Biochemistry, 4th Edition P55 ppt

16.5 How Do Bacterial Flagella Use a Proton Gradient to Drive Rotation? 503 hairpin loops of the six protein subunits form a spiral staircase, following the ssDNA as it threads through the central pore of the hexamer (Figure 16.26). Eric Enemark and Leemor Joshua-Tor have suggested that a central hairpin loop from one of the AAAϩ subunits coordinates each DNA nucleotide as it enters the helicase pore. Then, as each AAAϩ domain proceeds through the intermediate states of ATP binding and hydrolysis, its hairpin loop steps down through the six conformations of the staircase, maintaining continuous contact with its nucleotide, as it escorts it through the pore, finally releasing the nucleotide as it exits the pore. Following release, the hairpin moves back to the top of the staircase, picks up the next available nucleotide, and begins another journey down the staircase. For one full cycle of the hexamer, each subunit hydrolyzes one ATP, releases one ADP, and translocates one nucleotide through the central pore. A full cycle thus translocates six nucleotides with associated hydrolysis of six ATPs and release of six ADPs. 16.5 How Do Bacterial Flagella Use a Proton Gradient to Drive Rotation? Bacterial cells swim and move by rotating their flagella. The flagella of E. coli are helical filaments up to 15,000 nm (15 ␮m) in length and 15 nm in diameter. The direction of rotation of these filaments affects the movements of the cell. When the half-dozen flagella on the surface of the bacterial cell rotate in a counterclockwise (CCW) direction, they twist and bundle together in a left-handed helical structure and rotate in a concerted fashion, propelling the cell through the medium. Every few seconds, the flagellar motor reverses, the helical bundle of filaments (now turn- ing clockwise, or CW) unwinds into a jumble, and the bacterium somersaults or tumbles. Alternating between CCW and CW rotations, the bacterium can move toward food sources, such as amino acids and sugars. The rotations of bacterial flagellar filaments are the result of the rotation of motor protein complexes in the bacterial plasma membrane. (a) (b) FIGURE 16.26 The hairpin loops of the E1 helicase hexamer are arranged in a spiral staircase that winds around the DNA strand (a) side view; (b) axial view (pdb id ϭ 2GXA). As the helicase moves along the strand, the hairpin loop of one protein monomer binds each nucleotide as it enters the central cavity of the helicase.The loop adopts six conformations (a) as the helicase moves along the DNA, preserving the loop–nucleotide interaction until the nucleotide exits the cavity.The released protein loop then returns to the other end of the cavity to bind a new, incoming nucleotide. DNA is shown as a stick structure. His 507 of each hairpin loop is shown in space-filling mode. 504 Chapter 16 Molecular Motors LP rings FlgI, FlgH Outer membrane Cell membrane Cytoplasm Peptidogylcan layer (cell wall) Hook FlgE L P S M MS rings FliF, FliG C ring FliM, FliN and FliG Cap FliD Filament (Flagellin) The Flagellar Rotor Is a Complex Structure The flagellum is built from at least 25 proteins and comprises three parts: a rotary motor anchored in the bacterial inner membrane, a long filament that serves as a helical propellor, and a “hook” that functions as a universal joint that connects the motor with the filament (Figure 16.27). The rotary motor includes several rings of protein subunits, including the C ring, the MS ring, the P ring, and the L ring. The MS ring is built from 26 copies of the protein FliF. The C ring is attached to the MS ring and includes three “rotor” proteins—FliG, FliM, and FliN—involved in rotation of the motor. The C ring includes 26 copies of FliG, 34 copies of FliM, and 34 ϫ 4 ϭ 136 copies of FliN. The stationary portion of the motor—the “stator”—is formed from the proteins motA and motB. Eight motA 4 –motB 2 complexes are em- bedded in the bacterial inner membrane around the MS ring. Gradients of H ؉ and Na ؉ Drive Flagellar Rotors What energy source drives the flagellar motor? Gradients of protons and Na ϩ ions exist across bacterial inner membranes, typically with more H ϩ and Na ϩ outside the cell. In E. coli, spontaneous inward flow of protons through the motA–motB complexes drives the rotation of the motor (Figure 16.28). In Vibrio cholerae, inward Na ϩ ion flow powers the motor. Flagellar motors are thus energy conversion de- vices. In E. coli, each motA–motB complex passes 70 H ϩ per revolution of the motor. With a full complement of eight motA–motB complexes, a motor conducts about 560 protons per revolution. The H ϩ -driven flagellar rotors reach top rotational speeds of about 360 Hz (corresponding to 21,600 rpm). Thus, the overall rate of proton flow for the motor is approximately 200,000 H ϩ /sec! Flagellar motors driven by Na ϩ ions are even faster, with rotational rates of 1700 Hz (100,000 rpm) observed in Vibrio. The motA–motB complexes work with FliG in the C ring to transfer protons across the membrane. FliG contains 335 residues, and most of the FliG protein structure (residues 104 to 335) consists of two compact domains joined by an ␣-helix (Figure 16.28). A ridge on the C-terminal domain contains five charged residues that interact with motA and are important for motor rotation. Asp 32 of motB is essential for rotation of the motor and is probably involved in proton transfer. David Blair has proposed a model for creation of two membrane channels from the transmembrane segments of the motA 4 –motB 2 complex. Blair has suggested that each encounter of a motA–motB complex with a FliG subunit as the motor turns results in movement of one proton through each of these channels. The pas- sage of about 70 H ϩ through each motA–motB complex in one revolution of the FIGURE 16.27 A model of the E. coli flagellar motor.The motor is anchored by interactions of stationary motA and motB proteins in the M and S rings with the inner membrane. Spontaneous flow of protons through the motA–motB complexes and into the cell drives the rotation of the motor. Flow rates of 200,000 protons per second drive the motor at speeds approaching 22,000 rpm. (Adapted from Thomas, D. R., Morgan, D. G., Francis, N. R., and DeRosier, D. J., 2007. Bit by bit the structure of the complete flagellar hook/basal body complex. Microscopy and Microanalysis 13:34–35.Image provided by David J. DeRosier, Brandeis University.) 16.5 How Do Bacterial Flagella Use a Proton Gradient to Drive Rotation? 505 motor (which would involve encounters with 34 FliG subunits) is consistent with this suggestion (70/34 ϭ ϳ2). The Flagellar Rotor Self-Assembles in a Spontaneous Process Flagellar rotors are true masterpieces of biological self-assembly. The ring of FliF subunits, within the MS ring, is the first to assemble in the plasma membrane. Other proteins then attach to this ring one after another, from the base to the tip, to con- struct the motor structure. Once the motor has formed, the flexible hook and the flagellar filament are assembled. Precise recognition of the existing template structure allows this highly ordered self-assembly process to proceed without error. The flagellar filament is made from 20,000 to 30,000 copies of flagellin polymerized into a hollow helical tube structure. Each turn of the helical filament contains about 5000 flagellin subunits and is about 2300 nm long. A complete flagellum can have up to six full helical turns. Flagellin molecules are transported through the long, narrow, central channel of the motor and flagellum from the cell interior to the far end of the flagellum, where they self-assemble with the help of a pentameric complex of FliD, a capping protein (see Figure 16.27). The FliD complex has a plate and five leg domains. It rotates in a stepping fashion at the end of the filament, ex- posing one binding site at a time and guiding the binding of newly arriving flagellin molecules in a helical pattern. Flagellar Filaments Are Composed of Protofilaments of Flagellin Each cylindrical flagellar filament is composed of 11 fibrils or protofilaments that form the cylinder, with each fibril lying at a slight tilt to the cylinder axis (Figure 16.29a). An end-on view of the filament shows 11 subunits, each representing the end of a protofilament (Figure 16.29b). The flagellin protein of Salmonella typhi- ؊؊؉؊؊؉ MotA ؉؊؊؉؊؊ FliG middle domain Gly-Gly FliG C-terminal domain FliM MS ring MotA Membrane MotB K 262 D 288 D 289 R 297 R 281 FIGURE 16.28 Interactions between the stationary motA–motB complexes and the rotating FliG ring drive the flagellar motor. Proton flow through the motA–motB complexes is presumably coupled to conformation changes that alter ionic interactions between charged residues at the motA–motB and FliG interface, driving rotation of the FliG ring.Other conserved features include a hydrophobic patch (light green), a Gly-Gly motif (purple), and a EHPQR motif (blue, in the middle domain (pdb id ϭ 1LKV). 506 Chapter 16 Molecular Motors murium contains 494 residues and consists of four domains, denoted D0 through D3 (Figure 16.29c). D0 and D1 are composed of ␣-helices, whereas D2 and D3 consist primarily of ␤-strands. The N-terminus of the peptide chain lies at the base of D0. The peptide runs from D0 to D3 and then reverses and returns to D0, where the N- and C-termini are juxtaposed. The structure resembles a Greek capital gamma (⌫), with a height of 140 Å and a width of 110 Å. Each flagellin protein is arranged with D0 inside the filament and D3 facing the outside. The central pore, 20 Å in diameter, is lined by the ␣-helices of D0. Motor Reversal Involves Conformation Switching of Motor and Filament Proteins The flagellar motor reverses direction every few seconds so that the bacterium can change its swimming direction to seek better environments. Motor reversal involves conformation changes both in motor proteins and also in the filament itself. In the motor structure, the rotor proteins FliG, FliM, and FliN work together to control direction changes of the motor, and they are known collectively as the switch complex. FliN appears to lie at the base of the C ring, FliG lies at the top of the C ring, and FliM resides in the middle, contacting both FliN and FliG (Figure 16.30). Reversal of the flagellar motor causes the long filament to switch from a left- handed helical structure to a right-handed helical form. This makes the bundle of flagella fall apart, causing the bacterium to tumble. This left–to–right switch in the filament is caused by a conformational change that occurs in the flagellin subunits in some protofilaments. Interestingly, the driving force for these conformation changes is probably the torque applied to D0 and D1 of flagellin subunits along the filament when the motor itself reverses. D0 D0 D1 D1 (a) (b) D2 D2 D3 D3 S D0 D1 D2 D3 S D1 (c) N C D2 D3 D0 FIGURE 16.29 The E. coli flagellum is composed of 11 protofilaments that run the length of the flagellar filament.The filament is shown in cross section (a) and perpendicular to the filament (b). The protofilaments are long polymers of the flagellin protein (c), which consists of two ␣-helical domains (D0 and D1) that lie at a slight tilt to the filament axis and two ␤-sheet domains (D2 and D3) that extend outward from the filament. The N- and C-termini of the polypeptide are indicated (pdb id ϭ 1UCU). (Parts (a) and (b) courtesy of Keiichi Namba, Osaka University, Japan.) Summary 507 FliG FliM FliN FliG FliN FliM FIGURE 16.30 The switch complex that controls direction changes by the flagellar rotor consists of the rotor proteins FliG, FliM, and FliN. Interactions between these three proteins are presumed to control the direction of the rotor. Direction changes initiated here are communicated by protein conformation changes across the motor complex and throughout the length of the filament. Self-association of FliM subunits is mediated by hydrophilic residues of the ␣1 helix (red) on one subunit and on a short helix and loop on the adjacent subunit. Juxtaposed FliN subunits in the ring form a hydrophobic cleft (yellow). (FliG: pdb id ϭ 1LKV; FliM: pdb id ϭ 2HP7; FliN: pdb id ϭ 1YAB.) (Image on left courtesy of David J. DeRosier, Brandeis University.) SUMMARY 16.1 What Is a Molecular Motor? Motor proteins, also known as molecular motors, use chemical energy (ATP) to orchestrate different movements, transforming ATP energy into the mechanical energy of motion. In all cases, ATP hydrolysis is presumed to drive and control protein conformational changes that result in sliding or walking movements of one molecule relative to another. To carry out directed movements, molecular motors must be able to associate and dissociate reversibly with a polymeric protein array, a surface, or substructure in the cell. ATP hydrolysis drives the process by which the motor protein ratchets along the protein array or surface. Molecular motors may be linear or rotating. Linear motors crawl or creep along a polymer lattice, whereas rotating motors consist of a rotating element (the “rotor”) and a stationary element (the “stator”), in a fashion much like a simple elec- trical motor. 508 Chapter 16 Molecular Motors 16.2 What Is the Molecular Mechanism of Muscle Contraction? Exam- ination of myofibrils in the electron microscope reveals a banded or stri- ated structure. The so-called H zone shows a regular, hexagonally arranged array of thick filaments, whereas the I band shows a regular, hexagonal array of thin filaments. In the dark regions at the ends of each A band, the thin and thick filaments interdigitate. The thin filaments are composed primarily of three proteins called actin, troponin, and tropo- myosin. The thick filaments consist mainly of a protein called myosin. The thin and thick filaments are joined by cross-bridges. These cross- bridges are actually extensions of the myosin molecules, and muscle contraction is accomplished by the sliding of the cross-bridges along the thin filaments, a mechanical movement driven by the free energy of ATP hydrolysis. Myosin, the principal component of muscle thick filaments, is a large protein consisting of six polypeptides, including light chains and heavy chains. The heavy chains consist of globular amino-terminal myosin heads, joined to long ␣-helical carboxy-terminal segments, the tails. These tails are intertwined to form a left-handed coiled coil approximately 2 nm in diameter and 130 to 150 nm long. The myosin heads exhibit ATPase activity, and hydrolysis of ATP by the myosin heads drives muscle contraction. The free energy of ATP hydrolysis is translated into a conformation change in the myosin head, so dissociation of myosin and actin, hydrolysis of ATP, and rebinding of myosin and actin occur with stepwise movement of the myosin S1 head along the actin filament. The conformation change in the myosin head is driven by the hydrolysis of ATP. 16.3 What Are the Molecular Motors That Orchestrate the Mechano- chemistry of Microtubules? Microtubules are hollow, cylindrical structures, approximately 30 nm in diameter, formed from tubulin, a dimeric protein composed of two similar 55-kD subunits known as ␣-tubulin and ␤-tubulin. Tubulin dimers polymerize to form microtubules, which are es- sentially helical structures, with 13 tubulin monomer “residues” per turn. Microtubules are, in fact, a significant part of the cytoskeleton, a sort of intracellular scaffold formed of microtubules, intermediate filaments, and microfilaments. In most cells, microtubules are oriented with their minus ends toward the centrosome and their plus ends toward the cell periphery. This consistent orientation is important for mechanisms of intracellular transport. Microtubules are also the fundamental building blocks of eukaryotic cilia and flagella. Microtubules also mediate the intracellular motion of organelles and vesicles. 16.4 How Do Molecular Motors Unwind DNA? When DNA is to be replicated or repaired, the strands of the double helix must be unwound and separated to form single-stranded DNA intermediates. This separation is carried out by molecular motors known as DNA helicases that move along the length of the DNA lattice, sequentially destabilizing the hydrogen bonds between complementary base pairs. The movement along the lattice and the separation of the DNA strands are coupled to the hydrolysis of nucleoside 5Ј-triphosphates. The E. coli BCD helicase, which is involved in recombination processes, can unwind 33,000 base pairs before it dissociates from the DNA lattice. Processive movement is essential for helicases involved in DNA replication, where millions of base pairs must be replicated rapidly. Certain hexameric helicases form ringlike structures that completely encircle at least one of the strands of a DNA duplex. Other helicases, notably Rep helicase from E. coli, are homodimeric and move processively along the DNA helix by means of a “hand-over-hand” movement that is remarkably similar to that of kinesin’s movement along microtubules. 16.5 How Do Bacterial Flagella Use a Proton Gradient to Drive Rotation? Bacterial cells swim and move by rotating their flagella. The direction of rotation of these flagella affects the movements of the cell. When the half- dozen flagella on the surface of the bacterial cell rotate in a counterclockwise direction, they twist and bundle together and rotate in a concerted fashion, propelling the cell through the medium. The rotations of bacterial flagellar filaments are the result of the rotation of motor protein complexes in the bacterial plasma membrane. The flagellar motor consists of multiple rings (including the MS ring and the C ring). The rings are surrounded by a circular array of membrane proteins. In all, at least 40 genes appear to code for proteins involved in this magnificent assembly. One of these, the motB protein, lies on the edge of the M ring, where it interacts with the motA protein, located in the membrane protein array and facing the M ring. In contrast to the many other motor proteins described in this chapter, a proton gradient, not ATP hydrolysis, drives the flagellar motor. PROBLEMS Preparing for an exam? Create your own study path for this chapter at www.cengage.com/login. 1. The cheetah is generally regarded as nature’s fastest mammal, but another amazing athlete in the animal kingdom (and almost as fast as the cheetah) is the pronghorn antelope, which roams the plains of Wyoming. Whereas the cheetah can maintain its top speed of 70 mph for only a few seconds, the pronghorn antelope can run at 60 mph for about an hour! (It is thought to have evolved to do so in order to elude now-extinct ancestral cheetahs that lived in North America.) What differences would you expect in the muscle structure and anatomy of pronghorn antelopes that could account for their remarkable speed and endurance? 2. An ATP analog, ␤,␥-methylene-ATP, in which a OCH 2 O group re- places the oxygen atom between the ␤- and ␥-phosphorus atoms, is a potent inhibitor of muscle contraction. At which step in the contraction cycle would you expect ␤,␥-methylene-ATP to block contraction? 3. ATP stores in muscle are augmented or supplemented by stores of phosphocreatine. During periods of contraction, phosphocreatine is hydrolyzed to drive the synthesis of needed ATP in the creatine kinase reaction: Phosphocreatine ϩ ADP⎯⎯→creatine ϩ ATP Muscle cells contain two different isozymes of creatine kinase, one in the mitochondria and one in the sarcoplasm. Explain. 4. Rigor is a muscle condition in which muscle fibers, depleted of ATP and phosphocreatine, develop a state of extreme rigidity and cannot be easily extended. (In death, this state is called rigor mortis, the rigor of death.) From what you have learned about muscle contraction, explain the state of rigor in molecular terms. 5. Skeletal muscle can generate approximately 3 to 4 kg of tension or force per square centimeter of cross-sectional area. This number is roughly the same for all mammals. Because many human muscles have large cross-sectional areas, the force that these muscles can (and must) generate is prodigious. The gluteus maximus (on which you are probably sitting as you read this) can generate a tension of 1200 kg! Estimate the cross-sectional area of all of the muscles in your body and the total force that your skeletal muscles could generate if they all contracted at once. 6. Calculate a diameter for a tubulin monomer, assuming that the monomer MW is 55,000, that the monomer is spherical, and that the den- sity of the protein monomer is 1.3 g/mL. How does the number that you calculate compare to the dimension portrayed in Figure 16.12? 7. Use the number you obtained in problem 6 to calculate how many tubulin monomers would be found in a microtubule that stretched across the length of a liver cell. (See Table 1.2 for the diameter of a liver cell.) 8. The giant axon of the squid may be up to 4 inches in length. Use the value cited in this chapter for the rate of movement of vesicles Further Reading 509 and organelles across axons to determine the time required for a vesicle to traverse the length of this axon. 9. As noted in this chapter, the myosin molecules in thick filaments of muscle are offset by approximately 14 nm. To how many residues of a coiled-coil structure does this correspond? 10. Use the equations of Chapter 9 to determine the free energy dif- ference represented by a Ca 2ϩ gradient across the sarcoplasmic reticulum membrane if the luminal (inside) concentration of Ca 2ϩ is 1 mM and the concentration of Ca 2ϩ in the solution bathing the muscle fibers is 1 ␮M. 11. Use the equations of Chapter 3 to determine the free energy of hydrolysis of ATP by the sarcoplasmic reticulum Ca-ATPase if the concentration of ATP is 3 mM, the concentration of ADP is 1 mM, and the concentration of P i is 2 mM. 12. Under the conditions described in problems 10 and 11, what is the maximum number of Ca 2ϩ ions that could be transported per ATP hydrolyzed by the Ca-ATPase? 13. For each of the motor proteins in Table 16.2, calculate the force ex- erted over the step size given, assuming that the free energy of hydrolysis of ATP under cellular conditions is Ϫ50 kJ/mol. 14. When you go to the gym to work out, you not only exercise many muscles but also involve many myosins (and actins) in any given exercise activity. Suppose you lift a 10-kg weight a total distance of 0.4 m. Using the data in Table 16.2 for myosin, calculate the mini- mum number of myosin heads required to lift this weight and the number of sliding steps these myosins must make along their associated actin filaments. 15. In which of the following tissues would you expect to find smooth muscle? a. Arteries b. Stomach c. Urinary bladder d. Diaphragm e. Uterus f. The gums in your mouth 16. When an action potential (nerve impulse) arrives at a muscle membrane (sarcolemma), in what order do the following events occur? a. Release of Ca 2ϩ ions from the sarcoplasmic reticulum b. Hydrolysis of ATP, with release of energy c. Detachment of myosin from actin d. Sliding of myosin along actin filament e. Opening of switch 1 and switch 2 on myosin head 17. (Essay question.) You are invited by the National Science Founda- tion to attend a scientific meeting to set the agenda for funding of basic research related to molecular motors for the next 10 years. Only basic research will be funded, ruling out studies on human subjects. You are asked to suggest the research area most worthy of scientific research. Your presentation must include (1) a brief back- ground on what we currently know about the subject; (2) identifi- cation of a key research topic about which more needs to be known; and (3) a justification of why additional knowledge in this area is critical for advancing the field (that is, why investigations in this area are especially important). You are not being asked to provide the methods or experiments that might be used to address the problem—only the concept. Base your presentation on what you have learned in this chapter (you may consult and include refer- ences from the Further Reading section), and limit your presentation to 300 words. Preparing for the MCAT Exam 18. Consult Figure 16.17 and use the data in problem 8 to determine how many steps a kinesin motor must take to traverse the length of the squid giant axon. 19. When athletes overexert themselves on hot days, they often suffer immobility from painful muscle cramps. Which of the following is a reasonable hypothesis to explain such cramps? a. Muscle cells do not have enough ATP for normal muscle relax- ation. b. Excessive sweating has affected the salt balance within the muscles. c. Prolonged contractions have temporarily interrupted blood flow to parts of the muscle. d. All of the above. 20. Duchenne muscular dystrophy is a sex-linked recessive disorder associated with severe deterioration of muscle tissue. The gene for the disease: a. is inherited by males from their mothers. b. should be more common in females than in males. c. both a and b. d. neither a nor b. FURTHER READING Muscle Contraction Bagshaw, C. R., 2007. Myosin mechanochemistry. Structure 15:511–512. Coureux, P D., Sweeney, H. L., et al., 2004. Three myosin V structures delineate essential features of chemo-mechanical transduction. EMBO Journal 23:4527–4537. Fischer, S., Windshugel, B., et al., 2005. Structural mechanism of the re- covery stroke in the myosin molecular motor. 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Molecular motors and mechanisms of directional transport in neurons. Nature Reviews Neuro- science 6:201–214. Michie, K., and Lowe, J., 2006. Dynamic filaments of the bacterial cytoskeleton. Annual Review of Biochemistry 75:467–492. Styers, M., Kawalczyk, A. P., et al., 2007. Intermediate filaments and vesicular membrane traffic: The odd couple ’s first dance? Traffic 6:359–365. Tekotte, H., and Davis, I., 2002. Intracellular mRNA localization: Motors move messages. Trends in Genetics 18:636–642. Vale, R. D., 2003. The molecular motor toolbox for intracellular transport. Cell 112:467–480. Vale, R. D., and Milligan, R. A., 2000. The way things move: Looking under the hood of molecular motor proteins. Science 288:88–95. Verhey, K. J., and Gaertig, J., 2007. The tubulin code. Cell Cycle 6:2152–2160. Helicases Castella, S., Bingham, G., et al., 2006. Common determinants in DNA melting and helicase-catalyzed DNA unwinding by papillomavirus replication protein E1. 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Roles of charged residues of rotor and stator in flagellar rotation: Comparative study using H ϩ -driven and Na ϩ -driven motors in Escherichia coli. Journal of Bacteriology 188: 1466–1472. © Gray Hardel/CORBIS 17 Metabolism: An Overview 17.1 Is Metabolism Similar in Different Organisms? One of the great unifying principles of modern biology is that organisms show marked similarity in their major pathways of metabolism. Given the almost unlimited possibil- ities within organic chemistry, this generality would appear most unlikely. Yet it’s true, and it provides strong evidence that all life has descended from a common ancestral form. All forms of nutrition and almost all metabolic pathways evolved in early prokaryotes prior to the appearance of eukaryotes 1 billion years ago. For example, glycolysis, the metabolic pathway by which energy is released from glucose and captured in the form of ATP under anaerobic conditions, is common to almost every cell. It is believed to be the most ancient of metabolic pathways, having arisen prior to the appearance of oxygen in abundance in the atmosphere. All organisms, even those that can synthesize their own glucose, are capable of glucose degradation and ATP synthesis via glycolysis. Other prominent pathways are also virtually ubiquitous among organisms. Living Things Exhibit Metabolic Diversity Although most cells have the same basic set of central metabolic pathways, different cells (and, by extension, different organisms) are characterized by the alternative pathways they might express. These pathways offer a wide diversity of metabolic pos- sibilities. For instance, organisms are often classified according to the major metabolic pathways they exploit to obtain carbon or energy. Classification based on carbon requirements defines two major groups: autotrophs and heterotrophs. Autotrophs are organisms that can use just carbon dioxide as their sole source of carbon. Heterotrophs require an organic form of carbon, such as glucose, in order to synthesize other essential carbon compounds. Classification based on energy sources also gives two groups: phototrophs and chemotrophs. Phototrophs are photosynthetic organisms, which use light as a source of energy. Chemotrophs use organic compounds such as glucose or, in some instances, oxidizable inorganic substances such as Fe 2ϩ , NO 2 Ϫ , NH 4 ϩ , or elemental sulfur as sole sources of energy. Typically, the energy is extracted through oxidation–reduction reactions. Based on these characteristics, every organism falls into one of four categories (Table 17.1). Metabolic Diversity Among the Five Kingdoms Prokaryotes (the kingdom Monera—archaea and bacteria) show a greater metabolic diversity than all the Anise swallowtail butterfly (Papilio zelicans) with its pupal case. Metamorphosis of butterflies is a dra- matic example of metabolic change. All is flux, nothing stays still. Nothing endures but change. Heraclitus (c. 540–c. 480 B.C.) KEY QUESTIONS 17.1 Is Metabolism Similar in Different Organisms? 17.2 What Can Be Learned from Metabolic Maps? 17.3 How Do Anabolic and Catabolic Processes Form the Core of Metabolic Pathways? 17.4 What Experiments Can Be Used to Elucidate Metabolic Pathways? 17.5 What Can the Metabolome Tell Us about a Biological System? 17.6 What Food Substances Form the Basis of Human Nutrition? ESSENTIAL QUESTION The word metabolism derives from the Greek word for “change.” Metabolism repre- sents the sum of the chemical changes that convert nutrients, the “raw materials” necessary to nourish living organisms, into energy and the chemically complex finished products of cells. Metabolism consists of literally hundreds of enzymatic reactions organized into discrete pathways.These pathways proceed in a stepwise fashion, transforming substrates into end products through many specific chemical intermediates. Metabolism is sometimes referred to as intermediary metabolism to reflect this aspect of the process. What are the anabolic and catabolic processes that satisfy the metabolic needs of the cell? Create your own study plan for this chapter with tutorials, simulations, animations, and Active Figures at www.cengage.com/login. 512 Chapter 17 Metabolism: An Overview Classification Carbon Source Energy Source Electron Donors Examples Photoautotrophs Photoheterotrophs Chemoautotrophs Chemoheterotrophs TABLE 17.1 Metabolic Classification of Organisms According to Their Carbon and Energy Requirements CO 2 Organic compounds CO 2 Organic compounds Light Light Oxidation–reduction reactions Oxidation–reduction reactions H 2 O, H 2 S, S, other inorganic compounds Organic compounds Inorganic compounds: H 2 , H 2 S, NH 4 ϩ , NO 2 Ϫ , Fe 2ϩ , Mn 2ϩ Organic compounds (e.g., glucose) Green plants, algae, cyanobacteria, photosynthetic bacteria Nonsulfur purple bacteria Nitrifying bacteria; hydrogen, sulfur, and iron bacteria All animals, most microorganisms, nonphotosynthetic plant tissue such as roots, photosynthetic cells in the dark A DEEPER LOOK Calcium Carbonate—A Biological Sink for CO 2 A major biological sink for CO 2 that is often overlooked is the calcium carbonate shells of corals, molluscs, and crustacea. These invertebrate animals deposit CaCO 3 in the form of pro- tective exoskeletons. In some invertebrates, such as the sclerac- tinians (hard corals) of tropical seas, photosynthetic dinoflagel- lates (kingdom Protoctista) known as zooxanthellae live within the animal cells as endosymbionts. These phototrophic cells use light to drive the resynthesis of organic molecules from CO 2 released (as bicarbonate ion) by the animal’s metabolic activity. In the presence of Ca 2ϩ , the photosynthetic CO 2 fixation “pulls” the deposition of CaCO 3 , as summarized in the following coupled reactions: Ca 2ϩ ϩ 2 HCO 3 Ϫ 34 CaCO 3(s) ↓ ϩ H 2 CO 3 H 2 CO 3 34 H 2 O ϩ CO 2 H 2 O ϩ CO 2 ⎯⎯→carbohydrate ϩ O 2 CO 2 Photoautotrophic cells O 2 H 2 O Heterotrophic cells Glucose Solar energy FIGURE 17.1 The flow of energy in the biosphere is coupled primarily to the carbon and oxygen cycles. four eukaryotic kingdoms (Protoctista [previously called Protozoa], Fungi, Plants, and Animals) put together. Prokaryotes are variously chemoheterotrophic, photoautotrophic, photoheterotrophic, or chemoautotrophic. No protoctista are chemoautotrophs; fungi and animals are exclusively chemoheterotrophs; plants are characteristically photoautotrophs, although some are heterotrophic in their mode of carbon acquisition. Oxygen Is Essential to Life for Aerobes A further metabolic distinction among organisms is whether or not they can use oxygen as an electron acceptor in energy-producing pathways. Those that can are called aerobes or aerobic organisms; others, termed anaerobes, can subsist without O 2 . Organisms for which O 2 is obligatory for life are called obligate aerobes; humans are an example. Some species, the so-called facultative anaerobes, can adapt to anaerobic conditions by substituting other electron acceptors for O 2 in their energy-producing pathways; Escherichia coli is an example. Yet others cannot use oxygen at all and are even poisoned by it; these are the obligate anaerobes. Clostridium botulinum, the bacterium that produces botulin toxin, is representative. The Flow of Energy in the Biosphere and the Carbon and Oxygen Cycles Are Intimately Related The primary source of energy for life is the sun. Photoautotrophs utilize light energy to drive the synthesis of organic molecules, such as carbohydrates, from atmospheric CO 2 and water (Figure 17.1). Heterotrophic cells then use these organic products of photosynthetic cells both as fuels and as building blocks, or precursors, for the biosynthesis of their own unique complement of biomolecules. Ultimately, CO 2 is

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