Chemistry 301 Introduction to Biochemistry

Study Guide :: UNIT 3

Structure and Function

Overview

To understand how biological molecules function, it is important to understand their structure. Generally, molecules are classified based on their solubility in water. Hydrophobic molecules are not soluble in water, while hydrophilic molecules are. Amphiphilic molecules contain both hydrophobic and hydrophilic residues.

The main groups of biological molecules are proteins, nucleic acids, carbohydrates, and lipids. Each lesson in this unit will introduce the structures of these molecules and their general functions. The unit will also cover the structure of membranes and how transport of molecules across membranes occurs. Finally, the last lesson will cover the structure and function of vitamins, which play important roles in cellular processes

Amino acids are the building blocks that make up proteins. There are 20 common amino acids, and they can be joined together, much like beads on a string, to form proteins. Proteins are the “workers” in the cell. There are thousands of proteins, because amino acids can join together in various combinations and to almost any length. Lesson 1 will cover amino acid structure, ionic behaviour, and the structures and functions of proteins.

DNA and RNA are polymeric molecules, as are proteins and carbohydrates. The monomer units are deoxyribonucleotides and ribonucleotides respectively. Each deoxyribonucleotide or ribonucleotide consists of a sugar (deoxyribose or ribose), a phosphate group, and a nitrogenous base. There is less chemical variety in DNA or RNA than in proteins, because there are 20 common amino acids, and only 4 deoxyribonucleotides and 4 ribonucleotides. In Lesson 2, we will introduce the biochemical structure of genetic materials, and the biochemical processes involved in the creation of proteins from a genetic blueprint.

A simple sugar is an n-carbon polyhydroxyl chain (or ring), which also contains an aldehyde or ketone functional group. The value of n ranges from 3 to about 9, but 5-carbon sugars (pentoses) and 6-carbon sugars (hexoses) are the most common in biological systems. While amino acids are linked by peptide bonds, simple sugars are linked by ether bonds; and just as amino acids link to form proteins, simple 6-carbon sugars link to form polysaccharides. The 6-carbon sugars are the only ones that serve as building blocks for polysaccharides. Simple sugars are partially reduced molecules, and are the major biochemical fuel. Lesson 3 will explore the structure and chemistry of simple and complex sugars.

Lipids are a far more heterogenous class of biomolecules than are proteins, carbohydrates, or nucleic acids, but lipids do have some common features; for example, they are virtually insoluble in water, and generally have one or more unbranched hydrocarbon chains 10–20 carbons in length.

Lipids also have a polar or ionic functional group attached to one end of a hydrocarbon chain. This structure allows lipids to form organized aggregates with the carbon-hydrogen chains stacked together (excluding water) and the polar head groups interacting with water molecules. These lipid aggregates (membranes) can efficiently separate the aqueous compartments of the body.

Chemically, lipids differ from other biomolecules in that they are composed of reduced hydrocarbons. This property means that per unit mass, more chemical energy is available from lipids than from, say, carbohydrates.

Lesson 4 focuses on the structure and function of lipids and Lesson 5 will cover membrane structure and transport. Finally, in Lesson 6 we will consider vitamins—organic molecules that are not proteins, or carbohydrates, or lipids, but that are crucial to biological function. Vitamins may supply substances that are needed directly by the body, or they may function as “cofactors,” working with other molecules to induce chemical reactions.

The unit is divided into six lessons:

Learning Objectives

After completing this unit, you should be able to:

1. Differentiate between hydrophobic, hydrophilic, and amphipathic molecules.
2. Describe the structure of the following biological molecules: proteins, carbohydrates, lipids, and nucleic acids.
3. Identify and draw the structures of the 20 amino acids, and explain how amino acids are joined to form a protein.
4. Describe how the physiochemical properties of the amino acids determine the three-dimensional structure of a protein.
5. Demonstrate an understanding of the four levels of protein structure and the roles proteins play in biological systems.
6. Describe the chemical structures of a simple carbohydrate (sugar) and a complex carbohydrate (polysaccharide).
7. Explain how polysaccharides are synthesized from simple 6-carbon sugars.
8. Describe the roles of structural polysaccharides and storage polysaccharides, and explain how slight differences in their chemical structures enable them to fulfill their different roles.
9. Discuss the structures and functions of glycoproteins and proteoglycans.
10. Describe the structure of bacterial cell walls.
11. Identify the different classes of lipids and state their function.
12. Define “lipoprotein,” and explain their structure and function.
13. Describe the structure of biological membranes.
14. Discuss the dynamic nature of membranes, and describe some membrane transport mechanisms.
15. Identify the basic chemical structures of the nucleic acid bases, nucleotides, and nucleosides (both ribo- and deoxyribo- forms)
16. Describe the flow of genetic information (DNA to proteins)
17. Name the 3 types of RNA
18. Define the terms “chromosome,” “gene,” “virus,” and “plasmid.”

Glossary

A good definition identifies the thing described, and says something about where it is found and how it works. Each of the following definitions is incomplete. As you work through this and later units, you should develop a complete definition for the terms in each glossary.

Lesson 1: Proteins

Overview/Objectives

After completing this lesson, you should be able to:

1. Identify the 20 common amino acids.
2. Describe and draw a peptide bond.
3. Discuss how 20 amino acids can produce thousands of proteins with very marked differences in their physicochemical properties.
4. Determine the predominant ionic forms of each amino acid at pH 1, pH 7, and pH 11.
5. Define primary, secondary, tertiary, and quarternary protein structure.
6. Explain how a comparison of the primary structure of homologous proteins is used to determine both evolutionary relatedness and the importance of the sequence.
7. Explain why proteins evolve at different rates, even though mutation rates are constant with time.
8. Name the major features of secondary protein structure, and explain the role of H-bonds.
9. Explain why glycine and proline occur at positions where a polypeptide chain makes a reverse turn.
10. Describe how the physicochemical properties (solubility, polarity, charge, size, etc.) of its amino acids determine the tertiary structure of a protein.
11. List the forces that play a role in stabilizing tertiary structure.
12. Describe how X-ray crystallography is used in determining the tertiary structure of a protein.
13. Demonstrate an understanding of protein denaturation and list the conditions required.
14. Explain Chou and Fasman’s method to predict three-dimensional protein structure.
15. Explain how protein families can evolve from a single ancestor.
16. List the major role proteins play in vivo.

Readings and Activities

1. Read from “Proteins” to and including “Prions and Misfolding” (pages 43–58 of the textbook).

2. You can also view five video lectures:

   Amino Acids:

   • #03 Biochemistry Amino Acids Lecture for Kevin Ahern’s BB 450/550

     (A link to this is also provided on page 42 of the textbook.)

   Protein Structure:

   • #04 Biochemistry Protein Primary/Secondary Structure Lecture for Kevin Ahern’s BB 450/550

   • #05 Biochemistry Protein Tertiary/Quaternary Structure Lecture for Kevin Ahern’s BB 450/550

     (Links to these are also provided on pages 44 and 47 of the textbook.)

   Hemoglobin:

   • #08 Biochemistry Hemoglobin Lecture for Kevin Ahern’s BB 450/550

   • #09 Biochemistry Hemoglobin II/Enzymes I Lecture for Kevin Ahern’s BB 450/550

     (Links to these are also provided on page 52 of the textbook.)

3. To learn the structures of the 20 common amino acids, consult the Wikipedia page Amino acid.

Commentary

Amino Acids

The general structure of an α-amino acid can be drawn as follows:

Note: In the diagram above, the “R” denotes one of several different side chains. The R group is a distinguishing feature of each amino acid.

At pH 7, both the amino group and the carboxylic acid group are charged, so an amino acid is usually drawn in this way:

This type of molecule, in which there are opposing charges on either side, is known as a “zwitterion.”

The α-C has four different groups attached ($\ce{-}$H, $\ce{-}$R, $\ce{-}$NH₃⁺, $\ce{-}$COO⁻), and thus can exist as two different enantiomers (mirror images). Only the “L” enantiomer is found in proteins. It is often difficult to picture the spatial arrangement of an amino acid without using “ball-and-stick” molecules, as shown on page 43 of the text.

Several conventions have been developed to give a sense of the spatial organization of molecules when they are presented on paper. For example, in the structures below, the α-C is considered to be in the plane of the paper. The dotted lines indicate that $\ce{-}$COO⁻ and $\ce{-}$R are below the plane of the paper; the wedge-shaped lines indicate that $\ce{-}$NH₃⁺ and $\ce{-}$H are above the plane of the paper.

Twenty different R groups are found when the bulk of known proteins are analyzed, and four or five other R groups are found in selected proteins of specialized function. (Note: “R” is a term used in organic chemistry to represent the side-chain of a group. In organic chemistry, R refers to an alkyl functional group. In biochemistry, it has a broader meaning, referring to any small organic functional group attached to the α-C of an amino acid.) The R groups can be nonpolar or polar. Polar R groups can be charged.

As proteins are synthesized, the carboxylic group of one amino acid is bonded to the amino group of a second amino acid, and water is eliminated. The new bond (circled in the diagram below) is called a peptide bond.

When two amino acids are joined by a peptide bond, the result is called a “dipeptide.” Three amino acids joined by peptide bonds form “tripeptides.” When three to ten amino acids are linked, the structure is called an “oligopeptide,” and linkages of more than ten amino acids form “polypeptides.”

Note: The amino end of the polypeptide chain is called the “N-terminal residue,” and the carboxyl end is called the “C-terminal residue” (see page 44 of the text).

It is important that you remember that peptide bonds are made only one way: by linking amino groups and carboxylic acid groups that are attached to the α-C. Thus, in the long chain that we call a protein, the R groups stick out. Four amino acids contain either an amino group or a carboxylic acid group as their R-group (lys, arg, and glu, asp respectively). Peptide bonds are never made using an amino group or a carboxylic acid group in the R group of these amino acids. To repeat: a peptide bond is made only through the α-amino and α-carboxylic acid groups.

Proteins: Primary Structure

Amino acids are linked together by covalent peptide bonds to form proteins. Because of the charged nature of amino acids, the long polymer chain (the protein) will fold into a compact shape that is quite stable. In discussing the three-dimensional shape of a protein, we speak about its primary, secondary, tertiary, and quaternary structures. This lesson addresses the primary structure of proteins; you will see how proteins form globular molecules in subsequent lessons.

• “Primary structure” refers to the linear sequence of amino acids; for an example, see the figure on page 45 of the text. Each chain of the protein molecule consists of a line of amino acids joined by peptide bonds.
• “Secondary structure” refers to regular, easily recognized arrangements of the protein backbone; for example, α-helices (corkscrew shapes) and β-pleated sheets (zigzag shapes).
• “Tertiary structure” refers to the three-dimensional structure of an entire polypeptide.
• “Quaternary structure” describes the spatial arrangement of subunits in a protein with more than one polypeptide chain.

With one exception (discussed later), the forces that stabilize a protein’s secondary, tertiary, and quaternary structures are noncovalent: H-bonding, hydrophobic forces, ionic interactions, and van der Waals forces. We look at these bonds in more detail in subsequent lessons.

Earlier in this commentary, a peptide bond was shown to be a single bond between C and N. This representation is not strictly true. The peptide bond has some double bond character (remember electron delocalization from your organic chemistry course?) and can be drawn as shown below.

Because of the partial double bond character, completely free rotation around the peptide bond is not possible. However, adjacent amino acids will twist with respect to each other to minimize R-group interactions. These factors contribute to the rigid structure of peptides and to a natural α-helix. The twisting of peptide bonds and their relative rigidity build up secondary structures in proteins.

There are a number of chemical methods for analyzing and synthesizing proteins. It is important to think about why protein analysis and synthesis are so important in biochemistry and current research. Unit 9 will present methods for protein analysiss, but below we will discuss how these methods contribute to understanding how protein sequence can determine the relatedness of organisms and how they have evolved.

These techniques are powerful tools that enable us to determine the evolutionary history and relatedness of species. This determination can be made if the primary structures of a protein (common to all members of the species) are known. The phylogenetic tree derived from cytochrome c illustrates this technique. To construct a phylogenetic tree, a scientist determines not only how many amino acids differ between species, but also the likelihood that an amino acid mutated to a different amino acid directly. The analysis is typically done on a small protein. It would be useful to build several phylogenetic trees, based on several different proteins, for cross checking. Unfortunately, few proteins other than cytochrome c are sufficiently small and widespread.

Phylogenetic trees determined by protein analysis are not the be-all and end-all of evolutionary studies. The number of times a protein is expressed is at least as important as its amino acid sequence in determining the biochemical properties of a cell.

3-D Structure of Proteins I: Secondary Structure

What we call “secondary structure” in a protein consists of short stretches of the helical or zigzag structure of the protein backbone. These features (α-helices and β-sheets) arise because the R-groups in these regions of the protein are small enough not to interfere with the “natural” shape of the protein backbone. These features are both stabilized by H-bonds. Think back to your organic chemistry course. Large molecules adopt shapes that minimize the interactions between neighbouring atoms. Amino acids (and, hence, proteins) are made up of atoms that are roughly the same size: carbon, oxygen, and nitrogen. When amino acids are joined together in a peptide bond, the groups attached to the α-C will twist away from each other to minimize atomic interactions.

However, if minimization of atomic interactions were the only important force, all proteins would be composed of α-helices and would be as regular as DNA. As you know, DNA exists as a double helix—it is one of the most regular macromolecular structures known. It arises because the groups attached to the DNA backbone are all roughly the same size, shape, and charge, and so twist around the DNA backbone in a regular pattern to minimize atomic interactions.

β-sheets in proteins arise when the R-groups are very small (hydrogen or methyl) and uncharged. In β pleated sheets, H-bonds occur between polypeptide chains. (Remember that in α-helices, H-bonds occur within one polypeptide chain.) See page 46 of the text for a figure of a β-sheet.

Think about how the features of secondary structure might be important to the function of a protein. Consider the following two examples: silk fibroin (β-sheets) and collagen (α-helix). Silk fibres are strong because they cannot be stretched without breaking the covalent bonds of the backbone. However, silk fibres are also flexible, because neighbouring β-sheets associate only through van der Waals forces. Collagen has a rigid, triple-helical structure that confers great tensile strength. However, some collagens, like that in the Achilles tendon, can be stretched, because the α-helix is extensible. In an α-helix, stretching breaks only H-bonds, not covalent bonds. A schematic of collagen, a fibrous helical protein, is on page 46 of the text.

In summary, then, H-bonding and minimization of steric crowding yield the “natural” protein shape: α-helices. If the amino acid R-groups are very small, H-bonding forces between peptide chains can yield β-sheets.

3-D Structure of Proteins II: Tertiary and Quaternary Structures

Tertiary structures arise because some amino acid R groups are very large or are strongly charged, and thus interfere with the “natural” α-helix or β-sheet. Tertiary structures also form because of the hydrophobic (water-hating) or hydrophilic (water-loving) nature of the R-groups. Charged or polar R-groups (hydrophilic) are found most often on the outside surface of a protein. (Remember at pH 7 that amino groups and carboxylic acid groups are charged. Therefore asp, glu, lys, and arg will all have ionic R-groups at physiological pH.) Nonpolar R groups (hydrophobic) are found most often in the water-free interior of a protein.

Very few proteins possess only α-helices or β-sheets. Most do contain some helices and β-sheeting, but their overall shape can be called “globular”; that is, they consist of compact egg shapes. The overall shape (which includes α-helices, β-sheets, and seemingly disorganized regions) is called the “tertiary structure” of a protein. The tertiary structure comes about through

• H-bonding between various parts of the peptide chain
• hydrophobic interactions between the nonpolar R-groups to exclude water
• interactions between ionic and polar R-groups and water (i.e., forming the outside surface of the protein)
• the formation, by two cysteines, of a disulphide (covalent) bond which holds two parts of the peptide chain together. (If you have ever had a “perm” in your hair, you chemically broke existing disulphide bonds in hair proteins by using a reducing agent, and then formed different disulphide bonds using a neutralizing solution.)

Note that some proteins contain “cofactors” (the heme group, a divalent metal ion, or a small, non-amino acid organic group). Cofactors are discussed in more detail when enzymes are presented; they are mentioned in this lesson because their bonding to a protein (which may be covalent, ionic, or hydrophobic) also helps to determine the three-dimensional structure of a protein. Most of the information we have about protein structure comes from X-ray crystallography or nuclear magnetic resonance spectroscopy. A figure of the structure of myoglobin showing its tertiary structure is on page 48.

“Quaternary structure” describes the shape of a protein that consists of more than one polypeptide chain. In this case, there is usually an even number of polypeptide chains (two, four, or six), which are referred to as “subunits.” The forces holding the subunits together are noncovalent. Hemoglobin is an example. Hemoglobin consists of two sets of two identical polypeptide chains: α₂ β₂. Quaternary structure (held together by H-bonds, hydrophobic interactions, and in some cases disulphide bonds) can usually be disrupted by milder solution conditions than can tertiary structure. A figure illustrating the quaternary structure of hemoglobin is on page 52 of the text.

“Protein denaturation” refers to the unfolding or disruption of the tertiary and quaternary structures of proteins. The three-dimensional structure of a protein is a balance of H-bonds, hydrophobic interactions, and association with water. Anything that disrupts this balance (increased temperature, change in pH, change in solvent, or violent agitation) promotes denaturation. Using isopropanol (change of solvent) to clean the skin before injections denatures bacterial proteins—you hope. Boiling an egg denatures the albumin in the egg white. Beaten egg whites (meringues) are also denatured albumin. Raw fish soaked in lemon juice also displays denatured proteins. Denaturation and renaturation of ribonuclease is shown in the figure on page 56 of the text.

Protein Folding and Protein Dynamics

Suppose we know the primary structure of a protein, but do not have access to X-ray crystallography. Can we say anything about the secondary and tertiary structures? Yes, we can. Knowing the primary structure of a protein allows us to make rough predictions about the secondary structure. The most reliable method of making such predictions was devised by Peter Y. Chou and Gerald D. Fasman (1925–2003). They analyzed the structures of a large number of proteins whose X-ray crystal structures were known, and determined how frequently a given amino acid occurred in an α-helix, and how often in a β-sheet. When their classification scheme is applied to a protein of unknown three-dimensional structure, the secondary structure of the unknown protein can be predicted with reasonable accuracy.

Earlier in this unit, we noted that the structure of a protein is considered to be relatively rigid. This statement is both true and untrue. It is true that a protein will maintain a given shape, because all the bonding forces and the hydrophilic/ hydrophobic interactions are in equilibrium with the environment of the protein. (This statement means that if the salt concentration, pH, and temperature of a solution do not change, then the shape of the protein in this solution will not change either.) However, on a molecular level, there will always be minor changes in the environment of the protein (the “microenvironment”). Other molecules will interact with the protein, and the pH will change locally as a result of neighbouring reactions. Therefore, a protein will undergo minor and continuous changes in shape (these changes are called conformational flexibility or “breathing”).

Protein Function

Proteins vary in mass from about 10³ daltons (insulin) to 10⁶ daltons (immunoglobulins). Most proteins are water-soluble, because of the charged and polar R groups on their exterior surface. Most proteins are negatively charged at pH 7.0, because, in general, proteins contain more acidic than basic R groups. However, some proteins are also found in the non-aqueous membrane part of the cell. Because the more than 20 amino acids can be assembled in almost any combination to make up proteins, you can see that proteins are a very large and heterogeneous group, and in fact, proteins have many different functions. The following paragraphs provide a description of the major roles proteins play in vivo:

• Some proteins (called enzymes) act as catalysts. Virtually all reactions in the body are controlled by enzymes. For an idea of how efficient enzymes are, think about a time when you were suffering from “low blood sugar,” and how quickly you revived when you drank orange juice or ate a candy bar. Hundreds of reactions, all enzyme controlled, are involved in turning the sugar in the juice or candy bar into that feeling of well-being.
• Enzymes frequently contain sections of primary structure that are virtually identical across all species. These structures, known as “invariant regions,” usually are the portion of the enzyme which is the catalytic site.
• Because enzymes are such an important class of proteins, we discuss them in more detail in Unit 4.
• Hormones, which regulate enzyme activity, fall into two classes: proteins make up one of these classes; the other is based on cholesterol, a lipid.
• Growth factors are a group of protein molecules that induce growth and differentiation of specific cells.
• Some proteins, usually proteins with regularly repeating amino acid sequences, act as structural components. Silk fibre and collagen are examples of this class.
• Proteins act as transport molecules (e.g., the oxygen carriers, hemoglobin and myoglobin), and as storage molecules (e.g., the iron-containing protein, ferritin).
• The members of a globular class of proteins, called immunoglobulins, which constitute part of the immune system, recognize and bind foreign molecules for eventual removal from the body.
• Proteins make up the contractile elements involved in the movements of all non-stationary life forms; muscles are an example of such contractile elements.
• Positively charged proteins, called histones, closely surround the negatively charged DNA molecules to protect DNA from degradation and inappropriate reactions with other biomolecules. They also play an important role in the “packaging” of DNA.
• A very diverse, non-histone class of proteins interacts with DNA to inhibit RNA production, enhance RNA production, indicate where RNA production is to begin and to end, repair errors in DNA, and in general control all aspects of DNA action.

Study Questions

1. Draw the structure of each of the 20 amino acids and fill in the following table with the proper abbreviation for each.

   Amino acid	Abbreviation	Letter
   alanine
   arginine
   asparagine
   aspartic acid
   cysteine
   glutamic acid
   glutamine
   glycine
   histadine
   isoleucine
   leucine
   lysine
   methionine
   phenylalanine
   proline
   serine
   threonine
   tryptophan
   tyrosine
   valine

2. What is the R group in an amino acid basic structure?
3. Compare the arrangement of hydrophobic and hydrophilic amino acids in a protein in an aqueous environment to that of a membrane.
4. As you will see in the final lesson of this unit, a protein depends on its tertiary structure to perform its functions. A mutation in the interior of a given protein inactivated the protein. This mutation is ala → val. A second mutation, ile → gly, restored the activity of the protein. Explain these results.
5. Explain why proline is not common in α-helices, but can be found frequently in collagen.
6. Identify the six most abundant amino acids in proteins.
7. Why it is important to know the protein sequence in addition to the DNA sequence of a particular gene?
8. Wool consists of α-keratin. Pure wool sweaters exposed to hot water shrink alarmingly. After shrinkage, it is virtually impossible to stretch the sweater back to its original shape. What do you think is the biochemical cause of this phenomenon?
9. Describe the hydrogen-bonding pattern of an α-helix.
10. List the advantages of multiple subunits in proteins.
11. Explain how cooperativity between haemoglobin and oxygen is achieved.
12. Describe the energy and entropy changes that occur during protein folding.
13. Antibodies are proteins that recognize and help to inactivate foreign proteins (called antigens). Why do such antibodies rarely react to denatured proteins? A few antibodies do recognize denatured proteins. How can this occur?
14. What information is provided by comparing the sequences of proteins from different organisms?
15. How is a phylogenetic tree constructed?
16. Which side chains normally occur on a protein’s interior? On its surface?
17. Describe the forces that stabilize proteins.
18. List and describe two functions of proteins.
19. What is an invariant region and what does in mean in terms of enzymes?
20. What is a prion?

Lesson 2: Nucleic Acids

Overview/Objectives

After completing this lesson, you should be able to:

1. Identify the basic chemical structures of the nucleic acid bases, of nucleotides and of nucleosides (both ribo- and deoxyribo-forms).
2. Identify the standard abbreviations for the nucleic acid bases.
3. Describe the flow of genetic information (DNA → proteins).
4. Define the terms “chromosome,” “gene,” “virus,” and “plasmid.”
5. Explain why we know more about bacterial genetics than about mammalian genetics.
6. Name the three types of RNA.
7. Describe the chemical differences between DNA and RNA, and explain why DNA is the more stable molecule.
8. Describe the structure of B-DNA.
9. List the functions of DNA and RNA.
10. Describe the forces stabilizing nucleic acid secondary structures.
11. Explain denaturation and renaturation of complementary strands of DNA, RNA, or both, and define “TM.”
12. Describe the secondary structure of tRNA.

Readings and Activities

1. Read “Nucleic Acids,” “Superhelicity,” “RNA structures,” and “Denaturing Nucleic Acids” (pages 58–64 of the textbook).

Commentary

DNA and RNA are polymeric molecules, as are proteins and carbohydrates. The monomer units are deoxyribonucleotides and ribonucleotides. Each deoxyribonucleotide or ribonucleotide consists of a sugar (deoxyribose or ribose), a phosphate group, and a nitrogenous base. There is less chemical variety in DNA or RNA than there is in proteins, because there are 20 common amino acids, and only 4 deoxyribonucleotides and 4 ribonucleotides.

There is hydroxyl group at position 2 on ribose (2′-hydroxyl). This hydroxyl group chemically destabilizes RNA. This is relevant because RNA can exit the nucleus and send messages and make proteins and then be degraded when it is not needed.

DNA lacks a 2′-hydroxyl group. The absence of the OH group in DNA keeps DNA more stable and prevents it from being degraded. DNA remains in the nucleus.

The nitrogenous bases (N-bases) found in DNA are adenine (A), guanine (G), thymine (T), and cytosine (C).

The nitrogenous bases found in RNA are adenine, guanine, cytosine, and uracil (U). A figure showing how complementary bases pair with each other is on page 59 of the text. All of the nitrogenous bases are planar molecules, as are ribose and deoxyribose. The plane of the nitrogenous base is at right angles to that of the sugar. You can picture this by thinking of ribose as a boat and the nitrogenous base as a sail. When a deoxyribonucleotide or a ribonucleotide is free in solution, the base sits above the plane of the sugar, making a compact molecule (normal sailboat). When a deoxyribonucleotide or a ribonucleotide is found in DNA or or in RNA, the base points away from the sugar.

By pointing away from the sugar moiety (and thus also away from the phosphate group), the nitrogenous base is in position to hydrogen bond to another base. This positioning is the foundation of the double helical structure of DNA.

In DNA and RNA, the deoxyribonucleotides (and ribonucleotides) are linked together through phosphate-sugar ester bonds. Thus, DNA and RNA have sugar-phosphate backbones with nitrogenous bases extending away from the backbone. If you look at the figure below you can see that DNA is anti-parallel.

In complex forms of life and in many viruses, the flow of genetic information is as follows:

DNA → transcription RNA → translation proteins

That is, proteins are coded by RNA molecules and are produced under the control of RNA. RNA is always produced from a DNA template, except in RNA-viruses (also called “retroviruses”). These viruses contain reverse transcriptases, enzymes that use the viral RNA template to code for DNA. An example of a retrovirus is HIV.

DNA is relatively inert chemically; furthermore, repair mechanisms exist to help preserve the original structure of DNA. DNA molecules vary in length depending on the organism in question. However, DNA molecules are relatively large compared with all other biomolecules.

Most RNA molecules are very labile, with lifetimes of seconds to minutes. The exception is rRNA, which is complexed with proteins and is reasonably long lived.

Three types of RNA exist:

• messenger RNA (mRNA) molecules are the RNA species that carry the codes for protein primary structures (i.e., amino acid sequences).
• transfer RNA (tRNA) molecules bind specific amino acids as the first step in protein synthesis.
• ribosomal RNA (rRNA) molecules bind both mRNA and tRNA. Ribosomes are cellular structures consisting of rRNA and proteins; they are the platforms for protein synthesis.

One DNA molecule (i.e., one double helix) in association with certain proteins is called a chromosome. An organism may have only one chromosome (e.g., the bacterium E. coli), or it may have many chromosomes (e.g., human beings, who have 46). A gene is the portion of the chromosome that codes for one protein. Genes may follow one another closely on a chromosome or, more commonly, they may be separated by long sequences of bases. The intervening sections of DNA between genes are usually control points. These control points regulate how readily a gene is transcribed into RNA. The term “genome” refers to the total mass of DNA in a cell.

In addition to one or more chromosomes, some bacteria contain small circular DNA molecules called “plasmids.” As are chromosomes, plasmids are reproduced when a bacterium divides. In nature, plasmids can be transferred from one bacterium to another. Plasmids are probably the most important tool molecular geneticists have. Isolated genes can be inserted into plasmids. Small and relatively easy to manipulate, plasmids can then carry the new genetic information to a large population of bacteria.

Viruses are particles consisting of a capsid (protein coat) and genetic information, either DNA or RNA. Viruses can reproduce only by using the starting materials, reproductive machinery and metabolic energy of a host cell. So these are infective particles, inserting themselves into more biochemically sophisticated hosts. Viruses are studied because:

• with rudimentary structures and few genes, they are simple models for cellular development
• host-parasite relations and evolutionary changes can be more easily studied with viruses than with other life forms due to viral simplicity and rapid reproduction
• they are capable of causing disease and some animal cancers, and so also provide information on normal growth and development, as well as on the controls for these processes.

Bacteria are used extensively in molecular biology. Not only do they have simpler genomes than mammals, but they also reproduce rapidly. Thus, they are easy to study. Using plasmids, researchers have inserted many mammalian genes into bacteria for study. Drugs (e.g., insulin) can be produced in large amounts by insertion of the genes for mammalian insulin into bacteria. Information on specific mutations, viability, longevity, and production rate of mammalian proteins has been obtained from bacteria. More importantly, the effect of planned mutations on these properties can be studied using bacteria. However there are two drawbacks to using bacteria in this way:

• bacteria cannot glycosylate proteins. Most mammalian proteins are glycosylated in their natural state in vivo.
• the safety factor in inserting foreign genes into rapidly dividing hosts is not completely understood. As more and more powerful viral vectors and plasmid injection techniques are developed, it is theoretically possible to insert any gene into a bacterial host. How dangerous are these transformed cells? Scientists are beginning to understand the effects of foreign DNA, and to be able to engineer vectors to minimize any effect on the host.

We will cover recombinant DNA techniques in more detail in Unit 9.

RNA and DNA differ in these important aspects:

• RNA contains A, U, G, C; DNA contains the nitrogenous bases A, T, G, C.
• RNA is usually single stranded; DNA is double stranded.
• Normally the nitrogenous bases in RNA are chemically modified, while those in DNA are not.
• RNA contains an hydroxyl group at the 2 position of ribose. DNA does not.

The presence of a hydroxyl group destabilizes RNA, because the 2′-hydroxyl of ribose is close to a phosphate group in the phosphodiester backbone of RNA. Nucleophilic attack of the hydroxyl oxygen on the phosphorus atom cleaves the diester backbone.

DNA, lacking a 2′-hydroxyl group, is not prone to this cleavage. Consequently, DNA is far more stable than RNA. The biological advantage of this stability is that DNA, the information storage depot, is preserved from chemical attack. RNA (particularly mRNA) is very readily cleaved. Therefore, one or several copies of a protein can be made from a strand of mRNA, but protein production does not continue indefinitely.

DNA and RNA are linear phosphodiesters. There are nitrogenous bases attached to the sugar portions of the chain. Picture this molecule. The backbone is a negatively charged phosphate-sugar chain. A phosphate group bonds to the bottom of one sugar and the top of the next sugar, like a flight of stairs, with the plane of ribose being the step and phosphate the riser. To minimize steric crowding of the sugars and charge repulsion of phosphate groups, the backbone twists. (Picture a circular staircase.) Now consider the bases. These extend away from the backbone, at right angles to the plane of the sugars.

The linear model above is not the complete structure of DNA. Watson and Crick determined that DNA exists as a hydrogen-bonded dimer, not as a single linear phosphodiester molecule. It is because of the regularity of the sugar-phosphate backbone that the bases are in position to hydrogen bond so effectively to a partner strand. When the two complementary partners come together in double helix DNA, the bases stack regularly. The base stacking contributes a great deal to the stability of the DNA double helix structure.

As two strands of H-bonded DNA twist around each other, two grooves, which differ in size, are formed. These grooves, called the major groove and the minor groove (see the figure “DNA Double Helix” on page 59), allow replication and repair enzymes access to the inner bases.

Note that double-stranding places the bases inside a spiral, and the sugar-phosphate backbone on the outside. Thus, the bases are relatively protected from other reactants in solution. In addition, the phosphodiester backbone is negatively charged. This negative charge will further protect the bases (the crucial information part of DNA) from attack by negatively charged reactants.

The strain-free DNA structure described above is the B-form of DNA. Other structures exist, A-DNA and Z-DNA. All three forms of DNA are illustrated on page 60 of the text.

Real DNA deviates from the ideal B-form because of the sequence of bases within the DNA structure. Therefore, the major and minor grooves widen, decrease, and distort at various places on a chromosome. Hence, replication enzymes will not bind equally to all parts of a DNA duplex (double helix), and therefore, not all genes will be transcribed at the same rate.

RNA normally exists as a single chain. The exception is the RNA virus which can contain double-stranded RNA. (Note the information protection mechanisms discussed just above.)

RNA is produced from a DNA template; that is, from one side of a DNA duplex. The resulting structure will tend to be a regular spiral, as already described. Therefore, it is not surprising that sections of a single RNA strand will fold on themselves and form a duplex structure. This folding will happen when regular base pairing (discussed below) occurs.

tRNA is a good example of this phenomenon. tRNA is a set of single stranded molecules. All tRNA molecules consist of three regions of internal base pairing within the single chain. These regions twist just as the DNA double helix does. The overall shape of tRNA is a “T.” Each of the three arms of the “T” is an area of internal double helix. mRNA molecules are not as symmetrical as tRNA molecules. However, within the known mRNA structures, areas of double helix do occur.

Look back at the structures of the nitrogenous bases. They come in two sizes: pyrimidines, with a single ring structure (T, C, and U); and purines with with a double ring structure (A and G). For efficient base pairing to occur in a double helix, a purine on one side of the helix must be matched with a pyrimidine on the other. In DNA this matching is as follows:

adenine–thymidine
guanine–cytosine

Analysis of the total nitrogenous base content of DNA reveals that:

(A + G) always equals (T + C)

In RNA, U replaces T and so:

adenine–uracil

Study Questions

1. What difficulties can you see in producing proteins by inserting the mammalian genes into bacterial hosts?
2. In an isolated nucleotide/ribonucleotide, the nitrogenous base is positioned directly over the sugar ring. What prevents H-bonding between the two? (Note: Such H-bonding would prevent the base from twisting to H-bond to the partner strand of DNA.)
3. Draw the structures of adenine, cytosine, guanine, thymine, and uracil.
4. List and describe the major features of the Watson-Crick model of DNA structure.
5. What are the structural differences between DNA and RNA?
6. Would you expect DNA and RNA to be water-soluble? Explain.
7. Why would you expect RNA viruses to have double stranded RNA? While many RNA viruses are double stranded, not all are. What comments can you make on this latter class of viruses in light of your first answer?
8. It was stated that (A + G) always equals (T + C) for duplex DNA. Is this restriction also true for single strands of DNA? Justify your answer.
9. What differences would you expect between a duplex DNA which has more A–T base pairs and a duplex DNA which has more G–C pairs?
10. What are the forces that stabilize nucleic acids? What can be used to denature nucleic acids? Can nucleic acids renature?
11. Outline the role of the 2′ hydroxyl group in RNA. What is the reason this group is lacking in DNA?
12. Outline the flow of genetic information in a cell.
13. List and describe the three types of RNA.
14. What is a plasmid?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 3: Carbohydrates

Overview/Objectives

A simple sugar is an n-carbon, polyhydroxyl chain (or ring) that also contains an aldehyde or ketone functional group. The value of n ranges from 3 to about 9, but 5-carbon sugars (pentoses) and 6-carbon sugars (hexoses) are the most common in biological systems. As shown in the figures below, we can draw the structures of simple carbohydrates in linear or cyclic form.

While amino acids are linked by peptide bonds, simple sugars are linked by ether bonds; and just as amino acids link to form proteins, simple 6-carbon sugars link to form polysaccharides. The 6-carbon sugars are the only ones that serve as building blocks for polysaccharides. Simple sugars are partially reduced molecules, and are the major biochemical fuel. (Eventually C, H, and O atoms from the donuts you eat end up as the CO₂ and H₂O you expend.) Besides acting as biochemical fuel, sugars are also found linked to proteins (glycoproteins) and lipids (glycolipids), thereby increasing the water solubility of these molecules. The pattern of sugars linked to individual glycoproteins in cell membranes contributes to immunological identity. In this lesson, we explore the structure and chemistry of simple and complex sugars and then to the biochemistry of sugars; in Unit 6, we discuss how sugars are metabolized. As already discussed in Lesson 2, ribose (a pentose) is an important component of DNA and RNA.

After completing this lesson, you should be able to:

1. Describe the chemical structures of a simple carbohydrate (sugar) and a complex carbohydrate (polysaccharide).
2. Explain the chemistry involved in the cyclization of glucose.
3. Describe the difference between α-D-glucose and β-D-glucose.
4. Explain the difference between the chair and boat conformations of a hexose.
5. Relate the Haworth projection for a given sugar structure to its Fischer projection.
6. Explain how polysaccharides are synthesized from simple 6-carbon sugars.
7. Describe how a polysaccharide is formed from both α- and β-sugars.
8. Explain why polysaccharides are not always linear. (Remember that polypeptides are always linear.)
9. Name the different classes of polysaccharides.
10. List the structural differences between cellulose and glycogen.
11. Compare how the structures of cellulose and glycogen are stabilized, and relate the differences in stability to biological function.
12. Explain the difference between a homopolysaccharide and a heteropolysaccharide.
13. Describe the roles of structural polysaccharides and storage polysaccharides, and explain how slight differences in their chemical structures enable them to fulfill their different roles.
14. Define “glycoprotein,” and explain the roles glycoproteins play in vivo.
15. List the possible effects of glycosylation on the properties of a protein.
16. Describe a proteoglycan, and explain its function.
17. Describe the structure of bacterial cell walls.
18. Explain the action of penicillin and why some bacteria are resistant to the antibiotic.
19. Outline the differences between Gram-negative and Gram-positive bacteria.

Readings and Activities

1. Read “Carbohydrates,” “Monosaccharides,” “Stereoisomer Nomenclature,” “Boat/Chair Conformations,” “Disaccharrides,” “Oligosaccharides,” “Polysaccharides,” “Amylose/Amylopectin,” “Glycogen,” “Cellulose,” and “Glycosaminoglycans” (pages 64–70 of the textbook).
2. You can also watch two video lectures on carbohydrate structure:
   • #16 Biochemistry Blood Clotting/Carbohydrates I Lecture for Kevin Ahern’s BB 450/550

   • #17 Biochemistry Carbohydrates II Lecture for Kevin Ahern’s BB 450/550

     (Links to these are also provided on pages 65 and 68 of the textbook.)

Commentary

Monosaccharides

Look at the straight-line structures of the monosaccharides shown on pages 65–66 in the textbook reading. These stick figure representations do not give a clear idea of the three-dimensional shape of the simple sugars, but they do show how similar the sugars are. The aldoses display (n-2) chiral carbons (the end carbons are not chiral). The ketoses display (n-3) chiral carbons.

If we consider only D-hexoses (the stereoisomers found in vivo), there are eight possible aldohexoses and four possible ketohexoses. All of the eight D-aldohexoses and four D-ketohexoses are recognized as different compounds by the body. One way to test if this statement is true is to taste equal amounts of each of these sugars. They do not all taste equally sweet (a factor exploited in sugar substitutes). The three-dimensional structures of glucose in its linear form are shown below.

Note that a carbon atom with four groups attached to it will adopt a tetrahedral shape, not a planar one. As in the discussion of the three-dimensional shape of amino acids, solid lines indicate bonds in the plane of the paper; dotted lines indicate bonds extending below the plane of the paper; and wedge shapes indicate bonds coming out of the page toward the reader. Look at the way Structure A is drawn. There is steric hindrance among three of the hydroxyl groups—those marked with an asterisk (*). Structure B shows rotation around carbon$\ce{-}$carbon bonds to relieve this steric hindrance; hence, the bulky groups ($\ce{-}$OH and $\ce{-}$ CH₂OH) are as far from each other as possible.

Look closely at Structure B. One of the hydroxyl oxygens (which has a partial negative charge) is close to the aldehyde carbon (partial positive charge). This zone of proximity is the circled area in Structure B. When these atoms get close enough together, a new bond forms, and linear glucose becomes a cyclic molecule. This new molecule, formed by the reaction of an alcohol and an aldehyde, is called a “hemiacetal.” In aqueous solution, there is an equilibrium between the two forms of glucose, and over 90% of individual molecules are cyclic.

When glucose cyclizes, the anomeric hydroxyl group (i.e., the one on C_b which was an aldehyde in the linear structure of glucose) can be in one of two positions: on the opposite side of the ring from the free $\ce{-}$CH₂OH, giving α-D-glucose, or on the same side of the ring as the free $\ce{-}$CH₂OH, giving β-D-glucose. The cyclic form of glucose shown above is α-D-glucose. The molecular structures of α-D-glucose and β-D-glucose are shown on page 66 of the text.

This seemingly trivial difference (i.e., whether the anomeric hydroxyl group is above or below the glucose ring) is very important in determining polysaccharide solubility, because the position of the anomeric hydroxyl group will determine if water can penetrate the polysaccharide structure (starch, glycogen), or whether the sheets of linked glucoses can hydrogen bond together to exclude water (cellulose).

The cyclic structure of glucose can adjust to a chair or a boat conformation. Most often the chair conformation is the prevalent one, because in it steric crowding is minimized.

Glucose is by far the most common hexose in vivo, followed by mannose and galactose. The remaining hexoses are quite rare. One reason for the biochemical popularity of glucose is its ability to cyclize readily and without having to resort to complex reactions.

Polysaccharides

As it is part of a hemiacetal (or hemiketal) formation, the anomeric hydroxyl group of a cyclic sugar is unusually reactive, and can react with another alcohol group to form a new bond. An organic chemist would call this new bond an acetal; biochemists call it a glycosidic bond. Examples of glycosidic bond formation are shown below and in the textbook.

Note that a water molecule is released when a glycosidic bond is formed, just as a water molecule is released as a peptide bond is formed. The textbook figures indicate clearly the bond positions; the figure above emphasizes the three-dimensional shape of a disaccharide.

A glycosidic bond occurs between two sugar molecules to form a disaccharide (see the structure of sucrose on page 67 of the text), among three sugars to form a trisaccharide, 5–15 to form an oligosaccharide (page 68) and so on, until polysaccharides (also called glycans) are formed. Note the difference between an α- and a β-linked disaccharide (α and β refer to the original position of the anomeric hydroxyl group,: α means the new bond forms away from the free $\ce{-}$CH₂OH; β means the new bond forms on the same side as the free $\ce{-}$CH₂OH).

Homopolysaccharides occur if all monomeric sugar units are the same (usually glucose); heteropolysaccharides occur if more than one kind of hexose is used. Because the anomeric hydroxyl group can react with any other hydroxyl group, polysaccharides are not always linear, as proteins are; branched polysaccharides are well known. All biochemistry, and certainly all carbohydrate biochemistry, is governed by the need to minimize steric crowding between atoms and among chemical groups. So the number and types of glycosidic bonds, and the amount of glycosidic branching in vivo is limited. Amylopectin/amylose is a good example of a branched polysaccharide. The structure of amylose is shown in the textbook (page 69). Note how often branching occurs in this polysaccharide.

All of the common polysaccharides are polyglucoses. Some, such as cellulose and starch, are composed of unmodified glucose monosaccharide units; others, such as chitin and heparin, are composed of chemically modified glucose monosaccharide units.

The unmodified polyglucoses can be roughly divided into two groups: insoluble, structural polysaccharides; and hydrated, storage polysaccharides. In general, soluble polysaccharides are α-linked, and insoluble polysaccharides are β-linked.

The two most common examples of insoluble structural polysaccharides are cellulose (glucose units) and chitin (N-acetyl-D-glucosamine units). In both cases the monosaccharide units are joined by β(1 → 4) glycoside bonds. (Refer to page 70 in the text for information on how this has a large effect on the digestibility of this carbohydrate.) The polysaccharide chains of both cellulose and chitin stack in such a way that extensive hydrogen bonding can occur between the chains, and water is excluded.

In chitin, a hydroxyl group on each glucose has been substituted by an acetamido functional group; hence, chitin can hydrogen bond more efficiently to other chitin chains than cellulose can to other cellulose chains. Chitin, in fact, is one of the least soluble and toughest structural biochemical molecules known.

The storage polysaccharides, in contrast, are linked by α(1 → 4) glycoside bonds. This type of bonding places the terminal $\ce{-}$CH₂OH group of each glucose monomer on the same side of the polysaccharide chain. To minimize steric crowding, the glucose monomeric units bend away from each other in a large, left-handed helix. In this structural motif, the hydroxyl groups are all readily hydrated. This structure is the basis for the “thickening” properties of starch in cooking. The plant starch, α-amylose, is an example of this type of bonding. The polyglucose chain can also have side branches, usually α(1 → 6) linked, which further disorder the structure of the polysaccharide and allow for extensive hydration. Glycogen and amylopectin, storage polysaccharides found in animals and plants respectively, are examples.

Summarizing, then: α-linked polysaccharides have a more disordered, hydrated structure than do β-linked sugars. Therefore, the α-linked polysaccharides are the soluble ones. Branched polysaccharides (for the same reason) are also soluble.

Although the hydroxyl groups in glucose are polar, monomeric glucose does not bind metal cations in vivo. However, modified sugars are more polar than the parent glucose; some are even ionic. Some of these modified sugars bind cations, such as Na⁺, K⁺, and Ca²⁺ tightly. They are also more soluble than the parent sugar. Below, you will see that the solubility of sugars is used in vivo to increase the solubility of proteins and lipids.

Glycoproteins, Proteoglycans, and Ionic Polysaccharides

A glycoprotein is a protein to which one or more oligosaccharides is or are covalently linked. The amount of carbohydrate that can be linked to a protein varies widely, from 0% to about 90% carbohydrate (by weight). A surprisingly large number of proteins have some attached carbohydrate. Oligosaccharides are attached to newly formed proteins by enzymes, and enzymes are not nearly as efficient as DNA at producing identical copies. As a result, oligosaccharide structures may sometimes show subtle changes. (As you will learn later in the course, the protein comes from DNA, via RNA.) Furthermore, as a protein ages, its sugar content can vary. So we cannot be sure if we have the “right” compound when we analyze a glycoprotein. The good news in this uncertainty in the carbohydrate portion of glycoproteins is that the attached carbohydrates do not seem to affect the biochemical properties of proteins such as enzymes, transport proteins, and hormones. What we can say about the carbohydrate portion is that it:

• increases the aqueous solubility of a protein (the carbohydrate groups are always found on the outside of the three-dimensional structure of a protein).
• protects the protein portion from enzymatic degradation.
• can organize the structure of bulk water (ice-nucleation proteins), or conversely, can disorganize water structure (anti-freeze proteins).
• acts as a recognition marker in various biological processes (e.g., the carbohydrate portions of membrane-bound glycoproteins are always located on the external surface of the cell membrane).

Carbohydrates are attached to proteins through O-linkages or N-linkages
(refer to page 67).

• O-linkages resemble the glycoside bond, because the linkage is between the anomeric sugar carbon and the hydroxyl group of one of the alcohol amino acids (serine or threonine).
• N-linkages form between one of the sugar hydroxyls (not the anomeric one) and the amino group of asparagine.

An important subset of glycoproteins is the proteoglycans. Proteoglycans are a diverse group of glycoproteins made up from aggregates of proteins with glycosaminoglycans. They compose what is called the “ground substance” of the connective tissue. The proteoglycans can be very rigid (the cartilage in your nose), flexible (your Achilles tendon), or viscous (the lubricating fluid in your joints). The structure of the proteoglycans, together with their polyionic character, causes the complex to be highly hydrated. Proteins without attached sugars are not as highly hydrated as are those with bound sugars. The details of glycosaminoglycan structure are found in the textbook (page 70).

Bacterial Cell Walls

The bacterial cell wall is an example of a heteropolysaccharide.

Bacterial cell walls are a special subset of glycoproteins. Composed of linear carbohydrate chains cross-linked by repeating peptide chains, a bacterial cell wall forms a net enclosing the bacterial cell proper. The carbohydrate chains are alternating N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM) units that are β(1 → 4) linked as shown below:

Each NAM contains a lactic acid residue linked to a tetrapeptide: D-ala-isoglutamate-L-lys-D-ala. This structure is shown below.

The rest of the peptide linking (NAG-NAM)-chain 1 to (NAG-NAM)-chain 2 varies depending on the bacterial type. This framework is referred to as “peptidoglycan.”

The rigidity of their cell walls allows bacteria to live in a hypotonic environment (such as exists within a host) that would otherwise cause them to lyse (burst). Bacterial cell walls contribute to the ability of the organism to cause disease, and are of medical significance.

Bacterial cell walls contain D-amino acids; therefore, they cannot be cleaved by normal proteolytic enzymes. [Remember that the amino acids that make up proteins (and therefore proteolytic enzymes) are L-amino acids.] The bacterial cell wall can be cleaved, however, by lysozyme at a more vulnerable site: the glycosidic bond joining NAG with NAM. The biological method of choice for cleaving bacterial cell walls is through the use of penicillin, which inhibits the cross-linking of NAM-NAG chains.

Wikipedia’s page Peptidoglycan offers a schematic representation of a cross-linked bacterial cell wall and a description of the mechanism of action of peptidoglycan-specific antibiotics.

Study Questions

1. Why do ketoses have fewer chiral carbons than the equivalent aldoses?
2. Draw the three-dimensional linear structure of D-glucose (following the conventions shown in Structure A above for glucose). Determine if this molecule can cyclize, and if so, whether it is likely to cyclize.
3. Draw the boat and chair shapes for mannose. Determine which conformation provides the least amount of steric crowding.
4. What is the difference between the α anomer and the β anomer of D-glucose?

5. Look at the structures of lactose and sucrose in the textbook. Redraw them so that the rings are in the chair conformation and the glycoside bonds are correct.

   

   are illustrative of C$\ce{-}$O$\ce{-}$C ether bonds; but genuine bonds do not display kinks or right-angle bends. Which disaccharide displays the least steric crowding? Why?

6. What prevents branching from both sides of a glucose unit in a polysaccharide?
7. Microorganisms in the gut can hydrolyse the polysaccharides in beans, but not all humans can do so in the process of digestion. How does this fact explain the well-known production of gas that occurs after eating a meal of beans?
8. Why does paper (which is composed of unoriented cellulose fibres) lose its strength and structure when wet?
9. Chemically, how could you hydrolyze chitin to monosaccharides?
10. If cellulose forms such a compact, water-excluding structure, how can ruminants (e.g., cows) hydrolyze it?
11. Compare and contrast the structures and functions of cellulose, chitin, starch, and glycogen.
12. What common biochemical structure is found in cartilage and hyaluronic acid? Where are cartilage and hyaluronic acid found in the body? Why do they have such different physical properties?
13. If lysozyme is so effective in cleaving the heteropolysaccharide of the bacterial cell wall, why isn’t it used clinically to fight bacterial infection?
14. Outline the structure of peptidoglycan.
15. We noted earlier in this lesson that β-linked polysaccharides are normally insoluble. Does this insoluability apply to the β-linked bacterial cell wall? Describe the difference or similarity between cellulose (an insoluble β-linked polysaccharide) and the bacterial cell wall.
16. What is the mechanism of penicillin’s action on the bacterial cell wall?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 4: Lipids

Overview/Objectives

Lipids are a far more heterogenous class of biomolecules than are proteins, carbohydrates, or nucleic acids, but lipids do have some common features. For example, they are virtually insoluble in water, and generally have one or more unbranched hydrocarbon chains 10–20 carbons in length. Lipids also have a polar or ionic functional group attached to one end of a hydrocarbon chain. This structure allows lipids to form organized aggregates with the carbon-hydrogen chains stacked together (excluding water) and the polar head groups interacting with water molecules. These lipid aggregates (membranes) can efficiently separate the aqueous compartments of the body.

Chemically, lipids differ from other biomolecules in that they are composed of reduced hydrocarbons. This property means that per unit mass, more chemical energy is available from lipids than from, say, carbohydrates (in other words, French fries have more calories—energy—than plain potatoes do).

In Lesson 4, we discuss the structure and functions of lipids, including micelles, bilayers, liposomes, and lipoproteins.

After completing this lesson, you should be able to:

1. Identify the different classes of lipids.
2. Describe the properties of lipid aggregates.
3. Describe the structure of biological membranes.
4. Define “lipoprotein,” and explain the structure and functions of lipoproteins.
5. Discuss the dynamic nature of membranes, and describe some membrane transport mechanisms.
6. Describe the structure of a micelle, a bilayer, and a liposome.
7. Explain the forces that hold micelles, bilayers, and liposomes together.
8. List the basic physical properties of lipid bilayers.

Readings and Activities

1. Read “Lipids and Membranes,” “Fatty Acids,” and “Membrane Lipids,” (pages 70–74 of the textbook).
2. You can also view the video lectures on lipids and membranes:
   • #30 Biochemistry Lipids and Membranes Lecture for Kevin Ahern’s BB 451/551

   • #31 Biochemistry Membrane Transport Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on page 71 of the textbook.)

Commentary

Lipids are biomolecules characterized primarily by their lack of water solubility. Most lipids (but not all) contain unbranched hydrocarbon chains. Hydrocarbon chains with no C$\ce{=}$C double bonds stack well together. These “saturated lipids” are solid at room temperature. Unsaturated lipids do not stack as well, and so have a more disordered structure and a lower melting point. As you will recall from basic chemistry, the melting point of a substance is proportional to the amount of bonding between its molecules. Thus, the ordered, covalent stacking of saturated lipids would increase the melting point of these molecules.

The major roles lipids play in vivo are as

• the main components of membranes
• fuel molecules
• precursors of hormones and prostaglandins
• protective coatings on fur, feathers, fruits, etc.

Lipids are a very diverse group chemically. Based on their chemical structures they can be divided into five major groups:

1. fatty acids (building blocks of 2 and 3 below)
2. triacylglycerols (uncharged storage form of lipids in vivo)
3. glycerophospholipids (major lipid component of membranes)
4. sphingolipids (membrane components)
5. steroids.

There are three other classes of lipids, each of which has a distinctive structural type. These three classes contain substances with biological roles that are often very significant. However, they are quantitatively much less important than the first five classes. They include:

• waxes (fatty acids esterified to a sterol or fatty alcohol)
• terpenes (polymers of 2-methyl-1,3-butadiene; vitamins A, E, K, and certain plant compounds, such as menthol and camphor, are terpenes)
• prostaglandins (local hormones).

The structures of two fatty acids, palmitic acid (16:0) and palmitoleic acid (16:1^Δ9), are shown below. Here, (16:0) and (16:1^Δ9) refer to the total number of carbons in the fatty acid and to the number or numbers and position or positions of the double bonds. (The names and structures of other common fatty acids are shown on page 72 in the textbook.)

Note that there is one double bond in palmitoleic acid; it is between C₉ and C₁₀ (the carboxylic acid carbon is C₁). The most common polyunsaturated fatty acids possess double bonds on C₉, C₁₂, C₁₅, etc. This structure means that the double bonds are separated by two single C$\ce{-}$C bonds. Do you remember from organic chemistry that alternating C$\ce{=}$C double bonds (i.e., double, single, double, single) lead to delocalized electrons? If at all possible, in vivo systems avoid delocalized electrons. With two single C$\ce{-}$C bonds separating C$\ce{=}$C double bonds in lipids, there is no chance of delocalized electrons. In fact, many of the chemicals that are considered cancer-causing contain delocalized electrons in alternating bonds or in ring structures.

You may see the expression ω-3, referring to fatty acids found predominantly in fish. This designation means the double bond is between the third and fourth carbons, counting from the non-carboxylic acid end.

Triacylglycerols are found as lipid droplets in the “fat” cells making up adipose tissue. The three fatty acids (“triacyls”) esterified to glycerol differ from organism to organism, and vary with diet. These lipids are a backup source of biochemical fuel. The triacylglycerols do not contain a polar or ionic headgroup; therefore, they accumulate in very large aggregates, rather than forming micelles or liposomes. (We discuss these structures in the next lesson.) The poor water solubility of the triacylglycerols means that they are metabolized more slowly than are the water-soluble carbohydrates. A graphic demonstration of this difference occurs when a runner “hits the wall”; she has used up all her carbohydrate reserves, and must switch to lipid metabolism for energy.

Glycerophospholipids make up the bulk of biological membranes. The structure of a glycerophospholipid is shown below.

X is a highly polar or ionic group. The phosphoric acid group and X, together, are referred to as the head group, while the hydrocarbon portions are referred to as the tail. Most glycerophospholipids contain fatty acids (tails) with 16–18 carbon atoms. Because of this regularity, and because all the common head groups are approximately the same size, glycerophospholipids can stack to form very regularly shaped membranes.

Sphingolipids, also found in membranes, contain an amino alcohol and one fatty acid rather than two fatty acids. The amino alcohol is usually sphingosine or dihydrosphingosine, both C₁₈ compounds. Thus, the sphingosines stack well with the glycerophospholipids in membranes. The head groups of the gangliosides (a subclass of sphingolipids) can be large carbohydrate groups and can extend well beyond the surface of a cell membrane. These head groups are important recognition sites (“receptors”) for incoming molecules.

Cholesterol is the parent compound of steroids. As a component of animal membranes, cholesterol makes the membranes more rigid (saturated fatty acids have the same effect). Cholesterol is also a component of plasma lipoproteins; plasma is the straw-coloured fluid of the bloodstream and lymphatic system of humans and other animals. The most important steroids are bile acids (the detergent-like compounds that emulsify lipids in the intestine) and certain hormones. Plants do not contain cholesterol, but they do have sterol components in their membranes.

Study Questions

1. List the five classes of lipid, and outline their functions.
2. What are the names of the predominant fatty acids? How many carbons do they have?
3. Distinguish between unsaturated and saturated fatty acids.
4. What is the role of phospholipases?
5. What are the three types of steroid hormones? What responses do they evoke?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 5: Membranes and Transport

Overview/Objectives

After completing this lesson, you should be able to:

1. Describe the structure of a micelle, a bilayer, and a liposome.
2. Explain the forces that hold micelles, bilayers, and liposomes together.
3. List the basic physical properties of lipid bilayers.
4. Define “biological membrane.”
5. Name the two types of membrane protein, and describe how they are associated with the lipid bilayer.
6. List the features of the lipid bilayer that vary from membrane to membrane.
7. Explain what is meant by the “fluid mosaic model of membrane structure.”
8. Describe the effect of detergents on biological membranes.
9. Describe the basic structure of plasma lipoproteins.
10. List the categories of plasma lipoproteins, and state their functions.
11. Define “receptor-mediated endocytosis.”
12. Describe the role of lipoproteins in cardiovascular disease.
13. Describe a gap junction.
14. Explain the difference between mediated and non-mediated transport.
15. Define the terms “uniport,” “symport,” and “antiport.”
16. Define “electrogenic transport.”
17. Name the two sources of energy for active transport.

Readings and Activities

1. Read “Lipid Bilayers,” “Membrane Proteins,” “Membrane Transport,” “Na⁺/K⁺ ATPase,” and “Bacteriorhodopsin” (pages 74–78 of the textbook).
2. Consult these links for information on biological membranes:
   • Biological membrane on Wikipedia
   • Biological Membranes, Carnegie Mellon University, 2006.
3. Read the Wikipedia page Lipoprotein on plasma lipoproteins for information  not included in the text.
4. You can also watch two video lectures on membrane transport:
   • #31 Biochemistry Membrane Transport Lecture for Kevin Ahern’s BB 451/551 (This one was also provided earlier.)

   • #32 Biochemistry Nerve Transmission/Mitochondria Lecture for Kevin Ahern’s BB 451/551

     (Links to these are also provided on page 76 of the textbook.)

Commentary

The most important biochemical feature of lipids is the way they aggregate to exclude water from their hydrophobic (“water-hating”) portions. We live in a watery world, but only the hydrophilic (“water-loving”) head groups of lipids can interact with water.

For a lipid aggregate in water, the simplest conformation is a sphere with the polar head groups on the outside facing bulk water, and hydrophilic tails in the centre with water excluded; this structure is called a “micelle.” Micelles can be composed entirely of lipid or, more commonly, they can enclose non-lipid molecules such as proteins. In the latter case, the hydrophobic lipid tails interact with non-polar portions of the protein.

Lipids can also form bilayers. Here, the glycerophospholipids are arranged in two rows, with lipid tails facing each other and head groups facing bulk water. This is a very stable structure, since the tails are not crowded at their ends as they are in micelles. A micelle and a bilayer are shown below.

When a mixture of lipid bilayers in water is violently agitated, the bilayers twist to form self-sealing vesicles called liposomes, illustrated below. Liposomes have been used to deliver pharmaceutical agents or contrasting agents in clinical or veterinary applications.

Membranes, the lipid structures of most interest to biochemists, are complex lipid bilayers containing proteins and various pore structures. Two questions must be considered before we talk about membranes: “What forces hold bilayers (and liposomes) together?” and “How strong are those forces?”

Lipids aggregate to exclude water from the hydrophobic tails. These tails are unbranched and (usually) 16 to 20 carbons in length. If the lipids do not contain C$\ce{=}$C double bonds, then the chain shows a regular zigzag structure. These saturated chains stack very well together via van der Waals forces and hydrophobic forces.

Note that van der Waals forces, the directionless forces of attraction between any two particles or molecules, are very weak forces, weaker than H-bonds, which we discussed in Unit 1. However, like H-bonds, van der Waals forces are additive. A typical bilayer contains 32 to 40 carbons (i.e., 2 × 16–20 carbons) between the two outer head groups. Bilayers composed of saturated—that is, regularly stacking—hydrocarbon chains are quite stable and reasonably rigid. A cis-C$\ce{=}$C double bond, introduced into the middle of the chain, disrupts the regular stacking pattern of a bilayer, and has the effect of making the bilayer more fluid.

It is important to remember that, while bilayers are held together by noncovalent bonds, they are quite stable. The overall structure of a bilayer is determined by interaction with bulk water. The internal structure of a bilayer is determined by very weak van der Waals (directionless) forces. Also, bilayers can accept inclusions without destroying the entire bilayer structure. In essence, membranes are bilayers with protein inclusions.

Biological Membranes

Biological membranes are complex structures of proteins embedded in a lipid bilayer. It is important to keep in mind that the general shape and properties of any membrane—no matter how complex—are similar to those of the lipid bilayer.

Membranes have several important features, listed below. (While you are studying ths list, you may find it useful to refer to the diagram of the biological membrane in the links provided under “Readings and Activities” for this lesson.)

Features of the lipid portion of the membrane are:

• For any given membrane (e.g., rat liver nuclear membrane), the lipid composition in the two sides of the bilayer is different.
• The higher the ratio of saturated to unsaturated lipids, the more rigid the membrane.
• Some organisms (e.g., bacteria, hibernating animals) can change the ratio of saturated to unsaturated phospholipids in their membranes to maintain a constant membrane fluidity despite temperature change.
• Animal cell membranes contain cholesterol, while plant cell membranes do not.
• Cholesterol, because of its fixed structure, increases the rigidity of membranes.

Features of membrane proteins are:

• Integral proteins are held in the membrane by hydrophobic (van der Waals) interaction with the lipid hydrocarbon chains. The portion of the protein within the membrane is frequently in the form of one or more β-helices.
• Peripheral proteins are bound to the surface of a membrane by electrostatic forces or H-bonds. These proteins contain a higher proportion of polar or ionic amino acid residues than do the integral proteins.
• Integral membrane proteins frequently possess amino acid tails that extend beyond the membrane surface.
• Membrane proteins are usually glycoproteins. The carbohydrate portion or portions extend to the outside of the cell.
• Carbohydrates extending from a plasma membrane protein are recognized by the immune system as “self” or as “invader.”
• Membrane proteins function in shape stabilization, as receptors for circulating molecules, as enzymes, and to provide pore structures or active transport channels.

General features of membranes are:

• Membrane proteins and phospholipids can move laterally in a membrane, but do not “flip flop” within the membrane.
• Because the forces holding a membrane together are noncovalent, two cells can fuse by integrating their membranes, and portions of a given membrane can bud off as a vesicle.

Plasma Lipoproteins

As the name suggests, a lipoprotein is formed from a protein plus a fatty acid, a phospholipid, or a lipid-carbohydrate moiety. The attachment may be covalent or noncovalent. The class of proteins and lipids that are covalently linked is small and has not been widely studied. More attention has been paid to the noncovalent group of lipid-protein aggregates, as there is enormous medical interest in this class. In fact, the term “lipoprotein” is usually understood as referring to the protein-lipid complexes that are the transport system for lipids in the blood.

There are three general classes of plasma lipoproteins, described below.

• Chylomicrons, the least dense of the plasma lipoproteins, are complexes of proteins (minor component), and dietary lipids and cholesterol.
• Very low density (VLDL), intermediate density (IDL), and low density (LDL) lipoproteins are complexes that transport endogenous lipids and cholesterol from the liver to tissues.
• High density lipoproteins (HDL) transport endogenous cholesterol from tissues to the liver.

These complexes are categorized by their densities relative to one another (e.g., HDLs are more dense than LDLs). The density is almost always proportional to the protein content: the greater the density, the higher the proportion of protein in the aggregate. Analysis of lipoprotein complexes is complicated because the lipoproteins are not really dissolved in plasma, but are suspended (much as cream is in homogenized milk). When lipoproteins are studied, plasma is centrifuged at high speed over an aqueous solution of mixed density. Within the aqueous solution, the various lipoprotein fractions form bands that can be separated and analyzed.

The three classes of plasma lipoproteins function together to move lipids to the tissues from either the intestine or the liver, or to move lipids from tissues back to the liver. The lipoprotein complex is bound to a membrane protein at the receiving cell. Then, the lipid is dissociated from the aggregate in one of two ways:

1. The entire complex may be taken into the cell by endocytosis (e.g., LDL).
2. The lipid may be enzymatically separated from the protein (partially or completely); the lipid is then taken into the cell, and the protein is released back into the plasma (e.g., chylomicrons and VLDL).

LDL, which contains cholesterol, is the best example of the endocytosis process of lipid transport (review the Wikipedia page Lipoprotein, provided earlier in the readings). When LDL is bound to specific receptor proteins, these LDL receptors congregate on the cell surface. Clathrin, an intracellular protein, gathers on the opposite side of the membrane from the gathered LDL receptors, and helps to form “coated pits.” The coated pits deepen and fuse, forming closed vesicles which bud off inside the cell. Cholesterol is harvested for incorporation within the cell’s membranes, and the protein is degraded.

HDL removes cholesterol from a cell and transports it back to the liver, where this sterol is converted to a bile acid. The bile acids are the emulsifying agents that act in the digestion and absorption of fats and fat-soluble vitamins.

The lipoprotein complexes act in lipid transport, and so these aggregates constantly cycle from their apolipoprotein (i.e., protein alone) form to the lipid-protein aggregate form. There are at least nine different apolipoproteins, and all have large portions of α-helix in their secondary structure. The plasma lipoprotein complexes are not random aggregates; they have well-defined structures.

Elevated cholesterol levels have been implicated in the pathology of heart disease. However, it should also be recognized that the human body synthesizes a great deal of cholesterol. This synthesis occurs because of the need for cholesterol-based bile salts to function as digestive emulsifying agents. Small amounts (less than 1g/day) of bile acids escape recycling, and are metabolized in the large intestine and excreted. Synthesis of cholesterol is needed to replace these lost bile acids. In addition to scavenging by HDL, there are at least two feedback mechanisms to control cholesterol concentration in vivo: both LDL synthesis and cholesterol synthesis are decreased in the presence of high levels of cholesterol. Cholesterol is not an evil nutritional agent, but rather a necessary—and controlled—substance in mammalian systems. Problems result when the controlling mechanisms function incorrectly, as a result of poor diet, lack of exercise, environmental stresses, or heredity.

Cardiovascular Health and Disease

Cardiovascular health, in general, is associated with high levels of HDL (the cholesterol scavenger) and low levels of LDL (the cholesterol deliverer). Cardiac disease results when cholesterol delivery is increased, cholesterol removal diminished, or tissue uptake of cholesterol impaired. The most common cause of adult death in North America is heart disease or myocardial infarction (heart attack). Lipid deposition on arterial walls results in occluded vessels; a clot will eventually stop the blood flow in the occluded vessel. Of all the causes of heart disease the most well understood is the hereditary factor in familial hypercholesterolemia. In this disease, LDL receptors are absent or greatly reduced in number, preventing receptor-mediated endocytosis with the result that cholesterol-laden LDL accumulates in the plasma.

Membrane Transport

Membranes are dynamic structures. As we discussed earlier, there is constant movement of materials laterally within membranes. There is also dynamic movement of solutes across membranes. In this section, we consider methods of transporting water-soluble molecules across lipid barriers. There are two types of transport processes: nonmediated and mediated.

Nonmediated transport may involve diffusion through the lipid bilayer, or diffusion through “spaces” in the membrane.

Mediated transport depends on specific transport proteins. It may be passive or active. Passive mediated transport occurs when a specific molecule flows down its own concentration gradient; for example, transport via gated pores. Active mediated transport occurs when a specific molecule is transported against its own concentration gradient, using external energy; examples include ATP-driven transport and ion-driven transport.

Nonmediated Transport

Diffusion. Diffusion is one of the ways a small, noncharged species (such as ethanol) can enter a cell. The concentration gradient of the species determines whether the net flow is into or out of the cell. The oil-water partition coefficient of the substance (i.e., its ability to transfer readily from water to the lipid membrane) determines how quickly the substance will enter the cell (this is the way ethanol passes from stomach to bloodstream). Alternatively, a solute molecule may pass through a space in the membrane (this is the way RNA gets out of the nucleus).

Gap junctions. Gap junctions are hexagonal, tubular protein particles embedded in a plasma membrane. What is particularly important about the gap junctions is that they line up on adjacent cells to provide intercellular highways for the free flow of solutes. The junctions are channels of 16–20 Å inside diameter. This space allows some molecules, such as monosaccharides and amino acids, to pass between cells, but does not allow proteins or RNA to pass. Whether a gap junction is “open” or “closed” is controlled by the Ca²⁺ concentration. When the concentration of Ca²⁺ is less than 10⁻⁷, the normal situation in cells, the channels are open. A high concentration of Ca²⁺ closes a gap junction.

Mediated Transport (Passive)

Gated pores. A form of transport protein, gated pores are a special set of integral membrane proteins that span the membrane. A specific solute molecule (glucose is the best example) can bind to either side of its own transport protein. After the solute molecule is bound, a conformation change occurs in the transport protein, and the solute molecule is released on the other side of the membrane. This is mediated transport; however, as no external source of energy is necessary, it is not considered to be active transport. Net transport is from the area of high to the area of low concentration; the gated pore merely speeds up equilibration.

Mediated Transport (Active)

Many solutes are imported into or out of cells against a concentration gradient. Since this is equivalent to pushing uphill, energy is required, and the process is called active transport.

ATP-driven active transport. The ions Na⁺ and K⁺ are good examples of substances moved by active transport. Sodium ion concentration is approximately 4 mM inside cells and 100 mM outside. Potassium ion concentration is just the opposite: 4 mM outside and 100 mM inside the cell. To maintain these levels, potassium ions are pumped into cells as sodium ions are pumped out. This pumping is accomplished by an integral membrane protein, called (Na⁺-K⁺)-ATPase. The hydrolysis of ATP provides energy to pump Na⁺ and K⁺ ions against the concentration gradients noted.

Active transport across a membrane can result in the net movement of one solute species (uniport), two solute species moving in the same direction (symport), or two solute species moving in opposite directions (antiport). The transport of Ca²⁺ out of a cell is a uniport mechanism, while the Na⁺/K⁺ exchange is an antiport mechanism. Na⁺/glucose transport, discussed below, is a symport mechanism.

Ion-gradient-driven active transport. Ion gradients occur in vivo. These gradients are a source of electrochemical energy to move secondary species (i.e., species other than the ions themselves) across membranes. Glucose is transported from the lumen of the intestine by the movement of Na⁺ from areas of high to areas of low concentration (i.e., from outside to inside the cell).

Note: ATP-driven transport and ion-gradient-driven transport are similar, because both involve the use of stored energy to move a solute across a membrane against a concentration gradient. ATP-driven transport involves the use of chemical energy (ATP → ADP + P_i) while ion-gradient-driven transport involves the use of electrochemical energy.

Bacteriorhodopsin

Bacteriorhodopsin is a unique example of an integral membrane protein that is capable of generating ATP. It is a light-driven proton pump involved in photosynthesis. This protein is present in the bacterial genus Halobacterium. Vitamin A is associated with the protein and isomerizes between a cis and a trans form in the presence of light, which creates a proton gradient across the cell membrane and enables the bacterium to produce ATP.

Study Questions

1. Why is it virtually impossible for phospholipids and membrane proteins to flip-flop within the membrane, while it is relatively easy for these constituents to move laterally?
2. Explain the effect of a detergent on cell membrane structure.
3. List the six functions of a membrane according to the textbook.
4. List and describe the four types of membrane proteins.
5. Family hypercholesterolemia (FH) is a genetic disorder in humans in which cells cannot take up LDL. The cells are unable to internalize LDL, although they can bind LDL to their cell surface. Using your knowledge of receptor mediated endocytosis, suggest a reason for this anomaly.
6. Distinguish between passive and active transport, and indicate whether energy is required for each of these processes.
7. Why is the Na⁺/K⁺ ATPase considered to be involved in anti-port transport?
8. How does bacteriorhodopsin make ATP?

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.

Lesson 6: Vitamins and Cofactors

Overview/Objectives

In this brief unit, we consider vitamins: organic molecules that are not proteins, carbohydrates, or lipids, but are crucial to biological function. Vitamins may supply substances that are needed directly by the body, or they may function as “cofactors,” working with other molecules to induce chemical reactions.

After completing this lesson, you should be able to:

1. Define “vitamin” and “cofactor.”
2. Describe the function of vitamins and cofactors, and provide examples of each group.

Readings and Activities

1. Read “Fat-Soluble Vitamins” on page 78 of the textbook.

Commentary

Vitamins and cofactors are organic molecules contained in certain foods, and are essential for sustaining health. Only small amounts of vitamins are required to maintain the bodily integrity of an individual. Vitamins cannot be synthesized in the body, and must therefore be obtained from the diet. Some vitamins also serve as cofactors, in that they are associated with other molecules to induce a chemical reaction.

Vitamins are classified according to their properties in solution. Thus, two major groups, fat-soluble and water-soluble, have been identified. Should there be a deficiency in a vitamin, the body may succumb to a disease that is specific to that deficiency. The table below summarizes typical water-soluble and fat-soluble vitamins, their role in human health, and their deficiency effects.

Study Questions

1. Define the terms “vitamin” and “cofactor.”
2. What does folate do? What does a deficiency in this vitamin result in?
3. Name one function of vitamin A. Is this vitamin fat-soluble or water-soluble?
4. What is the function of vitamin D? How is this vitamin synthesized?
5. What vitamin is essential for blood clotting and how is this achieved?

Answers to Unit 3 Study Questions

Lesson 1 Proteins

1. Please consult the website given in the reading on amino acid structures to check your drawing of each. The abbreviations and the letter designations for the amino acids are:

   Amino acid	Abbreviation	Letter
   Alanine	Ala	A
   Arginine	Arg	R
   Asparagine	Asn	N
   Aspartic Acid	Asp	D
   Cysteine	Cys	C
   Glutamic Acid	Glu	E
   Glutamine	Gln	Q
   Glycine	Gly	G
   Histadine	His	H
   Isoleucine	Ile	I
   Leucine	Leu	L
   Lysine	Lys	K
   Methionine	Met	M
   Phenylalanine	Phe	F
   Proline	Pro	P
   Serine	Ser	S
   Threonine	Thr	T
   Tryptophan	Trp	W
   Tyrosine	Tyr	Y
   Valine	Val	V

2. The R group in an amino acid structure is denotes the amino acid side chain. This is where amino acids differ as each individual amino acid has a unique side chain or R-group.
3. In an aqueous environment the amino acids on a protein that have hydrophilic side chains will face the outside of the protein to interact with water and the hydrophobic side chains will be on the interior of the protein. In a membrane amino acids with hydrophobic side chains will face the outside where they interact with hydrophobic fatty acids and the hydrophilic side chains will be on the inside or any place they can interact with water.
4. The first mutation substituted a larger amino acid for a smaller one, thus distorting the tightly packed protein interior. The second mutation reversed this change by substituting a smaller amino acid for a larger one.
5. Proline is the least flexible amino acid because of the attachment of the side chain to the alpha-amino group and so it is too rigid and does not “bend” or fit with other amino acid side chains to allow the formation of a helix structure. In alpha helices, the side chains of the amino acids all point out from the axis and the structure of proline cannot accommodate this. Proline is found in collagen though because of the abundance of glycine which has the smallest side chain. The presence of many glycine residues would allow for proline to fit into this type of helical structure.
6. The six most abundant amino acids in proteins are: leu, ala, gly, set, val, and glu.
7. It is important to know the protein sequence in addition to the DNA sequence of a particular gene because it reveals information about the protein structure that is not possible with nucleic acid sequencing alone. One example is the positioning of disulphide bonds in proteins. In addition to this, many proteins are modified after they are synthesized. This modification can include cleavage of the protein to form the “mature” protein, or the addition of carbohydrates, phosphate groups, or acetyl groups to amino acid side chains.
8. The biochemical cause of this phenomenon is that the α-keratins are a large group of proteins which range from horn and fingernails (high cysteine content) to wool, hair, and skin (low cysteine content). The latter group is characterized by extensive α-helix formation. When wool is processed (“spun”), the fibres are teased to an elongated shape. Exposure to steam or hot water will cause the processed fibres to revert to their original, tightly H-bonded α-helices. Test whether and how much α-helices stretch by experimenting with a strand of your hair.
9. The regular coding of an α-helix is caused by H-bonding of the $\ce{-}$C$\ce{=}$O group of the first amino acid residue to the $\ce{-}$N$\ce{-}$H group of the fifth amino acid residue.
10. There are two major advantages of multiple subunits in proteins. The first is that having subunits that are assembled rather than one large polypeptide facilitates repair of a defective protein. It is easier and more efficient to repair one subunit rather than one very large protein. The second advantage is that having subunits rather than one large polypeptide allows better manipulation and regulation of the protein. This is true for enzymes. Since each subunit has an active site, having subunits makes it easier to increase the size of the enzyme through association of subunits instead of increasing the length of the polypeptide. In addition, the presence of subunits allows for more effective regulation of enzyme activity.
11. Hemoglobin binds oxygen by the iron of the heme group in one of the subunits. The heme iron is attached to a histadine side chain. This binding changes the conformation of that subunit as well as the other subunits of the hemoglobin protein which facilitates more binding of oxygen. This cooperativity allows hemoglobin to bind oxygen readily in areas of the body where oxygen is abundant and take that oxygen to areas of low oxygen. When oxygen is low the hemoglobin releases the oxygen.
12. This question refers to the energy and entropy changes that occur during protein folding. Experiments have shown that proteins fold to their native conformations quickly through directed pathways. The increase in conformational stability of the folded protein results in a decrease in free energy. A protein that is folding proceeds from high-energy, high-entropy to low-energy, low-entropy as the number of possible conformations that the folding protein can assume decreases. This energy-entropy relationship is referred to as the “folding funnel.”

13. Antibodies bind to the surface of an antigen. Therefore, part of the “recognition” is the specific arrangement of amino acids at a particular spot on the surface of an antigen. Denaturation destroys the spatial arrangement of amino acids which form the surface of a protein.

    However, in two instances antibodies may bind to denatured antigens. In the first case, the antibody may have been raised toward a denatured antigen; this type of bonding would occur in laboratory situations more commonly than in vivo. In the second case, the area of recognition on an antigen may be held together, at least partially, by disulphide bonds. Disulphide bonds are covalent bonds and will not break under the conditions of temperature, pH, solvent, and agitation which will dissociate H-bonds and hydrophobic interactions.

14. This question refers to the process of comparing protein sequences among different organisms. Several important steps are involved:
    1. Identify a protein that is common to the organisms (e.g., cytochrome c).
    2. Sequence the amino acids of the protein.
    3. Compare the amino acid sequences, and note any differences (changes) among them.

    Comparing sequences of related proteins between different organisms can yield information related to possible functional differences or binding specificities, whether the protein is a membrane protein or secreted protein, evolutionary relationships between organisms, and conserved regions or sequences, which can be very useful for predictions of protein function. Databases enable the functional prediction of unknown or previously uncharacterized proteins by comparing the sequence to those of characterized proteins.

15. A phylogenetic tree is constructed by:
    1. classifying the organisms under study according to similarities (e.g., are they mammals, insects, reptiles, etc.).
    2. noting the amino acid sequence of a protein common to all of the organisms.
    3. Grouping the organisms according to common alterations in an amino acid sequence.

    An organism that contains a subtle change in the sequence may be considered ancestral to other organisms in the group.

16. This question is looking for which side chains typically occur on the surface of a protein and which ones occur in the interior of a protein. To answer this question, one must consider the polarity of the amino acid side chain. Typically the nonpolar residues Val, Leu, Ile, Met, and Phe occur most often in the interior. The charged polar residues Arg, His, Lys, Asp, and Glu are located on the protein surface. Uncharged polar side chains, Ser, Thr, Asn, Gln, and Tyr, are typically on the surface, but may be found in the interior of a protein H-bonded to other groups (this neutralizes their polarity). To check the polarity of the amino acids check the link provided in the readings. The figure of the structures on this link indicates their polarity as well as the table titled, “Table of standard amino acid abbreviations and properties.
17. The forces that stabilize proteins are:
    • Hydrophobic effect – stabilizes proteins by causing nonpolar groups to minimize their association with water.
    • Electrostatic interactions, such as Van der Waals forces, hydrogen bonding, and ionic associations stabilize proteins by allowing side chains to bond or associate.
    • Chemical cross-linking through the formation of disulphide bonds or by the cross-linking of metal ions to a protein add stability to the folding of the protein.
18. Two functions of proteins are:
    • Enzymes – catalyse reactions in the cell.
    • Growth factors – induce growth and differentiation of specialized cells.
    • Other functions are listed in the commentary of this lesson.
19. An invariant region is a section of primary structure that is identical or conserved across all species. For enzymes, this is usually in the catalytic region.
20. A prion is an infectious protein that is responsible for mad cow disease and Creutzfeldt-Jacob disease. Infection with a prion causes misfolding of the PrP protein which is found in the brain. The misfolded PrP does two things: it 1) causes aggregations called amyloid plaques that damage nerve cells in the brain and 2) causes more of the PrP proteins in the brain to misfold as well.

Lesson 2 Nucleic Acids

1. The proteins produced by mammalian genes inserted into bacterial hosts will lack sugar groups, and so may not be as effective as the authentic proteins. This is because bacteria cannot glycosylate proteins like mammalian cells do.
2. There are no polar hydrogen atoms nor oxygen atoms on the side of the ribose ring which is closer to the base. Therefore no H-bonding is possible.
3. Refer to the structures shown in the commentary for Lesson 2 Nucleic Acids.
4. The Watson-Crick model has the following four features:
   • two polynecleotide chains that form a double helix structure.
   • anti-parallel DNA strands and each forms a right-handed helix.
   • The double helix contains major and minor grooves as a result of the bases on the periphery.
   • Each base is hydrogen bonded to a base in the opposite strand, A pairs with T and G with C. This is called complementary base pairing.
5. DNA is double-stranded and RNA is single-stranded. DNA contains deoxyribonucleotides as the pentose sugar is deoxyribose. RNA contains ribonucleotides as the pentose sugar is ribose. The nitrogenous bases in DNA are A-T and G-C, but in RNA they are A-U and G-C, whereby A pairs with uracil instead of thymine.
6. With both an anion (phosphate) and a sugar on every monomer unit, DNA and RNA are readily hydrated; therefore, solubility is not a problem.
7. RNA viruses have just as great a need to preserve their genomes from generation to generation as other species do. Double stranding would protect the base composition from chemical attack (mutation, destruction, or both), just as duplex DNA is protected. Single strand RNA viruses may mutate faster than duplex RNA viruses; they many have protective protein coats surrounding the RNA; or the RNA may double back on itself to give internal double stranding (see next lesson).
8. The restriction is true for double stranded DNA because every purine is H-bonded to a pyrimidine. Therefore, the sum of the purines (A + G) must equal the sum of the pyrimidines (T + C). There is no such restriction for each individual strand of DNA.
9. A and T are joined by two H-bonds, while G and C are joined by three H-bonds. Therefore, the duplex with more G-C pairs will have a higher melting temperature; and it may replicate more slowly.
10. The forces that stabilize nucleic acids are H-bonding. In addition a stabilizing force for DNA is stacking interactions between bases in a strand. Nucleic acids can be denatured by heat (to boiling) and urea. Yes, nucleic acids can renature.
11. The role of the 2′ OH group in RNA is to render it sensitive to degradation so that RNA can send messages and be involved in protein synthesis, but be broken down shortly after. The lack of this group in DNA makes the DNA more stable and resistant to degradation in order to protect the genetic information of the cell.
12. The flow of genetic information is DNA to RNA to protein. DNA is transcribed to mRNA in a process called transcription which encodes the DNA to make messenger RNA and the mRNA is then used to make proteins with ribosomes in a process called translation.
13. The three types of RNA are:
    • messenger RNA (mRNA): carries the code for amino acids to make proteins.
    • transfer RNA (tRNA): binds specific amino acids to start protein synthesis.
    • ribosomal RNA (rRNA): binds mRNA and tRNA and associate with proteins in ribosomes which conduct protein synthesis.
14. A plasmid is a small, circular, mobile genetic element that is common in bacteria. A plasmid has a small number of genes and can replicate on its own without the chromosomal DNA replicating. They can be transferred to bacteria and replicate. That is why they are used for cloning. Genes of interest can be inserted into a plasmid which is then transferred to a bacteria to make many copies of that gene

Lesson 3 Carbohydrates

1. An aldose with n carbons will have an aldehyde group as C₁ and a $\ce{-}$CH₂OH as C_n. Neither group contains a chiral carbon: the aldehyde group has only three substituents on carbon, and the terminal carbon does not have four different substituents on carbon. However, all the intervening carbons have four different groups attached, and so will be chiral (n-2 chiral carbons). A ketose possesses two terminal $\ce{-}$CH₂OH groups (non-chiral) and an internal ketone group (non-chiral). A ketose has n-3 chiral carbons.

2. 

   D-glucose can cyclize because it contains an aldehyde group and an appropriate hydroxyl group to form a hemiacetal. However, D-glucose is less likely to cyclize than is D-glucose, because there is less steric hindrance between hydroxyl groups in the extended form of D-glucose.

3. The chair form of mannose contains less steric crowding.

   

4. The α anomer and the β anomer of D-glucose differ in the position of the hydroxyl (OH) group at the anomeric carbon. In the α anomer, the OH substituent of the anomeric carbon is on the opposite side of the ring from the CH₂OH group at the chiral centre. In the β anomer, the OH group is on the same side as the CH₂OH group.

5. Both disaccharides are stable because the bulky side groups (except C₄–OH on galactose) are in the equatorial position.

   

6. In theory, nothing prevents branching from both sides of a glucose unit in a polysaccharide. Steric crowding is probably what prevents one sugar from being attached to more than three others (i.e., the branch and the two parts of the main chain). In fact, branches usually protrude from every sixth to tenth unit of the polysaccharide chain, rather than from every monosaccharide unit.
7. Microorganisms, after hydrolysing the polysaccharides to monosaccharides, will continue to process the monosaccharides before the host organism can absorb them. The end product of metabolism is frequently a gas, so that metabolism can proceed smoothly.
8. Water will penetrate areas of disorder in the cellulose structure. The water molecules will then disrupt some H-bonds by competing with interchain H-bonds.
9. You could hydrolyse chitin to monosaccharides by boiling it with a concentrated base. Base hydrolysis is the most efficient way to disrupt most polymeric biochemical molecules, because OH⁻ is such an effective nucleophile.
10. Time is the key to ruminant use of cellulose. Humans get one pass at hydrolysing food as it passes through their systems. Ruminants process foods in a more leisurely manner (which is what they are doing while chewing the cud). Continual mechanical processing and exposure to water and enzymes will hydrolyze the cellulose molecule from the outside in.
11. This question refers to the functions of the following polysaccharides: cellulose, chitin, starch, and glycogen. Cellulose and chitin are both structural polysaccharides which mean that they function to provide strength and rigidity. Cellulose is a linear polymer of up to 15, 000 glucose residues that provides strength to plant cells due to its highly cohesive and hydrogen-bonded structure. Chitin is the principal structural component of the exoskeletons of invertebrates. It is also found in the cell walls of most fungi and many algae. Chitin is a homopolymer of N-acetyl-D-glucosamine residues. Both cellulose and chitin have similar functions, and their structures differ only in that the OH group on the second carbon in cellulose has a acetamido functional group in chitin. Starch and glycogen are both composed of glucose residues and are storage polysaccharides that function as an energy source. Starch is composed of α-amylose and amylopectin. α-amylose is a linear polymer of several thousand glucose residues linked by α (1 → 4) bonds, and amylopectin consists of α(1 → 4)-linked glucose residues but with α(1 → 4) branch points every 24–30 glucose residues. Starch is the primary energy reserve for plants. Glycogen has the same structure as amylopectin, but is more highly branched. Glycogen is the primary energy reserve for animals.
12. The structures are glycoproteins. The flexibility of the various glycoproteins is related to the degree of hydration of the individual glycoprotein. The degree of hydration of the glycoprotein, in turn, is related to the amount (and accessibility) of the sugar residues. Joint lubricating fluid (hyaluronic acid) consists of a gel-like matrix of modified, highly charged polysaccharides. This matrix is more highly hydrated and more flexible than any of the other glycoproteins. Cartilage is the least hydrated and thus the least flexible of the structures.
13. Lysozyme is fairly indiscriminate in cleaving polysaccharides. It cleaves the host’s polysaccharides as well as the bacterial cell wall polysaccharides.
14. The structure of peptidoglycan is composed of linear chains of alternating β(1 → 4)-linked N-acetylglucosamine and N-acetylmuramic acid. The lactic group of N-acetylmuramic acid forms an amide bond with a D-amino acid-containing tetrapeptide to form the peptidoglycan repeating unit (e.g., L-Ala-D-isoglutamyl-L-Lys-D-Ala in the bacterium Staphylococcus aureus.
15. Cellulose is insoluble because of interchain hydrogen bonding, which excludes water. NAM and NAG both contain N-acetyl groups side chains. These N-acetyl groups interfere with the close packing between polysaccharide chains that is a necessary condition for interchain H-bonding. (It is not β-linking per se that causes insolubility, but the packing allowed by the symmetry of β-linking.) In addition, the heteropolysaccharide chains of the bacterial cell wall are linked by polypeptide chains. The overall structure of the bacterial cell wall is a covalently bonded net, overlying the plasma membrane, which protects the hypertonic bacterium without impeding passage of biomolecules in and out of the bacterium.
16. Penicillin specifically binds to and inactivates enzymes that cross-link the peptidoglycan strands of the bacterial cell wall, thus preventing growing cells from synthesizing the cell wall needed for protection from the hypotonic environment. As a result, the bacterial cells lyse.

Lesson 4: Lipids

1. The five classes of lipids and their functions are:
   • Fatty acids: compose all of the other lipids except for some steroids and fat-soluble vitamins. They are composed of a carboxyl group linked to an aliphatic tail.
   • Triacylglycerols: primary form of energy storage for metabolic activity. Consist of a glycerol molecule esterified to 3 fatty acids.
   • Glycerophospholipids: major lipid components of biological membranes. They are derived from phosphatidic acid.
   • Sphingolipids: major membrane components as well, and gangliosides that act as specific receptors for hormones and are also involved in cell-cell recognition. They are similar in structure to glycerophospholipids, but may have a simple or complex carbohydrate attached.
   • Steroids: component of animal plasma membranes to provide rigitidy. It is a sterol or modified sterol that has a complex ringed-structure.
2. The names of the predominant fatty acids are palmitic, oleic, linoleic, and stearic acids. They all contain 18 carbon chains.
3. Unsaturated fatty acids contain double bonds and saturated fatty acids are fully saturated with hydrogen and therefore do not contain double C$\ce{=}$C bonds. As a result saturated fatty acids have the least amount of steric interference and a higher melting point.
4. Phospholipases are enzymes that hydrolyze glycerolphospholipids. Hydrolytic cleavage of glycerolphospholipids by phospholipase A₂ results in a lysophospholipid, which is a strong detergent capable of lysing cells by disrupting membranes.
5. The three types of steroid hormones and the responses they evoke are:
   • glucocorticoids: e.g., cortisol; affect metabolism and inflammation
   • aldosterone: regulates the excretion of salt and water by the kidneys
   • androgens and estrogens: affect sexual development and function.

Lesson 5: Membranes and Transport

1. To flip-flop, the polar portion of a lipid must pass through the hydrophobic interior of the bilayer to get to the other side. This movement is thermodynamically unfavourable. Moving laterally is relatively easy, because the same polar portions stay in contact with water while hydrophobic areas of the moving components remain within the hydrophobic core of the bilayer. Remember that the interior forces in a bilayer are directionless and relatively weak. Moving within the bilayer would be a little like moving through a field of tall grass.
2. Detergents can interact with membrane protein, principally by van der Waals bonds. The hydrophobic portion of the detergent interacts with the protein while the hydrophilic portion of the detergent interacts with water. Detergent molecules generally do not have as ordered a structure as do glycerophospholipids; hence, the detergent-protein complex is more likely to form a micelle than a pseudo-membrane (see the next lesson). Triton X-100 does not contain any charged groups; it is a neutral detergent. Therefore Triton X-100 is less likely to bind to water-soluble proteins than is a detergent with a charged head group (e.g., sodium dodecyl sulphate).
3. The six functions of lipid bilayers (membranes) are:
   • Obtain food for energy for the cell.
   • Export materials out of the cell.
   • Maintain osmotic balance.
   • Create gradients for secondary transport.
   • Provide an electromotive force for nerve signalling.
   • Store energy in electrochemical gradients for ATP production (oxidative phosphorylation or photosynthesis).
4. The four types of membrane proteins are:
   • Integral membrane proteins: embedded in the membrane and pass through the membrane, emerging on each side.
   • Peripheral membrane proteins: embedded or tightly associated with part of the membrane, but do not project completely through both sides.
   • Associated membrane proteins: found near membranes, but are not embedded. These proteins may associate through interactions with other proteins or molecules in the lipid bilayer.
   • Anchored membrane proteins: not embedded in the bilayer, but are attached to a molecule that is embedded in the membrane, typically a fatty acid.
5. Endocytosis is a co-operative process. Normal LDL receptors with bound LDL molecules congregate in exactly the area where internal clathrin molecules pull the membrane in to form an internal vesicle. Therefore, there must be a physical connection linking clathrin with the receptors. Binding of LDL causes the congregation in one area and signals the process of vesicle production. In the genetically distinct form of familial hypercholesterolemia mentioned in the question, the physical connection does not occur.
6. Passive transport does not require ATP or energy. It is driven by diffusion whereby molecules move from a higher concentration to a lower concentration. This can occur directly across a membrane (oxygen, carbon dioxide, water) or through special transport proteins (glucose transport proteins of red blood cells). Active transport requires energy, ATP as well as other sources (sodium-glucose transporter uses a sodium gradient for the force to transport glucose). Ions are usually transported in active transport.
7. The Na⁺/K⁺ ATPase transporter is an example of anti-port transport because it transports Na⁺ ions out (3 atoms) of the cell and K⁺ (2 atoms) ions into the cell. This means it transports ions in two different directions. This transporter uses ATP energy to maintain cellular osmotic pressure and ion gradients needed for transmission of signals in nerve cells.
8. Bacteriorhodopsin has three identical polypeptide chains each one with a transmembrane component and a molecule of retinal (Vitamin A). When light reaches the Vitamin A it isomerizes from the cis to trans form. This change in conformation allows the transport of protons into and out of the cell (bacterium) which creates a proton gradient across the cell membrane that provides the energy to synthesize ATP.

Lesson 6: Vitamins and Cofactors

1. A vitamin is an organic compound that is essential for normal growth and nutrition. They typically cannot be synthesized by the body and so are required in small quantities in the diet. A cofactor is a non-protein compound or metallic ion that is essential for the activity of an enzyme. Vitamins can be cofactors.
2. Folate, or folic acid, is a B vitamin. It is water-soluble so not stored by the body. It is necessary for DNA and RNA synthesis as well as protein synthesis. A deficiency in folate results in anemia and in fetal neural tube and brain defects.
3. One of the functions of vitamin A is in vision. Vitamin A is a fat-soluble vitamin.
4. The function of vitamin D is to assist in the intestinal absorption of calcium and phosphate to maintain bones. Vitamin D is synthesized from cholesterol in a reaction catalysed by UV light.
5. Vitamin K is needed for blood clotting. It functions as a cofactor for the enzyme that modifies prothrombin to bind calcium and allow it to be positioned closer to the site of a wound.

If you wish to discuss any of these questions or need assistance with the material, please contact your Academic Expert (AE) by emailing the Student Success Centre at fst_success@athabascau.ca.