Lb 3 - MOLECULES OF LIVING SYSTEMS

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 Element Symbol Black Carbon C White Hydrogen H Red Oxygen O Blue Nitrogen N Green, Orange, or Purple Any other element or group -

1 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

1. 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 alcohol- can be toxic). Make a diagram of this molecule below.

O ║ 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 (NH3). Remove one of the hydrogen atoms from the nitrogen to create the non-ionized version of the amino or amine functional group.

2 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 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? ______. 3 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. 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 alpha- or 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.

4 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 ║ ▼ 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 Tub Solution Color Color Is Color Color Is starch e No before after sugar before after present? present 1 10 drops DI water

2 10 drops glucose

5 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

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.

6 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. 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 O ║ 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 Solution Color Before Color after Result- is there No. adding adding protein? 1 20 drops DI water 7 2 20 drops egg albumin

3 20 drops milk

4 20 drops unknown A or B

5 20 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 UPID STRUCTURE O ║ 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

8 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

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? ______

9 ______

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 one sharp knife for cutting 100 ml of liquid dishwashing two test tubes with caps onion soap/salt solution one test tube rack funnel 10 ml pineapple juice one 250 ml beaker coffee filter (or cheesecloth) 5 ml of 95% or higher ethanol 2 disposable pipettes ice water bath 60°C water bath one spatula distiled water in a container food processor or blender

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 10 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.