Test Four Chapter 16

When a signal needs to be sent to most cells throughout a multicellular organism, the signal most suited for this is a ___________. (a) neurotransmitter. (b) hormone. (c) dissolved gas. (d) scaffold.

(b)

Acetylcholine binds to a GPCR on heart muscle, making the heart beat more slowly. The activated receptor stimulates a G protein, which opens a K+ channel in the plasma membrane, as shown in Figure Q16-32. Which of the following would enhance this effect of the acetylcholine? Figure Q16-32 (a) addition of a high concentration of a nonhydrolyzable analog of GTP (b) addition of a drug that prevents the α subunit from exchanging GDP for GTP (c) mutations in the acetylcholine receptor that weaken the interaction between the receptor and acetylcholine (d) mutations in the acetylcholine receptor that weaken the interaction between the receptor and the G protein

Choice (a) is correct. The heart is induced to beat more slowly by the binding of acetylcholine to a GPCR, activating a G protein whose βγ complex binds to and opens K+ channels. The addition of high concentrations of a nonhydrolyzable analog of GTP will increase the length of time that the G protein βγ complex remains free of the α subunit and able to activate the K+ channel; this will therefore enhance the effect of acetylcholine [choice (a)]. All the other choices will make it more difficult for the signal to proceed from the GPCR to the K+ channel.

Match the class of cell-surface receptor with the best description of its function. Not all descriptors will be used. G-protein-coupled receptors ______ ion-channel-coupled receptors ______ enzyme-coupled receptors ______ A. alter the membrane potential directly by changing the permeability of the plasma membrane B. signal by opening and closing in a ligand-independent manner C. must be coupled with intracellular monomeric GTP-binding proteins D. all receptors of this class are polypeptides with seven transmembrane domains E. discovered for their role in responding to growth factors in animal cells

G-protein-coupled receptors ___D___ ion-channel-coupled receptors ___A___ enzyme-coupled receptors ___E___

Which of the following statements is true? (a) Extracellular signal molecules that are hydrophilic must bind to a cell- surface receptor so as to signal a target cell to change its behavior. (b) To function, all extracellular signal molecules must be transported by their receptor across the plasma membrane into the cytosol. (c) A cell-surface receptor capable of binding only one type of signal molecule can mediate only one kind of cell response. (d) Any foreign substance that binds to a receptor for a normal signal molecule will always induce the same response that is produced by that signal molecule on the same cell type.

Only choice (a) is true. A hydrophilic molecule cannot diffuse across the membrane and it can therefore only affect a cell if it binds to a cell-surface receptor [choice (a)]. Most signal molecules remain bound to the extracellular domain of the receptor, whereas the intracellular domain mediates signal transduction; although many signal molecules are endocytosed with their receptor, they remain inside membrane-bounded compartments and are therefore not transported into the cytosol [choice (b)]. A cell-surface receptor capable of binding only one type of signal molecule can stimulate more than one kind of cell response, depending on the types of intracellular signaling pathway it activates [choice (c)]. Foreign substances that bind to a receptor for a normal signal molecule can sometimes induce the same response as the natural signal molecule, but in other cases they can block the binding of the natural signal molecule without activating the receptor [choice (d)].

Intracellular steroid hormone receptors have binding sites for a signaling molecule and a DNA sequence. How is it that the same steroid hormone receptor, which binds to a specific DNA sequence, can regulate different genes in different cell types?

The specific genes regulated in response to an activated steroid hormone receptor depends not only on the genes having the appropriate DNA sequence for binding the receptor but also on a variety of other nuclear proteins that influence gene expression, some of which vary between different cell types.

Which of the following mechanisms is not directly involved in inactivating an activated RTK? (a) dephosphorylation by serine/threonine phosphatases (b) dephosphorylation by protein tyrosine phosphatases (c) removal of the RTK from the plasma membrane by endocytosis (d) digestion of the RTK in lysosomes

(a) RTKs are phosphorylated on tyrosines by their dimerization partner, which is also a tyrosine kinase, and thus the reversal of these phosphorylations involves protein tyrosine phosphatases, and not protein serine/threonine phosphatases. Endocytosis of the receptor and its ultimate digestion in the lysosome are other methods that the cell uses to down- regulate active receptors.

Figure Q16-18 shows the pathway through which nitric oxide (NO) triggers smooth muscle relaxation in a blood-vessel wall. Which of the following situations would lead to relaxation of the smooth muscle cells in the absence of acetylcholine? Figure Q16-18 (a) a smooth muscle cell that has a defect in guanylyl cyclase such that it cannot bind NO (b) a muscle cell that has a defect in guanylyl cyclase such that it constitutively converts GTP to cyclic GMP (c) a muscle cell that has cyclic GMP phosphodiesterase constitutively active (d) a drug that blocks an enzyme involved in the metabolic pathway from arginine to NO

(b) A constitutively active guanylyl cyclase will produce cyclic GMP (cGMP) even in the absence of a signal and thus will lead to relaxation of smooth muscle cells in the absence of acetylcholine. Choice (a) would lead to a block in the production of cGMP such that even if NO were to reach the smooth muscle cells, relaxation would not occur. Choice (c) would not lead to muscle-cell relaxation independently of acetylcholine, because cyclic GMP phosphodiesterase is involved in the degradation of cGMP. Choice (d) will lead to a block in the production of NO.

You are interested in further understanding the signal transduction pathway that controls the production of Pig1, a protein important for regulating cell size. Activation of the TRK receptor leads to activation of the GTP-binding protein, Ras, which then activates a protein kinase that phosphorylates the SZE transcription factor. SZE only interacts with the nuclear transport receptor when it is phosphorylated. SZE is a gene activator for the Pig1 gene. This pathway is diagrammed in Figure Q16-50. Normal cells grown under standard conditions (without ligand) are 14 μm in diameter while normal cells exposed to TRK ligand are 10.5 μm in diameter. Given this situation, which of the following conditions do you predict will more likely lead to smaller cells? (a) addition of TRK ligand and a drug that stimulates the GTPase activity of Ras (b) addition of TRK ligand and a drug that inhibits the activity of the phosphatase that acts on SZE (c) addition of TRK ligand and a drug that stimulates the degradation of Pig1 (d) addition of TRK ligand and a drug that inhibits Pig1 binding to DNA

(b) The activation of the TRK receptor and its downstream signaling pathway leads to smaller cells. Thus, by inhibiting the dephosphorylation of SZE, SZE will likely activate the expression of Pig1 longer than it would in normal cells, leading to smaller cells. All other scenarios interfere with TRK receptor signaling, and should lead to cells that are not as small.

The local mediator nitric oxide stimulates the intracellular enzyme guanylyl cyclase by ________________. (a) activating a G protein. (b) activating a receptor tyrosine kinase. (c) diffusing into cells and stimulating the cyclase directly. (d) activating an intracellular protein kinase.

(c)

Figure Q16-63 shows how normal signaling works with a Ras protein acting downstream of an RTK. You examine a cell line with a constitutively active Ras protein that is always signaling. Which of the following conditions will turn off signaling in this cell line? Figure Q16-63 (a) addition of a drug that prevents protein X from activating Ras (b) addition of a drug that increases the affinity of protein Y and Ras (c) addition of a drug that blocks protein Y from interacting with its target (d) addition of a drug that increases the activity of protein Y

(c) If protein Y cannot interact with its target, signaling will not occur. Increasing the activity of protein Y [choice (d)] or increasing the affinity of protein Y and Ras [choice (b)] would not turn off signaling. Preventing protein X from activating Ras [choice (a)] would have no effect in a cell line with a constitutively active Ras protein, because Ras is already active.

A protein kinase can act as an integrating device in signaling if it ___________________. (a) phosphorylates more than one substrate. (b) catalyzes its own phosphorylation. (c) is activated by two or more proteins in different signaling pathways. (d) initiates a phosphorylation cascade involving two or more protein kinases.

(c) Integrating devices are able to relay signals from more than one signaling pathway. Being activated by two or more proteins in different signaling pathways allows a kinase (or any other signaling molecule) to be affected by more than one upstream signal. Choices (a), (b), and (d) affect the output signal that a kinase is able to produce, not its ability to integrate upstream signals from more than one signaling pathway.

The length of time a G protein will signal is determined by _______. (a) the activity of phosphatases that turn off G proteins by dephosphorylating Gα. (b) the activity of phosphatases that turn GTP into GDP. (c) the degradation of the G protein after Gα separates from Gβγ. (d) the GTPase activity of Gα.

(d)

Adrenaline stimulates glycogen breakdown in skeletal muscle cells by ultimately activating glycogen phosphorylase, the enzyme that breaks down glycogen, as depicted in Figure Q16-36. Figure Q16-36 Which of the following statements is false? (a) A constitutively active mutant form of PKA in skeletal muscle cells would lead to a decrease in the amount of unphosphorylated phosphorylase kinase. (b) A constitutively active mutant form of PKA in skeletal muscle cells would not increase the affinity of adrenaline for the adrenergic receptor. (c) A constitutively active mutant form of PKA in skeletal muscle cells would lead to an excess in the amount of glucose available. (d) A constitutively active mutant form of PKA in skeletal muscle cells would lead to an excess in the amount of glycogen available.

(d) A constitutively active mutant form of PKA in skeletal muscle cells would lead to a decrease in the amount of glycogen available, because active PKA stimulates enzymes that are responsible for the breakdown of glycogen so that glucose can be produced. All the other statements are true.

Which of the following statements about G-protein-coupled receptors (GPCRs) is false? (a) GPCRs are the largest family of cell-surface receptors in humans. (b) GPCRs are used in endocrine, paracrine, and neuronal signaling. (c) GPCRs are found in yeast, mice, and humans. (d) The different classes of GPCR ligands (proteins, amino acid derivatives, or fatty acids) bind to receptors with different numbers of transmembrane domains.

(d) Although it is true that many types of ligands can bind to and activate GPCRs, all GPCRs have a similar structure with seven transmembrane domains.

When the neurotransmitter acetylcholine is applied to skeletal muscle cells, it binds the acetylcholine receptor and causes the muscle cells to contract. Succinylcholine, which is a chemical analog of acetylcholine, binds to the acetylcholine receptor on skeletal muscle cells but causes the muscle cells to relax; it is therefore often used by surgeons as a muscle relaxant. Propose a model for why succinylcholine causes muscle relaxation. What might be the mechanism to explain the different activities of acetylcholine and succinylcholine on the acetylcholine receptor?

Although succinylcholine can bind to the acetylcholine receptor, it does not activate the receptor and therefore does not cause the muscle cell to contract. Instead, succinylcholine blocks the ability of acetylcholine to bind to the receptor and thereby prevents acetylcholine from stimulating muscle contraction.

Cells can signal to each other in various ways. A signal that must be relayed to the entire body is most efficiently sent by __________________ cells, which produce hormones that are carried throughout the body through the bloodstream. On the other hand, __________________ methods of cell signaling do not require the release of a secreted molecule and are used for very localized signaling events. During __________________ signaling, the signal remains in the neighborhood of the secreting cell and thus acts as a local mediator on nearby cells. Finally, _________________ signaling involves the conversion of electrical impulses into a chemical signal. Cells receive signals through a __________________, which can be an integral membrane protein or can reside inside the cell. amplification G protein phosphorylation contact-dependent K+ channel receptor endocrine neuronal target epithelial paracrine

Cells can signal to each other in various ways. A signal that must be relayed to the entire body is most efficiently sent by endocrine cells, which produce hormones that are carried throughout the body through the bloodstream. On the other hand, contact-dependent methods of cell signaling do not require the release of a secreted molecule and are used for very localized signaling events. During paracrine signaling, the signal remains in the neighborhood of the secreting cell and thus acts as a local mediator on nearby cells. Finally, neuronal signaling involves the conversion of electrical impulses into a chemical signal. Cells receive signals through a receptor, which can be an integral membrane protein or can reside inside the cell.

Activated GPCRs activate G proteins by reducing the strength of binding of GDP to the α subunit of the G protein, allowing GDP to dissociate and GTP (which is present at much higher concentrations in the cell than GDP) to bind in its place. How would the activity of a G protein be affected by a mutation that reduces the affinity of the α subunit for GDP without significantly changing its affinity for GTP?

The mutant G protein would be constantly active. Each time the α subunit hydrolyzed GTP to GDP, the GDP would dissociate spontaneously, allowing GTP to bind and reactivate the α subunit, especially because the intracellular concentration of GTP is higher than that of GDP. Normally, GDP is tightly bound by the α subunit, which keeps the G protein in its inactive state until interaction with an appropriate activated GPCR stimulates the release of GDP.

Explain why the signal molecules used in neuronal signaling work at a longer range than those used in contact-dependent signaling.

The neurotransmitter released from a neuron in neuronal signaling must diffuse across the synaptic cleft to reach receptors on the target cell. In contrast, in contact-dependent signaling, the signal molecule is attached to the plasma membrane of the signaling cell and interacts with receptors located on the plasma membrane of the receiving cell; thus, the cells must be in direct contact for this type of signaling to occur.

Match the target of the G protein with the appropriate signaling outcome. adenylyl cyclase ________ A. cleavage of inositol phospholipids ion channels _________ B. increase in cAMP levels phospholipase C _________ C. changes in membrane potential

adenylyl cyclase ___B_____ ion channels ___C_____ phospholipase C ___A_____

Name the three main classes of cell-surface receptor.

ion-channel-coupled receptors; G-protein-coupled receptors; enzyme-coupled receptors

Activated protein kinase C (PKC) can lead to the modification of the membrane lipids in the vicinity of the active PKC. Figure Q16-38 shows how G proteins can indirectly activate PKC. You have discovered the enzyme activated by PKC that mediates the lipid modification. You call the enzyme Rafty and demonstrate that activated PKC directly phosphorylates Rafty, activating it to modify the plasma membrane lipids in the vicinity of the cell where PKC is active; these lipid modifications can be detected by dyes that bind to the modified lipids. Cells lacking Rafty do not have these modifications, even when PKC is active. Which of the following conditions would lead to signal-independent modification of the membrane lipids by Rafty? Figure Q16-38 (a) the expression of a constitutively active phospholipase C (b) a mutation in the GPCR that binds the signal more tightly (c) a Ca2+ channel in the endoplasmic reticulum with an increased affinity for IP3 (d) a mutation in the gene that encodes Rafty such that the enzyme can no longer be phosphorylated by PKC

(a) A constitutively active phospholipase C will lead to the constitutive production of IP3 and diacylglycerol, leading to activation of PKC in a signal-independent manner; thus, Rafty activation and the lipid modification will be signal-independent. Choices (b) and (c) will increase activity of the signal transduction pathway in a signal-dependent manner. Choice (d) will prevent PKC from activating Rafty and will thus prevent the lipid modifications.

When Ras is activated, cells will divide. A dominant-negative form of Ras clings too tightly to GDP. You introduce a dominant-negative form of Ras into cells that also have a normal version of Ras. Which of the following statements is true? (a) The cells you create will divide less frequently than normal cells in response to the extracellular signals that typically activate Ras. (b) The cells you create will run out of the GTP necessary to activate Ras. (c) The cells you create will divide more frequently compared to normal cells in response to the extracellular signals that typically activate Ras. (d) The normal Ras in the cells you create will not be able to bind GDP because the dominant-negative Ras binds to GDP too tightly.

(a) Dominant-negative Ras is considered "dominant" because it prevents normal Ras from doing its job. This mutant protein, when overexpressed in a cell, binds to—and essentially monopolizes—other signaling partners in the pathway, but it cannot activate the target proteins that lie downstream. In this way, dominant-negative Ras mutants block Ras signaling and inhibit cell proliferation. The creation of GTP in the cell is independent of Ras and its associated signaling pathways [choice (b)]. An activated Ras mutant will cause cells to divide more frequently. In this situation, the Ras pathway does not work and cells should not divide more frequently [choice (c)]. The binding of GDP to dominant-negative Ras should not affect the ability of normal Ras to bind GDP, as the concentration of GTP in the cell is quite high, and the normal Ras is inactive and bound to the GDP it created from GTP the last time it was active.

You are interested in cell-size regulation and discover that signaling through a GPCR called ERC1 is important in controlling cell size in embryonic rat cells. The G protein downstream of ERC1 activates adenylyl cyclase, which ultimately leads to the activation of PKA. You discover that cells that lack ERC1 are 15% smaller than normal cells, while cells that express a mutant, constitutively activated version of PKA are 15% larger than normal cells. Given these results, which of the following treatments to embryonic rat cells should lead to smaller cells? (a) addition of a drug that causes cyclic AMP phosphodiesterase to be hyperactive (b) addition of a drug that prevents GTP hydrolysis by Gα (c) addition of a drug that activates adenylyl cyclase (d) addition of a drug that mimics the ligand of ERC1

(a) Hyperactivating cyclic AMP phosphodiesterase will degrade the cAMP, terminating the signal more quickly than usual. All other answers will lead to larger cells.

The growth factor RGF stimulates proliferation of cultured rat cells. The receptor that binds RGF is a receptor tyrosine kinase called RGFR. Which of the following types of alteration would be most likely to prevent receptor dimerization? (a) a mutation that increases the affinity of RGFR for RGF (b) a mutation that prevents RGFR from binding to RGF (c) changing the tyrosines that are normally phosphorylated on RGFR dimerization to alanines (d) changing the tyrosines that are normally phosphorylated on RGFR dimerization to glutamic acid

(b) Binding of a ligand to RTKs leads to their dimerization, and thus a mutation that prevents RGFR from binding to RGF will prevent dimerization. A mutation that increases the affinity of RGFR for RGF will increase dimerization in the presence of ligand [choice (a)]. RTKs become phosphorylated on dimerization. However, changing the relevant tyrosines to alanine will block receptor activation but should not cause or prevent dimerization [choice (c)]. Because glutamic acid is negatively charged, it can mimic the addition of a phosphate to an amino acid; thus, changing the relevant tyrosines to glutamic acid may mimic receptor activation, but it should not cause or prevent receptor dimerization [choice (d)].

Male cockroaches with mutations that strongly decrease the function of an RTK called RTKX are oblivious to the charms of their female comrades. This particular RTK binds to a small molecule secreted by sexually mature females. Most males carrying loss-of-function mutations in the gene for Ras protein are also unable to respond to females. You have just read a paper in which the authors describe how they have screened cockroaches that are mutant in RTKX for additional mutations that partly restore the ability of males to respond to females. These mutations decrease the function of a protein that the authors call Z. Which of the following types of protein could Z be? Explain your answer. (a) a protein that activates the Ras protein by causing Ras to exchange GDP for GTP (b) a protein that stimulates hydrolysis of GTP by the Ras protein (c) an adaptor protein that mediates the binding of the RTKX to the Ras protein (d) a transcriptional regulator required for the expression of the Ras gene

(b) Mutations that increase the activity of Ras should mimic the effect of stimulating RTKX in a receptor-independent fashion. Because the intracellular concentration of GTP is higher than that of GDP, some proportion of the Ras molecules is expected to be GTP- bound and active; ridding the cells of a protein that stimulates GTP hydrolysis will increase this pool of active Ras. Mutants that cannot stimulate Ras to exchange GDP for GTP will have the same phenotype as mutants lacking Ras [choice (a)], as will mutants lacking a transcriptional regulator required for expression of the Ras gene [choice (d)]. Defects in an adaptor protein that mediates the binding of receptor X to Ras will have no further effect on a mutant already lacking the receptor [choice (c)].

The activation of the serine/threonine protein kinase Akt requires phosphoinositide 3-kinase (PI 3-kinase) to _________. (a) activate the RTK. (b) create phosphorylated lipids that serve as docking sites that localize Akt to the plasma membrane. (c) directly phosphorylate Akt. (d) to create DAG.

(b) PI 3-kinase activity causes the localization of Akt to the plasma membrane, where it is phosphorylated by another protein kinase. The RTK is activated before PI 3-kinase is activated [choice (a)]. PI 3-kinase phosphorylates lipids, not proteins such as Akt [choice (c)]. DAG is created by phospholipase C and is not involved in this particular signaling pathway [choice (d)].

The growth factor Superchick stimulates the proliferation of cultured chicken cells. The receptor that binds Superchick is a receptor tyrosine kinase (RTK), and many chicken tumor cell lines have mutations in the gene that encodes this receptor. Which of the following types of mutation would be expected to promote uncontrolled cell proliferation? (a) a mutation that prevents dimerization of the receptor (b) a mutation that destroys the kinase activity of the receptor (c) a mutation that inactivates the protein tyrosine phosphatase that normally removes the phosphates from tyrosines on the activated receptor (d) a mutation that prevents the binding of the normal extracellular signal to the receptor

(c) RTKs are usually activated by signal-induced dimerization, which allows the receptors to phosphorylate themselves and activate intracellular signaling proteins that are stimulated by the phosphorylated receptor. After it is activated, the receptor is dephosphorylated, and thereby inactivated, by a protein tyrosine phosphatase. Therefore, a mutation in the gene that encodes the protein tyrosine phosphatase will inappropriately increase the activity of the receptor and promote uncontrolled cell proliferation. Mutations that prevent dimerization of the receptor (including mutations that prevent ligand binding) or autophosphorylation (which requires the kinase activity of the receptor) will inactivate the receptor.

During nervous-system development in Drosophila, the membrane-bound protein Delta acts as an inhibitory signal to prevent neighboring cells from developing into neuronal cells. Delta is involved in ______________ signaling. (a) endocrine (b) paracrine (c) neuronal (d) contact-dependent

(d)

The rod photoreceptors in the eye are extremely sensitive to light. The cells sense light through a signal transduction cascade involving light activation of a GPCR that activates a G protein that activates cyclic GMP phosphodiesterase. How would you expect the addition of the following drugs to affect the light-sensing ability of the rod cells? Explain your answers. A. a drug that inhibits cyclic GMP phosphodiesterase B. a drug that is a nonhydrolyzable analog of GTP

A. A drug that inhibits cyclic GMP phosphodiesterase would decrease any light response in the rod cell. Normally, cyclic GMP is continuously being produced in the eye. The perception of light by a rod cell normally leads to the activation of cyclic GMP phosphodiesterase, which then hydrolyzes cyclic GMP molecules. This causes Na+ channels to close, which changes the membrane potential and alters the signal sent to the brain. If cyclic GMP phosphodiesterase were blocked, levels of cyclic GMP would remain high and there would be no cellular response to light. B. A drug that is a nonhydrolyzable analog of GTP would lead to a prolonged response to light. This is because a nonhydrolyzable analog of GTP would prevent the G protein from turning itself off by hydrolyzing its bound GTP to GDP. Continued activation of the G protein would keep cyclic GMP phosphodiesterase levels higher than normal, leading to a prolonged period of lowered levels of cyclic GMP. This in turn would cause Na+ channels to be closed for longer than normal, leading to a prolonged change in the membrane potential and an extended light response.

Receipt of extracellular signals can change cell behavior quickly (for example, in seconds or less) or much more slowly (for example, in hours). A. What kind of molecular changes could cause quick changes in cell behavior? B. What kind of molecular changes could cause slow changes in cell behavior? C. Explain why the response you named in A results in a quick change, whereas the response you named in B results in a slow change.

A. Any answer that involves the modification of existing cell components is correct. Protein phosphorylation, protein dephosphorylation, protein ubiquitylation, lipid phosphorylation, and lipid cleavage are all examples of correct answers. B. Responses that involve alterations in gene expression occur slowly. C. Modification of existing cell components can happen quickly, whereas responses that depend on changes in gene expression take much longer, because the genes will need to be transcribed, the mRNAs will need to be translated, and the proteins need to accumulate to high-enough levels to instigate change.

A calmodulin-regulated kinase (CaM-kinase) is involved in spatial learning and memory. This kinase is able to phosphorylate itself such that its kinase activity is now independent of the intracellular concentration of Ca2+. Thus, the kinase stays active after Ca2+ levels have dropped. Mice completely lacking this CaM-kinase have severe spatial learning defects but are otherwise normal. A. Each of the following mutations also leads to similar learning defects. For each case explain why. (1) a mutation that prevents the kinase from binding ATP (2) a mutation that deletes the calmodulin-binding part of the kinase (3) a mutation that destroys the site of autophosphorylation B. What would be the effect on the activity of CaM-kinase if there were a mutation that reduced its interaction with the protein phosphatase responsible for inactivating the kinase?

A. Because a complete lack of the CaM-kinase causes a learning defect, we can assume that mutations leading to inactivation of the kinase would also have a similar effect. (1) Protein kinases have a binding site for ATP, which is the source of the phosphate used for phosphorylating their target proteins; if the kinase cannot bind ATP, it will be inactive. (2) Because binding to calmodulin in the presence of Ca2+ activates CaM-kinases, deletion of the calmodulin-binding portion would inactivate the kinase. (3) A mutation that destroys the site of autophosphorylation will also impair the normal function of the kinase, because the kinase will become inactive as soon as Ca2+ levels decrease. B. The kinase would stay active for longer after a transient increase in intracellular Ca2+ concentration.

Your friend is studying a segment of a newly discovered virus that carries an enhancer of gene expression that confers responsiveness to glucocorticoid (a hormone) on genes that are linked to it. He constructs two versions of a reporter gene: one has only a minimal promoter linked to it (which contains sites for RNA polymerase binding); the other reporter gene has both this minimal promoter plus the viral enhancer attached to it. The reporter gene allows him to measure the amount of transcription that occurs from each construct. Your friend puts each of these constructs into two different cell lines and examines the expression of the reporter gene in each cell line, as shown in Figure Q16-25. He is puzzled by these findings and asks for your help in interpreting them. Figure Q16-25 A. From these data, can you tell whether both cell lines contain glucocorticoid receptors? Why? B. What might account for the difference in the transcription of the reporter gene in cell lines 1 and 2 after introduction of the construct containing the viral enhancer in the presence of glucocorticoid?

A. Both cell lines seem to contain the glucocorticoid receptor, because both cell lines demonstrate increases in reporter gene expression in response to the addition of glucocorticoid. With the introduction of the construct containing the glucocorticoid-responsive viral enhancer, cell line 1 showed a 1000-fold increase and cell line 2 showed an 80-fold increase. B. There are four reasonable explanations for this difference in glucocorticoid responses in the two cell lines (any one would be correct). 1. Cell line 1 contains a protein that does not exist in cell line 2. This protein cooperates with the glucocorticoid receptor to activate the transcription of the reporter gene, leading to higher levels of the reporter gene in cell line 1 than in cell line 2. 2. Cell line 2 contains a slightly defective glucocorticoid receptor that does not bind DNA as strongly as the normal glucocorticoid receptor in cell line 1, thus leading to lower levels of reporter gene activity in cell line 2 than in cell line 1. 3. Cell line 1 contains a mutant glucocorticoid receptor that binds DNA much more strongly than the normal glucocorticoid receptor found in cell line 2. Therefore, higher levels of reporter gene activity are seen in cell line 1 than in cell line 2. 4. Cell line 2 contains a protein that does not exist in cell line 1. This protein acts as an antagonist to the glucocorticoid receptor to decrease the level of transcription from the reporter gene, leading to lower levels of reporter gene expression in cell line 2 than in cell line 1.

Acetylcholine acts at a GPCR on heart muscle to make the heart beat more slowly. It does so by ultimately opening K+ channels in the plasma membrane (as diagrammed in Figure Q16-32), which decreases the cell's excitability by making it harder to depolarize the plasma membrane. Indicate whether each of the following conditions would increase or decrease the effect of acetylcholine. A. addition of a drug that stimulates the GTPase activity of the Gα subunit B. mutations in the K+ channel that keep it closed all the time C. modification of the Gα subunit by cholera toxin D. a mutation that decreases the affinity of the βγ complex of the G protein for the K+ channel E. a mutation in the acetylcholine receptor that prevents its localization on the cell surface F. adding acetylcholinesterase to the external environment of the cell

A. Decrease. An increase in the GTPase activity of the Gα subunit will decrease the length of time that the G protein is active. B. Decrease. If the K+ channel remains closed, acetylcholine will not slow the heart. C. Increase. Cholera toxin inhibits the GTPase activity of the Gα subunit, keeping the subunit in an active state for a longer time. D. Decrease. The activated βγ complex binds to and activates the K+ channel; decreasing their affinity for each other will decrease the time that the K+ channel is open, effectively decreasing the effect of acetylcholine. E. Decrease. If there is no receptor on the cell surface, cells will be unable to respond to acetylcholine. F. Decrease. Acetylcholinesterase degrades acetylcholine and thus will decrease the effect of acetylcholine.

Indicate by writing "yes" or "no" whether amplification of a signal could occur at the particular steps described below. Explain your answers. A. An extracellular signaling molecule binds and activates a GPCR. B. The activated GPCRs cause Gα to separate from Gβ and Gγ. C. Adenylyl cyclase produces cyclic AMP. D. cAMP activates protein kinase A. E. Protein kinase A phosphorylates target proteins.

A. No. Each signaling molecule activates only one receptor molecule. B. Yes. Each activated GPCR activates many G-protein molecules. C. Yes. Each activated adenylyl cyclase molecule can generate many molecules of cAMP. D. No. In unstimulated cells, protein kinase A is held inactive in a protein complex. Binding of cAMP to the complex induces a conformational change, releasing the active protein kinase A. Therefore, one cAMP cannot activate more than one molecule of protein kinase A. E. Yes. Each activated protein kinase A molecule can phosphorylate many molecules of each type of target protein.

Two protein kinases, PK1 and PK2, work sequentially in an intracellular signaling pathway. You create cells that contain inactivating mutations in the genes that encode either PK1 or PK2 and find that these cells no longer respond to a particular extracellular signal. You also create cells containing a version of PK1 that is permanently active and find that the cells behave as though they are receiving the signal even when the signal is not present. When you introduce the permanently active version of PK1 into cells that have an inactivating mutation in PK2, you find that these cells also behave as though they are receiving the signal even when no signal is present. A. From these results, does PK1 activate PK2, or does PK2 activate PK1? Explain your answer. B. You now create a permanently active version of PK2 and find that cells containing this version behave as though they are receiving the signal even when the signal is not present. What do you predict will happen if you introduce the permanently active version of PK2 into cells that have an inactivating mutation in PK1?

A. Normally, PK2 activates PK1. We are told that PK1 and PK2 normally work sequentially in an intracellular signaling pathway. If PK1 is permanently activated, a response is seen independently of whether or not PK2 is present. If PK1 activated PK2, no response should be seen if PK1 were activated in the absence of PK2. B. You would predict that no response to the signal would be observed. This is because PK2 normally needs to activate PK1 for the cells to respond to the signal. When PK2 is permanently activated in the absence of PK1, PK1 is not there to relay the signal.

When activated by the extracellular signal protein platelet-derived growth factor (PDGF), the PDGF receptor phosphorylates itself on multiple tyrosines (as indicated in Figure Q16-55A by the circled Ps; the numbers next to these Ps indicate the amino acid number of the tyrosine). These phosphorylated tyrosines serve as docking sites for proteins that interact with the activated PDGF receptor. These proteins are indicated in the figure, and include the proteins A, B, C, and D. One of the cell's responses to PDGF is an increase in DNA synthesis, which can be measured by the incorporation of radioactive thymidine into the DNA. To determine which protein or proteins—A, B, C, or D—are responsible for the activation of DNA synthesis, you construct mutant versions of the PDGF receptor that retain one or more tyrosine phosphorylation sites. You express these mutant versions in cells that do not make their own PDGF receptor. In these cells, the various mutant versions of the PDGF receptor are expressed normally, and, in response to PDGF binding, become phosphorylated on whichever tyrosines remain. You measure the level of DNA synthesis in cells that express the various mutant receptors and obtain the data shown in Figure Q16-55B. Figure Q16-55 A. From these data, which, if any, of proteins A, B, C, and D are involved in the stimulation of DNA synthesis by PDGF? Explain your answer. B. Which, if any, of these proteins inhibit DNA synthesis? Explain your answer. C. Which, if any, of these proteins seem to have no detectable role in DNA synthesis? Explain your answer.

A. Proteins A and D stimulate DNA synthesis. PDGF receptors that can bind to only A or D (see experiments 2 and 5) can stimulate DNA synthesis to about 50% of normal amounts (which is represented by experiment 1). Proteins A and D are both needed and are used in an additive fashion; this is evident from experiment 6: when a PDGF receptor can bind both A and D, the DNA synthesis level is close to that obtained with the normal receptor. B. Protein B is an inhibitor of DNA synthesis. Consequently, receptors with binding sites for B and D (see experiment 7) stimulate a lower DNA synthesis rate than do receptors that bind only D (experiment 5). C. Protein C has no detectable role in DNA synthesis. Receptors that can bind only C (experiment 4) activate DNA synthesis about as much as receptors that do not bind any of the four proteins (experiment 9; the negative control). Furthermore, the binding of protein C does not affect the response mediated by protein D when the receptor can bind both C and D (experiment 8).

You are interested in cell-size regulation and discover that signaling through an enzyme-coupled receptor is important for the growth (enlargement) of mouse liver cells. Activation of the receptor activates adenylyl cyclase, which ultimately leads to the activation of PKA, which then phosphorylates a transcription factor called TFS on threonine 42. This phosphorylation is necessary for the binding of TFS to its specific sites on DNA, where it then activates the transcription of Sze2, a gene that encodes a protein important for liver cell growth. You find that liver cells lacking the receptor are 15% smaller than normal cells, whereas cells that express a constitutively activated version of PKA are 15% larger than normal liver cells. Given these results, predict whether you would expect the cell's size to be bigger or smaller than normal cells if cells were treated in the following fashion. A. You change threonine 42 on TFS to an alanine residue. B. You create a version of the receptor that is constitutively active. C. You add a drug that inhibits adenylyl cyclase. D. You add a drug that increases the activity of cyclic AMP phosphodiesterase. E. You mutate the cAMP-binding sites in the regulatory subunits of PKA, so that the complex binds cAMP more tightly.

A. Smaller. This mutation will make a TFS that cannot be phosphorylated by PKA. B. Bigger. C. Smaller. D. Smaller. cAMP phosphodiesterase is involved in converting cAMP to AMP and will down-regulate this signaling pathway. E. Bigger. Higher affinity of the PKA complex for cAMP will increase its activity, and thus cells will be bigger.

Circle the phrase in each pair that is likely to occur more rapidly in response to an extracellular signal. A. changes in cell secretion / increased cell division B. changes in protein phosphorylation / changes in proteins being synthesized C. changes in mRNA levels / changes in membrane potential

A. changes in cell secretion B. changes in protein phosphorylation C. changes in membrane potential

When adrenaline binds to adrenergic receptors on the surface of a muscle cell, it activates a G protein, initiating an intracellular signaling pathway in which the activated α subunit activates adenylyl cyclase, thereby increasing cAMP levels in the cell. The cAMP molecules then activate a cAMP-dependent kinase (PKA) that, in turn, activates enzymes that result in the breakdown of muscle glycogen, thus lowering glycogen levels. You obtain muscle cells that are defective in various components of the signaling pathway. Referring to Figure Q16-36, indicate how glycogen levels would be affected in the presence of adrenaline in the following cells. Would they be higher or lower than in normal cells treated with adrenaline? A. cells that lack adenylyl cyclase B. cells that lack the GPCR C. cells that lack cAMP phosphodiesterase D. cells that have an α subunit that cannot hydrolyze GTP but can interact properly with the β and γ subunits

A. higher B. higher C. lower D. lower

An extracellular signal molecule can act to change a cell's behavior by acting through cell-surface __________________ that control intracellular signaling proteins. These intracellular signaling proteins ultimately change the activity of __________________ proteins that bring about cell responses. Intracellular signaling proteins can __________________ the signal received to evoke a strong response from just a few extracellular signal molecules. A cell that receives more than one extracellular signal at the same time can __________________ this information using intracellular signaling proteins. __________________ proteins can act as molecular switches, letting a cell know that a signal has been received. Enzymes that phosphorylate proteins, termed ___________, can also serve as molecular switches; the actions of these enzymes are countered by the activity of __________________. acetylase decouple GTP-binding AMP-binding decrease neurotransmitter amplify effector protein kinases autocrine esterases protein phosphatases cleavage integrate receptors convolute GMP-binding sterols

An extracellular signal molecule can act to change a cell's behavior by acting through cell-surface receptors that control intracellular signaling proteins. These intracellular signaling proteins ultimately change the activity of effector proteins that bring about cell responses. Intracellular signaling proteins can amplify the signal received to evoke a strong response from just a few extracellular signal molecules. A cell that receives more than one extracellular signal at the same time can integrate this information using intracellular signaling proteins. GTP-binding proteins can act as molecular switches, letting a cell know that a signal has been received. Enzymes that phosphorylate proteins, termed protein kinases, can also serve as molecular switches; the actions of these enzymes are countered by the activity of protein phosphatases.

Can signaling via a steroid hormone receptor lead to amplification of the original signal? If so, how?

Because the interactions of the signal molecule with its receptor and of the activated receptor with its gene are both one-to-one, there is no amplification in this part of the signaling pathway. The signal can, however, be amplified when the target genes are transcribed, because each activated gene produces multiple copies of mRNA, each of which is used to make multiple copies of the protein that the gene encodes.

Ca2+ can trigger biological effects in cells because an unstimulated cell has an extremely __________________ concentration of free Ca2+ in the cytosol, compared with its concentration in the __________________ space and in the __________________, creating a steep electrochemical gradient. When Ca2+ enters the cytosol, it interacts with Ca2+-responsive proteins such as __________________, which also binds diacylglycerol, and __________________, which activates CaM-kinases. adenylyl cyclase endoplasmic reticulum nuclear average extracellular peroxisome Ca2+ high phospholipase C calmodulin intracellular protein kinase A colorful low protein kinase C

Ca2+ can trigger biological effects in cells because an unstimulated cell has an extremely low concentration of free Ca2+ in the cytosol, compared with its concentration in the extracellular space and in the endoplasmic reticulum, creating a steep electrochemical gradient. When Ca2+ enters the cytosol, it interacts with Ca2+-responsive proteins such as protein kinase C, which also binds diacylglycerol, and calmodulin, which activates CaM-kinases.

Cells signal to one another in various ways. Some use extracellular signal molecules that are dissolved gases, such as __________________, which can diffuse easily into cells. Others use cytokines, which bind to cytokine receptors. Cytokine receptors have no intrinsic enzyme activity but are associated with cytoplasmic tyrosine kinases called __________________s, which become activated on the binding of cytokine to its receptor and go on to phosphorylate and activate cytoplasmic transcriptional regulators called __________________s. Some intracellular signaling pathways involve chains of protein kinases that phosphorylate each other, as seen in the __________________ signaling module. Lipids can also relay signals in the cell, as we observe when phospholipase C cleaves the sugar-phosphate head off a lipid molecule to generate the two small messenger molecules __________________ (which remains embedded in the plasma membrane) and __________________ (which diffuses into the cytosol). cyclic GMP MAP kinase STAT diacylglycerol NO TGFβ IP3 Ras JAK SMAD

Cells signal to one another in various ways. Some use extracellular signal molecules that are dissolved gases, such as NO, which can diffuse easily into cells. Others use cytokines, which bind to cytokine receptors. Cytokine receptors have no intrinsic enzyme activity but are associated with cytoplasmic tyrosine kinases called JAKs, which become activated on the binding of cytokine to its receptor and go on to phosphorylate and activate cytoplasmic transcriptional regulators called STATs. Some intracellular signaling pathways involve chains of protein kinases that phosphorylate each other, as seen in the MAP kinase signaling module. Lipids can also relay signals in the cell, as we observe when phospholipase C cleaves the sugar-phosphate head off a lipid molecule to generate the two small messenger molecules diacylglycerol (which remains embedded in the plasma membrane) and IP3 (which diffuses into the cytosol).

Which of the following statements about molecular switches is false? (a) Phosphatases remove the phosphate from GTP on GTP-binding proteins, turning them off. (b) Protein kinases transfer the terminal phosphate from ATP onto a protein. (c) Serine/threonine kinases are the most common types of protein kinase. (d) A GTP-binding protein exchanges its bound GDP for GTP to become activated.

Choice (a) is false. GTP-binding proteins themselves hydrolyze their bound GTP to GDP, using their own intrinsic GTPase activity.

The ethylene response in plants involves a dimeric transmembrane receptor. When the receptor is not bound to ethylene, the receptor binds to and activates a protein kinase, which activates an intracellular signaling pathway that leads to the degradation of a transcriptional regulator important for transcribing the ethylene- responsive genes (see Figure Q16-60). You discover a phosphatase that is important for ethylene signaling, and you name it PtpE. Plants lacking PtpE never turn on ethylene-responsive genes, even in the presence of ethylene. You find that PtpE dephosphorylates serine 121 on the transcriptional regulator. Furthermore, plants lacking PtpE degrade the transcriptional regulator in the presence of ethylene. Figure Q16-60 Which of the following statements is inconsistent with your data? (a) When the transcriptional regulator is phosphorylated, it activates transcription of the ethylene-responsive genes. (b) When the transcriptional regulator is not phosphorylated, it binds to DNA. (c) Activation of the protein kinase that binds to the ethylene receptor leads to inactivation of PtpE. (d) Binding of ethylene to its receptor leads to the activation of PtpE.

Choice (a) is inconsistent. Cells lacking PtpE do not transcribe the ethylene-reponsive genes, suggesting that dephosphorylation of the transcriptional regulator is required before it can activate transcription. From your data, PtpE is either constitutively active or inactivated by the active protein kinase downstream of the ethylene receptor. The latter scenario would permit maximum repression in the absence of ethylene. All the other choices are consistent with your data.

Which of the following statements is true? (a) Because endocrine signals are broadcast throughout the body, all cells will respond to the hormonal signal. (b) The regulation of inflammatory responses at the site of an infection is an example of paracrine signaling. (c) Paracrine signaling involves the secretion of signals into the bloodstream for distribution throughout the organism. (d) The axons of neurons typically signal target cells using membrane-bound signaling molecules that act on receptors in the target cells.

Choice (b) is correct. Choice (a) is not true because only the cells with a receptor for the hormone will respond to the signal. Choice (c) is not true because paracrine signaling involves signaling in a more local fashion, unlike endocrine signaling where signals are sent through the bloodstream. Choice (d) is untrue because axons typically signal using diffusible neurotransmitters that are released at the synapse.

Which of the following statements is true? (a) MAP kinase is important for phosphorylating MAP kinase kinase. (b) PI 3-kinase phosphorylates a lipid in the plasma membrane. (c) Ras becomes activated when an RTK phosphorylates its bound GDP to create GTP. (d) STAT proteins phosphorylate JAK proteins, which then enter the nucleus and activate gene transcription.

Choice (b) is correct. MAP kinases are phosphorylated by MAP kinase kinases [choice (a)]. Ras exchanges its GDP for GTP when activated [choice (c)]. JAK proteins are receptor-associated kinases that phosphorylate the transcriptional regulators called STAT proteins [choice (d)].

During the mating process, yeast cells respond to pheromones secreted by other yeast cells. These pheromones bind GPCRs on the surface of the responding cell and lead to the activation of G proteins inside the cell. When a wild-type yeast cell senses the pheromone, its physiology changes in preparation for mating: the cell stops growing until it finds a mating partner. If yeast cells do not undergo the appropriate response after sensing a pheromone, they are considered sterile. Yeast cells that are defective in one or more components of the G protein have characteristic phenotypes in the absence and presence of the pheromone, which are listed in Table 16-34. Table Q16-34 Which of the following models is consistent with the data from the analysis of these mutants? Explain your answer. (a) α activates the mating response but is inhibited when bound to βγ (b) βγ activates the mating response but is inhibited when bound to α (c) the G protein is inactive; either free α or free βγ complex is capable of activating the mating response (d) the G protein is active; both free α and free βγ complex are required to inhibit the mating response

Choice (b) is correct. Single mutations in the β and γ subunits of the G protein permit growth in the absence of pheromone and display a sterile phenotype in the presence of pheromone, whereas loss of the α subunit causes growth arrest in either the presence or the absence of pheromone. Because arrested growth is a normal response that occurs when the cells sense a pheromone, it must be the βγ complex that normally activates the cellular mating response when released from the α subunit. In cells lacking α, the βγ complex causes growth arrest inappropriately because α normally inhibits the action of the complex until it is activated by the binding of pheromone to the yeast cell. This interpretation is consistent with the analysis of the double mutants in which α is deleted together with either β or γ. In these double mutants, normal growth is seen in the absence of pheromone because β and γ act together as a complex to activate target proteins; a lack of either β or γ leads to no response to pheromone, even when α is not present to inhibit the response.

Cell lines A and B both survive in tissue culture containing serum but do not proliferate. Factor F is known to stimulate proliferation in cell line A. Cell line A produces a receptor protein (R) that cell line B does not produce. To test the role of receptor R, you introduce this receptor protein into cell line B, using recombinant DNA techniques. You then test all of your various cell lines in the presence of serum for their response to factor F, with the results summarized in Table Q16-1. Table Q16-1 Which of the following cannot be concluded from your results above? (a) Binding of factor F to its receptor is required for proliferation of cell line A. (b) Receptor R binds to factor F to induce cell proliferation in cell line A. (c) Cell line A expresses a receptor for factor F. (d) Factor F is not required for proliferation in cell line B.

Choice (b) is false. Expressing receptor R in cell line B causes proliferation in the absence of factor F, suggesting that something else in the serum can bind to receptor R in cell line B. We do not know the effect of eliminating receptor R from cell line A, and thus we cannot say whether binding of receptor R to its ligand is required for cell line A to proliferate. The data support all the other answers. Because cell line A proliferates only in the presence of factor F, binding of factor F must be required for cell line A to proliferate [choice (a)]. Because cell line A proliferates in response to factor F, it must express a receptor for factor F since it responds to factor F [choice (c)]. Because cell line B can proliferate in the absence of factor F once R is expressed, factor F is not required for cell line B to proliferate [choice (d)].

Acetylcholine is a signaling molecule that elicits responses from heart muscle cells, salivary gland cells, and skeletal muscle cells. Which of the following statements is false? (a) Heart muscle cells decrease their rate and force of contraction when they receive acetylcholine, whereas skeletal muscle cells contract. (b) Heart muscle cells, salivary gland cells, and skeletal muscle cells all express an acetylcholine receptor that belongs to the transmitter-gated ion channel family. (c) Active acetylcholine receptors on salivary gland cells and heart muscle cells activate different intracellular signaling pathways. (d) Heart muscle cells, salivary gland cells, and skeletal muscle cells all respond to acetylcholine within minutes of receiving the signal.

Choice (b) is not true. Only skeletal muscle cells express an acetylcholine receptor that belongs to the transmitter-gated ion channel family; salivary gland cells and heart muscle cells express a different receptor. The other choices are all true.

The last common ancestor to plants and animals was a unicellular eukaryote. Thus, it is thought that multicellularity and the attendant demands for cell communication arose independently in these two lineages. This evolutionary viewpoint accounts nicely for the vastly different mechanisms that plants and animals use for cell communication. Fungi use signaling mechanisms and components that are very similar to those used in animals. Which of the phylogenetic trees shown in Figure Q16-59 does this observation support?

Choice (b) is the correct answer (see Figure Q16-59). The similarities in signaling mechanisms between animals and fungi support the phylogenetic tree in which fungi branched from the animal lineage after plants and animals separated. This branching order is supported by a wide variety of other data, including genomic sequence comparisons.

Foreign substances like nicotine, morphine, and menthol exert their initial effects by _____. (a) killing cells immediately, exerting their physiological effects by causing cell death. (b) diffusing through cell plasma membranes and binding to transcription factors to change gene expression. (c) interacting with cell-surface receptors, causing the receptors to transduce signal inappropriately in the absence of the normal stimulus. (d) removing cell-surface receptors from the plasma membrane.

Choice (c) is correct. Although some foreign substances will remove cell-surface receptors from the plasma membrane [choice (d)], this is a long-term response and not part of the initial response.

The following happens when a G-protein-coupled receptor activates a G protein. (a) The β subunit exchanges its bound GDP for GTP. (b) The GDP bound to the α subunit is phosphorylated to form bound GTP. (c) The α subunit exchanges its bound GDP for GTP. (d) It activates the α subunit and inactivates the βγ complex.

Choice (c) is correct. Choice (d) is incorrect because the βγ complex can also activate downstream targets. The other statements are simply untrue.

You are interested in how cyclic-AMP-dependent protein kinase A (PKA) functions to affect learning and memory, and you decide to study its function in the brain. It is known that, in the cells you are studying, PKA works via a signal transduction pathway like the one depicted in Figure Q16-35. Furthermore, it is also known that activated PKA phosphorylates the transcriptional regulator called Nerd that then activates transcription of the gene Brainy. Which situation described below will lead to an increase in Brainy transcription? Figure Q16-35 (a) a mutation in the Nerd gene that produces a protein that cannot be phosphorylated by PKA (b) a mutation in the nuclear import sequence of PKA from PPKKKRKV to PPAAAAAV (c) a mutation in the gene that encodes cAMP phosphodiesterase that makes the enzyme inactive (d) a mutation in the gene that encodes adenylyl cyclase that renders the enzyme unable to interact with the α subunit of the G protein

Choice (c) is correct. cAMP phosphodiesterase is important for converting cAMP into AMP and thus down-regulating the activity of PKA. Without cAMP phosphodiesterase, transcription of Brainy will be increased. All the other choices will lead to inactivation of the signaling pathway and a decrease in Brainy transcription. A mutant form of Nerd that cannot be phosphorylated will not be active [choice (a)]. If PKA cannot be imported into the nucleus, it will be unable to phosphorylate Nerd [choice (b)]. Adenylyl cyclase interaction with the α subunit of the G protein is important for the G protein's activation [choice (d)].

Figure Q16-61 shows that intracellular signaling pathways can be highly interconnected. Figure Q16-61 From the information in Figure Q16-61, which of the following statements is incorrect? (a) The GPCR and the RTK both activate phospholipase C. (b) Activation of either the GPCR or the RTK will lead to activation of transcriptional regulators. (c) CaM-kinase is only activated when the GPCR is active and not when the RTK is active. (d) Ras is activated only when the RTK is active and not when the GPCR is active.

Choice (c) is incorrect. CaM-kinase is activated by calmodulin, which is ultimately activated by phospholipase C. Either the GPCR or the RTK activates phospholipase C. All the other statements are correct.

The lab you work in has discovered a previously unidentified extracellular signal molecule called QGF, a 75,000-dalton protein. You add purified QGF to different types of cells to determine its effect on these cells. When you add QGF to heart muscle cells, you observe an increase in cell contraction. When you add it to fibroblasts, they undergo cell division. When you add it to nerve cells, they die. When you add it to glial cells, you do not see any effect on cell division or survival. Given these observations, which of the following statements is most likely to be true? (a) Because it acts on so many diverse cell types, QGF probably diffuses across the plasma membrane into the cytoplasm of these cells. (b) Glial cells do not have a receptor for QGF. (c) QGF activates different intracellular signaling pathways in heart muscle cells, fibroblasts, and nerve cells to produce the different responses observed. (d) Heart muscle cells, fibroblasts, and nerve cells must all have the same receptor for QGF.

Choice (c) is most likely to be true. Because heart muscle cells, fibroblasts, and nerve cells all respond to QGF with different outcomes, it is likely that QGF activates different effector proteins in these different cell types, leading to the diversity of outcomes observed in the experiment. QGF is unlikely to diffuse across the cell membrane, given that it is a large protein [choice (a)]. Although glial cells do not die or divide in response to QGF, they could have a receptor for QGF, as receptor activation could lead to some other response [choice (b)]. A signal molecule can often bind to different types of receptor on different cell types, so choice (d) may or may not be correct.

All members of the steroid hormone receptor family __________________. (a) are cell-surface receptors. (b) do not undergo conformational changes. (c) are found only in the cytoplasm. (d) interact with signal molecules that diffuse through the plasma membrane.

Choice (d) is correct. All members of the steroid hormone receptor family are intracellular proteins [thus choice (a) is not correct] that interact with signal molecules that can diffuse through the plasma membrane. Once activated, steroid hormone receptors regulate gene transcription in the nucleus [choice (c)]. The binding of the signal molecule induces a large conformational change in the receptor protein [choice (b)]. This conformational change activates the steroid hormone receptors, allowing them to promote or inhibit the transcription of the appropriate genes.

Akt promotes the survival of many cells by affecting the activity of Bad and Bcl2, as diagrammed in Figure Q16-58. Figure Q16-58 Which of the following statements is false? (a) In the presence of a survival signal, Akt is phosphorylated. (b) In the absence of a survival signal, Bad inhibits the cell-death inhibitor protein Bcl2. (c) In the presence of a survival signal, the cell-death inhibitory protein Bcl2 is active. (d) In the absence of a survival signal, Bad is phosphorylated.

Choice (d) is false. Bad is phosphorylated in the presence of a survival signal. When the survival signal is not present, Bad binds to the cell-death inhibitor protein Bcl2, promoting cell death. All the other statements are correct.

Which of the following statements is false? (a) Nucleotides and amino acids can act as extracellular signal molecules. (b) Some signal molecules can bind directly to intracellular proteins that bind DNA and regulate gene transcription. (c) Some signal molecules are transmembrane proteins. (d) Dissolved gases such as nitric oxide (NO) can act as signal molecules, but because they cannot interact with proteins they must act by affecting membrane lipids.

Choice (d) is not true. NO can diffuse across the plasma membrane and directly activate intracellular proteins such as the enzyme guanylyl cyclase.

Bacteria undergo chemotaxis toward amino acids, which usually indicates the presence of a food source. Chemotaxis receptors bind a particular amino acid and cause changes in the bacterial cell that induce the cell to move toward the source of the amino acid. Four types of chemotaxis receptor that mediate responses to different amino acids have been identified in a bacterium. The receptors are called ChrA, ChrB, ChrC, and ChrD. Each receptor specifically senses serine, aspartate, glutamate, or glycine, although you do not know which receptor senses which amino acid. You have been given a wild-type bacterial strain that contains all four receptors, as well as various mutant bacterial strains that are lacking one or more of the receptors. To figure out which receptor senses which amino acid, you conduct experiments in which you fill a capillary tube with an amino acid to attract the bacteria, dip the capillary tube into a solution containing bacteria, remove the capillary tube after 5 minutes, and count the number of bacteria in the capillary tube. Your results are shown in Table Q16-65. Table Q16-65 From these results, indicate which receptor is used for which amino acid.

ChrA senses glycine, ChrB senses aspartate, ChrC senses glutamate, and ChrD senses serine. To figure this out, you must match the pattern of intact receptors with the pattern of responses to the various amino acids. For example, ChrD is missing in strain 2, which does not sense serine. Therefore, ChrD is the receptor used to sense serine. Since we know that one of the receptors senses glutamate, but all strains respond to glutamate, ChrC must be the sensor for glutamate because it is present in all strains.

G-protein-coupled receptors (GPCRs) all have a similar structure with __________________ transmembrane domains. When a GPCR binds an extracellular signal, an intracellular G protein, composed of __________________ subunits, becomes activated. __________________ of the G-protein subunits are tethered to the plasma membrane by short lipid tails. When unstimulated, the α subunit is bound to __________________, which is exchanged for __________________ on stimulation. The intrinsic __________________ activity of the α subunit is important for inactivating the G protein. __________________ inhibits this activity of the α subunit, thereby keeping the subunit in an active state. adenylyl cyclase cholera toxin GTPase AMP diacylglycerol phosphodiesterase ATP five seven ATPase four three Ca2+ GDP twelve cAMP GTP two

G-protein-coupled receptors (GPCRs) all have a similar structure with seven transmembrane domains. When a GPCR binds an extracellular signal, an intracellular G protein, composed of three subunits, becomes activated. Two of the G-protein subunits are tethered to the plasma membrane by short lipid tails. When unstimulated, the α subunit is bound to GDP, which is exchanged for GTP on stimulation. The intrinsic GTPase activity of the α subunit is important for inactivating the G protein. Cholera toxin inhibits this activity of the α subunit, thereby keeping the subunit in an active state.

Antibodies are Y-shaped molecules that have two identical binding sites. Suppose that you have obtained an antibody that is specific for the extracellular domain of an RTK. When the antibody binds to the RTK, it brings together two RTK molecules. If cells containing the RTK were exposed to the antibody, would you expect the kinase to be activated, inactivated, or unaffected? Explain your reasoning.

The RTK will probably become activated on binding of the antibody molecule. This is because signal-induced dimerization usually activates RTKs. When RTK molecules are brought together, their cytoplasmic kinase domains become activated and each receptor phosphorylates the other.

Your friend is studying mouse fur color and has isolated the GPCR responsible for determining its color, as well as the extracellular signal that activates the receptor. She finds that, on addition of the signal to pigment cells (cells that produce the pigment determining fur color), cAMP levels rise in the cell. She starts a biotech company, and the company isolates more components of the signaling pathway responsible for fur color. Using transgenic mouse technology, the company genetically engineers mice that are defective in various proteins involved in determining fur color. The company obtains the following results. Normal mice have beige (very light brown) fur color. Mice lacking the extracellular signal have white fur. Mice lacking the GPCR have white fur. Mice lacking cAMP phosphodiesterase have dark brown fur. Your friend has also made mice that are defective in the α subunit of the G protein in this signaling pathway. The defective α subunit works normally except that, once it binds GTP, it cannot hydrolyze GTP to GDP. What color do you predict that the fur of these mice will be? Why?

These mice will have dark brown fur. The inability to hydrolyze GTP to GDP will lead to inappropriate activation of the signaling pathway that makes pigment. Too much pigment will be produced, as seen in the mice lacking cAMP phosphodiesterase (which lack the ability to damp the signal), and the mice will end up with dark brown fur.

Given the generic signaling pathway in Figure Q16-12, write the number corresponding to the item on the line next to the descriptor below. Figure Q16-12 _________ receptor protein _________ effector proteins _________ intracellular signaling proteins _________ ligand

____2____ receptor protein ____4____ effector proteins ____3____ intracellular signaling proteins ____1___ ligand

Rank the following types of cell signaling from 1 to 4, with 1 representing the type of signaling in which the signal molecule travels the least distance and 4 the type of signaling in which the signal molecule travels the largest distance. ______ paracrine signaling ______ contact-dependent signaling ______ neuronal signaling ______ endocrine signaling

____3__ paracrine signaling ____1__ contact-dependent signaling ____2__ neuronal signaling ____4__ endocrine signaling