Signal Transduction Pathways
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AP Biology › Signal Transduction 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?
Increasing intracellular ATP concentration to provide more substrate
Deleting the catalytic subunits of PKA while leaving cAMP levels unchanged
Adding a competitive inhibitor that binds the active site of enzyme E
Overexpressing PDE so cAMP is rapidly hydrolyzed even with inhibitor present
Mutating receptor R so H cannot bind its extracellular domain
Explanation
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, cells are exposed to ligand L that activates receptor R and increases cytosolic cAMP. A FRET sensor reports PKA activity. With L alone, PKA activity peaks at 30 s and returns to baseline by 3 min. With L plus a drug that blocks receptor internalization, PKA activity remains elevated for 10 min. Ligand binding affinity is unchanged. Which explanation is best supported by these data?
Receptor internalization increases ligand concentration by importing L into the cytosol
Blocking internalization prevents cAMP synthesis but extends PKA activity by feedback
Receptor internalization is required to activate adenylyl cyclase at the membrane
Receptor internalization normally reduces signaling by removing active receptors from the membrane
Blocking internalization increases cytosolic ATP, which directly activates PKA
Explanation
This question assesses the skill of analyzing a signal transduction pathway. Receptor internalization normally reduces signaling by removing active receptors from the membrane, so blocking it prolongs PKA activity as receptors continue activating the pathway despite unchanged binding. In the pathway, L activates R to increase cAMP and PKA, with normal transient response, but blocking internalization extends it. This supports desensitization via internalization. Choice B is tempting but wrong because it suggests internalization activates cyclase, which is a misconception as the prolonged signal indicates internalization terminates signaling. A transferable strategy is to manipulate endocytosis to study desensitization mechanisms in GPCR pathways.
In a cell-free system, activated receptor fragments phosphorylate adaptor protein Ad on tyrosines. Ad recruits enzyme PI3K, which converts membrane lipid PIP$_2$ to PIP$_3$. PIP$_3$ recruits kinase Akt to the membrane, where Akt becomes phosphorylated and then phosphorylates cytosolic target X. When a lipid phosphatase that converts PIP$_3$ back to PIP$_2$ is added, Akt phosphorylation decreases and X phosphorylation drops. Which change would most likely counteract the effect of the lipid phosphatase?
Block ATP binding to Akt so it stays phosphorylated and continues phosphorylating X
Remove membrane lipids so Akt remains cytosolic and is phosphorylated more efficiently
Add extra PIP$_2$ so the phosphatase has more substrate and produces more PIP$_3$
Increase PI3K activity to raise PIP$_3$ levels despite phosphatase conversion
Inhibit Ad phosphorylation so PI3K binds more strongly to the receptor fragment
Explanation
This question assesses the skill of analyzing signal transduction pathways by countering phosphatase effects in PI3K-Akt signaling. The lipid phosphatase converts PIP3 to PIP2, reducing PIP3, Akt membrane recruitment, phosphorylation, and X phosphorylation. Increasing PI3K activity raises PIP3 levels, opposing the phosphatase and restoring Akt activation and X phosphorylation. This boosts production to overcome degradation, maintaining the lipid signal. A tempting distractor is choice E, adding PIP2 to increase PIP3 via phosphatase, but phosphatases degrade, not produce PIP3, confusing reaction direction. For balancing enzymes, enhance the opposing activity to shift equilibrium toward the desired product.
A ligand binds receptor R and activates enzyme AC, producing cAMP. cAMP activates PKA, which phosphorylates channel Ch, increasing ion flux. A regulatory protein Reg binds PKA’s catalytic subunit and reduces its activity when cAMP levels fall. In cells treated with a phosphodiesterase activator, cAMP drops quickly and channel phosphorylation decreases rapidly. Which change would most likely maintain channel phosphorylation despite rapid cAMP breakdown?
Decrease receptor number so AC is activated more strongly per receptor
Increase phosphodiesterase activity further to stabilize cAMP by faster turnover
Express a PKA catalytic subunit variant that is active without cAMP binding
Block ion flux through Ch so phosphorylation remains detectable for longer
Add a ligand antagonist to prevent receptor activation and reduce channel dephosphorylation
Explanation
This question assesses the skill of analyzing signal transduction pathways by selecting ways to sustain phosphorylation despite rapid cAMP breakdown. Rapid cAMP breakdown by phosphodiesterase activation causes quick loss of PKA activity and channel Ch dephosphorylation, as Reg inhibits PKA when cAMP falls. Expressing a PKA catalytic subunit active without cAMP bypasses cAMP dependence, maintaining phosphorylation even with low cAMP. This variant escapes Reg inhibition, sustaining the signal. A tempting distractor is choice B, increasing phosphodiesterase further, but this would worsen cAMP loss, not sustain it, confusing turnover with stabilization. To maintain signals, introduce components insensitive to the degrading factor.
A receptor activates a small GTPase (G) that cycles between inactive GDP-bound and active GTP-bound forms. A guanine nucleotide exchange factor (GEF) promotes GDP release so GTP can bind, activating G. A GTPase-activating protein (GAP) increases G’s GTP hydrolysis, returning it to the GDP-bound form. Active G-GTP activates effector enzyme E, increasing E activity. Cells expressing a mutant G that cannot hydrolyze GTP show high E activity even without ligand stimulation. Which change would most likely restore regulation of E activity in cells expressing the mutant G?
Increasing GEF activity to promote more frequent exchange of GDP for GTP on G
Adding more ligand so receptor activation can override constitutive G activity
Inhibiting effector enzyme E so it cannot be activated by G-GTP
Reducing cytosolic GDP concentration to favor spontaneous GTP binding to G
Overexpressing GAP to accelerate GTP hydrolysis and shorten G activity duration
Explanation
This question assesses the skill of analyzing a signal transduction pathway. The pathway activates GTPase G via receptor-stimulated GEF to promote GDP-GTP exchange, with GAP accelerating hydrolysis for inactivation and G-GTP activating effector E. Choice C restores regulation in mutant G cells by inhibiting E, preventing its activation despite constitutive G-GTP from impaired hydrolysis, thus controlling E activity without ligand. This counters the high basal E activity in mutants lacking GTPase function. A tempting distractor is A, overexpressing GAP, but this misconception assumes GAP can accelerate hydrolysis in a mutant that cannot hydrolyze GTP, whereas it fails to inactivate the mutant. A transferable strategy is to target downstream effectors when upstream regulators are constitutively active in signaling mutations.
A toxin covalently modifies the G$$ subunit of a heterotrimeric G protein, preventing GTP hydrolysis but not affecting GTP binding. In cells where a GPCR normally activates adenylyl cyclase transiently, toxin-treated cells show prolonged cAMP elevation after a brief ligand pulse. Which explanation best accounts for the prolonged cAMP signal?
cAMP elevation persists because the receptor moves into the cytosol where it produces more ligand
The toxin prevents ligand from dissociating, so receptors stay occupied and increase cAMP permanently
G$$ remains active longer because it cannot hydrolyze GTP, sustaining adenylyl cyclase activation
cAMP remains elevated because phosphodiesterase converts cAMP into ATP more slowly
Adenylyl cyclase becomes a ligand-gated channel that stays open after covalent modification
Explanation
This question assesses the skill of analyzing a signal transduction pathway. The toxin prevents GTP hydrolysis by the Gα subunit, which normally deactivates the G protein after activation, leading to prolonged activation of adenylyl cyclase and sustained cAMP elevation even after the ligand is removed. In the pathway, ligand binding to the GPCR causes Gα to exchange GDP for GTP and activate adenylyl cyclase, but without hydrolysis, Gα remains in the active GTP-bound state. This explains the prolonged cAMP signal as the pathway fails to terminate properly. Choice B is tempting but wrong because it assumes the toxin affects ligand dissociation from the receptor, which is a misconception about the toxin's target being the G protein rather than the receptor-ligand interaction. A transferable strategy is to map out the sequence of activation and deactivation steps in G protein-coupled pathways to identify where disruptions occur.
A signaling pathway begins when ligand X binds receptor R, activating kinase A. Kinase A phosphorylates kinase B, which phosphorylates kinase C. Kinase C phosphorylates cytosolic enzyme Y, increasing Y activity. A phosphatase PP removes phosphates from kinase C, decreasing its activity. In an experiment, cells treated with a PP inhibitor show normal kinase A activation after X addition but exhibit a larger increase in Y activity than untreated cells. Which explanation best accounts for the increased Y activity in PP-inhibited cells?
Inhibiting PP directly phosphorylates enzyme Y, bypassing kinases A, B, and C
Inhibiting PP prevents ligand X from binding receptor R, increasing kinase A activation
Inhibiting PP increases kinase C phosphorylation, extending kinase C activity toward enzyme Y
Inhibiting PP reduces ATP availability, slowing dephosphorylation and increasing Y activity
Inhibiting PP decreases kinase B phosphorylation, causing kinase C to activate enzyme Y faster
Explanation
This question assesses the skill of analyzing a signal transduction pathway. The pathway activates kinase A upon ligand X binding receptor R, leading to sequential phosphorylation of kinases B and C, which then activates enzyme Y, with phosphatase PP normally dephosphorylating C to reduce its activity. Choice A accounts for increased Y activity in PP-inhibited cells because inhibiting PP sustains kinase C phosphorylation, extending its activity and enhancing Y phosphorylation despite normal A activation. This results in a larger Y response, as seen in the experiment. A tempting distractor is C, suggesting inhibited PP decreases B phosphorylation, but this misconception confuses dephosphorylation targets, as PP acts on C to prolong, not accelerate, downstream activation. A transferable strategy is to assess how phosphatase inhibition affects phosphorylation persistence in multi-step kinase cascades.
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?
Synthesis of IP$_3$ from PIP$_2$, preventing second messenger formation
Activation of PKC by $\mathrm{Ca^{2+}}$ binding, preventing phosphorylation of transporter T
Autophosphorylation of receptor R, preventing recruitment of PLC$\gamma$
Binding of growth factor G to receptor R at the cell surface
Insertion of transporter T into the membrane by vesicle trafficking
Explanation
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 muscle cells, ligand L binds membrane receptor R and increases intracellular $\mathrm{Ca^{2+}}$. R activates G protein G\alpha, 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?
Remove channel C from the membrane so phosphorylation has no cellular purpose
Increase cytosolic ribosomes to enhance synthesis of PKA catalytic subunits
Block ATP production so adenylyl cyclase cannot generate any second messenger
Mutate receptor R to prevent ligand L binding at the extracellular surface
Increase PDE activity to accelerate cAMP breakdown after PKA activation
Explanation
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.
A ligand-gated receptor channel opens upon ligand binding and allows Na$^+$ influx. The increased Na$^+$ activates an intracellular Na$^+$-sensitive enzyme N, which produces a second messenger M. M activates kinase K, and K phosphorylates cytosolic protein V, increasing V activity. When extracellular Na$^+$ is replaced with an impermeant cation, ligand binding still occurs but M levels and V activity do not increase. Which step is most immediately prevented by removing extracellular Na$^+$?
Kinase K phosphorylation of protein V to increase V activity
Second messenger M activation of kinase K in the cytosol
Ligand binding to the extracellular domain of the receptor channel
Opening of the receptor channel to permit Na$^+$ influx into the cytosol
Na$^+$-sensitive enzyme N synthesis to increase intracellular enzyme concentration
Explanation
This question assesses the skill of analyzing a signal transduction pathway. The pathway involves ligand binding opening a receptor channel for Na⁺ influx, activating enzyme N to produce messenger M, which stimulates kinase K to phosphorylate and activate protein V. Choice B is most immediately prevented by removing extracellular Na⁺, as ligand binding occurs but without Na⁺ influx, N remains inactive, blocking M production and downstream V activation. This is evidenced by unchanged M levels and V activity in the absence of Na⁺. A tempting distractor is A, ligand binding to the receptor, but this misconception assumes binding requires Na⁺, whereas the results show binding persists without influx. A transferable strategy is to pinpoint the first disrupted step in ion-dependent pathways by examining immediate downstream outputs.