Chapter_02_Solutions

Chapter_02_Solutions - Chapter 2 Cells Chemistry and...

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Unformatted text preview: Chapter 2 Cells Chemistry and Biosynthesis 2 THE CHEMICAL COMPONENTS OF A CELL In This Chapter A13 DEFINITIONS THE CHEMICAL COMPONENTS OF A CELL CATALYSIS AND THE USE OF ENERGY BY CELLS A27 HOW CELLS OBTAIN ENERGY FROM FOOD A34 2–1 Avogadro’s number 2–2 Hydrophobic force 2–3 Molecule 2–4 Atomic weight 2–5 Hydrogen bond 2–6 Acid 2–7 van der Waals attraction TRUE/FALSE 2–8 True. With each half-life, half the remaining radioactivity decays. After 10 half-lives (1/2)10, about 1/1000 (exactly 1/1024) of the original radioactivity will remain. (It is useful to remember that 210 is about 1000.) 2–9 False. The pH of the solution will be very nearly neutral, essentially pH 7, because the few H+ ions contributed by HCl will be outnumbered by the H+ ions from dissociation of water. No matter how much a strong acid is diluted it can never give rise to a basic solution. In fact, calculations that take into account both sources of H+ ions and also the effects on the dissociation of water give a pH of 6.98 for a 10–8 M solution of HCl. 2–10 False. Strong acids bind protons weakly and give them up readily in a water environment. 2–11 False. Many of the functions that macromolecules perform rely on their ability to associate and dissociate readily, which would not be possible if they were linked by covalent bonds. By linking their macromolecules noncovalently, cells can, for example, quickly remodel their interior when they move or divide, and easily transport components from one organelle to another. It should be noted that some macromolecules are linked by covalent bonds. This occurs primarily in situations where extreme structural stability is required, such as in the cell walls of many bacteria, fungi, and plants, and in the extracellular matrix that provides the structural support for most animal cells. THOUGHT PROBLEMS 2–12 Organic chemistry in laboratories—even the very best—is rarely carried out in a water environment because of low solubility of some components and because water is reactive and usually competes with the intended reaction. A13 A14 Chapter 2: Cell Chemistry and Biosynthesis The most dramatic difference, however, is the complexity. It is critical in laboratory organic chemistry to use pure components to ensure a high yield of the intended product. By contrast, living cells carry out thousands of different reactions simultaneously with good yield and virtually no interference between reactions. The key, of course, is that cells use enzyme catalysts, which bind substrate molecules in an active site, where they are isolated from the rest of the environment. There the reactivity of individual atoms can be manipulated to encourage the correct reaction. It is the ability of enzymes to provide such special environments—miniature reaction chambers—that allows the cell to carry out an enormous number of reactions simultaneously without cross talk between them. 2–13 The atomic weights of elements represent the average for the element as isolated from nature. Elements in nature include a mixture of isotopes. For most elements one isotope represents that vast majority; those elements have atomic weights that are nearly integers. Chlorine, however, has two abundant isotopes (75% 35Cl and 25% 37Cl), which average to an atomic weight of 35.5. 2–14 A. The atomic number of carbon, which equals the number of protons, is six. The atomic weight, which equals the number of protons plus neutrons, is 12. B. The number of electrons, which equals the number of protons, is six. C. The first shell can accommodate two electrons and the second shell eight. Carbon therefore needs four additional electrons (or would have to give up four electrons) to obtain a full outermost shell. Carbon is most stable when it shares four additional electrons with other atoms (including other carbon atoms) by forming four covalent bonds. D. Carbon 14 has two additional neutrons in its nucleus. Because electrons determine the chemical properties of an atom, carbon 14 is chemically identical to carbon 12. 2–15 A. The identity of the element specifies the number of protons; thus, 14C and 12C, for example, both have 6 protons. The difference in atomic weight is due to differences in numbers of neutrons. 14C has 2 more neutrons than 12C; 3H has 2 more neutrons than 1H; 35S has 3 more neutrons than 32S; and 32P has 1 more neutron than 31P . B. The change in identity of the element (from P to S) indicates that the number of protons has increased by 1 (from 15 for P to 16 for S). Because the atomic weight has remained the same, the number of neutrons must have decreased by 1 (from 17 in 32P to 16 in 32S). The conversion of a neutron into a proton is accompanied by emission of an electron. C. The equations for decay are indicated below with the number of protons shown as subscripts to make the relationships clearer. This process is illustrated for decay of 14C in Figure 2–38. 14C 6 3H 1 Æ 14N7 + e – Æ 3He2 + e – 35S 16 Æ 35Cl17 + e – + + antineutrino ‘b particle’ 14C 6 14N 7 Figure 2–38 Radioactive decay of 14C6 to 14N (Answer 2–15). 7 THE CHEMICAL COMPONENTS OF A CELL Among the product atoms, only the isotope of helium, 3He2, is not the most common (the most common isotope is 4He2). D. In each case the product atom will initially be missing an electron and thus will be positively charged. It will have one extra proton, but the same number of electrons it started with in its electron shell (remember the emitted electron came from the nucleus and is long gone). A positive charge on these product atoms would be highly unstable, and the atoms will steal an electron from some other molecule in their environment, returning the atom to electrical neutrality. But the theft of an electron from another molecule will initiate a free-radical cascade as that molecule, in turn, scrambles to replace its missing electron. Such free-radical cascades are one source of the biological damage caused by radioactivity. 2–16 You would expect the DNA backbone to break. The chemistry of an atom is determined by its number of protons, which establishes the number and reactivity of electrons in the outer electron shell. A phosphorus atom is comfortable, chemically speaking, bonded to the four oxygen atoms in the arrangement that makes up the phosphodiester bond in DNA, whereas a sulfur atom (which is what the phosphorus atom would become) is not. 2–17 The time at which an individual radioactive atom will decay is impossible to predict. For a sufficiently large population of radioactive atoms, however, it is possible to calculate very accurately the fraction that will decay over a defined period of time. 2–18 There is a good reason for the inverse relationship between half-life and maximum specific activity: the shorter the half-life, the greater the number of atoms that will decay per unit time, hence the greater the number of dpm or curies. If the radioactive atoms were present on an equimolar basis, then those with shorter half-lives would give more dpm; that is, they would have more Ci/mmol—a higher specific activity. 2–19 No. It is a coincidence that the ratio of C, H, and O in living organisms is the same as that for sugars. Much of the H and O (70%) is due to water, and the rest is from a mixture of sugars, amino acids, nucleotides, and lipids—the whole variety of small and large molecules that make up living organisms. 2–20 A. From the innermost to the outermost, the first three electron shells can carry 2, 8, and 8 electrons. B. H can gain one electron to fill the first shell, or it can lose one electron to leave a completely empty first shell. C can gain or lose four electrons to generate a filled outer shell. N and P will gain three electrons to fill their outer shells. O and S will gain two electrons to fill their outer shells. In general, it is energetically most favorable for an atom to lose or gain the fewest number of electrons required to generate a filled outer shell. C. The valences for these atoms equal the number of electrons that must be gained or lost to complete the outer shell. Thus, the valences are the same as the numbers in part B. 2–21 B (less than 1 kcal/mole), E (1–3 kcal/mole), A (12 kcal/mole), C (84 kcal/ mole), and D (675 kcal/mole). 2–22 Permanent dipoles are critical in biology because they allow molecules to interact through electrical forces. Any large molecule with many polar groups will have a pattern of partial positive and negative charges on its surface. When such a molecule encounters a second molecule with a complementary set of charges, the two molecules will be attracted to one another by interactions between their permanent dipoles. Such interactions resemble ionic bonds but are weaker. 2–23 A. Hydrogen bonds form between specific groups; one is always hydrogen linked in a polar bond to a nitrogen or an oxygen, and the other is usually a A15 A16 Chapter 2: Cell Chemistry and Biosynthesis nitrogen or an oxygen atom. Van der Waals attractions are weaker and occur between any two atoms that are in close enough proximity. Both hydrogen bonds and van der Waals attractions are short-range interactions that come into play only when two molecules are already close. Hydrogen bonds are directional, whereas van der Waals attractions are not. Both types of bonds can be thought of as ways to ‘fine-tune’ an interaction; that is, helping to position two molecules correctly with respect to each other once they have been brought together by diffusion. B. Van der Waals attractions would occur in all three examples. A hydrogen bond would occur only in example 3; that is, between a nitrogen atom and a hydrogen bound to an oxygen atom. 2–24 Because of its larger size, the outermost electrons in a sulfur atom are not as strongly attracted to the nucleus as they are in an oxygen atom. Consequently, the hydrogen–sulfur bond is much less polar than the hydrogen–oxygen bond. Because of the reduced polarity, the sulfur in H2S is not strongly attracted to hydrogen atoms in adjacent H2S molecules, and hydrogen bonds do not form. It is the lack of hydrogen bonds in H2S that allows it to be a gas, and the presence of strong hydrogen bonds in water that makes it a liquid. 2–25 Although the symbol ‘p’ in common usage denotes the ‘negative logarithm of,’ what it stands for is unclear. In the original 1909 paper in which the concept of pH was developed, the author—Danish chemist Soren P.L. Sorensen—was not explicit. In textbooks where it is commented on at all, it is most commonly reputed to stand for the French or German words for power or potential. Close examination of the original paper reveals that the ‘p’ in pH is likely a consequence of the author’s arbitrary choice to call two solutions by the letters ‘p’ and ‘q’. The q solution had the known H+ concentration of 1, the p solution had the unknown H+ concentration. If the solutions had been switched, do you think qH would ever have caught on? Reference: Norby JG (2000) The origin and the meaning of the little p in pH. Trends Biochem. Sci. 25, 36–37. 2–26 A solution of sodium chloride will be neutral. Neither the sodium ion nor the chloride ion binds H+ or OH– and thus neither influences the dissociation of water. A solution of potassium acetate (the salt of a weak acid) will be basic because the acetate ion will steal sufficient numbers of protons from water to satisfy the equilibrium CH3COO– + H2O CH3COOH + OH– The increase in hydroxyl ions will cause the number of protons to decrease, satisfying the equilibrium for water ionization ([OH–][H+] = 10–14) and making the solution basic. A solution of ammonium chloride (the salt of a weak base) will be acidic because the ammonium ion will dissociate sufficiently to satisfy the equilibrium NH4+ + H2O NH3 + H3O+ The increase in hydronium ions lowers the pH and makes the solution more acidic. 12 10 9.60 (4) NH3CH2COO– & NH2CH2COO– + 8 5.97 pH 6 A. The dissociation expression for a carboxylate group is –COOH H+ + –COO–. The dissociation expression for an amine group is –NH3+ H+ + –NH2. B. The acidic carboxyl group gives up its proton much more readily than the protonated amine group (that’s why the amine group is basic—it tends to pick up a proton from water). Thus, as shown in Figure 2–39, the pK for the (3) +NH3CH2COO– 4 2.34 (2) +NH3CH2COOH & +NH3CH2COO– 2 2–27 NH2CH2COO– (5) 11.30 (1) +NH3CH2COOH 0 0 0.5 1.0 1.5 NaOH added (equivalents) Figure 2–39 Titration of a solution of glycine (Answer 2–27). 2.0 A17 THE CHEMICAL COMPONENTS OF A CELL carboxyl group corresponds to the point at which 0.5 equivalents of OH– have been added, which is pH 2.3. The pK for the amine group corresponds to the point at which 1.5 equivalents of OH– have been added, which is pH 9.6. C. The predominant ionic species of glycine are shown in Figure 2–39. At point (2), the pK for the carboxyl group, two species +H3NCH2COOH and +H NCH COO– are present in equal concentrations. Similarly, at point (4), 3 2 the pK for the amine group, two species are present at equal concentrations. D. The isoelectric point occurs when 1.0 equivalents of OH– have been added (Figure 2–39). At that point—point (3) on the curve—the predominant ionic species is +H3NCH2COO–, which carries no net charge. The isoelectric point for glycine occurs at pH 5.97, which is exactly halfway between the pK values for the carboxyl group (2.34) and the amine group (9.60). At this pH all the other minor ionic species of glycine are present in exactly balancing amounts so that there is no net charge on the solute. 2–28 The structures of these three forms of glycine are shown in Figure 2–40. 2–29 The titration of histidine is shown in Figure 2–4A and that for glutamate is shown in Figure 2–4B. The requirement for three equivalents of OH– in both cases indicates that three ionizable groups are involved. Estimating the pK values from the points on the curves at which 0.5, 1.5, and 2.5 equivalents of OH– were added allows a match to be made with the amino acids listed in Table 2–2. 2–30 The rank order for pK values is expected to be 4, 1, 2, 3. It is convenient to discuss the rank order starting with the aspartate side chain carboxyl group on the surface of a protein with no other ionizable groups nearby (1). The side chain would be expected to have a pK around 4.5, somewhat higher than observed in the free amino acid because of the absence of the influence of the positively charged amino group. If the side chain were buried in a hydrophobic pocket on the protein (2), its pK would be higher because the presence of a charge in a hydrophobic environment (without the easy bonding to water) would be disfavored. If there were another negative charge in the same hydrophobic environment (3), the pK of the aspartate side chain would be elevated even further (even more difficult to give up a proton and become charged) because of electrostatic repulsion. If there were a positively charged group in the same environment (4), then the favorable electrostatic attraction would make it very easy for the proton to come off, lowering the pK even below that of the side chain on the surface (1). 2–31 You should advise the runner to breathe rapidly just before the race. Since a sprint will cause a lowering of blood and cell pH, the object of the pre-race routine would be to raise the pH with the idea that the runner could then sprint longer before feeling fatigue. Holding your breath or breathing rapidly both temporarily affect the amount of dissolved CO2 in the bloodstream. Holding your breath will increase the amount of CO2 and push the equilibrium to the right, leading to an increase in [H+] and a lower pH. By contrast, breathing rapidly will reduce the concentration of CO2 and pull the equilibrium to the left, leading to a decrease in [H+] and a higher pH. 2–32 The majority of aspirin is absorbed into the bloodstream through the lining of the stomach. At the low pH in the stomach, which is below the pK of aspirin, most of the aspirin will be uncharged and will therefore diffuse through the plasma membranes of the cells that line the stomach. + H3 N C H H free glycine COO– Na + COOH COO– Cl– +H3 N C H H glycine hydrochloride H2 N C H H glycine sodium salt Figure 2–40 Three forms of glycine (Answer 2–28). A18 Chapter 2: Cell Chemistry and Biosynthesis Figure 2–41 The functional groups in 1,3-bisphosphoglycerate, pyruvate, and cysteine (Answer 2–35). O O O– P – O C 2–33 2–34 The statement is correct. The hydrogen–oxygen bond in water molecules is polar; thus, the oxygen atom carries a partial negative charge and the hydrogen atoms carry partial positive charges. The partial negative charges on the oxygen atoms are attracted to the positive charges on the sodium ions but are repelled by the negative charges on the chloride ions. O hydroxyl HO C carboxylicphosphoric acid anhydride H Although individually weak, several such noncovalent interactions can, in aggregate, provide sufficient stability to hold a pair of molecules together. The situation is analogous to objects held together with VelcroTM: a small bit holds them together weakly, whereas a large bit holds them together tightly. Such fastenings are easy to peel apart because the links can be broken a few at a time rather than all at once. CH2 O O P – O phosphoryl O– 1,3 bisphosphoglycerate O O– carboxylate C 2–35 The functional groups on the three molecules are indicated and named in Figure 2–41. C 2–36 The drawings are accurate. The smaller hydrogen atoms are linked to oxygen atoms, and the larger ones are linked to carbon atoms. The difference in size reflects the polarity of the respective bonds: the H–C bond is nonpolar, whereas the H–O bond is polar. As a result, oxygen draws the shared electrons away from the hydrogen more strongly, resulting in a smaller radius of the electron cloud around the hydrogen atom. pyruvate Both amylose and cellulose are polymers of glucose. Amylose is a polymer of a-D-glucose, linked together by a1 Æ 4 glycosidic bonds. Cellulose is a polymer of b-D-glucose linked together by b1 Æ 4 glycosidic bonds. It is more difficult to discern this for cellulose, as every other monomer is flipped through 180° about an axis that passes through carbons 1 and 4. The structures of aD-glucose and b-D-glucose and the linked dimers in amylose and cellulose, respectively, are shown in Figure 2–42. CH 2–37 2–38 HO CH2 O HO OH OH HO b-D-glucose HO CH2 HO a-D-glucose CH2 OH HO HO CH2 O a-D-glucose O SH sulfhydryl CH2 CH2 amino NH3 + O carboxylate C O– cysteine b HO OH O OH 1 HO OH 4 CH2 HO O cellulose: b-1,4 glycosidic linkages HO O OH CH2 HO b-D-glucose rotated O OH HO OH HO OH HO HO CH3 The structures of the sedative (R)-thalidomide and the teratogenic (S)thalidomide differ at a single chiral center (Figure 2–43). After the teratogen had been identified, it was assumed that if thalidomide had been synthesized as the pure, correct optical isomer, it would have caused no problems. Recent experiments, however, have shown that thalidomide is rapidly racemized (converted to a mixture of optical isomers) in animals. Thus, a protocol designed to synthesize the correct isomer would not have made a difference in the end. HO HO O carbonyl OH HO HO CH2 O OH HO CH2 4 1 O a HO O OH amylose: a-1,4 glycosidic linkages Figure 2–42 The structures of a-D-glucose and b-D-glucose and their linkage into dimers (Answer 2–37). Arrows point to ring oxygens to show alignment of monomers that are joined into dimers. The OH curved arrow indicates a 180° rotation of the sugar. A19 THE CHEMICAL COMPONENTS OF A CELL (A) (R)-THALIDOMIDE (B) (S)-THALIDOMIDE O O N N O O O N O O H N O H sedative 2–39 Figure 2–43 The structures of the teratogenic (S) and sedative (R) forms of thalidomide (Answer 2–38). teratogen A molecule is amphiphilic if its hydrophobic and hydrophilic portions are segregated into two distinct regions of the molecule. As indicated in Figure 2–44, fatty acids and phospholipids are amphiphilic because each of these molecules has one well-defined portion that is hydrophilic and another that is hydrophobic. By contrast, triacylglycerols are relatively hydrophobic throughout. Because fatty acids and phospholipids are amphiphilic, collections of these molecules can form distinctive kinds of structures, including lipid monolayers (at an air interface), lipid bilayers (the essence of membrane structure), and micelles (aggregates with a hydrophilic surface and a hydrophobic center). Triacylglycerols, which are predominantly hydrophobic, separate from a water solution, forming a lipid droplet (the storage form of fat in adipose cells). (A) FATTY ACID O – O C CH2 CH2 CH2 CH2 (B) TRIACYLGLYCEROL O H2 C O C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 O HC O C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 O H2 C O C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 hydrophobic (C) PHOSPHOLIPID O H2 C O C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 O HC O C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 O H2 C O P O CH2 CH2 NH3 O– hydrophilic + Figure 2–44 A fatty acid, a triacylglycerol, and a phospholipid (Answer 2–39). The hydrophilic and hydrophobic portions of the molecules are indicated by light and dark shading, respectively. A20 Chapter 2: Cell Chemistry and Biosynthesis side chain side chain NH3 + H N CH2 CH2 CH2 side chain H N-terminus + H H N CH a H G glycine side chain O– O C H C O CH2 O N CH a C W tryptophan N H CH a K lysine side chain CH3 S CH2 CH2 H CH2 C CH2 O N CH a O Figure 2–45 The components of the polypeptide (Answer 2–40). The shaded atoms indicate peptide bonds. C E glutamate CH2 N H CH a O C C-terminus O– M methionine 2–40 The components of the polypeptide are shown in Figure 2–45. 2–41 The synthesis of a macromolecule with a unique structure requires that a single stereoisomer is used in each position. Changing one amino acid from its L- to its D-form would result in a different protein. Thus, if a random mixture of the D- and L-forms were used to build a protein, its amino acid sequence would not specify a single structure, but rather many different structures (2N different structures would be formed, where N is the number of amino acids in the protein). Why L-amino acids were selected in evolution as the exclusive building blocks of proteins is a mystery; we could easily imagine a cell in which certain (or even all) amino acids were used in the D-forms to build proteins, as long as these particular stereoisomers were used exclusively. 2–42 A major advantage of condensation reactions is that they are readily reversible by hydrolysis (and water is readily available in the cell). This allows cells to break down their macromolecules (or macromolecules of other organisms ingested as food) and to recover the subunits intact so that they can be ‘recycled’ to build new macromolecules. 2–43 The components of the oligonucleotides are shown in Figure 2–46. DNA is identifiable by the use of deoxyribose in the sugar-phosphate backbone and thymine (T) as one of the pyrimidine bases. By contrast, RNA uses ribose in the sugar-phosphate backbone and uracil (U) in place of thymine (T). CALCULATIONS 2–44 A. Although we do not know the molecular weight of cellulose (because the molecules contain variable numbers of C6H12O6 subunits), we know it is 40% carbon {(6 ¥ 12)/[(6 ¥ 12) + (12 ¥ 1) + (6 ¥ 16)]}. Thus 2 g of carbon atoms, which corresponds to 1023 atoms of carbon, are contained in the cellulose that makes up a page. 23 C atoms = 2 g ¥ 6 ¥ 10 d ¥ C atom g 12 d = 1023 C atoms B. The product of the number of carbon atoms in each dimension equals 1023 (X ¥ Y ¥ Z = 1023). The number of carbon atoms in each dimension will be in the same ratio as the lengths. Thus, if Z is the number of carbon atoms in the thickness of the page and X is the number in the width, the ratio of X/Z is [(21 ¥ 10–2 m)/(0.07 ¥ 10–3 m)], and X = [(21 ¥ 10–2)/(0.07 ¥ 10–3)] Z. Similarly, the ratio of Y/Z is [(27.5 ¥ 10–2 m)/(0.07 ¥ 10–3 m)], and Y = [(27.5 ¥ 10–2)/(0.07 ¥ 10–3)] Z. Substituting for X and Y, 21 ¥ 10–2 Z ¥ 27.5 ¥ 10–2 Z ¥ Z = 1023 0.07 ¥ 10–3 0.07 ¥ 10–3 1023 ¥ (0.07 ¥ 10–3)2 Z3 = (21 ¥ 10–2) ¥ (27.5 ¥ 10–2) 3 = 8.48 ¥ 1015 Z Z = 2.04 ¥ 105 A21 THE CHEMICAL COMPONENTS OF A CELL (B) DNA 5¢ end (A) RNA 5¢ end O O – O O P O Figure 2–46 Components of oligonucleotides (Answer 2–43). (A) RNA oligonucleotide. (B) DNA oligonucleotide. Shaded components from top to bottom are bases, nucleosides, and nucleotides. – NH O CH2 O N O BASES uracil O P O N O O CH2 NH N O O deoxyribose OH O O P – N CH2 – NH2 O N N O NH2 N ribose O guanine O P O NH2 N O adenine CH2 N N N O adenine N NUCLEOSIDES OH O – O O P O NH2 O – N CH2 N O O P O O H3 C O cytosine O CH2 NH N O thymine O NUCLEOTIDES O O – P OH O O – O P O O 3¢ end O 3¢ end The suggested shortcut makes the calculation a little more straightforward. The volume of the page is 4 ¥ 10–6 m3 [(21 ¥ 10–2 m) ¥ (27.5 ¥ 10–2 m) ¥ (0.07 ¥ 10–3 m)], which equals a cube with a side of 1.6 ¥ 10–2 m. The presence of 1023 carbon atoms in this volume corresponds to 4.6 ¥ 107 carbon atoms per side (1023)0.33. This corresponds to about 200,000 carbon atoms to span the thickness of the page [(4.6 ¥ 107 atoms/1.6 ¥ 10–2 m) ¥ (0.07 ¥ 10–3 m)]. C. With a diameter of 0.4 nm each it would take 175,000 carbons atoms to span the thickness of the page [(0.07 ¥ 106 nm)/0.4 nm], if they were laid end to end at their van der Waals contact distance. D. At first glance, it might seem strange that it takes more carbon atoms in cellulose, where they account for only 40% of the mass, than it takes as free, pure carbon atoms to span the thickness of the page. The key is that the atoms in cellulose are covalently bound to one another and therefore are much closer together than their van der Waals radii. The nuclei of covalently linked carbon atoms are separated by 0.15 nm, whereas those in van der Waals contact are separated by 0.4 nm. 2–45 Glucose (C6H12O6) has a molecular weight of 180 [(6 ¥ 12) + (12 ¥ 1) + (6 ¥ 16)] and therefore a mass of 180 g/mole. A concentration of 90 mg/dL corresponds to 5 ¥ 10–3 M or 5 mM. g [glucose] = 90 mg ¥ mole ¥ 10 dL ¥ 3 dL 180 g L 10 mg = 5 ¥ 10–3 moles/L, which is 5 ¥ 10–3 M, or 5 mM 2–46 The values of l for each of the isotopes are shown in Table 2–7. At t© 2.303 log (0.5) = –lt© A22 Chapter 2: Cell Chemistry and Biosynthesis Table 2–7 Decay constants for radioactive isotopes (Answer 2–46). RADIOACTIVE ISOTOPE HALF-LIFE 14C 5730 years 12.3 years 87.4 days 14.3 days 3H 35S 32P DECAY CONSTANTS (l) 10–4/year 1.21 ¥ 5.63 ¥ 10–2/year 7.93 ¥ 10–3/day 4.85 ¥ 10–2/day 2.30 ¥ 10–10/min 1.07 ¥ 10–7/min 5.51 ¥ 10–6/min 3.37 ¥ 10–5/min Rearranging, l = –2.303 log (0.5) = –2.303 (–0.301) t© t© 0.693 l= t© For 14C, 0.693 5730 years l = 1.21 ¥ 10–4/year l= Since there are 5.26 ¥ 105 min/year, l = 2.30 ¥ 10–10/min. 2–47 A. 14C atoms correspond to 1.3 ¥ 10–12 of the total carbon in living organisms. A disintegration rate of 15.3 dpm corresponds to 6.65 ¥ 1010 atoms of 14C. dpm = lN N= 15.3 dpm 2.30 ¥ 10–10 min–1 = 6.65 ¥ 1010 In one gram of carbon there are a total of 5.0 ¥ 1022 atoms [1 g ¥ (6 ¥ 1023 d/g) ¥ (C atom/12 d)]. Thus, 14C atoms are 1.3 ¥ 10–12 (6.65 ¥ 1010/5.0 ¥ 1022) of the total carbon in living organisms. B. A 70-kg human contains about 9 ¥ 10–8 Ci, or 0.09 mCi of 14C. A 70-kg human contains 13,000 g of carbon (70 kg ¥ 0.185 ¥ 1000 g/kg). At 15.3 dpm per gram of carbon, this corresponds to 1.98 ¥ 105 dpm (13,000 g ¥ 15.3 dpm/g), which is 8.9 ¥ 10–8 Ci [(1.98 ¥ 105 dpm) ¥ (Ci/2.22 ¥ 1012 dpm)]. A little over twice this amount of radioactivity (0.2 mCi) is present in humans due to the natural abundance of 40K, which has a half-life of 1.3 ¥ 109 years (a decay constant of 1.01 ¥ 10–15 min–1) and constitutes 0.012% of the potassium on Earth. (A human is about 0.35% potassium by weight.) C. The sample is 13,500 years old. Its age can be calculated from the equation for first-order decay (see Problem 2–46). 2.3 log N = –lt N0 –2.3 N t= log l N0 Since the dpm per gram of carbon is proportional to the numbers of carbon atoms present initially (N0) and at the current time (N), dpm values can be substituted for them. –2.3 ¥ log 3.0 15.3 1.21 ¥ 10–4 year –1 = –1.9 ¥ 104 years ¥ –0.71 t = 1.35 ¥ 104 years, or 13,500 years t= 2–48 The specific activity of glycine would be 0.12 Ci/mmol if all carbons were 14C. The number of radioactive atoms in 1 mmol of glycine is 12 ¥ 1020 [(6 ¥ 1020 molecules/mmol) ¥ (2 14C/molecule)]. With a decay constant, l, of 2.30 ¥ 10–10/min, this number of 14C atoms corresponds to 2.76 ¥ 1011 dpm [(12 ¥ A23 THE CHEMICAL COMPONENTS OF A CELL 1020) ¥ (2.30 ¥ 10–10)], which is 0.124 Ci [(2.76 ¥ 1011 dpm)/(2.22 ¥ 1012 dpm/Ci)]. At a specific activity of 200 mCi/mmol, about 1/300 molecules of glycine would carry a 14C atom. 200 mCi/mmol is 1/600 of the maximum specific activity [(200 ¥ 10–6 Ci/mmol)/(0.12 Ci/mmol)], and since there are 2 carbons per glycine, about 1/300 molecules will have a 14C atom. 2–49 A. Ethanol in 5% beer is 0.86 M. Pure ethanol is 17.2 M [(789 g/L) ¥ (mole/46 g)], and thus 5% beer would be 0.86 M ethanol (17.2 M ¥ 0.05). B. At a legal limit of 80 mg/100 mL, ethanol will be 17.4 mM in the blood [(80 mg/0.1 L) ¥ (mmol/46 mg)]. C. At the legal limit (17.4 mM), ethanol in 5% beer (0.86 M) has been diluted 49.4-fold (860 mM/17.4 mM). This dilution represents 809 mL in 40 L of body water (40 L/49.4). At 355 mL per beer, this equals 2.3 beers (809 mL/355 mL). D. It would take nearly 4 hours. At twice the legal limit the person would contain 64 g of ethanol [(0.16 g/0.1 L) ¥ (40 L)]. The person would metabolize 8.4 g/hr [(0.12 g/hr kg) ¥ (70 kg)]. Thus, to metabolize 32 g of ethanol (the amount in excess of the legal limit) would require 3.8 hours [(32 g) ¥ (hr/8.4 g)]. 2–50 A. Hydronium (H3O+) ions result from water dissociating into protons and hydroxyl ions, each proton binding to a water molecule to form a hydronium ion (2 H2O Æ H2O + H+ + OH– Æ H3O+ + OH–). At neutral pH the concentrations of H3O+ ions and OH– ions are equal. We know that at neutrality the pH is 7.0, and therefore, the H+ (H3O+) concentration is 10–7 M. B. The molecular weight of water is 18 and thus has a mass of 18 g/mole. The mass of 1 liter of water is 1000 g. Thus, pure water is 55.6 M [(1000 g/L) ¥ (mole/18 g)]. (Because the mass of water is actually 18.015 g/mole, pure water is 55.5 M.) C. The ratio of H3O+ ions (10–7 M) to H2O molecules (55.5 M) is 1.8 ¥ 10–9. Thus, at neutral pH only about 2 water molecules in a billion are dissociated. (A) TITRATION OF HCl 12 10 pH at equivalence point 8 2–51 A. A solution is said to be neutral when the concentrations of H+ and OH– are exactly equal. This occurs when the concentration of each ion is 10–7 M, so that their product is 10–14 M2. B. In a 1 mM solution of NaOH, the concentration of OH– is 10–3 M. Thus, the concentration of H+ is 10–11 M, which is pH 11. [H+] 7.0 pH 4 2 K = w– [OH ] –14 2 = 10 –3 M = 10–11 M 10 M 0 20 40 60 80 100 (B) TITRATION OF ACETIC ACID 12 10 A. It would take 50 mL of KOH to neutralize the protons in the HCl solution and in the solution of acetic acid. Both solutions contain 50 mmol of titratable H+ (0.5 L ¥ 0.1 mole/L = 0.05 mole or 50 mmol). The titratable H+ will be exactly neutralized when 50 mL of 1 M KOH (0.05 L ¥ 1 mole/L = 0.05 mole or 50 mmol) has been added, as indicated in Figure 2–47. B. At the equivalence point, titration of the HCl solution by KOH will give a pH 7 solution of slightly less than 0.1 M KCl (0.091 M because of the 10% increase in volume due to added KOH) (Figure 2–47A). At the equivalence point the solution is the same as would be generated by dissolving the appropriate amount of KCl in water, which gives a neutral pH (pH 7) as discussed in Problem 2–26. At the equivalence point, titration of acetic acid by KOH will give a 0.091 M solution of potassium acetate (K+ CH3COO–). As discussed in Problem 2–26, a solution of potassium acetate will be basic. The pH changes rapidly in the equivalence point 0 C. A pH of 5.0 corresponds to an H+ concentration of 10–5 M. Thus, the OH– concentration is 10–9 M (10–14 M2/10–5 M). 2–52 pH changes rapidly 6 pH at equivalence point 8.9 8 pH changes rapidly pH 6 pH = pK 4.8 equivalence point half titrated [HA] = [A–] 4 2 0 0 20 40 60 80 1 M KOH added (mL) Figure 2–47 Titration curves for 500 mL solutions of 0.1 M HCl and acetic acid (Answer 2–52). 100 A24 Chapter 2: Cell Chemistry and Biosynthesis Table 2–8 Dissociation of a weak acid at pH values above and below the pK (Answer 2–53). – log [A ] [HA] pH pK +4 pK +3 pK +2 pK +1 pK pK –1 pK –2 pK –3 pK –4 [A–] [HA] ‘RULE-OF-THUMB’ % DISSOCIATION 104 103 102 101 100 10–1 10–2 10–3 10–4 4 3 2 1 0 –1 –2 –3 –4 % DISSOCIATION 99.99 99.9 99 91 50 9.1 0.99 0.099 0.0099 99.99 99.9 99 90 50 10 1 0.1 0.01 region of the equivalence point, making it difficult to estimate the pH from the curve in that region, but it is somewhere around pH 9 (Figure 2–47B). Knowing the pK for acetic acid and the concentration allows an exact calculation of the pH at the equivalence point, which is 8.86. As discussed below, the titration curve does give a good estimate of the pK for acetic acid, which would allow the calculation of pH at the equivalence point. C. The pK is equal to the pH when the concentrations of CH3COO– and CH3COOH are equal, which occurs when half the ionizable H+ has been titrated; that is, when 25 mL of 1 M KOH has been added. From the graph in Figure 2–47B, the pH of the solution at that point can be estimated to be between 4.6 and 4.9. The actual pK of acetic acid is 4.76. 2–53 A. The values for log [A–]/[HA], [A–]/[HA], and the percentage of the acid that has dissociated, are shown in Table 2–8. Included in the table are a set of ‘rule-of-thumb’ values that may be easier to remember, and are handy to have mentally available for estimating answers. B. A plot of pH versus percentage dissociation of the weak acid, HA, is shown in Figure 2–48. All weak acids, regardless of pK, yield titration curves that are identical to this one. The curves for different weak acids are shifted along the pH scale depending on their pK values. +4 +3 +2 +1 pH pK 0 all HA 10 20 30 40 50 60 percent dissociated 70 80 90 100 all A– Figure 2–48 Percentage dissociation of a weak acid as a function of pH (Answer 2–53). THE CHEMICAL COMPONENTS OF A CELL This titration curve is fundamentally similar to protein–ligand binding curves and to enzyme activity curves. As pointed out in Problem 3–103, all three phenomena—titration of weak acids, protein–ligand binding, and enzyme activity—generate identical curves. Most importantly, the ‘rule-ofthumb’ values pertain to each, allowing rapid estimates in all three situations. 2–54 A. Estimating the percentages of the four forms of phosphate at pH 7 is straightforward from the rule-of-thumb values derived in Problem 2–53 (see Table 2–8). Since the cytosol is about 5 pH units above the pK for dissociation of H3PO4, it will be about 99.999% ionized; thus, H3PO4 will account for only about 0.001% of the total. Since the cytosol is just slightly above the pK for dissociation of H2PO4–, H2PO4– would be slightly less than 50% and HPO42– would be slightly greater than 50% of the total. Since the cytosol is more than 5 pH units below the pK for dissociation of HPO42–, it will be less than 0.001% ionized; thus, PO43– will account for less than 0.001% of the total. B. The ratio of [HPO42–] to [H2PO4–] ([A–]/[HA]) in the cytosol at pH 7 can be calculated using the Henderson–Hasselbalch equation – pH = pK + log [A ] [HA] Substituting, 2– 7.0 = 6.9 + log [HPO4 – ] [H2PO4 ] 2– 0.1 = log [HPO4 –] [H2PO4 ] [HPO42–] = 1.26 [H2PO4–] Because these two forms of phosphate sum to 100% (the other two forms are negligible), the ratio can be used to calculate the percentage of each form that is present (44% for H2PO4– and 56% for HPO42–). Thus, if the cytosolic concentration of phosphate is 1 mM, then the concentration of H2PO4– is 0.44 mM and that of HPO42– is 0.56 mM. 2–55 You could start with any of the individual solutions of phosphate and make a 1 mM solution by adding 10 mL to a liter of water. (You would want to start with a little less than a liter of water so that you could adjust the pH by addition of HCl or KOH, as needed. When the pH was adjusted to 6.9, you would then bring the solution to a final volume of 1 liter by addition of water.) If you started with H3PO4, you would need to add 1.5 mL of 1 M KOH (1.5 equivalents of OH–) to bring the solution to pH 6.9. If you started with KH2PO4, you would need to add 0.5 mL of KOH (0.5 equivalents of OH–) to bring the solution to pH 6.9. If you started with K2HPO4, you would need to add 0.5 mL of 1 M HCl (0.5 equivalents of H+) to bring the pH to 6.9. If you started with K3PO4, you would need to add 1.5 mL of HCl (1.5 equivalents of H+) to bring the pH to 6.9. In all these cases you are moving phosphate along its titration curve to reach 6.9 (the pK for H2PO4– H+ + HPO42–) at which point [H2PO4–] equals [HPO42–]. You could reach that same point by mixing equal amounts of the species directly. Thus, you would get a 1 mM solution at pH 6.9 by mixing 5 mL aliquots of the KH2PO4 and K2HPO4 solutions in 1 liter of water. In practice you would still measure the pH to make sure it was 6.9 in case one of your original solutions was not exactly ‘as advertised.’ Similarly, you could also achieve the same end by mixing 5 mL aliquots of the H3PO4 and K3PO4 solutions. The solutions will not all be the same. Although they will all be 1 mM phosphate, they can differ in the amount of K+ and Cl– ions. Solutions that are brought up to pH 6.9 by addition of KOH and those that are obtained by direct mixing of different phosphate solutions will be 1.5 mM K+. A solution A25 A26 Chapter 2: Cell Chemistry and Biosynthesis of K2HPO4 that is adjusted downward to pH 6.9 with HCl will be 2 mM K+ and 0.5 mM Cl–. A solution of K3PO4 that is adjusted to pH 6.9 with HCl will be 3 mM K+ and 1.5 mM Cl–. Depending on the uses for which the buffer is intended, these differences could be crucial. It is especially important to note that these calculations assume that the pH adjustments with KOH or HCl were precise. If one is sloppy, and overshoots the pH and then undershoots it several times before hitting it exactly, the concentrations of K+ and Cl– can be arbitrarily high—even though the phosphate is exactly 1 mM and the pH is exactly 6.9. This can lead to very puzzling results…and has! 2–56 Since the pK values of the two buffering systems are not very different, the key consideration is overall concentration of the buffers. The concentration of globin chains in red blood cells is 6.7 mM. g [globin] = 100 mg ¥ mole ¥ ¥ 1000 mL mL 15,000 g 1000 mg L = 6.7 ¥ 10–3 mole/L = 6.7 mM If all ten histidines can interact with the cytosol (that is, are not tied up in ionic bonds, for example), then the total concentration of histidines in globin is 67 mM (10 ¥ 6.7 mM). Thus, the potential buffering capacity of the histidines in globin, 67 mM, is much greater than that of phosphate at 1 mM. 2–57 A. The ratio of [HCO3–] to [CO2(dis)] at pH 7.4 is 20. – pH = pK ¢ + log [HCO3 ] [CO2(dis)] – 7.4 = 6.1 + log [HCO3 ] [CO2(dis)] [HCO3–] = 1.3 and [HCO3–] = 20 log [CO2(dis)] [CO2(dis)] Since the total carbonate is 25 mM, HCO3– is 23.8 mM [(20/21) ¥ 25 mM] and CO2(dis) is 1.2 mM [(1/21) ¥ 25 mM]. B. Addition of 5 mM of H+ would drive 5 mM of HCO3– to CO2(dis), thereby maintaining the equilibrium for hydration and dissociation of CO2. Thus, addition of 5 mM H+ would reduce [HCO3–] to 18.8 mM, and increase [CO2(dis)] to 6.2 mM. At these concentrations the pH would be 6.6. – pH = pK ¢ + log [HCO3 ] [CO2(dis)] = 6.1 + log 18.8 6.2 pH = 6.1 + 0.48 = 6.6 In a closed system bicarbonate/CO2 would provide a very weak buffering system with a very small buffering capacity. C. In an open system, addition of 5 mM H+ would cause the same changes as above except that the excess CO2 would be removed by exhalation, maintaining its concentration at 1.2 mM. Under these conditions the pH would be 7.3. – pH = pK ¢ + log [HCO3 ] [CO2(dis)] = 6.1 + log 18.8 1.2 pH = 6.1 + 1.19 = 7.3 Thus, in an open system the pH decreases by only about 0.1 pH unit. The beauty of this buffering system is that HCO3– is constantly being added back to the system through metabolism, which generates CO2 that is then hydrated to HCO3–. Moreover, the two components of the system are independently regulated: CO2 exhalation by the lungs can be controlled by the rate of breathing, and HCO3– can be excreted or retained by the kidneys. 2–58 The concentration of protein is about 200 mg/mL (0.18 ¥ 1.1 g/mL = 198 mg/ CATALYSIS AND THE USE OF ENERGY BY CELLS mL). Note that if you are given the density of the cell, you don’t need to know its volume to calculate the concentration of protein. DATA HANDLING 2–59 The effects on pK values are due to electrostatic interactions between the carboxyl and amino groups. In alanine a large electrostatic attraction between the –NH3+ and the –COO– is present at pH 7. This favorable interaction makes it more difficult to remove a proton from –NH3+, raising its pK, and more difficult to add a proton to –COO–, lowering its pK. The electrostatic attraction decreases as the amino and carboxyl groups are moved farther and farther away from one another in the oligomers of alanine, virtually disappearing by Ala4, as reflected in the changes in pK values. Reference: Cantor CR & Schimmel PR (1980) Biophysical Chemistry, Part 1: The Conformation of Biological Macromolecules, pp 42–46. San Francisco: WH Freeman and Company. 2–60 Assuming that the change in enzyme activity is due to the change in protonation state of histidine, the enzyme must require histidine in the protonated, charged state. The enzyme is active only below the pK of histidine (which is typically around 6.5 to 7.0 in proteins), where the histidine is expected to be protonated. CATALYSIS AND THE USE OF ENERGY BY CELLS DEFINITIONS 2–61 Activation energy 2–62 Standard free-energy change (DG°) 2–63 Oxidation 2–64 Substrate 2–65 Diffusion 2–66 Enzyme 2–67 Coupled reaction 2–68 Equilibrium 2–69 Free energy (G) TRUE/FALSE 2–70 True. The difference between plants and animals is in how they obtain their food molecules. Plants make their own using the energy of sunlight, plus CO2 and H2O, whereas animals must forage for their food. 2–71 True. Oxidation–reduction reactions refer to those in which electrons are removed from one atom and transferred to another. Since the number of electrons is conserved (no loss or gain) in a chemical reaction, oxidation—removal of electrons—must be accompanied by reduction—addition of electrons. 2–72 False. The equilibrium constant for the reaction A B remains unchanged; it’s a constant. Linking reactions together can convert an unfavorable reaction into a favorable one, but it does so not by altering the equilibrium constant, but rather by changing the concentration ratio of products to reactants. A27 A28 Chapter 2: Cell Chemistry and Biosynthesis THOUGHT PROBLEMS 2–73 Catabolic pathways break down larger molecules (often derived from food) into smaller molecules, and abstract energy in a useful form in the process. Anabolic pathways (or biosynthetic pathways) construct larger molecules from smaller ones. The small molecules generated by catabolic pathways are used as starting points and intermediates in anabolic pathways, and the energy from catabolic pathways, harnessed in the form of activated carriers, is used to drive the energetically unfavorable process of biosynthesis. 2–74 The second law of thermodynamics applies to closed systems, which could be a chamber in a scientist’s laboratory, for example, or the entire universe. Closed systems do not exchange matter or energy with their surroundings. Living organisms such as cells and human beings are not closed systems; they continually exchange matter and energy with their surroundings. It is perfectly permissible for a portion of a closed system—a human being in the universe—to increase its order, provided that the rest of the system (the rest of the universe) becomes disordered to a greater extent. This is what living organisms do: they take in food and use the energy to increase their order. But to do so they release waste products that are less complex (less ordered) than the food they took in, and much of the energy in the food is released in its most disordered form—as heat. Whatever order is created within a cell or an organism is more than paid for by the disorder introduced into its environment. 2–75 The overall reaction of photosynthesis is unlikely to be carried out by a single enzyme because of the large number of covalent bonds that must be made to convert CO2 to glucose. Typically, enzymes alter one or a pair of covalent bonds in a reaction. The conversion of CO2 to glucose requires a large number of proteins to carry out the individual steps in the overall pathway. Because sugars are more complicated molecules than CO2 and H2O, the reaction generates a more ordered state inside the cell. As demanded by the second law of thermodynamics, heat is generated at many steps along the pathway of these reactions, as summarized in the equation for photosynthesis. 2–76 If the reaction is rewritten as its two half reactions, it is then clear that Na is oxidized and Cl is reduced. NET: 2 Na Æ 2 Na+ + 2 e – 2 e – + Cl2 Æ 2 Cl– 2 Na + Cl2 Æ 2 Na+ + 2 Cl– Electrons are removed from sodium; therefore it is oxidized. Electrons are added to chlorine; therefore it is reduced. 2–77 A. For polymerization to be favored at high temperature and depolymerization to be favored at low temperature, DH and DS must both be positive. At high temperature, where polymerization is favored, the –TDS term becomes large enough to overcome the positive DH term, yielding a negative, favorable DG for polymerization. At low temperature, where depolymerization is favored, the –TDS becomes small enough that it is outweighed by the positive DH term, giving rise to a positive, unfavorable DG for polymerization. B. It seems counterintuitive that polymerization of free tubulin subunits into highly ordered microtubules should occur with an overall increase in entropy (decrease in order). But it is counterintuitive only if one considers the subunits in isolation. Remember that thermodynamics refers to the whole system, which includes the water molecules. The increase in entropy is due largely to the effects of polymerization on water molecules. The surfaces of the tubulin subunits that bind together to form microtubules are fairly hydrophobic, and constrain (order) the water molecules in their immediate vicinity. Upon polymerization, these constrained water CATALYSIS AND THE USE OF ENERGY BY CELLS molecules are freed up to interact with other water molecules. Their newfound disorder much exceeds the increased order of the protein subunits, and thus the net increase in entropy (disorder) favors polymerization. 2–78 A. Reaction rates could be limited by any combination of the following factors: (1) the frequency of collision with the active site of the enzyme; (2) the proportion of molecules that are energetic enough to undergo reaction; or (3) the rate of release of the products from the active site. B. The 107–fold rate enhancement corresponds to the ratio of the areas under the curve to the right of the thresholds in Figure 2–22. The number of molecules with sufficient energy to undergo a catalyzed reaction (the area to the right of threshold A) divided by the number of molecules with sufficient energy to undergo an uncatalyzed reaction (the area to the right of threshold B) is equal to 107. 2–79 The statement is correct. A reaction with a negative DG°, for example, would not proceed spontaneously under conditions where there is already an excess of products over those that would be present at equilibrium. Conversely, a reaction with a positive DG° would proceed spontaneously under conditions where there is an excess of substrates compared to those present at equilibrium. 2–80 At the same concentrations, the DG values for the forward and reverse reactions will be the same magnitude but differ in sign. Thus, the DG for C + D Æ A + B is 4.5 kcal/mole. 2–81 Absolutely none. These values provide information about how energetically favorable a reaction is under standard conditions (DG°) and under actual conditions (DG). They provide no information about the rate at which a favorable reaction will occur. Rates depend on other factors: in the cell, most commonly on the existence of enzymes and their properties. 2–82 The cell links these two reactions by providing an enzyme that catalyzes the net reaction directly. Thus, in the cell, phosphorylation of glucose does not occur as the sum of two reactions but as a single reaction. The enzyme hexokinase, for example, binds glucose and ATP and catalyzes the transfer of a phosphate directly from ATP to glucose. 2–83 A positive DG for D Æ E and a negative DG for E Æ F is an unstable situation. A positive DG for D Æ E means that E will be converted to D. This will reduce the concentration of E and increase that of D. Meanwhile more D will be added from the upstream reaction, and E will be removed by the downstream reaction. These effects, which increase D and decrease E, continue until the concentration ratio [E]/[D] is sufficient to drive the reaction in the forward (D Æ E) direction, at which point the DG becomes negative. 2–84 The free energy DG (–11 to –13 kcal/mole) derived from ATP hydrolysis depends on both DG° (–7.3 kcal/mole) and the concentrations of the substrates and products. DG = DG ° + 1.41 kcal/mole log [ADP][Pi] [ATP] Given that DG° is –7.3 kcal, the ratio of [ADP][Pi]/[ATP] in cells must range from a little more than 10–3 (DG = –11.5) to a little less than 10–4 (DG = –12.9 kcal/mole), as they do under different cellular conditions. 2–85 Enzyme A is beneficial. It allows the interconversion of two energy carrier molecules, both of which are required in the triphosphate form for many metabolic reactions. Any ADP that is formed is quickly converted to ATP by oxidative phosphorylation, and thus the cell maintains a high [ATP]/[ADP] ratio. Because of enzyme A, called nucleotide phosphokinase, some of the ATP is used to keep the [GTP]/[GDP] ratio similarly high. A29 A30 Chapter 2: Cell Chemistry and Biosynthesis Enzyme B would be highly detrimental to the cell. Cells use NAD+ as an electron acceptor in catabolic reactions and must maintain a high [NAD+]/[NADH] ratio to support the breakdown of glucose and fats to make ATP By contrast, NADPH is used as an electron donor in biosynthetic reac. tions; cells thus maintain a high [NADPH]/[NADP+] ratio to drive the synthesis of various biomolecules. Since enzyme B would bring both ratios to 1, it would reduce the rates of both catabolic and anabolic reactions. 2–86 The two lists match up as follows: A with 1; B with 5; C with 6; D with 2; E with 3; and F with 4. 2–87 Reactions B, D, and E all require coupling to other, energetically favorable reactions. In each case, molecules are made that have higher-energy bonds. By contrast, in reactions A and C simpler molecules (A) or lower-energy bonds (C) are made. CALCULATIONS 2–88 A. The oxidation states for each of the carbons in the various two-carbon molecules are shown in Figure 2–49. B. The molecules are ordered from most reduced (ethane) to most oxidized (acetate and acetamide) in Figure 2–49. C. In all cases the differences in oxidation state correspond to a pair of electrons. Virtually all redox chemistry between organic molecules in cells involves transfers of pairs of electrons (usually as a hydride ion—a proton with two electrons). Transfers of pairs of electrons are favored because molecular orbitals are most stable with even numbers of electrons. If a molecule acquires (or gives up) a single electron, it will have an unpaired electron, an extremely unstable and reactive state known as a free radical. Free-radical chemistry is rarely used in biological reactions, although it is critically important, for example, in electron transport during respiration (see Chapter 14) and in the reduction of ribonucleotides to deoxyribonucleotides. D. A carbon–carbon double bond (H2C=CH2), an alcohol (H3C–CH2OH), an amine (H3C–CH2NH2), a thiol (H3C–CH2SH), and a phosphate ester (H3C–CH2PO42–) are all at the same oxidation state. Thus any reaction that converts one into the other is not a redox reaction. Similarly, carboxyls (H3C–COO–) and amides (H3C–CONH2) are at the same oxidation state. Carbonyls (H3C–CHO) are intermediate, and carbon–carbon single bonds are the most reduced form. If you remember the relationships among these molecules, you will be able to handle about 95% of biological redox reactions at a glance. For the rest you can apply the rules in this problem. 2–89 A. The first and third reactions in the series (succinate Æ fumarate and malate Æ oxaloacetate) are redox reactions. In the first, the two central carbons of succinate are oxidized by the introduction of a double bond—they have OXIDATION STATES MOLECULE C2 C1 C2 SUM 1 ethane H3 C CH3 –3 –3 –6 2 ethene H2 C CH2 –2 –2 –4 3 ethanol H3 C CH2 OH –3 –1 –4 –3 –1 –4 –3 –1 –4 4 phosphoethanol C1 H3 C CH2 PO4 2– + 5 ethylamine H3 C CH2 NH3 6 thioethane H3 C CH2 SH –3 –1 –4 7 acetaldehyde H3 C CHO –3 +1 –2 8 acetate H3 C COO –3 +3 0 9 acetamide H3 C CONH2 –3 +3 0 most reduced – most oxidized Figure 2–49 A series of two-carbon molecules arranged in order of increasing oxidation state (Answer 2–88). A31 CATALYSIS AND THE USE OF ENERGY BY CELLS each lost one electron (their oxidation states have increased from –2 to –1). In the third reaction, the carbon attached to the hydroxyl group has been oxidized by conversion to a carbonyl group—it has lost a pair of electrons (its oxidation state has increased from 0 to +2). The second reaction is not a redox reaction, since the oxidation states of the two central carbons have changed in compensating ways: the upper one from –1 to 0 and the lower one from –1 to –2. B. Since the molecules in the pathway have been oxidized, the unseen electron carriers must have been reduced; that is, they must have accepted a pair of electrons from pathway intermediates. It is instructive to examine the reactions a little closer. In addition to losing a pair of electrons, both succinate and malate also lose a pair of hydrogens. One of the hydrogens and both electrons travel together as a hydride ion (H–) to the electron carrier; the other hydrogen is released as a proton. The electron carrier in the oxidation of succinate is FAD, which picks up both the hydride ion and the proton to become FADH2. The electron carrier in the oxidation of malate is NAD+, which picks up the hydride ion to become NADH, but leaves the proton in solution. 2–90 The enzyme catalyzes events at the rate of 1 event/3.2 ¥ 10–5 sec, or 1 event/32 msec [(1014 events/100 yr) ¥ (yr/365 days) ¥ (days/24 hr) ¥ (hr/60 min) ¥ (min/60 sec)]. 2–91 The instantaneous velocities are H2O = 3.8 ¥ 104 cm/sec, glucose = 1.2 ¥ 104 cm/sec, and myoglobin = 1.3 ¥ 103 cm/sec. The calculation for a water molecule, which has a mass of 3 ¥ 10–23 g [(18 g/mole) ¥ (mole/6 ¥ 1023 molecules)], is shown below. v = (kT/m)© 1 1.38 ¥ 10–16 g cm2 v= ¥ 310 K ¥ K sec2 3 ¥ 10–23 g v = 3.78 ¥ 104 cm/sec ( © ( When these numbers are converted to km/hr the results are fairly astounding. Water moves at 1,360 km/hr, glucose at 428 km/hr, and myoglobin at 47 km/hr. Thus, even the largest (slowest) of these molecules is moving faster than the swiftest human sprinter! And water molecules are traveling at Mach 1.1! Unlike a human sprinter, or a jet airplane, these molecules make forward progress only slowly because they are constantly colliding with other molecules in solution. Reference: Berg HC (1993) Random Walks in Biology, Expanded Edition, pp 5–6. Princeton, NJ: Princeton University Press. 2–92 It would take glucose an average of 0.13 second and myoglobin an average of 1.3 seconds to diffuse 20 mm. The calculation for glucose is t = x 2/6D t = (20 mm)2 ¥ t = 0.13 sec (cm)2 sec ¥ (104 mm)2 6 ¥ (5 ¥ 10–6 cm2) Reference: Berg HC (1993) Random Walks in Biology, Expanded Edition, pp 5–6. Princeton, NJ: Princeton University Press. 2–93 A. At equilibrium DG is zero (for any reaction); there is no tendency for the reaction to proceed in one direction over the other direction. Substituting K for [F6P]/[G6P] gives 0 = DG° + 2.3 RT log K DG° = –2.3 RT log K, or –1.41 kcal/mole log K B. At equilibrium DG is zero and DG° is 0.42 kcal/mole DG ° = –2.3 RT log [F6P] [G6P] A32 Chapter 2: Cell Chemistry and Biosynthesis Substituting 1.41 kcal/mole for 2.3 RT, = –1.41 kcal log (0.5) mole = 0.42 kcal/mole C. Since DG° relates to the equilibrium, it is unchanged; that is, DG° = 0.42 kcal/mole. At DG = –0.6 kcal/mole, the ratio of [F6P] to [G6P] is 0.19. DG = DG ° + 2.3 RT log [F6P] [G6P] –0.6 kcal 0.42 kcal 1.41 kcal log [F6P] = + [G6P] mole mole mole [F6P] = –1.02 kcal/mole = –0.72 log [G6P] 1.41 kcal/mole [F6P] = 0.19 [G6P] 2–94 A. The equilibrium constant K = 4.6 ¥ 10–3 M–1. DG ° = –2.3 RT log K 3.3 kcal/mole = –1.41 kcal/mole log K 3.3 kcal/mole = –2.34 log K = –1.41 kcal/mole K = 4.57 ¥ 10–3 = 4.57 ¥ 10–3 M–1 (You will note that this expression gives the numerical value of K because it enters the equation as a log value, where units are meaningless. Because of this and because the units derived from the equilibrium expression can vary so widely (for example, M2, M1, no units, M–1, M–2, etc., depending on the reaction), some sources treat the equilibrium constant as a unitless number by convention. Throughout this book we have added back the appropriate units so that they can be used as a guide in other calculations such as the one below.) B. The equilibrium concentration of G6P would be 1.1 ¥ 10–7 M, or 0.11 mM, if [GLC] and [Pi] were at 5 ¥ 10–3 M (5 mM). Remember that in the expression for equilibrium all concentrations are M. K = [G6P] [GLC][Pi] [G6P] = (4.57 ¥ 10–3 M–1) ¥ (5 ¥ 10–3 M) ¥ (5 ¥ 10–3 M) [G6P] = 1.14 ¥ 10–7 M = 0.11 mM This would be a low concentration for an intermediate in glucose metabolism, most of which are in the range 10 to 100 mM or so. However, given that one intermediate in the glycolytic pathway, 1,3-bisphosphoglycerate, is present at about 1 mM, perhaps this concentration for G6P is not out of the question. Nevertheless, cells have devised a different way to carry out this reaction. C. The value of DG° for the net reaction is obtained by adding together the DG° values for the individual reactions; thus, DG° for the net reaction is –4.0 kcal/mole (3.3 kcal/mole – 7.3 kcal/mole). The equilibrium constant K is 6.9 ¥ 102. DG ° = –2.3 RT log K –4.0 kcal/mole = –1.41 kcal/mole log K –4.0 kcal/mole log K = = 2.84 –1.41 kcal/mole K = 6.87 ¥ 102 D. The equilibrium [G6P] would be 10.3 M, which is an improbably high concentration for a cell. It far exceeds the amount of available phosphate and would be more viscous than maple syrup. CATALYSIS AND THE USE OF ENERGY BY CELLS A33 K = [G6P][ADP] [GLC][ATP] (6.87 ¥ 102) ¥ (5 ¥ 10–3 M) ¥ (3 ¥ 10–3 M) [G6P] = (10–3 M) [G6P] = 10.3 M E. The cellular [G6P] of 200 mM is about 50,000-fold lower than the equilibrium concentration (10.3 M) calculated in part D. Obviously, in cells the phosphorylation of glucose to glucose 6-phosphate is not at equilibrium. In cells DG for the reaction is –6.6 kcal/mole. DG = DG ° + 2.3 RT log [G6P][ADP] [GLC][ATP] (200 ¥ 10–6)(10–3) (5 ¥ 10–3)(3 ¥ 10–3) = –4.0 kcal/mole + (1.41 kcal/mole)(–1.88) DG = –4.0 kcal/mole – 2.64 kcal/mole = –6.64 kcal/mole = –4.0 kcal/mole + 1.41 kcal/mole log 2–95 The whole population of ATP molecules in the body would turn over (cycle) 1800 times per day, or a little more than once a minute. Conversion of 3 moles of glucose to CO2 would generate 90 moles of ATP [(3 moles glucose) ¥ (30 moles ATP/mole glucose)]. The whole body contains 5 ¥ 10–2 mole ATP [(2 ¥ 10–3 mole/L) ¥ 25 L]. Since the concentration of ATP doesn’t change, each ATP must cycle 1800 times per day [(90 moles ATP/day)/(5 ¥ 10–2 mole ATP)]. DATA HANDLING 2–96 Addition of the final subunit into the ring involves two sets of bonds, one to each of its neighbors (Figure 2–50). If the bonds were equally strong, then you might expect a squaring of the equilibrium constant (a doubling of DG). In reality the situation is somewhat more complex and can give rise to equilibrium constants several orders of magnitude higher than the square. This rather simple treatment, however, serves to illustrate the stability that can be gained by closure. Icosahedral viruses, for example, are very stable as a consequence of closure in three dimensions. Reference: Howard J (2001) Mechanics of Motor Proteins and the Cytoskeleton, pp 151–163. Sunderland, MA: Sinauer Associates, Inc. 2–97 The theoretical underpinnings of the assertion that all DG values must be negative are so strong that the error must lie with the experiment. One potential source of error is that the concentrations of the intermediates have not been measured precisely enough. That is the most likely explanation given the formidable experimental challenges to such precise, instantaneous measurements of concentration. The other possibility is that the DG° values (equlibrium values) are slightly off relative to their true values under physiological conditions. References: Berg JM, Tymoczko JL & Stryer L (2002) Biochemistry, Fifth Edition, pp 436–437. New York: WH Freeman and Co. Minakami S, Suzuki C, Saito T & Yoshikawa H (1965) Studies on erythrocyte glycolysis. I. Determination of the glycolytic intermediates in human erythrocytes. J. Biochem. (Tokyo) 58, 543–550. Figure 2–50 Closure of the pentameric ring with two bonds (Answer 2–96). A34 Chapter 2: Cell Chemistry and Biosynthesis HOW CELLS OBTAIN ENERGY FROM FOOD DEFINITIONS 2–98 Citric acid cycle 2–99 Fat 2–100 Glycogen 2–101 Oxidative phosphorylation 2–102 Electron-transport chain 2–103 Glycolysis TRUE/FALSE 2–104 False. Glycolysis is the only metabolic pathway that can generate ATP in the absence of oxygen. There are many circumstances in which cells are temporarily exposed to anoxic conditions, during which time they survive by glycolysis. For example, in an all-out sprint the circulation cannot deliver adequate oxygen to leg muscles, which continue to power muscle contraction by passing large amounts of glucose (from glycogen) down the glycolytic pathway. Similarly, there are several human cell types that do not carry out oxidative metabolism; for example, red blood cells, which have no mitochondria, make ATP via glycolysis. Thus, glycolysis is critically important, but it’s sort of like insurance: it’s not so important until you need it, and then it’s hard to do without. 2–105 True. Oxygen is not a substrate (or a product) for any reaction in the citric acid cycle. Thus, the reactions can occur in the absence of oxygen. In cells, however, the reactions cannot proceed for very long in the absence of oxygen because NADH and FADH2 cannot be converted back to NAD+ and FAD by oxidative phosphorylation (which does depend on oxygen). In the absence of NAD+ and FAD four separate reactions of the cycle (see Figure 2–35) will cease to operate. THOUGHT PROBLEMS 2–106 The two lists match up as follows: A with 2 and 3, B with 4, and C with 1. 2–107 A. One way to balance the equation for a pathway is to write down each reaction and sum them all up, as done below for the first two reactions of glycolysis Glucose + ATP Æ G6P + ADP + H+ G6P Æ F6P SUM: Glucose + ATP + G6P Æ G6P + F6P + ADP + H+ (1) (2) Note that in the sum, the intermediate G6P appears on both sides and thus cancels out. Because the pathway intermediates always drop out of such a balanced equation, there is a less tedious way to balance the equation: all molecules at the blunt ends of the arrows are reactants and all molecules at the pointed ends are products. Ignore the intermediates, but pay careful attention to stoichiometry, since it is usually just the flow that is indicated. Using this method the equation for the first stage of glycolysis is Glucose + 2 ATP Æ 2 G3P + 2 ADP + 2 H+ B. The equation for the second stage of glycolysis is A35 HOW CELLS OBTAIN ENERGY FROM FOOD G3P + Pi + NAD+ + 2 ADP Æ pyruvate + NADH + 2 ATP + H2O C. The overall equation is just the sum of these two equations, after doubling the numbers in the second equation to get the stoichiometry right. Glucose + 2 ATP Æ 2 G3P + 2 ADP + 2 H+ 2 G3P + 2 Pi + 2 NAD+ + 4 ADP Æ 2 pyruvate + 2 NADH + 4 ATP + 2 H2O SUM: Glucose + 2 ADP + 2 Pi + 2 NAD+ Æ 2 pyruvate + 2 ATP + 2 NADH + 2 H2O + 2 H+ 2–108 The extreme conservation of glycolysis is one form of evidence that all present cells are derived from a common ancestor. In this view the elegant reactions of glycolysis would have evolved only once, and then they would have been inherited as cells evolved. The later invention of oxidative phosphorylation allowed 15 times more energy to be captured than is possible by glycolysis alone. This remarkable efficiency is close to the theoretical limit and hence virtually eliminates the opportunity for further improvements. The generation of alternative pathways would result in no obvious growth advantage that could have been selected in evolution. 2–109 Under anaerobic conditions, cells are unable to make use of pyruvate—the end product of the glycolytic pathway—and NADH. The electrons carried in NADH are normally delivered to the electron transport chain for oxidative phosphorylation, but in the absence of oxygen the carried electrons are a waste product, just like pyruvate. Thus, in the absence of oxygen, pyruvate and NADH accumulate. Fermentation combines these waste products into a single molecule, either lactate or ethanol, which is shipped out of the cell. The flow of material through the glycolytic pathway could not continue in the absence of oxygen in cells that cannot carry out fermentation. Because NAD+ + NADH is present in cells in limited quantities, anaerobic glycolysis in the absence of fermentation would quickly convert the pool largely to NADH. The change in the ratio NAD+/NADH would stop glycolysis at the step in which glyceraldehyde 3-phosphate (G3P) is converted to 1,3-bisphosphoglycerate (1,3BPG), a step with only a small negative DG normally (see Table 2–4). The purpose of fermentation is to regenerate NAD+ by transferring the pair of carried electrons in NADH to pyruvate and excreting the product. Thus, fermentation allows glycolysis to continue. 2–110 2–111 2–112 1-arseno-3-phosphoglycerate O In the absence of oxygen the energy needs of the cell must be met by fermentation to lactate, which requires a high rate of flow through glycolysis to generate sufficient ATP. When oxygen is added, the cell can generate ATP by oxidative phosphorylation, which generates ATP much more efficiently than glycolysis. Thus, less glucose is needed to supply ATP at the same rate. In the presence of arsenate, 1-arseno-3-phosphoglycerate is formed instead of 1,3-bisphosphoglycerate (Figure 2–51). Because it is sensitive to hydrolysis in water, the arsenate high-energy bond is destroyed before the molecule that contains it can diffuse to the next enzyme. The product of the hydrolysis, 3-phosphoglycerate, is the same product normally formed, but because it is formed nonenzymatically, the reaction is not coupled to ATP formation. Arsenate wastes metabolic energy by uncoupling many phosphotransfer reactions by the same mechanism, and that is why it is so poisonous. The reverse of the forward reaction is simply not a possibility under physiological conditions. Recall from Problems 2–83 and 2–97 that flow of material through a pathway requires that the DG values for every step must be negative. Thus, for a flow from liver glycogen to serum glucose the step from glucose 6-phosphate to glucose must have a negative DG. To simply reverse the forward reaction (that is, G6P + ADP Æ GLC + ATP, DG° = 4.0 kcal/mole) would require that the concentration ratio [GLC][ATP]/ [G6P][ADP] be less than 10–2.84 (0.0015) in order to bring the reaction to equilibrium (DG = 0). O O C As O– O– HOCH CH2 O O – P O O – H2O spontaneous O O– C HOCH CH2 + HAsO42– + H+ O – O P O O – 3-phosphoglycerate Figure 2–51 Hydrolysis of 1-arseno3-phosphoglycerate (Problem 2–111). A36 Chapter 2: Cell Chemistry and Biosynthesis DG = DG ° + 1.41 kcal/mole log [GLC][ATP] [G6P][ADP] 0 = 4.0 kcal/mole + 1.41 kcal/mole log [GLC][ATP] [G6P][ADP] [GLC][ATP] = –4.0 kcal/mole = –2.84 log [G6P][ADP] 1.41 kcal/mole [GLC][ATP] = 0.0015 [G6P][ADP] Inside a functioning cell, such as a liver cell exporting glucose, the concentration of ATP always exceeds that of ADP, but just for illustrative purposes let’s assume the ratio is 1. Under these conditions the ratio [GLC]/[G6P] must be 0.00145; that is the concentration of G6P must be nearly 700 times higher than that of GLC. Since the circulating concentration of glucose is maintained at between 4 and 5 mM, this corresponds to about 3 M G6P, an impossible concentration given that the total concentration of cellular phosphate is less than about 25 mM. 2–113 The statement is incorrect. The oxygen atoms that are part of CO2 do not come from the oxygen atoms that are consumed as part of the oxidation of glucose (or of any other food molecule). The electrons that are abstracted from glucose at various stages in its oxidation are finally transferred to oxygen to produce water during oxidative phosphorylation. Thus, the oxygen used during oxidation of food in animals ends up as oxygen atoms in H2O. One can show this directly by incubating living cells in an atmosphere that contains molecular oxygen enriched for the isotope, 18O, instead of the naturally occurring isotope, 16O. In such an experiment one finds that all the CO2 released from cells contains only 16O. Therefore, the oxygen atoms in the released CO2 molecules do not come directly from the atmosphere but rather from the organic molecules themselves and from H2O. 2–114 The carbon atoms in a sugar molecule are already partially oxidized, in contrast to all but the very first carbon in a fatty acid. Thus, more electrons and more energy can be abstracted per carbon from a fatty acid than from a sugar. One consequence of this can be seen just by looking at what fraction of a fatty acid versus a sugar enters the citric acid cycle as acetyl CoA. Two carbon atoms are lost from glucose in its conversion to acetyl CoA; thus, only two-thirds of its carbons enter the citric acid cycle. By contrast, all of the carbons of fatty acids enter the citric acid cycle as acetyl CoA. 2–115 You make net amounts of protein only after a meal, when a mixture of essential amino acids is circulating. Depending on the composition of the meal, amino acids circulate for about 1 to 4 hours after eating, during which time they’re consumed by protein synthesis, oxidation, and conversion to fatty acids and glycogen. Synthesis of increased amounts of protein is confined to this window because you must have the essential amino acids, which come from the diet, to carry out synthesis of net amounts of protein. Once the influx of amino acids from the diet abates, you can only recycle the amino acids you already have, which means that you cannot increase the amount of proteins. 2–116 The citric acid cycle continues because intermediates are replenished as necessary by reactions leading to the citric acid cycle (instead of away from it). One of the most important reactions of this kind is the conversion of pyruvate to oxaloacetate by the enzyme pyruvate carboxylase (Figure 2–52). pyruvate + CO2 + ATP + H2O Æ oxaloacetate + ADP + Pi + 2 H+ 2–117 Darwin exhaled the carbon atom, which therefore must be the carbon atom of a CO2 molecule. After spending some time in the atmosphere, the CO2 molecule likely entered a plant cell, where it became ‘fixed’ by photosynthesis and converted into part of a sugar molecule. While it is certain that these early steps must have happened this way, there are many different paths from there to the carbon atom in your hemoglobin. A37 HOW CELLS OBTAIN ENERGY FROM FOOD CO2 PYR pyruvate carboxylase AcCoA aspartate asparagine pyrimidines glucose OAA fatty acids cholesterol CIT MAL ICIT aKG FUM SUC ScCoA glutamate glutamine proline arginine purines heme chlorophyll The sugar could have been broken down by the plant cell into pyruvate or acetyl CoA, for example, which then could have entered biosynthetic reactions to build an amino acid. The amino acid might have been incorporated into a plant protein, maybe an enzyme or a protein that builds the cell wall. You might have eaten the delicious leaves of the plant in your salad, and digested the protein in your gut to produce amino acids again. After circulating in your bloodstream, the amino acid might have been taken up by a developing red blood cell to make its own protein, such as the hemoglobin in question. The food chain scenario can be much more complicated, of course. The plant, for example, might have been eaten by an animal, which in turn was consumed by you on your lunch break. Moreover, because Darwin died more than 100 years ago, the carbon atom could have traveled from animals to plants and back many times. In each round it would have started again as fully oxidized CO2 gas and entered the living world following its reduction during photosynthesis. CALCULATIONS 2–118 At this rate of ATP regeneration, the cell will consume oxygen at 6.7 ¥ 10–15 L/min [(0.9 ¥ 109 ATP/min) ¥ (1 O2/5 ATP) ¥ (22.4 L/6 ¥ 1023 O2) = 6.72 ¥ 10–15 L/min]. The volume of the cell is 10–12 L [(1000 mm3) ¥ (cm/104 mm)3 ¥ (mL/cm3) ¥ (L/1000 mL)]. Dividing the cell volume by the rate of consumption of O2 [(10–12 L)/(6.7 ¥ 10–15 L/min)] indicates that the cell will consume its own volume of oxygen in 149 minutes, or about 2.5 hours. (Since air contains only 20% oxygen, a cell would consume its own volume of air in about 30 minutes.) 2–119 The human body operates at about 70 watts—about the same as a light bulb. 3 watts = 109 ATP ¥ 5 ¥ 1013 cells ¥ mole ¥ 12 kcal ¥ 4.18 ¥ 10 J body 60 sec cell body 6 ¥ 1023 ATP kcal mole 69.7 J/sec = 69.7 watts = body body 2–120 You would need to expend 496 kcal in climbing from Zermatt to the top of the Matterhorn, a vertical distance of 2818 m. Substituting into the equation for work J work = 75 kg ¥ 9.8 m ¥ 2818 m ¥ ¥ kcal 3 sec2 kg m2/sec2 4.18 ¥ 10 = 495.5 kcal This is equal to about 1.5 Snickers™ (496 kcal/325 kcal), so you would be well advised to plan a stop at Hörnli Hut to eat another one. Figure 2–52 An important reaction for replenishing citric acid cycle intermediates that are removed for biosynthesis (Answer 2–116). PYR = pyruvate, AcCoA = acetyl coenzyme A, CIT = citrate, ICIT = isocitrate, aKG = a-ketoglutarate, ScCoA = succinyl coenzyme A, SUC = succinate, FUM = fumarate, MAL = malate, and OAA = oxaloacetate. A38 Chapter 2: Cell Chemistry and Biosynthesis In reality the human body does not convert chemical energy into external work at 100% efficiency, as assumed in this answer, but rather at an efficiency of around 25%. Moreover, you will be walking laterally as well as uphill. Thus you would need more than 6 Snickers™ to make it all the way. Reference: Frayn KN (1996) Metabolic Regulation: A Human Perspective, p 179. London: Portland Press. 2–121 A. Since there is only one reactant and one product, the ratio of equilibrium concentrations would not change. K = [2PG] [3PG] K [3PG]= [2PG] Thus the concentration of 2PG is fixed by the equilibrium constant. Whatever amount of 3PG is added, a constant fraction of it is converted to 2PG, so that the equilibrium ratio of concentrations would remain the same. The value of K is 0.20 (1.7 mM/8.3 mM). DG° for the reaction is DG ° = –1.41 kcal/mole log [2PG] [3PG] = –1.41 kcal/mole log 1.7 8.3 DG ° = 0.97 kcal/mole B. For the reaction PEP Æ 2PG the equilibrium constant is K = [2PG] [PEP] and the standard free-energy change is DG ° = –1.41 kcal/mole log [2PG] [PEP] 2.9 DG ° = –1.41 kcal/mole log 7.1 DG ° = 0.55 kcal/mole For the reverse reaction, DG° has the same magnitude but the opposite sign; thus, DG° = –0.55 kcal/mole. C. DG° for the overall reaction is the sum of the DG° values for the linked reactions; however, the DG° values must refer to the proper direction (that is, the directions of the linked reactions and the overall reaction must be the same). For the conversion of 3PG to PEP , DG°3PG Æ PEP = DG°3PG Æ 2PG + DG°2PG Æ PEP = 0.97 kcal/mole – 0.55 kcal/mole DG°3PG Æ PEP = 0.42 kcal/mole 2–122 A. The DG° for conversion of 3PG to pyruvate (PYR) and phosphate is the sum of the DG° values for the individual steps in the reaction: DG°3PG Æ PYR = DG°3PG Æ PEP + DG°PEP Æ PYR = 0.42 kcal/mole – 14.8 kcal/mole DG°3PG Æ PYR = –14.4 kcal/mole A39 HOW CELLS OBTAIN ENERGY FROM FOOD B. The DG° for conversion of 3PG to pyruvate and phosphate is independent of the pathway for the conversion. Thus, the DG° is –14.4 kcal/mole. The DG° for conversion of glycerate to pyruvate is obtained by subtracting the DG° for 3PG to glycerate (GLY) from the overall DG°: DG°GLY Æ PYR = DG°3PG Æ PYR – DG°3PG Æ GLY = –14.4 kcal/mole + 3.3 kcal/mole DG°GLY Æ PYR = –11.1 kcal/mole C. The analysis above indicates that a very large standard free-energy change occurs between glycerate and pyruvate. Removal of water (DG° = –0.5 kcal/ mole) does not account for very much of this free-energy change. Thus it appears that the conversion of enolpyruvate to pyruvate is accompanied by a large standard free-energy change of around –10.6 kcal/mole. This reasoning suggests that the majority of the standard free-energy change associated with the conversion of PEP to pyruvate (–10.6 kcal/mole out of –14.8 kcal/ mole) comes from the conversion of enolpyruvate to pyruvate and not from the hydrolysis of the phosphate bond. In fact, the standard free-energy change for PEP to pyruvate (–14.8 kcal/ mole) is close to the sum of the enolpyruvate to pyruvate step (about –11 kcal/mole) and the standard free-energy change for hydrolysis of a simple phosphate ester bond (about –3.0 kcal/mole). Thus, the phosphate bond in PEP is a high-energy bond because its hydrolysis is linked to the very favorable conversion of enolpyruvate to pyruvate. D. The DG° for the linked conversion of PEP to pyruvate and of ADP to ATP is –7.5 kcal/mole. The DG° for the linked reaction can be obtained by adding together the DG° values for the individual reactions. NET: PEP Æ PYR + Pi ADP + Pi Æ ATP PEP + ADP Æ PYR + ATP DG° = –14.8 kcal/mole DG° = 7.3 kcal/mole DG° = –7.5 kcal/mole Reference: Lipmann F (1941) Metabolic generation and utilization of phosphate bond energy. Adv. Enzymol. 1, 99–162. 2–123 A. The DG in resting muscle is –0.2 kcal/mole. DG = DG ° + 1.41 kcal/mole log [C][ATP] [CP][ADP] (13 ¥ 10–3)(4 ¥ 10–3) (25 ¥ 10–3)(0.013 ¥ 10–3) = –3.3 kcal/mole + 1.41 kcal/mole log 160 = –3.3 kcal/mole + 3.1 kcal/mole = –3.3 kcal/mole + 1.41 kcal/mole log DG = –0.2 kcal/mole This very small negative value should not surprise you; it says that this energy buffering system is nearly at equilibrium, which you should expect in the absence of heavy ATP usage. B. If the concentration of ATP decreases to 3 mM and that of ADP increases to 1 mM, then the DG for the reaction will be –3.0 kcal/mole. DG = DG ° + 1.41 kcal/mole log [C][ATP] [CP][ADP] –3 –3 = –3.3 kcal/mole + 1.41 kcal/mole log (13 ¥ 10–3)(3 ¥ 10–3) (25 ¥ 10 )(1 ¥ 10 ) = –3.3 kcal/mole + 1.41 kcal/mole log 1.56 = –3.3 kcal/mole + 0.27 kcal/mole DG = –3.0 kcal/mole A40 Chapter 2: Cell Chemistry and Biosynthesis Thus, as soon as exercise begins, the reaction will become highly favored and creatine phosphate will drive the conversion of ADP to ATP. In reality the enzyme that catalyzes this reaction is efficient enough to keep the reaction nearly at equilibrium (DG = 0) so that ATP levels remain high as creatine phosphate levels fall, fulfilling its role as an energy buffer. C. If ATP (4 mM) could sustain a sprint for 1 second, then creatine phosphate (25 mM) could sustain a sprint for an additional 6 seconds by regenerating an equal amount of ATP. This is not long enough to allow a sprinter to finish 200 meters; thus, there must be another source of energy. The additional energy comes from the breakdown of muscle glycogen, which is processed through anaerobic glycolysis, producing lactate and ATP. Typical stores of muscle glycogen are sufficient for about 80 seconds of sprinting. 2–124 A. The mitochondrion obtains a flow through this reaction by maintaining high concentrations of substrates and low concentrations of products. The concentration ratio of products to substrates, [OAA][NADH]/[MAL][NAD+], must be sufficiently small to overcome a positive DG° value, as calculated in part B. B. The minimum ratio of [MAL] to [OAA] is the ratio that will make DG zero, which is 1.1 ¥ 104. DG = DG ° + 1.41 kcal/mole log [OAA][NADH] [MAL][NAD+] = 7.1 kcal/mole + 1.41 kcal/mole log [OAA](1) [MAL](10) = 7.1 kcal/mole + 1.41 kcal/mole log 0.1 + 1.41 kcal/mole log [OAA] [MAL] if DG = 0, –7.1 kcal/mole – 1.41 kcal/mole log 0.1 log [OAA] = 1.41 kcal/mole [MAL] [OAA] = –5.69 kcal/mole = –4.04 log [MAL] 1.41 kcal/mole [OAA] = 9.2 ¥ 10–5, and thus [MAL]/[OAA] = 1.1 ¥ 104 [MAL] DATA HANDLING 2–125 Knoop’s result was surprising at the time. One might have imagined that the obvious way to metabolize fatty acids to CO2 would be to remove the carboxylate group from the end as CO2 and then oxidize the newly exposed carbon atom until it too could be removed as CO2. However, removal of singlecarbon units is not consistent with Knoop’s results, since it predicts that odd- and even-number fatty acids would generate the same final product. Similar inconsistencies crop up with removal of fragments containing more than two carbon atoms. For example, removal of three-carbon fragments would work for the eight-carbon and seven-carbon fatty acids shown in Figure 2–34, but would not work for six-carbon and five-carbon fatty acids. Removal of two-carbon fragments from the carboxylic acid end is the only scheme that accounts for the consistent difference in the metabolism of odd- and even-number fatty acids. One might ask why the last two-carbon fragment is not removed from phenylacetic acid. It turns out that the benzene ring interferes with the fragmentation process, which involves modification of the third carbon from the carboxylic acid end. Since that carbon is part of the benzene ring in phenylacetic acid, it is protected from modification and further metabolism of phenylacetic acid is blocked. Knoop’s results also specify the direction of degradation. If the nonacidic end of the chain were attacked first, either the benzene ring would make the fatty acids resistant to metabolism, or the same benzene compound would always be excreted, independent of the length of the fatty acid fed to the dogs. HOW CELLS OBTAIN ENERGY FROM FOOD Reference: Knoop F (1905) Der Abbau aromatischer Fettsäuren im Tierkörper. Beitr. Chem. Physiol. 6, 150–162. 2–126 A. The complete oxidation of citrate to CO2 and H2O occurs according to the balanced chemical reaction. C6H8O7 + 4.5 O2 Æ 6 CO2 + 4 H2O Thus each molecule of citrate would require 4.5 molecules of oxygen for its complete oxidation. The results in Table 2–5 were surprising to Krebs and others at the time because much more oxygen is consumed (40 mmol) than could be accounted for by oxidation of citrate itself. Only 13.5 mmol of oxygen would be required to oxidize 3 mmol of citrate completely (3 mmol ¥ 4.5). This calculation shows that citrate is acting catalytically in the oxidation of carbohydrates (which in these experiments were endogenous in the minced pigeon breasts). Although others were aware of the catalytic nature of other intermediates, Krebs was the first person to complete the circle of chemical reactions that constitute the citric acid cycle. Krebs’s experimental rationale is clearly laid out in the paper: “Since citric acid reacts catalytically in the tissue, it is probable that it is removed by a primary reaction but regenerated by a subsequent reaction. In the balance sheet no citrate disappears and no intermediate products accumulate. The first object of the study of intermediates is therefore to find conditions under which citrate disappears in the balance sheet.” B. The consumption of oxygen is low in the presence of the metabolic poisons because citrate is prevented from acting catalytically. The balanced equations for the conversion of citrate to a-ketoglutarate and succinate show that the amount of oxygen consumed is approximately what is expected. For citrate conversion to a-ketoglutarate, half a molecule of oxygen is consumed. C6H8O7 + 0.5 O2 Æ C5H6O5 + CO2 + H2O For citrate conversion to succinate, one molecule of oxygen is consumed. C6H8O7 + O2 Æ C4H6O4 + 2 CO2 + H2O Thus the observed stoichiometry of oxygen consumption matches the expectations. C. The absence of oxygen is crucial for demonstrating an accumulation of citrate from an intermediate in the cycle. In the presence of oxygen, citrate acts catalytically—is consumed and then regenerated—so that it does not accumulate no matter what intermediate is added. In the absence of oxygen, however, the conversion of citrate to a-ketoglutarate is blocked, since that conversion requires oxygen, as described below. Under these conditions citrate will accumulate if an appropriate intermediate is present. Of all the intermediates, only conversion of oxaloacetate to citrate does not require oxygen. The immediate precursor of oxaloacetate is malate. Since the conversion of malate to citrate requires oxygen, all other intermediates also must require oxygen to be converted to citrate. The requirement for oxygen is indirect and is mediated through the cofactors NAD+ and FAD; they accept electrons from the substrates and transfer them to the electron-transport chain and ultimately to oxygen. In the absence of oxygen, all of the NAD+ will quickly be converted to NADH, and all of the FAD will quickly be converted to FADH2. In the absence of NAD+ and FAD, the reactions of the cycle cannot proceed. D. E. coli and yeast do indeed use the citric acid cycle. Krebs got this point wrong because he did not realize (nor did anyone for a long time) that citrate cannot get into these cells. Therefore, when he added citrate to intact E. A41 A42 Chapter 2: Cell Chemistry and Biosynthesis coli and yeast, he found no stimulation of oxygen consumption. Passage of citrate across a membrane requires a transport system, which is present in mitochondria but is absent from yeast and E. coli plasma membranes. References: Krebs HA & Johnson WA (1973) The role of citric acid in intermediate metabolism in animal tissues. Enzymologia 4, 148–156. Stare FJ & Baumann CA (1936) The effect of fumarate on respiration. Proc. R. Soc. Lond. B121, 338–357. Szent-Györgyi AV (1924) Über den mechanismus de Succin- und Paraphenylendiaminoxydation. Ein Betrag der Zellatmung. Biochem Z. 150, 141–149. See also Albert Szent-Györgyi’s Nobel Lecture (1937) at www.nobel.se/medicine/laureates/1937/szent-gyorgyi-lecture.pdf 2–127 A. The cross-feeding experiments indicate that the three steps controlled by the products of the TrpB, TrpD, and TrpE genes are arranged in the order X TrpE TrpD TrpB Y Z tryptophan where X, Y, and Z are undefined intermediates in the pathway. The ability of the TrpE– strain to be cross-fed by the other two strains indicates that the TrpD– and TrpB– strains accumulate intermediates that are farther along the pathway than the step controlled by the TrpE gene. The ability of the TrpD– strain to be cross-fed by the TrpB– strain but not the TrpE– strain places it in the middle. The inability of the TrpB– strain to be cross-fed by either of the other strains is consistent with its controlling the step closest to tryptophan. B. The patterns of growth on minimal medium supplemented with known intermediates in the tryptophan biosynthetic pathway are consistent with the order deduced from cross-feeding: chorismate TrpE anthranilate TrpD indole TrpB tryptophan In reality, of course, the intermediates for the pathway were unknown (or not fully known) at the time the cross-feeding experiments were done. The intermediates were worked out by a combination of educated guesses at the likely intermediates, which could then be tested on mutant strains, and of analysis of the compounds that accumulated in the mutants. Reference: Yanofsky C (2001) Advancing our knowledge in biochemistry, genetics, and microbiology through studies on tryptophan metabolism. Annu. Rev. Biochem. 70, 1–37. 2–128 A. The ProA– strain is a deletion; it doesn’t grow at any temperature. The ProB– strain is temperature-sensitive; it doesn’t grow at 42°C, but grows at the other temperatures. The ProC– strain is cold-sensitive; it doesn’t grow at 22°C, but grows at the other temperatures. B. At 22°C, the ProC– strain cross-feeds the ProA– strain, indicating that the intermediate that accumulates in the ProC– strain comes after the block in the ProA– strain. Similarly, at 42°C the ProA– strain cross-feeds the ProB– strain, indicating that the intermediate that accumulates in the ProA– strain comes after the block in the ProB– strain. These two results—C after A, and A after B—suggest the order shown below. intermediate 1 ProB intermediate 2 ProA intermediate 3 ProC proline C. The identification of three genes by the cross-feeding experiments shown here indicates that there are very likely to be at least three steps in the pathway. But A43 HOW CELLS OBTAIN ENERGY FROM FOOD O – O O C CH2 ATP ADP H2 N C O O glutamate O CH2 CH2 C H2 N ProA CH C H2O H H CH2 CH2 C N spontaneous C O C O– H NADPH NADP+ CH2 ProB CH Pi CH2 CH2 H2 N O– O O O– C P O– H2 C CH2 CH2 H2 N+ CH ProC C O NADPH NADP+ O– CH C O O– O– g-glutamyl phosphate glutamate g-semialdehyde D-pyrroline-5-carboxylic acid there could be many more steps than identified genes. For example, mutations in the genes controlling some steps in the pathway may not be present in your collection of mutant strains. Conversely, there could be many more genes than steps. For example, multiple genes could encode different subunits of an enzyme that controls a single step. It is important to confirm the pathway by isolation of the enzymes and by detailed biochemical studies. Such studies have elucidated the complete pathway for proline biosynthesis (Figure 2–53). D. The lack of cross-feeding of the ProA– strain by the ProC – strain at 30°C or 42°C, or between the wild-type bacteria and the mutant strains under any conditions, indicates that neither intermediates nor end products accumulate under normal growth conditions. The proline that is produced is rapidly used for protein synthesis and is prevented from being synthesized in excess of needs by careful regulation of the pathway, in this case through a feedback inhibition of proline acting directly on the ProB enzyme. proline Figure 2–53 The proline biosynthetic pathway (Answer 2–128). ...
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