32.7 How Do Neurotransmission Pathways Control the Function of Sensory Systems? 1053 been used illegally as a hallucinogenic drug, under the street name of angel dust. Sadly, it has caused many serious, long-term psychological problems in its users. ␥-Aminobutyric Acid and Glycine Are Inhibitory Neurotransmitters Certain neurotransmitters, acting through their conjugate postsynaptic receptors, inhibit the postsynaptic neuron from propagating nerve impulses from other neu- rons. Two such inhibitory neurotransmitters are ␥-aminobutyric acid (GABA) and glycine. These agents make postsynaptic membranes permeable to chloride ions and cause a net influx of Cl Ϫ , which in turn causes hyperpolarization of the post- synaptic membrane (making the membrane potential more negative). Hyperpolar- ization of a neuron effectively raises the threshold for the onset of action potentials in that neuron, making the neuron resistant to stimulation by excitatory neuro- transmitters. These effects are mediated by the GABA and glycine receptors, which are ligand-gated chloride channels (Figure 32.60). GABA is derived by a decar- boxylation of glutamate (Figure 32.61) and appears to operate mainly in the brain, whereas glycine acts primarily in the spinal cord. The glycine receptor has a specific affinity for the convulsive alkaloid strychnine (Figure 32.62). The effects of ethanol on the brain arise in part from the opening of GABA receptor Cl Ϫ channels. FIGURE 32.60 GABA (␥-aminobutyric acid) and glycine are inhibitory neurotransmitters that activate chloride channels. Influx of Cl Ϫ causes a hyperpolar- ization of the postsynaptic membrane. GABA recep- tors are similar in many respects to nicotinic acetyl- choline receptors, with an ␣ 2 ␥␦ stoichiometry. (a) Top view; (b) side view. (Images courtesy of Donald Weaver, University of Nova Scotia, and Valerie Campagna- Slater, University of Toronto.) (a) (b) Glutamate + H 3 NCCH 2 H CH 2 COO – COO – + H 3 NCH 2 CH 2 COO – CH 2 ␥-Aminobutyrate (GABA) CCH 2 CH 2 COO – O H Succinate semialdehyde – OOC CH 2 CH 2 COO – Succinate Glutamate decarboxylase GABA-glutamate transaminase Succinate semialdehyde dehydrogenase FIGURE 32.61 Glutamate is converted to GABA by glutamate decarboxylase. GABA is degraded by the action of GABA–glutamate transaminase and succinate semialdehyde dehydrogenase to produce succinate. N O H O N H H Strychnine FIGURE 32.62 Glycine receptors are distinguished by their unique affinity for strychnine. 1054 Chapter 32 The Reception and Transmission of Extracellular Information HUMAN BIOCHEMISTRY The Biochemistry of Neurological Disorders Defects in catecholamine processing are responsible for the symp- toms of many neurological disorders, including clinical depression (which involves norepinephrine [NE]) and parkinsonism (involv- ing dopamine [DA]). Once these neurotransmitters have bound to and elicited responses from postsynaptic membranes, they must be efficiently cleared from the synaptic cleft (see accompanying figure, part a). Clearing can occur by several mechanisms. NE and DA transport or reuptake proteins exist both in the presynaptic mem- brane and in nearby glial cell membranes. On the other hand, cat- echolamine neurotransmitters can be metabolized and inactivated by two enzymes: catechol-O-methyl-transferase in the synaptic cleft and monoamine oxidase in the mitochondria (see figure, part b). Catecholamines transported back into the presynaptic neuron are accumulated in synaptic vesicles by the same H ϩ -ATPase/H ϩ -ligand exchange mechanism described for glutamate. Clinical depression has been treated by two different strategies. Monoamine oxidase inhibitors act as antidepressants by increasing levels of cate- cholamines in the brain. Another class of antidepressants, the tri- cyclics, such as desipramine (see figure, part c), act on several classes of neurotransmitter reuptake transporters and facilitate more prolonged stimulation of postsynaptic receptors. Prozac is a more specific reuptake inhibitor and acts only on serotonin reup- take transporters. Parkinsonism is characterized by degeneration of dopaminergic neurons, as well as consequent overproduction of postsynaptic dopamine receptors. In recent years, Parkinson’s patients have been treated with dopamine agonists such as bromocriptine (see figure, part d) to counter the degeneration of dopamine neurons. Catecholaminergic neurons are involved in many other interest- ing pharmacological phenomena. For example, reserpine (see fig- ure, part e), an alkaloid from a climbing shrub of India, is a power- ful sedative that depletes the level of brain monoamines by inhibiting the H ϩ –monoamine exchange protein in the membranes of synaptic vesicles. Cocaine (see figure, part f), a highly addictive drug, binds with high affinity and specificity to reuptake transporters for the mono- amine neurotransmitters in presynaptic membranes. Thus, at least one of the pharmacological effects of cocaine is to prolong the synaptic effects of these neurotransmitters. Postsynaptic neuron Norepinephrine Norepi- nephrine receptor Receptor Mitochondrion H + Presynaptic neuron Monoamine oxidase Desipramine Tranylcypromine Reserpine (a) ᮣ (a) The pathway for reuptake and vesicular repackaging of the catecholamine neurotransmit- ters. The sites of action of desipramine, tranyl- cypromine, and reserpine are indicated. 32.7 How Do Neurotransmission Pathways Control the Function of Sensory Systems? 1055 Catechol- O-methyltransferase Monoamine oxidase HO C C H NH 3 + H H 3 CO H Methyl group donor OH CH 3 3-O-Methylepinephrine HO C C H NH 3 + H HO H OH Norepinephrine NH 4 + HO C C HO H OH 3,4-Dihydroxyphenylglycolaldehyde O H (b) NH 2 Tranylcypromine (c) N CH 2 CH 2 CH 2 NH CH 3 O CH CH 2 CF 3 CH 2 NH CH 3 HCl . Desipramine Prozac ® HN Br H H N CH 3 C O H N N O O HO CH(CH 3 ) 2 HCH 2 CH(CH 3 ) 2 O N H Bromocriptine (d) CH 3 O N H N H H H CH 3 O H 2 O C O OC OCH 3 O OCH 3 OCH 3 OCH 3 Reserpine (e) OCH 3 H N CH 3 Cocaine OC O C O (f) ᮡ (b) Norepinephrine can be degraded in the synaptic cleft by catechol-O-methyltransferase or in the mitochondria of presynaptic neurons by monoamine oxidase. (c) The structures of tranylcypromine, desipramine, and Prozac. (d) The structure of bromocriptine. (e) The struc- ture of reserpine. (f) The structure of cocaine. SUMMARY 32.1 What Are Hormones? Many different chemical species act as hor- mones. Steroid hormones, all derived from cholesterol, regulate metab- olism, salt and water balances, inflammatory processes, and sexual func- tion. Several hormones are amino acid derivatives. Among these are epinephrine and norepinephrine (which regulate smooth muscle con- traction and relaxation, blood pressure, cardiac rate, and the processes of lipolysis and glycogenolysis) and the thyroid hormones (which stimu- late metabolism). Peptide hormones are a large group of hormones that appear to regulate processes in all body tissues, including the release of yet other hormones. Hormones and other signal molecules in biological 1056 Chapter 32 The Reception and Transmission of Extracellular Information The Catecholamine Neurotransmitters Are Derived from Tyrosine Epinephrine, norepinephrine, dopamine, and L-dopa are collectively known as the catecholamine neurotransmitters. These compounds are synthesized from tyrosine (Figure 32.63), both in sympathetic neurons and in the adrenal glands. They func- tion as neurotransmitters in the brain and as hormones in the circulatory system. However, these two pools operate independently, thanks to the blood–brain barrier, which permits only very hydrophobic species in the circulatory system to cross over into the brain. Hydroxylation of tyrosine (by tyrosine hydroxylase) to form 3,4-dihydroxyphenylalanine ( L-dopa) is the rate-limiting step in this pathway. Dopa- mine, a crucial catecholamine involved in several neurological diseases, is synthe- sized from L-dopa by a pyridoxal phosphate-dependent enzyme, dopa decarboxy- lase. Subsequent hydroxylation and methylation produce norepinephrine and epinephrine (Figure 32.63). The methyl group in the final reaction is supplied by S-adenosylmethionine. Each of these catecholamine neurotransmitters is known to play a unique role in synaptic transmission. The neurotransmitter in junctions between sympathetic nerves and smooth muscle is norepinephrine. On the other hand, dopamine is in- volved in other processes. Either excessive brain production of dopamine or hy- persensitivity of dopamine receptors is responsible for psychotic symptoms and schizophrenia, whereas lowered production of dopamine and the loss of dopamine neurons are important factors in Parkinson’s disease. Various Peptides Also Act as Neurotransmitters Many relatively small peptides have been shown to possess neurotransmitter activity (see Table 32.3). One of the challenges of this field is that the known neuropeptides may represent a very small subset of the neuropeptides that exist. Another chal- lenge arises from the small in vivo concentrations of these agents and the small number of receptors that are present in neural tissue. Physiological roles for most of these peptides are complex. For example, the endorphins and enkephalins are natural opioid substances and potent pain relievers. The endothelins are a family of homologous regulatory peptides, synthesized by certain endothelial and epithelial cells that act on nearby smooth muscle and connective tissue cells. They induce or affect smooth muscle contraction; vasoconstriction; heart, lung, and kidney func- tion; and mitogenesis and tissue remodeling. Vasoactive intestinal peptide (VIP) produces a G protein–adenylyl cyclase–mediated increase in cAMP, which in turn triggers a variety of protein phosphorylation cascades, one of which leads to con- version of phosphorylase b to phosphorylase a, stimulating glycogenolysis. More- over, VIP has synergistic effects with other neurotransmitters, such as norepineph- rine. In addition to increasing cAMP levels through -adrenergic receptors, norepinephrine acting at ␣ 1 -adrenergic receptors markedly stimulates the increases in cAMP elicited by VIP. Many other effects have also been observed. For example, injection of VIP increases rapid eye movement (REM) sleep and decreases waking time in rats. VIP receptors exist in regions of the central nervous system involved in sleep modulation. Tyrosine HO CH 2 C COO – NH 3 + H Dopa HO CH 2 O 2 CO 2 C COO – NH 3 + H HO Dopamine HO CH 2 CH NH 3 + H HO HO C C H NH 3 + H HO H OH CH 3 HO C C H + H 2 N H HO H Methyl group donor OH CH 3 Epinephrine (Adrenaline) Norepinephrine (Noradrenaline) ᮤ FIGURE 32.63 The pathway for the synthesis of catecholamine neurotransmitters.Dopa, dopamine, noradren- aline, and adrenaline are synthesized sequentially from tyrosine. Problems 1057 PROBLEMS Preparing for an exam? Create your own study path for this chapter at www.cengage.com/login. 1. Compare and contrast the features and physiological advantages of each of the major classes of hormones, including the steroid hormones, polypeptide hormones, and the amino acid–derived hormones. 2. Compare and contrast the features and physiological advantages of each of the known classes of second messengers. 3. Nitric oxide may be merely the first of a new class of gaseous second messenger/neurotransmitter molecules. Based on your knowledge of the molecular action of nitric oxide, suggest another gaseous mol- ecule that might act as a second messenger and propose a molecular function for it. 4. Herbimycin A is an antibiotic that inhibits tyrosine kinase activity by binding to SH groups of cysteine in the src gene tyrosine kinase and other similar tyrosine kinases. What effect might it have on normal rat kidney cells that have been transformed by Rous sarcoma virus? Can you think of other effects you might expect for this interesting antibiotic? 5. Monoclonal antibodies that recognize phosphotyrosine are com- mercially available. How could such an antibody be used in studies of cell signaling pathways and mechanisms? 6. Explain and comment on this statement: The main function of hor- mone receptors is that of signal amplification. 7. Synaptic vesicles are approximately 40 nm in outside diameter, and each vesicle contains about 10,000 acetylcholine molecules. Calcu- late the concentration of acetylcholine in a synaptic vesicle. 8. GTP␥S is a nonhydrolyzable analog of GTP. Experiments with squid giant axon synapses reveal that injection of GT␥S into the pre- synaptic end (terminal) of the neuron inhibits neurotransmitter re- lease (slowly and irreversibly). The calcium signals produced by presynaptic action potentials and the number of synaptic vesicles docking on the presynaptic membrane are unchanged by GTP␥S. Propose a model for neurotransmitter release that accounts for all of these observations. 9. A typical hormone binds to its receptor with an affinity (K D ) of ap- proximately 1 ϫ 10 Ϫ9 M. Consider an in vitro (test-tube) system in which the total hormone concentration is approximately 1 nM and the total concentration of receptor sites is 0.1 nM. What fraction of the receptor sites is bound with hormone? If the concentration of receptors is decreased to 0.033 nM, what fraction of receptor is bound with the hormone? 10. (Integrates with Chapter 24.) All steroid hormones are synthesized in the human body from cholesterol. What is the consequence for steroid hormones and their action from taking a “statin” drug, such as Zocor, which blocks the synthesis of cholesterol in the body? Are steroid hormone functions compromised by statin action? 11. Given that -strands provide a more genetically economical way for a polypeptide to cross a membrane, why has nature chosen ␣-helices as the membrane-spanning motif for G-protein–coupled receptors? That is, what other reasons can you imagine to justify the use of ␣-helices? 12. Write simple reaction mechanisms for the formation of cAMP from ATP by adenylyl cyclase and for the breakdown of cAMP to 5Ј-AMP by phosphodiesterases. 13. (Integrates with Chapter 9.) Consider the data in Figure 32.49a. Recast Equation 9.2 to derive a form from which you could cal- culate the equilibrium electrochemical potential at which no net systems bind with very high affinities to their receptors, displaying K D val- ues in the range of 10 Ϫ12 to 10 Ϫ6 M. Hormones are produced at concen- trations equivalent to or slightly above these K D values. Once hormonal effects have been induced, the hormone is usually rapidly metabolized. 32.2 What Is Signal Transduction? Hormonal regulation depends upon the transduction of the hormonal signal across the plasma mem- brane to specific intracellular sites, particularly the nucleus. Signal trans- duction pathways consist of a stepwise progression of signaling stages: receptor⎯→transducer⎯→effector. The receptor perceives the signal, trans- ducers relay the signal, and the effectors convert the signal into an intra- cellular response. Often, effector action involves a series of steps, each of which is mediated by an enzyme, and each of these enzymes can be con- sidered as an amplifier in a pathway connecting the hormonal signal to its intracellular targets. 32.3 How Do Signal-Transducing Receptors Respond to the Hormonal Message? Steroid hormones may either bind to plasma membrane receptors or exert their effects within target cells, entering the cell and migrating to their sites of action via specific cytoplasmic receptor pro- teins. The nonsteroid hormones, which act by binding to outward- facing plasma membrane receptors, activate signal transduction path- ways that mobilize various second messengers—cyclic nucleotides, Ca 2ϩ ions, and other substances—that activate or inhibit enzymes or cas- cades of enzymes in very specific ways. 32.4 How Are Receptor Signals Transduced? Receptor signals are transduced in one of three ways, to initiate actions inside the cell: 1. Exchange of GDP for GTP on GTP-binding proteins (G proteins), which in turn leads to generation of second messengers, including cAMP, phospholipid breakdown products, and Ca 2ϩ . 2. Receptor-mediated activation of phosphorylation cascades that in turn trigger activation of various enzymes. 3. Conformation changes that open ion channels or recruit proteins into nuclear transcription complexes. 32.5 How Do Effectors Convert the Signals to Actions in the Cell? Transduction of the hormonal signal leads to activation of effectors— usually protein kinases and protein phosphatases—that elicit a variety of actions that regulate discrete cellular functions. Of the thousands of mammalian kinases and phosphatases, the structures and functions of a few are representative, including protein kinase A (PKA), protein kinase C (PKC), and protein tyrosine phosphatase SHP-2. 32.6 How Are Signaling Pathways Organized and Integrated? All sig- naling pathways are organized in time and space in the cell, they are care- fully regulated, and they are integrated with one another. PIDs modulate and control the association of signaling molecules with one another, of- ten in large signalsomes; signaling molecules are switched on and off by covalent modifications such as phosphorylations; and signaling often in- volves amplification and cooperative effects. GPCR signaling can occur through G-protein-independent pathways and is sometimes modulated by RGS/GAPs. Responses of signaling receptors can be coordinated by trans- activation, and signals from multiple pathways can be integrated. 32.7 How Do Neurotransmission Pathways Control the Function of Sensory Systems? Nerve impulses, which can be propagated at speeds up to 100 m/sec, provide a means of intercellular signaling that is fast enough to encompass sensory recognition, movement, and other phys- iological functions and behaviors in higher animals. The generation and transmission of nerve impulses in vertebrates is mediated by an in- credibly complicated neural network that connects every part of the or- g anism with the brain—itself an interconnected array of as many as 10 12 cells. Despite their complexity and diversity, the nervous systems of ani- mals all possess common features and common mechanisms. Physical or chemical stimuli are recognized by specialized receptor proteins in the membranes of excitable cells. Conformational changes in the receptor protein result in a change in enzyme activity or a change in the perme- ability of the membrane. These changes are then propagated through- out the cell or from cell to cell in specific and reversible ways to carry in- formation through the organism. 1058 Chapter 32 The Reception and Transmission of Extracellular Information flow of potassium would occur. This is the Nernst equation. Cal- culate the equilibrium potential for K ϩ and also for Na ϩ , assum- ing T ϭ 37°C. 14. The calculation of the actual transmembrane potential difference for a neuron is accomplished with the Goldman equation: ⌬ ϭ ᎏ R Ᏺ T ᎏ ln where [C] and [A] are the cation and anion concentrations, re- spectively, and P C and P A are the respective permeability coefficients of cations and anions. Assume relative permeabilities for K ϩ , Na ϩ , and Cl Ϫ of 1, 0.04, and 0.45, respectively, and use this equation to calculate the actual transmembrane potential difference for the neuron whose ionic concentrations are those given in Figure 32.49a. 15. Use the information in problems 13 and 14, together with Fig- ure 32.50, to discuss the behavior of potassium, sodium, and chloride ions as an action potential propagates along an axon. 16. Review the cell signaling pathway shown in Figure 32.4. With the rest of the chapter as context, discuss all the steps of this pathway that involve signal amplification. 17. In the cell signaling pathway shown in Figure 32.4, what would be the effect if Ras were mutated so that it had no GTPase activity? 18. One of the topics discussed in this chapter is the ability of GPCRs to exert signaling effects without the involvement of G proteins. Using ⌺P C [C] outside ϩ⌺P A [A] inside ᎏᎏᎏ ⌺P C [C] inside ϩ⌺P A [A] outside the pathway shown in Figure 32.40, and considering everything you have learned in this chapter, suggest some reasons that would ex- plain why this G-protein–independent signaling was difficult to ver- ify experimentally. Preparing for the MCAT Exam 19. Malathion (Figure 32.58) is one of the secrets behind the near- complete eradication of the boll weevil from cotton fields in the United States. For most of the 20th century, boll weevils wreaked havoc on the economy of states from Texas to the Carolinas. When boll weevils attacked cotton fields in a farming community, the de- struction of cotton plants meant loss of jobs for farm workers, bankruptcies for farm owners, and resulting hardship for the en- tire community. Relentless application of malathion to cotton crops and fields has turned the tide, however, and agriculture ex- perts expect that boll weevils will be completely gone from cotton fields within a few years. Remarkably, malathion-resistant boll wee- vils have not emerged despite years of this pesticide’s use. Consider the structure and chemistry of malathion and suggest what you would expect to be the ecological consequences of chronic malathion application to cotton fields. 20. Consult the excellent review article “Assembly of Cell Regulatory Sys- tems Through Protein Interaction Domains” (Science 300:445–452, 2003, by Pawson and Nash) and discuss the structural requirements for a regulatory protein operating in a signaling network. FURTHER READING Signal Transduction and Signaling Pathways Delcourt, N., Bockaert, J., et al., 2007. GPCR-jacking: From a new route in RTK signaling to a new concept in GPCR activation. Trends in Pharmacological Sciences 28:602–607. Ferguson, S. S. 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Refined structure of the nicotinic acetylcholine recep- tor at 4 Å resolution. Journal of Molecular Biology 346:967–989. Webb, T. I., and Lynch, J. L., 2007. Molecular pharmacology of the glycine receptor chloride channel. Current Pharmacological Design 13: 2350–2367. This page intentionally left blank 20 m on a side. Maximal information content of liver-cell DNA ϭ 3 ϫ 10 9 bp, which, expressed in proteins 400 amino acids in length, could encode 2.5 ϫ 10 6 proteins. 6. The amino acid side chains of proteins provide a range of shapes, polarity, and chemical features that allow a protein to be tailored to fit almost any possible molecular surface in a comple- mentary way. 7. Biopolymers may be informational molecules because they are constructed of different monomeric units (“letters”) joined head to tail in a particular order (“words, sentences”). Polysac- charides are often linear polymers composed of only one (or two repeating) monosaccharide unit(s) and thus display little information content. Polysaccharides with a variety of monosac- charide units may convey information through specific recogni- tion by other biomolecules. Also, most monosaccharide units are typically capable of forming branched polysaccharide struc- tures that are potentially very rich in information content (as in cell surface molecules that act as the unique labels displayed by different cell types in multicellular organisms). 8. Molecular recognition is based on structural complementarity. If complementary interactions involved covalent bonds (strong forces), stable structures would be formed that would be less re- sponsive to the continually changing dynamic interactions that characterize living processes. 9. Two carbon atoms interacting through van der Waals forces are 0.34 nm apart; two carbon atoms joined in a covalent bond are 0.154 nm apart. 10. Slight changes in temperature, pH, ionic concentrations, and so forth may be sufficient to disrupt weak forces (H bonds, ionic bonds, van der Waals interactions, hydrophobic interactions). 11. Living systems are maintained by a continuous flow of matter and energy through them. Despite the ongoing transformations of matter and energy by these highly organized, dynamic systems, no overt changes seem to occur in them: They are in a steady state. 12. The fraction of the M. genitalium genes encoding proteins ϭ 0.925. Genes not encoding proteins encode RNA molecules. (0.925)(580,074 base pairs) ϭ 536,820 base pairs devoted to protein-coding genes. Since 3 base pairs specify an amino acid in a protein, 369 amino acids are found in the average M. genitalium protein. If each amino acid contributes on average 120 D to the mass of a protein, the mass of an average M. genitalium protein is 44,280 D. 13. (0.925)(206) ϭ 191 proteins. Assuming its genes are the same size as M. genitalium, the minimal genome would be 228,480 base pairs. For detailed answers to the end-of-chapter problems as well as addi- tional problems to solve, see The Student Solutions Manual, Study Guide and Problems Book by David Jemiolo and Steven Theg that accompanies this textbook. Chapter 1 1. Because bacteria (compared with humans) have simple nutri- tional requirements, their cells obviously contain enzyme systems that allow them to convert rudimentary precursors (even inorganic substances such as NH 4 ϩ , NO 3 Ϫ , N 2 , and CO 2 ) into complex biomolecules—proteins, nucleic acids, polysaccha- rides, and complex lipids. On the other hand, animals have an assortment of different cell types designed for specific physiolog- ical functions; these cells possess a correspondingly greater repertoire of complex biomolecules to accomplish their intri- cate physiology. 2. Consult Figures 1.20 to 1.22 to confirm your answer. 3. a. Laid end to end, 250 E. coli cells would span the head of a pin. b. The volume of an E. coli cell is about 10 Ϫ15 L. c. The surface area of an E. coli cell is about 6.3 ϫ 10 Ϫ12 m 2 . Its surface-to-volume ratio is 6.3 ϫ 10 6 m Ϫ1 . d. 600,000 molecules. e. 1.7 nM. f. Because we can calculate the volume of one ribosome to be 4.2 ϫ 10 Ϫ24 m 3 (or 4.2 ϫ 10 Ϫ21 L), 15,000 ribosomes would occupy 6.3 ϫ 10 Ϫ17 L, or 6.3% of the total cell volume. g. Because the E. coli chromosome contains 4600 kilobase pairs (4.6 ϫ 10 6 bp) of DNA, its total length would be 1.6 mm— approximately 800 times the length of an E. coli cell. This DNA would encode 4300 different proteins, each 360 amino acids long. 4. a. The volume of a single mitochondrion is about 4.2 ϫ 10 Ϫ16 L (about 40% the volume calculated for an E. coli cell in problem 3). b. A mitochondrion would contain on average fewer than eight molecules of oxaloacetate. 5. a. Laid end to end, 25 liver cells would span the head of a pin. b. The volume of a liver cell is about 8 ϫ 10 Ϫ12 L (8000 times the volume of an E. coli cell). c. The surface area of a liver cell is 2.4 ϫ 10 Ϫ9 m 2 ; its surface-to- volume ratio is 3 ϫ 10 5 m Ϫ1 , or about 0.05 (1/20) that of an E. coli cell. Cells with lower surface-to-volume ratios are limited in their exchange of materials with the environment. d. The number of base pairs in the DNA of a liver cell is 6 ϫ 10 9 bp, which would amount to a total DNA length of 2 m (or 6 feet of DNA!) contained within a cell that is only Abbreviated Answers to Problems A-2 Abbreviated Answers to Problems 14. Given 1109 nucleotides (or base pairs) per gene, the minimal virus, with a 3500-nucleotide genome, would have only 3 genes; the maximal virus, with a 280,000-bp genome, would have 252 genes. 15. Fate of proteins synthesized by the rough ER: a. Membrane proteins would enter the ER membrane, and, as part of the membrane, be passed to the Golgi, from which vesicles depart and fuse with the plasma membrane. b. A secreted protein would enter the ER lumen and be trans- ferred as a luminal protein to the Golgi, from which vesicles depart. When the vesicle fuses with the plasma membrane, the protein would be deposited outside the cell. 16. Increasing kinetic energy increases the motions of molecules and raises their average energy, which means that the difference between the energy to disrupt a weak force between two mole- cules and the energy of the weak force is smaller. Thus, in- creases in kinetic energy may break the weak forces between molecules. 17. Informational polymers must have “sense” or direction, and they must be composed of more than one kind of monomer unit. Chapter 2 1. a. 3.3; b. 9.85; c. 5.7; d. 12.5; e. 4.4; f. 6.97. 2. a. 1.26 mM; b. 0.25 mM; c. 4 ϫ 10 Ϫ12 M; d. 2 ϫ 10 Ϫ4 M; e. 3.16 ϫ 10 Ϫ10 M; f. 1.26 ϫ 10 Ϫ7 M (0.126 M). 3. a. [H ϩ ] ϭ 2.51 ϫ 10 Ϫ5 M; b. K a ϭ 3.13 ϫ 10 Ϫ8 M; pK a ϭ 7.5. 4. a. pH ϭ 2.38; b. pH ϭ 4.23. 5. Combine 187 mL of 0.1 M acetic acid with 813 mL of 0.1 M sodium acetate. 6. [HPO 4 Ϫ2 ]/[H 2 PO 4 Ϫ ] ϭ 0.398. 7. Combine 555.7 mL of 0.1 M Na 3 PO 4 with 444.3 mL of 0.1 M H 3 PO 4 . Final concentrations of ions will be [H 2 PO 4 Ϫ ] ϭ 0.0333 M; [HPO 4 2Ϫ ] ϭ 0.0667 M; [Na ϩ ] ϭ 0.1667 M; [H ϩ ] ϭ 3.16 ϫ 10 Ϫ8 M. 8. Add 432 mL of 0.1 N HCl to 1 L 0.05 M BICINE. [BICINE] total ϭ 0.05 M/1.432 L ϭ 0.0349 M [protonated form] ϭ 0.0302 M. 9. a. Fraction of H 3 PO 4 : @pH 0 ϭ 0.993; @pH 2 ϭ 0.58; @pH 4 ϭ 0.01; negligible @pH 6. b. Fraction of H 2 PO 4 Ϫ : @pH 0 ϭ 0.007; @pH 2 ϭ 0.41; @pH 4 ϭ 0.986; @pH 6 ϭ 0.94; @pH 8 ϭ 0.14; negligible @pH 10. c. Fraction of HPO 4 2Ϫ : negligible @pH 0, 2, and 4; @pH 6 ϭ 0.06; @pH 8 ϭ 0.86; @pH 10 Ϸ 1.0; @pH 12 ϭ 0.72. d. Fraction of PO 4 3Ϫ : negligible at any pH Ͻ10; @pH 12 ϭ 0.28. 10. At pH 5.2, [H 3 A] ϭ 4.33 ϫ 10 Ϫ5 M; [H 2 A Ϫ ] ϭ 0.0051 M; [HA 2Ϫ ] ϭ 0.014 M; [A 3Ϫ ] ϭ 0.0009 M. 11. a. pH ϭ 7.02; [H 2 PO 4 Ϫ ] ϭ 0.0200 M; [HPO 4 2Ϫ ] ϭ 0.0133 M. b. pH ϭ 7.38; [H 2 PO 4 Ϫ ] ϭ 0.0133 M; [HPO 4 2Ϫ ] ϭ 0.0200 M. 12. [H 2 CO 3 ] ϭ 2.2 M; [CO 2(d) ] ϭ 0.75 mM. When [HCO 3 Ϫ ] ϭ 15 mM and [CO 2(d) ] ϭ 3 mM, pH ϭ 6.8. 13. Titration of the fully protonated form of anserine will require the addition of three equivalents of OH Ϫ . The pK a values lie at 2.64 (COOH); 7.04 (imidazole-N ϩ H); and 9.49 (NH 3 ϩ ). Its iso- electric point lies midway between pK 2 and pK 3 , so pH I ϭ 8.265. To prepare 1 L of 0.04 M anserine buffer, add 164 mL of 0.1 M HCl to 400 mL of 0.1 M anserine at its isoelectric point, and make up to 1 L final volume. 14. Add 410 mL of 0.1 M NaOH to 250 mL of 0.1 M HEPES in its fully protonated form and make up to 1 L final volume. 15. 166.7 g/mole. 16. Add 193 mL of water and 307 mL of 0.1 M HCl to 500 mL of 0.1 M triethanolamine. 17. Combine 200 mL of 0.1 M Tris-H ϩ with 732 mL water and 68 mL of 0.1 M NaOH. 18. a. b. c. The relevant pK a for this calculation is 8.3. Combine 400 mL of 0.1 M bicine at its pH I (pH 5.3) with 155 mL of 0.1 M NaOH and 345 mL of water. d. The relevant pK a for this calculation is 2.3. The concentra- tion of fully protonated form of bicine at pH 7.5 is 2.18 ϫ 10 Ϫ7 M. 19. 0.063 M 20. 5.2 mM 21. 22. 0.0153 nmol/mL и sec 23. A drop in blood pH would occur. 24. c. Chapter 3 1. K eq ϭ 613 M; ⌬G° ϭϪ15.9 kJ/mol. 2. ⌬G° ϭ 1.69 kJ/mol at 20°C; ⌬G° ϭϪ5.80 kJ/mol at 30°C. ⌬S° ϭ 0.75 kJ/mol и K. 3. ⌬G ϭ Ϫ24.8 kJ/mol. 4. State functions are quantities that depend on the state of the system and not on the path or process taken to reach that state. Volume, pressure, and temperature are state functions. Heat and all forms of work, such as mechanical work and electrical work, are not state functions. 5. ⌬G°Ј ϭ⌬G° Ϫ 39.5 n (in kJ/mol), where n is the number of H ϩ produced in any process. So ⌬G° ϭ⌬G°Јϩ39.5 n ϭ Ϫ30.5 kJ/mol ϩ 39.5(1) kJ/mol. ⌬G° ϭ 9.0 kJ/mol at 1 M [H ϩ ]. 0 2 0.2 0.4 0.6 0.8 1.0 4 6 8 pH equivalents OH – pK a = 4.7 NCH 2 COO – CH 2 CH 2 OH CH 2 CH 2 OH 0 2 0.5 1.0 1.5 2.0 4 6 8 10 12 pH equivalents OH – pK a = 2.3 pK a = 8.3