Chapter_14_Solutions

Chapter_14_Solutions - Chapter 14 Energy Conversion:...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 14 Energy Conversion: Mitochondria and Chloroplasts THE MITOCHONDRION DEFINITIONS 141 142 143 144 145 146 147 148 149 Electron-transport chain Intermembrane space Citric acid cycle Respiratory chain ATP synthase Chemiosmotic coupling Oxidative phosphorylation Electrochemical proton gradient Outer membrane 14 In This Chapter THE MITOCHONDRION A307 ELECTRON-TRANSPORT A315 CHAINS AND THEIR PROTON PUMPS CHLOROPLASTS AND PHOTOSYNTHESIS A323 THE GENETIC SYSTEMS A331 OF MITOCHONDRIA AND PLASTIDS TRUE/FALSE 1410 False. The intermembrane space is chemically equivalent to the cytosol with respect to small molecules, not because of the presence of any specialized transport proteins in the mitochondrial outer membrane, but because this membrane contains many copies of the channel-forming protein porin, which forms large aqueous channels. These channels convert the outer membrane into a sieve that allows free passage of all molecules of molecular mass less than 5000 daltons, which includes all the ions and metabolites in a cell and some small proteins. True. When the citric acid cycle is operating as a cycle of reactions, its sole products are CO2 and reduced forms of the electron carriers NADH and FADH2. The electrons in NADH and FADH2 are passed into the electrontransport chain to generate ATP via oxidative phosphorylation. True. The natural (thermodynamic) tendency of electrons is to move from low-affinity to high-affinity carriers. True. Because submitochondrial particles are inside out, electron transport initiated by the addition of oxygen will cause protons to be pumped into the vesicle, causing the medium to become more basic. 1411 1412 1413 THOUGHT PROBLEMS 1414 The characteristic organization of mitochondria in liver cells and cardiac muscle cells is likely to reflect functional differences in these cells. Most of the energy needs of liver cells derive from metabolic processes in the cytosol, whereas the energy needs of cardiac muscle cells result primarily A307 A308 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts from contractions of the myofibrils. Mitochondria are thus positioned in these cells at the sites where energy, in the form of ATP, is most required. 1415 The inner mitochondrial membrane is the site of oxidative phosphorylation, and it produces most of the cell's ATP. Cristae are portions of the mitochondrial inner membrane that are folded inward. Mitochondria that have a higher density of cristae have a larger area of inner membrane and therefore a greater capacity to carry out oxidative phosphorylation. Heart muscle expends a lot of energy during its continuous contractions, whereas skin cells have a lesser energy demand. An increased density of cristae therefore increases the ATP-production capacity of the heart muscle cell. This is a remarkable example of how cells adjust the abundance of their individual components according to need. In the presence of oxygen, yeast can generate about 15 times more ATP from each glucose molecule than they can in the absence of oxygen. Thus, to meet their energy needs they need to process about 15-fold fewer glucose molecules; hence the dramatic drop in glucose consumption when O2 is introduced. The sprinter generates sufficient ATP to power muscle contraction during a sprint by the simple expedient of running enormous amounts of glucose through the glycolytic pathway. Thus, a sprinter metabolizes 15 molecules of glucose to harvest the same amount of energy that a marathon runner gets from a single molecule, and metabolizes even more glucose to meet the increased per-second energy demand of sprinting over marathon running. One measure of this difference is the length of time that the muscle's glycogen reserve--the intracellular source of glucose--will last during the two types of running. At maximum power output, muscle glycogen would last a sprinter less than 80 seconds, whereas the glycogen reserve (supplemented by circulating glucose and fatty acid oxidation) lasts the marathon runner more than 2 hours. The citric acid cycle uses NAD+ and FAD to make NADH and FADH2. Oxidative phosphorylation converts NADH and FADH2 back to NAD+ and FAD. The cycling of these cofactors between the citric acid cycle and oxidative phosphorylation is critical because they are present in very small amounts relative to the quantity of acetyl CoA that is metabolized in the cycle. In the absence of oxygen, electron transport and ATP synthesis do not occur; thus, NADH and FADH2 cannot be converted back to NAD+ and FAD. As a result, the citric acid cycle quickly converts the available NAD+ and FAD into NADH and FADH2 and then grinds to a halt. The requirement for cycling of cofactors is a general principle of metabolic regulation: it provides a rough coordination of flow through pathways that are not otherwise connected. In the intact mitochondrion NADH donates its electrons to the respiratory chain from the matrix side of the inner membrane. Likewise, ATP is made from ADP and phosphate on the matrix side of the inner membrane. With intact mitochondria it is difficult for a scientist to manipulate the concentrations of these small molecules in the matrix because the inner membrane encloses it. By contrast, in submitochondrial particles with the matrix side exposed to the surroundings, it is easy to add different concentrations of NADH, ATP ADP, and phosphate (as well as inhibitors and artificial electron , donors and acceptors) and follow the consequences for electron transport and ATP synthesis. As soon as protons enter the intermembrane space, they move rapidly throughout the cytosol. They are not confined to the intermembrane space because of the large aqueous channels that are present in the outer mitochondrial membrane. 1416 1417 1418 1419 1420 1421 A. You would expect ATP to be synthesized, as shown in Figure 1431. When the vesicles are exposed to light, protons will be pumped inside by bacterio- THE MITOCHONDRION rhodopsin, creating a pH difference across the vesicle membranes. The proton-motive force represented by the pH difference would drive the protons back out of the vesicles through the ATP synthase, causing ATP to be synthesized from ADP and phosphate in the external medium. B. If the vesicles were leaky to protons, no pH difference would be generated, hence no ATP could be synthesized. The protons pumped into the vesicles by bacteriorhodopsin would immediately leak back out without generating a pH difference. C. If the ATP synthase molecules were randomly oriented, you would still expect ATP to be synthesized, although at about half the rate. The molecules that were oriented correctly would make ATP; the oppositely oriented ATP synthase molecules would be inert. If bacteriorhodopsin were randomly oriented, you would expect much less ATP to be synthesized. In vesicles with equal numbers of oppositely oriented bacteriorhodopsin molecules, no pH difference would be generated upon exposure to light because the proton pumping in both directions would be equal. In vesicles with an excess of outwardly directed proton pumps, the pH difference would be in the wrong direction to be utilized by ATP synthase and, thus, no ATP would be made. In vesicles with an excess of inwardly directed proton pumps, a pH difference of the right orientation would be generated; thus, those vesicles would be capable of synthesizing some ATP. D. Using components from widely divergent organisms can be a very powerful experimental tool. Because the two proteins come from such different sources, it is very unlikely that they form a direct functional interaction. The experiment therefore strongly suggests that the pumping of protons (normally carried out by the respiratory chain) and the synthesis of ATP are separate events. The notion that electron transport and ATP synthesis are separate is now clearly established (this experiment was one of the earliest demonstrations); thus, this approach is a valid one. Reference: Racker E & Stoeckenius W (1974) Reconstitution of purple membrane vesicles catalyzing light-driven proton uptake and adenosine triphosphate formation. J. Biol. Chem. 249, 662663. 1422 A. DNP will collapse the electrochemical proton gradient completely. Protons that are pumped across the inner membrane will be carried back across the membrane by DNP, and therefore no energy can be stored across the membrane. B. The electrochemical proton gradient is made up of two components: an H+ concentration difference and an electrical potential. If the membrane is made permeable to K+ with valinomycin, K+ ions will be driven into the matrix by the electrical potential of the inner membrane (negative inside, positive outside). The influx of positively charged K+ ions will abolish the membrane's electrical potential. By contrast, the concentration component of the proton gradient (the pH difference) is unaffected by valinomycin. As a result, only part of the driving force that makes it energetically favorable for protons to flow back into the matrix will be lost. 1423 The stator plays the crucial role of anchoring the a3b3 complex to the rotor component of the ATP synthase. Fixing the a3b3 complex allows the g subunit to turn inside it and drive the set of conformational changes that are responsible for ATP synthesis. In the absence of the stator, the a3b3 complex would rotate along with the g subunit, and there would be no conformational changes. Thus, ATP would not be synthesized in the absence of the stator. The antiport of the amino acids Asp and Glu (V) is electrically neutral and doesn't involve a proton; thus, it is unaffected by the electrochemical proton gradient. All the rest are affected by the gradient. The symport of pyruvate and H+ (I) is electrically neutral, but because a proton is transported into the matrix, it is driven by the DpH component of the electrochemical gradient. The antiport of citrulline and ornithine (II), which are components of the LIGHT H+ sealed vesicle A309 H+ H+ H+ H+ H+ ADP + Pi ATP Figure 1431 ATP synthesis in reconstituted lipid vesicles containing bacteriorhodopsin and ATP synthase (Answer 1421). 1424 A310 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts urea cycle, is driven by the membrane-potential component of the electrochemical gradient. The symport of phosphate with two protons (III) is driven by the DpH component of the electrochemical gradient. The antiport of citrate plus a proton with malate (IV) occurs against the DpH component of the electrochemical proton gradient. CALCULATIONS 1425 The number of protons in an actively respiring liver mitochondrion at pH 7.5 (3.16 108 M H+) is about 10. 8 + 23 + 3 H+ = 3.16 10 mole H 6 10 H (4/3)(3.14)(0.5 mm) 15L 3 10 mm mitochondrion L mole H+ mitochondrion = 9.9 If the matrix of the mitochondrion started at pH 7 (107 M H+), it originally held about 31 protons (31.4). Thus, to reach pH 7.5, about 21 protons would need to be pumped out. These are remarkable results. Regardless of the particulars of mitochondrial size and exact pH values, it is clear that only a few tens of protons are normally involved in establishing the proton-motive force. More than anything, these results emphasize the dynamic nature of proton pumping and ATP synthesis. 1426 T. ferrooxidans can make ATP for free in its acidic environment. In the absence of a membrane potential, the free-energy change available from inward proton transport is DG = 2.3RT log [H+]in [H+]out 3 6.5 = 2.3 1.98 10 kcal 310 K log 102.0 K mole 10 3 = 2.3 1.98 10 kcal 310 K log 104.5 K mole DG = 6.4 kcal/mole If DG for ATP synthesis is 11 kcal/mole, it would take at least 2 moles of protons (2 6.4 kcal/mole = 12.8 kcal/mole), to drive ATP synthesis. Thus, each molecule of ATP would require that two protons be transported into the cell. The energy of transport would have to be coupled to the synthesis of ATP, but thermodynamic calculations give no clue as to the actual mechanism of coupling. The protons could enter singly or together, depending on the mechanism. If they entered one at a time, the energy from the first proton would have to be stored in such a way that the energy of the second proton could be added to it. ATP synthase in normal cells accomplishes much the same task: it couples the flow of 34 protons to the synthesis of one molecule of ATP . Reference: Ingledew JW (1982) Thiobacillus ferrooxidans: the bioenergetics of an acidophilic chemolithotroph. Biochim. Biophys. Acta 683, 89117. 1427 It would take the heart 6 seconds to consume its steady-state levels of ATP. Because each pair of electrons reduces one atom of oxygen, the 12 pairs of electrons generated by oxidation of one glucose molecule would reduce 6 O2. Thus, 30 ATP are generated per 6 O2 consumed. At steady state, the rate of ATP production equals its rate of consumption. The time in seconds required to consume the steady-state level of ATP is time = 5 mmol ATP 6 O2 60 sec min g 30 ATP 10 mmol O2 g min = 6 sec 1428 A. When the concentrations of the reactants and products are all 1 M, the reaction is at standard conditions and DG equals DG, which is 7.3 kcal/mole. THE MITOCHONDRION [ADP][Pi] [ATP] 3 = 7.3 kcal/mole + 2.3 1.98 10 kcal 310 K log 1 1 1 K mole A311 DG = DG + 2.3RT log Since the log of 1 is 0, DG = 7.3 kcal/mole When the concentrations of the reactants and products are all 1 mM, DG equals 11.5 kcal/mole. (At 37C, 2.3RT is 1.41 kcal/mole.) 3 3 DG = 7.3 kcal/mole + (1.4 kcal/mole) log 10 10 103 = 7.3 kcal/mole + (1.4 kcal/mole)(3) = 7.3 kcal/mole 4.2 kcal/mole DG = 11.5 kcal/mole B. At the given concentrations of ATP ADP, and Pi, DG for ATP hydrolysis is , 11.1 kcal/mole. 3 2 DG = 7.3 kcal/mole + (1.4 kcal/mole) log 10 10 5 103 = 7.3 kcal/mole + (1.4 kcal/mole)(2.7) = 7.3 kcal/mole 3.8 kcal/mole DG = 11.1 kcal/mole C. At equilibrium DG is 0. At equilibrium there is no longer any tendency for a reaction to proceed. If [Pi] is 10 mM at equilibrium, then the ratio of [ATP] to [ADP] will be 6.1 108. 0 = 7.3 kcal/mole + (1.4 kcal/mole) log [ADP] 10 [ATP] 7.3 kcal/mole = (1.4 kcal/mole)(log102 + log[ADP]/[ATP]) = (2)(1.4 kcal/mole) + (1.4 kcal/mole) log[ADP]/[ATP] log[ADP]/[ATP]= 10.1 kcal/mole 1.4 kcal/mole = 7.2 log[ATP]/[ADP] = 7.2 [ATP]/[ADP] = 6.1 108 2 D. At a constant [Pi], every 10-fold change in the ratio of [ATP] to [ADP] will alter DG by 1.4 kcal/mole. As shown below, a 10-fold increase in [ATP]/[ADP] will decrease DG by 1.4 kcal/mole. DG = DG + 1.4 kcal/mole log [ADP][Pi] [ATP] = DG + 1.4 kcal/mole log[Pi] + 1.4 kcal/mole log [ADP] [ATP] A 10-fold increase in [ATP]/[ADP], which is equal to a 10-fold decrease in [ADP]/[ATP], causes the log of the ratio in the expression above to decrease by 1. Thus each 10-fold increase in the ratio causes 1.4 kcal/mole to be subtracted from the right-hand side, thereby decreasing DG by 1.4 kcal/mole. A 100-fold increase in the ratio of [ATP]/[ADP] decreases DG by 2.8 kcal/mole; a 1000-fold increase in the ratio decreases DG by 4.2 kcal/mole. 1429 At an [ATP]/[ADP][Pi] ratio of 104, three protons would be required to drive synthesis of one ATP. The cost of synthesis of ATP under the specified conditions is DG = DG + 2.3RT log [ATP] [ADP][Pi] 7.3 kcal + 2.3 1.98 103 kcal 310 K log (104) = mole K mole = 12.9 kcal/mole A312 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts At least three protons (3 4.6 kcal/mole = 13.8 kcal/mole) would be required to balance the cost of ATP synthesis under these conditions. (A) ONE-STEP MECHANISM S ADP O P 16 18 H O + 17 O H DATA HANDLING 1430 A. This experiment distinguishes very nicely between mechanisms involving one or two phosphate transfers. Since each phosphate transfer results in inversion of the configuration around the phosphate atom, a one-step mechanism results in inversion and a two-step mechanism results in retention (inversion followed by inversion gives retention). As illustrated in Figure 1432A, direct attack of water on ATP to generate ADP and phosphate is a one-step mechanism and therefore produces inversion of configuration. Hydrolysis of ATP via an intermediate phosphorylated substance is a two-step mechanism, and therefore the configuration is retained (Figure 1432B). (Note that the result does not distinguish between a one-step mechanism and a mechanism involving three--or any odd number of--phosphate transfers.) B. Inversion of configuration during ATP hydrolysis by ATP synthase indicates that the hydrolysis reaction does not occur via two phosphate transfers and therefore argues against the involvement of a single phosphorylated intermediate. Thus, hydrolysis of ATP by ATP synthase probably involves the direct attack of H2O on ATP If hydrolysis of ATP is the reverse of the synthetic reac. tion (as it is thought to be), then synthesis of ATP from ADP and phosphate must also occur directly and not through a phosphorylated intermediate. Reference: Webb MR, Grubmeyer C, Penefsky HS & Trentham DR (1980) The stereochemical course of phosphoric residue transfer catalyzed by beef heart mitochondrial ATPase. J. Biol. Chem. 255, 1163711639. 1431 A. Presumably the hydrolysis of an individual ATP molecule provides the driving force for the 120 rotation of the g subunit, hence the corresponding revolution of the actin filament. Since a low concentration of ATP was used in these experiments, the pauses represent the variable times it takes for the next molecule of ATP to bind. Rotation through 120 corresponds to one ab dimer, the unit of ATP hydrolysis (or of synthesis in the normal direction). B. If three ATP molecules must be hydrolyzed to drive one complete rotation of the g subunit, then in its normal operation ATP synthase must synthesize three molecules per rotation of the g subunit. Reference: Masaike T, Mitome N, Noji H, Muneyuki E, Yasuda R, Kinosita K & Yoshida M (2000) Rotation of F1-ATPase and the hinge residues of the b subunit. J. Exp. Biol. 203, 18. 1432 The crystal structure of the a3b3 complex, with ADP and an analog of ATP bound, shows clearly that the correct arrangement is the one in Figure 145B. This arrangement can also be deduced from the counterclockwise revolution of the actin filament when ATP synthase hydrolyzes ATP. When ATP synthase is operating in reverse to hydrolyze ATP, the sequence of conformational changes is also reversed, as shown for arrangement B in Figure 1433. An actin filament is shown arbitrarily between the ATP-binding conformation and the empty conformation in step I in Figure 1433. After ATP hydrolysis to step II, a 120 counterclockwise revolution of the actin filament leaves it in the same relative position: between an ATP-binding conformation and an empty conformation. Likewise, after hydrolysis of another ATP to step III, a 120 counterclockwise revolution of the actin filament once again places it between an ATP-binding and an empty conformation. (If you try the same analysis for the arrangement in Figure 145A, you'll find that the neighbors of the actin filament change at each step, indicating that its revolution does not match the sequence of conformation changes.) Another way you might recognize that the arrangement in Figure 145B is the correct H 17 O ADP S 18 O 16 P O 17 O inversion (B) TWO-STEP MECHANISM S ADP O P 16 18 O + X-ENZYME O ADP S O 18 O 16 P O X-ENZYME H S 17 O P 16 18 O O retention Figure 1432 Stereochemical analysis of ATP hydrolysis (Answer 1430). (A) Inversion by a one-step mechanism. (B) Retention by a two-step mechanism. THE MITOCHONDRION A313 I P AT em pt y Figure 1433 The correct arrangement of ab conformations (Answer 1432). Roman numerals indicate successive steps in the cycle of conformational changes driven by hydrolysis of ATP. ATP ADP + Pi ATP ADP Pi ADP + Pi AD P AT P III P II P AD i y pt em Pi empty one is to notice that a particular ab conformation, for example the ATPbinding conformation, moves counterclockwise--the same direction as filament rotation--in going from step I to II to III. References: Noji H, Yasuda R, Yoshida M & Kinosita K (1997) Direct observation of the rotation of F1-ATPase. Nature 386, 299302. Abrahams JP, Leslie AG, Lutter AGW & Walker JE (1994) Structure at 2.8 resolution of F1-ATPase from bovine heart mitochondria. Nature 370, 621628. 1433 Such a result is entirely reasonable; mechanical force has just been substituted for the proton-motive force to turn the axle-like g subunit. This experiment would suggest a two-step model for ATP synthase: (1) proton flow causes rotation of the g subunit and (2) rotation of the g subunit inside the a3b3 complex causes synthesis of ATP. In this experiment the authors have succeeded in uncoupling these two steps: mechanically rotating the g subunit inside the a3b3 complex is sufficient to cause synthesis of ATP. This would be a very exciting experiment, indeed, because it would directly demonstrate the relationship between mechanical movement and enzymatic activity. There is no doubt that it should be published and that it would become a `classic.' The key to understanding these results is to recognize that an exchange of ADP for ATP is not electrically neutral. ATP carries four negative charges; ADP carries only three (see Figure 146). As a result, each exchange of ADP for ATP increases the negative charge on the side of the membrane that receives the ATP. In a respiring mitochondrion there is an electrochemical proton gradient across the mitochondrial inner membrane such that the outside of the membrane is positive. The resulting proton-motive force drives the exchange of an external ADP for an internal ATP, thereby reducing the positive charge on the outside of the membrane. In your results, the exchange of external ADP was favored over that of external ATP under conditions in which the membrane was charged: when substrate was present without an inhibitor and when substrate was present with an inhibitor of ATP synthase (see Table 141, experiments 2 and 4). When the electrochemical proton gradient was absent, that is, when there was no substrate or when the membrane was made permeable to protons, ADP and ATP were exchanged at equal rates (see experiments 1 and 3). 1434 ATP ADP + Pi ATP A314 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts The electrogenic nature of the ADPATP antiporter has important physiological consequences. In essence, the exchange harnesses the energy of the electrochemical proton gradient to drive transport so that the cytosolic ratio of ATP/ADP remains high (up to 50). This concentration difference provides up to one-third of the free-energy change (DG) for ATP hydrolysis in the cytoplasm. References: Nicholls DG & Ferguson SJ (1992) Bioenergetics 2, pp 215218. London: Academic Press. Tzagoloff A (1982) Mitochondria, pp 212213. New York: Plenum Press. 1435 A. The rate of oxygen consumption is determined by the rate of electron transport down the respiratory chain. Electron transport generates an electrochemical proton gradient, which opposes the flow of electrons. In the complete absence of a way to dissipate the gradient, the flow of electrons ultimately would stop when the electron pressure balances the opposing electrochemical proton gradient. In the experiment in Figure 147, the electrochemical proton gradient is dissipated at a slow background rate, which accounts for the slow background rate of oxygen consumption. Addition of ADP and its subsequent conversion to ATP allows protons to flow back into the mitochondria, dramatically decreasing the electrochemical proton gradient and permitting the rapid transport of electrons to oxygen. The increased rate of electron transport produces an increased rate of oxygen consumption. When all the ADP is converted to ATP, proton flow again slows to the background rate, and the increased electrochemical proton gradient once again decreases the flow of electrons. B. The slow background rate of oxygen consumption by mitochondria in the absence of added ADP indicates that electrons continue to flow down the electron-transport chain to oxygen in the absence of ATP synthesis. Such a flow can continue only if the electrochemical proton gradient is slowly being dissipated. If the mitochondrial inner membrane were completely impermeable to protons, the rate of oxygen consumption would drop to zero when proton pumping due to electron transport was balanced by the back pressure of the electrochemical proton gradient. Thus, the protons must be crossing the membrane in the absence of ATP synthesis. Several processes other than ATP synthesis from added ADP might account for the slow passage of protons across the membrane and the slow background rate of oxygen consumption. (1) The mitochondrial inner membrane is not completely impermeable to protons, which can slowly cross the membrane even in the absence of ATP synthesis. (2) The internal mitochondrial supply of ATP may be hydrolyzed to ADP and then reconverted to ATP using the proton-motive force. (3) If some mitochondria in the preparation are damaged so that their inner membranes are not intact, they will transport electrons to oxygen continuously because there will be no electrochemical proton gradient to oppose electron flow. C. Since each pair of electrons that flows down the respiratory chain from NADH to oxygen reduces one oxygen atom, the P/2e ratio is equivalent to the P/O ratio. The P/O ratio, as calculated below, is between 2.5 and 2.8 molecules of ATP per O atom. Uncertainty in the P/O (P/2e) ratio arises from the uncertainty in how much oxygen is consumed during the conversion of 500 nmol ADP to ATP. If oxygen consumption is calculated as the difference between the horizontal gray lines in Figure 147, which is 100 nmol O2, then the P/O ratio is 2.5 (500 nmol ATP/200 nmol O). On the other hand, if oxygen consumption is calculated as the difference between the slanted gray lines in Figure 147, which is 90 nmol O2, then the P/O ratio is 2.8 (500 nmol ATP/180 nmol O). The latter calculation makes the implicit assumption that the background rate of oxygen consumption continues during the conversion of ADP to ATP which is a perfectly reasonable assumption. It turns , out, however, that the natural slow flow of protons across intact inner membranes is quite sensitive to the size of the electrochemical proton gradient. ELECTRON-TRANSPORT CHAINS AND THEIR PROTON PUMPS The slight decline in the proton-motive force during ATP synthesis may reduce the leakage to nearly zero, in which case the larger value for oxygen consumption may be the more valid one (giving a P/O ratio of 2.5). A P/O ratio of 2.5 is now accepted as the value for the number of ATP molecules formed by flow of a pair of electrons down the electron-transport chain from NADH to O2. D. Several processes in these kinds of experiments, in addition to ATP synthesis, are driven by the electrochemical proton gradient. The uptake of substrate (b-hydroxybutyrate) into mitochondria may require symport with protons. The import of phosphate into mitochondria also requires symport with a proton. Finally, the membrane potential--one component of the electrochemical proton gradient--drives the exchange of internal ATP for external ADP Given that several processes are driven by the electrochemical . proton gradient, it is not surprising that the P/O ratio is not an integer. Before the chemiosmotic theory, when chemical coupling hypotheses were fashionable, integral values were expected and values of 2.5 or 2.8 were assumed to `really' mean 3. Reference: Nicholls DG & Ferguson SJ (1992) Bioenergetics 2, pp 65104. London: Academic Press. A315 ELECTRON-TRANSPORT CHAINS AND THEIR PROTON PUMPS DEFINITIONS 1436 1437 1438 1439 1440 1441 1442 Cytochrome oxidase complex Respiratory control Cytochrome Ironsulfur center Redox reaction Respiratory enzyme complex Redox potential TRUE/FALSE 1443 True. The flow of electrons down the respiratory chain moves from carriers with lower affinity for electrons to ones with higher affinity and finally to oxygen, which has the highest affinity of all. Thus, the relative positions of cytochromes and ironsulfur centers make sense in terms of the natural flow of electrons to oxygen. False. The three respiratory enzyme complexes exist as independent entities in the mitochondrial inner membrane. The ordered transfers of electrons between complexes are mediated by random collisions between the complexes and two mobile carriers--ubiquinone and cytochrome c--with electron transfers occurring when the appropriate components meet. False. Lipophilic weak acids act as uncoupling agents that dissipate the proton-motive force and stop ATP synthesis; however, they increase the flow of electrons through the respiratory chain by eliminating the respiratory control imposed by the electrochemical proton gradient. Normally, the electrochemical proton gradient exerts a backpressure that restricts the flow of electrons down the electron-transport chain. In the absence of the gradient electron transport runs unchecked at the maximum rate. 1444 1445 A316 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts THOUGHT PROBLEMS 1446 H+ ions move much faster than Ca2+ ions through an aqueous solution because their `movement' is virtual; the H+ ion that appears on one side of the cell is not the same one that started on the other side. H+ ion movement, instead, depends on exchanges of hydrogen bonds for covalent bonds in a chain of water molecules. By contrast, a Ca2+ ion must actually diffuse through the medium. The difference in the nature of the movements of these two ions is nowhere better illustrated than by their behavior in ice. As expected, the rate of diffusion of Ca2+ ions decreases considerably. Surprisingly, H+ ions move even more rapidly. That is because H+ movement depends on chains of water molecules. In ice most of the water molecules are linked into chains, so that H+ ions can move very rapidly over long distance, whereas in liquid water the chains involve only a few water molecules, which means that there are periodic delays as H3O+ ions connect up with a new chain. A hydrogen atom is a proton plus one electron; it is neutral. A proton is a hydrogen atom that has lost its single electron and thus is positively charged. A hydride ion is a hydrogen atom that has gained an extra electron and thus is negatively charged. A hydrogen molecule is a pair of hydrogen atoms that share their two electrons in a covalent bond; it is neutral. The E0 values for the electron carriers in Table 142 suggest that they appear in the respiratory chain in the same order as they appear (top to bottom) in the table. Remember that E0 values refer to standard conditions (reactants and products at 1 M). You would really need to know the E values to know the order of the carriers in the chain. The concentration terms might be especially important in distinguishing the order of ubiquinone and cytochrome b, and the orders of cytochromes c1, c, and a. Despite these caveats, the order of carriers in the respiratory chain is as listed in Table 142. Only in mixture E would electron transfer occur, with cytochrome c becoming reduced. In this mixture a portion of the respiratory chain has been reconstituted, so that electrons can flow in the energetically favored direction from reduced ubiquinone to the cytochrome b-c1 complex to cytochrome c. Although energetically favorable, the transfer in mixture A cannot occur spontaneously in the absence of the cytochrome b-c1 complex, which catalyzes this reaction. No electron flow occurs in the other mixtures, whether the cytochrome b-c1 complex is present or not: in mixtures B and F both ubiquinone and cytochrome c are oxidized; in mixtures C and G both are reduced; and in mixtures D and H electron flow is energetically disfavored because reduced cytochrome c has a lower free energy than oxidized ubiquinone. Treatment with sodium nitrite oxidizes the Fe2+ in a proportion of hemoglobin molecules to Fe3+. Because there is so much more hemoglobin than cytochrome oxidase, the Fe3+ form of hemoglobin competes effectively with cytochrome oxidase for the binding of cyanide. Treatment with sodium nitrite also impairs the oxygen-carrying function of hemoglobin, which will bind oxygen only when the heme group carries an Fe2+ ion. The treatment is therapeutically effective because there is a useful middle range where sufficient hemoglobin has been converted to the Fe3+ form to bind up the cyanide, but adequate Fe2+ hemoglobin remains to carry oxygen. It is unlikely that the same diffusible carrier could be used productively for both steps. Such a carrier would need to have a redox potential and a binding specificity that would allow it to accept electrons from the NADH dehydrogenase complex and pass them on to the cytochrome b-c1 complex, and to accept electrons from the cytochrome b-c1 complex and pass them on to the cytochrome oxidase complex. Leaving aside the tricky problem of donating and receiving electrons from the cytochrome b-c1 complex, such an 1447 1448 1449 1450 1451 ELECTRON-TRANSPORT CHAINS AND THEIR PROTON PUMPS arrangement would be counterproductive because it would allow frequent bypass of the cytochrome b-c1 complex: electrons could be accepted from the NADH dehydrogenase complex and passed directly to the cytochrome oxidase complex. This short-circuit of the standard flow would waste the energy normally harvested by the cytochrome b-c1 complex, releasing it as heat instead. 1452 An uncoupler promotes weight loss by decreasing the efficiency of oxidative phosphorylation. For example, if sufficient uncoupler were ingested to reduce the efficiency of oxidative phosphorylation to 50%, twice as many calories (from food or internal stores, mainly fat) would have to be burned to generate the same amount of ATP. Dinitrophenol is no longer prescribed because its use led to several deaths; if oxidative phosphorylation is too efficiently compromised, not enough ATP will be generated to support essential cell functions and death is the result. In the presence of an artificially large electrochemical gradient, you would expect electrons to flow in the reverse of their normal direction. This is a straightforward expectation of thermodynamics and the principle of microscopic reversibility. The imposed large electrochemical gradient provides the thermodynamic driving force; that is, the free-energy change (DG) of the gradient is sufficiently negative to overcome the normally positive freeenergy change of moving electrons from carriers with high affinity to carriers with lower affinity (the reverse of normal). The principle of microscopic reversibility provides a mechanism for doing so. If electron flow down the respiratory chain is coupled to proton pumping out of the matrix, then the same mechanism must be able to operate in reverse to couple proton movement into the matrix (backwards through the proton pumps) to electron flow up the respiratory chain. For all ATP synthases that use a proton-motive force to generate ATP, the flow of protons is to the same side of the membrane on which ATP is made. For a bacterium that lives in an alkaline environment, the natural flow of protons would be from the inside of the cell to the outside. Thus, the proton flow is in the wrong direction to allow a normally oriented ATP synthase to make ATP; in fact, outward proton flow would cause ATP hydrolysis. Even if the ATP synthase were reversed in the membrane, so that it could use the proton flow, it would then make ATP on the outside of the cell--not a very useful arrangement. Bacteria of this kind have adapted to use a gradient of another ion such as Na+ to drive their ATP synthases, suggesting that using another ion probably is the simplest adaptation accessible to evolution. As a purely theoretical engineering problem, other solutions can be imagined. Structurally, the ATP synthase consists of a proton-powered motor embedded in the membrane, an ADP to ATP converter, and a drive shaft that connects the two. Normally, an inward proton flow through the motor drives a counterclockwise rotation of the drive shaft inside the converter to cause the normal sequence of conformational changes that converts ADP to ATP. (An outward flow drives a clockwise rotation of the shaft that causes the reverse sequence of conformational changes, converting ATP to ADP.) With this scheme in mind, you might consider re-engineering the motor so that an outward flow drives a counterclockwise rotation of the drive shaft. Alternatively, you might re-engineer the converter so that a clockwise rotation of the drive shaft inside it would drive the normal sequence of conformational changes. Either of these changes would allow the re-engineered ATP synthase to make ATP from the proton gradient in an alkaline environment. A317 1453 1454 CALCULATIONS 1455 A. With all reactants and products at 1 M, the concentration term reduces to zero; thus, DE is equal to DE0, and DG is A318 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts DG = nF DE = 2 23 kcal 0.13 V V mole = 6 kcal/mole B. When the ratio of lactate to pyruvate equals 1 and the ratio of NAD+ to NADH equals 1, the overall concentration term reduces to zero; thus, DE is equal to DE0, and DG equals 6 kcal/mole. C. In order for DG to be zero, the concentration term must equal 15,600. When DG is zero, DE is also zero, thus + DE0 = 2.3 RT log [lactate][NAD ] nF [pyruvate][NADH] + 0.13 V = 0.031 V log [lactate][NAD ] [pyruvate][NADH] [lactate][NAD+] 4.19 = log [pyruvate][NADH] [lactate][NAD+] 15,600 = [pyruvate][NADH] D. At an NAD+ to NADH ratio of 1000, and a lactate to pyruvate ratio of 5.1 (0.77/0.15) DE is DE = DE0 2.3 RT log (5.1 1000) nF = 0.13 V 0.031 V log (5100) = 0.13 V 0.115 V = 0.015 V And DG is DG = nF DE = 2 23 kcal 0.015 V V mole = 0.7 kcal/mole Thus, under normal conditions in vascular smooth muscle the reaction is slightly in favor of reduction of pyruvate to lactate. Reference: Barron JT, Gu L & Parrillo JE (2000) NADH/NAD redox state of cytoplasmic glycolytic compartments in vascular smooth muscle. Am. J. Physiol. Heart Circ. Physiol. 279, H2872H2878. 1456 A. Reversing the fumarate/succinate half reaction and summing gives DE0 values of 0.02 V for FAD and 0.35 V for NAD+. Since a positive DE0 value corresponds to a negative DG value, only the reaction using FAD would yield a favorable (negative) DG value. B. The concentration ratio needed to yield a DG value of zero is 4.4 for FAD and 5.1 1012 for NAD+. When DG is zero, DE is also zero. Thus, for FAD, [fumarate][FADH2] =0 DE = DE0 2.3 RT log nF [succinate][FAD] [fumarate][FADH2] DE0 = 2.3 RT log nF [succinate][FAD] [fumarate][FADH2] 0.02 V = 0.031 V log [succinate][FAD] [fumarate][FADH2] 0.65 = log [succinate][FAD] [fumarate][FADH2] 4.4 = [succinate][FAD] With NAD+ as the cofactor, the analogous calculation gives 5.1 1012 as the concentration ratio where DG is zero. C. From these calculations it is clear that FAD is a more appropriate cofactor for the oxidation of succinate than NAD+. Using FAD, the concentration ratio ELECTRON-TRANSPORT CHAINS AND THEIR PROTON PUMPS would have to be maintained at the reasonable value of less than 4.4 to allow the reaction to proceed. By contrast, with NAD+ the concentration ratio would have to be maintained at less than 5.1 1012. Such a ratio could be achieved, for example, if the succinate concentration were 4.4 105 times the fumarate concentration, and the FAD concentration were 4.4 105 times the FADH2 concentration. Such an excess of reactants over products is highly unlikely, if not impossible, for a biological reaction pathway. 1457 A. The balanced equation for the reduction of O2 by Fe2+ is 4 Fe2+ + O2 + 4 H+ 4 Fe3+ + 2 H2O Reversing the half reaction for Fe3+/Fe2+ and summing gives a DE0 value of 0.05 V (0.77 V + 0.82 V). Under standard conditions DE = DE0 and DG is DG = nF DE = 4 23 kcal 0.05 V V mole DG = 4.6 kcal/mole or as sometimes stated, since this is a four-electron reaction, DG = 1.15 kcal/mole for each electron Although the flow of electrons from Fe2+ to O2 is thermodynamically favorable, the free-energy change for each electron is fairly small. Fortunately, T. ferrooxidans does not depend on this redox reaction as a source of energy, but rather as a way of detoxifying entering protons (from which it makes ATP; see Problem 1426) and as a source of electrons for reducing NADP+. B. The balanced reaction for reduction of NADP+ + H+ by Fe2+ is NADP+ + H+ + 2 Fe2+ NADPH + 2 Fe3+ Reversing the half reaction for Fe3+/Fe2+ and summing give a DE0 value of 1.09 V (0.77 V 0.32 V). Because DE = DE0 under standard conditions, DG is DG = nF DE = 2 23 kcal 1.09 V V mole DG = 50.1 kcal/mole (or 25 kcal/mole for each electron) At the concentrations specified in the problem, 2+ 2 DE = DE0 2.3 RT log [NADPH][Fe3+ ]2 [NADP+][Fe ] nF A319 Note the use of exponents for Fe2+ and Fe3+, in accord with the balanced equation. Since the concentrations of Fe2+ and Fe3+ are equal they cancel out, and 3 DE = 1.09 V 2.3 1.98 10 kcal 310 K V mole log 10 2 K mole 23 kcal 1 = 1.09 V 0.03 V DE = 1.12 V At these concentrations DG = 2 23 kcal 1.12 V V mole DG = 51.5 kcal/mole (or 26 kcal/mole for each electron) A320 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts These calculations make it very clear that reduction of NADP+ by Fe2+ is extremely unfavorable. Yet T. ferrooxidans manages to do exactly that! Problem 1488 shows how. Reference: Ingledew JW (1982) Thiobacillus ferrooxidans: the bioenergetics of an acidophilic chemolithotroph. Biochim. Biophys. Acta 683, 89117. 1458 Reversing the NAD+/NADH half reaction and summing E0 values give a DE0 value of 1.14 V (0.82 V + 0.32 V). Noting that the balanced equation is a fourelectron reaction, the standard free-energy change is DG = nFDE0 = 4 23 kcal 1.14 V V mole = 105 kcal/mole The balanced equation that uses O2 has the same DE0 value of 1.14 V, but because it is a two-electron reaction, the standard free-energy change is only 52.4 kcal/mole, just half the value calculated above. When the values from these calculations are expressed per mole of electrons, they agree: 26.2 kcal/mole of electrons. DATA HANDLING 1459 A. The reduced (electron-rich) forms of the cytochromes give rise to the bands. Oxygen accepts electrons from the electron-transport chain and is reduced to H2O. Therefore in the presence of oxygen the cytochromes would be drained of their electrons; that is, oxidized. Since the absorption bands do not show up in the presence of oxygen, the oxidized forms must not absorb light. The reduced forms of the cytochromes absorb light and are therefore responsible for the characteristic absorption patterns that Keilin obeserved. In the absence of oxygen the cytochromes pick up electrons from substrates (become reduced) but cannot get rid of them by transfer to oxygen. In the presence of oxygen, the electrons are transferred efficiently, leaving the cytochromes in their electron-deficient or oxidized state. B. Keilin's observations indicate that the order of electron flow through the cytochromes is reduced cytochrome cytochrome cytochrome O 2 substrates b c a This order can be deduced from Keilin's results. Since the bands become visible in the absence of oxygen, they represent the reduced (electron-rich) forms of the cytochromes. When oxygen is added, they are all converted to the oxidized (electron-poor) form. When cyanide is added, all the cytochromes are reduced, indicating that cyanide blocks the flow of electrons from the cytochromes to oxygen; that is, all the cytochromes are `upstream' of oxygen (in the sense of electron flow). When urethane is added, cytochrome b remains reduced but cytochromes a and c become oxidized. Thus, urethane interrupts the flow of electrons from cytochrome b to cytochromes a and c, indicating that cytochrome b is `upstream' of cytochromes a and c. These results indicate that either cytochrome a or c transfers electrons to oxygen. The inability of oxygen to oxidize a preparation of cytochrome c suggests, by elimination, that cytochrome a is responsible for the transfer of electrons to oxygen. This ordering of cytochromes a and c is weak since it is based on a negative result (which could have other interpretations). Keilin himself confirmed this order by observing subtle spectral shifts in the cytochrome a band in the presence of cyanide under reducing conditions; he named the active component cytochrome a3. We now know that cytochrome a is a large complex with several redox centers, one of which reacts with molecular oxygen. ELECTRON-TRANSPORT CHAINS AND THEIR PROTON PUMPS C. The rapid oxidation of glucose to CO2 prevents the disappearance of the absorption bands by providing a source of reduced substrates (ultimately NADH and FADH2) that transfer electrons into the electron-transport chain faster than oxygen can remove them. Under these conditions the cytochromes remain reduced (electron rich) and therefore continue to absorb light. Reference: Keilin D (1966) The History of Cell Respiration and Cytochrome. Cambridge, U.K.: Cambridge University Press. 1460 The rates of oxidation of the electron carriers, if measured rapidly enough, reveal their order in the respiratory chain. The carriers closest to oxygen will be oxidized first, and those farthest from oxygen will be oxidized last. This rationale allows you to deduce the order of electron flow through the carriers: cytochrome cytochrome cytochrome cytochrome O 2 b c1 c (a + a3) 1461 A. Malonate, cyanide, and butylmalonate all give the oxygen trace shown in Figure 1411A. Each of these inhibitors blocks the flow of electrons to oxygen, thereby stopping oxygen consumption. Butylmalonate and malonate block the uptake and oxidation of succinate, respectively, thereby eliminating the flow of electrons at the source. Cyanide stops the flow of electrons at cytochrome oxidase, which normally transfers the electrons to oxygen. Atractyloside and oligomycin give the oxygen trace shown in Figure 1411B. By blocking exchange of ADP for ATP, atractyloside prevents entry of ADP into mitochondria and subsequent ATP synthesis. This `removal' of ADP returns the rate of respiration to the rate before ADP was added. By inhibiting ATP synthase, oligomycin prevents synthesis of ATP, which has the same effect as eliminating ADP, and returns the rate of respiration to the rate before ADP was added. FCCP gives the oxygen trace shown in Figure 1411C. By making the inner membrane permeable to protons, FCCP uncouples electron transport from oxidative phosphorylation, thereby allowing the maximum rate of electron flow through the respiratory chain to oxygen (since it is not opposed by the electrochemical proton gradient). B. The oxygen traces expected for the three combinations of inhibitors are shown in Figure 1434. 1. FCCP uncouples the flow of electrons from oxidative phosphorylation, allowing a maximum rate of electron flow and oxygen consumption. The subsequent addition of cyanide stops the flow of electrons directly and, thereby, stops oxygen consumption (Figure 1434A). 2. Since FCCP uncouples electron flow from oxidative phosphorylation, oligomycin (which inhibits ATP synthase) has no effect on the rate of oxygen consumption (Figure 1434B). 3. Oligomycin slows the rate of oxygen consumption by blocking ATP synthase. This block is bypassed by FCCP because it uncouples electron transport from oxidative phosphorylation (Figure 1434C). Reference: Nicholls DG & Ferguson SJ (1992) Bioenergetics 2, pp 8287. London: Academic Press. 1462 The experiments in which individual lysines were modified indicate that a common set of lysines is important for electron transfers to and from cytochrome c. One way that such modifications might inhibit electron transfer is by interfering with the binding of cytochrome c to the two complexes. This interpretation is supported by the protection experiments, which show that the same lysines are shielded from modification when cytochrome c is bound either to the cytochrome b-c1 complex or to the cytochrome oxidase complex. A321 (A) mitochondria ADP FCCP cyanide (B) mitochondria ADP FCCP oligomycin (C) mitochondria ADP oligomycin FCCP Figure 1434 Oxygen traces expected for three combinations of inhibitors (Answer 1461). (A) FCCP followed by cyanide. (B) FCCP followed by oligomycin. (C) Oligomycin followed by FCCP. A322 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts Together, the two sets of experiments indicate that many of the same lysines on cytochrome c interact with the cytochrome b-c1 complex and with the cytochrome oxidase complex. If the same lysines are used to interact with both complexes, cytochrome c cannot be bound to both complexes simultaneously (and therefore cannot link the two complexes like a wire). Since cytochrome c must interact separately with the two complexes, it must serve as a mobile carrier of electrons that binds to the two complexes alternately. References: Smith HT, Ahmed AJ & Millet F (1981) Electrostatic interaction of cytochrome c with cytochrome c1 and cytochrome oxidase. J. Biol. Chem. 256, 49844990. Capaldi RA, Darley-Usmar V, Fuller S & Millet F (1982) Structural and functional features of the interaction of cytochrome c with complex III and cytochrome c oxidase. FEBS Lett. 138, 17. 1463 An inhibitor of electron transport (such as cyanide in mitochondria) would be expected to block electron flow, which would rapidly lead to collapse of the proton gradient (as it was used up to make ATP) and a drop in ATP levels (as it was consumed in cellular processes). Neither TCS nor DCCD causes all three of these effects in methanogenic bacteria. An inhibitor of ATP synthase (such as oligomycin in mitochondria) would be expected to stop ATP synthesis, leading to a drop in intracellular ATP levels. Blocking ATP synthase, which would prevent the flow of protons across the membrane, would lead to a backpressure that would restrict the flow of electrons down the respiratory chain. As shown in Figure 1413B, DCCD behaves like an inhibitor of ATP synthase: ATP levels fall, the proton gradient remains high, and the rate of production of CH4 (the indicator of electron flow) decreases substantially. An inhibitor that uncouples electron transport from ATP synthesis (such as dinitrophenol in mitochondria) would be expected to collapse the proton gradient, which would prevent ATP synthesis. In addition, collapse of the proton gradient would allow electrons (unopposed by the proton gradient) to flow more readily down the electron-transport chain. As shown in Figure 1413A, TCS behaves like an uncoupler: the proton gradient collapses, ATP levels fall, and CH4 production accelerates. The sequential addition of DCCD and TCS (see Figure 1413C) confirms these assignments. Addition of an uncoupler (TCS) to a system in which ATP synthase is inhibited (by DCCD) would be expected to collapse the proton gradient and restore electron flow (CH4 production). Reference: Nicholls DG & Ferguson SJ (1992) Bioenergetics 2, pp 149154. London: Academic Press. 1464 A. Each of the observations with ionophores is consistent with the idea that the movement of protons down the electrochemical proton gradient powers the flagella, as explained below for each observation. 1. During oxidation of glucose, bacteria pump protons out of the cell, establishing an electrochemical proton gradient, which is the sum of a proton gradient and a membrane potential. Addition of FCCP makes the membranes permeable to protons, thereby collapsing both the proton gradient and the membrane potential. In the absence of an electrochemical proton gradient to drive protons across the membrane back into the cell, the flagellar motor cannot function. 2. In a medium containing K+, valinomycin collapses the membrane potential specifically by allowing an influx of K+ to balance the efflux of protons. Under these circumstances it is entirely the proton gradient (which is larger than usual because it is not opposed by the membrane potential) that drives protons through the flagellar motor. The ability of bacteria to swim normally in the presence of the proton gradient alone is strong evidence that the flagellar motor is proton-powered. CHLOROPLASTS AND PHOTOSYNTHESIS 3. In the absence of glucose (or any other substrate) for oxidation, there is no electrochemical proton gradient. In the presence of external K+, valinomycin facilitates a flow of K+ into the cell; this results in a membrane potential that is positive inside. Although there are protons available in the medium (from H2O), this membrane potential is in the wrong orientation to drive the protons into the cell. As a result, the bacteria remain motionless. 4. In the absence of glucose, there is no electrochemical proton gradient. In the absence of external K+ (that is, when Na+ is in the medium), addition of valinomycin allows internal K+ to move out of the cell (down its concentration gradient), creating a membrane potential that is positive outside. This membrane potential can drive protons into the cell for a while. Each proton that enters the cell lessens the membrane potential until the membrane potential is dissipated, at which point the cells stop swimming. B. At first glance it seems peculiar that normal bacteria can swim in the absence of oxygen. Without oxygen, there is no electron flow down the electron-transport chain and no linked proton translocation across the membrane. What, then, is the source of protons to power the motor under anaerobic conditions? The mutant strain provides the essential clue. In the absence of the ATP synthase, bacteria cannot swim. This observation suggests that the ATP synthase in some way generates the proton gradient. In normal bacteria in the absence of oxygen, ATP that is generated anaerobically is used to drive the ATP synthase in reverse, causing protons to flow out of the cell. The resulting electrochemical proton gradient drives the protons back through the flagellar motor, allowing the bacteria to swim. The mutant bacteria cannot swim in the absence of oxygen because they have no ATP synthase and therefore cannot create an electrochemical gradient in the absence of electron flow. Reference: Manson MD, Tedesco P, Berg HC, Harold FM & van der Drift C (1977) A protonmotive force drives bacteria flagella. Proc. Natl Acad. Sci. USA 74, 30603064. A323 CHLOROPLASTS AND PHOTOSYNTHESIS DEFINITIONS 1465 1466 1467 1468 1469 1470 1471 1472 1473 Photosynthetic electron-transfer Antenna complex Carbon fixation Photochemical reaction center Chloroplast Sucrose Chlorophyll Noncyclic photophosphorylation Carbon-fixation cycle (Calvin cycle) TRUE/FALSE 1474 True. The thylakoids and cristae have the same arrangement of membrane proteins--ATP synthase, for example, with its lollipoplike head protruding into the matrix or stroma. A324 1475 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts False. When an electron in a chlorophyll molecule in the antenna complex is excited, it transfers its energy--not the electron--from one chlorophyll molecule to another by resonance energy transfer. THOUGHT PROBLEMS 1476 Protons are pumped across the inner membrane into the intermembrane space in mitochondria. By contrast, they are pumped across the thylakoid membrane into the thylakoid space in chloroplasts. Mitochondria have no structures that are analogous to the thylakoid membrane or the thylakoid space of chloroplasts. ATP is synthesized in the corresponding compartments in the two organelles: in the matrix in mitochondria and in the stroma in chloroplasts. Choice A in Table 145 gives the true balanced equation for the Calvin cycle. You can decide which is balanced by seeing whether the oxygen and hydrogen atoms on each side of the equation are equal. For equation A there are 12 Hs from NADPH and 24 Hs from H2O for 36 total on the left, and 12 Hs from glucose, 18 from Pi, and 6 from H+ for 36 total on the right; thus, H atoms balance. There are 12 Os from CO2 and 12 from H2O for 24 total on the left, and 6 Os from glucose and 18 from Pi for 24 total on the right; thus, O atoms balance. For none of the other equations do both the H atoms and O atoms balance. (A difficulty in deciding on the balanced equation is the form of Pi that the authors intend in their equations. The pK for H2PO4 HPO42 + H+ is 6.9, which means that at intracellular pH--slightly above 7--most of the Pi will be in the HPO42 form, the form you were directed to use in this problem. If H2PO4 is used instead, then equation B in Table 145 is the balanced equation. Because cells contain a mixture of HPO42 and H2PO4 at intracellular pH, the actual balanced equation is somewhere between A and B.) In the absence of light, but the presence of CO2, ribulose 1,5-bisphosphate will decrease and 3-phosphoglycerate will increase. The presence of CO2 allows ribulose 1,5-bisphosphate to be converted to 3-phosphoglycerate but in the absence of light (and therefore reduced amounts of NADPH and ATP) 3-phosphoglycerate will accumulate because subsequent reactions require NADPH and ATP. In the absence of CO2, ribulose 1,5-bisphosphate will accumulate (and 3phosphoglycerate will decrease) because its conversion to 3-phosphoglycerate is dependent on CO2. The corn plant (C4) eventually will kill the geranium (C3). Because both plants fix CO2, the concentration of CO2 in the chamber will fall. At low CO2 concentrations, the corn plant has a distinct advantage since its enzyme for carbon fixation has a high affinity for CO2. By contrast, the geranium depends on ribulose bisphosphate carboxylase, which has a lower affinity for CO2. At low CO2 concentrations, O2 also competes with CO2 for addition to ribulose 1,5-bisphosphate, ultimately liberating CO2 in the process known as photorespiration. Not only will the geranium give up CO2 in an abortive attempt at photosynthesis, it will continue to respire (using its mitochondria), thereby providing even more CO2 for the corn plant. The corn plant will continue to fix CO2 until the geranium wastes away and dies. Plants are green because they absorb light efficiently at the blue and red ends of the spectrum but poorly at the green and yellow wavelengths (see Figure 1417A). Thus the light that reaches our eyes after striking a plant is enriched for green and yellow wavelengths; hence, plants appear green. If a chemical reagent that accepts electrons from plastoquinone can restore O2 evolution in DCMU-treated chloroplasts, then the components from photosystem II to plastoquinone cannot be affected by DCMU. Because DCMU blocks photophosphorylation, it must act before the sites at which 1477 1478 1479 1480 1481 CHLOROPLASTS AND PHOTOSYNTHESIS protons are pumped within the cytochrome b6-f complex. These two arguments restrict the site of action of DCMU to a component after plastoquinone but before the first site of proton pumping in the cytochrome b6-f complex (see Figure 1414). DCMU kills plants because it blocks their photosynthetic production of ATP and NADPH. Deprived of their ability to fix CO2 and generate energy, plants die. 1482 Protons pumped across the inner mitochondrial membrane into the intermembrane space equilibrate with the cytosol, which serves as a huge H+ sink. Both the mitochondrial matrix and the cytosol house many metabolic reactions that require a pH around neutrality. The largest H+ concentration difference between mitochondrial matrix and cytosol that is compatible with function is therefore relatively small (less than 1 pH unit). Much of the energy stored in the mitochondrial electrochemical proton gradient is instead due to the electrical potential of the membrane (about 140 mV of the 200 mV potential difference is due to the membrane potential). By contrast, chloroplasts have a smaller, dedicated compartment--the thylakoid space--into which H+ ions are pumped. Much higher concentration differences can be achieved (more than 3 pH units), and nearly all of the energy stored in the thylakoid electrochemical proton gradient is due to the pH difference between the stroma and the thylakoid space. Even during daylight hours in chloroplast-containing cells, it is the mitochondria that supply the cell with ATP, which is produced by oxidative phosphorylation. Glyceraldehyde 3-phosphate made by photosynthesis in chloroplasts moves to the cytosol and is eventually used as a source of energy to drive ATP production in mitochondria. A325 1483 CALCULATIONS 1484 A. The energy of a mole of photons at any particular wavelength is the energy of one photon times Avogadro's number (N). Therefore, the energy of a mole of photons at a wavelength of 400 nm is E = Nhc/l 23 37 17 1 = 6 10 photons 1.58 10 kcal/sec 3 10 nm mole photon sec 400 nm E = 71 kcal/mole for 400-nm light This calculation for 680-nm and 800-nm light gives E = 42 kcal/mole for 680-nm light E = 36 kcal/mole for 800-nm light B. If a square meter receives 0.3 kcal/sec of 680-nm light, which is worth 42 kcal/mole of photons, then the time it will take for a square meter to receive one mole of photons is sec 42 kcal 0.3 kcal mole time = 140 sec/mole time = C. If it takes 140 seconds for a square meter of tomato leaf to receive a mole of photons and eight photons are required to fix a molecule of CO2, it will take just under 2 hours to synthesize a mole of glucose: 140 sec 8 mole photons 6 mole CO2 mole glucose mole photons mole CO2 time = 6720 seconds or 112 minutes time = A326 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts The actual efficiency of photon capture is considerably less than 100%. Under optimal conditions for some rapidly growing plants, the efficiency of utilization of photons that strike a leaf is about 5%. However, even this value greatly exaggerates the true efficiency of utilization of the energy in sunlight. For example, a field of sugar beet converts only about 0.02% of the energy that falls on it during the growing season. Several factors limit the overall efficiency, including saturation of the photosystems far below maximum sunlight, availability of water, and low temperatures. D. In contrast to the very low overall efficiency of light utilization, the efficiency of conversion of light energy to chemical energy after photon capture is 33%. mole CO2 mole photons 112 kcal 8 mole photons 42 kcal mole CO2 = 0.33 or 33% When a photon is absorbed by P700, 90% of the energy of 700-nm light is captured. The energy in a photon of 700-nm light is efficiency = E = hc/l 37 17 1 = 1.58 10 kcal/sec 3 10 nm 700 nm photon sec = 6.8 1023 kcal/photon Reversing the P700+/P700* half cell and summing gives a standard redox potential (DE0) of 1.6 V. Under standard conditions, DG is DG = nFDE0 = 1 23 kcal 1.6 V V mole = 36.8 kcal/mole Dividing by Avogadro's number (6 1023 molecules/mole) gives 6.1 1023 kcal/molecule. Thus, 90% (6.1 1023/6.8 1023) of the energy of 700-nm light is captured when a photon is absorbed by P700. 1485 1486 Eight photons are required: four by photosystem II and four by photosystem I. Four electrons (ultimately from water, generating O2) are excited by absorption of four photons in photosystem II and then re-excited by absorption of four photons in photosystem I. The four electrons from photosystem I are used to reduce two molecules of NADP+ to NADPH. The standard free-energy change for the reaction is 105 kcal/mole, which is extremely unfavorable. (It is the energy from the absorbed photons that makes this reaction feasible.) Reversing the O2/H2O half reaction and summing gives the balanced equation 2 H2O + 2 NADP+ O2 + 2 NADPH + 2 H+ and a DE0 of 1.14 V. (Remember that adjusting the number of atoms and electrons in order to balance the chemical reaction does not affect E0 or DE0 values.) Using the relationship between DE0 and DG and remembering that the balanced equation involves a four-electron transfer, DG = nFDE0 = 4 23 kcal 1.14 V V mole = 105 kcal/mole 1487 1488 A minimum of eight protons (8 6.4 kcal/mole = 51.2 kcal/mole) would be required to drive electrons from Fe2+ to NADP+, a process with an unfavorable free-energy change of 50.6 kcal/mole. The free-energy change for transfer of two electrons from Fe2+ to NADP+ (DE = 1.1 V) is DG = nFDE = 2 23 kcal 1.1 V V mole = 50.6 kcal/mole CHLOROPLASTS AND PHOTOSYNTHESIS These thermodynamic considerations say nothing about the mechanism of coupling. Reasoning from what is known in other organisms, it is likely that an electron-transport chain links the Fe2+/Fe3+ redox pair to the NADP+/NADPH redox pair. In other organisms the flow of electrons in the favorable direction is coupled to the pumping of protons across a membrane. This is likely to be the arrangement in T. ferrooxidans as well; however, the naturally imposed pH difference across the membrane drives an inward proton flow that reverses the proton pumps to cause an upward (otherwise unfavorable) flow of electrons from Fe2+ to NADP+ (see Problem 1453). 1489 Noncyclic photophosphorylation, which generates one NADPH per electron pair, must process 12 pairs of electrons to generate the 12 NADPH required in the Calvin cycle. At 1 ATP per electron pair, it also generates 12 of the required 18 ATP. Cyclic photophosphorylation generates the remaining 6 ATP by processing an additional 6 electron pairs. Thus, the ratio of electron pairs processed via noncyclic versus cyclic photophosphorylation is 2:1. Noncyclic photophosphorylation is run twice as often as cyclic photophosphorylation to meet the needs of the Calvin cycle. The noncyclic pathway involves both photosystems I and II, whereas the cyclic pathway involves only photosystem I. Thus, photosystem I will be used 18 times and photosystem II will be used 12 times--a ratio of 3:2--to meet the needs of the Calvin cycle. The free-energy change for ATP synthesis in chloroplasts under these conditions is 12.2 kcal/mole, about the same as in mitochondria: DG = DG + 2.3RT log [ATP] [ADP][Pi] 3 3 = 7.3 kcal/mole + 2.3 1.98 10 kcal 310 K log 3 4 10 2 K mole (10 )(10 ) = 7.3 kcal/mole + 4.9 kcal/mole = 12.2 kcal/mole CACTUS CO2 A327 1490 DATA HANDLING 1491 A. Starch formation requires light in the cactus and in C4 plants (as well as in C3 plants). The synthesis of starch via CO2 fixation in the Calvin cycle requires ATP and NADPH. These molecules are present in cells in small amounts; they are not stored. In order for starch synthesis to continue, ATP and NADPH must be continuously regenerated by photophosphorylation. B. CO2 fixation in the cactus is outlined in Figure 1435. Reactions shown with thick lines occur at night; reactions shown with thin lines occur during the day. The carbon-fixation reactions in the cactus are essentially the same as those in C4 plants. The key difference in the metabolic pathways of CO2 fixation is that the cactus uses starch as a source of phosphoenolpyruvate (PEP) (via glycolysis) in the CO2 pumping cycle. However, the common reactions of the pumping cycle are distributed differently in both space and time in C4 plants and the cactus. In C4 plants the reactions involve different cells but occur all at the same time. By contrast, in the cactus they all occur in the same cell but at different times. C. Starch is not required for CO2 pumping in C4 plants because the molecules that constitute the pump are used catalytically. In principle, a few molecules of PEP could pump an unlimited amount of CO2, because PEP is regenerated at the end of each pumping cycle. By contrast, the reactions in the cactus are stoichiometric: each CO2 molecule that is stored as malate requires one molecule of PEP. Starch in the cactus is used as the source of PEP molecules: the number of CO2 molecules that can be fixed is limited by the amount of starch. D. The principal advantage of this method of CO2 fixation is that the cactus can seal itself off (close its stomata) during the heat of the day, thereby preventing water loss. Yet, it can still provide a rich supply of CO2 (from stored STARCH PEP PYR OAA MAL CO2 Rb15BP 3PG G3P GLC STARCH Figure 1435 CO2 fixation in the cactus (Answer 1491). Night reactions are shown as thick lines; day reactions are shown as thin lines. Only the carbon pathways are indicated; the cofactor requirements are not shown. A328 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts malate) for sugar synthesis during the day, when the production of ATP and NADPH are maximal due to photophosphorylation. At night, when there is less risk of water loss, it can open its stomata and fix CO2. Reference: Foyer CH (1984) Photosynthesis, pp 176195. New York: Wiley. 1492 One measure of photosynthesis is the evolution of O2. In 1882 none of the sensitive devices now used for measuring O2 were available. Instead, Engelmann made use of bacteria that grow best in the presence of O2 and actively seek it. When the alga was illuminated with a spectrum of light, only those portions that received light at the blue or red ends of the spectrum were able to carry out photosynthesis and evolve O2. Bacteria that use O2 tend to collect around those portions of the alga that give off O2. Thus the density of bacteria is a crude measure of the rate of O2 evolution. The action spectrum (the rate of O2 evolution at different wavelengths) can be approximated by the density of the bacteria at different places in the spectrum (Figure 1436). The burst of oxygen production when the illumination was switched to 650 nm suggests that this wavelength stimulates photosystem II, which accepts electrons directly from water and generates oxygen. Similarly, the dip in oxygen production when the illumination was switched to 700 nm suggests that this wavelength stimulates photosystem I, which accepts electrons from the electron-transport chain and is thus farther removed from the reactions that generate oxygen. This interpretation is supported by the more detailed analysis of the chromatic transients below. The chromatic transients result because the two photosystems are out of balance with one another. Each separate wavelength preferentially (but not absolutely) stimulates one of the two photosystems. Thus, when photosystem I is stimulated (by 700-nm light), it pulls electrons out of the electrontransport chain that links the two photosystems, leaving them in a relatively oxidized state, primed to accept electrons from photosystem II. When the light is switched to 650 nm (which stimulates photosystem II), there is an initial rush of electrons (from H2O) into the cytochrome chain that causes a burst of O2 evolution. The flow of electrons through the cytochromes is quickly limited by the ability of the electrons to be transferred to photosystem I, which is suboptimally stimulated, and O2 evolution slows. When the light is switched back to 700 nm, the electron pressure from photosystem II (which is now suboptimally stimulated) is insufficient to push electrons into the relatively reduced (electron-rich) cytochromes. As a result, O2 evolution is depressed transiently while electrons are bled off from the cytochromes. Once the cytochromes have been partially drained of their electrons, they can accept new electrons from photosystem II, thereby reestablishing the normal level of oxygen production. References: Emerson R (1958) The quantum yield of photosynthesis. Annu. Rev. Plant Physiol. 9, 124. Lawlor DW (1987) Photosynthesis: Metabolism, Control, and Physiology. New York: Wiley. 1493 ultraviolet 450 blue 500 green 600 yellow 700 red infrared Figure 1436 Action spectrum of a filamentous green alga (Answer 1492). CHLOROPLASTS AND PHOTOSYNTHESIS 1494 A. Since stimulation by 680-nm light removes electrons from the cytochromes, causing their oxidation, 680-nm light must preferentially stimulate photosystem I, which transports electrons from the cytochromes to NADP+ (Figure 1437). The subsequent stimulation by 562-nm light causes electrons to flow into the cytochromes at a faster rate than before, thereby causing them to become more reduced. Consequently, 562-nm light must stimulate photosystem II, which transfers electrons from water to the cytochromes (Figure 1437). Thus, in these algae, as in most plants, the longer wavelength preferentially stimulates photosystem I, and the shorter wavelength preferentially stimulates photosystem II. B. These results support the Z scheme of photosynthesis in several ways. First, the different effects at the two wavelengths suggest that there are at least two components that differ in their responses to these wavelengths of light. Second, the two wavelengths have opposite effects on the redox poise of the cytochromes--680-nm light causing oxidation and 562-nm light causing reduction. Finally, the effects at the two wavelengths could be separated by DCMU, which indicates that the two photosystems communicate through the cytochromes (Figure 1437). C. These results indicate that DCMU blocks electron transport through the cytochromes on the upstream side; that is, on the side nearer photosystem II (Figure 1437). When photosystem I is stimulated by 680-nm light in the presence of DCMU, it transfers what electrons are available out of the cytochromes, causing their oxidation. In addition, in the presence of DCMU electrons cannot be transferred into the cytochromes by stimulation of photosystem II by 562-nm light (see Figure 1419B). These two effects indicate that DCMU blocks electron transport very near the beginning of the cytochrome chain. Reference: Duysens LNM, Amesz J & Kamp BM (1961) Two photochemical systems in photosynthesis. Nature 190, 510511. 1495 A. These results support a gear-wheel connection between the abstraction of electrons from water and their activation in photosystem II reaction centers. The periodicity of O2 evolution in response to light flashes rules out the possibility that four photons must be delivered simultaneously to the reaction center. If four photons were needed simultaneously, each flash should yield an equal burst of O2. The periodicity also argues against cooperation among four reaction centers to produce a molecule of O2. At saturating light intensities, most of the reaction centers should be stimulated during each flash; if they could cooperate, they would generate O2 at each flash. Furthermore, the results of the DCMU experiment definitely eliminate the possibility of cooperation. If four reaction centers were required to cooperate, one might expect a fourthpower dependence on the concentration of active centers. However, a 30fold reduction in active centers (DCMU inhibited 97% of the active centers) gave only a 30-fold reduction in O2 evolution (peaks of oxygen production were 3% of those in the absence of DCMU) instead of the enormous reduction (304) expected from a fourth-power dependence. A periodicity in O2 evolution is exactly what one would expect from a gearwheel link between extraction of multiple electrons from water and photon excitation of single electrons in photosystem II reaction centers. Furthermore, each gear wheel must service a single reaction center. If one gear wheel could interact with four reaction centers, for example, then it could donate its four electrons from water during each flash, which would allow it to evolve O2 during each flash, eliminating the periodicity. B. The four-flash periodicity in the evolution of O2 argues strongly that the gear wheel picks up four electrons at a time from two water molecules and passes them on to the photosystem II reaction center one at a time. The timing of the appearance of the first burst of O2 says something about the darkDCMU A329 cytochromes NADP + PSI (680 nm) PSII (562 nm) Figure 1437 Simplified Z scheme of photosynthesis showing the relationship of the two photosystems, the cytochromes, and the point of action of DCMU (Answer 1494). A330 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts adapted state of the gear wheel; namely, that it holds three electrons. The first three flashes transfer those electrons. The gear wheel can then pick up four new electrons from water (in a reaction that depends on light), generating a molecule of oxygen in the process. (Actually, about a quarter of the gear wheels carry four electrons in the dark-adapted state, which is why there is significant oxygen evolution on the fourth flash.) C. The periodicity is gradually damped out with increasing flash number because the multiple photosystems fall out of phase with one another. During a single flash most of the photosystem reaction centers capture one photon, but some capture two photons, and others capture no photons. Those reaction centers that capture zero or two photons are out of step with the majority. After several flashes the number of out-of-step reaction centers increases sufficiently to obscure any periodicity. The period of dark adaptation at the beginning of the experiment is required to bring the majority of the reaction centers to the same state so that periodicity can be observed at all. Reference: Forbush B, Kok B & McGloin M (1971) Cooperation of charges in photosynthetic oxygen evolution II. Damping of flash yield, oscillation and deactivation. Photochem. Photobiol. 14, 307321. 1496 A. The switch in solutions creates a pH gradient across the thylakoid membrane. The flow of protons down their electrochemical potential drives ATP synthase, which converts ADP to ATP . B. No light is needed because the proton gradient is established artificially without a need for the light-driven electron-transport chain. C. No ATP would be synthesized because the proton gradient would be in the wrong direction for ATP synthase to make ATP. In fact, one might expect that ATP would be hydrolyzed because the backwards proton gradient would drive ATP synthase in reverse, causing hydrolysis of ATP. D. These experiments provided early supporting evidence for the chemiosmotic model by showing that a pH difference alone is sufficient to drive ATP synthesis. 1497 A. To promote ATP synthesis, the protons must get to the inside of the thylakoid vesicle to lower its pH, so that in the alkalinization part of the experiment an electrochemical proton gradient will be generated that can drive ATP synthesis. Thylakoid membranes, like mitochondrial inner membranes, are relatively impermeable to protons. As a result, HCl, which dissociates completely into ions, is minimally effective in lowering the internal pH of the vesicles, because H+ ions have great difficulty penetrating the membrane. Succinic acid, on the other hand, is less than half dissociated at pH 4 (which is below its lower pK), and in its nonionized form it can cross the membrane. Once across the membrane, it can dissociate to provide a supply of protons on the inside of the vesicle. Thus, acidification with succinic acid yields more ATP because it is more effective at delivering protons to the interior of the vesicles. The yield of ATP increases with increasing concentration of succinic acid because more succinic acid enters the vesicles. The higher concentration of succinic acid inside the vesicle provides a greater supply of protons to flow through the ATP synthase and, therefore, a higher yield of ATP. B. The yield of ATP depends on the pH of the alkaline stage of the experiment because the second pH determines the magnitude of the pH gradient: the larger the electrochemical proton gradient, the more ATP that can be synthesized. C. Only the addition of FCCP has an effect on the synthesis of ATP. FCCP carries protons across the membrane, thereby collapsing the proton gradient and eliminating ATP synthesis. Addition of DCMU, which stops the flow of electrons through the electron-transport chain, does not affect ATP synthesis. Stopping the electron flow has no effect on the size of the electrochemical proton gradient because its magnitude is set by the acidbase treatment. THE GENETIC SYSTEMS OF MITOCHONDRIA AND PLASTIDS Reference: Jagendorf AT & Uribe E (1966) ATP formation caused by acidbase transition of spinach chloroplasts. Proc. Natl Acad. Sci. U.S.A. 55, 170177. A331 THE GENETIC SYSTEMS OF MITOCHONDRIA AND PLASTIDS DEFINITIONS 1498 1499 Endosymbiont hypothesis Maternal inheritance 14100 Mitotic segregation TRUE/FALSE 14101 False. Mitochondria and chloroplasts divide throughout interphase, out of phase with the division of the cell or with each other. Similarly, replication is not limited to S phase but occurs throughout the cell cycle. The process is regulated, so that the total number of organellar DNA molecules doubles in every cell cycle. The mechanism of this regulation is not known. 14102 False. The mitochondrial genetic code differs slightly from the nuclear code, but it also varies slightly from species to species. 14103 False. The presence of introns in organellar genes is surprising precisely because corresponding introns are so uncommon in related bacterial genomes. 14104 True. Inheritance of organellar genomes is very different from the inheritance of nuclear genes, which is governed by Mendelian rules. A pattern of inheritance that does not obey Mendelian rules is unlikely to be due to a nuclear gene, which leaves the organellar genomes--the only other genomes in a cell. 14105 True. Petite mutants in yeast have been shown to lack a mitochondrial genome, but they retain readily identifiable mitochondria that maintain their number for as long as the cells are grown. Such mitochondria do not carry out oxidative phosphorylation, the key function (partially) encoded by mitochondrial genes; thus, the cells are restricted to growth by fermentation. THOUGHT PROBLEMS 14106 It is likely that your organism is derived from an ancient eucaryote that once possessed an endosymbiont. Transfer of DNA from the endosymbiont to the nuclear genome occurred, giving rise to the precursors of the scattered bits of `bacterial' DNA you found in your organism's genome. At some later point the endosymbiont-derived organelles (mitochondria) were lost, perhaps in adaptation to the anaerobic niche in which it now lives. An alternative hypothesis, which is difficult to rule out, is that the organism never possessed any mitochondria and that the bits of bacterial DNA were picked up by lateral gene transfer directly from other bacteria; that is, that there was no endosymbiont stage. This hypothetical organism resembles Giardia, one of the rare eucaryotes without mitochondria, which is thought to have been derived from a more typical eucaryotic cell that lost its mitochondria (see Problem 145). 14107 What you have neglected in your scheme is a mechanism to get the ATP out of the proto-Paracoccus. An adenine nucleotide carrier is absent from all A332 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts free-living bacteria, as expected, since they must retain ATP inside if it is to do them any good. All mitochondria have an ATPADP antiporter in their membranes to allow free exchange of ATP and ADP between the cytoplasm and the mitochondrial matrix. Only after the acquisition of this carrier would ATP synthesized by the endosymbiont be available to the host. 14108 Four. Two of the three in the standard code (UAA and UAG) plus two (AGA and AGG) that encode arginine in the standard code. 14109 Variegation occurs because the plants have a mixture of normal and defective chloroplasts. These sort out by mitotic segregation to give patches of green and yellow in leaves. Many of the green patches have cells that still retain defective chloroplasts in addition to the normal ones. As such patches grow, they can segregate additional cells that have only defective chloroplasts, giving rise upon cell division to an island of yellow cells in a sea of green ones. By contrast, yellow patches are due to cells that retain only defective chloroplasts. Thus, yellow cells cannot give rise to green cells by mitotic segregation; hence, there are no green islands surrounded by yellow. 14110 Pedigree A is for an autosomal recessive mutation. Both the mother and the father must be heterozygotes who carry one copy of the mutant chromosome. One quarter of their children, irrespective of gender, are expected to get a defective copy of the autosome from each parent, thereby becoming homozygous (and affected). Pedigree B is for an X-linked recessive mutation. The mother carries the mutation on one copy of her two X chromosomes, and is thus unaffected. She passes the X chromosome with the mutation randomly to half her offspring. The males who get that X chromosome are affected because it's their only X chromosome. Females who get the mutant X chromosome are unaffected because they have a normal X chromosome from their father. Note that without additional information (a larger pedigree) it would be difficult to be certain that this pedigree truly resulted from an X-linked recessive mutation and not from an autosomal recessive mutation. The distinguishing feature--a small number of affected individuals who are all male--could have resulted by chance in a small pedigree for an autosomal recessive mutation. Pedigree C is for an autosomal dominant mutation. The mother is heterozygous, but affected, because one copy of a pair of autosomal chromosomes carries a dominant mutation. She will pass the affected chromosome to half her children, regardless of gender, all of whom will be affected because the mutation is dominant. Pedigree D is for a mitochondrial mutation. The mother is affected and passes on her defective mitochondria to all the children because the fertilized egg contains only her mitochondria. In reality, mitochondrial inheritance is rarely as clear-cut as this example would indicate. The mother's mitochondria are rarely all of the defective type; thus, mitotic segregation can give rise to a range of mixtures of mutant and normal mitochondria in the children, who may display phenotypes that range from unaffected to severely affected. Real pedigrees for mitochondrial mutations are thus sometimes difficult to distinguish from pedigrees for autosomal dominant mutations. CALCULATIONS 14111 On average there are about 3.9 mitochondrial genomes per mitochondrion in human liver cells. If mitochondrial DNA is 1% of the nuclear DNA, its total amount is equivalent to 6.4 107 bp (0.01 6.4 109 bp), which is equal to 3.9 103 genomes (6.4 107 bp/16,569 bp) per cell. If there are 1000 mitochondria per liver cell, there are 3.9 genomes per mitochondrion (3.9 103 genomes/1000 mitochondria). THE GENETIC SYSTEMS OF MITOCHONDRIA AND PLASTIDS A333 DATA HANDLING 14112 A. The results in Figure 1426 are exactly what you would expect if mitochondrial DNA were replicated at random times throughout the cell cycle. Regardless of the length of the chase, a constant fraction of the labeled DNA should be triggered to replicate. The fraction of the DNA that is shifted (about 10%) is what you expect, since the labeling time with BrdU (2 hours) is about 10% of the 20-hour cell cycle. If replication were confined to a specific part of the cell cycle, then DNA that was labeled with 3H-thymidine in the first pulse should not be replicated a second time until the critical phase of the cell cycle came around again. As a result, very little of the labeled DNA would be shifted in density until that time. The critical time in the cell cycle would show up in this experiment as a high fraction of labeled DNA that was density shifted at a particular chase time. B. It is true that the cell population is asynchronous, but asynchrony has no bearing on the interpretation of the experiment. If the mitochondrial DNA is replicating at random times, then the synchrony of the cell population is irrelevant. Your colleague's concern is directed at the possibility that an asynchronous cell population might obscure your ability to detect a timed replication of mitochondrial DNA. Your elegant experimental design nicely gets around that potential objection. If mitochondria replicated at a specific time during the cell cycle, the brief pulse of 3H-thymidine would label only those cells in that portion of the cycle. Since in the remainder of the experiment you follow only the radioactive mitochondrial DNA molecules, the brief labeling period has, in essence, synchronized the cell population--you are blind to what happens in any cells that were outside the critical labeling period. C. A similar analysis of nuclear DNA would be expected to show a peak of density shifting between 15 and 20 hours, with very little shifting of radioactive DNA at shorter chase periods (Figure 1438). A pulse of 3H-thymidine would label nuclear DNA only in S phase. The labeled DNA would not replicate again until the next S phase. If the DNA were radioactively labeled at the end of one S phase, it would come to the beginning of the next S phase after 15 hours or so--at which point its density would be shifted by exposure to BrdU. If the DNA were labeled at the beginning of S phase, it would not enter S phase again for nearly 20 hours. In actual experiments, the peak of density shifting of labeled nuclear DNA was indeed observed between 15 and 20 hours. D. If mitochondrial DNA molecules were replicated at all times, but individual molecules had to wait one entire cell cycle between replication events, there would be a peak of density shifting after 18 to 20 hours, when the labeled molecules returned to their critical time in the cell cycle. These results would resemble those for nuclear DNA replication, since nuclear DNA also has to wait one full cycle, but the timing would be slightly different because mitochondrial DNA replicates in 2 hours, whereas nuclear DNA takes 5 hours. DNA shifted to heavy density (%) Reference: Bogenhagen D & Clayton DA (1977) Mouse L cell mitochondrial DNA molecules are selected randomly for replication throughout the cell cycle. Cell 11, 719727. 14113 A. Chloroplast DNA, whether it is pure or a contaminant in the mitochondrial DNA preparation, will be cut into the same restriction fragments. Therefore any band that shows up at the same position in the chloroplast and mitochondrial lanes is likely to be due to contamination. Thus, the middle band in the zucchini digest and both bands in the pea digest are likely to have resulted from contamination. Only the fragments that are uniquely represented in the mitochondrial DNA lanes are candidates to have resulted from transfer of DNA from chloroplast to mitochondrion. 20 nuclear DNA 10 0 0 mitochondrial DNA 5 10 15 20 chase time (hours) 25 Figure 1438 Peak of density-shifted nuclear DNA (Answer 14112). A334 Chapter 14: Energy Conversion: Mitochondria and Chloroplasts B. Only zucchini and corn show evidence of bands that are likely to have arisen by transfer of DNA from chloroplasts to mitochondria, since they alone have bands that are uniquely represented in the mitochondrial DNA lanes. Reference: Stern DB & Palmer JD (1984) Extensive and widespread homologies between mitochondrial DNA and chloroplast DNA in plants. Proc. Natl Acad. Sci. U.S.A. 81, 19461950. 14114 A. Initiation of protein synthesis in mitochondria differs from that in the cytoplasm in two distinct ways. The first is straightforward: the codon AUA in mitochondria can serve as an initiation codon and encodes methionine (see Figure 1429, mRNA 13). In cytoplasmic protein synthesis, AUA encodes isoleucine and does not serve as an initiator of protein synthesis. The second difference is more subtle: the encoded protein can begin immediately at the 5 end of the mRNA (see Figure 1429, mRNAs 7 and 16). Cytoplasmic (and bacterial) mRNAs typically have a short stretch of untranslated nucleotides at their 5 ends that are thought to help guide ribosomes onto the mRNA. In bacteria there is a short sequence in front of the start codon to which a ribosomal RNA hybridizes. In the cytoplasm, ribosomes bind to a 5 cap (a modified G attached post-transcriptionally to the end of the mRNA) and thread onto the mRNA for some distance before reaching the first start codon. Neither of these features is present at the 5 ends of mitochondrial mRNA. B. The termination codons for protein synthesis in mitochondria are unusual in two ways. First, the termination codon in mRNA 16 is AGA (see Figure 1429), which in the nucleus encodes arginine. Second, the termination codons in mRNAs 7 and 13 are not completely encoded in the DNA; instead, they are generated by addition of the poly-A tail (see Figure 1429). In mRNA 7 only the initial U of the UAA stop codon is encoded; in mRNA 13 only the initial UA of the stop codon is encoded. C. The presence of tRNA genes at the exact boundaries of the mRNA genes suggests that they might be involved in processing the mRNAs out of the single, long primary transcript. The tRNAs are thought to serve as structural signposts for the processing of the primary transcript. The folding of the tRNAs into cloverleaf structures would place distinctive structures at the ends of the mRNAs. It is thought that the tRNA structures are recognized and cleaved at their ends to remove them from the primary transcript. The mRNAs are then the remains of tRNA processing. This scheme is referred to as the tRNA punctuation model of RNA processing. References: Montoya J, Ojala D & Attardi G (1981) Distinctive features of the 5-terminal sequences of the human mitochondrial mRNAs. Nature 290, 465470. Ojala D, Montoya J & Attardi G (1981) tRNA punctuation model of RNA processing in human mitochondria. Nature 290, 470474. 14115 The abnormal patterns of cytochrome absorption suggest that both poky and puny affect mitochondrial function. The genetic analysis is consistent with a cytoplasmic mode of inheritance for poky, but a nuclear mode of inheritance for puny. Crosses 7, 8, and 9 in Table 146 are control crosses, which show that wild type always yields fast-growing progeny and that mutants always yield slowgrowing progeny. Crosses 1 and 2 show the cytoplasmic mode of inheritance of poky. When poky was present in the protoperithecial parent (cytoplasmic donor), all the spores grew slowly (cross 1); when it was in the fertilizing parent, the spores grew rapidly (cross 2). This result is expected if the cytoplasmic donor determines the type of mitochondria present in the spores. In cross 1 poky was the cytoplasmic donor and the spores grew slowly. In cross 2 wild type was the cytoplasmic donor and the spores grew rapidly. Crosses 3 and 4 show the nuclear mode of inheritance of puny. In both crosses puny contributes a mutant gene to the fusion and the wild type con- THE GENETIC SYSTEMS OF MITOCHONDRIA AND PLASTIDS tributes a normal gene to the fusion. These genes are divided up equally (in a Mendelian fashion) among the spores so that half the progeny grow rapidly and half the progeny grow slowly. Crosses 5 and 6 are slightly more complicated because they involve the interplay of the two mutations. In cross 5 poky is the cytoplasmic donor, and since all spores receive `poky' mitochondria, all spores grow slowly. Some spores (about half) will also carry the puny mutation in their nuclei (the other half will have wild-type nuclei--from poky), but it makes no difference whether the nuclei are normal or mutant because the mitochondria are already compromised by the poky mutation. In cross 6 poky is the nuclear donor (puny is the cytoplasmic donor); therefore, the poky mutation is present in none of the spores. Once again, half the spores will carry the puny mutation and half will be wild type; however, in the absence of `poky' mitochondria, the nuclear phenotypes are expressed. Thus, half the spores will grow rapidly and half will grow slowly. It was this sort of distortion from the expected Mendelian behavior of genes that led ultimately to the realization that mitochondria (and later chloroplasts) carried genetic material. References: Mitchell MB & Mitchell HK (1952) A case of "maternal" inheritance in Neurospora crassa. Proc. Natl Acad. Sci. U.S.A. 38, 442449. Mitchell MB, Mitchell HK & Tissieres A (1953) Mendelian and nonMendelian factors affecting the cytochrome system in Neurospora crassa. Proc. Natl Acad. Sci. U.S.A. 39, 605613. 14116 A. The deletions that suppress the Pet494 mutation do not affect the coding portion of the mRNA. Therefore, they do not alter the CoxIII gene product and cannot affect its stability. By contrast, the replacement of the normal 5 untranslated region with one from any of several other genes is perfectly consistent with an alteration in translation. These results suggest that the normal Pet494 gene product promotes the translation of CoxIII from the wild-type mitochondrial mRNA. B. The rearrangements of the 5 end of the CoxIII gene result in the deletion of essential mitochondrial genes, which normally would produce a cytoplasmic petite strain of yeast. Yet these strains have all the usual mRNAs and grow perfectly well. These observations suggest that the deleted DNA must be present somewhere else in the mitochondria. It turns out that the Pet494 suppressor strains contain both wild-type and deleted mitochondrial genomes. Although it is not uncommon for mitochondria to contain more than one DNA molecule, the mixture of DNAs is quite unusual. This socalled heteroplasmic state is normally unstable, and the individual mitochondrial genomes segregate rapidly. By demanding that the cells grow on glycerol, it is possible to maintain the heteroplasmic state indefinitely. In the presence of the nuclear Pet494 mutation, both mitochondrial genomes are required for growth on glycerol: the deleted genome provides translatable CoxIII mRNA and the wild-type genome provides all other essential gene products. Reference: Fox TD (1986) Nuclear gene products required for translation of specific mitochondrially coded mRNAs in yeast. Trends Genet. 2, 97100. A335 ...
View Full Document

This note was uploaded on 01/07/2011 for the course BIOLOGY 7.012 taught by Professor Ericlander during the Spring '04 term at MIT.

Ask a homework question - tutors are online