ch12 - Biosignaling 1. Hormone Experiments in CellmFree...

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Unformatted text preview: Biosignaling 1. Hormone Experiments in CellmFree Systems in the i9503, Earl W. Sutherland, Jr, and his coileagues carried out pioneering experiments to elucidate the mechanism of action of epinephrine and glucagon. Given what you have learned in this chapter about hormone action, interpret each of the experiments described beiow. Identify substance X and indicate the sigmficance oi the results. (3.) Addition of epinephrine to a homogenate of normal liver resuited in an increase in the activity of giycogen phosphorylase. However, if the homogenate was first centrifuged at a high speed and epinephrine or glucagon was added to the clear supernatant fraction that contains phosphory— Ease, no increase in the phosphoryiase activity occurred. (b) When the particulate fraction from the centrifugation in (a) was treated with epinephrine, substance X was produced. The substance was isolated and purified. Unlike epinephrine, substance X activated glycogen phosphorylase when added to the clear supernatant fraction of the centrifuged homogenate. (c) Substance X was heat—stable; that is, heat treatment did not affect its capacity to activate phos- phorylase. (Hint: Wouid this be the case if substance X were a protein?) Substance X was nearly identical to a compound obtained when pure ATP was treated with barium hydroxide. (Fig. 8—6 will be helpful.) Answer Substance X is cyclic AMP. Epinephrine stimulates glycogen phosphorylase by activating the enzyme adenylyl cyclase, which catalyzes formation of CAMP, the second messenger. (a) Adenylyl cyciase is a membrane—bound protein; centrifugation sediments it into the particulate fraction. (b) Cyclic AMP directly stimulates glycogen phosphorylase. (c) Cyclic AMP is heat-stabie; it can be prepared by treating ATP with barium hydroxide. 2. Effect of Dibutyryl CAMP versus CAMP on Intact Cells The physiological effects of epinephrine shonid in principle be mimicked by addition of cAMP to the target celis. in practice, addition of CAMP to intact target ceils elicits only a minimal physioiogical response. Why? When the structuraliy related derivative dibutyryi CAMP (shown below) is added to intact cells, the expected physiological response is readily apparent. Explain the basis for the difference in ceilular response to these two substances. Dibutyryl CAMP is widely used in studies of CAMP function. 5-132 Chapter12 Biosignaling O \ \cli—(cngizons NH N/ I N\> K N N 0—0112 0 H H H n o=P~———o o\ Cl)- //’C—(CH2)20H3 o DibutyryloAMP (N 6,02’-Dibutyryl adenosine 3',5’-cyc1io monophosphate) Answer Unlike CAMP, dibutyryi cAMP passes readily through the plasma membrane. 3. Effect of Cholera Toxin on Adenylyi Cyclase rI‘he gram-negative bacterium thm‘o cholerae produces a protein, cholera toxin (MJr 90,000), that is responsible for the characteristic symptoms of cholera: extensive loss of body water and Na+ through continuous, debilitating diarrhea. If body fluids and Na+ are not replaced, severe dehydration results; untreated, the disease is often fatal. When the cholera toxin gains access to the human intestinai tract it binds tightly to specific sites in the plasma membrane of the epitheliai ceils fining the small intestine, causing adenylyl cyclase to undergo pro- ionged activation (hours or days). (a) What is the effect of cholera toxin on {CAMP} in the intestinal cells? (b) Based on the information above, suggest how CAMP normaily functions in intestinal epithelial cells. (c) Suggest a possible treatment for cholera. Answer (a) it increases {CAMP}. (b) The observations suggest that CAMP reguiates Na‘“ permeability. (c) Replace iost body fluids and electrolytes. 4. Mutations in PKA Explain how mutations in the R or 0 subunit of CAMPmdependent protein kinase (PM) might lead to (a) a constantiy active PKA or (b) a constantly inactive PKA. Answer (a) If a mutation in the R subunit makes it unable to bind to the C subunit, the 0 subunit is never inhibited; it is constantly active. (b) If a mutation prevents the binding of CAMP to the R subunit but stilt ailows normal R—O interaction, the inhibition of C by R cannot be relieved by elevated [CAMP], so the en— zyme is constantiy inactive. . Therapeutic Effects of Albuterol The respiratory symptoms of asthma result from constriction of the bronchi and bronchioles of the lungs caused by contraction of the smooth muscle of their wails. This constriction can be reversed by raising the {CAMP} in the smooth muscle. Explain the therapeutic effects of aibuteroi, a B-adrenergic agonist taken (by inhaiation) for asthma. Would you expect this drug to have any side effects? How might one design a better drug that did not have these effects? Answer By mimicking the actions of epinephrine on smooth muscle, aibuterol raises [CAMP], leading to reiaxation and eniargement (dilation) of the bronchi and bronchioles. Because B-adrenergic receptors also control many other processes, drugs that act as 6~adrenergic Chapter 12 Biosignaling 3-133 agonists generally have other, undesirable effects. To minimize such side effects, the goal is to find an agonist that is specific for the subtype of fi-adrenergic receptors found in bronchial smooth muscle. 6. Termination of Hormonal Signals Signais carried by honnones must eventualiy be terminated. Describe several different mechanisms for signal termination. Answer A hormone can be degraded by extraceilular enemas (such as acetyl- cholinesterase). ri‘he GTP bound to a G protein can be hydrolyzed to GDP. A second messen- ger can be degraded (cAMP, cGMP), further metabolized (1:33), or resequestered (Ca2+, in the endoplasmic reticulum). A receptor can be desensitized (acetylcholine receptor/channel), phosphoryiated/inactivated, bound to an arrestin, or removed from the surface (B-adrenergic receptor, rhodopsin). 7. Using FRET to Explore Protein-Protein Interactions in Vivo Figure 12—8 shows the interaction between Bwarrestin and the B-adrenergic receptor. How would you use FRET (see Box 12—8) to demonstrate this interaction in living cells? Which proteins would you fuse? Which wavelengths wouid you use to illuminate the cells, and which would you monitor? What would you expect to observe if the interaction occurred? if it did not occur? How might you explain the failure of this approach to demonstrate this interaction? Answer Fuse CFP to B-arrestin and YFP to the cytoplasmic domain of the ,G-adrenergic re- ceptor, or vice versa. in either case, illuminate at 433 nm and observe at both 476 and 52? run. If the interaction occurs, emitted light intensity wiil decrease at 476 nm and increase at 527 nm on addition of epinephrine to celis expressing the fusion proteins. if the interaction does not occur, the waveiength of the emitted light will remain at 476 nm. There are several reasons why this might fail; for exampie, the fusion proteins (1) are inactive or otherwise unable to interact, {2) are not translocated to their normal subcellular location, or (3) are not stable to proteolytic breakdown. 8. EGTA Injection EGTA (ethyiene glycol—bis (,Gmaminoethyl ether)—N,N, N ’,N ’—tetraacetic acid) is a chelating agent with high affmity and Specificity for Ca2+. By microinjecting a cell with an appropriate Ca2+-EGTA solution, an experimenter can prevent cytosolic {0912+} from rising above 10‘7 M. How wouid EGTA nucroinjection affect a cell’s response to vasopressin (see Tabie 12—4)? To glucagon? Answer Vasopressin acts through a PLO—coupled GPCR. The 1P3 released by PLC normaliy ei— evates cytosolic [Ca3+] to 10”“6 M, activating (with diacylgiycerol) protein kinase C. Preventing this elevation of {0212+} by using EGTA to “buffer” the internal [Ca2+i would block vasopressin action, but shouid not directiy affect the response to glucagon, which uses CAMP, not Ca2+, as its intraccliular second messenger. 9. Amplification of Hormonal Signals Describe alt the sources of amplification in the insuiin receptor system. Answer The amplification results from catalysts activating cataiystsminciuding protein kinases that act in enzyme cascades. Two moiecuies of insuiin activate an insulin receptor . dimer for a finite period, during which the receptor phosphorylates many moiecules of [1284. Through a series of interactions with other proteins (including Grbéi, Sos, Rae), IRS—1 activates Raf, which phosphorylates and activates many molecules of MEK, each of which phosphorylates and activates many molecules of ERK Each activated ERK phosphorylates and activates several moiecules of a tronscrtpttonfactoi; and each or" these stimulates the transcription of multiple copies of mRNA for specific genes. Each mRNA can direct the synthesis of many copies of the protein it encodes. (See Fig. 12—15.) 3-134 Chapter 3.2 Biosignaliag Answer When active, Ras activates the protein kinase Raf, which initiates the MAPK cascade that leads to the phosphorylation of nuclear proteins. A Rae protein without GTPase activity would, once activated by the binding of GTP, remain active, continuing to produce an insulin response. 11. Differences among G Proteins Compare the G proteins G5, which acts in transducing the signal from fi-adrenergic receptors, and Rae. What properties do they share? How do they differ? What is the functional difference between GS and (if? Answer Shoer properties of Bus and GS: both can bind either GDP or GTP; both are activated by GTP; both can, when active, activate a downstream enzyme; both have intrinsic GTPase activity that shuts them off after a short period of activation. Diggerences between Res and GS: Ras is a small, monomeric protein; Gs is heterotrimeric. Functional dfierences between Gs and Gd: GS activates adenylyl cyclase, Gi inhibits it. 12. Mechanisms for Regulating Protein Kinases Identify eight general types of protein kinases found in eukaryotic cells, and explain what factor is directly responsible for activating each type. Answer Kinase (factor(s)): PKA (cAMP); PKG (CGMP); PKG (Ca2+, DAG); Ca2+/CaM kinase (Ca2+, CaM); cyclin—dependent kinase (cyclin); protein Tyr kinase (ligand for the receptor, such as insulin); MAPK (Rat); Raf (Res); glycogen phosphorylase ldnase (PKA). 13. Nonhydrolyzable G'I'P Analogs Many enzymes can hydrolyze GTP between the .8 and y phosphates. The GTP analog liq-irnidoguanosine 5’—triphosphate Gpp(NH)p, shown below, cannot be hydrolyzed between the B and 7 phosphates. Predict the effect of microinjection of Gpp£Nl-l)p into a myocyte on the cell’s response to fi~adrenergic stimulation. OH OH GppCNH)p (,8, ry-imidoguanosine 5’-triphosphate) Answer Ncnhydrolyzabie analogs of GTP have the effect of keeping the stimulatory G protein (G5) in its activated form once it has encountered the receptor-hormone complex; it cannot shut itself off by converting the bound GTP (analog) to GDP. Injection of the analog would therefore be expected to prolong the effect of epinephrine on the injected cell. 14. Use of Toxin Binding to Purify a Channel Protein a—Bungarotoxin is a powerful neurotoxin found in the venom of a poisonous snake (Bungdrus multict'nctus). it binds with high specificity to the nicotinic acetylcholine receptor (AChR) protein and prevents the ion channei from opening. This interaction was used to purify AChR from the electric organ of torpedo fish. (a) Outline a strategy for using ambungarotofin covalently bound to chromatography beads to purify the AChR protein. (Hint: See Fig. 3—17c.) (b) Outline a strategy for the use of {lgfilla-bungarotoxin to purify the AChR protein. Chapter 12 Biosignaling 3-135 Answer (3) Use the (a~bungarotoxin—abound beads for affinity purification (see Fig. 3—17c, p. 87) of AChR. Extract proteins from the electric organs and pass the mixture through the Chro- matography column; the AChR binds selectively to the beads. Elute the AChR with a soiu- tion of N301 or a solvent of lower pH, which weakens its interaction with oa—bungarotoxin. (i1) Use binding of [1251}or-bungarotoxin as a quantitative (assay for AChR during purifica- tion by various techniques. At each step, assay AChR by measuring [125lia—bungarotoxin binding to the proteins in the sample. Optimize purification for the highest specific activity of AChR (counts/min of bound {igfilla-bungarotoxin per mg of protein) in the final material. £5. Resting Membrane Potential A variety of unusuai invertebrates, including giant clams, mussels, and polychaete worms, live on the fringes of deep—sea hydrothermai vents, where the temperature is 60 °C. (a) The adductor muscle of a giant clam has a resting membrane potential of “95 mV. Given the intracellular and extracellular ionic compositions shown below, would you have predicted this membrane potentiai? Why or Why not? Mememvmtgmmnwna «wumwww “Mira-"F _ “w- -- - - - Concentration (mm) - ._amv.sr.-m=cr.m>€ 440 E W 20 3| g Cl“ 560 I C 2+ 1 :1 a a {1 : fl m" m-----:--:_-.-- _ .- v --; -. w--- v-_ .«m _mu-.m;—u—~-1=;mr;::_r.~wrm—-mq:m«s: (b) Assume that the adductor muscle membrane is permeable to only one of the ions listed above. Which ion could determine the Vm? Answer (a) In most myocytes at rest, the plasma membrane is permeable primarily to K“ ions. Vm is a function of the distribution of K+ ions across the membrane. If Vm in the clam adductor muscle is determined primarily by EC, then Vm at rest would be predicted by the Nernst equation Eh“ = (RT/Zr?) 1n (Cent/Cm) using the values for [K43 given in the table. EK+ m [(8.315 J/moi ‘ K}(333 K)/(l) (96,480 J/V ' 11101)] ln (20/400) m 0.028? V X (—3.0) m -0.09 V, 01‘ —90 HIV Because the experimentaliy observed Vm is ~95 mV, the plasma membrane in the adduc- tor muscle must be permeable to some other ion or combination of ions. (b) Use the Nernst equation to calculate E for each ion. The ion with an E vaiue closest to the membrane potential is the permeant ion that influences Vm. EK+ = ~90 mV (see above) ENa+ = 0.0287 V X In (440/50) 5 0.06 V, or 60 mV Ear a (0.0287 V/—1)>< 1n (560/21) = —0.094 V, or ~94 mV (note that Z for Cl“ is M1) E03“ a (0.0287 VIZ) x in (EU/0.4) m 0.05 V, or 50 mV (Z for Ca2+ is 2) 3-135 Chapteri2 Biosignaling Thus, because E01" = —94 mV is very close to the resting Vm of —95 mV, it is likely that the membrane is permeable only to Cl“ ions at rest. You could verify this experimentally by changing the extracellular [Ci'], then measuring the effect on resting membrane potential. If this potential does depend only on Cl“ ions, the Nernst equation shouid predict how the membrane potential will change. 16. Membrane Potentials in Frog Eggs Fertilization of a frog oocyte by a sperm cell triggers ionic changes similar to those observed in neurons (during movement of the action potential) and initiates the events that result in cell division and development of the embryo. Oocytes can be stimulated to divide without fertilization by suspending them in 80 mM K01 (normal pond water contains 9 mM KCI). (3.) Calculate how much the change in extracellular {K01} changes the resting membrane potential of the oocyte. (flint: Assume the oocyte contains 120 mM K“ and is permeable only to K‘“.) Assume a temperature of 20 °C. (b) When the experiment is repeated in Ca2+«free water, elevated EKCl] has no effect. What does this suggest about the mechanism of the K01 effect? Answer (a) v... a g a( Egg) 2 {(8.815 i/moi - 19(293 K)/(1)(96,480 J/V - moi); In ([KWouc/[Kflkg = (0.025 V) In Gwen/{Km Vm in pond water = 0.025 v ln (9/120) = —0.06 v, or has mV V,“ in so mM KCI = 0.025 v in (80/120) 3 -—o.01 v, or —10 mV The membrane of the oocyte has been depolarized—the resting membrane potential has become less negative—by exposure to elevated extracellular 1K+}. (b) This observation suggests that the effect of increased {K01} depends on an influx of Ca2+ from the extracellular medium, which is required to stimulate cell division. High [KCIl treatment must depolarize the oocyte sufficiently to open voltage-dependent Ca2+ channels in the plasma membrane. 17. Excitation Triggered by Hyper-polarization In most neurons, membrane depolarization leads to the opening of voltage-dependent ion channels, generation of an action potential, and ultimately an in- flux of Ca2+, which causes release of neurotransmitter at the axon terminus. Devise a cellular strategy by which hyperpolam‘zotion in rod cells could produce excitation of the visual pathway and passage of visual signals to the brain. (Hint: The neuronal signaling pathway in higher organisms consists of a series of neurons that relay information to the brain (see Fig. 12—35). The signal released by one neuron can be either excitatory or inhibitory to the following, postsynaptic neuron.) Answer Hyperpolarization of rod cells in the retina occurs when the membrane potential, Vm, becomes more negative. ri‘his results in the closing of voltage—dependent Caz+ channels in the presynaptic region of the rod cell. The resulting decrease in intracellular [Ca2+] causes a corresponding decrease in the release of neurotransmitter by exocytosis. The neurotransmitter released by rod cells is actually an inhibitory neurotransmitter, which leads to suppression of activity in the next neuron of the visual circuit. When this inhibition is removed in reaponse to a light stimulus, the circuit becomes active and visual centers in the brain are excited. 18. Genetic “Channelopathies” There are many genetic diseases that result from defects in ion channels. For each of the following, explain how the molecular defect might lead to the symptoms described. (a) A lossnof-function mutation in the gene encoding the a subunit of the cGMP-gated cation channel of retinal cone cells leads to a complete inability to distinguish colors. I9. 20. 21. Chapter12 Biosignafing 8—13? (b) Loss-of~function alleles of the gene encoding the or subunit of the ATP—gated K+ channel shown in Figure 23m29 lead to a condition known as congenital hyperinsulinismwpersistently high lev- els of insulin in the blood. (c) Mutations affecting the ,8 subunit of the ATP—gated Kl" channel that prevent ATP binding lead to neonatal diabetesmpersistentiy low levels of insulin in the blood in newborn babies. Answer (a) Loss of function of the cGMP—gated channel prevents influx of Na“ and Ca2+ into cone cells in response to light; consequently, the cells fail to signal the brain that light had been received. Because red cells are unaffected, the individual can see but does not have color vision. A loss»of~function mutation in the ATP-gated cation channel prevents efflux of 16* through these channels, leading to continuous depolarization of the B—cell membrane and constitu— tive release of insulin into the blood. ATP is responsible for closing this channel, so in an individual with the mutant protein, the channels will remain open, preventing depolarization of the B—cell membrane and thereby preventing release of insulin, resulting in diabetes. ('0) Visual Desensitization Ognchi’s disease is an inherited form of night blindness. Affected individuals are slow to recover vision after a flash of bright light against a dark background, such as the headlights of a car on the freeway. Suggest what the molecular defectCs) might be in Oguchi‘s disease. Explain in molecular terms how this defect would account for night blindness. Answer Some individuals with Oguchi’s disease have a defective rhodopsin kinase that slows the recycling of rhodopsin after its conversion to the all—trans form on illumination. This defect leaves retinal rod and cone cells insensitive for some time after a bright flash. Other individuals have genetic defects in arrestin that prevent it from interacting with phosphorylated rhodopsin to trigger the process that leads to replacement of all-trans-retinal with ll~ois~retinal Effect of a Permeant cGMP Analog on Rod Cells An analog of cGlVlP, 8-Br—cGMP, will permeate cellular membranes, is only slowly degraded by a rod cell's PDE activity, and is as effective as cGMP in opening the gated channel in the cell’s outer segment. If you suspended rod cells in a buffer containing a relatively high [S—Br-cGMPl, then illuminated the cells while measuring their membrane potential, what would you observe? Answer Rod cells would no longer show any change in membrane potential in response to light. This experiment has been done. illumination did activate PDE, but the enzyme could not significantly reduce the 8-Br—cGMP level, which remained well above that needed to keep the gated ion channels open. Thus, light had no impact on membrane potential. Hot and Cool Taste Sensations The sensations of heat and cold are transduced by a group of temperaturewgated cation channels. For example, TRPVl, TRPVB, and TRPMS are usually closed, but open under the following conditions: TRPVl at 243 °C; TRPV3 at 238 °C; and TRPMS at <25 °C. These channels are expressed in sensory neurons known to be responsible for temperature sensation. (a) Propose a reasonable model to explain how exposing a sensory neuron containing TRPVl to high temperature leads to a sensation of heat. Capsaicin, one of the active ingredients in “hot” peppers, is an agonist of TRPVl. Capsaicin shows 50% activation of the TRPVl response at a concentration (i.e., it has an E050) of 32 nM. Explain why even a very few drops of hot pepper sauce can taste very “hot” without actually burning you. Menthol, one of the active ingredients in mint, is an agonist of TRPMS (E050 m 30 ,LLM) and TRPVS (E1050 2 20 ptM}. What sensation would you expect from contact with low levels of menthol? With high levels? (1)) (C) 3-138 ChapterlZ Biosignaling Answer (a) On exposure to heat, TRPVl channels open, causing an influx of Na+ and Ca2+ into the sensory neuron. This depolarizes the neuron, triggering an action potential. When the action potentiai reaches the axon terminus, neurotransmitter is released, signaling the nervous system that heat has been sensed. (b) Capsaicin mimics the effects of heat by binding to and opening the TRPVl channel at low temperature, leading to the false sensation of heat. The extremely low E050 indi- cates that even very small amounts of capsaicin will have dramatic sensory effects. (c) At low levels, menthol should Open the TRPMS channel, ieading to a sensation of cool; at high levels, both TRPMS and TRPVS will open, leading to a mixed sensation of cool and heat, such as you may have experienced with very strong peppermints. 22. Oncogenes, Tumor-Suppressor Genes, and Tumors For each of the following situations, provide a plausible explanation for how it could lead to unrestricted cell division. (a) Colon cancer cells often contain mutations in the gene encoding the prostaglandin E3 receptor. PGEZ is a growth factor required for the division of cells in the gastrointestinal tract. (b) Kaposi sarcoma, a common tumor in people with untreated AIDS, is caused by a virus carrying a gene for a protein similar to the chemokine receptors CXCR1 and CXCR2. Chemokines are cell- specific growth factors. ' (c) Adenovirus, a tumor virus, carries a gene for the protein ElA, which binds to the retinobiastoma protein, pr. (Hint: See Fig. 12418.) (d) An important feature of many oncogenes and tumor suppressor genes is their celi~type speci— ficity. For example, mutations in the PGE2 receptor are not typically found in lung tumors. Explain this observation. (Note that PGEZ acts through a GPCR in the plasma membrane.) Answer (a) These mutations might lead to permanent activation of the PGEg receptor. The mutant cells would behave as though stimulatory levels of PGEZ were always present, leading to unregulated cell division and tumor formation. (1)) The viral gene nught encode a constitutively active form of the receptor, such that the cells send a constant signal for cell division. This unrestrained division would lead to tumor formation. (c) ElA protein might bind to pr and prevent E2]? from binding, so E2]? is constantly active as a transcription factor. It constantly activates genes that trigger cell division, so cells divide uncontrofiably. (d) Lung cells do not normally respond to PGEZ because they do not express the PGEg receptor; mutations resulting in a constitutiveiy active PGEZ receptor do not affect lung cells. 23. Mutations in Tumor Suppressor Genes and Oncogenes Explain why mutations in tumor suppressor genes are recessive (both copies of the gene must be defective for the regulation of cell division to be defective), whereas mutations in oncogenes are dominant. Answer A tumor suppressor gene in its normal cellular form encodes a protein that re- strains cell division. Mutant forms of the protein fail to suppress cell division, but if either of the two alleles of the gene present in the individual encodes a normal protein, normal function will continue. Only if both alleles are defective will the suppression of cell division fail, leading to unregulated division. An oncogene in its normal form encodes a regulatory protein that sig~ mate the cell to divide, but only when other, external or internal factors (such as growth fac- tors) signal cell division. If a defective oncogene product is formed by either of the two alleles, unregulated cell growth and division will occur: the mutant protein sends the signal for ceil division, whether or not growth factors are present. Chapter 12 Biosignaiing 34.39 24. Retinoblastoma in Children Explain why some chiidren with retinoblastoma develop multiple tn~ more of the retina in both eyes, whereas others have a singie tumor in oniy one eye. Answer Chiidren who develop multiple tumors in both eyes were born with a defective copy of the Rb gene, occurring in every cell of the retina. Early in their lives, as retinai Ceils divided, one or several celis independently underwent a second mutation that damaged the remaining good c0py of the Rb gene Each celi with two defective Rb alleles develops into a tumor. In the later onset, 'singietumor form of the disease, children were born with two good copies of the Rb gene. A tumor develops when mutation in a single retinal cell damages one aileie of the Rb gene, then a second mutation damages the second allele in the same cetl. Two mutations in the same gene in the same cell are extremely rare, and when this does happen, it occurs in only one ceil and develops into a single tumor. 25. Specificity of 3. Signal for a Single Cell Type Discuss the validity of the following proposition. A signaling molecule (hormone, growth factor, or neurotransmitter) elicits identical responses in differ- ent types of target cells if they contain identicai receptors. Answer ri‘he proposition is invalid. Two cells expressing the same surface receptor for a given hormone may have different complements of target proteins for phosphorylation by protein kinases, resulting in different physiological and biochemicai responses in different ceils. Data Analysis Problem 26. Exploring Taste Sensation in Mice Figure 1241 shows the signai—transduction pathway for sweet taste in mammals. Pleasing tastes are an evolutionary adaptation to encourage animals to consume nutritious foods. Zhao and coauthors {2003} examined the two major pleasurable taste sensations: sweet and umami. Urnami is a “distinct savory taste” triggered by amino acids, especiaily aspartate and giutamate, and probably encourages animals to consume protein-rich foods. Monosodium giutamate (MSG) is a flavor enhancer that expioits this sensitivity. At the time the article was published, specific taste receptor proteins (labeled SR in Fig. 1241) for sweet and urnami had been tentatively characterized. Three such proteins were knOW~T1Ri, T1R2, and TiR3~which function as heterodimeric receptor complexes: TlRl-Tl R3 was tentatively identified as the umami receptor, and TlRZ—T1R3 as the sweet receptor. It was not clear how taste sensation was encoded and sent to the brain, and two possible modeis had been suggested. In the cell- based model, individual taste-sensing cells express oniy one kind of receptor; that is, there are “sweet cells," “bitter belts," “umami cells,” and so on, and each type of cell sends its information to the brain via a different nerve. The brain “knows” which taste is detected by the identity of the nerve fiber that transmits the message. In the receptor-based model, individual taste-sensing coils have severai kinds of receptors and send different messages along the same nerve fiber to the brain, the message depend ing on which receptor is activated. Aiso unclear at the time was whether there was any interaction be- tween the different tests sensations, or whether parts of one taste-sensing system were required for other taste sensations. (a) Previous work had shown that different taste receptor proteins are expressed in nonoverlapping sets of taste receptor cells. Which model does this support? Explain your reasoning. Zhao and colleagues constructed a set of “knockout nuce"————mice homozygous for ioss-of—function alleles for one of the three receptor proteins, TlRl, TlRZ, or T1R3w~and double~knockout mice with nonfunctioning TlRZ and TlR3. The researchers measured the taste perception of these mice by mea- suring their “lick rate” of solutions containing different taste molecules. Mice will lick the spout of a feeding bottie with a pleasant—tasting solution more often than one with an unpleasant-tasting soiution. The researchers measured relative lick rates: how often the mice licked a sample soiution compared with water. A relative tick rate of i indicated no preference; <1, an aversion; and >1, a preference (b) All four types of knockout strains had the same responses to salt and bitter tastes as did wild~ type mice. Which of the above issues did this experiment address? What do you conciude from these results? 5440 Chapter 12 Biosignaling The researchers then studied umarni taste reception by measuring the relative tick rates of the dif— ferent mouse strains with different quantities of MSG in the feeding solution. Note that the solutions aiso contained inosine monophosphate (IMP), a strong potentiator of umami taste reception {and a conunon ingredient in ramen soups, along with MSG), and ameloride, which suppresses the pleasant salty taste imparted by the sodium of MSG. The results are shown in the graph. 10 Wild type and T1R2 knockout Relative lick rate TIRl knockout TIRB knockout 1 10 MSG + IMP + ameloride (mM) 100 (c) Are these data consistent with the umami taste receptor censisting of a heterodimer of TiRl and TERB? Why or why not? ((1) Which modele) of taste encoding does this result support? Explain your reasoning. Zhao and coworkers then performed a series of similar experiments using sucrose as a sweet taste. These results are shown below. 20 Wild type and T1R1 knockout _ g .. g 'I‘lRZJI‘lRS E double knockout _ T.le knockout 2"; ’I‘lRS knockout £ 1 _. . , i 1 100 1000 Sucrose (mM} (e) Are these data consistent with the sweet taste receptor consisting of a heterodimer of T1R2 and T1R3? Why or why not? (f) There were some unexpected responses at very high sucrose concentrations. How do these com- plicate the idea of a heterodirneric system as presented above? In addition to sugars, hmnans also taste other compounds (eg, the peptides monellin and asparw tame) as sweet; mice do not taste these as sweet. Zhao and coworkers inserted into T1R2 knockout mice a copy of the human T1R2 gene under the control of the mouse T1R2 promoter. These modified mice now tasted monellin and saccharin as sweet. The researchers then went further, adding to THE knockout mice the RASSL protein—a G proteinmiinked receptor for the synthetic opiate spiradoline; the RASSL gene was under the control of a promoter that could be induced by feeding the mice tetracyciine. These mice did not prefer spiradoline in the absence of tetracycline; in the presence of tetracycline, they showed a strong preference for nanomolar concentrations of spiradoline. ' (g) How do these results strengthen Zhao and coauthors’ conciusions about the mecharnsm of taste sensation? Chapter12 Biosignaiing 3-141 Answer (a) The cell—based model, which predicts different receptors present on different coils. (in) This experiment addresses the issue of the independence of different taste sensations. Even though the receptors for sweet and/or umami are missing, the animais’ other taste sensations are normal; thus, pieasant and unpleasant taste sensations are independent. (c) Yes. Loss of either Tlitl or TIRB subunits abolishes umami taste sensation. ((1) Both modeis. With either modei, removing one receptor would abolish that taste sensation. (e) Yes. Loss of either the TIRE or TiRS subunits almost completeiy abolishes the sweet taste sensation; complete elimination of sweet taste requires deletion of both subunits. (f) At very high sucrose concentrations, TIRZ and, to a iesser extent, T1 R3 receptors, as homodimers, can detect sweet taste. (g) The results are consistent with either model of taste encoding, but do strengthen the researchers’ conclusions. Ligand binding can be completeiy separated from taste sensa- tion. if the ligand for the receptor in “sweettasting cells” binds a molecule, mice prefer that molecuie as a sweet compound. Reference Zhao, G.Q., Zhang, Y., H0011, Meet, Chandrashekar, J ., Erlenbach, L, Ryba, N.J.P., 8; Zuker, C. (2003) The receptors for mam- malian sweet and umarai taste. Cell 115, 255—266. ...
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ch12 - Biosignaling 1. Hormone Experiments in CellmFree...

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