AP Biology Quiz: 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?

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?

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?

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?

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?

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?

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?

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?

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$^+$?

Cells are stimulated with ligand L, activating a receptor that triggers a kinase cascade leading to phosphorylation of cytosolic substrate S. Adding a broad phosphatase inhibitor increases both the peak level and duration of S phosphorylation. Adding a kinase inhibitor eliminates S phosphorylation. No changes in ligand binding are detected under any condition. Which change would most likely decrease the duration of S phosphorylation without affecting its initial peak?

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?

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?

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?

In photoreceptor cells, light activates rhodopsin (a GPCR), which activates transducin (G protein). Transducin activates phosphodiesterase (PDE), which decreases cGMP. Lower cGMP closes cGMP-gated $\mathrm{Na^+}$ channels, hyperpolarizing the cell. A guanylyl cyclase restores cGMP levels, reopening channels. In a mutant with reduced PDE activity, light causes a much smaller hyperpolarization. Which change would most likely increase hyperpolarization in the mutant during light exposure?

In an experiment, cells are stimulated with ligand L that activates receptor R and leads to phosphorylation of protein P within 10 seconds. A kinase inhibitor blocks P phosphorylation, but a phosphatase inhibitor increases both the magnitude and duration of P phosphorylation. No changes occur in ligand binding. Which interpretation best explains the effect of the phosphatase inhibitor?

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?

A membrane receptor activates two parallel pathways after ligand binding. Pathway 1: G protein activates adenylyl cyclase, increasing cAMP and activating PKA, which phosphorylates substrate S1. Pathway 2: the same receptor activates PLC, producing IP$_3$ and raising cytosolic Ca$^{2+}$, activating kinase CaK, which phosphorylates substrate S2. A selective adenylyl cyclase inhibitor blocks S1 phosphorylation but does not change S2 phosphorylation. Which statement best explains the inhibitor's specificity?

In a receptor signaling pathway, ligand L activates receptor R, which activates kinase K. K phosphorylates inhibitor I, causing I to release enzyme E. Free E then catalyzes conversion of substrate to product. In cells expressing a mutant I that cannot be phosphorylated, L binding to R is normal but product formation does not increase. Which step is most directly disrupted by the mutation?

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?

A membrane receptor activates JAK kinase, which phosphorylates a cytosolic protein STAT. Phosphorylated STAT (STAT-P) forms dimers in the cytosol within 1 minute. A phosphatase dephosphorylates STAT-P, returning STAT to monomers. When the phosphatase is inhibited, STAT remains phosphorylated longer and dimer levels stay high. Which change would most likely decrease STAT dimer levels without altering receptor activation?