Signal Transduction Pathways - AP Biology

A bacterial two-component system uses membrane sensor kinase S and response regulator R. Ligand L binds S, triggering autophosphorylation on histidine and transfer of the phosphate to aspartate on R. Phosphorylated R binds enzyme E and increases its catalytic rate. S also has phosphatase activity that dephosphorylates R when L is absent. A mutant S lacks phosphatase activity but retains kinase activity. After brief L exposure, phosphorylated R remains high longer in the mutant than in wild type. Which outcome is most likely in the mutant after L removal?
Answer: Enzyme E activity stays elevated longer because phosphorylated R persists without dephosphorylation This question assesses the skill of analyzing a signal transduction pathway. In the mutant, enzyme E activity stays elevated longer because phosphorylated R persists without dephosphorylation, as S lacks phosphatase activity but retains kinase function, prolonging the signal after L removal. In the two-component system, L activates S to phosphorylate R, which binds and activates E, and normally S dephosphorylates R. The prolonged phosphorylated R in mutants supports this. Choice C is tempting but wrong because it assumes phosphatase is required for kinase function, which is a misconception since kinase activity is retained. A transferable strategy is to assess phosphorylation dynamics in bacterial systems to understand regulatory balances between kinases and phosphatases.
In a pathway, ligand L activates receptor R, leading to activation of small GTPase Q. Q-GTP activates enzyme Z, which produces second messenger M. M activates kinase K, which phosphorylates target T. A GTPase-activating protein (GAP) increases GTP hydrolysis on Q, promoting Q-GDP. When GAP is overexpressed, M levels and phosphorylated T are reduced after L stimulation. Which intervention would most likely restore phosphorylated T despite GAP overexpression?
Answer: Add membrane-permeable M to activate kinase K downstream of Q and enzyme Z This question assesses the skill of analyzing signal transduction pathways by evaluating interventions to restore downstream signaling when an upstream regulator is perturbed. The pathway involves ligand L activating receptor R, which then activates GTPase Q to its GTP-bound form, enabling enzyme Z to produce second messenger M, which in turn activates kinase K to phosphorylate target T. Overexpression of GAP accelerates GTP hydrolysis on Q, shifting it to the inactive GDP form, thereby reducing M production and subsequent phosphorylation of T despite L stimulation. Adding a membrane-permeable form of M directly bypasses the reduction in Q-GTP and Z activity, allowing activation of kinase K and restoration of phosphorylated T levels. A tempting distractor is choice B, which suggests decreasing ligand L concentration to conserve Q-GTP, but this reflects the misconception that reducing input signal preserves limited active components, when it actually diminishes overall pathway activation. A transferable strategy is to identify the point of disruption in a pathway and consider downstream interventions that circumvent the block to restore function.
In a yeast cell, mating factor F binds GPCR R, activating G protein and a MAPK cascade: kinase M1 activates M2, which activates M3. M3 phosphorylates target protein T, changing its activity within minutes. A scaffold protein holds M1, M2, and M3 in a complex near the receptor. In cells lacking the scaffold, receptor binding and G protein activation occur, but phosphorylation of T is much lower and slower. Which mechanism best explains the scaffold effect?
Answer: The scaffold increases signaling efficiency by colocalizing kinases to promote sequential phosphorylation This question assesses the skill of analyzing a signal transduction pathway. The scaffold increases signaling efficiency by colocalizing M1, M2, and M3 near the receptor, promoting rapid sequential phosphorylation in the MAPK cascade despite normal receptor and G protein activation. In the yeast pathway, F activates R and G protein to the cascade, phosphorylating T, but without the scaffold, efficiency drops, leading to lower and slower T phosphorylation. This explains the scaffold's role in organization. Choice D is tempting but wrong because it suggests the scaffold transports T to the nucleus, which is a misconception as the effect is on phosphorylation rate, not localization. A transferable strategy is to consider protein complexes and scaffolds when analyzing speed and efficiency in multi-step kinase pathways.
In muscle cells, epinephrine binds a β-adrenergic GPCR, activating Gαs and adenylyl cyclase to raise cAMP. cAMP activates PKA, which phosphorylates glycogen phosphorylase kinase (GPK). Activated GPK phosphorylates glycogen phosphorylase (GP), increasing glucose-1-phosphate release. A student adds a nonhydrolyzable GTP analog that locks Gαs in its GTP-bound state, then adds epinephrine. Compared with epinephrine alone, GP activity increases more and remains elevated longer. Which step is most directly affected by the nonhydrolyzable GTP analog?
Answer: Hydrolysis of GTP on Gαs, prolonging adenylyl cyclase stimulation and cAMP production This question analyzes how a nonhydrolyzable GTP analog affects the β-adrenergic signal transduction pathway. The pathway involves epinephrine → GPCR → Gαs-GTP → adenylyl cyclase → cAMP → PKA → GPK → GP. Normally, Gαs hydrolyzes GTP to GDP, terminating its activation of adenylyl cyclase. The nonhydrolyzable GTP analog prevents this hydrolysis, keeping Gαs locked in its active GTP-bound state, which prolongs adenylyl cyclase stimulation and sustained cAMP production (choice B). Students often mistakenly think the analog affects ligand binding (choice A), but it specifically affects the G protein's GTPase activity. When analyzing G protein pathways, remember that GTP hydrolysis is the key termination mechanism.
In a plant cell assay, peptide P binds a membrane receptor kinase and triggers phosphorylation of a cytosolic relay protein K. Phosphorylated K activates an NADPH oxidase in the plasma membrane, producing reactive oxygen species (ROS). ROS oxidize cysteines on a nearby protein phosphatase (PP), decreasing PP activity. PP normally dephosphorylates K, limiting signal duration. When P is added, K phosphorylation rises quickly and then plateaus. A mutant line expresses PP that cannot be oxidized by ROS but retains normal catalytic activity. Compared with wild-type cells, what is most likely in the mutant after P addition?
Answer: Lower K phosphorylation because PP remains active and continues dephosphorylating K This question examines how feedback regulation affects signal transduction dynamics. In wild-type cells, ROS production leads to phosphatase (PP) oxidation and inactivation, reducing K dephosphorylation and allowing K phosphorylation to plateau at a higher level. In the mutant, PP cannot be oxidized by ROS, so it remains fully active throughout signaling. This continued PP activity constantly dephosphorylates K even as the receptor kinase phosphorylates it, resulting in lower steady-state K phosphorylation compared to wild-type cells where PP is partially inactivated. Choice A incorrectly assumes the mutation permanently inactivates PP, when actually it prevents ROS-mediated inactivation while maintaining normal catalytic function. When analyzing feedback loops, consider how preventing regulatory modifications affects the balance between activating and deactivating processes.
A ligand (L) binds a G protein–coupled receptor (GPCR) in the plasma membrane, causing the Gα subunit to exchange GDP for GTP and dissociate. Gα-GTP activates adenylyl cyclase, increasing intracellular cAMP. cAMP binds protein kinase A (PKA), releasing active catalytic subunits that phosphorylate a cytosolic enzyme E, increasing E activity. In a cell-free assay, adding L increases E activity 8-fold. When a nonhydrolyzable GTP analog is provided, E activity remains high even after L is removed. When a phosphodiesterase that degrades cAMP is added during L exposure, E activity increases only 1.5-fold. Which change would most likely prevent amplification of the signal from receptor binding to enzyme E activation?
Answer: Preventing cAMP from binding PKA regulatory subunits to release active kinase This question assesses the skill of analyzing a signal transduction pathway. The pathway involves ligand L binding to a GPCR, activating Gα to increase cAMP via adenylyl cyclase, which then activates PKA to phosphorylate and enhance enzyme E activity, creating amplification through multiple steps. Choice B prevents amplification by blocking cAMP from binding PKA, halting the release of active kinase subunits and thus stopping phosphorylation of E despite upstream activation. This directly interrupts the cascade, as evidenced by the cell-free assay where manipulations like nonhydrolyzable GTP or phosphodiesterase affect signal duration and magnitude but do not block the core amplification. A tempting distractor is D, which provides a nonhydrolyzable GTP analog, but this misconception assumes sustained Gα activation prevents amplification, whereas it actually prolongs signaling and increases E activity. A transferable strategy is to map each intervention to its specific step in the pathway to determine if it blocks, enhances, or sustains signal propagation.
In yeast cells, pheromone P binds a GPCR that activates a heterotrimeric G protein. The released Gβγ subunit activates a MAP kinase cascade (M1→M2→M3) through sequential phosphorylation. Activated M3 phosphorylates a cytosolic target protein X, changing its activity within minutes. A specific phosphatase removes phosphate groups from M3, decreasing M3 activity and limiting phosphorylation of X. When the phosphatase is overexpressed, pheromone P produces less phosphorylated X. Which manipulation would most likely increase phosphorylated X levels in cells overexpressing the phosphatase?
Answer: Use an M3 variant lacking the phosphatase recognition site, reducing dephosphorylation This question tests the ability to analyze components and interventions in a signal transduction pathway. The pathway involves pheromone P activating a GPCR and MAP kinase cascade where M1 activates M2, M2 activates M3, and M3 phosphorylates X, with a phosphatase limiting M3 activity by dephosphorylation. Using an M3 variant lacking the phosphatase recognition site would reduce dephosphorylation, prolonging M3 activation and increasing phospho-X despite phosphatase overexpression. This directly protects M3 from excessive negative regulation. A tempting distractor is choice A, increasing phosphatase activity, but this misinterprets the goal by assuming more dephosphorylation helps, when it would further decrease M3 activity and phospho-X. A transferable strategy is to alter enzyme recognition sites to mitigate overactive negative regulators in phosphorylation cascades.
A bacterial chemotaxis receptor in the membrane activates a cytosolic kinase when attractant is absent. The kinase phosphorylates response regulator RR, and RR-P binds flagellar motor proteins to increase tumbling frequency. When attractant binds the receptor, kinase activity decreases, RR phosphorylation drops, and tumbling decreases within seconds. A mutant strain has normal receptor–attractant binding but maintains high RR phosphorylation even with attractant present. Which defect most likely explains the mutant phenotype?
Answer: Loss of a phosphatase that normally dephosphorylates RR when kinase activity decreases This question assesses the skill of analyzing signal transduction pathways by deducing defects causing persistent signaling in bacterial chemotaxis. In the mutant, attractant binds normally but kinase activity remains high, keeping RR phosphorylated and tumbling frequency elevated, unlike wild type where attractant decreases kinase and RR-P. Loss of a phosphatase that dephosphorylates RR would cause RR-P to persist, explaining sustained high phosphorylation despite attractant. This defect impairs signal termination, leading to prolonged tumbling. A tempting distractor is choice C, a mutation preventing RR from binding flagellar proteins, but this would reduce tumbling rather than maintain it, misunderstanding the role of RR-P in promoting tumbling. When inferring mutations, compare mutant phenotypes to wild-type responses and pinpoint where regulation fails.
A steroid-like ligand X is membrane-impermeable and binds a cell-surface receptor that activates a heterotrimeric G protein. The G subunit activates adenylyl cyclase, increasing cAMP. cAMP activates a cyclic nucleotide–gated channel, allowing Ca $^{2+}$ influx; Ca $^{2+}$ activates a cytosolic protease that cleaves protein M, changing its activity within 90 s. When adenylyl cyclase is inhibited, Ca $^{2+}$ influx and M cleavage are not detected. Which change would most likely bypass the need for adenylyl cyclase in this pathway?
Answer: Add a cAMP analog that opens cyclic nucleotide–gated channels This question assesses the skill of analyzing signal transduction pathways by finding ways to bypass adenylyl cyclase inhibition in GPCR-cAMP signaling. Adenylyl cyclase inhibition prevents cAMP production after receptor and G protein activation, stopping cAMP-gated Ca2+ channel opening, Ca2+ influx, protease activation, and M cleavage. Adding a cAMP analog that opens cyclic nucleotide-gated channels provides the cAMP signal directly, allowing Ca2+ influx and downstream protease activity without needing adenylyl cyclase. This intervention mimics the second messenger's effect on the channel, restoring the pathway beyond the inhibited enzyme. A tempting distractor is choice A, which suggests adding a ligand that prevents G protein activation, but this would further inhibit signaling rather than bypass it, misunderstanding antagonism versus compensation. For bypassing defects, target the immediate downstream effector of the blocked component with a mimic or activator.
A growth factor binds an RTK, leading to receptor dimerization and phosphorylation. The phosphorylated RTK recruits PI3K, which converts PIP $_2$ to PIP $_3$ in the inner leaflet of the plasma membrane. PIP $_3$ recruits kinase Akt to the membrane, where Akt becomes activated and phosphorylates cytosolic target S, increasing S activity. PTEN is a lipid phosphatase that converts PIP $_3$ back to PIP $_2$ . In cells with reduced PTEN activity, growth factor exposure produces a larger and longer-lasting increase in S activity than in control cells. Which change best explains the altered signaling output?
Answer: Reduced PTEN increases PIP $_3$ persistence, prolonging Akt recruitment and activation This question assesses the skill of analyzing a signal transduction pathway. The pathway involves growth factor binding RTK, recruiting PI3K to produce PIP₃, which activates Akt to phosphorylate and enhance S activity, with PTEN normally converting PIP₃ back to PIP₂ to terminate signaling. Choice A explains the larger, longer S activity in reduced PTEN cells because PIP₃ persists, prolonging Akt membrane recruitment and activation. This aligns with the observed amplified response, as PTEN reduction removes a key negative regulator. A tempting distractor is E, suggesting reduced PTEN blocks Akt binding by removing PIP₃, but this misconception reverses PTEN's role, as lower PTEN actually increases PIP₃ and enhances docking. A transferable strategy is to evaluate how altering regulators affects signal duration and amplitude in kinase cascades.
In smooth muscle cells, ligand L binds receptor tyrosine kinase (RTK) and causes receptor dimerization and autophosphorylation on cytosolic tyrosines. An adaptor protein binds the phosphotyrosines and recruits a Ras-GEF, which converts Ras-GDP to Ras-GTP. Ras-GTP activates a kinase cascade: Raf phosphorylates MEK, and MEK phosphorylates ERK. Activated ERK phosphorylates a cytosolic enzyme, increasing its catalytic rate within minutes. A mutant cell line expresses Ras that can bind GDP but cannot exchange GDP for GTP; all other proteins are normal and L binding to RTK is unchanged. Which outcome is most likely when mutant cells are treated with ligand L?
Answer: The cytosolic enzyme shows little increase in activity because ERK activation is reduced This question requires analyzing how a mutation in a signal transduction pathway affects downstream signaling events. The RTK pathway normally proceeds: ligand binding → RTK dimerization/autophosphorylation → adaptor protein recruitment → Ras-GEF activation → Ras-GTP formation → Raf/MEK/ERK cascade → enzyme phosphorylation. The mutant Ras cannot exchange GDP for GTP, blocking its activation despite normal RTK function upstream. Without active Ras-GTP, the Raf/MEK/ERK cascade cannot be initiated, preventing ERK activation and subsequent enzyme phosphorylation. Choice A incorrectly assumes ERK activation depends only on RTK autophosphorylation, ignoring the requirement for Ras-GTP in activating the downstream kinase cascade. To analyze pathway disruptions, identify where the block occurs and predict effects on all downstream components.
Immune cells respond to ligand S by activating a receptor that stimulates phospholipase C (PLC). PLC cleaves membrane PIP $_2$ into DAG and IP $_3$ . IP $_3$ binds an ER channel, releasing Ca $^{2+}$ into the cytosol; DAG and Ca $^{2+}$ together activate protein kinase C (PKC). PKC phosphorylates substrate Q, measurable after 60 s. When cells are treated with an IP $_3$ receptor blocker, Ca $^{2+}$ release and Q phosphorylation are both reduced, even though DAG levels increase normally. Which result is best supported by the pathway described?
Answer: Blocking the IP $_3$ receptor reduces PKC activation by limiting cytosolic Ca $^{2+}$ This question assesses the skill of analyzing signal transduction pathways by interpreting the effects of blocking IP3 receptors in PLC-mediated signaling. The IP3 receptor blocker prevents Ca2+ release from the ER, reducing cytosolic Ca2+ needed alongside DAG to activate PKC, which phosphorylates Q, even though DAG increases normally. This indicates that blocking the IP3 receptor limits PKC activation by restricting Ca2+ availability, explaining the reduced Q phosphorylation. The pathway logic shows both second messengers are required for full PKC activity, so impairing one reduces the output despite the other being present. A tempting distractor is choice A, suggesting DAG alone suffices for maximal PKC, but this is wrong because the data show reduced phosphorylation without Ca2+, overlooking the cooperative role of Ca2+ and DAG. To analyze such results, compare observed outcomes to pathway requirements and identify which components are essential for activation.
In immune cells, cytokine C binds a receptor that lacks intrinsic kinase activity but is associated with a cytosolic kinase J. Ligand binding causes receptor dimerization, bringing two J molecules together so they phosphorylate each other and then phosphorylate the receptor tails. A latent cytosolic protein S binds the phosphotyrosines and becomes phosphorylated by J; phosphorylated S rapidly forms dimers and activates a cytosolic enzyme by direct binding. A small molecule is added that prevents receptor dimerization but does not block cytokine binding. Which step is most directly disrupted by the small molecule?
Answer: Autophosphorylation of J due to reduced proximity of kinase molecules This question tests identification of which step in a signal transduction pathway is directly affected by an inhibitor. The JAK-STAT pathway requires receptor dimerization to bring two JAK (J) kinases close enough to phosphorylate each other through trans-autophosphorylation. The small molecule prevents receptor dimerization, which keeps the J kinases separated and unable to phosphorylate each other. Without J autophosphorylation, the kinases cannot become active to phosphorylate the receptor or STAT (S) proteins, blocking all downstream signaling. Choice D incorrectly suggests S dimers directly activate the enzyme without mentioning that S must first be phosphorylated by active J kinases. When analyzing pathway inhibitors, identify the most proximal step affected and recognize that all downstream events will be blocked.
In cultured muscle cells, ligand L binds membrane receptor R and increases intracellular $\mathrm{Ca^{2+}}$ . R activates G protein Gα, which activates adenylyl cyclase to produce cAMP. cAMP activates protein kinase A (PKA), which phosphorylates channel C, increasing $\mathrm{Ca^{2+}}$ influx. A phosphodiesterase (PDE) hydrolyzes cAMP to AMP. When cells are treated with L plus a PDE inhibitor, cAMP rises 4× higher than with L alone, and $\mathrm{Ca^{2+}}$ influx remains elevated longer. Which change would most likely reduce the duration of the $\mathrm{Ca^{2+}}$ response to L without preventing initial receptor binding?
Answer: Increase PDE activity to accelerate cAMP breakdown after PKA activation This question assesses the skill of analyzing a signal transduction pathway. Increasing PDE activity would accelerate cAMP breakdown, reducing the duration of PKA activation and thus shortening the Ca²⁺ influx response without affecting initial receptor binding by ligand L. In the pathway, L activates the G protein and adenylyl cyclase to produce cAMP, which activates PKA to phosphorylate channel C, but faster cAMP hydrolysis by PDE would limit this. The PDE inhibitor experiment shows that slowing cAMP breakdown prolongs the response, supporting that enhancing PDE would have the opposite effect. Choice B is tempting but wrong because mutating receptor R to prevent binding would block the initial response entirely, which is a misconception about targeting downstream steps to modulate duration without preventing initiation. A transferable strategy is to identify negative regulators like degradative enzymes in pathways and consider how altering their activity affects signal timing.
In immune cells, ligand L binds a cytokine receptor lacking intrinsic kinase activity. Binding causes receptor clustering and activation of an associated cytosolic kinase J, which phosphorylates a docking protein D. Phosphorylated D recruits and activates enzyme E, which generates many molecules of second messenger M from a membrane lipid. M binds and activates protein kinase K, which phosphorylates cytosolic substrate S, altering its activity rapidly. A lipid phosphatase converts M back to the original lipid, limiting signaling. Which change would most likely increase the magnitude of substrate S phosphorylation after L is added?
Answer: Inhibiting the lipid phosphatase that removes M, allowing M to persist and activate K longer This question analyzes a cytokine receptor signal transduction pathway where signal magnitude depends on second messenger duration. The pathway shows: L binding → receptor clustering → kinase J activation → D phosphorylation → enzyme E activation → second messenger M production → kinase K activation → substrate S phosphorylation. A lipid phosphatase normally converts M back to the original lipid, terminating the signal. Inhibiting this phosphatase (choice C) would allow M to persist longer, keeping kinase K active for an extended period and increasing S phosphorylation. Students often confuse increasing versus decreasing activities (choice A increases phosphatase, which would decrease signaling). When analyzing pathways, consider how the duration of second messengers affects overall signal strength.
In epithelial cells, growth factor G binds receptor tyrosine kinase R, causing autophosphorylation and recruitment of enzyme PLC $\gamma$ . PLC $\gamma$ generates IP $_3$ , which releases $\mathrm{Ca^{2+}}$ from the ER. $\mathrm{Ca^{2+}}$ activates protein kinase C (PKC) together with DAG at the membrane. PKC phosphorylates transporter T, increasing ion transport rate. When cells are treated with a drug that chelates cytosolic $\mathrm{Ca^{2+}}$ , PKC activation and increased transport do not occur even though IP $_3$ levels rise. Which step is directly blocked by the chelator?
Answer: Activation of PKC by $\mathrm{Ca^{2+}}$ binding, preventing phosphorylation of transporter T This question assesses the skill of analyzing a signal transduction pathway. The Ca²⁺ chelator directly blocks activation of PKC by preventing Ca²⁺ binding, which is required alongside DAG for PKC to phosphorylate transporter T and increase ion transport. In the pathway, G activates R autophosphorylation, PLCγ recruitment, IP₃ production, and Ca²⁺ release, but without free Ca²⁺, PKC cannot activate despite IP₃ rising. This step is disrupted downstream of IP₃ but upstream of transport. Choice C is tempting but wrong because it describes IP₃ synthesis as preventing second messenger formation, which is a misconception reversing the role of PLCγ in generating messengers. A transferable strategy is to identify cofactor requirements like Ca²⁺ in kinase activation to predict blocks in RTK pathways.
In cultured liver cells, hormone H binds membrane receptor R and activates G protein Gα, which stimulates adenylyl cyclase to convert ATP to cAMP. cAMP binds protein kinase A (PKA), releasing active catalytic subunits that phosphorylate enzyme E, increasing its activity 6-fold within 30 s. A phosphodiesterase (PDE) hydrolyzes cAMP to AMP. When cells are treated with a PDE inhibitor, the same dose of H produces a larger and longer-lasting increase in E activity; however, without H the inhibitor causes little change. Which change would most likely eliminate the effect of the PDE inhibitor on E activity?
Answer: Mutating receptor R so H cannot bind its extracellular domain This question assesses the skill of analyzing signal transduction pathways by identifying changes that abolish the impact of a modulator on downstream enzyme activity. The pathway begins with hormone H binding receptor R to activate G protein Gα, stimulating adenylyl cyclase to produce cAMP from ATP, which activates PKA to phosphorylate and enhance enzyme E activity, while PDE hydrolyzes cAMP to limit the signal. The PDE inhibitor prolongs cAMP elevation only when H is present, leading to a larger and sustained increase in E activity, but has minimal effect without H due to low basal cAMP production. Mutating receptor R to prevent H binding blocks pathway initiation, eliminating the inhibitor's ability to enhance E activity since no cAMP is generated in response to H. A tempting distractor is choice E, which suggests overexpressing PDE to rapidly hydrolyze cAMP despite the inhibitor, but this overlooks that the inhibitor's effect depends on upstream activation, not just PDE levels, reflecting a misconception about signal dependency. A transferable strategy is to dissect how inhibitors affect pathway dynamics and test upstream blocks to confirm dependency on initial signal transduction steps.
In an experiment, ligand Q activates a membrane receptor that triggers production of second messenger M. M activates kinase A, which phosphorylates kinase B; kinase B phosphorylates enzyme E, increasing product formation. When kinase A is inhibited, neither kinase B nor E becomes phosphorylated after Q addition. When kinase B is inhibited, kinase A is still phosphorylated but E is not phosphorylated. Which conclusion is best supported by these results?
Answer: Kinase A is upstream of kinase B in the transduction pathway leading to E phosphorylation This question assesses the skill of analyzing signal transduction pathways by ordering components based on inhibitor effects. Inhibiting kinase A prevents phosphorylation of both B and E, while inhibiting B allows A phosphorylation but not E, indicating kinase A is upstream of B, which is upstream of E in the pathway from Q to product formation. This sequential logic places A before B in activating E. The results show dependency where A activates B, and B activates E. A tempting distractor is choice A, suggesting B upstream of A, but this contradicts the data where A inhibition blocks B, reversing the order misconception. For ordering pathways, use selective inhibitors to see which steps block downstream events.
A ligand L binds receptor R and activates kinase K, leading to phosphorylation of protein P. Phosphorylated P opens a chloride channel, changing membrane potential. A second protein, U, binds phosphorylated P and targets it for dephosphorylation by recruiting phosphatase D. In cells with a mutation that prevents U from binding P, channel opening persists longer after L removal. Which change would most likely reverse the persistence in the mutant cells?
Answer: Increase phosphatase D activity so P is dephosphorylated efficiently without recruitment by U This question assesses the skill of analyzing signal transduction pathways by determining modifications that can alter signal termination in a mutant context. The normal pathway features ligand L binding receptor R to activate kinase K, which phosphorylates protein P, opening a chloride channel to change membrane potential, while protein U recruits phosphatase D to dephosphorylate P and terminate the signal. In the mutation preventing U from binding P, phosphorylated P persists, leading to prolonged channel opening after L removal because dephosphorylation is impaired. Increasing phosphatase D activity allows efficient dephosphorylation of P even without U-mediated recruitment, reversing the persistence by enabling faster signal shutoff in mutant cells. A tempting distractor is choice E, which proposes increasing U transcription to overcome the mutation by mass action, but this misunderstands that the binding defect prevents U from interacting with P regardless of quantity. A transferable strategy is to trace how mutations disrupt feedback or termination steps in pathways and evaluate changes that compensate by enhancing alternative regulatory mechanisms.
In cardiac cells, epinephrine binds a β-adrenergic GPCR, activating Gs and increasing cAMP. cAMP activates PKA, which phosphorylates a Ca2+ channel, increasing Ca2+ influx. Elevated Ca2+ binds a regulatory subunit of an enzyme complex, increasing the complex's catalytic activity. A second protein, arrestin, binds phosphorylated GPCRs and prevents further G protein activation without blocking epinephrine binding. Cells are treated with a kinase inhibitor that prevents GPCR phosphorylation but does not affect PKA. Which effect is most likely during continuous epinephrine exposure?
Answer: More sustained cAMP because arrestin binding is reduced, prolonging G protein activation This question analyzes how preventing receptor desensitization affects signal duration. Normally, activated GPCRs are phosphorylated by specific kinases (like GRKs), allowing arrestin to bind and block further G protein activation, thus terminating signaling. The kinase inhibitor prevents GPCR phosphorylation, so arrestin cannot bind to terminate Gs activation. This allows the epinephrine-bound receptor to continue activating Gs proteins, producing more sustained cAMP levels and prolonged PKA activity compared to normal desensitization. Choice A incorrectly suggests unphosphorylated GPCRs cannot activate G proteins, confusing the role of phosphorylation in desensitization versus initial activation. When analyzing receptor regulation, distinguish between mechanisms for initial activation (ligand binding) and subsequent inactivation (phosphorylation and arrestin binding).

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