Exercise 4: Biologically Important Molecules______________________________________________________________________________

OBJECTIVES: Learn the structure and importance of the four types of biological molecules. Understand hydrolysis and dehydration synthesis. Become familiar with different tests used to identify macromolecules. Recognize the difference between a negative and a positive control, and the purpose of

each type in an experimental design. ______________________________________________________________________________

INTRODUCTION:

Biological organisms need and use four classes of macromolecules: carbohydrates, proteins, lipids and nucleic acids. These organic compounds (i.e., carbon-based molecules) are important for proper cellular functioning and each plays a different role within the cell. Carbohydrates provide the primary source of energy or “fuel” for cells and are used to support cell walls of bacteria, fungi and plants. Proteins function as “building blocks” or structural elements within the cell and aid in the transport of molecules across membranes. Proteins also regulate cellular activities (enzymes and hormones) and are important components of the immune system (antibodies). Lipids are used for “storage” of excess fuel and are an integral part of cell membranes. Nucleic acids comprise our genes (DNA, RNA), regulate cell function, and are involved in energy transfer. In addition, both DNA and RNA participate in cellular replication.

Many biological molecules are polymers comprised of small subunits (monomers) held together by covalent bonds. Carbohydrates, for example, are composed of varying combinations of monosaccharides (e.g. glucose) that can be joined to form disaccharides (e.g. sucrose) and polysaccharides (e.g. starch, glycogen). Similarly, proteins are made from unique combinations of amino acids. Monomers are linked together by dehydration synthesis (condensation) reactions in which energy is used to remove a water molecule, resulting in the covalent bonding of two subunits (Fig. 1a). Conversely, the bond between monomers can be broken by the addition of a water molecule and the release of energy - hydrolysis (Fig. 1b).

Although all macromolecules are characterized by the presence of a carbon backbone, the four classes vary in their elemental structure and chemical properties. The disparate functional groups impart different solubilities and polarities to each type of macromolecule. For instance, lipids, which are made of fatty acids and have very little oxygen, are nonpolar and insoluble in water. Proteins, on the other hand, are polymers of amino acids covalently linked by peptide bonds. Because amino acids are polar, non-polar, charged or aromatic, the properties of the resulting proteins vary in accordance with the type of amino acids that comprise them.

In today’s lab, you will perform several biochemical tests to detect the presence of organic molecules in known substances. Throughout this process, you will learn about the use of controls as standards for comparison and their role in identifying unknown solutions. Controls are an essential component of every experiment because they help eliminate alternate

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explanations of experimental results. In general, a control is any variable kept constant during the experiment, and it is compared to the experimental sample(s) being tested. There are two types of controls: negative and positive. A negative control helps minimize false positives by providing a known negative result for a given experimental treatment. Thus, a negative control provides an example of what the results should look like if the experimental manipulation had no effect on the variable of interest. A positive control helps minimize false negatives by demonstrating what a positive result should look like if the experimental manipulation produces a change. For example, when testing for the presence of salt in a substance, a salt solution would serve as the appropriate positive control and distilled water as the negative control.

Figure 1. Dehydration synthesis and hydrolysis reactions

Task 1: CARBOHYDRATES

Carbohydrates are molecules made of Carbon (C), Hydrogen (H) and Oxygen (O) in a ratio of 1:2:1. Monosaccharides are made of single sugar molecules while disaccharides and polysaccharides are composed of two or more sugar molecules, respectively.

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Monosaccharides contain either aldehyde (-CHO) or ketone (-C=O) side groups that reduce oxidizing compounds. A molecule is oxidized if it loses an electron or hydrogen atom and is reduced when it gains an electron or hydrogen atom (Fig. 2). Collectively, the two processes are referred to as a redox reaction because when one molecule is oxidized, another is reduced.

Figure 2. Redox reactions

I. Examine Reducing SugarsBenedict’s reagent can be used to identify the presence

of reducing sugars and is a good indicator for the presence of some carbohydrates. At basic/alkaline pHs (8-14) the copper ions (Cu2+) in Benedict’s reagent are reduced by the monosaccharide functional groups (i.e. -CHO or –C=O) to form cuprous oxide. In the Benedict’s test for reducing sugars, the Benedict’s reagent is reduced while the reducing sugar is oxidized. This redox reaction results in a tractable color change going from a light blue solution to a green/reddish orange one. The intensity of the color change is indicative of the amount of reducing sugar present (Fig. 5).

Figure 3. Benedict’s test results

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A = negative (no reducing sugars present) – light blue

B = positive (small amount of reducing sugars present) – green

C = positive (larger amount of reducing sugars present) – orange

D = positive (abundance of reducing sugars present) – red orange

Procedure:

1. Obtain seven test tubes and label them 1-7 using tape wrapped near the top of tube.

2. Add the materials listed in Table 1 to each of your tubes.

3. Half fill a 250mL beaker with water. Place it on the hot plate at your station and allow it to come to a gentle boil. Never let glass boil dry.

4. In the meantime, predict the color changes you expect to occur in each tube and record them in Table 1 in the “Benedict’s Test Results Expected (color)” column. Also mark which tube you think is the positive control and which is the negative control.

5. Add 2mL of Benedict’s reagent to each tube.

6. Place all 7 tubes in the gently boiling water bath for 3 minutes. Observe for change in color.

7. After 3 minutes, remove the tubes and allow them to cool to room temperature. Record the color of each tube in Table 1 in the “Benedict’s Test Results Observed (color)” column.

Table 1:

Tube # SolutionBenedict’s Test Results Iodine Test Results

Expected (color)

Observed (color)

Expected (color)

Observed (color)

1 10 drops onion juice

2 10 drops potato juice

3 10 drops sucrose

4 10 drops glucose

5 10 drops distilled water

6 10 drops reducing sugar

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7 10 drops starch

Questions:

a. How do mono-, di- and polysaccharides differ?

b. Explain, in your own words, what are negative and positive controls? Why are they used in experiments?

o Which of the solutions in Table 1 were your positive and negative controls? Explain.

c. What chemical reactions does Benedict’s test for, and what does this have to do with carbohydrates? Will it be able to detect ALL sugars?

II. Examine StarchStarch is a polysaccharide often used by organisms for storage of metabolic energy.

Unlike the simpler mono- and disaccharides, starch is a structurally complex polymer (Fig. 4). Iodine (iodine-potassium iodide, I2KI) reacts with the three-dimensional (3D) structure of this molecule, resulting in a color change (going from yellow to blue-black, Fig. 5). In this exercise, we will test the substances previously examined for the presence of reducing sugars for starch.

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Figure 4. Starch Molecule

Figure 5. Expected test results for Iodine test for starch.Procedure:

1. Obtain seven test tubes and label them 1-7.

2. Add the materials listed in Table 1 to each of your tubes.

3. Predict the color changes you expect to occur in each tube and record them in Table 1 in the “Iodine Test Results Expected (color)” column. Also mark which tube you think is the positive control and which is the negative control.

4. Add 8 drops of iodine to each tube.

5. Record the color of each tube in Table 1 in the “Iodine Test Results Observed (color)” column.

Questions:

a. What type of carbohydrate are you testing for when you use the Iodine test? Is this type of carbohydrate a mono-, di- or polysaccharide?

b. Based on both your answer to part a, as well as your test results, what is the predominant carbohydrate in onion juice? What about potato juice?

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A = negative (no starch present) – orange

B = positive (starch present) - black-blue

c. Which of the solutions in Table 1 were your positive and negative controls for starch? Explain.

Task 2: PROTEINS  

Proteins are composed of amino acids covalently linked by peptide bonds (Fig. 9). All amino acids contain an amino group (-NH2), a carboxyl group (-COOH), and a variable side chain (R-group, Fig. 7) by which they are categorized. Peptide bonds (C-N) form when the amino group of one amino acid reacts with the carboxyl group of another (Fig. 7). The Biuret reagent, regularly colored blue, is used to identify proteins. When the copper ions (Cu2+) in the reagent interact with peptide bonds, a violet color is produced (Fig. 8). In order for the interaction between Cu2+ and the peptide bonds to result in a color change, a minimum of 4-6 peptide bonds is required. In general, the longer the protein chain, the greater the intensity of the reaction.

Figure 7. Protein composed of two amino acids and linked by a peptide bond (red).

Figure 8. Expected test results for Biuret test for protein.

Procedure:

1. Obtain 5 test tubes and label them 1-5.

2. Add the substances listed in Table 2 to each test tube.

A B C

Fig. 3.20Fig. 3.20Fig. 3.20

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A = negative (no protein containing more than 4-6 peptide bonds present) – light blue

B = positive (small amount of protein containing more than 4-6 peptide bonds present) – light tan

C = positive (large amount of protein containing more than 4-6 peptide bonds present) – purple

3. Predict the color changes you expect to occur in each tube and record them in Table 2 in the “Expected Results (color)” column. Also mark which tube you think is the positive control and which is the negative control.

4. Add 2mL of 2.5% sodium hydroxide followed by 3 drops of Biuret reagent. 5. Record the color of each tube in Table 1 in the “Observed Results (color)” column.

Table 2:Tube # Solution Expected Results

(color)Observed Results

(color)

1 2mL egg albumen1

2 2mL honey

3 2mL amino acid solution

4 2mL distilled water

5 2mL protein solution

1Albumen = clear liquid inside of an egg (“egg white”)

Questions:

a. What monomer comprises a polypeptide? What molecular structures indicate a molecule is a protein?

b. Using the Biuret test, what would a positive and a negative result indicate?

c. Hypothetically, you test two different substances for the presence of protein. One test provides a strong positive result and the other a weak, but still positive result. What does this indicate about the two samples?

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Task 3: LIPIDS

Lipids, which include triglycerides (fats), steroids, waxes, and oils, vary in function. Similar to carbohydrates, fatty acids bond to glycerol with the input of energy and the formation of water. While triglycerides and oils serve as energy-storage molecules, phospholipids aggregate to form cellular membranes which are important sources of cholesterol, a necessary component of steroid hormones. All lipids share one characteristic; they are insoluble in water (i.e. hydrophobic) because they have a high proportion of non-polar carbon-hydrogen bonds and can only dissolve in non-polar solvents such as ether, ethanol and acetone. This property can be used to test unknown solutions for the presence of lipids. One indicator commonly used is Sudan IV, a fat-soluble dye that binds to lipids when added to a solution (Fig. 11).

Figure 9. Expected test results for Sudan IV test for lipids

Examine Lipid Solubility

Procedure:

1. Obtain two test tubes and label them 1 and 2.

2. In this exercise, you will assess the solubility of lipids in polar and non-polar solvents. Predict what you expect to occur in each tube and record your predictions in Table 3 in the “Expected Results” column.

3. Add 1mL of vegetable oil to each tube followed by the solutions listed in Table 3.

4. Record your observations in Table 3 in the “Observed Results” column.

Table 3:Tube # Solution Expected Results Observed Results

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A. B.

A = positive (lipids present)

B = negative (no lipids present)

1 5mL water

2 5mL acetone

Questions:a. Can a lipid dissolve in water? Why or why not?

b. How did this test incorporate what you know about a lipid’s structure and its solubility properties?

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