms are H, the
black atoms are C, and the atoms with vertical bars are 0). Before proceeding,
please make sure you read the paragraph below.
The numbers next to the carbon atoms represent their positions in the ring structure (note: they do
not represent the number of carbon atoms at each position). The positions of the -H and -OH
groups at each carbon also show their true relations to one another (the -OH groups alternate
between being above the ring and below the ring). Since the H at C) is below the OH. this
should also be true of your constructed model.
1. How is C6 different from the other carbon atoms? __________________________.
This ring is a hexagon, but what’s peculiar about it?___________________________________
What does C1 share with C5? _________________________.
How many hydroxyl groups are present in the glucose molecule? __________________________.
How many hydrogen atoms are attached directly to carbon atoms? ________________________.
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2. Change beta-glucose to galactose. Remove the hvdrogen and the hvdroxyl of your beta-glucose
molecules and reverse their positions. The hydroxyl group should now point upward and the
hydrogen downward. This minor change results in the conversion of glucose into galactose,
another monosaccharide. These two molecules are isomers since they have the same types and
quantity of atoms but different arrangement. This should illustrate an important concept:
SMALL CHANGES CAN MAKE BIG DIFFERENCES IN MOLECULES.
3. Alpha-glucose vs. beta-glucose. Beta-glucose can be easily converted to another version of
glucose called alpha-glucose. This can be done by rearranging the hydroxyl and hydrogen on
C1. These two forms of glucose are highly interconvertible; if we start with either a pure alphaor pure beta- solution of glucose, we will, within minutes, have a mixture of the two. Construct
alpha-glucose as shown to the right.
Glucose molecules can join together in long chains to form polysaccharide molecules. They are
joined in the region of the hydroxyl on C1 of one molecule and the hydroxyl of C4 of another
molecule. The reaction that takes place between the two monosaccharides is a dehydration
synthesis, which results in the removal of a molecule of water. Starch (amylose) is a very important
polysaccharide as a fuel storage molecule in plants. It is formed in plants by linking alpha-glucose
molecules together. Cellulose is also an important polysaccharide in plants that is formed by
linkages between beta-Glucose molecules. Cellulose is an important structural component of the
cell walls of plant cells. The simple differences between starch and cellulose result in very dramatic
differences in the ability of animals to digest these two compounds. Animals have the enzyme
amylase, which breaks down starch but not cellulose. Observe the diagram of these linkages
below.
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III. PROCEDURE FOR THE IDENTIFICATION OF MONOSACCHARIDES
Benedict's solution is a simple indicator that detects the presence of simple monosaccharides such as
glucose and fructose. If Benedict's solution is heated in the presence of a monosaccharide, it will
turn from a deep blue color to a reddish orange color. If no monosaccharide is present then the
color change does not occur.
Benedict's Solution (blue) + Unknown Monosaccharide Solution (clear)
Benedict’s Reagent (blue) + Unknown Monosaccharide Solution (clear)
+ HEAT
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Reddish Orange Solution
1. Obtain 7 test tubes and label them 1-7
2. Add to each test tube the materials to be tested as listed in the second column of Table 1
of the following page.
3. Next, add 20 drops of Benedict's solution to each tube and MIX well by carefully swirling
or flicking the test tube- do not splash liquids.
4. Place all seven tubes in a boiling water bath for 3 minutes and record any color changes
that occur during this time in Table 1 under "Observations of Benedict's Test."
5. After 3 minutes, use test tube holders to remove the tubes from the baths and let them
cool. Again, note their color for differences from your last observation.
IV PROCEDURE FOR IDENTIFICATION OF POLYSACCHARIDES
Iodine solution, which is yellow brown in color, reacts with starch to produce a dark blue or
purple color. This distinguishes starch from monosaccharides, disaccharides and other
polysaccharides, which do not react with iodine in the same way.
1. Wash and dry the labeled tubes from the previous experiment.
2. Add to each test tube the materials to be tested as listed in Table 1 of the following page.
3. Add 3-5 drops of iodine solution to each tube and MIX well well by carefully swirling or
flicking the test tube- do not splash liquids.
4. Record color changes in Table 1 under "Obsevations of Iodine Test."
Table 1. Solutions and color reactions for Benedict's Test and iodine test
Observations of Benedict’s Test
Observation of Iodine Test
Tube
Solution
Color
Color
Is sugar
Color
Color
Is starch
No
before
after
present
before
after
present?
1
10 drops DI water
2
10 drops glucose
3
10 drops sucrose
4
10 drops starch
5
10 drops milk
6
10 drops
unknown A or B
7
10 drops
unknown C or D
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QUESTIONS FOR BOTH BENEDICT'S TESTAND THE IODINE TEST
1. Which of the 7 solutions that you used was a positive control for the benedict’s test? Which was
a negative control? ____________________________________________________________________
2. Which of the 7 solutions that you used was a positive control for the iodine test? Which was
a negative control? ____________________________________________________________________
3. Did the positive and negative controls come out as you predicted? _________________________
___________________________________________________________________________________
4. What can you say about the unknown solutions? ________________________________________
___________________________________________________________________________________
5. Name four foods that you might expect to react positively with iodine._____________________
_____________________________________________________________________________________
PROTEINS
Protein macromolecuies are complex assemblages of amino acids. The smallest proteins contain as
few as six amino acids and the largest over 30,000! Proteins are the major structural molecules of
animals, and also include entities such as enzymes, blood proteins, antibodies, & many hormones.
V. PROCEDURE FOR STRUCTURE OF PROTEINS
1. To a carbon atom attach two hydrogen atoms. Now, to one of the free spaces on the carbon
attach an amine functional group, and to the other attach a carboxyl group (-COOH). You have
just constructed the amino acid glycine, the simplest amino acid. Sketch it below:
• Now replace one of the hydrogen atoms on the central carbon with a green molecule. This
green molecule will stand for any one of twenty R-groups, which represent the variable groups
on the twenty biologically important amino acids. Are the R-groups the same as functional
groups? ________________________. What does having a different R-group do to each amino
acid? _______________________________________________________________________________
____________________________________________________________________________________
2. Now, keeping your old amino acid, construct another amino acid with an orange or purple
R-group. You will now proceed to link these two amino acids.
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3. Remove a hydrogen from the amine group on one amino acid and the -OH group from the
carboxyl group of the other amino acid. Place a bond between the two holes you just created
to make a peptide linkage. What molecule do you think is formed from the atoms you removed
just now from the amino acids? ______________________________________________________.
This is another example of___________________________________________________________
VI. PROCEDURE FOR IDENTIFICATION OF PROTEINS
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The amino acids of proteins are made up of an amino group (-NH2) & a carboxyl group ( C─OH ).
The bond between these two groups found on adjacent amino acids is called a peptide bond (a
type of covalent bond). This type of bond can be identified by Biuret's test. Biuret's solution, which
is specific for polypeptides and not free amino acids, will turn from a blue color to a violet color
upon reacting with polypeptides.
1. Obtain 5 test tubes and label them 1-5.
2. Add the materials listed in Table 2 and then add the materials listed in steps 3 and 4.
3. Add 1 mL (10 drops) of 2.5%s sodium hydoxide (NaOH) to each tube.
4. Add 2 mL (20 drops) of Biuret's solution to each tube and mix.
Table 2. Solutions and color reactions for Biuret's test
Tube
No.
1
Solution
20 drops DI water
2
20 drops egg albumin
3
20 drops milk
4
20 drops unknown A or B
5
20 drops unknown C or D
Color Before
adding
Color after
adding
Result- is there
protein?
QUESTIONS
1. What can you say about the chemical composition of egg albumin?
2. Would free amino acids react positively with Biuret's solution? Explain your answer.
3. What was the positive control in this experiment? What was the negative control?
4. Would you expect Biuret's solution to change color if you spilled some on your hair? Explain
your answer.
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LIPIDS
The lipids (biological fats) are a fairly diverse group of macromolecules with varied functions in
organisms; many are energy-storage compounds, but they are also integral in membrane structure.
Biological waxes serve in water regulation, while sterols make up some of the most important
hormones, such as the sex hormones. The major building blocks of lipids are the fatty acids, which
are very long chains of hydrocarbons with a carboxyl group at one end
VII. PROCEDURE FOR THE UPID STRUCTURE
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1. Construct a fatty acid chain by building a carboxyl functional group ( C─OH ) and then
attaching 7 carbon atoms to it. Add hydrogen atoms to all available sites on the carbon atoms
and diagram your molecule below.
The fatty acid that you have constructed is a saturated fatty acid, which is typical of the fats
found in animal products. What is this fatty acid saturated with? ___________________________
2. Insert a double bond between carbons #3 and #4. How does this alter the shape of the fatty
acid? ___________________________________________. Because the C-H bond of fatty acids
contain energy (expressed as kJ/mol), what has inserting a double bond in the chain done to
the energy content of the chain? ____________________________________________.
By inserting the double-bond you have created an unsaturated fatty acid, which is typical of the
fats found in plants.
Does margarine or butter contain more calories? ________________________________________.
Remember that butter is an saturated fat.
Average Bond Energies
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NUCLEIC ACIDS
The nucleic acid macromolecules provide information for biological systems. The best known of
these molecules is DNA, which makes up the genetic information systems in organisms. Another
nucleic acids is RNA, which provides for the flow of genetic information from DNA to the cell;
ATP, which functions in energy transfer within the cell; and a variety of energy-carrier molecules in
the cell also contain nucleotide components. The basic building blocks of nucleic acids are
nucleotides.
IX. PROCEDURE FOR NUCLEIC ACID STRUCTURE
Identify the three major components of an nucleotide: sugar, phosphate group, and amino base.
Look at the DNA molecule model on the instructor’s desk and at diagrams in your text book to
make more observation on how nucleotides link to make DNA strands.
Where on the DNA model do nucleotides occur? _____________________
______________________________________________________________________________________
How are these molecules paired and held together on the model? __________________________
__________________________________________________________________________________
X. DNA EXTRACTION FROM ONION
(Modified from the Office of Biotechnology, Iowa State University and from other sources.)
DNA is present in the cells of all living organisms. This procedure is designed to extract DNA from
onion in sufficient quantity to be seen and spooled. One of the goals of this procedure is to
introduce you to the concept of chemical extraction and to show you that purifying biochemicals
can be fun and easy!
The process of extracting DNA from a cell is the first step for many laboratory procedures in
biotechnology. The scientist must be able to separate DNA from the unwanted substances of the
cell gently enough so that the DNA is not broken up. An onion is used because it has a low starch
content, which allows the DNA to be seen clearly.
MATERIALS
You will work in groups of 4 for this procedure. Obtain the following materials from the central
table and take them back to your stations (the following materials are per group)
blender
two test tubes with caps
one test tube rack
one 250 ml beaker
2 disposable pipettes
one spatula
one sharp knife for cutting
onion
funnel
coffee filter (or cheesecloth)
ice water bath
distiled water in a container
100 ml of liquid dishwashing
soap/salt solution
10 ml pineapple juice
5 ml of 95% or higher ethanol
60°C water bath
food processor or blender
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PROCEDURE
1. Cut an inch square out of the center of one medium onion. Coarsely chop the onion and place
it in a blender and blend on very low for no more than 10 seconds. IT IS CRITICAL THAT YOU
DO NOT BLEND TOO LONG OR TOO AGGRESSIVELY; OTHERWISE THE DNA GETS ALL
CUT UP. The size of the pieces should be like those used in making spaghetti. It is better to have
the pieces too large than too small.
2. Place the onion in a beaker and cover it with 100 ml of the soap/salt solution.
The liquid detergent causes the cell membrane to break down and dissolves the lipids and proteins
of the cell by disrupting the bonds that hold the ceil membrane together. Salt enables nucteic acids
to precipitate out of an alcohol solution because it shields the negative phosphate end of DNA,
causing the DNA strands to come closer together and band together.
3. Put the beaker in a hot water bath at 60°C for 10 minutes. Use the back of a spoon or spatula to
press the onion against the sides of the beaker.
The heat treatment softens the phospholipids in the cell membrane and denatures the DNAse
enzymes, which, if present, would cut the DNA into small fragments so that it could not be
extracted.
4. Cool the mixture in an ice water bath for 5 minutes. During this time, press the chopped onion
mixture against the side of the measuring cup with the back of the spoon.
This step slows the breakdown of DNA
5. Filter the mixture through a coffee filter that is fitted into a funnel with the funnel placed in the
100 ml graduated cylinder. When you filter the onion mixture, try to keep the foam from
getting into the filtrate. It sometimes filters slowly, so you might want to gently stir the material
in the filter.
6. Dispense the filtered onion solution into test tubes. Each test tube should be about 1/3 full.
7. The solution in the test tube should be very gently mixed before the next procedure.
8. Add 10 ml of pineapple juice to the test tube and mix it gently by inversion.
The pineapple juice contains enzymes that destroy unwanted proteins that could interfere with our
analysis.
9. VERY SLOWLY add cold alcohol to the test tube by SLOWLY pouring the alcohol down the
inside of the test tube with a disposable pipette or medicine dropper to a depth of about 1 cm.
For best results, the alcohol should be as cold as possible.
DNA is not soluble in alcohol. When alcohol is added to the mixture, all the components of the
mixture, except for DNA, stay in solution while the DNA precipitates out into the alcohol layer.
10. Let the solution sit for 2-3 minutes without disturbing it. It is important not to shake the test
tube. You can watch the white DNA precipitate out into the alcohol layer. When good results
are obtained, there wilt be enough DNA to spool on to a glass rod, a Pasteur pipette that has
been heated at the tip to form a hook, or similar device. DNA has the appearance of white
mucus (snot-like appearance).
If you are not able to isolate DNA, look at the DNA of another lab group. If you were continuing
the process the DNA would be purified, cleaved with enzymes, ran on a gel for separation, and
viewed as a banding pattern on a gel.
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