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APPENDIX A

Answers to Review Quest ions

Chapter 1

1. The answer is C. In a steady state, the amount or con-centration of a substance in a compartment does notchange with respect to time. Although there may beconsiderable movements into and out of the compart-ment, there is no net gain or loss. Steady states in thebody often do not represent an equilibrium condition,but they are displaced from equilibrium by the con-stant expenditure of metabolic energy.

2. The answer is D. The increase in plasma insulin lowersthe plasma glucose concentration back to normal andis an example of negative feedback. Negative feedbackopposes change and results in stability. Positive feed-back would produce a further increase in plasma glu-cose concentration. Chemical equilibrium indicates acondition in which the rates of reactions in forwardand backward directions are equal. End-product inhi-bition occurs when the products of a chemical reactionslow the reaction (for example, by inhibiting an en-zyme) that produces them. Feedforward control in-volves a command signal and does not directly sensethe regulated variable (plasma glucose concentration).

3. The answer is E. The EGF receptor is a tryosine kinasereceptor; therefore, an inhibitor of tyrosine kinasesshould have the desired effect. An adenylyl cyclasestimulator or phosphodiesterase inhibitor would in-crease intracellular levels of cAMP, but this is not thesecond messenger problem here. An EGF agonistwould increase signaling along the EGF pathway andwould increase the problem, causing an undesired ef-fect. Likewise, a phosphatase inhibitor would slow thehydrolysis of phosphorylated intermediates and main-tain the activated state of the EGF pathway.

4. The answer is D. Second messengers are a class of sig-naling molecules generated inside cells in response tothe activation of a receptor. If second messengers werealways available, signal transduction pathways couldnot be regulated. The response to a second messengervaries depending on the cell type because each celltype differs with respect to the number and comple-ment of receptors, effectors, and downstream targets.Second messengers include nucleotides such as cAMPor cGMP, ions such as calcium, and gases such as nitricoxide. Many different plasma membrane receptors, notonly tyrosine kinase receptors, are coupled to secondmessenger generating systems.

5. The answer is B. Cyclic nucleotides are generated by

the action of either adenylyl or guanylyl cyclase onATP or GTP, respectively. Cyclic AMP and cGMP ac-tivate distinct signaling pathways. For example, cAMPcan activate protein kinase A, which will phosphory-late its substrates; cGMP activates protein kinase G,which phosphorylates a different set of substrates. Al-though signal transduction in sensory tissues involvesboth cAMP and cGMP, cGMP has a more importantrole in signal transduction than cAMP. PhospholipaseC activation is coupled to the activation of a G protein(Gq), not to cAMP or cGMP.

6. The answer is D. Steroid hormone receptors are tran-scriptional regulators found in the cytoplasm or in thenucleus. These receptors are activated by the bindingof steroid ligands that diffuse through lipid bilayersand enter the cytosol. Activated steroid receptors me-diate their effects by direct interaction with gene reg-ulatory elements and do not activate G proteins orcause binding of IP3 to the IP3-gated calcium releasechannel in the sarcoplasmic reticulum. Steroid hor-mone receptors do not have tyrosine kinase activityand do not cause the phosphorylation of tyrosineresidues in these receptors. Steroid hormone receptorsare not linked to activation of the MAP kinase path-way. Estrogen receptors are located in the cytoplasmof cells; upon binding of estrogen, they move to thenucleus to bind to estrogen response elements to acti-vate gene transcription.

7. The answer is E. Cardiac muscle cells have many gapjunctions that allow the rapid transmission of electricalactivity and the coordination of heart muscle contrac-tion. Gap junctions are pores composed of paired con-nexons that allow the passage of ions, nucleotides, andother small molecules between cells.

8. The answer is E. Inositol trisphosphate (IP3) and dia-cylglycerol (DAG) are generated by the action ofphospholipase C (PLC) on PIP2, phosphatidylinositol4,5-bisphosphate. IP3 and DAG are second messen-gers, not first messengers. DAG is important for the ac-tivation of protein kinase C, not PLC. Tyrosine kinasereceptors are activated by the binding of ligands, suchas hormones or growth factors, not by IP3 or DAG. IP3

can indirectly activate calcium-calmodulin-dependentprotein kinases by causing the release of calcium fromintracellular stores; DAG has no direct effect on thesekinases.

9. The answer is C. The activation of tyrosine kinase re-ceptors often results in a cellular response that is in-volved in growth or differentiation. Tyrosine kinase re-ceptors do not have constitutively active receptors; if

708 APPENDICES

this were true, there could be no regulation of signal-ing. The activation of ras occurs indirectly by the acti-vation of adapter molecules (Grb2 and SOS) that asso-ciate with phosphorylated tyrosine residues in thecytoplasmic tail of the receptor. The activation of ty-rosine kinase receptors usually involves multimeriza-tion into dimers or trimers.

10. The answer is B. If it is unable to hydrolyze GTP, theG�s subunit remains in its active form and results in in-creases in adenylyl cyclase activity, intracellularcAMP, and release of growth hormone. If G�i is acti-vated, adenylyl cyclase activity will be decreased. Alack of GHRH receptors should produce decreased,not increased, growth hormone secretion.

Chapter 2

1. The answer is D. Phospholipids have both a polar hy-drophilic head group and a hydrophobic region be-cause of the long hydrocarbon chains of the two fattyacids. Since the hydrophilic and hydrophobic regionsare present within the same molecule, phospholipidsare described as amphipathic. Phospholipids are notsoluble in water and do not have a steroid structure.

2. The answer is B. Phospholipid molecules can rotateand move laterally within the plane of the lipid bilayer,but movement from one half of the bilayer to the otheris slow because it is an energetically unfavorableprocess. Cholesterol is an example of a separate class oflipids that do not contain fatty acids. Many phospho-lipids are distributed unequally between the two halvesof the lipid bilayer. In the red blood cell membrane, forexample, most of the phosphatidylcholine is in theouter half. Both ion channels and symporters are mem-brane proteins, not phospholipids.

3. The answer is A. Membrane-spanning segments of in-tegral proteins frequently adopt an �-helical confor-mation because this structure maximizes the opportu-nities for the polar peptide bonds to form hydrogenbonds with one another in the hydrophobic interior ofthe lipid bilayer. These segments are composed largelyof amino acids with nonpolar hydrophobic side chainsthat interact with the surrounding lipids. There are nocovalent bonds with cholesterol or phospholipids, andthe peptide bonds are not unusually strong.

4. The answer is D. The Nernst equation calculates themembrane potential that develops when a single ion isdistributed at equilibrium across a membrane. TheGoldman equation gives the value of the membranepotential when all permeable ions are accounted for.The van’t Hoff equation calculates the osmotic pres-sure of a solution, and Fick’s law refers to the diffu-sional movement of solute. The permeability coeffi-cient accounts for several factors that determine theease with which a solute can cross a membrane.

5. The answer is B. Intracellular K� is high comparedwith all other intracellular ions.

6. The answer is C. Active transport always moves soluteagainst its electrochemical gradient. All the other op-tions are shared by both active transport and equili-brating carrier-mediated transport systems.

7. The answer is A. Cyclic GMP is the ligand that opensthe ion channel in this example. Ion pumps and Na�

solute-coupled transporters are examples of activetransport systems, not Na� channels. The process isunrelated to receptor-mediated endocytosis.

8. The answer is C. K� is the major intracellular ion; ef-flux of K� will produce an osmotic flow of water out ofthe cell. Water exit will lead to a decrease in cell vol-ume. An influx of Na� and synthesis of sorbitol do notoccur during this process because both processeswould increase intracellular osmolytes and drive waterinto the cell, increasing cell volume.

9. The answer is D. By substituting in the Nernst equa-tion (equation 7):

Ei � Eo � 61/–1 � log10 120/8� �61 � 1.176� �71.7 mV inside the cell.

Note that for Cl�, the value for z (valence) is �1.

10. The answer is E. By using the van’t Hoff equation:

� � n R T �? C � 3 � 0.0821 � 300 � 0.86 � 0.1� 6.35 atm

Chapter 3

1. The answer is B. Potassium ion concentration is highin the intracellular fluid relative to the extracellularfluid. The opposite is true for sodium ion concentra-tion; therefore, there is a strong driving force for potas-sium to leave the cell and sodium to enter the cell. Ifthe permeability to K� increases, more potassiumwould leave the cell, and the cell would become morenegative (hyperpolarize). If the permeability to Na�

decreases, less sodium would enter the cell, and the cellwould become less positive (hyperpolarize).

2. The answer is C. Voltage-gated potassium channelsopen with a delay relative to voltage-gated sodiumchannels in response to depolarization. Concomi-tantly, there is a delay in their closing relative to thesodium channels. During the afterhyperpolarizationphase of the action potential, the sodium channels areclosed, but the potassium channels remain open. Be-cause there is a strong driving force for K� to leave thecell, the cell hyperpolarizes. An outward calcium cur-rent, an outward sodium current, or an inward chloridecurrent could conceivably hyperpolarize the cell, butthese currents are not the basis for this phase of the ac-tion potential.

3. The answer is D. A specialization occurs in myelinatedaxons in which the voltage-gated sodium channels arepreferentially distributed to the axonal membrane be-neath the nodes of Ranvier. Since these channels arerequired for the generation of an action potential, theaction potential jumps from node to node. Thisprocess is facilitated by an increased membrane resist-ance and a decreased capacitance associated with themyelinated regions of the axon, both of which pro-mote the electrotonic spread of the positive chargethat accumulates beneath one node of Ranvier at the

peak of the action potential. Nongated ion channelsare not involved in the generation of action potential.

4. The answer is C. Myelin contributes substantially tothe effective membrane resistance, Rm. The space con-stant increases as Rm increases because it is more diffi-cult for ions to flow across the membrane relative tothe ease with which they flow within the axon. Whenan axon demyelinates, its space constant decreases andconduction velocity is slowed. This slowing of theconduction velocity is the basis for the neurologicaldeficits associated with multiple sclerosis.

5. The answer is C. SNARES are the group of proteins re-sponsible for docking and binding synaptic vesicles tothe presynaptic membrane to prepare them for release.If the vesicles cannot dock, they cannot fuse with themembrane to release their neurotransmitter. Disrup-tion of SNARES has no direct effect on other compo-nents of neurochemical transmission, including actionpotential propagation, transmitter-receptor interac-tion, or uptake mechanisms.

6. The answer is A. Spatial summation of synaptic poten-tials can occur if they are close enough that the spaceconstant spans the two synapses; therefore, propertiesof the cell that increase the space constant would opti-mize the effectiveness of the two synapses. The spaceconstant increases with increasing membrane resist-ance or decreasing cytoplasmic resistance. Cytoplas-mic resistance decreases as the cross-sectional area in-creases. Temporal summation could also increase theeffectiveness of the two synapses; this would be facili-tated by a large time constant.

7. The answer is C. Acetylcholinesterase is the enzymethat breaks acetylcholine down into acetate andcholine. The acetate diffuses away, and choline istaken back up into the presynaptic nerve terminal forthe synthesis of more ACh. Blocking the function ofacetylcholinesterase would prevent the breakdown ofACh, which would accumulate in the cleft becausethere is no uptake mechanism for ACh and it diffusesaway more slowly than acetate.

8. The answer is C. Catecholaminergic transmission is ef-ficient, in part, because there is a significant reuptakeof the catecholamines for repackaging into synapticvesicles to use again. MAO and COMT are not foundin the cleft and do not aid in the removal of the cate-cholamines from the cleft. The postsynaptic cell mayhave an uptake mechanism (not endocytosis) for thecatecholamines, but the efficacy of this mechanism issubstantially lower than the reuptake mechanism.

9. The answer is A. Dopamine plays a major role in twofunctional systems of the brain, the motor system andthe limbic system. Within the limbic system, DA is as-sociated with affect. Too much dopaminergic trans-mission can result in psychotic disorders, such as schiz-ophrenia. A blockade of dopaminergic transmissionameliorates psychosis. Cholinergic transmission is in-volved in cognitive function and motor control. Therole of nitrergic transmission in cognition and behav-ior is unknown.

10. The answer is D. Most neurotransmitters are synthe-sized locally within the axon terminals from precursors

available at the terminals, using enzymes that reside inthe terminals. However, peptides must be synthesizedby ribosomes, which are not found in axons or termi-nals. The supply of peptide transmitters in the axonterminal must be continuously replenished via axoplas-mic transport from the cell body. Microtubules are anessential component of axoplasmic transport; disrupt-ing their integrity would diminish axonal transport anddeplete the peptide transmitter from the terminal.

11. The answer is B. GABA is the major inhibitory trans-mitter in the brain. The activation of GABA receptorshyperpolarizes neurons. Activity of the GABA systemis widespread in the brain, and a disruption of GABAsignaling results in a hyperexcitability of neurons thatcan lead to seizures.

12. The answer is B. The acute onset of symptoms in bothpeople suggests food poisoning and not a chronic dis-order or a stroke. A toxin that blocked nerve-muscletransmission would produce muscle paralysis or weak-ness and no sensory disturbances. The tingling feelingsuggests abnormally high excitability and firing of sen-sory nerves. Ciguatera toxin, the product of a dinofla-gellate that sometimes contaminates red snapper andother reef fishes, is probably the cause of the sensoryabnormality and gastrointestinal symptoms. Ciguateratoxin binds to voltage-gated sodium channels and re-sults in their persistent activation.

13. The answer is C. Loss of myelin will result in a lowerconduction velocity because the action potential willno longer “jump” from node to node. The compoundaction potential (the sum of many individual action po-tentials) will be more spread out and will have a slowerrate of rise than normal. The afterhyperpolarizationwill last longer.

14. The answer is C. Release of transmitter depends onopening of voltage-gated calcium channels and entryof extracellular calcium into the nerve terminals. Defi-cient acetylcholine release by motor nerve terminalscould explain muscle weakness. Nerve conduction ve-locity is not dependent on calcium channels. The re-polarization phase of the nerve action potential de-pends on voltage-gated potassium channels. Theupstroke of the nerve action potential depends on volt-age-gated sodium channels. Nerve excitability (and,hence, nerve firing) is affected by extracellular calciumconcentration (hypocalcemia results in increased ex-citability), but this is because of an effect on sodiumchannels, not calcium channels.

Chapter 4

1. The answer is A. The intensity of sensory informationis encoded in the action potential frequency. Cessationof the stimulus would lead to a rapid decrease in the ac-tion potential frequency, and adaptation of the recep-tor would also lead to a decrease in frequency. With aconstant and maintained stimulus, at least some adap-tation would take place, and the frequency would fallsomewhat (and certainly not increase). The action po-tential velocity is a property of the nerve—not the re-ceptor—and it would not be affected.

APPENDIX A Answers to Review Questions 709

710 APPENDICES

2. The answer is C. Rods and cones are absent from thearea of the retina where the optic nerve exits. Theblind spot is of appreciable size, but because its loca-tion is off-center and the eyeballs are mirror images,each eye fills in the information missing from theother, even when the gaze is fixed at a point. There areno connections from lateral cells to the blind spot be-cause nerves are exiting there and do not makesynapses.

3. The answer is C. Presbyopia is the age-related inabil-ity of the eye to focus on close objects. The decreasedcompliance of the lens prevents it from assuming a suf-ficiently curved state, and the focal point is behind theretina. As a person ages, only minor changes occur inthe shape of the eyeball. Age-related changes in theopacity of structures through which light must pass,while they can impair vision, have little effect on thefocal point of the light rays.

4. The answer is B. A myopic eyeball is too long, andlight rays coming from great distances focus in front ofthe retina. The range of motion available to the lens isnot sufficient to provide accommodation regardless ofthe effort made. A negative lens placed in front of theeye corrects the eye’s refractive power, and the rayswill now focus on the retina. A positive lens would onlyworsen matters, and a cylindrical lens (which has twofoci, depending on the orientation considered) wouldnot compensate satisfactorily. Reducing the light in-tensity also would not help; the pupils would dilate andadmit more peripheral rays that would be further out offocus.

5. The answer is A. The frequency response of the basi-lar membrane changes steadily from high to low alongits length, so that high frequencies are detected closeto the oval window and low frequencies are detected atthe other end, near the helicotrema.

6. The answer is C. Relative motion between the en-dolymph and the cupulae of the semicircular canals isdue to the inertia of the endolymph, whether the bodymotion is starting or stopping. As the fluid continues tomove when the head has stopped moving, the cupulaewill be stimulated, producing the sensation of rotarymotion. Moving in a straight line, without accelera-tion, will produce no fluid movement and no sensation.The sensation of static body position is accomplishedby the maculae, which are sensitive to gravity but notendolymph motion.

7. The answer is A. When a receptor adapts, the sensa-tion decreases although the stimulus may be un-changed. Adaptation is largely a result of the fall inmagnitude of the generator potential and is not due tofatigue. Sensory responses are graded in response tochanges in stimulus intensity regardless of the level ofadaptation, and the phenomenon of compression al-lows a wide range of environmental intensities to betranslated into a much more narrow range of sensoryresponses.

8. The answer is B. Rapidly adapting sensory receptorsare best suited for detecting motion and change. Ac-tions such as holding a steady weight and sensing theresting position of the body or the position of an ex-

tended limb are best sensed by receptors that adaptslowly. Likewise, sensors that adapt quickly would notbe well suited for detecting the continued presence ofa chemical stimulus. (Our rapidly adapting olfactorysensors can sometimes fail to provide informationabout a continuing hazard.)

9. The answer is D. Reduction in the intensity of a sensa-tion is largely the result of a decline in the generatorpotential. In this sense, it mimics the effects of a re-duction in the stimulus intensity. Because the actionpotentials arising in a sensory nerve are all-or-none,their velocity of conduction, amplitude, and durationof depolarization are not affected by the stimulus in-tensity; rather, they are properties of the nerve cell.

10. The answer is C. The transfer of energy through themiddle ear from the relatively large eardrum to thesmaller oval window by the ossicular chain increases theefficiency of the mechanical transduction process. Thebones do not support the membrane structures but allowthem to move relatively freely. Interference with the os-sicular transmission process by external influences (as bythe stapedius and tensor tympani muscles) or by diseaseprocesses, acts to reduce the vibration transfer effi-ciency, a change that can be either protective or harm-ful. The function of the eustachian tube is independentof the ossicles. While the bones themselves are passive,they are essential to the process of sound conduction.

11. The answer is B. The cone cells, which are responsiblefor color vision, are located at the point of sharpest fo-cus, but they do not function if the light intensity toolow. In such cases, the single-pigment rod cells (withgreater sensitivity, but with less advantageous locationand interconnection) provide monochromatic but dif-fuse vision. The color composition of light does notdepend on its intensity, and dark adaptation does notchange the spectral sensitivity of the individual pig-ments. While focusing mechanisms may be less effec-tive with low light, they still function.

Chapter 5

1. The answer is A. The maintenance of posture requirescontinuous muscle action. The low threshold for acti-vation, fatigue-resistant motor units are the type activein postural control. Intrafusal muscle fibers do not con-tribute to force generation.

2. The answer is C. The Golgi tendon organ senses theforce of muscular contraction. The nuclear chain andbag fibers, along with type Ia endings, are all compo-nents of the muscle spindle which reports musclelength and velocity of muscle shortening.

3. The answer is D. Motor neurons controlling axial mus-cles are positioned most medially in the ventral hornarea. An enlarged central canal would impinge on thatpool of motor neurons first.

4. The answer is C. Muscle spindles monitor musclelength. If the muscle is suddenly stretched, the spindleproduces action potentials that activate homonymousmotor neurons to contract the stretched muscle and re-sist the length change.

5. The answer is E. The spinal cord has the intrinsic cir-

cuitry in the form of central pattern generators to pro-duce the basic motions of walking. All the other listedareas may influence the local pattern generators.

6. The answer is D. The rubrospinal tract descends in thelateral spinal cord and influences distal muscle func-tion. This is also the function of the corticospinal tract.The vestibulospinal and reticulospinal tracts descendmedially and influence proximal muscle action. Thespinocerebellar tract is an ascending pathway.

7. The answer is C. The primary motor area is locatedalong the precentral gyrus. The supplementary motorarea is located on the medial aspect of the hemisphere.The other areas have sensory and association functionsthat influence the motor areas.

8. The answer is E. The supplementary motor area tendsto produce bilateral motor responses when stimulated.The other areas would tend to produce unilateral re-sponses.

9. The answer is C. The neurons of the primary motorcortex contribute about one-third of the axons thatmake up the corticospinal tract. Other tracts, such asthe rubrospinal, do not sprout additional axons. Thealpha motor neurons do not atrophy if deprived of cor-ticospinal input.

10. The answer is C. Decreased inhibitory input to theGPi from the putamen, would enhance inhibitory out-put from the GPi to the thalamus. The result is inhibi-tion of excitatory output from the thalamus back to thecortex.

11. The answer is B. Spinal input, such as from the spin-ocerebellar tracts, enters the cerebellum on the mossyfibers. The climbing fibers originate from the inferiorolivary nucleus of the medulla. The other componentsare intrinsic to the cerebellum.

Chapter 6

1. The answer is D. Pupillary dilation is a function of thesympathetic innervation that originates from the upperthoracic spinal cord. The preganglionic axons pass upthe paravertebral sympathetic chain to the superiorcervical ganglion from which the postganglionic axonsarise. These axons then ascend in the pericarotidplexus to the eye.

2. The answer is D. Destruction of the lumbar paraverte-bral ganglia would impair sympathetic function to theleg on that side. This would result in skin dryness fromthe absence of sweating and warmth from persistentvasodilation. There should be no alteration of sensa-tion or skeletal muscle function.

3. The answer is D. Muscarine is an exogenous agonistthat acts at postganglionic synapses. All the otheragents are neurotransmitters used at autonomicsynapses.

4. The answer is A. The muscarinic parasympathetic andadrenergic sympathetic receptors are both G protein-linked and share a seven-membrane-spanning segmentconfiguration. The parasympathetic and sympatheticpreganglionic synapses are both of the direct ligand-gated type, which is similar in configuration to the re-ceptor at the neuromuscular junction.

5. The answer is E. Sweat glands are controlled by thesympathetic nervous system.

6. The answer is B. Postsynaptic neurons are about a 100-fold more numerous than the presynaptic neurons.This divergence is why presynaptic neuron activationcan produce a widespread sympathetic response.

7. The answer is D. Bright light would cause constrictionof the pupil as a result of parasympathetic activation.Medications that inhibit the action of acetylcholine(anticholinergic drugs) could impair pupillary con-striction. Inhibition of adrenergic action might assist inpupillary constriction, but the primary constrictor ac-tion is cholinergic.

8. The answer is D. Muscarinic receptors at the synapsebetween the postganglionic axon and the target tissueare of the indirect ligand-gated type, which utilizes a Gprotein. This type synapse alters the function ofadenylyl cyclase and produces changes in cyclic AMPlevels. The preganglionic to postganglionic sympa-thetic and parasympathetic synapses are directly gatedby acetylcholine. Curare blocks the receptor at theneuromuscular junction but not at the direct ligand-gated cholinergic autonomic synapses.

9. The answer is E. The medullary reticular formation isthe anatomic site for coordination of cardiac sympa-thetic and parasympathetic activity.

Chapter 7

1. The answer is A. Alpha waves are noted in the EEG ina relaxed, awake person whose eyes are closed. An alertstate is indicated by beta waves. Theta and delta wavesare noted during sleep. Variability in the wave formsmight indicate a seizure or damage locus.

2. The answer is D. Dreams are associated with REMsleep. Normally, a person does not “act out” his or herdream because all of the motor neurons to musclesother than those for respiration, the middle ear, andthe extraocular eye muscles are inhibited, abolishingmuscle tone. If this inhibition does not occur, a personexhibits marked and often dangerous movement dur-ing dreaming. Muscle tone is reduced but not abol-ished during slow-wave sleep; however, movementsare not an issue, presumably because dreaming doesnot occur. (Note, however, that sleepwalking occurs inslow-wave sleep.) Increased activity in motor areas ofthe cortex (or other areas) during REM sleep normallywould not cause movement because motor neurons areinhibited.

3. The answer is D. Melatonin is secreted by the pinealgland. Adrenaline is secreted by the adrenal medulla,leptin by adipocytes, and melanocyte-stimulating hor-mone and vasopressin by the pituitary.

4. The answer is D. The basal forebrain nuclei and the pe-dunculopontine nuclei are major sources of distributedcholinergic innervation in the CNS. These cell groupsare functionally dissimilar. Neither is a major input tothe striatum or involved in language construction.Only the basal forebrain nuclei receive input from thecingulate gyrus. Although not known for certain, it isunlikely that either of these cell groups is atrophied in

APPENDIX A Answers to Review Questions 711

712 APPENDICES

schizophrenia, which appears to be a disorder ofdopaminergic function.

5. The answer is A. Circulating leptin levels are sensed byneurons in the arcuate nucleus, which does not possessa blood-brain barrier. While some other regions of thehypothalamus also lack a blood-brain barrier, these re-gions do not contain leptin-sensing cells.

6. The answer is B. Circadian rhythms are entrained bythe SCN to the external day/night cycle. This externalinformation reaches the SCN directly by an opticnerve projection from the retina. The internal clock re-sides in the SCN and regulates the production of mela-tonin by the pineal gland. Reticular formation and vi-sual cortical inputs are not directly involved in theregulation of circadian rhythms.

7. The answer is C. Magnocellular neurons of the par-aventricular and supraoptic nuclei of the hypo-thalamus, whose axons reach the posterior pituitary viathe hypothalamo-hypophyseal tract, secrete the poste-rior pituitary hormones. The portal capillary systemfrom the hypothalamus to the pituitary gland is associ-ated with the anterior pituitary. While pituitary func-tion may be altered by the fight-or-flight response, thereticular activating system, or emotional state, none ofthese directly mediates posterior pituitary hormone se-cretion.

8. The answer is E. The arcuate fasciculus is the fiber bun-dle connecting Broca’s and Wernicke’s areas. Thefornix connects the hippocampus with the hypothala-mus and basal forebrain. The thalamocortical tractconnects the thalamus with the cortex and the reticu-lar activating system connects the brainstem with thethalamus and cortex. The prefrontal cortex is not afiber bundle.

9. The answer is B. Neuroleptics drugs ameliorate thesymptoms of psychosis in disorders such as schizophre-nia. While the etiology of schizophrenia is far from un-derstood and many transmitter systems may be in-volved, all neuroleptics block dopamine receptors.

10. The answer is A. The intralaminar nuclei of the thala-mus receive input from the brainstem reticular activat-ing system and convey information to the cortex.These nuclei are critical for the maintenance of arousaland consciousness. Without the intralaminar nuclei,beta rhythms and attention would be severely compro-mised. Both slow-wave and REM sleep would be af-fected because the regulation of sleep is also driven bythe reticular activating system and its input.

11. The answer is D. Somatic sensory information fromthe left hand would be perceived in the right cortex,which does not generate language. To verbally explainwhat the object is, the information must cross to theleft hemisphere. This crossing occurs through the cor-pus callosum. The fornix and hippocampus would beinvolved in storing memories about particular items,not in retrieving the memory. Neither the primary so-matic sensory cortex on the left side nor the visual cor-tex on either side plays a role in identifying an objectplaced in the left hand by tactile cues.

12. The answer is D. The hippocampus is crucial for theformation of long-term (declarative) memory. With-

out the hippocampus, short-term memory is intact butthe conversion to long-term does not take place. Theretrieval of stored declarative memory does not requirethe hippocampus. The hippocampus is not needed forthe formation or retrieval of procedural memory.

13. The answer is E. Wernicke’s area is responsible for therecognition and construction of words and language;when it is damaged, the individual speaks but the con-tent is nonsensical. Damage to Broca’s area results in aninability to speak clearly because it controls the motorpatterns required to speak; the little speech that is pro-duced is grammatically and syntactically correct. Thehippocampus and corpus callosum are not involved inthe generation of speech. Damage to the arcuate fasci-culus would result in a loss of speech because languagegenerated in the Wernicke’s area would not be con-veyed to Broca’s area.

14. The answer is D. Mania is an affective disorder char-acterized by increased transmission through noradren-ergic pathways. Other transmitter systems may play arole, but effective treatments are targeted at the nora-drenergic system.

15. The answer is A. Acetylcholine is critical for cognitivefunction because of the cholinergic neurons in thebasal forebrain that relay hippocampal information tothe rest of the cortex. Nicotine activates cholinergicreceptors. The only effective drugs for the treatment ofcognitive deficits in Alzheimer’s disease are choliner-gic, although cognition clearly involves neurons inmany regions of the brain that utilize a variety of trans-mitters.

Chapter 8

1. The answer is C. In all muscle types, the interaction be-tween actin and myosin provides the forces that resultin shortening. Skeletal and cardiac muscle have repeat-ing sarcomeres, but smooth muscle does not. Smoothand cardiac muscles have small cells, whereas skeletalmuscle has large cells.

2. The answer is B. The width of the I band changes be-cause the thin filaments enter farther into the A band.The Z lines move closer together. The decrease in Iband width and the moving of the Z lines together areproportional, but there is no change in A band dimen-sions.

3. The answer is B. ATP must bind to the myosin headsto allow the crossbridges to detach and the cycle tocontinue.

4. The answer is A. Relaxed skeletal muscle is in a state ofinhibited contraction. The enzymatic activity of myosinis greatly enhanced by its interaction with actin. Therole of calcium is as an activator, not an inhibitor; at rest,the concentration of free calcium is low.

5. The answer is C. Removal of calcium from the myofil-ament space into the sarcoplasmic reticulum (not theextracellular space) is an absolute requirement for nor-mal relaxation. A reduction of ATP would promoterigor, not relaxation.

6. The answer is B. When the myofilament overlap is de-creased above the optimal length, fewer crossbridges

(borne on the myosin filaments) are able to interactwith actin, and there is a proportionate decrease in theforce produced. The filaments actually come closer to-gether as the muscle becomes thinner.

7. The answer is C. The sarcoplasmic reticulum releasescalcium rapidly and in close proximity to the myofila-ments. Calcium is not stored in the T tubules and is notinvolved in action potential events at the sarcolemmain skeletal muscle.

8. The answer is A. Calcium diffuses away from the tro-ponin complex because the intracellular concentrationhas become low and the gradient favors dissociation.Calcium does not bind to active sites on myosin mole-cules, and individual actin molecules do not have en-zymatic activity.

9. The answer is C. ATP is the immediate source of en-ergy. The other substances are in metabolic pathwaysthat provide energy, via several routes, into the ATPpool. They are not used directly in the crossbridge cy-cle.

10. The answer is B. The condition called rigor mortis de-velops after death because the processes that generateATP stop. ADP does not contain energy usable to sup-port contraction.

11. The answer is B. By shifting its metabolism to anaero-bic pathways (glycolysis), the muscle keeps function-ing at the expense of generating end products that willeventually require oxygen consumption for their fur-ther processing. This condition is called oxygendeficit.

12. The answer is C. A reduction in the calcium-pumpingability of the sarcoplasmic reticulum would leave ahigher concentration of calcium ions in the myofila-ment space for a longer time. The diffusion of calciumaway from the regulatory proteins would be slower,and crossbridges would detach less rapidly; conse-quently, the muscle would relax more slowly. Activa-tion processes would not be as affected because theydo not directly depend on the effectiveness of the cal-cium pump.

Chapter 9

1. The answer is B. This is a chemically gated channelwithout a highly selective filter and voltage sensormechanism. Both sodium and potassium pass throughit simultaneously down their respective electro-chemical gradients.

2. The answer is B. The endplate potential and the actionpotential are based on changed ionic permeabilities,but the postsynaptic channels in the endplate regionare not voltage-sensitive. This means that the endplatepotential cannot regenerate and be propagated. Be-cause the channels do not select between sodium andpotassium, the endplate potential is close to zero. Assuch, it can never assume a large inside-positive value.

3. The answer is A. The postsynaptic membrane channelsare blocked by the bound curare molecules and will notallow ions to pass; therefore, this membrane will notdepolarize. The other choices are all presynaptic

events and will not be affected by the blocked postsy-naptic membrane.

4. The answer is B. The contraction will be twitch-like,but it will have increased amplitude, reflecting the ad-ditional calcium released from the SR in response tothe second stimulus. Its duration will also be increasedfor the same reason.

5. The answer is A. This is the definition of isometric.The three other responses address factors that mightchange the size of the contraction but have nothing todo with whether it is isometric.

6. The answer is B. As long as the muscle is actually lift-ing the afterload, this is the only factor that determinesthe force. The other factors may make the muscleshorten faster or slower, but they do not affect theforce produced.

7. The answer is A. This is a statement of relationship thatis graphically represented in the force-velocity curve.Regarding choice D, note that it is force that deter-mines velocity and not the other way around.

8. The answer is C. This is a point at the maximum of thepower output curve. Fmax and Vmax represent points ofzero power. A velocity of two-thirds Vmax correspondsto a force too small to produce maximal power.

9. The answer is C. The forearm/biceps combination, be-cause of the proportions involved, operates at a me-chanical disadvantage with regard to force, trading de-creased hand force for increased hand velocity.

10. The answer is D. This mixture of fiber types is ideal forthe stated exercise because it can mobilize energyquickly. Choice A is a possibility, but almost all mus-cles have some mixture of fiber types.

11. The answer is B. Isometric contraction is possiblewhen the volume of the organ is prevented fromchanging, as by a closed sphincter. Any shortening ofsmooth muscle in a hollow organ would be againstsome sort of load.

12. The answer is C. The level of phosphorylation woulddecline because the myosin light chains dephosphory-lated by the phosphatase could not be rephosphory-lated by MLCK because it would no longer be calmod-ulin-activated, as a result of the lowered cellularcalcium. Choice A represents the skeletal muscle con-dition.

13. The answer is B. This emphasizes the primary role ofmyosin-based regulation in smooth muscle. Choice Arepresents the skeletal muscle condition, while choicesC and D do not reflect the actual physiological effects;in particular, choice D is the reverse of the truth.

14. The answer is C. While smooth and skeletal musclecan exert about the same amount of force per cross-sectional area, smooth muscle does it much more eco-nomically. It is capable of extreme shortening whenconditions external to the muscle allow.

15. The answer is B. The crossbridge cycle of smoothmuscle is similar to that of skeletal muscle, with theadded complexity (in some smooth muscles) of thelatch state of crossbridges.

16. The answer is C. Smooth muscle membrane receptorsperform a wide variety of functions and are involved in

APPENDIX A Answers to Review Questions 713

714 APPENDICES

both chemical and electrical activities at the mem-brane.

Chapter 10

1. The answer is C. Cardiac muscle has small cells thatmust be coupled electrically for communication to oc-cur. Because it receives no motor innervation, it mustbe spontaneously active.

2. The answer is B. The intercalated disk is the site ofelectrical coupling and mechanical linkage, both ofwhich are necessary for the tissue to behave as a syn-cytium.

3. The answer is B. Cardiac muscle is similar to skeletalmuscle in both the structure of its contractile apparatusand the means by which it is regulated. It is similar tosmooth muscle in its small cell size and syncytial be-havior.

4. The answer is C. Until the relative refractory period isover, the muscle cannot be restimulated. By this time,it has begun to relax, so a smooth (fused) tetanus cannot occur.

5. The answer is B. Because the afterload is removed atthe end of the isotonic portion of the contraction, it isnot available to reextend the muscle and relaxation oc-curs isometrically at the shortened length.

6. The answer is C. Although the relationship is not lin-ear, the muscle is extended in proportion to the pre-load. In the intact heart, when the heart is filled moreat rest, the muscle will shorten a greater distance whenit contracts.

7. The answer is D. Because the force is high, the muscleis nearer the limit set by the length-tension curve.

8. The answer is C. As is the case for skeletal muscle,when the muscle is shortening isotonically, the onlyfactor that controls the force is the afterload. Theother factors mentioned will affect the velocity or ex-tent of the shortening, not the force.

9. The answer is C. Inotropic interventions of manytypes, including heart rate changes, epinephrine, anddigitalis-like drugs, all affect the availability of calciumto the contractile proteins.

10. The answer is C. The force-velocity curve states thebasic relationship between the speed of shortening andthe afterload at a given level of contractility. This rela-tionship can be modified (up or down) by changes incontractility.

Chapter 11

1. The answer is C. Adult erythrocytes normally do notcontain any carboxyhemoglobin, which is formedwhen hemoglobin binds carbon monoxide. Adult ery-throcytes possess two distinct types of hemoglobin,HbA and HbA2. These hemoglobin molecules may besaturated with oxygen (HbO2 ) or reduced to Hb whenoxygen is released to cells within tissues.

2. The answer is D. Superoxide anion is generated whenoxygen is reduced by cytoplasmic NADPH. The re-duction is carried out by the enzyme NADPH oxidase,which is not a reactant but a catalyst activated in cells

responding to bacteria. The hexose monophosphateshunt is an enzyme cascade (not a reactant) that func-tions to provide high levels of reduced NADPH todrive this reaction. G proteins are not reactants, butplay an essential role in the activation of this cellularcascade. Similarly, the enzyme myeloperoxidase is nota reactant; it enhances the ability of reactants, such ashydrogen peroxide, to exert a lethal effect on invadingbacteria.

3. The answer is A. T cells are infected by HIV in indi-viduals who have AIDS. B cells, like T cells, are lym-phocytes, but they are not targets for HIV. Neu-trophils are not lymphocytes and are not infected bythe AIDS virus. Monocytes and basophils similarly arenot targets for the virus that causes AIDS.

4. The answer is B. Umbilical cord blood, derived fromthe circulating blood of newborn infants, possesseshigh levels of hematopoietic progenitors. Levels of cir-culating progenitors rapidly decrease after birth, de-pleting the progenitor content within the circulatingblood of adults. The spleen of adult humans functionsas a hematopoietic organ in certain disease states, suchas leukemia. However, in other animals and in devel-oping human fetuses, the spleen plays an importantrole in the hematopoietic response. While the liver andthe thymus are important in hematopoiesis and im-mune reconstitution prior to birth, these organs are notinvolved in hematopoiesis in adult humans.

5. The answer is E. When specifically programmed Tcells or B cells of the adaptive immune system first rec-ognize specific antigens, they begin to divide rapidly,generating several copies of cells similarly pro-grammed against the inciting stimulus. Hematopoiesisinvolves the nonspecific generation of all cells inblood, including leukocytes, erythrocytes, andplatelets. Hematotherapy is a therapeutic process inwhich specifically amplified cells are infused in pa-tients to increase resistance to infection or to restorehematopoiesis. Inflammation is not a specific responseagainst individual antigenic determinants and does notrequire T cell or B cell amplification. Similarly, innateimmunity does not require amplification of T cells or Bcells as a result of interaction with an invading stimulusbut is affected by cells present and programmed to re-spond to specific stimuli.

6. The answer is B. Delayed-type hypersensitivity reac-tions to PPD and other specific antigens developslowly over 24 to 48 hours as T cells become activatedand secrete factors that effect the skin response. B cellsplay no role in this type of reaction; instead, they pro-duce antibodies involved in more immediate re-sponses. Neutrophils do not arrive at sites of delayed-type hypersensitivity in large numbers. Eosinophilsplay a role in immediate hypersensitivity to many anti-gens that cause symptoms of allergy, such as sneezingand stuffy nose, but do not participate in the delayedresponse. Finally, the response to PPD is driven bycells programmed to respond specifically to this anti-gen derived from the bacteria that cause tuberculosis,and not by a metabolite of this protein.

7. The answer is C. Antibody specificity is dictated by the

sequence of amino acids within the variable regions ofthe light and heavy chains. The Fc region is a site for an-tibody docking to effector cells and does not play a rolein antigen binding. The constant region has a similarstructure in antibodies of widely divergent specificityand, therefore, does not dictate specificity. Fc receptorsare sites on immune effector cells that interact with theFc region of the antibody molecule and do not define anantibody’s specificity. The J chain is a unique portion ofsecreted IgA molecules that allows the molecule tomove from the circulation through mucous membranes.

8. The answer is D. The extrinsic coagulation pathway isactivated when tissue thromboplastin (tissue factor) isreleased from injured tissues. Activation of factor X oc-curs later and is a step involved in the activation ofboth the intrinsic and the extrinsic pathways. Activa-tion of factor XII is the first step in activation of the in-trinsic coagulation pathway. Conversion of prothrom-bin to thrombin and conversion of fibrinogen to fibrinare the final steps that lead to clot formation by eitherthe intrinsic or the extrinsic pathway.

Chapter 12

1. The answer is C. See equation 3 in the text.2. The answer is C. See equation 6 in the text. Changes

in transmural pressure can be caused by changes insideor outside of a vessel (see equation 5). The viscosity ofblood does not directly affect transmural pressure. Re-sistance, not transmural pressure, is proportional to thelength of a tube.

3. The answer is D. When the heart stops, blood contin-ues to flow from the arteries to the veins until the pres-sures in the two sides of the circulation are equal. Thatpressure is mean circulatory filling pressure. Hemody-namic pressure is the potential energy that causesblood to flow. Mean arterial pressure is the averagepressure in the aorta or a large artery over the cardiaccycle. Transmural pressure is the difference betweenthe pressure inside and outside a blood vessel. Hydro-static pressure is the pressure caused by the force ofgravity acting on a fluid.

4. The answer is D. Although flow velocity, viscosity,and tube diameter all influence turbulence, it is thecombination of these variables (plus the density ofblood), expressed as the Reynolds number (equation 4in the text), that determines whether flow is turbulentor laminar.

5. The answer is E.

Compliance � �V/�P� 30 mL/40 mm Hg� 0.75 mL/mm Hg

6. The answer is B. See equation 1 in text. The tube isanalogous to the systemic circulation in which thereare many branches. The overall resistance can be cal-culated from the sum of the flows through the individ-ual branches and �P, provided it is the same for allbranches.

R � �P/Q

where Q� 95 � 5 � 100 mL/minand �P� 75 � 25 � 50 mm Hg.

R � 50/100 � 0.5 mm Hg/(mL/min) � 0.5 PRU

Chapter 13

1. The answer is B. Voltage-gated Na� channels are re-sponsible for phase 0 in ventricular muscle. Voltage-gated Ca2� channels are responsible for phase 0 innodal cells. The potassium channels mentioned do notplay a role in mediating depolarization.

2. The answer is D. The form of the QRS will be normalbecause electrical excitation of the ventricles occursover essentially the normal pathway (i.e., AV node tobundle branches to Purkinje system to myocardium).The T wave will be normal as well. With completeheart block, P waves and QRS complexes are com-pletely independent of each other. Some PR intervalscould be shortened by chance, others will be very long;that is, there is no predictable PR interval. There willnot be a consistent ratio of P waves to QRS complexesbecause the two are disassociated, but the average ra-tio would be 80/40 or 2:1.

3. The answer is B. The shape of the QRS complex willbe significantly different from normal because depolar-ization now originates in the right ventricle and prop-agates in a retrograde fashion. Because the right side ofthe heart depolarizes before the left, the configurationof the QRS may resemble that seen with left bundlebranch block, another situation in which the right sideof the heart depolarizes before the left. The duration ofthe QRS complex will be increased because the spe-cialized conducting system of the ventricles is not fullyemployed: Depolarization moves through more slowlyconducting muscle instead of the rapidly conductingPurkinje system. Retrograde conduction through theAV node is extremely unlikely, so P waves will not fol-low each QRS complex. Because excitation of the atriaand ventricles is still independent, there will be no pre-dictable PR interval.

4. The answer is D. Voltage-gated Ca2� channels are pri-marily responsible for the upswing of the action poten-tial (phase 0) of nodal cells. Voltage-gated Na� chan-nels are inactivated because the resting membranepotential in these cells never becomes sufficiently nega-tive to allow reactivation. Acetylcholine-activated K�

channels are important only in mediating the effect ofACh on the pacemaker potential of nodal cells. Inwardrectifying K� channels are responsible for maintainingthe resting membrane potential in nonnodal cells buthave a less important role in cells with a pacemaker po-tential.

5. The answer is B. Atrial repolarization normally occursduring the QRS complex. A dipole is created by atrialrepolarization but it is not observed on the ECG be-cause the dipole created by ventricular depolarizationis much larger.

6. The answer is D. Depolarization of the ventricles pro-ceeds from subendocardium to subepicardium, but this

APPENDIX A Answers to Review Questions 715

716 APPENDICES

does not result in the P wave. In lead I, when the ECGelectrode attached to the right arm is positive relativeto the electrode attached to the left arm, a downward de-flection is recorded. AV nodal conduction is slowerthan atrial conduction, but this does not cause the Pwave. When cardiac cells are depolarized, the inside ofthe cells is positive or neutral relative to the outside ofthe cells.

7. The answer is C. Stimulation of the sympathetic nervesto the normal heart decreases the duration of the ven-tricular action potential and, therefore, decreases theQT interval. As heart rate increases, the duration of di-astole and, therefore, the TP interval decreases. In-creased conduction velocity in the AV node decreasesthe duration of the PR interval. Fewer P waves thanQRS complexes are indicative of AV block. On thecontrary, sympathetic stimulation may reverse AVblock. The frequency of QRS complexes increaseswith the heart rate.

8. The answer is D. The drug could act on �1-adrenergicreceptors to increase the rate of depolarization ofsinoatrial nodal cells. An adrenergic receptor antago-nist would have the opposite effect, as would a cholin-ergic receptor agonist and the closing of voltage-gatedCa2� channels. Opening of acetylcholine-activatedK� channels would slow pacemaker depolarization bykeeping the membrane potential closer to the K� equi-librium potential.

9. The answer is C. Excitation of the ventricles does notordinarily lead to excitation of the atria because retro-grade conduction in the AV node is unusual. Norepi-nephrine modulates the ventricular force of contrac-tion and conduction velocity and lowers the thresholdfor excitation, but it does not, by itself, initiate excita-tion. Excitation of the ventricles is initiated by phase 0of the action potential. Normal ventricular cells do notexhibit pacemaker potentials.

10. The answer is C. AV nodal cells exhibit action poten-tials characterized by slow depolarization (phase 0)because fast voltage-gated Na� channels do not par-ticipate. This is because the diastolic potential of thesecells does not become sufficiently negative to allow re-activation of Na� channels. Acetylcholine slows andnorepinephrine speeds conduction velocity. AV nodalcells are capable of pacemaker activity but at a rate ofapproximately 25 to 40 beats/min.

11. The answer is C. When stimulation of the parasympa-thetic nerves to the normal heart leads to complete in-hibition of the SA node for several seconds, nodal es-cape usually occurs. In this situation, pacemakeractivity usually is taken over by cells in the AV node orbundle of His. QRS complexes are normal because thepacemaker activity is high enough in the conductingsystem to lead to a normal pattern of ventricular exci-tation. T waves would be normal for the same reason.Because at least one beat begins without atrial excita-tion, there would be fewer P waves than either QRScomplexes or T waves.

12. The answer is B. The R wave in lead I of the ECG re-flects a net dipole associated with ventricular depolar-ization. Repolarization causes the T wave. The R wave

is smallest when the mean axis is directed perpendicu-lar to a line drawn between the two shoulders becauseboth electrodes are equally influenced by the negativeand positive sides of the dipole.

13. The answer is C. The ST segment of the normal ECGoccurs during a period when both ventricles are com-pletely depolarized. It is present in all leads.

Chapter 14

1. The answer is C. Loop B shows increased contractilitybecause stroke volume is increased at a constant pre-load and afterload. When loop B is compared to loopA, preload is not increased or decreased because thereis no change in the pressure or volume at which the mi-tral valve closes and isovolumetric contraction begins.Afterload is not changed because there is no change inthe pressure or volume at which the aortic valve opensand ejection begins. The evidence that stroke volumeis increased is the larger volume difference between thepoint at which the aortic valve opens and closes—thatis, between isovolumetric contraction and relaxation.

2. The answer is A. The aortic and mitral valves are neveropen at the same time. This is the basic principle of thecardiac pump. The first heart sound is caused by clo-sure of the mitral and tricuspid valves. The mitral valveis open throughout diastole except isovolumetric relax-ation. Left ventricular pressure is less than aortic pres-sure during diastole and isovolumetric contraction butis greater than aortic pressure during a substantial pe-riod of ventricular ejection. Ventricular filling occursduring diastole.

3. The answer is C. Aortic pressure reaches its lowestvalue during the isovolumetric contraction phase ofventricular systole. The second heart sound is associ-ated with closure of the aortic valve. Left atrial pressureis less than left ventricular pressure during ventricularsystole and isovolumetric relaxation. The ventricleseject blood during all of systole except isovolumetriccontraction. Ventricular end-diastolic volume isgreater than end-systolic volume.

4. The answer is D. Increased ventricular filling meansa larger end-diastolic volume. Of the three pointsrepresenting increased end-diastolic volume, onlychoice D is on a higher ventricular function curve,signaling increased contractility. If you chose choiceA, you recognized that the upper curve representedincreased contractility, but missed the fact that end-diastolic volume would be increased as well. If youselected choices C or D, you recognized increasedend-diastolic volume, but did not understand that in-creased contractility means that the ventricular func-tion curve would be higher. Point B is the graphicaldefinition of decreased contractility at an unchangedend-diastolic volume.

5. The answer is B. Drug B increases the internal work ofthe left ventricle more than drug A because it increasesexternal work by increasing pressure. Drug A increasesthe external work of the left ventricle the same as drugB. External work is stroke volume multiplied by meanarterial pressure, so equivalent increases in stroke vol-

ume and pressure yield equivalent increases in strokework. Because drug B increases internal work morethan drug A, total work is more increased. For this rea-son, drug B increases the oxygen consumption of theheart more than drug A. The “double product” (aorticpressure times heart rate) is greater for drug B than fordrug A. Cardiac efficiency is higher with drug A thanwith drug B because efficiency is a measure of the oxy-gen cost of external work. Because of the greater inter-nal work, drug B increases oxygen consumption morethan drug A. The ratio of external work to oxygen con-sumption would be higher for drug A than drug B.

6. The answer is D.

CO � HR � SV � HR � (EDV � ESV)� 70 beats/min � (130 � 60) mL� 4,900 mL/min

SW � SV � MAP� 70 mL � 90 mm Hg� 6,300 mL � mm Hg

7. The answer is B.

CO � V̇O2/(aO2 � v�O2)� 4,000 mL O2/min/(190 � 30 mL O2/L)� 25 L/min

SV � CO/HR� 25 (L/min)/(180 beats/min)� 139 mL/beat

8. The answer is D. Electrical pacing to a heart rate of200 beats/min would decrease time for filling and re-duce end-diastolic volume. A reduction in afterloadwould make it easier for the ventricle to eject bloodand would raise stroke volume. An increase in end-diastolic pressure will increase end-diastolic fiberlength and increase the force of contraction andstroke volume. Stimulation of the vagus nerves slowsthe heart, increases the time for ventricular filling,and increases stroke volume. Stimulation of sympa-thetic nerves to the heart increases heart rate andcontractility. Despite the decreased filling accompa-nying an increase in heart rate, stroke volume willstay the same or increase because of the increasedcontractility.

Chapter 15

1. The answer is C. Strictly speaking, mean arterial pres-sure minus right atrial pressure equals cardiac outputtimes systemic vascular resistance. Right atrial pressureis often ignored because it is so much smaller thanmean arterial pressure that it does not have much effecton the calculation.

2. The answer is B. A stroke volume change with nochange in heart rate means that cardiac output ischanged. If we assume that mean arterial pressure is de-termined by CO and SVR and SVR is constant, thenmean arterial pressure must have changed. Heart ratechanges with no changes in cardiac output or SVR willhave no effect on mean arterial pressure. A doubling of

heart rate with no change in stroke volume gives a dou-bling of cardiac output; if SVR is halved at the sametime, then mean arterial pressure will not change. Ar-terial compliance influences pulse pressure but notmean arterial pressure.

3. The answer is C. If the cuff is too small, it takes a falselyhigh pressure in the cuff to transmit sufficient pressureto the vessel wall for total occlusion of the artery.Blood pressure may be falsely high in patients withbadly stiffened arteries because of the extra pressureneeded to compress the arteries. The measurementgives an indirect reading of systolic and diastolic pres-sure; mean arterial pressure must be calculated. Themeasurement depends on the appearance of sound to sig-nal systolic pressure

4. The answer is C. Vessel radius is the most importantvariable influencing vascular resistance. Resistancechanges occur primarily in small arteries and arterioles.Blood viscosity and length are important determinantsof underlying vascular resistance, but ordinarily do notchange enough to be influential in altering vascular re-sistance.

5. The answer is E. Standing up causes a shift in bloodfrom the chest to the periphery, lowering central bloodvolume. The diameter of the leg veins increases be-cause of increased transmural pressure caused by thecolumn of blood in the vessels above them. Right atrialpressure decreases and, therefore, decreases filling ofthe ventricles and stroke volume.

6. The answer is B. By convention, the first of the twonumbers is the systolic pressure and the second is thediastolic pressure.

Pulse pressure � 125 � 75 mm Hg� 50 mm Hg

Mean arterial pressure � 75 mm Hg � 50 mm Hg/3� 92 mm Hg

7. The answer is B. Mean arterial pressure both beforeand after the tricuspid valve becomes incompetent is110 mm Hg. The pressure gradient before tricuspid in-sufficiency is P�a � Pra � 107 mm Hg. The pressuregradient after the valve becomes incompetent is 110 �13 � 97 mm Hg. If all other hemodynamic factors re-main unchanged (which would be unlikely in this situ-ation), systemic blood flow will fall in proportion tothe decrease in pressure gradient.

8. The answer is A. Pulse pressure is determined by strokevolume and arterial compliance. Stroke volume is un-changed, and if arterial compliance were to remainconstant, the pulse pressure would not change. How-ever, an increase in mean arterial pressure will tend tostretch the aorta and decrease its compliance. Ejectingthe same stroke volume into a less compliant aorta willresult in an increased pulse pressure.

9. The answer is A. The increase in transmural pressureexerted by the column of blood above the veins wouldhave little effect on their volume if they were as stiff asthe arteries. In this situation, relatively little bloodwould accumulate in the veins and little would be dis-placed from the central blood volume.

APPENDIX A Answers to Review Questions 717

718 APPENDICES

Chapter 16

1. The answer is B. Although small arteries do have a sig-nificant resistance, arterioles dominate the total resist-ance.

2. The answer is B. Molecular-sized openings within thetight junctions are the most influential sites sieving themolecules that diffuse through the capillary wall. Largedefects are highly permeable areas, but their occur-rence is too infrequent to affect the total amount ofmaterial moved.

3. The answer is A. All the other possibilities include oneminor force for filtration or absorption.

4. The answer is C. Myogenic mechanisms seem to in-volve only the physical loading of vascular muscle cellsin the form of stretch and increased tension or in justincreased tension.

5. The answer is A. Each of the choices is a function ofthe microcirculation, but its most important functionby far is to provide tissue with nutrients and removethe wastes.

6. The answer is A. Lipids are not particularly water-sol-uble and must primarily diffuse through the lipid layersof cell membranes. A small amount of lipid does movethrough water-filled channels.

7. The answer is B. The cardiovascular system is designedto support a much higher metabolic rate than exists atrest. Only a fraction of the available blood flow is nec-essary for functioning at rest, and the remainder movesslowly through the venules and smallest veins.

8. The answer is C. The interstitial space consists of al-ternating gel and liquid areas with a low plasma proteinconcentration. It is permeable compared to the capil-lary wall.

9. The answer is D. Both adenosine diphosphate (ADP)and acetylcholine cause the release of NO from en-dothelial cells. The other choices involve mechanismsthat function without endothelial cells.

10. The answer is A. Although all of the choices are eventsthat happen in lymph vessels, the first key event islowering the lymphatic hydrostatic pressure to enabletissue fluid to enter the lymphatic vessel.

11. The answer is C. Nerve fibers, not vascular smoothmuscle, release norepinephrine. The norepinephrinefrom the sympathetic nerves simply diffuses from theaxons and binds to specific receptors on smooth mus-cle cells.

12. The answer is A. More capillaries in use at a constantblood flow actually slows the flow velocity in individ-ual capillaries. The distances between capillaries aredecreased. The perfusion of additional capillaries doesnot influence the permeability of the individual capil-laries.

13. The answer is B. The amount of oxygen exchanged isequal to the product of the blood flow and the arterial-venous oxygen content difference: 200 mL/min � (20mL/100 mL � 15 mL/100 mL) � 10 mL/min.

14. The answer is D. Fluid will be filtered at a net pressureof �4 mm Hg. The balance of hydrostatic pressures is22 mm Hg (capillary hydrostatic pressure � tissue hy-drostatic pressure) and is greater than the balance ofcolloid osmotic pressures, which is 18 mm Hg [reflec-

tion coefficient � (plasma colloid osmotic pressure �tissue colloid osmotic pressure)].

Chapter 17

1. The answer is A. Increased arterial blood pressure orthe increased cardiac output of exercise imposes an in-creased workload on the heart, and the coronary ves-sels dilate to improve oxygen delivery. When theblood pressure falls or the blood lacks oxygen, the au-toregulatory mechanisms of the heart vasculature di-late the microvessels to maintain the blood flow.

2. The answer is C. Resting after a meal is associated withreduced sympathetic nervous system activity and re-duced arterial pressure, but the expected increase inblood flow would not meet the substantially increasedmetabolic needs of the intestine during nutrient pro-cessing. Much more potent mechanisms are needed toincrease blood flow, such as increased NO production.Although parasympathetic nervous system activity in-creases during food absorption, the effect on bloodflow is minor.

3. The answer is C. The hepatic arterial and portal venousblood mix in the capillaries of the hepatic acinus.

4. The answer is C. Brain blood flow is constant despitelarge changes in the arterial blood pressure becausevascular resistance usually changes in the same direc-tion as the arterial pressure and by almost the same per-centage.

5. The answer is D. The skeletal muscle vasculature has a20-fold or greater range of blood flows, from minimalperfusion at rest tovery high blood flows during in-tense exercise. No other organ system has appreciablymore than a 4- to 5-fold change in blood flow from restto maximum flow.

6. The answer is D. See Figure 17.6.7. The answer is D. The autoregulatory range is shifted to

higher pressures because the arteries and arterioles in-crease their resistance. The functional and structuralchanges increase the arterial pressure at which au-toregulation of blood flow occurs, but increase thelowest pressure at which blood flow can be main-tained.

8. The answer is B. Oxygenated blood from the placentadoes not become fully mixed with blood returning forthe superior vena cava and is diverted through the fora-men ovale into the left atrium. Consequently, the oxy-gen content of blood in the ascending aorta is signifi-cantly greater than that in the ductus arteriosus. Theupper body, brain, and coronary arteries are suppliedby vessels that branch from the aorta before the ductusarteriosus. The ductus carries blood with lower oxygencontent into the aorta, to perfuse the lower body andfetal placenta.

Chapter 18

1. The answer is B. Norepinephrine, the sympatheticpostganglionic neurotransmitter, causes constrictionof blood vessels in the skin. Increased sensitivity to NEwould greatly reduce skin blood flow, which would

cause the skin to be cold and painful. Epinephrine con-stricts skin blood vessels; abnormally low epinephrinein the blood would allow skin vessels to dilate. An in-sensitivity of blood vessels to epinephrine would havethe same effect. As there are no parasympatheticnerves to skin blood vessels, parasympathetic activitydoes not affect blood flow in the skin. Although acetyl-choline causes nitric oxide release from skin blood ves-sels, this would cause vasodilation.

2. The answer is B. Activation of parasympathetic nervesto the heart would lower the heart rate below its in-trinsic rate. However, with all effects of norepineph-rine and epinephrine blocked, the sympathetic nervoussystem cannot raise the heart rate above its intrinsicrate. The withdrawal of parasympathetic nerve tonecould only raise the heart rate to the intrinsic rate. (SeeChapter 13 for a discussion of intrinsic heart rate.)

3. The answer is D. The cold pressor response is initiatedby the stimulation of pain receptors by exposing thesurface of the skin to ice water.

4. The answer is A. The release of acetylcholine fromparasympathetic nerves to the sinoatrial node results ina slowing of diastolic depolarization of pacemaker cellsand a slowing of the heart rate. ACh slows conductionvelocity, inhibits NE release from sympathetic termi-nals, enhances NO release from endothelial cells, and di-lates blood vessels of the external genitalia (via NO)—all by binding to muscarinic receptors.

5. The answer is A. The function of these baroreceptorsis the rapid short-term regulation of arterial bloodpressure. The receptors start firing at a pressure of ap-proximately 40 mm Hg. They completely adapt over 1to 2 days, not weeks. In general, changes in barorecep-tor activity have little effect on cerebral blood flow.The sympathetic activity following a fall in blood pres-sure results in increased heart rate and contractility,which raises myocardial metabolism and coronaryblood flow.

6. The answer is D. Peripheral chemoreceptor activationplays a significant role in enhancing the diving responseby enhancing peripheral vasoconstriction and brady-cardia. Activation is increased by a decrease in pH andby a lowering of arterial PO2, not oxygen content. Pe-ripheral chemoreceptors are located in the aortic andcarotid bodies.

7. The answer is B. The fight-or-flight response and ex-ercise are characterized by increased sympathetic toneand decreased parasympathetic tone. The diving re-sponse is associated with increased parasympatheticand sympathetic tone. The cold pressor response ischaracterized by increased sympathetic activity to theheart and blood vessels.

8. The answer is D. The hemorrhage has decreased arte-rial pressure below normal. The fall in blood volumewould result in a fall in central blood volume, rightventricular end-diastolic volume, and cardiopulmonaryreceptor activity. Carotid baroreceptor activity wouldbe lowered in the presence of a low mean arterial pres-sure. The resulting sympathetic activity would causevasoconstriction in the splanchnic bed, and especiallywith a lowered arterial pressure, splanchnic blood flow

would be decreased. The heart rate would be elevatedby the increased sympathetic activity and decreasedparasympathetic activity caused by the baroreceptorreflex.

9. The answer is B. Standing up increases the transmuralpressure in the veins of the legs. Because the veins arehighly compliant, their volume increases at the ex-pense of central blood volume. A lower central bloodvolume means reduced cardiac filling pressure (pre-load). Within seconds, the decrease in preload de-creases stroke volume, cardiac output, and arterialpressure. However, within the first minute, the arterialbaroreflex and the cardiopulmonary reflex work to-gether to increase sympathetic activity and decreaseparasympathetic activity. As a result, cardiac contrac-tility and heart rate increase, and cardiac output de-creases less than it would have without compensation.“Noncritical” vascular regions, such as the splanchnicarea and skin, constrict in response to increased sym-pathetic nervous system activity. Brain blood flowchanges little because sympathetic nerve activationcauses little vasoconstriction in the brain and autoreg-ulation of blood flow prevents a fall in brain bloodflow, even if mean arterial pressure decreases.

10. The answer is C. Pressure diuresis lowers arterial pres-sure by lowering blood volume and, thereby, loweringcardiac output. All of the other choices do lower arte-rial pressure, but are not caused by pressure diuresis.

11. The answer is C. By increasing the excretion of saltand water, blood volume is decreased. Central bloodvolume participates in this decrease, reducing ventric-ular filling, cardiac output, and venous return as well(remember that cardiac output equals venous return inthe steady state). Both the muscle pump and the respi-ratory pump increase the pressure gradient toward theheart and increase venous return and central blood vol-ume. Lying down increases venous return and centralblood volume by reducing venous volume of the lowerextremities. Going into space is similar to lying down,in that the force of gravity on blood is removed andblood is not held in the leg veins when a person is up-right. Therefore, central blood volume is increased.

Chapter 19

1. The answer is D. Pleural pressure is the most negativeat total lung capacity because of the elastic recoil of thelungs pulling inward. At residual volume, pleural pres-sure would be the least negative. Choice E is not an op-tion because pleural pressure is positive during a forcedvital capacity maneuver.

2. The answer is A. Transpulmonary pressure is equal toalveolar pressure minus pleural pressure.

3. The answer is B. The outward recoil of the chest walland the inward recoil of the lungs reach equilibrium atFRC. At residual volume, the outward recoil of thechest is the greatest and the inward recoil of the lungis the smallest. At total lung capacity, the inward recoilof the lung is the greatest.

4. The answer is B. Transairway pressure is pressureacross the airways and is measured by subtracting pleu-

APPENDIX A Answers to Review Questions 719

720 APPENDICES

ral pressure from airway pressure (Pta � Paw � Ppl).Transairway pressure is most negative in the condi-tions described in choice B. Transairway pressure is themost positive in the conditions described in choice D.

5. The answer is D. The ratio for FEV1/FVC is 0.80 (80%)for healthy adults, including trained athletes. Thisvalue tends to decrease with age.

6. The answer is C. Emphysema is an obstructive disor-der that leads to highly compliant lungs, while pul-monary edema, fibrosis, congestion, and respiratorydistress syndrome are restrictive disorders that lead tostiff lungs with decreased compliance.

7. The answer is D. An increase in airway diameter low-ers airway resistance, which has the greatest effect onforced expiration. Total lung capacity, inspiratory ca-pacity, and tidal volume would not appreciablychange. FRC is high with asthma and would decreasewith a bronchodilator.

8. The answer is C. A restrictive lung disease causes a de-crease in FEV1, FVC, FRC, and RV. However, the ra-tio of FEV1/FVC is likely to be increased.

9. The answer is A. Minute ventilation is equal to expiredair per minute, tidal volume times frequency of breath-ing, or alveolar ventilation plus dead space ventilation.

10. The answer is D. Tidal volume � minute ventilation(8 L/min) � frequency (10 breaths/min) � 0.8L/breath.

11. The answer is E. Fibrosis leads to stiff lungs, resultingin reduced compliance and the need for more work toinflate the lungs. Stiffer lungs also have greater elasticrecoil, so the lungs will deflate easier.

12. The answer is D. There is no airflow during breathholding with an open glottis. Under these conditions,alveolar pressure equals atmospheric pressure.

13. The answer is E. CL � �V/�P � 0.5 L/5 cm H2O �0.1 L/cm H2O

14. The answer is C. PTP � PA � Ppl � �1 � (�7) cmH2O � � 6 cm H2O

15. The answer is D. TLC � VC � RV � 5.0 L � 1.2 L� 6.2 L

16. The answer is B. VD � V̇E � V̇A � 7.0 L/min � 5.0L/min � 2.0 L/min

Chapter 20

1. The answer is E. The pulmonary circulation is a high-flow, low-pressure, low-resistance, and high-compli-ance system.

2. The answer is D. Pulmonary vascular resistance de-creases with an increase in pulmonary arterial pressure.The primary reason is capillary recruitment, but it isalso due to capillary distension. Pulmonary vascular re-sistance is increased at low and high lung volumes (seeFig. 20.6) and by hypoxia.

3. The answer is D. The pulmonary and the systemic cir-culations both receive all of the cardiac output andhave the same flow. Pressure, resistance, and compli-ance are different.

4. The answer is B. The gravitational effect on the pul-monary circulation causes blood flow to be greatest at

the base. Vascular resistance is high at the apex be-cause alveolar pressure exceeds capillary pressure.

5. The answer is C. In the supine position, the heart is inthe middle of the chest. Pulmonary arterial pressure atthe top of the chest is 15 cm H2O minus 7.5 cm H2O� 7.5 cm H2O. Therefore, arterial pressure exceedsvenous pressure (7 cm H2O). Since alveolar pressure isless than venous pressure in a healthy individual, wehave the situation that Pa Pv PA, or a zone 3.There is no zone 4.

6. The answer is C. A drop in venous pressure has thegreatest effect in zone 3 because the pressure gradientfor flow is determined by the arterial-venous pressuredifference. In zone 1 there is no flow, and the pressuregradient for flow in zone 2 is the arterial-alveolar pres-sure difference.

7. The answer is A. At the base, both airflow and bloodflow are higher; however, blood flow exceeds airflowat the base, which results in a low V̇A/Q̇ ratio. At theapex, blood flow and airflow are lower than at the base,but airflow is greater than blood flow, which leads to ahigh V̇A/Q̇ .

8. The answer is C. The regional differences in bloodflow and airflow are the result of gravity.

9. The answer is C. The ventilation-perfusion ratio ishighest at the apex and lowest at the base of the lung.As a result, the lungs are overventilated at the apex rel-ative to blood flow; PO2 is high and PCO2 is low at theapex.

10. The answer is B. R � �P/Q̇ � (20 � 5 mm Hg)/5 Lper min � 3 mm Hg/L per min.

11. The answer is C. 20 cm H2O � 1.36 cm H2O per mmHg � 15 mm Hg.

12. The answer is C. �P � R � Q̇ � 4 mm Hg/L/min �5 L/min � 20 mm Hg.

Chapter 21

1. The answer is D. The A-aO2 gradient in a healthy per-son is due to both a low V̇A/Q̇ ratio at the base of thelungs and a small shunt from the bronchial circulation.

2. The answer is A. A decrease in the diffusion distancewill lead to an increase in DL. A decrease in capillaryblood volume, surface area, cardiac output, and bloodhemoglobin concentration will decrease DL.

3. The answer is D. The equilibrium curves are not simi-lar; that for CO2 is steeper and more linear. The bloodcarries more CO2 than O2. The presence of CO2 willincrease the P50. Although red cells carry most of the O2,the plasma carries the majority of the CO2 (mainly asbicarbonate).

4. The answer is A. A decrease in hemoglobin concentra-tion will decrease the O2 content, but will not affectthe oxygen saturation or PO2.

5. The answer is B. All will favor the unloading of oxygenfrom hemoglobin except a rise in pH.

6. The answer is D. A low V̇A/Q̇ ratio will cause hypox-emia, but it will have little effect on arterial PCO2 be-cause of the linearity of the CO2 equilibrium curve.Also, a low PaO2 stimulates ventilation, which pro-motes CO2 loss.

7. The answer is E. In a normal resting condition, theblood leaving the lungs is 98% saturated with oxygen,and the blood returning to the lungs is 75% saturatedwith oxygen. With vigorous exercise, blood leavingthe lungs is still 98% saturated, but blood returning isusually less than 75% saturated because more oxygenis unloaded from hemoglobin in exercising muscles.

8. The answer is B. Carbon monoxide will lower oxygencontent and saturation, but the arterial PO2 is un-changed. Airway obstruction or pulmonary edema willlower O2 saturation and arterial PO2.

9. The answer is D. A shunt, low V̇A/Q̇ ratio, and diffu-sion impairment all cause an increase in the A-aO2 gra-dient. The reason the A-aO2 gradient is normal withgeneralized hypoventilation is that both alveolar oxy-gen and arterial oxygen tension decrease together.

10. The answer is C. The alveolar gas equation is requiredto obtain the A-aO2 gradient. The alveolar gas equationis PAO2 � 150 mm Hg � 1.2 � PaCO2.

11. The answer is D. DL � V̇CO/PACO � 10 mL/0.5 mmHg � 20 mL/min per mm Hg.

Chapter 22

1. The answer is D. The basic rhythm exists in the ab-sence of the pontine respiratory group, afferent vagalinput to the pons and medulla, or an intact spinal cord.These can modify the rhythm of breathing but are notrequired.

2. The answer is A. A brief early burst by the inspiratoryneurons occurs with expiration.

3. The answer is D. An inverse relationship exists be-tween hypoxia-induced hyperventilation and oxygencontent. Hypoxia-induced hyperventilation is depend-ent on PaCO2 and more on carotid than aorticchemoreceptors.

4. The answer is D. Stimulation of lung C fibers will causebronchoconstriction, apnea, rapid shallow breathing,and skeletal muscle relaxation.

5. The answer is E. CSF and plasma differ in protein con-centration, PCO2, and electrolyte composition (includ-ing the [H�]).

6. The answer is B. Slow-wave sleep is characterized byperiodic breathing, hypercapnia, and a decreased sen-sitivity to hypoxia. The cough reflex is suppressed, andskeletal muscle relaxation is less than in REM sleep.

7. The answer is B. During sleep, airway irritation will notevoke a cough, but will evoke apnea and arousal. Air-way occlusion or hypercapnia will evoke arousal.

8. The answer is E. Negative-feedback systems are notnecessarily the most stable.

9. The answer is C. The control of ventilation by PaCO2

works primarily through the central chemoreceptors.However, the central effects are mediated indirectlythrough a change in CSF [H�], and the sensitivity isinversely related to PaO2.

10. The answer is B. Minute ventilation is inversely relatedto SaO2 and increases in linear fashion as SaO2 de-creases.

Chapter 23

1. The answer is C. Renal clearance is measured in vol-ume of plasma per unit time.

2. The answer is D. Na� reabsorption by collecting ductprincipal cells occurs via a Na� channel called ENaC(epithelial sodium channel). Na� reabsorption in prox-imal tubule cells is coupled to transport of solutes viacotransport (e.g., Na-glucose) and antiport (e.g.,Na�/H� exchanger) mechanisms. Na� reabsorption inthe thick ascending limb involves a Na-K-2Cl cotrans-porter and, in the distal convoluted tubule, a Na-Cl co-transporter. Collecting duct intercalated cells are pri-marily concerned with acid-base, not Na�, transport.

3. The answer is B. Amount � concentration � volumeor volume � amount/concentration. Volume � 570mosm/day �1,140 mosm/kg H2O � 0.5 kg H2O/day(or, 0.5 L/day because urine is mostly water and a literof urine weighs about 1 kg).

4. The answer is D. Long loops of Henle are associatedwith a steep gradient in the medulla because there ismore opportunity for countercurrent multiplication. Adrug that inhibits Na� reabsorption by the thick as-cending limb will reduce the single effect, resulting ina loss of the medullary gradient. A very low GFR resultsin inadequate input of solute into the medulla and a di-minished ability to concentrate the urine. Excess waterintake causes the medullary gradient to fall because toomuch water is added to the medulla. A protein-defi-cient diet results in less urea production by the liverand less urea accumulation in the kidney medulla.

5. The answer is A. The decrease in vascular resistanceleads to an increase in glomerular blood flow.Glomerular capillary pressure will fall, however, andconsequently, GFR will fall. The filtration fraction(GFR/RPF) will fall because GFR falls and RPF rises.Less fluid is filtered into the space of Bowman’s cap-sule, so the hydrostatic pressure there should fall.

6. The answer is B. Active reabsorption of Na�, poweredby the Na�/K�-ATPase, is the main driving force forwater reabsorption. Reabsorption of amino acids andwater is secondary to active Na� reabsorption. Thereis no active water reabsorption, and pinocytosis is toosmall to account for appreciable water reabsorption.The high colloid osmotic pressure in peritubular capil-laries favors uptake of reabsorbed fluid from the renalinterstitial fluid, but does not cause the removal of fluidfrom the proximal tubule lumen.

7. The answer is B. The percentage excretion is equal to100 � excreted Na�/filtered Na� � 100 � (UNa � V̇)

(PNa � GFR) � 100 � (UNa � V̇) � PNa � (UIN �V̇/PIN) � 100 � UNa/PNa � UIN/PIN � 7,000/140 �10/1 � 5.

8. The answer is D. In the autoregulatory range, vascularresistance falls when arterial blood pressure falls.Changes in vessel caliber primarily occur in vessels up-stream to the glomeruli (cortical radial arteries and af-ferent arterioles). Because autoregulatory range ex-tends from an arterial blood pressure of about 80 to180 mm Hg, renal blood flow is not maintained whenblood pressure is low; in fact, the sympathetic nervoussystem will be activated and cause intense vasocon-

APPENDIX A Answers to Review Questions 721

722 APPENDICES

striction in the kidneys. Renal autoregulation does notdepend on nerves.

9. The answer is B. When the kidney is producing maxi-mally concentrated urine, fluid in the cortical collectingduct becomes isosmotic with the surrounding cortical in-terstitial fluid. Therefore, the osmolality will be about300 mosm/kg H2O; it cannot go above this value becausehyperosmotic values (compared to systemic bloodplasma) can be produced only in the kidney medulla.

10. The answer is B. The patient is older and severely dehy-drated; the GFR can be expected to be low. Conse-quently, the proximal tubules may be able to reabsorb allof the filtered glucose (because the filtered load is re-duced), even though the plasma [glucose] is elevated. Ifsplay is increased, glucose Tm is low, or threshold islow, glucose should be present (not absent) from theurine. An abnormally high glucose Tm would reduceglucose excretion; however, in the scenario presented,this is not a likely cause of the absence of glucose in theurine.

11. The answer is D. Excretion of phenobarbital is pro-moted by increasing urine output and making the urinemore alkaline. The latter would keep phenobarbital inits anionic form, which is not reabsorbed by the kidneytubules.

12. The answer is C. Inulin clearance is the standard formeasuring GFR. PAH clearance is used to measure re-nal plasma flow, not GFR.

13. The answer is C. Liddle’s syndrome is due to excessiveactivity of the Na� channel in collecting duct principalcells, leading to salt retention and hypertension. Bart-ter and Gitelman syndromes are salt-wasting disorders;blood pressure would tend to be low, not high. Dia-betes insipidus and renal glucosuria produce excessivefluid loss and would not be likely causes of the patient’shypertension.

14. The answer is A. In the absence of arginine vaso-pressin, the kidneys produce a large volume of osmot-ically dilute urine.

15. The answer is C. The renal clearance of PAH is thehighest (it is nearly equal to the renal plasma flow) be-cause PAH is not only filtered by the glomeruli but isalso secreted vigorously by proximal tubules. Creati-nine is filtered and secreted, to a small extent only, inthe human kidney. Inulin is only filtered. Urea is fil-tered and variably reabsorbed; its clearance is alwaysbelow the inulin clearance in people. Na� has the low-est clearance of all because filtered Na� is extensivelyreabsorbed.

16. The answer is A. The filtered load of the substance isPx � GFR � 2 mg/mL � 100 mL/min � 200 mg/min.The rate of excretion is Ux � � 10 mg/mL � 5 mL/min� 50 mg/min. Hence, more substance X was filteredthan was excreted, and the difference, 200 mg/min �50 mg/min � 150 mg/min, gives the rate of tubular re-absorption of substance X.

17. The answer is C. The true renal plasma flow (RPF) �CPAH/EPAH � UPAH � V̇/PPAH � (Pa

PAH � PrvPAH)/Pa-

PAH � 0.60 � 5.0/0.02 � (0.02 � 0.01)/0.02 � 300mL/min. The renal blood flow � RPF/(1 � hematocrit)� 300/(1 � 0.40) � 500 mL/min.

18. The answer is C. There is an inverse hyperbolic rela-tionship between plasma [creatinine] and GFR and,therefore, a rise in plasma [creatinine] is associatedwith a fall in GFR (see Fig. 23.7). The greatest absolutechange in GFR occurs when plasma [creatinine] dou-bles starting from a normal GFR and plasma [creati-nine].

19. A is the answer. Granular cells (also known as juxta-glomerular cells) are located primarily in the wall of af-ferent arterioles and are the major site of renin synthe-sis and release.

20. The answer is E. GFR � Kf (PGC � PBS � COP).Therefore, Kf � 42 nL/min � (50 � 12 � 24) mm Hg� 3.0 nL/min per mm Hg.

Chapter 24

1. The answer is B. ICF volume is calculated by subtract-ing ECF volume from the total body water. The otherfluid volumes can be determined from the volume ofdistribution of a single indicator, such as radioactivesulfate for ECF volume, radioiodinated serum albuminfor plasma volume, and deuterium oxide for total bodywater.

2. The answer is A. Cardiac failure results in a decrease ineffective arterial blood volume, which stimulates thirst.Because angiotensin stimulates thirst, a low plasmalevel would have the opposite effect. Distension of theatria (increased blood volume) or stomach inhibitsthirst. Volume expansion and a low plasma osmolalityboth inhibit thirst.

3. The answer is B. AVP is synthesized in the cell bodiesof nerve cells located in the supraoptic and paraven-tricular nuclei of the anterior hypothalamus.

4. The answer is D. From the indicator dilution method,the plasma volume � 10 Ci � 4 Ci/L� 2.5 L. If thehematocrit ratio is 0.4, then the blood volume � 2.5 Lplasma � 0.6 L plasma per L blood � 4.17 L.

5. The answer is A. An increase in central blood volumewill stretch the atria, cause the release of atrial natri-uretic peptide, and result in diminished Na� reabsorp-tion. All other choices produce increased tubular Na�

reabsorption.6. The answer is B. The loop of Henle (mostly the thick

ascending limb) reabsorbs about 65% of the filteredMg2�.

7. The answer is C. Infusion of isotonic saline tends toraise blood pressure, decrease renal sympathetic nerveactivity, and increase fluid delivery to the maculadensa; all of these changes suppress renin release. Allother choices result in increased renin release.

8. The answer is E. Skeletal muscle cells contain largeamounts of K�; injury of these cells can result in addi-tion of large amounts of K� to the ECF. Insulin, epi-nephrine, and HCO3

� promote the uptake of K� bycells. Hyperaldosteronism causes increased renal ex-cretion of K� and a tendency to develop hypokalemia.

9. The answer is B. PTH inhibits tubular reabsorption ofphosphate, stimulates tubular reabsorption of Ca2�,and increases bone resorption. PTH secretion is in-

creased in patients with chronic renal failure. Its secre-tion is stimulated by a fall in plasma ionized Ca2�.

10. The answer is D. Aldosterone increases K� secretionand Na� reabsorption by cortical collecting ducts. Itdoes not affect water permeability.

11. The answer is B. Autoregulation refers to the relativeconstancy of renal blood flow and GFR despitechanges in arterial blood pressure. Mineralocorticoidescape refers to the fact that the salt-retaining action ofmineralocorticoids does not persist but is overpoweredby factors that promote renal Na� excretion. Satura-tion of transport occurs when the maximal rate of tu-bular transport is reached. Tubuloglomerular feedbackresults in afferent arteriolar constriction when fluid de-livery to the macula densa is increased; it contributes torenal autoregulation.

12. The answer is C. Nephrogenic diabetes insipidus ischaracterized by increased output of dilute urine.Plasma AVP is elevated because of the volume deple-tion. Plasma osmolality is on the high side of the nor-mal range because of the loss of dilute fluid in theurine. The increased urine output is not due to diabetesmellitus because there is no glucose in the urine andthe urine is very dilute. Diuretic drug abuse should notproduce very dilute urine because Na� reabsorption isinhibited. Neurogenic diabetes insipidus is unlikely be-cause the plasma AVP level is reduced in this case. Pri-mary polydipsia produces output of a large volume ofdilute urine, but plasma osmolality and AVP levels aredecreased.

13. The answer is E. Na� is the major osmotically activesolute in the ECF and is the major determinant of theamount of water in and, hence, volume of this com-partment.

14. The answer is A. Although the plasma osmolality is ex-traordinarily high, the plasma Na�, glucose, and BUNare normal. This indicates the presence of anothersolute (it could be ethanol) in the plasma. The calcu-lated osmolality � 2 � [Na�] � [glucose]/18 �[BUN]/2.8 � 280 � 5.6 � 5.4 � 291 mosm/kg H2O,a lot less than the measured osmolality (370 mosm/kgH2O). Simple dehydration would cause a rise inplasma [Na�]. Diabetes insipidus or diabetes mellituscannot explain the high osmolality. The normal BUNdoes not support the existence of renal failure.

15. The answer is B. The inhibitor will block the conver-sion of angiotensin I to angiotensin II, and therefore,the plasma angiotensin I level will rise and the plasmaangiotensin II and aldosterone levels will fall. Theplasma bradykinin level will rise because the convert-ing enzyme catalyzes the breakdown of this hormone.The plasma renin level will rise because (1) the fall inblood pressure stimulates renin release, and (2) an-giotensin II directly inhibits renin release by acting onthe granular cells of afferent arterioles, so that this in-hibition is removed when less angiotensin II is present.

16. The answer is D. In response to an increase in dietaryK� intake, the cortical collecting duct principal cellsincrease the rate of K� secretion, accounting for mostof the K� excreted in the urine.

17. The answer is D. The subject in choice D has a low

plasma osmolality but inappropriately concentratedurine. The subject in choice A may have diabetes in-sipidus. The subject in choice B has a low plasma os-molality, but the urine osmolality is appropriately low.The subjects in choices C and E are normal, althoughthe subject in choice E is producing concentrated urineand may be water-deprived.

18. The answer is A. The low blood pH and hyper-glycemia (or hyperosmolality) would tend to raiseplasma [K�], yet the plasma [K�] is normal. Thesefindings suggest that the total body store of K� is re-duced. Remember that most of the body’s K� is withincells. In uncontrolled diabetes mellitus, the osmotic di-uresis (increased Na� and water delivery to the corti-cal collecting ducts), increased renal excretion ofpoorly reabsorbed anions (ketone body acids), and el-evated plasma aldosterone level (secondary to volumedepletion) would all favor enhanced excretion of K�

by the kidneys. The subject has normokalemia, not hy-pokalemia or hyperkalemia.

19. The answer is D. Isotonic saline does not change cellvolume. The plasma AVP level will fall because of vol-ume expansion and cardiovascular stretch receptor in-hibition of its release. The plasma aldosterone levelwill be low because of inhibited release of renin andless angiotensin II formation. The plasma ANP levelwill be increased from stretch of the cardiac atria. Alarge part of the infused isotonic saline will be filteredthrough capillary walls into the interstitial fluid.

20. The answer is A. ECF volume and blood volume are in-creased, but these should promote Na� excretion, notlead to Na� retention by the kidneys. A decrease in ef-fective arterial blood volume is the best explanation forrenal Na� retention.

Chapter 25

1. The answer is D. Using the Henderson-Hasselbalchequation, 6.0 � 9.0 � log ([NH3]/[NH4

�]),[NH3]/[NH4

�] � 10�3.0 � 1:1,000.2. The answer is B. The plasma [HCO3

�] is easily calcu-lated from the formula: [H�] � 24 � PCO2 /[HCO3

�],so [HCO3

�] � 24 � 24/48 � 12 mEq/L. Alternatively,the Henderson-Hasselbalch equation could be used,but it requires the use of logarithms.

3. The answer is E. The collecting duct is lined by a tightepithelium and can lower the urine pH to 4.5 (a tubulefluid/plasma [H�] ratio of 102.9/1 or about 800/1 ifplasma pH is 7.4). The proximal convoluted tubule islined by a leaky epithelium and can lower tubule fluidpH to about 6.7 (a tubule fluid/plasma [H�] ratio of100.7/1 or 5/1 when plasma pH is 7.4). Other nephronsegments beyond the proximal convoluted tubule donot lower tubular fluid pH as much as the collectingducts.

4. The answer is A. The kidneys filter about 4,320mEq/day of HCO3

� and usually reabsorb all but a fewmEq/day; reabsorption of HCO3

� occurs via H� se-cretion and consumes the bulk of secreted H�. A typ-ical excretion rate for NH4

� is about 50 mEq/day; fortitratable acid about 25 mEq/day. The quantity of free

APPENDIX A Answers to Review Questions 723

724 APPENDICES

H� in a typical urine sample (e.g., 2 L/day, pH 6.0) isnegligible (e.g., 0.002 mEq/day).

5. The answer is C. Net acid excretion is calculated from:urinary titratable acid � urinary NH4

� � urinaryHCO3

� excretion � 30 � 60 � 2 � 88 mEq/day, inthis case.

6. The answer is E. In the process of excreting titratableacid and ammonia, the kidneys generate and add to theblood an equivalent amount of new HCO3

�. There-fore, the answer is 200 � 500 � 700 mEq.

7. The answer is E. When Na� reabsorption is stimu-lated, Na�/H� exchange is increased, resulting ingreater H� secretion in the proximal tubule and loopof Henle. Additionally, increased Na� reabsorption inthe collecting ducts renders the duct lumen more neg-ative, which favors H� secretion. All of the other fac-tors result in decreased H� secretion.

8. The answer is C. The subject has a severe metabolicacidosis. The anion gap (140 � 105 � 6 � 19 mEq/L)is high. Methanol intoxication (see Table 25.5) pro-duces this type of acid-base disturbance, resultingmainly from formic acid production. Acute renal fail-ure would also produce a high anion gap metabolic aci-dosis, but because the BUN is normal, this is unlikely.Uncontrolled diabetes mellitus also produces a highanion gap metabolic acidosis, but because the plasmaglucose is normal, this is unlikely. Diarrhea produces ametabolic alkalosis. A drug that depresses breathingproduces retention of CO2 and respiratory acidosis.

9. The answer is C. Because of the low ambient baromet-ric pressure and oxygen tension at high altitude, hy-poxia develops. Therefore, we can immediately ruleout choices A and D. Choice B is a subject with hy-poxia that resulted from inadequate ventilation; thissubject has CO2 retention and a respiratory acidosis.Hypoxia stimulates ventilation and results in a lowPCO2 and respiratory alkalosis. Choice E shows valuesfor an acute respiratory alkalosis; the plasma [HCO3

�]has been lowered by 4 mEq/L, corresponding to the 20mm Hg decrease below normal in PCO2 (see Table25.4). Choice C shows typical values for a chronic(one week) respiratory alkalosis; the kidneys have fur-ther lowered the plasma [HCO3

�] and reduced theseverity of the alkalemia.

10. The answer is D. Aspirin (salicylate) intoxication pro-duces a mixed acid-base disturbance—respiratory alka-losis (as a result of stimulation of the respiratory cen-ter) and metabolic acidosis (as a result of inhibition ofoxidative metabolism and accumulation of lactic andketone body acids). The respiratory alkalosis predom-inates during the first several hours in adults; metabolicacidosis occurs at the same time and becomes over-whelming late in the course of the intoxication. ChoiceD shows the predominant respiratory alkalosis; the re-duction in plasma [HCO3

�] is accounted for by the ac-cumulation of organic acids in the blood and is tooearly to reflect significant renal compensation. ChoiceA represents metabolic acidosis with normal respira-tory compensation. Choice B represents respiratoryacidosis as a result of alveolar hypoventilation or a mis-match between alveolar ventilation and pulmonary

capillary blood flow; note the abnormally low PO2.Choice C represents the normal condition for arterialblood. Choice E represents simple acute respiratory al-kalosis.

Chapter 26

1. The answer is B. Successive small intestinal structuresbetween the serosa and mucosa are longitudinal mus-cle, myenteric plexus, a network of interstitial cells ofCajal network, circular muscle, submucous plexus, andmuscularis mucosae.

2. The answer is D. Interstitial cells of Cajal are pace-maker cells that generate electrical slow waves. Theother cell types do not generate electrical slow waves.

3. The answer is C. Inhibitory motor neurons determinewhen electrical slow waves trigger contractions. Dam-age to the enteric nervous system, including the in-hibitory motor neurons, frees the musculature from in-hibition. In the absence of inhibition, the musclecontracts continuously in a disorganized manner. Ef-fective propulsion is impossible in the absence of theENS.

4. The answer is C. Of the possible choices, only cellbodies in the dorsal vagal nucleus have axons ending inthe wall of the stomach.

5. The answer is A. Fast EPSPs in the ENS are mediatedmainly by nicotinic receptors for ACh. Hyperpolariz-ing after-potentials reduce excitability. Metabotropicreceptors stimulate adenylyl cyclase. Fast EPSPs arenot hyperpolarizing potentials.

6. The answer is C. Suppression of EPSPs by NE could bethrough an action at the presynaptic site of ACh re-lease or an action at the postsynaptic membrane. Thefinding that NE does not affect the action of exoge-nously applied ACh, blocking the fast EPSP indicatesthat the mechanism of suppression of the EPSPs is sup-pression of ACh release at the synapse.

7. The answer is D. Once triggered by the stimulus, theaction potential travels from muscle fiber to musclefiber as the ionic current travels across gap junctions.Gap junctions account for the functional electricalsyncytial properties of smooth muscle. Nerve fibersand the release of neurotransmitter cannot account forthe spread of the action potential and associated con-traction because tetrodotoxin blocked all neural func-tion. Interstitial cells of Cajal is not correct because theaction potential traveled from cell to cell in the bulk ofthe smooth muscle. Electrical slow waves are not cor-rect because the action potential was triggered by astimulus applied at one point, not slow waves originat-ing along the segment of intestine.

8. The answer is E. Rapid transit is not likely because theloss of inhibitory motor neurons results in delayedtransit (i.e., pseudoobstruction). Accelerated gastricemptying does not occur mainly because pseudoob-struction in the duodenum presents a high resistance toinflow from the stomach. Gastroesophageal reflux isnot correct because in the absence of inhibition, thelower esophageal sphincter remains contracted and is abarrier to reflux. Diarrhea is unlikely because diarrhea

requires intestinal propulsion and this is compromisedby the loss of inhibitory motor neurons. Inhibitory mo-tor neurons are necessary for the relaxation of sphinc-ters.

9. The answer is D. Longitudinal muscle is relaxed andcircular muscle is contracted in the propulsive seg-ment. Longitudinal muscle contracts and circular mus-cle is inhibited in the receiving segment.

10. The answer is A. Choice A is correct because sphinc-ters function to prevent reflux; therefore, flow across asphincter is generally unidirectional. Choice B is notcorrect because tone in the lower esophageal sphincteris increased during the MMC in the stomach. ChoiceC is incorrect because the sphincter cannot be relaxedafter blockade of the inhibitory innervation by a localanesthetic. Choice D is incorrect because pressure inthe sphincter is higher than in the two compartmentsit separates. Choice E is incorrect because inhibitoryneurons fire to relax the sphincter during a swallow.

11. The answer is D. Physiological ileus is defined as theabsence of contractile activity. It is a significant behav-ior pattern, requiring a functional ENS. Each of theother neurally programmed patterns involves contrac-tile behavior and motility.

12. The answer is D. Gastric emptying of particles greaterthan about 7 mm does not occur during the digestivestate. The lag phase is the time required for the stom-ach to grind large particles into smaller particles in thissize range. Choice A is not correct because conversionfrom interdigestive to digestive states occurs immedi-ately upon the first few swallows of a meal. Choice B isincorrect because cephalic and gastric phases of acidsecretion reach maximum near the onset of the lagphase. Choice C is incorrect because the lag phase is atthe beginning of the emptying curve, not at the end.Choice E is incorrect because the lag phase does notapply for a liquid meal.

13. The answer is A. The plateau phase of the gastric ac-tion potential and the associated trailing contractionincrease in direct relation to the amount of ACh re-leased by excitatory motor neurons to the antral mus-culature. The higher the firing frequency of the excita-tory motor neurons, the more ACh is released. ChoiceB is not correct because the release of NE from sympa-thetic postganglionic neurons decreases the amplitudeof the plateau phase of the gastric action potential.Choice C is incorrect because the firing frequency ofthe pacemaker does not affect the amplitude of theplateau phase. Opening of the pylorus cannot affectthe trailing contraction; nevertheless, the pylorus isclosed as the trailing contraction approaches. Choice Eis incorrect because firing of the motor neurons to thegastric reservoir does not directly influence the inner-vation of the antral pump.

14. The answer is D. Lipids (fats) have the greatest effectin slowing gastric emptying because they have thehighest caloric content. Decreased pH in the duode-num is also a powerful suppressant of gastric emptying.However, the question asks about an ingested meal,not conditions in the duodenum.

15. The answer is A. As the gastric reservoir fills during a

meal, mechanoreceptors signal the CNS. When thelimits of adaptive relaxation in the reservoir arereached, signals from the stretch receptors in the reser-voir’s walls account for the sensations of fullness andsatiety. Overdistension is perceived as discomfort.Adaptive relaxation appears to malfunction in theforms of functional dyspepsia characterized by thesymptoms described in this question. If adaptive relax-ation is compromised (e.g., by an enteric neuropathy),mechanoreceptors are activated at lower distendingvolumes and the CNS wrongly interprets the signals asif the gastric reservoir were full. None of the otherchoices would be expected to activate mechanosen-sory signaling of the state of fullness of the gastricreservoir.

16. The answer is E. Power propulsion is the pattern ofmotility for defense of the intestinal tract. It occurs inthe retrograde direction during emetic responses thatempty the lumen of threatening material in the uppersmall intestine. It occurs in the orthograde direction inthe lower small intestine and in the large intestinewhere it also functions to quickly eliminate threateningsubstances or organisms from the intestine. In the largeintestine, secretion flushes the material from the mu-cosa and holds it in suspension in the lumen. This isfollowed by power propulsion, which rapidly clearsthe lumen of the material. This form of behavior is de-fensive but has the adverse effects of diarrhea and ab-dominal pain. None of the other choices evokes con-scious sensations during daily occurrence.

17. The answer is D. Observations on the transit of mark-ers after instillation in the human cecum show that themarkers remain for the longest time in the transversecolon. Transit is significantly faster in the other partsof the large intestine.

18. The answer is D. Examination of older patients oftenreveals weakness in the pelvic floor musculature.Weakness in the puborectalis muscle allows theanorectal angle to straighten and lose its barrier func-tion to the passage of feces into the anorectum. ChoiceA is incorrect because the rectoanal reflex (i.e., relax-ation of the internal anal sphincter in response to dis-tension of the rectum) does not weaken significantly inolder persons. Choice B is incorrect because a deficit insensory detection, not elevated sensitivity, can be afactor in fecal incontinence. Choice C is incorrect be-cause adult Hirschsprung’s disease results in constipa-tion, not incontinence. The myopathic form ofpseudoobstruction is not associated with fecal inconti-nence because propulsive motility is absent as a resultof weakening of the intestinal smooth muscle.

Chapter 27

1. The answer is D. Salivary secretion is exclusively un-der neural control. The others need both neural andhormonal stimulation and are, therefore, only partiallystimulated by the sight, smell, and chewing of food(cephalic phase). The sight, smell, and chewing offood stimulate the parasympathetic nervous system,which stimulates salivary secretion.

APPENDIX A Answers to Review Questions 725

726 APPENDICES

2. The answer is C. The uptake of bile acid by hepato-cytes is sodium-dependent and is not dependent oncalcium, iron, potassium, or chloride.

3. The answer is A. Intrinsic factor is critical for the ab-sorption of vitamin B12 by the ileum. None of the othersubstances is secreted by parietal cells. Gastrin, so-matostatin, and CCK are secreted by specialized GIendocrine cells, whereas chylomicrons are producedby enterocytes.

4. The answer is C. Although the cephalic and intestinalphases stimulate gastric secretion, the gastric phase is,by far, the most important.

5. The answer is B. Carbonic anhydrase catalyzes the for-mation of carbonic acid from carbon dioxide and wa-ter. It is not involved in the formation of carbon diox-ide from carbon and oxygen, bicarbonate ion fromcarbonic acid, hydrochloric acid, or hypochlorousacid.

6. The answer is B. Parasympathetic stimulation inducesthe release of kallikrein by the salivary acinar cells,which converts kininogen to form lysyl-bradykinin (apotent vasodilator). Bradykinin is a vasoactive peptide.Kininogen is the precursor for kinins. Kinins includebradykinin and lysyl-bradykinin. Aminopeptidase re-leases amino acids from the amino end of peptides andis found in the brush border membrane and cytoplasmof enterocytes.

7. The answer is D. Intrinsic factor is secreted by theparietal cells of the stomach and is not secreted by thesalivary glands. Lactoferrin, amylase, mucin, and mu-ramidase are found in saliva.

8. The answer is B. In the fasting state, the pH of thestomach is low, between 1 and 2.

9. The answer is A. Salivary secretion is inhibited by at-ropine. Atropine is an anticholinergic drug that com-petitively inhibits ACh at postganglionic sites, inhibit-ing parasympathetic activity. Pilocarpine actuallystimulates salivation because of its muscarinic action.Cimetidine is an antagonist for the histamine H2 re-ceptor. Aspirin is the most widely used analgesic (painreducer), antipyretic (fever reducer), and anti-inflam-matory drug. Omeprazole inhibits the H�/K�-ATPaseand, thus, inhibits acid secretion.

10. The answer is C. The chief cells of the stomach secretepepsinogen, and the parietal cells of the stomach se-crete hydrochloric acid and intrinsic factor. Gastrinand CCK are secreted by specialized endocrine cells.

11. The answer is B. Histamine interacts with its receptorin parietal cells to increase the intracellular cAMP. His-tamine does not cause an increase in intracellularsodium or cGMP or a decrease in intracellular calcium.

12. The answer is D. When the pH of the stomach fallsbelow 3, the antrum secretes somatostatin, which actslocally to inhibit gastrin release; therefore, somato-statin inhibits gastric secretion. Enterogastrones arehormones produced by the duodenum that inhibit gas-tric secretion and motility. Intrinsic factor is involvedin the absorption of vitamin B12 and is not involved inthe release of gastrin. Secretin is a hormone secretedby the duodenal and jejunal mucosa when exposed toacidic chyme and is responsible for stimulating pancre-

atic secretion rich in bicarbonate. CCK stimulates thegallbladder to contract and the pancreas to secrete ajuice rich in enzymes.

13. The answer is B. Secretin stimulates secretion of a bi-carbonate-rich pancreatic juice. Somatostatin, gastrin,and insulin do not. CCK stimulates a pancreatic secre-tion rich in enzymes and potentiates the action of se-cretin.

14. The answer is C. Excessive production of gastrin re-sults in acid hypersecretion and peptic ulcer disease.Patients with Zollinger-Ellison syndrome do not sufferfrom excessive acid reflux, excessive secretion of CCK,failure of the liver to secrete VLDLs, or failure to se-crete a bicarbonate-rich pancreatic juice.

15. The answer is B. Lactase hydrolyzes lactose to formboth glucose and galactose. None of the other combi-nations is correct.

16. The answer is A. Maltase hydrolyzes maltose to formglucose. Because maltose does not contain galactose orfructose, none of the other choices is correct.

17. The answer is C. Fructose is taken up by enterocytesby facilitated diffusion. Both glucose and galactose aretaken up by enterocytes through a sodium-dependenttransporter. Xylose and sucrose are not taken up by en-terocytes.

18. The answer is D. Pancreatic lipase hydrolyzes triglyc-eride to form 2-monoglyceride and two fatty acids.The hydrolysis of phosphatidylcholine, not triglyc-eride, results in the formation of lysophosphatidyl-choline. Although diglyceride is an intermediate in thehydrolysis of triglyceride by pancreatic lipase, the hy-drolysis continues until 2-monoglyceride and fattyacids are formed. Pancreatic lipase does not hydrolyzetriglyceride totally to form glycerol and fatty acids.

19. The answer is C. The small intestine transports dietarytriglyceride as chylomicrons in lymph. VLDLs are se-creted by the small intestine during fasting. Althoughsome dietary fatty acids are transported in the portalblood bound to albumin, it is not the predominantpathway for the transport of dietary lipids to the circu-lation by the small intestine. The intestine does not se-crete LDLs, and although it does secrete HDLs, theyare not used as a vehicle for transporting dietary lipidsto the blood by the small intestine.

20. The answer is C. Amino acids, as well as dipeptidesand tripeptides, use different brush border transportersfor their uptake. Dipeptides and tripeptides are nottaken up passively by any part of the GI tract.

21. The answer is A. Dietary protein is transported in theportal blood as free amino acids. Although dipeptidesand tripeptides are taken up by enterocytes, they arehydrolyzed by the brush border membrane, as well asby cytoplasmic peptidase to form free amino acids.

22. The answer is D. Vitamin B1 is a water-soluble vitamin.Vitamins A, D, E, and K are all fat-soluble vitamins.

23. The answer is B. Vitamin D plays an important indi-rect role in the absorption of calcium by the GI tract.The other vitamins listed are not involved in the ab-sorption of calcium.

24. The answer is C. Vitamin A is transported in chylomi-crons as ester. Vitamins D, E, and K are transported in

the free form associated with chylomicrons. VitaminB12, a water-soluble vitamin, is transported in the bloodbound to transcobalamin.

25. The answer is C. Potassium is passively absorbed bythe jejunum. The other choices do not apply to the ab-sorption of potassium by the small intestine.

26. The answer is C. Ascorbic acid enhances iron absorp-tion mostly by its reducing capacity, keeping iron inthe ferrous state. Ascorbic acid does not enhance hemeiron absorption, nor does it affect heme oxygenase ac-tivity or the production of ferritin or transferrin.

Chapter 28

1. The answer is D. Alcohol dehydrogenase catalyzes theconversion of alcohol to acetaldehyde, which is thenconverted to acetate. Acetate is then metabolized byhepatocytes. Cytochrome P450 is a primary compo-nent of the oxidative enzyme system involved in themetabolism of drugs. NADPH-cytochrome P450 re-ductase is an enzyme involved in phase I reactions ofdrug metabolism. There is no such enzyme as alcoholoxygenase. Glycogen phosphorylase is an enzyme in-volved in glycogen breakdown, not alcohol metabo-lism.

2. The answer is C. Unlike patients who have diabetes,healthy humans are capable of keeping their blood glu-cose within a relatively narrow range after a meal, 120to 150 mg/dL. Blood levels of 30 to 50 mg/dL and 50to 70 mg/dL indicate hypoglycemia, and blood levelsof 220 to 250 mg/dL and 300 to 350 mg/dL indicatehyperglycemia.

3. The answer is A. The liver has the enzyme glucose-6-phosphatase, but muscle does not. Consequently, mus-cle is incapable of releasing glucose from glucose 6-phosphate. Glucose undergoes reactions other thanglycolysis. Both liver and muscle have glucose-1-phos-phatase and glycogen phosphorylase enzymes. Thesynthesis of glucose, called gluconeogenesis, is carriedout mostly in the liver and, to some extent, in the kid-neys.

4. The answer is A. Fatty acid synthesis occurs only in thecytoplasm. Mitochondria are involved in fatty acid ox-idation rather than synthesis. Fatty acid synthesis doesnot occur in the nucleus. Endosomes and the Golgi ap-paratus are not involved in fatty acid synthesis.

5. The answer is B. Although both chylomicrons andVLDLs are triglyceride-rich lipoproteins, the liver, un-like the small intestine, produces only VLDLs. LDLsand HDLs are not triglyceride-rich lipoproteins. Chy-lomicron remnants are generated in the circulation bythe metabolism of chylomicrons.

6. The answer is C. Both urea and glutamine play an im-portant role in the storage and transport of ammonia inthe blood. Histidine, phenylalanine, methionine, andlysine are not involved in ammonia transport.

7. The answer is C. The liver is one of the major sites forthe removal of hormones, including glucagon. Conse-quently, patients with a portacaval shunt have highlevels of circulating glucagon and other hormones be-cause portal blood bypasses the liver. Choice A is in-

correct because the pancreas does not produce moreglucagon in portacaval shunt patients. Choice B is in-correct because the kidneys are capable of removingglucagon in these patients. However, the kidneys arenot nearly as important as the liver in removingglucagon in healthy individuals. Choice D is incorrectbecause the small intestine does not produce glucagon.Choice E is incorrect because blood flow to the smallintestine is not compromised in portacaval shunt pa-tients.

8. The answer is C. The liver makes transferrin to carryiron in the blood. Hemosiderin is an intracellular com-plex of ferric hydroxide, polysaccharides, and proteins.Haptoglobin binds free hemoglobin in the blood.Ceruloplasmin is a circulating plasma protein involvedin the transport of copper. Lactoferrin is an iron-bind-ing glycoprotein found in secretions (e.g., milk, saliva)and in neutrophil granules; it appears to contribute toantimicrobial host defenses.

9. The answer is A. Smokers inhale polycyclic aromatichydrocarbons, which stimulate drug-metabolizing en-zymes. Therefore, smokers have higher levels of he-patic drug-metabolizing enzymes than nonsmokers.The level of drug-metabolizing enzymes in the liver islowered by malnutrition and is lower in the newborn.

10. The answer is C. Phase I reactions of drug metabolismrefer to the addition of one or more polar groups to thedrug molecule. Hydrophilic, not hydrophobic, groupsare introduced into the drug molecule in a phase I re-action. The conjugation of drugs with glucuronic acid,glycine, taurine, or sulfate is a phase II reaction.

11. The answer is C. A healthy liver converts vitamin D(cholecalciferol) to form 25-hydroxycholecalciferol,but a diseased liver has a reduced capacity to do so.The kidney, not the liver, is responsible for the con-version of 25-hydroxycholecalciferol to 1,25-dihy-droxycholecalciferol. Vitamin D, not 1,25-hydroxyc-holecalciferol, is absorbed by the small intestine.

12. The answer is A. LDLs are removed from the blood bythe liver by binding to LDL receptors, followed by en-docytosis of the LDL-receptor complex. LDLs do notbind to HDL receptors, albumin, transferrin, or cerulo-plasmin.

Chapter 29

1. The answer is C. Antipyretics, such as aspirin, inhibitthe synthesis of prostaglandin E2, which mediates theelevation of the thermoregulatory set point duringfever. Antipyretics cannot prevent the increase in coretemperature during exercise because that increase isnot produced by an elevated thermoregulatory setpoint (see Fig. 29.11). Therefore, considerations of thepossible harm or benefits as a result of the increase, asin choices A and B, are irrelevant. As antipyretics donot directly stimulate heat loss responses, choices Dand E are not applicable.

2. The answer is D. Blood vessels in the skin have a dualnervous control, but both vasoconstriction and activevasodilation are mediated by sympathetic fibers. Thenerve endings that control sweating are also part of the

APPENDIX A Answers to Review Questions 727

728 APPENDICES

sympathetic nervous system, although they releaseacetylcholine. Sympathectomy abolishes sweating,vasoconstriction, and active vasodilation.

3. The answer is C. The core temperature of a resting per-son shows a circadian rhythm and is higher at 4:00 PM

than at 4:00 AM. This rhythm in core temperature is theresult of an underlying circadian rhythm in the ther-moregulatory set point. Because a change in the ther-moregulatory set point affects core temperature at restand the thresholds for sweating and vasodilation all inthe same way, these thresholds are also higher at 4:00PM.

4. The answer is E. Acclimatization to cold produces sev-eral different (and contrasting) sets of changes, de-pending on the acclimatizing environment (and, per-haps, on characteristics of the population beingacclimatized).

5. The answer is B. Fever enhances the body’s defenses,partly by magnifying the responses of leukocytes andmacrophages to the other stimuli that are operativeduring an immune response. Choice E reflects whatwas widely believed until the 1970s. Although a fewpathogens do not flourish at temperatures above 37�C,they are so much the exception that A is not the bestchoice.

6. The answer is C. The classic changes observed in heatacclimatization are lower heart rate during exercise; anincreased sweating response; and a lower core tempera-ture during exercise, which is due to both the increasedsweating response and a lower thermoregulatory setpoint. In addition, salt is conserved by a reduced saltconcentration in sweat.

7. The answer is B. Before the hike, the total osmotic con-tent of the body was 40 L � 280 mosm/L � 11,200mosm (assuming that plasma osmolarity � 2 � plasma[Na�]). The subject lost 400 mmol (50 mmol/L � 8 L)of NaCl or 800 mosm from the ECF in sweat and re-placed all of his water losses. His plasma osmolarity af-ter the hike and rest is 260 mosm/L [(11,200 � 800mosm) � 40 L], or plasma [Na�] � 130 mmol/L. Thereduced plasma osmolality causes water to move fromthe ECF into the cells until a new osmotic equilibriumis established. The initial Na� content of the subject’sECF was 15 L � 140 mmol/L � 2,100 mmol. He lost400 mmol Na� during the hike, and his ECF [Na�] waslowered to 130 mmol/L. His new ECF volume �(2,100 � 400 mmol) � 130 mmol/L � 13.1 L.

8. The answer is B. His metabolic rate is 800 W; however,he is performing external work at a rate of 140 W andneeds to dissipate 660 W (� 800 W � 140 W) as heat.(It is true that he requires a higher metabolic rate to gouphill than if he were going on a level road, but wehave already specified his metabolic rate.) His skintemperature is 14�C above air temperature, and theconvective heat transfer coefficient is 15 W/(m2��C),so he loses heat by convection at a rate of 210 W/m2 ofsurface area. Because his body surface area is 1.8 m2,convection accounts for a loss of 378 W, leaving 282W � 16,920 J/min to be dissipated by evaporation ofsweat. Because it takes evaporation of 1 g of sweat to

remove 2,425 J of heat, he must secrete and evaporate7 g of sweat per min.

Chapter 30

1. The answer is C. A maximal voluntary contraction in-volving the identical muscles in an identical form ofcontraction provides the most readily quantified andaccurate basis for normalization of isometric exerciseintensity. Choice A is incorrect because the basis forcomparison involves rhythmic, dynamic exercise. Theother choices also contradict the principle that exer-cise can only be compared with other exercise involv-ing the same muscles and the same types of musclecontraction.

2. The answer is A. The physiological responses to dy-namic exercise are predictable when healthy individu-als differing in endurance exercise capacity are com-pared at matched levels of relative oxygen transportdemand. Exercise at 75% of the maximal oxygen up-take will lead to exhaustion in typically 1 to 2 hours,rendering choice B incorrect. The more highly trainedperson will show increased work output despite fatigu-ing at about the same time as the person with lower ca-pacity, rendering choices C and D incorrect. Traininglowers lactic acid production at any matched fractionaluse of the maximal oxygen uptake, making choice E in-correct.

3. The answer is D. Active muscle vasodilation during dy-namic exercise is quantitatively much greater than thenet vasoconstriction in the gut, skin, kidneys, and in-active muscle. Choices B, C, and D contradict this an-swer. Total systemic vascular resistance can be meas-ured, albeit indirectly, from measurements of systemicarterial pressure and cardiac output.

4. The answer is C. This answer presumes that vasocon-striction occurs in these vascular beds and that its ef-fect is to help balance vasodilation in active skeletalmuscles and prevent exercise-induced systemic hy-potension. This effect is ubiquitous across all individu-als during all forms of dynamic exercise, makingchoices B, C, and E incorrect. Cerebral blood flow isheld constant during all forms of exercise, unlike renalor splanchnic blood flow.

5. The answer is A. Even highly trained and heat-accli-matized individuals are at risk for heat-related illness ifexercise is sufficiently prolonged and if environmentalconditions are sufficiently severe. In healthy personsduring exercise, coronary vasodilatory capacity is ade-quate, renal blood flow reductions in health are en-tirely safe, and gastric mucosal blood flow reductionsare easily tolerated. In long-term exercise in a warm en-vironment, hypotension, not hypertension, is the pos-sible cardiovascular risk.

6. The answer is E. During dynamic exercise, the balanceof active muscle vasodilation and sympatheticallydriven vasoconstriction in other organs provides thehighest systemic arterial pressure when the involvedmuscle mass size is intermediate. Isometric exercise al-ways causes blood pressure to increase more thanmatched dynamic exercise. Prolongation of work low-

ers blood pressure. The state of training, fatigue, andprolongation of activity are largely irrelevant or non-specific as factors affecting blood pressure during exer-cise.

7. The answer is E. The baroreceptor blood pressure setpoint is increased during exercise, depending on exer-cise mode, intensity, and duration. Blood pressure onlyfalls during exercise when there is preexisting cardiacdisease or during prolonged work in the heat. Traininghas no apparent effects on the baroreflex.

8. The answer is E. In the broadest terms, the changes incholesterol transport in response to chronically in-creased physical activity occur from prolonged en-hancement of fat metabolism. The increase in HDLand decrease in LDL occur, at least in part, in responseto enhanced lipoprotein lipase activity and increasedapo A-I synthesis. These effects of long-term, regular,dynamic exercise are largely independent of diet andweight loss.

9. The answer is B. A normal or reduced arterial PCO2 isa respiratory response that regulates arterial blood pHduring exercise. Oxygen partial pressure significantlydeclines in arterial blood during exercise only whenthere is preexisting lung disease (choice A), while therespiratory control system allows ventilation to match,but not exceed, levels of CO2 production (choice C).Exercise in healthy persons does not result in respira-tory acidosis or dizziness resulting from decreasedcerebral perfusion.

10. The answer is B. Exercise training has no effect on lungtissues other than the respiratory muscles. The incor-rect choices represent aspects of lung function that aredetermined by lung tissues unaltered by any form ofphysical activity. Decreases in ventilation and dyspneaduring exercise do occur with chronic increases in dy-namic exercise, but these arise from adaptations local-ized in the active skeletal muscles (including the respi-ratory muscles).

11. The answer is D. Weight-bearing exercise and in-creased muscle strength reduce osteoporosis by in-creasing the forces applied to bone. These changes areaugmented by exercise-linked improvements in motorcoordination that reduce the risk of falls. These factorsin combination sharply reduce the incidence of hipfracture in older persons. Activities that decrease grav-itational forces on bone (e.g., water immersion), whilevaluable, decrease forces applied to bone and are lessuseful in the prevention of osteoporosis.

12. The answer is D. Eccentric contractions cause delayedmuscle soreness. This muscle inflammatory response isa result of the greater force per active motor unit foundduring eccentric as compared with concentric exerciseat the same force development. Soreness is not foundafter isometric exercise (choice A), solely on the basisof ischemia (which occurs in many forms of musclecontraction; choice B), in response to increased en-durance (choice C), or in direct proportion to the per-centage usage of the maximum voluntary contractileforce (which is defined in terms of isometric contrac-tions; choice E).

13. The answer is C. Motor unit rotation allows frequent

rest and recovery for activated cells, delaying fatigue.Inactive muscle cells undergo atrophic changes that re-duce cell cross-sectional area, reducing strength andincreasing the number of mobilized cells and motorunits required for a fixed external force development.These facts contradict choice A. Atrophy causes allsystems required for force production to down-regu-late in parallel, contradicting choice B, and lack of ac-tivity reduces, rather than increases, oxidative capac-ity, rendering choice D erroneous. Choice E is falsebecause the form of exercise must be standardized formeaningful comparisons of strength or endurance.

14. The answer is E. Relative to weight-bearing activity,estrogen plays a more important role in the mainte-nance of bone mass in women. Reductions in bonemass, which increase the risk of fracture and are invari-ably associated with reduced body weight, occur de-spite increased dynamic exercise endurance and intra-muscular adaptations that are appropriate for the highlevel of dynamic exercise training.

15. The answer is D. Increases in both insulin-dependentand insulin-independent glucose uptake in active mus-cles during exercise enhance the measured insulin sen-sitivity. These effects reduce the requirements for ei-ther insulin itself or for oral antiglycemic agents inpersons with type 2 diabetes. In contrast, in type 1 di-abetes, these same effects increase the risk for hypo-glycemia, leading to requirements for careful monitor-ing of activity, as well as food intake, insulinadministration, and stress in persons with this illness.All of the other choices directly contradict this princi-ple, other than choice B, which is incorrect because in-creased sympathetic activity during exercise directlyreduces pancreatic insulin release and blood insulinlevels.

16. The answer is C. Maternal activity reduces the risk ofmaternal gestational diabetes as a result of the samemechanisms (increased insulin-dependent and insulin-independent muscle glucose uptake) that reduce therisk and severity of type 2 diabetes in all persons. Thereare no known negative effects of maternal exercise oneither the course or pregnancy or its outcome, and ma-ternal exercise does not alter the duration of gestationor fetal weight at term.

Chapter 31

1. The answer is C. Right or left shifts in dose-responsecurves indicate changes in sensitivity. Changes in max-imal biological response indicate changes in respon-siveness. Because there is no change in maximal re-sponse, the correct answer must relate to a change insensitivity only. A right shift indicates decreased sensi-tivity.

2. The answer is C. Hormones produce their effects ontarget cells by interacting with specific receptors. Hor-mone binding to its receptor generally initiates a cas-cade of events that lead to biological effects in the tar-get cells.

3. The answer is A. Aldosterone is a steroid and the pri-mary mineralocorticoid in the body. Testosterone,

APPENDIX A Answers to Review Questions 729

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progesterone, and cortisol are steroid hormones hav-ing primarily androgen, progestin, and glucocorticoidactivities, respectively. Prostaglandin E2 is a local sig-naling molecule derived from arachidonic acid.

4. The answer is B. Scatchard plots of hormone-receptorbinding data give information regarding the number ofreceptors and the affinity of the hormone for its recep-tor. The x-intercept provides data regarding total re-ceptor number, and the slope is equal to the negativeof the association constant (�Ka).

5. The answer is D. Preprohormones are the gene prod-ucts for most peptide and protein hormones. These arerapidly cleaved to form prohormones. POMC and pro-pressophysin are two examples of specific prohor-mones.

6. The answer is E. Cortisol, like other steroid hormones,is carried in the blood largely bound to carrier proteins,although a small percentage exists free in solution. Themajority of cortisol is bound to a specific carrier pro-tein, corticosteroid-binding globulin (CBG), whilesmaller amounts are bound nonspecifically to albumin.Few, if any, cortisol receptors would be expected in theplasma and transthyretin binds primarily thyroxine.

7. The answer is D. Hormones generally circulate at con-centrations from 10�9 to 10�12 M. They producemuch larger changes in a variety of biological parame-ters as a result of signal amplification, in which therather weak hormonal signal is amplified into a largerbiological response.

Chapter 32

1. The answer is C. Destruction of the neurons in the par-aventricular nuclei of the hypothalamus decreasesCRH release, which causes decreased synthesis and se-cretion of ACTH. Hyperosmolality of the bloodwould lead to an increase in portal blood AVP, whichincreases ACTH secretion by corticotrophs. Physicalor emotional stress increases ACTH release. Glucocor-ticoids feed back to the hypothalamus and anterior pi-tuitary to decrease ACTH synthesis and secretion. Pri-mary adrenal insufficiency is characterized by a lack ofglucocorticoids in the blood, resulting in an increase inACTH synthesis and secretion. Increased PKA activityin corticotrophs increases ACTH synthesis and secre-tion.

2. The answer is D. Thyroid hormones stimulate the ex-pression of the GH gene in somatotrophs. Thyroidhormones exert a negative-feedback signal on the hy-pothalamic-pituitary-thyroid axis to inhibit their ownsynthesis and secretion. Therefore, thyroid hormonesdecrease the sensitivity of thyrotrophs to TRH, decreasethe formation of IP3 in thyrotrophs, inhibit the expres-sion of the genes for the � and � subunits of TSH inthyrotrophs, and decrease the secretion of TSH by thy-rotrophs. Thyroid hormones have no effect on ACTHrelease.

3. The answer is B. Galactorrhea is commonly associatedwith pituitary tumors secreting excess PRL. Prolactin isimportant in maintaining breast milk production afterbirth. Galactorrhea is diagnosed if present longer than

6 months postpartum in a nonnursing mother. Thecombination of both galactorrhea and amenorrhea isdiagnostic of a PRL-secreting pituitary tumor. TSHgenerally has little effect on PRL secretion. GH haslactogenic activity when high, not low. Hypothalamicdopamine is an inhibitor of PRL release.

4. The answer is A. Neurophysin is the other productgenerated when the prohormone for AVP or oxytocinis cleaved. A decrease in blood volume would result inthe release of AVP and neurophysin from magnocellu-lar neurons. The hormones oxytocin, �-lipotropin,ACTH, and somatostatin are not involved in the regu-lation of blood volume.

5. The answer is C. Growth hormone deficiency in adultsis characterized by decreased muscle strength and ex-ercise intolerance and a reduced sense of well-being(including effects on libido). Lean body mass (muscle)is lost, and excess body fat deposition occurs in the ab-dominal region. GH replacement can reverse these ef-fects. Thyroid dysfunction is ruled out by the normalthyroid hormone levels. Glucocorticoid deficiencyusually results from primary adrenal insufficiency, as inAddison’s disease. Clinical symptoms include a de-creased sense of well-being, GI disturbances, and ab-normal glucose metabolism. Primary adrenal insuffi-ciency is also characterized by high blood levels ofACTH, which can result in hyperpigmentation as a re-sult of the melanocyte-stimulating activity of theamino terminal portion of ACTH. Adrenal insuffi-ciency is not usually associated with a redistribution ofbody fat to central stores. Prolactin does not appear tohave a major physiological effect in human males.Acromegaly results from excessive GH secretion in anadult; the symptoms are not consistent withacromegaly.

6. The answer is C. The data demonstrate a higher aver-age ACTH and higher average cortisol concentrationin the evening hours. This is opposite the usual diurnalpattern in which ACTH and cortisol are high in themorning. It is possible that the subject works nightsand has a reversed but normal diurnal rhythm ofACTH and cortisol release. There is no adrenal disease(primary or secondary) because both ACTH and cor-tisol are higher at the same time and then are lower atthe same time. The diurnal pattern rules out an ACTH-secreting tumor because ACTH release would tend tobe constant.

7. The answer is B. Somatostatin, given as a long-actinganalog octreotide, is effective in reducing excess secre-tion of GH. It can also reduce tumor size, if one is pres-ent. Glucocorticoid would feed back to inhibit the hy-pothalamic-pituitary-adrenal axis but have little effecton GH release. Because acromegaly is characterized byexcessive GH secretion, the administration of GHwould be inappropriate. Insulin could be given tocounter the diabetogenic effects of excess GH, but itwould have little effect on tumor size (if present), bonethickening, or hypertrophy of the liver. GHRH andthyroid hormone would stimulate GH release in a situ-ation of high GH.

8. The answer is B. GHRH increases cAMP and stimu-

lates GH synthesis and secretion; somatostatin de-creases cAMP and inhibits GH synthesis and secretionfrom somatotrophs. TRH stimulates TSH secretionand the synthesis of the � and � subunits of TSH byincreasing inositol trisphosphate and calcium in thy-rotrophs. cAMP has no effect on AVP release, and itstimulates ACTH synthesis in corticotrophs.

Chapter 33

1. The answer is A. TSH stimulates the endocytosis ofcolloid by the apical membrane of the follicular cell.Thyroglobulin in the colloid is hydrolyzed in the lyso-somal vesicles to release thyroid hormones. T4 and T3

are stored in thyroglobulin in the colloid, not in secre-tory vesicles in the follicular cell. TSH stimulates theuptake of iodide from the blood, not the colloid. It hasno effect on blood flow to the thyroid gland and no di-rect effect on the binding of T4 and T3 to thyroxine-binding globulin. TSH stimulates an increase in cAMP,not an increase in the hydrolysis of this second mes-senger.

2. The answer is C. Thyroid hormones are important fornormal development of the CNS and for body growth.TSH stimulates the synthesis and release of thyroidhormones, as well as the growth of the thyroid gland.In a disorder in which the thyroid gland does not re-spond to TSH, thyroid hormone production would bedecreased, resulting in poor development of the CNSand poor body growth. TSH would also not be able tostimulate the growth of the thyroid, resulting in a smallgland.

3. The answer is A. Giving thyroid hormones to the childwould improve body growth but not mental ability be-cause thyroid hormones are most important for CNSdevelopment in utero. Therefore, giving thyroid hor-mones after birth would be too late. Thyroid gland sizewould remain smaller than normal because thyroidhormones have no trophic effect on the gland; onlyTSH has a trophic effect.

4. The answer is C. Uncoupling proteins allow protons toflow down their electrochemical gradient across themitochondrial membrane, uncoupled from the synthe-sis of ATP. The resulting energy generated is releasedas heat, and ATP is not synthesized. Uncoupling pro-teins are increased by thyroid hormones. The noveluncoupling proteins are found in many tissues, includ-ing muscle and adipose tissue. Oxidation of fatty acidsand glucose is not coupled in mitochondria, and theuncoupling proteins are not the switch between oxida-tion of these two substrates. Uncoupling proteins havenot been demonstrated to be essential to the mainte-nance of body temperature in mammals. However,UCP-1 is important in the ability of small mammals,such as rodents, to tolerate cold temperatures.

5. The answer is F. T3 is produced from T4 by 5�-deiodi-nase (type 2) in the anterior pituitary. The major thy-roid hormone product of the thyroid gland is T4. Thethyroid hormone receptor (TR) is located in the nu-cleus. A 5�-deiodinase acts on T4 to make reverse T3.The half-life of T3 in the bloodstream is about 1 day

because of the protective actions of the thyroid hor-mone-binding proteins. Thyroid peroxidase catalyzesthe iodination of thyroglobulin to form MIT and DIT,precursor molecules for T3.

6. The answer is D. The patient’s symptoms of chronic fa-tigue, aching muscles, occasional numbness in the fin-gers, and weight gain are consistent with a hypothy-roid state. The high TSH rules out a defect in thehypothalamic-pituitary axis and suggests an unrespon-sive thyroid gland, most likely a result of autoimmunethyroid disease. The presence of antibodies to thyroidperoxidase or thyroglobulin would confirm the diag-nosis. The absence of a goiter rules out hypothy-roidism as a result of iodine deficiency; low serum thy-roid hormone levels would result in elevated TSH withsubsequent trophic effects on thyroid growth. There isno growth of the thyroid in this patient because of theautoimmune attack on the gland.

7. The answer is B. Thyroid peroxidase catalyzes the cou-pling of two adjacent iodotyrosine residues in the thy-roglobulin precursor to form iodothyronine and dehy-droalanine. Thyroid peroxidase uses hydrogenperoxide produced by mitochondria to iodinate tyro-sine residues and to couple adjacent iodotyrosineresidues. Thyroid peroxidase is localized to the apicalmembrane of the follicular cell and catalyzes all reac-tions in this location. The release of thyroid hormoneis mediated by lysosomal degradation of thyroglobu-lin. Thyroid peroxidase iodinates tyrosine residues inthe thyroglobulin molecule to form MIT and DIT. De-hydroalanine is derived from the free-radical re-arrangement of 2 DIT residues to form thyroxine. Thy-roid peroxidase forms the free radicals necessary forthis reaction.

8. The answer is F. A TSH secreting tumor of the pitu-itary would result in elevated thyroid hormone levelsand symptoms of thyrotoxicosis. Graves’ disease ischaracterized by elevated thyroid hormone levels andanti-TSH receptor antibodies. TSH would be low be-cause of feedback inhibition of its release. Resistanceto thyroid hormone action could result in elevated thy-roid hormone levels but would not cause symptoms ofthyrotoxicosis. Thyroid gland adenomas commonlyresult from a point mutation in the TSH receptor, re-sulting in chronic activation of signaling. This wouldincrease thyroid hormones but should result in a re-duction in TSH. A deficiency in 5�-deiodinase couldresult in increased thyroid hormone levels and symp-toms of thyrotoxicosis, but would not be associatedwith elevated TSH. Early in the progression ofHashimoto’s disease, symptoms of thyrotoxicosis maybe present, but the absence of antithyroid antibodiesrules out this condition.

Chapter 34

1. The answer is B. Cholesterol esters in LDL are themost important source of cholesterol for sustaining ad-renal steroidogenesis when it occurs at a high rate overa long time. This cholesterol can be used directly afterrelease from LDL and not stored. De novo synthesis of

APPENDIX A Answers to Review Questions 731

732 APPENDICES

cholesterol from acetate is a minor source of choles-terol in humans. Cholesterol from the plasma mem-brane or endoplasmic reticulum is not used forsteroidogenesis. Cholesterol esters in lipid dropletswithin adrenal cortical cells would be used first and de-pleted during periods of high adrenal steroid hormonesynthesis.

2. The answer is C. The increase in body weight with lit-tle linear growth suggests that the patient has Cush-ing’s disease rather than general obesity because lineargrowth usually continues in obesity syndromes. Labo-ratory findings in Cushing’s disease include elevatedACTH, serum cortisol, urinary cortisol, and serum in-sulin (as a result of the cortisol-induced resistance toinsulin action in skeletal muscle and adipose tissue).

3. The answer is A. Congenital adrenal hyperplasia is theresult of genetic defects that affect adrenal steroido-genic enzymes, producing an impaired formation ofcortisol. Low serum cortisol is a stimulus for ACTH re-lease from the hypothalamus. The increase in ACTHhas a proliferative effect on the adrenal gland, resultingin hyperplasia. Addison’s disease is the result of patho-logical destruction of the adrenal glands by microor-ganisms or autoimmune disease and would, therefore,not result in adrenal hyperplasia. ACTH stimulates thegrowth of the adrenal gland. A reduction in ACTH inthe blood would result in atrophy of the adrenal gland.Corticosteroid-binding globulin noncovalently bindssteroid hormones in plasma; defects in this protein arenot associated with adrenal hyperplasia. Cushing’s dis-ease results from a pituitary ACTH-secreting tumor;adrenal hyperplasia is secondary, not congenital, inthis disease. Aldosterone synthesis is regulated by therenin-angiotensin system. Defective aldosterone syn-thesis would, therefore, not lead to increased ACTHand adrenal hyperplasia.

4. The answer is E. Catecholamines stimulateglycogenolysis and gluconeogenesis in the liver, caus-ing glucose to be synthesized and released into theblood. Catecholamines stimulate glycogen phospho-rylase in muscle to free glucose for use by the muscle.Muscle cannot release glucose to the circulation be-cause it lacks glucose-6-phosphatase. However, themuscle can release lactate, which can be used in gluco-neogenesis by the liver. Catecholamines inhibit the re-lease of insulin from the pancreas. Insulin would becounterproductive to attempts to increase blood glu-cose. Catecholamines increase the release of fatty acidsfrom the adipose tissue, to be used in gluconeogenesisby the liver.

5. The answer is F. Patients on long-term glucocorticoidtherapy should have the dose increased prior to under-going surgery to minimize the effects of surgical stress.These patients cannot mount their own stress responsebecause of the lack of adrenal cortisol release. Gluco-corticoid-induced hypoglycemia or interactions withanesthetics are unlikely, and these concerns would besecondary to stimulating the response to surgicalstress. Glucocorticoids inhibit ACTH release and theimmune response. Glucocorticoids increase the re-sponse of the vasculature to catecholamines.

6. The answer is F. IP3 is one of the second messengers inthe cells of the zona glomerulosa that signals for al-dosterone release. A decrease in IP3 would result in lesssignal for aldosterone synthesis and release. The rate ofaldosterone secretion would increase in response to anincrease in renin release from the kidney. Renin cat-alyzes the rate-limiting step in the conversion of an-giotensinogen to angiotensin II, which is a stimulus foraldosterone synthesis and release. A rise in serumpotassium or renal sympathetic nerve activity, a fall inblood pressure in the kidney, or a decrease in tubulefluid sodium concentration at the macula densa wouldstimulate aldosterone synthesis and release.

7. The answer is C. The first and rate-limiting step in allsteroid biosynthesis is catalyzed by cholesterol side-chain cleavage enzyme, resulting in pregnenolone andisocaproic acid. 17�-hydroxylase, 3�-hydroxysteroiddehydrogenase, 21-hydroxylase, and 11�-hydroxylaseare all involved in the synthesis of cortisol, but are notrate-limiting. 3-Hydroxy-3-methylglutaryl CoA re-ductase catalyzes the rate-limiting step in de novo cho-lesterol synthesis.

8. The answer is B. Addison’s disease results from thepathological destruction of the adrenal glands by mi-croorganisms or by an autoimmune response; it is char-acterized by glucocorticoid and aldosterone deficiency.Hyperpigmentation is caused by a lack of cortisol pro-duction, which results in increased ACTH production.Hyponatremia and hyperkalemia occur in the absenceof aldosterone, which normally stimulates sodium re-tention and potassium excretion by the kidneys. Cush-ing’s disease produces excessive cortisol release fromthe adrenals, secondary to excessive anterior pituitarysecretion of ACTH; patients with this disease do nothave the symptoms of aldosterone deficiency. Primaryhypoaldosteronism is characterized by a lack of aldos-terone secretion. The hyperpigmentation indicates amore severe disease with lack of cortisol production aswell. Congenital adrenal hyperplasia is the result of ge-netic defects that affect adrenal steroidogenic enzymes,resulting in impaired formation of cortisol. Low serumcortisol is a stimulus for ACTH release and hyperpig-mentation. Congenital adrenal hyperplasia is usually as-sociated with hypertension as a result of the excess pro-duction of steroidogenic intermediates such asdeoxycorticosterone, which has substantial mineralo-corticoid activity. Hypopituitarism is a condition inwhich pituitary function is suppressed, resulting in re-duced ACTH secretion; this is not applicable becausethe patient presents with hyperpigmentation as a resultof excess ACTH release. Patients with glucocorticoid-suppressible hyperaldosteronism are hypertensive.

9. The answer is E. Glucocorticoids maintain the tran-scription of genes and, therefore, the intracellular con-centrations of many of the enzymes needed to carryout gluconeogenesis in the liver and kidneys. Gluco-corticoids maintain the liver and kidneys in a state thatmakes them capable of accelerated gluconeogenesiswhen fasting occurs. Glucocorticoids inhibit insulin re-lease. Insulin inhibits gluconeogenic enzymes in theliver. The glucocorticoid-induced inhibition of glu-

cose utilization by skeletal muscle does not stimulategluconeogenesis but provides glucose for utilization bythe CNS. Inhibition of glycogenolysis by glucocorti-coids does not occur in fasting. Glucocorticoids do notinhibit, but actually permit, lipolysis and the release offatty acids from adipose tissue.

Chapter 35

1. The answer is D. Epinephrine stimulates glucagon se-cretion but inhibits insulin secretion. Amino acids andacetylcholine both stimulate insulin and glucagon se-cretion.

2. The answer is C. Insulin inhibits protein degradationand stimulates amino acid uptake in skeletal muscle. Itstimulates glucose uptake in many, but not all, tissues.It inhibits hormone-sensitive lipase in adipose tissue.

3. The answer is C. Glucagon stimulates gluconeogene-sis and ureagenesis in the liver. Under certain condi-tions, glucagon can actually stimulate insulin secretion.Glucagon does not have its primary actions in adiposetissue. Somatostatin does not play a role in ketogene-sis.

4. The answer is B. Persons with type 1 diabetes are in-sulin-deficient, not insulin resistant; they are treatedwith exogenous insulin. Persons with type 2 diabetesare treated with sulfonylureas. Secondary complica-tions are difficult to avoid in any form of diabetes.

5. The answer is A. The development of type 2 diabeteshas a strong genetic component. Persons with type 2diabetes often have normal or elevated insulin levels.Although there is an association of type 2 diabetes andobesity, it is not true that it only occurs in obese indi-viduals. Type 1 diabetes is a disease of insulin defi-ciency, and type 2 is a disease of insulin resistance.

6. The answer is D. Neuropathy, nephropathy, andretinopathy are chronic complications of type 2 dia-betes. Ketoacidosis is an acute complication seen intype 1 diabetes.

7. The answer is D. Delta cells produce somatostatin.8. The answer is D. The I/G ratio is highest after feeding

and decreases progressively during fasting.

Chapter 36

1. The answer is B. Half (50%) of the total plasma cal-cium is free or ionized.

2. The answer is C. Most of the ingested calcium is notabsorbed by the GI tract and leaves the body via the fe-ces.

3. The answer is A. The majority of ingested phosphateis absorbed by the GI tract and leaves the body via theurine.

4. The answer is A. Skin, kidney, and liver can all be in-volved in forming the active metabolite of vitamin D,1,25-dihydroxycholecalciferol. Bone does not formthis hormone, but is a target for its actions.

5. The answer is C. The kidneys are the site of formationof 1,25-dihydroxycholecalciferol from 25-hydroxyc-holecalciferol, a reaction catalyzed by the 1�-hydrox-ylase enzyme. 7-Dehydrocholesterol is converted to

cholecalciferol in the skin. Vitamin D3 is not convertedto vitamin D2. Calcium is incorporated into hydroxya-patite in bone.

6. The answer is C. Osteoporosis is characterized by anequivalent loss of bone mineral and organic matrix.Paget’s disease is characterized by disordered bone re-modeling; rickets and osteomalacia are characterizedby inadequate bone mineralization.

7. The answer is D. PTH stimulates bone resorption andrenal calcium reabsorption and, via stimulated synthe-sis of 1,25-dihydroxycholecalciferol, intestinal calciumabsorption, raising plasma calcium concentration.PTH inhibits renal phosphate reabsorption, leading tophosphaturia and hypophosphatemia.

Chapter 37

1. The answer is A. Reduced secretion of GnRH will re-sult in extremely low levels of circulating LH and FSH,causing testicular atrophy, as in Kallmann’s syndrome.Hypersecretion of LH and FSH, increased activin, andan increased number of FSH receptors all lead to hy-perfunction of the testis, not hypofunction. A failure ofthe hypothalamus to respond to testosterone increasesLH, leading to increased Leydig cell androgens andtesticular hypertrophy.

2. The answer is E. Follistatin is a binding protein for ac-tivin. Activin cannot increase FSH secretion when fol-listatin is bound to it, so FSH secretion decreases. Fol-listatin does not bind FSH, does not inhibit seminalfluid production and Leydig cell testosterone secretion,and does not stimulate the production of spermatogo-nia.

3. The answer is A. The epididymis and vas deferens aremajor storage sites of spermatozoa. Spermatozoa de-velop in the in the seminiferous tubules. Sertoli cells,not the epididymis, secrete estrogens and inhibin. Theprostate gland, seminal vesicles, and bulbourethralglands secrete the seminal fluids.

4. The answer is B. It takes approximately 65 to 70 daysto develop spermatozoa from the earliest stages ofspermatogonia. Because the production of sperm de-pends on LH and FSH, a lack of GnRH (Kallmann’ssyndrome) will reduce the production of LH, FSH, andsperm. Temperature is important in regulating spermproduction. Optimal sperm production occurs at 2 to3�C lower than body temperature.

5. The answer is A. The initial reaction and the rate-lim-iting step in the production of testosterone is the con-version of cholesterol to pregnenolone, which is regu-lated by

6. The answer is LH-stimulated cAMP in the Leydig cells.The cholesterol side-chain cleavage enzyme is locatedin mitochondria. All other sex hormone synthesis oc-curs outside of the mitochondria. Aromatization is thelast reaction, the conversion of testosterone to estra-diol. Pregnenolone is the immediate derivative of cho-lesterol, not progesterone. The initial reaction is stim-ulated by LH, not FSH. LH receptors are on Leydigcells, the site of testosterone synthesis.

7. The answer is C. The enzyme 5�-reductase is found in

APPENDIX A Answers to Review Questions 733

734 APPENDICES

the prostate and converts testosterone to dihy-drotestosterone. Testosterone does not bind HDL;HDL is a source of cholesterol. Activin does not bindtestosterone. Testosterone cannot be converted di-rectly to 17-hydroxyprogesterone, which is derivedfrom progesterone and is converted to androstene-dione. The side-chain cleavage enzyme converts cho-lesterol to pregnenolone.

8. The answer is A. Sex hormone-binding globulin bindsto both testosterone and estradiol, but it binds withhigher affinity to testosterone. The bioactivity oftestosterone is reduced by SHBG because testosteronecannot bind to its receptor when bound by SHBG.SHBG increases the circulating half-life of testosteroneby slowing the clearance and metabolism of testos-terone. SHBG does not alter the secretion of inhibin orandrogen-binding protein.

9. The answer is D. The production of estradiol requiresLeydig cells, under the influence of LH, which stimu-lates androgen production. The androgen diffuses toSertoli cells, which contain aromatase, the enzymethat converts androgens to estrogens under the influ-ence of FSH. Therefore, Leydig cells, Sertoli cells, LH,and FSH are required. Follistatin binds activin andwould reduce FSH secretion, an essential componentfor estradiol production. Estradiol is not produced byLeydig cells. Activin would increase the secretion ofFSH, which is a necessary component for estradiol, butother cells and hormones are required. Similarly, Ley-dig cells would need LH to stimulate the production ofthe androgen precursor of estrogen. Sertoli cells, underthe influence of FSH, are needed to aromatize andro-gen from Leydig cells.

10. The answer is C. Androgens and estrogens are knownto stimulate the closure of the epiphyses at puberty.Because eunuchs are castrated, they have no testicularsource of androgen and estrogen, and the closure ofthe epiphyses is delayed. In eunuchs, long bones con-tinue to grow, resulting in a tall stature. Estrogens dohave a positive effect in maintaining bone; however,eunuchs have little or no estrogen because the testesare absent. Choice B is incorrect, although eunuchsmay have elevated circulating LH (as a result of thelack of androgen negative feedback). LH has no effecton bone. The absence of testes delays the closure ofthe epiphyses, and androgen levels are low in eunuchsbecause of the lack of testes.

Chapter 38

1. The answer is B. Aromatase is present only in granu-losa cells and is regulated mainly by FSH. AlthoughLH may stimulate aromatase in granulosa cells, granu-losa cells do not produce androgens. Estradiol synthe-sis in the graafian follicle is unrelated to progesteronesynthesis in the corpus luteum and does not increaseLH during this phase. Estradiol increases LH secretionduring the LH surge. There is no evidence for synergybetween FSH and progesterone in regulating estradiolsecretion by the graafian follicle.

2. The answer is A. Granulosa cells do not have the en-

zyme called 17�-hydroxylase, which converts proges-terone to 17�-hydroxyprogesterone. Aromatase is theenzyme that converts androgens to estrogens. 5�-Re-ductase converts testosterone to dihydrotestosterone.Sulfatase is an enzyme that conjugates steroids withsulfate for subsequent excretion in the urine. Steroido-genic acute regulatory protein transports cholesterolfrom the outer to the inner mitochondrial membrane.

3. The answer is B. One of the first clinical measures formenopause is an increase in the serum concentration ofFSH (and LH), indicative of the lack of ovarian func-tion. Menses starts at age 12, not age 50, and its onsetat this time would not indicate menopause. Excessivecorpora lutea would likely indicate multiple ovulationsor a failure of luteal regression. Increased vaginal corni-fication is an indicator of estrogen secretion, whichdoes not occur in menopause. Menstrual cycles be-come irregular at menopause.

4. The answer is D. Progesterone has a thermogenic ef-fect on the hypothalamus, increasing the basal bodytemperature for a few days after ovulation. Womenwho, because of ovulatory problems, are having trou-ble getting pregnant are sometimes asked to recordtheir daily oral temperatures and look for the increasein basal body temperature, indicating an increase inprogesterone (which indicates ovulation). Proges-terone induces a secretory type of endometrium,whereas estrogens induce a proliferative type. Duringthe luteal phase, when progesterone is increasing,graafian follicles are not present. Progesterone levelsdecrease during luteal regression. FSH decreases whenprogesterone is rising.

5. The answer is A. Theca interna cells produce andro-gens under the influence of LH, whereas granulosacells do not produce androgens. Theca interna cells docontain cholesterol side-chain cleavage enzyme,which converts cholesterol to pregnenolone. Becausetheca cells do not express aromatase, they cannot con-vert testosterone to estradiol. The theca interna has arich blood supply. Granulosa cells produce inhibin.

6. The answer is A. Disruption of the hypothalamic-pitu-itary portal system leads to a lack of dopamine andGnRH reaching the pituitary. Because dopamine in-hibits PRL secretion, PRL levels will increase. In addi-tion, the lack of GnRH will lead to reduced secretionof LH and FSH, reduced ovarian function, and even-tual ovarian atrophy. PRL will have no effect on theovary or inhibit ovarian follicle development. Disrup-tion of the hypothalamic-pituitary axis will lead to re-duced follicular development, lack of ovulation, andlow circulating progesterone. Inhibin levels will de-crease, but FSH will not increase because there is noGnRH reaching the pituitary from the disrupted axis.Excessive ovarian androgen usually occurs in the pres-ence of excessive LH secretion or an androgen tumorin the ovary. LH secretion is reduced by the lack ofGnRH.

7. The answer is B. Inhibin is produced by granulosa cellsand inhibits the secretion of FSH. Inhibin does not in-hibit the secretion of LH and PRL. Although inhibincan have local ovarian effects, it has profound in-

hibitory effects on FSH secretion. Inhibin has twoforms, A and B; the � subunits are the same, whereasthe � subunits are different. Inhibin binds activin anddecreases FSH secretion.

8. The answer is D. Estrogen induces the formation of astringy vaginal secretion that is called spinnbarkeit,observed in the late follicular phase. The secretory en-dometrium is under the influence of progesterone;therefore, spinnbarkeit would not be present.Spinnbarkeit is not produced in response to proges-terone, androgen, or prolactin.

9. The answer is A. Fertilization occurs in the oviduct.The oocyte must have entered a second meiotic divi-sion to reduce the chromosome number of the oocyteto a haploid state (n) so that it may fuse with the sperm(also haploid), producing a 2n zygote. Fertilizationdoes not occur in the uterus, especially not after thefirst meiotic division when the chromosome number is2n. In the adult ovary, oocytes do not undergo mitosis.Graafian follicles do not enter the oviduct and are notfertilized. Fertilization does not occur in the uterus,and the oocyte does not implant. The blastocyst willimplant in the uterus. In addition, extrusion of the po-lar body is associated with fertilization, but this eventoccurs within the oviduct.

10. The answer is B. 5�-Reductase is the enzyme that con-verts testosterone to dihydrotestosterone. 5�-Reduc-tase is associated with increasing the most potent an-drogen, dihydrotestosterone, and reducing LHsecretion. Estrogens are associated with female sec-ondary sex characteristics, although some androgensregulate pubic hair development.

Chapter 39

1. The answer is D. Suckling involves hormonal and neu-ronal components, but the hormonal component is ef-ferent and the neuronal component is afferent. Whenthe baby suckles, neural signals from the nipple travelvia nerves to the spinal cord and up to the brain (affer-ent component), which triggers the release of oxytocinfrom the supraoptic nuclei (efferent component). Oxy-tocin enters the circulation, enters the breasts, andcauses contraction of the myoepithelial cells. Placentallactogen is no longer present after parturition; it is aplacental hormone. Dopamine release is decreased bysuckling, and as a result, PRL secretion is increased.

2. The answer is D. Under normal circumstances, theuterus must be primed with both progesterone and es-trogen for successful implantation. Implantation oc-curs on day 7 after fertilization. The decidual reactionoccurs as the result of the implanting blastocyst. Theembryo is in the blastocyst stage of development at thetime of implantation. A morula does not implant. Thedeveloping embryo enters the uterus on day 3 or 4, itremains suspended in the uterus for 3 or 4 more days,and implantation occurs on day 7.

3. The answer is A. The acrosome reaction causes a fusionof the plasma membrane and the acrosomal membraneof the sperm, with subsequent release of proteolyticenzymes that help the sperm enter the ovum. The zona

reaction and pronuclei formation occur after the spermhas entered the ovum. Sperm enter the perivitellinespace after penetration; there is no evidence that thisspace has any role in penetration. Cumulus expansionassists in movement of the sperm through the mass ofgranulosa cells for the sperm to get to the surface of thezona pellucida. However, the cumulus cells do not as-sist in actual penetration of the zona.

4. The answer is B. The production of hCG by tro-phoblast cells stimulates the corpus luteum to continueto produce progesterone so that luteal regression doesnot occur at the end of the anticipated cycle. AlthoughPRL levels increase throughout pregnancy, PRL is notresponsible for maintenance of the corpus luteum ofpregnancy. Prostaglandins are generally luteolytic,causing regression of the corpus luteum, termination ofthe luteal phase, and return to the next cycle; they donot prolong the cycle or postpone it. Oocytes are notdepleted after implantation. In fact, pregnancy tendsto preserve oocytes, as ovulation ceases during preg-nancy. Plasma progesterone levels are high duringpregnancy as a result of activation of the corpus luteumand placental production of progesterone. Elevatedprogesterone blocks follicular development and theensuing LH surge; low levels of progesterone wouldpermit a return to cyclicity.

5. The answer is A. Fertilization by more than one spermis prevented by the cortical reaction. Cortical granulescontaining proteolytic enzymes fuse just beneath theentire surface of the oolemma. The proteolytic en-zymes are released to the perivitelline space, destroythe sperm receptors, and harden the zona, preventingother sperm from penetrating the fertilized ovum. En-zyme reaction is a nonspecific term with little meaningfor polyspermy. The acrosome reaction allows thesperm to penetrate the zona. The decidual reaction isan inflammatory-like reaction that occurs simultane-ously with implantation of the blastocyst into the uter-ine endometrium.

6. The answer is A. Oral steroidal contraceptives gener-ally contain progesterone and estrogen-like molecules,which feed back negatively on the hypothalamic-pitu-itary axis and reduce the secretion of LH and FSH; thisis the primary mechanism of action in preventing preg-nancy. Choices B, C, and D are not the best answers,although oral contraceptives do alter the uterine envi-ronment, thicken the cervical mucus, and reduce spermmotility. If ovulation were not blocked, the other pa-rameters would not be effective in blocking pregnancy.Oral contraceptives block the LH surge; they do notinduce a premature surge.

7. The answer is D. hCG is produced by trophoblast cellsprior to implantation of the embryo and binds to lutealLH receptors, signaling them to produce progesterone,which is necessary for the maintenance of pregnancy.Therefore, hCG signals the mother that she is preg-nant via stimulation of luteal LH receptors. Placentallactogen is not produced until after pregnancy is wellestablished. Progesterone is a common hormone asso-ciated with the menstrual cycle and pregnancy. In hu-mans, the inflammatory reaction at implantation does

APPENDIX A Answers to Review Questions 735

736 APPENDICES

not signal the mother that she is pregnant and it fol-lows secretion of hCG.

8. The answer is C. The placenta cannot make androgensfrom progestin precursors because it lacks 17�-hy-droxylase. DHEAS from the fetal adrenal glands isconverted to 16OH-DHEAS by the fetal liver andthen to estriol by the placenta; this reaction is substan-tial and is an indicator of fetal stress (estriol low) orwell-being (estriol high). The mother’s adrenal canalso make DHEAS, which can be converted to 16OH-DHEAS by 16-hydroxylase in the fetal liver, but thisreaction is limited (10%). Androgens cannot be pro-duced from cholesterol in the placenta; the placentalacks 17�-hydroxylase. Estradiol is generally not con-verted to estriol. Androgens from the ovary are gener-ally not converted to estriol.

9. The answer is C. Insulin resistance is associated withreduced utilization of glucose by the mother and thisglucose is spared for the fetus. Plasma glucose is notlower but higher with insulin resistance. Insulin movesglucose into cells. During pregnancy, the developmentof insulin resistance may be a predictor of diabeteslater in life. Reduced pituitary function occurs becauseof the high levels of steroids and PRL, all independent

of insulin resistance. Progesterone, not insulin, in-creases appetite during pregnancy.

10. The answer is C. The female phenotype can developin an XY male if the biological action of testosterone isabsent. This absence can be due to a lack of testos-terone secretion caused by enzyme deficiencies or alack of the testosterone (DHT) receptor. In thisprocess, called testicular feminization, a phenotypicfemale develops in the presence of an XY karyotype.There is a lack of pubic and axillary hair, well-devel-oped breasts (as a result of the conversion of testos-terone to estrogen), with inguinal or abdominal testes,no uterus (because AMH is secreted), underdevelopedmale accessory ducts (lack of testosterone action), andthe vagina ends in a blind pouch. Progesterone has noeffect on phenotype. There is no evidence that adrenalinsufficiency (low cortisol and androgens from theadrenals) have any effect on inducing female pheno-type in a male. Inhibin would reduce FSH secretionand ultimately reduce adult testis size, but in the fetusthere is no effect on the development of the femalephenotype. AMH will prevent formation of theoviducts, uterus, and upper vagina; it does not increasefemale characteristics in the male.