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Handout Lab 3 - 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, ormonomers, 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 a dehydration synthesis. In this process, water is removed from two monomers, which become linked together by various types of linkages. 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 / Element / Symbol
Black / Carbon / C
White / Hydrogen / H
Red / Oxygen / O
Blue / Nitrogen / N
Green, Orange, or Purple / Any other element or group / -

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. One of our first tasks will be to construct some of these functional groups and discuss their properties.

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

  1. Construct one of the simplest organic molecules, methane (CH4), by inserting four bonds with hydrogens into a carbon atom. Notice that the mode 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 alcohol). Make a diagram of this molecule below.

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  1. Another simple organic molecule is called formic acid ( H-C-OH ). 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.
  2. 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.
  3. Construct an ammonia molecule (NH3). 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.

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 targetsolution (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)n, where “n” may indicate almost any number. For example, in glucose (C6H1206)n, n = 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 CARBOHYDRATES

1. Construct a molecule of beta-glucose as follows: (note: the white atoms are H, theblack atoms are C, and the atoms with vertical bars are O). 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. Thisshould also be true of your constructed model. Have your instructor check your work when you are finished.

1. How is C6 different from the other carbon atoms? ______.

What feature 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? ______.

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. 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 alpha-

or pure beta- solution of glucose, we will, within minutes, have a mixture of the two. Construct

alpha-glucose as shown below.

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.

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 (note: keep the labels on the tubes for other experiments).
  2. Add to each test tube the materials to be tested as listed in the second column of Table 1.
  3. Next, add 10 drops of Benedict's solution to each tube and MIX well.
  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 "Result of Benedict's Test."
  5. After 3 minutes, remove the tubes from the baths and let them cool. Again, note their color for differences from your last observation.

Table 1. Solutions and color reactions for Benedict's Test and iodine test

Tube No. / Solution / Results of Benedict’s Test / Result of Iodine Test
1 / 10 drops deionized water
2 / 10 drops glucose solution
3 / 10 drops sucrose solution
4 / 10 drops starch solution
5 / 10 drops milk
6 / 10 drops unknown A or B
7 / 10 drops unknown C or D

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.
  3. Add 3-5 drops of iodine solution to each tube and MIX well.
  4. Record color changes in Table 1 under "Result of Iodine Test."

QUESTIONS FOR BOTH BENEDICT'S TESTAND THE IODINE TEST

1. Which of the 7 solutions that you used was a positive control for this test? Which was a

negative control? ______

2. Did the positive and negative controls come out as you predicted? ______

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3. What can you say about the unknown solutions? ______

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4. Name four foods that you might expect to react positively with iodine.

PROTEINS

Protein macromolecules 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 (NH2), 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? ______

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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.

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 what type of reaction? ______

VI. PROCEDURE FOR IDENTIFICATION OF PROTEINS O

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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 (10 drops) of Biuret's solution to each tube and mix.

Table 2. Solutions and color reactions for Biuret's test

Tube No. / Solution / Result of Biuret’s Test
1 / 10 drops deionized water
2 / 10 drops egg albumin
3 / 10 drops milk
4 / 10 drops unknown A or B
5 / 10 drops unknown C or D

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.

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 LIPID 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 a saturated fat.

Average Bond Energies

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. Other nucleic acids include RNA, which provided 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. The basic building blocks of nucleic acids are nucleotides.

IX. PROCEDURE FOR NUCLEIC ACID STRUCTURE

Construct the nucleotide unit pictured below (use a green module to represent the entire phosphate group) and compare your results to the model of DNA on display. Attempt to locate on the DNA model a nucleotide like the one you just constructed.