Structure and Function

Introduction

In the course description for AP^® Biology, the College Board lists structure and function as one of its eight major themes. Use this set of three demonstrations to engage your students in hands-on review activities and to assess their understanding of this fundamental theme in biology.

The set of three demonstrations includes:

1. Isolating DNA—Begin with this classic activity then use the accompanying questions to review enzymes, molecular structure and structural changes in DNA during the cell cycle.
2. Permeability of Cell Membranes—Continue the review of structure and function with a demonstration of the selective permeability of a cell membrane.
3. Protein Denaturation—Finally, adjust the pH of the solution surrounding a protein to visually demonstrate the ability of some proteins to rebound from denaturation.

The series of demonstrations may be presented in a variety of ways. Each activity may be used to review a specific AP test topic or all the activities may be performed together to evaluate students’ ability to explain the functions of biological molecules and higher levels of biological organization in terms of their structure and properties. A student worksheet is included as an optional assessment tool for the instructor.

Materials Included In Kit

Additional Materials Required

Experiment Overview

Isolating DNA

With its unique structure and essential functions, DNA is an excellent biomolecule for describing the mutually dependent structure and function. The structure of individual nucleotides predicts how they will join together to form a polymer, and also shows how the tertiary structure of the double helix is possible. The structure of the double helix, in turn, explains the ability of DNA to be replicated with few errors. In addition, the structure has a built-in mechanism for controlling gene expression to make proteins which function throughout the entire organism.

Permeability of Cell Membranes

The cell membrane plays a vital role in regulating what goes into and out of the cell. Discover the characteristics of cell membranes that make this possible.

Protein Denaturation

Any change in the pH of a protein’s environment will cause observable changes in the solubility of the protein. These changes reflect changes in the three-dimensional structure of the protein. The effect of pH on protein solubility explains why most enzymes function well at an optimum pH, and why their activity decreases substantially at pH values other than the optimum. This demonstration examines the effect of pH on the solubility and structure of casein, a milk protein.

Materials

Safety Precautions

Ethyl alcohol is a flammable liquid and a dangerous fire risk—keep away from flames and other sources of ignition. The denatured alcohol is toxic by ingestion. Sodium dodecyl sulfate solution may be irritating to skin. Dilute hydrochloric acid, ammonium hydroxide, potassium hydroxide and sodium hydroxide solutions are skin and body tissue irritants and slightly toxic by ingestion. Neutral red will stain skin and clothing. Avoid contact of all chemicals with eyes and skin. Hydrochloric acid is toxic by ingestion or inhalation and is severely corrosive to skin and eyes. Sodium hydroxide solution is a corrosive liquid and is very dangerous to eyes; it causes skin burns. Any food-grade items that have been brought into the lab are considered laboratory chemicals and are for lab use only. Do not taste or ingest any food in the lab and do not remove any remaining food items. Wear chemical splash goggles, chemical-resistant gloves and a chemical-resistant apron. Wash hands thoroughly with soap and water before leaving the laboratory. Please follow all laboratory safety guidelines. Please review current Safety Data Sheets for additional safety, handling and disposal information.

Disposal

Please consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. For Isolating DNA, the solids may be disposed of in the regular trash according to Flinn Suggested Disposal Method #26a. Move the leftover solution remaining at the end of the Procedure to a chemical fume hood. Allow the ethyl alcohol to evaporate according to Flinn Suggested Disposal Method #18a. After the ethyl alcohol has evaporated the remaining solution may be disposed of down the drain with plenty of excess water according to Flinn Suggested Disposal Method #26b. For Permeability of Cell Membranes, all of the products and excess sodium bicarbonate may be disposed of down the drain with excess water according to Flinn Suggested Disposal Method #26b. Excess hydrochloric acid may be neutralized with base and then poured down the drain with plenty of excess water according to Flinn Suggested Disposal Method #24b. Excess sodium hydroxide, potassium hydroxide, and ammonium hydroxide may be neutralized with acid and then poured down the drain with an excess water according to Flinn Suggested Disposal Method #10. For Protein Denaturation, the solutions may be rinsed down the drain with excess water according to Flinn Suggested Disposal Method #26b.

Prelab Preparation

Isolating DNA

1. Use the balance and a weighing dish or wax paper to mass 25 g of banana chips.
2. Place the banana chips and 150 mL of 1 M sodium chloride solution into a resealable bag overnight. Note: Check the resealable bag for leaks.
3. The next day, gently knead the bag, breaking the soaked banana chips into pea-sized or smaller pieces.
4. Prepare an ice-water bath and chill the ethyl alcohol for use in step 6 of the Procedure.

1. Prepare 0.02% neutral red solution: Dilute 1 mL of 1% neutral red to 50 mL with DI water.
2. Prepare 1% sodium bicarbonate solution: Dilute 25 mL of the 10% sodium bicarbonate solution to 250 mL with DI water.
3. Prepare 0.01 M solution of ammonium hydroxide, potassium hydroxide and sodium hydroxide. Add 90 mL of deionized water to each bottle. Cap and shake. The pH of each solutions should be 11–12.
4. Prepare the yeast suspension just prior to the lab. Add one entire package of dry yeast to 240 mL of the 1% sodium bicarbonate solution. Mix thoroughly and swirl when dispensing.
5. Prepare the boiling water baths for use in step 11.

a. Fill a 250-mL beaker about one-third full with tap water.
b. Add several boiling stones.
c. Heat the water on a hot plate until the water just begins to boil.

Procedure

Isolating DNA

1. Layer four layers of cheesecloth on top of one another and place into a funnel. Place the funnel into a clean, 400-mL beaker.
2. Pour the banana mixture through four layers of cheesecloth so that the liquid drains into the beaker.
3. Gently squeeze the cheesecloth to remove most of the banana solution leaving the solids trapped in the cheesecloth. Typically, about 100 mL of banana solution is recovered from the original 150 mL of salt solution.
4. Use a clean, graduated cylinder to add 10 mL of the 0.1 M EDTA to the beaker. Stir with a stirring rod.
5. Use a clean, graduated cylinder to add 10 mL of the 10% SDS solution to the beaker. Stir with a stirring rod. Note: A cloud of precipitated proteins may form during this step.
6. Holding the beaker at a slight angle, use a clean, graduated cylinder to carefully transfer 15 mL of ice-cold 95% ethyl alcohol down the side of the beaker so that the ethyl alcohol forms a layer on top of the banana solution in the beaker (see Figure 1).

7. Carefully place the beaker back on the tabletop, making sure the two layers do not mix.
8. Allow the beaker to sit for one minute and observe the DNA precipitating out of the banana solution at the interface between the cold ethyl alcohol and the aqueous banana solution layers. Note: DNA appears as a white cloud with numerous tiny bubbles attached to the precipitate.
9. Use the stirring rod or inoculating loop to gently lift the DNA from the interface.
10. Answer the questions on the DNA Worksheet.

1. Place six test tubes in a test tube rack. Use a marker to label them 1 through 6.
2. Using a graduated pipet, add 1 mL of neutral red solution into test tube 1.
3. Use a clean graduated pipet to add 1% sodium bicarbonate solution one drop at a time to test tube 1 until the color changes. Record the new color in Data Table 1 on the Permeability Worksheet.
4. Use a clean, graduated pipet to add 0.01 M hydrochloric acid solution one drop at a time to test tube 1 until the color changes again. Record the color in the Data Table on the Permeability Worksheet.
5. Use a stirring rod to stir the yeast suspension. Using the graduated cylinder, transfer 25 mL of the yeast suspension to a 100-mL beaker. Record the initial color of the yeast suspension in the Data Table on the Permeability Worksheet.
6. Using the graduated cylinder, add 25 mL of the 0.02% neutral red solution to the yeast suspension in the beaker. Immediately record the color of the yeast–neutral red suspension in the Data Table on the Permeability Worksheet.
7. After 5 minutes, observe and record the color of the suspension in the Data Table on the Permeability Worksheet.
8. Place a funnel into test tube 2.
9. Fold a piece of filter paper in half and then in half again. Open the filter paper to form a filter paper cone. Place the filter paper cone into the funnel.
10. Filter 10 mL of the yeast–neutral red suspension from step 6 into test tube 2. Observe and record the color of (a) the yeast cells in the filter paper and (b) the liquid in the test tube in the Data Table on the Permeability Worksheet.
11. Pour 10 mL of the yeast–neutral red suspension from step 6 into test tube 3. Using a clamp, place the test tube into a boiling water bath for 5 minutes. Record the initial and final color of the yeast–neutral red suspension in the Data Table on the Permeability Worksheet.
12. Pour 10 mL of the yeast–neutral red suspension from step 6 into test tubes 4–6.
13. Use a clean, graduated pipet to add 1 mL of 0.01 M sodium hydroxide to test tube 4. Observe and record the color of the yeast cells in the Data Table on the Permeability Worksheet.
14. Use a clean, graduated pipet to add 1 mL of 0.01 M potassium hydroxide to test tube 5. Observe and record the color of the yeast cells in the Data Table on the Permeability Worksheet.
15. Use a clean, graduated pipet to add 1 mL of 0.01 M ammonium hydroxide to test tube 6. Observe and record the color of the yeast cells in the Data Table on the Permeability Worksheet.

1. Add 25 mL of 0.1 M sodium hydroxide and a stirring bar to a 600-mL beaker or flask and place the beaker on a magnetic stirrer. Add 225 mL of distilled or deionized water and stir at moderate speed. Add 1 g of casein and stir to dissolve. (The solution will be slightly cloudy or translucent.)
2. Add 1–2 mL of universal indicator to observe pH changes, if desired. Consult the universal indicator color chart for pH values.
3. With rapid stirring, add 2 M hydrochloric acid in 0.5-mL increments using a graduated Beral-type pipet. (The solution will turn cloudy, but will clear up again as the hydrochloric acid is dispersed. After 1–2 mL of acid has been added, the cloudiness will reach a maximum—this is the isoelectric point. The pH at the isoelectric point is 4–5.)
4. Once the isoelectric point has been reached, pause just long enough to record observations (15–20 seconds). Continue adding 2 M hydrochloric acid in 0.5-mL increments with stirring until the solution is clear again. (The cloudiness will fade and the precipitate will redissolve after the addition of another 2–3 mL of acid, when the pH drops below the isoelectric point, pH ≤2.)
5. Continue to stir the solution. Reverse the process by adding 2 M sodium hydroxide in 0.5-mL increments using a clean, graduated Beral-type pipet. (After 2–3 mL of sodium hydroxide has been added, the protein will precipitate out again at the isoelectric point.)
6. Continue adding sodium hydroxide in 0.5-mL increments with stirring until the solution is clear again. (The solution will clear up after an additional 1–2 mL of sodium hydroxide has been added and the pH >10–12.)
7. The process may be repeated using the same casein solution. This will give time to explain the observations.

Student Worksheet PDF

10986_Student1.pdf

Sample Data

Permeability of the Cell Membrane
(Student answers will vary.)

Answers to Questions

Isolating DNA

1. In the procedure for the isolation of DNA, EDTA is added to remove DNase, a natural cellular enzyme that catalyzes the breakdown of DNA. What are some prossible reasons that a cell would need this type of function?

The ability to break down DNA is important in cell death, the immune response and in the repair of damaged DNA.

2. DNA is precipitated by adding ethyl alcohol, a solvent that is less polar than water. Based on its structure, explain why DNA is insoluble in alcohol.

The phosphate backbone is negatively charged, as a highly charged ionic compound it requires a polar solvent such as water.

3. Compare the structure and function of a eukaryotic chromosome to that of a prokaryotic chromosome.

Eukaryotic chromosomes are have a complex linear structure with intron and extron sections within the genes. Eukaryotic chromosomes are bound to proteins called histones which help with packaging the chromosomes and help regulate replication and gene transcription. Typically there are more than one chromosome in a eukaryotic cell. Prokaryotic chromosomes are a single, circular chromosome without histones. There are no introns within the chromosome and replication starts at a single point on the chromosome.

4. How does the structure of DNA lend itself to being the genetic blueprint for life?

The directionality of DNA, the unit repeatability, the bonding conformations and the availability of raw materials all contribute to DNA serving as the molecule of choice as the genetic blueprint of life.

5. Describe the structural differences between chromatin and a chromosome and explain which structural form of DNA is dominant a) during mitosis and b) in nondividing cells.

The tightly bound chromosome form found during interphase allows for tight control of gene expression, whereas the looser, partially unwound chromatin allows room for replication of the entire strand.

Permeability of the Cell Membrane 

1. Use the results from test tube 1 to explain the slow color change observed in the yeast–neutral red mixture in the Filtered Yeast–Neutral Red Suspension section of the data table.

The indicator color changed from yellow to red as the neutral red was transported across the yeast cell membrane and encountered the acidic cytoplasm. Neutral red is yellow in basic solutions, red in acidic.

2. Explain the color change observed in the boiled yeast–neutral red solution in the Boiled Yeast–Neutral Red Suspension section of the data table.

The yeast suspension was initially prepared in sodium bicarbonate solution and was therefore basic. The suspended cells however were red due to acidic cytoplasm. Boiling altered the permeability of the yeast cell membrane, allowing the sodium bicarbonate to enter the cells and causing the cells to become basic. The neutral red stain turned yellow in the presence of the base.

3. Based upon the observations in the Yeast–Neutral Red Suspension with Bases section of the data table, is there any evidence for transport of the bases across the yeast cell membrane? Explain.

Transport of the ammonium ion into the yeast cells caused the neutral red to turn yellow in the presence of the base.

4. Discuss the structure and function of the following components of eukaryotic cell membrane.

Protein Denaturation 

1. Like most proteins, casein has both acidic (−CO₂H) and basic (−NH₂) side chains in its structure. What is the overall charge on a protein at a) high pH when the acidic side chains are ionized; and b) at low pH when the basic side chains are protonated? c) What is the relationship between the charge on a protein and its solubility in water? Explain.

Proteins are a) negatively charged when the acidic side chains are ionized to −CO₂⁻ groups; and b) positively charged when the basic side chains are protonated to −NH₃⁺ groups. c) In general, proteins will be most soluble when they are either positively or negatively charged due to hydration or solvation of the ionic groups with water molecules. Proteins will be least soluble when they have a net charge of zero, because large nonpolar polymer molecules will be hydrophobic and insoluble in water.

2. The isoelectric point of a protein is defined as the pH at which the protein has a net charge of zero. Predict the approximate isoelectric point of casein based on the results of this demonstration.

Casein was most soluble in water at pH <2 and pH >10. It precipitated out to give a very cloudy solution at pH 4−5. Since proteins are least soluble when they have no net charge, the isoelectric point of casein must be approximately pH 4−5.

3. Most enzymes are proteins. The pH value at which an enzyme is most active is called its optimum pH. Using the basic model of enzyme function, explain at least two main ways in which structural changes accompanying pH changes can influence the function of enzymes.

Enzyme–substrate binding is the first step in the basic model for how enzymes work. Substrate binds to the active site in the enzyme, and further reaction of the substrate within the active site is facilitated by acidic and basic side changes that are nearby in the overall three-dimensional or tertiary structure of the enzyme. Changes in pH affect the ionization of side chains, which in turn can alter the interactions between amino acids that give rise to the unique three-dimensional folded structure of the protein or enzyme. Changing the tertiary structure may alter the active site so that is no longer accessible for binding to the substrate or the substrate does not bind as strongly. If nearby side chains are no longer charged, they may not be able to catalyze the reaction of the substrate within the enzyme–substrate complex.

4. Describe the general structural features of fibrous and globular proteins, give one example of each, and relate the unique functional role of the protein to its structure.

Fibrous proteins have rigid, elongated structures defined primarily by the secondary structure, either alpha helix chains or beta-pleated sheet, of the polypeptide chain. The helical chains may further intertwine, like strands of yarn, to form very strong fibers. Examples of fibrous proteins include collagen and keratin. The strength of protein fibers or sheets makes fibrous proteins well-suited to structural roles within the body—in connective tissue (skin and nails), hair, muscles, etc. Fibrous proteins tend to be insoluble in water or in cytoplasm. Globular proteins have highly folded, compact, spherical three-dimensional tertiary structures stabilized by intermolecular forces between amino acid side chains. Many globular proteins also have several subunits joined together in a stable quaternary arrangement. Most enzymes, as well as receptor and transport proteins, are globular proteins. The unique compact structures of globular proteins give rise to a variety of binding sites for small molecules for catalytic activity or transport to take place. Hemoglobin is an example of a globular protein. It binds oxygen molecules and catalyzes their transport. Globular proteins tend to be soluble in water so that they can be carried throughout the bloodstream.

Discussion

Isolating DNA

Deoxyribonucleic acid (DNA) is considered the molecular “blueprint” which the body uses for creating new proteins. All organisms from bacteria to plants to animals have DNA. DNA is normally found in the cell nucleus where it is complexed with proteins to form chromosomes. The structure of DNA is universal—it is a biological polymer consisting of multiple repeating units with three main components: deoxyribose (a sugar), nitrogen bases, and phosphate groups. One deoxyribose molecule bonds to one of four nitrogenous bases to form a nucleoside. The deoxyribose unit of the nucleoside complex bonds with a phosphate group forming a completed nucleotide. The phosphate of one nucleotide bonds with the deoxyribose of another nucleotide forming a repeating chain of nucleotides. The nitrogenous bases of the polynucleotide chain are hydrophobic, whereas the phosphate groups are ionic. The polynucleotide chain of DNA has directionality conferred by the deoxyribose sugar. DNA tends to form a double strand with opposing strands forming the two halves. It is the pairing of the nitrogen bases which confers stability to the macromolecule.

Four nitrogen bases are present in DNA—adenine, cytosine, thymine and guanine. The bases on opposite strands in the double-stranded structure of DNA are complementary, meaning that adenine only pairs with thymine and cytosine only pairs with guanine in DNA. The base pairs are stabilized by hydrogen bonding between the complementary bases. This specific pairing assists in creating the famous helical structure of DNA, known as a double helix.

Within the double helix model of DNA, the phosphate–sugar groups appear to make up the “backbone” of the structure, and the nitrogen bases project inward and appear to be “stacked” on top of one another within the interior. The double-helix structure of DNA provides an obvious explanation for the essential functions of DNA, namely the process of both replication and transcription. During replication, the process by which DNA is duplicated for cell division, the two strands of DNA separate and each serves as a template for the synthesis of a complementary strand. The process of messenger RNA synthesis, transcription, occurs on one strand of unwound DNA, referred to as template strand. An enzyme moves along the template DNA strand, joining together nucleotides that are complementary to the bases on the DNA template.

Permeability of Cell Membranes

Proteins are an important component of the cell membrane. Carrier proteins and ion channel proteins are integral proteins that act as channels allowing ions, like chloride, and large polar molecules, like glucose, to completely cross the phospholipid bilayer. These channels may selectively close as necessary based upon the cell’s requirements. Other proteins act as part of the active transport system to move specific molecules across the phospholipid bilayer. Additional proteins function as receptors which bind molecules, like hormones. Hormones provide the interior of the cell with information about the external environment of the cell. The position of a membrane protein depends upon its function. Receptor proteins are peripheral proteins—they adhere to the outer or inner surface of the phospholipid bilayer. Integral proteins that help transport molecules across the phospholipid bilayer cross through the bilayer with ends that often protrude into the cell or out from the cell (see Figure 3).

Yeast cells are used in this experiment. Yeast cells are hardy, unicellular, eukaryotic organisms that divide quickly. Since yeast cells are eukaryotes they contain the same organelles as humans and their cell membranes function in a similar way. One difference between human cells and yeast cells is the pH of their cytoplasm. Human cytoplasm has an average pH of about 7.4. Yeast cytoplasm has a pH of about 5.8. In order to study the transport of molecules across the cell membrane, the cytoplasm must be stained with an indicator stain that does not immediately kill the cell. Stains that do not immediately kill living cells and unicellular organisms are called vital stains. Neutral red is a vital stain that also acts as an acid–base indicator. Acid–base indicators are chemicals that change color based on the pH of the substance to which they are added. Neutral red appears red in solutions with a pH less than 6.8. However, neutral red appears yellow in solutions with a pH greater than 8.0.

Protein Denaturation

Proteins are composed of amino acid molecules joined together in chain-like fashion via peptide linkages. Amino acids are thus often referred to as the “building blocks” of protein structure. The number of amino acids in a single protein can vary from around 50 amino acid residues in insulin to more than 500 in hemoglobin and more than 5,000 in some viruses. When fewer than 50 amino acids are joined together, the resulting compounds are called polypeptides.

All amino acids have two structural groups in common—they contain a carboxylic acid group (–COOH) on one end and an amine group (–NH₂) on the other end. Peptide linkages are created when the carboxyl group of one amino acid reacts with the amino group of the next amino acid in the sequence. As each amino acid is added to the growing polypeptide chain, a molecule of water is formed as a byproduct, as shown in Figure 4.

All proteins are made from about 20 different, naturally occurring amino acids, which can be arranged in an almost infinite number of ways. The primary structure of a protein is determined by the number and identity of amino acids and the order in which they are joined together. Higher levels of protein structure (called secondary, tertiary and quaternary structure) result as the polypeptide chains form ribbons, sheets and coils that then fold in on themselves to form more stable three-dimensional arrangements.

In addition to these reactive amine and carboxylic acid functional groups, amino acids contain a third group of atoms, called the side chain (shown as “R” groups in Figure 4). Although not involved in peptide bond formation, the side chains may contain other functional groups that influence both the structure and function of proteins. Hydrophobic amino acids contain nonpolar (“water-fearing”) side chains, such as large hydrocarbon groups. Protein molecules often fold in on themselves so that the hydrophobic amino acids are tucked away in the interior of the structure. This reduces unfavorable contact between the nonpolar side chains and polar water molecules within cells. Amino acids are classified as hydrophilic if they contain polar (“water-loving”) side chains that are able to form strong hydrogen bonds. Hydrophilic amino acids are often found at the “active” sites in enzymes, where they bind to small molecules and catalyze chemical reactions. Finally, ionic amino acids contain extra acidic and basic groups in their side chains; at physiological pH these side chains exist in charged, ionic forms. Oppositely charged side chains form so-called “salt bridges” that stabilize the three-dimensional structure of proteins.

Casein is the principal protein in milk (80% of the total protein content). Casein, like other proteins, is an ionic species containing amino groups and carboxyl groups on its terminal amino acids. It also contains a variety of other acidic and basic groups on the side chains of its non-terminal amino acids. Casein is a phosphoprotein—it contains a large number of phosphate groups attached to the amino acid side chains in its polypeptide structure. The negatively charged phosphate groups are balanced by positively charged calcium ions and are responsible for the highly nutritional calcium content in milk. Casein is almost completely insoluble in water at neutral pH (pH = 7). The effect of pH on the solubility of casein reflects the ionization of the acidic and basic groups in its structure. At high pH, casein will have a net negative charge due to ionization of all the acidic side chains (—CO₂^–) in its structure. Because casein is ionized at high pH values, it is soluble in dilute sodium hydroxide solution. At low pH, casein will have a net positive charge due to protonation of all basic side chains (—NH₃⁺) in its structure. Because casein is ionized at low pH values, casein is also soluble in strongly acidic solutions. At intermediate pH values, casein will contain roughly equal numbers of positively and negatively charged groups and the protein will have a net charge of zero. Casein is insoluble in neutral solutions because it is not charged under these conditions. The solubility of a protein is usually at a minimum at its isoelectric point. The isoelectric point is defined as the pH at which a protein has a net charge of zero. For casein, due to the attached phosphate groups, the isoelectric point is close to pH = 4.