Introduction to Signal Transduction
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AP Biology › Introduction to Signal Transduction
A signaling ligand G binds to a receptor on the plasma membrane and quickly increases cytosolic cGMP. Cells pretreated with a drug that locks heterotrimeric G proteins in the GDP-bound state show no cGMP increase after ligand addition, even though ligand binding is unchanged. When a membrane-permeable cGMP analog is added, the intracellular response occurs despite the drug. Which of the following best explains the early transduction step normally required for cGMP production?
Ligand G is transported into the cytosol, where it directly synthesizes cGMP from GTP.
Ligand G increases transcription of guanylyl cyclase, leading to rapid accumulation of cGMP.
Ligand G binds the receptor and is converted into cGMP on the extracellular surface of the membrane.
Ligand G blocks phosphodiesterase export, causing cGMP to accumulate outside the cell and diffuse inward.
Ligand G activates a receptor that promotes GTP binding to a G protein, enabling activation of guanylyl cyclase.
Explanation
This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because locking G proteins in GDP-bound state prevents cGMP increase despite ligand binding, indicating that receptor activation promotes GTP exchange on G proteins, which then activate guanylyl cyclase to produce cGMP. The permeable cGMP analog restoring the response shows the block is upstream of cGMP production. This fits basic signaling principles of GPCRs where ligand binding activates G proteins by facilitating GDP-GTP exchange, enabling effector activation. A tempting distractor is C, which is wrong due to the misconception that rapid second messenger increases involve transcription, whereas cGMP production occurs enzymatically within seconds. For signal transduction questions, use drugs affecting G protein states to identify their role in linking receptors to second messenger-generating enzymes.
A cell expresses Receptor R, a transmembrane protein. When Ligand L binds, an intracellular protein becomes rapidly phosphorylated on tyrosine residues. If a mutation deletes the receptor’s cytosolic tail but leaves the extracellular ligand-binding domain intact, Ligand L still binds but phosphorylation of the intracellular protein does not occur. Which of the following best explains this result in terms of early signal transduction?
Phosphorylation fails because the mutation reduces ATP production in mitochondria
Tyrosine phosphorylation requires ligand entry into the nucleus, which is blocked by mutation
Deleting the cytosolic tail prevents Ligand L from binding to the extracellular domain
The cytosolic tail is required to couple ligand binding to activation of intracellular signaling proteins
The cytosolic tail normally blocks ligand binding, so deletion should increase phosphorylation
Explanation
This question assesses the skill of analyzing signal transduction pathways, focusing on the transduction step across the membrane. The correct answer is A because the cytosolic tail of Receptor R is essential for transducing the signal from ligand binding to intracellular events, as its deletion prevents phosphorylation of the downstream protein despite intact extracellular binding, indicating the tail couples reception to signaling proteins like kinases. This is evidenced by the rapid tyrosine phosphorylation normally occurring after Ligand L binding, which requires intracellular domain interactions. Basic signaling principles highlight that transmembrane receptors use cytosolic domains to activate cascades, such as phosphorylation events. A tempting distractor is B, which is incorrect because it assumes the cytosolic tail affects extracellular binding, a misconception ignoring that the mutation leaves the binding domain intact. In signal transduction questions, consider how receptor domains (extracellular, transmembrane, cytosolic) contribute to each pathway stage to interpret mutation effects.
A cultured animal cell is exposed to signaling molecule X, which cannot cross the plasma membrane. Within seconds, intracellular cAMP levels increase. When researchers add a competitive antagonist that binds the extracellular side of receptor R, the cAMP increase is eliminated. However, when cells are treated with a membrane-permeable cAMP analog, the intracellular response occurs even in the presence of the antagonist. Receptor R spans the membrane and is present at similar levels in all treatments. Which of the following best explains how molecule X initiates the intracellular response?
Molecule X binds receptor R to supply ATP for cAMP synthesis in the extracellular fluid.
Molecule X increases transcription of the adenylyl cyclase gene, causing higher cAMP within seconds.
Molecule X diffuses through the membrane and directly activates adenylyl cyclase in the cytosol.
Molecule X binds receptor R, triggering a conformational change that activates a G protein leading to cAMP production.
Molecule X binds receptor R and is transported into the cell by receptor-mediated endocytosis to raise cAMP.
Explanation
This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because molecule X cannot cross the plasma membrane, indicating it must bind to an extracellular receptor site, and the rapid increase in cAMP within seconds suggests activation of a second messenger pathway rather than slower processes like transcription. The competitive antagonist blocking the extracellular side of receptor R eliminates the cAMP increase, confirming that X binds R to initiate signaling, while the membrane-permeable cAMP analog bypassing this block shows the issue is upstream of cAMP production. This aligns with basic signaling principles where ligand binding to a G protein-coupled receptor (GPCR) causes a conformational change that activates an associated G protein, which then stimulates adenylyl cyclase to produce cAMP. A tempting distractor is D, which is wrong due to the misconception that transcriptional changes can occur within seconds, whereas gene expression typically takes minutes to hours. For signal transduction questions, always consider the ligand's ability to cross the membrane and the response timeline to differentiate between direct intracellular actions and receptor-mediated pathways.
Immune cells exposed to lipid signal L show an increase in cytosolic Ca$^{2+}$ within seconds. L is hydrophobic and crosses the plasma membrane. A receptor protein R is found primarily in the cytosol and binds L with high affinity. When R is mutated so it cannot bind L, Ca$^{2+}$ does not increase after L addition, even though L still enters the cell. Which of the following is the most consistent prediction about the earliest step in this signaling pathway?
L binds a cell-surface receptor that directly pumps Ca$^{2+}$ into the cytosol.
R binding to L initiates an intracellular interaction that triggers Ca$^{2+}$ release from internal stores.
L binding to R causes immediate synthesis of new Ca$^{2+}$ channels in the plasma membrane.
L must be converted into a peptide outside the cell before any Ca$^{2+}$ response can occur.
R mutation increases membrane permeability, preventing L from entering the cytosol.
Explanation
This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because L's hydrophobicity allows it to cross the membrane and bind cytosolic receptor R, and the mutation preventing binding eliminates the Ca²⁺ increase, indicating R's role in initiating intracellular signaling. In basic signaling principles, intracellular receptors like those for hydrophobic ligands can trigger rapid responses such as Ca²⁺ release from stores upon ligand binding. The evidence that L still enters the cell but no Ca²⁺ response occurs with mutated R supports that the pathway starts with R-L interaction leading to downstream Ca²⁺ mobilization. A tempting distractor is C, which is wrong because it assumes the receptor directly pumps Ca²⁺, reflecting the misconception that cytosolic receptors act as membrane transporters rather than initiators of intracellular cascades. A transferable strategy for signal transduction questions is to consider ligand properties and receptor location to predict whether signaling is membrane-bound or intracellular.
In heart muscle cells, ligand Z binds a membrane receptor and causes a rapid decrease in cytosolic cAMP. When cells are treated with pertussis toxin, which prevents certain G proteins from interacting with receptors, Z still binds but cAMP levels no longer decrease. Which of the following best explains the earliest intracellular event normally triggered by Z binding?
Z binding activates a G protein whose $ $ subunit inhibits adenylyl cyclase, reducing cAMP production.
Z binding blocks cAMP diffusion out of the cell by closing membrane pores.
Z binding increases transcription of a cAMP inhibitor, lowering cAMP within seconds.
Z binding directly degrades cAMP by acting as a phosphodiesterase enzyme.
Z binding causes receptor endocytosis, which immediately consumes cAMP as ATP is hydrolyzed.
Explanation
This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is A because the rapid cAMP decrease after Z binding and its prevention by pertussis toxin, which affects inhibitory G proteins, indicate activation of a Gi protein that inhibits adenylyl cyclase. In basic signaling principles, ligand-bound GPCRs can activate Gi, whose α subunit reduces cAMP production by inhibiting the cyclase enzyme. The evidence that Z binds but cAMP does not decrease with toxin treatment supports this as the earliest event, distinguishing it from stimulatory Gs pathways. A tempting distractor is C, which is wrong because it assumes the receptor directly degrades cAMP, reflecting the misconception that receptors act as enzymes rather than activators of G protein-mediated transduction. A transferable strategy for signal transduction questions is to differentiate between stimulatory and inhibitory pathways using specific toxins and second messenger changes.
A ligand binds a receptor and triggers rapid activation of Protein S by phosphorylation. When cells are treated with a phosphatase inhibitor, Protein S phosphorylation becomes higher and persists longer after ligand removal. Ligand binding kinetics to the receptor are unchanged by the inhibitor. Which of the following best explains the role of phosphatases in early signal transduction here?
Phosphatases degrade the ligand outside the cell, reducing receptor occupancy over time
Phosphatases transport phosphate into the cytosol, enabling phosphorylation of Protein S
Phosphatases are required to add phosphate groups to receptors to initiate ligand binding
Phosphatases increase signaling by synthesizing ATP from ADP at the plasma membrane
Phosphatases remove phosphate groups from signaling proteins, opposing kinase-driven activation states
Explanation
This question assesses the skill of analyzing signal transduction pathways, focusing on the transduction step regulating phosphorylation. The correct answer is A because phosphatases dephosphorylate proteins like Protein S, terminating signals, as evidenced by higher, prolonged phosphorylation with inhibition, without affecting binding. This aligns with basic signaling principles of kinase-phosphatase balance for dynamic control. The rapid activation supports enzymatic regulation. A tempting distractor is B, which is wrong because it reverses phosphatase roles, a misconception of their deactivating function. For signal transduction questions, consider opposing enzymes to explain signal duration and intensity.
A receptor is predicted to be a seven-transmembrane protein. Ligand binding leads to rapid activation of a cytosolic enzyme and increased cAMP. When the receptor’s cytosolic loops are mutated, ligand binding remains normal but cAMP does not increase. Which of the following best explains the function of the receptor’s cytosolic loops in early signaling?
The cytosolic loops transport cAMP into the extracellular space, so mutation increases intracellular cAMP
The cytosolic loops are required for ribosome binding, enabling rapid synthesis of cAMP enzymes
The cytosolic loops form the ligand-binding pocket, so mutation should prevent ligand binding
The cytosolic loops interact with and activate a G protein, linking receptor binding to cAMP production
The cytosolic loops reduce cAMP by converting it to ATP, so mutation should lower ATP instead
Explanation
This question assesses the skill of analyzing signal transduction pathways, focusing on the transduction step in G protein-coupled receptors. The correct answer is A because the cytosolic loops interact with G proteins to facilitate activation and cAMP production, and their mutation disrupts this without affecting binding. This is supported by the receptor's seven-transmembrane structure typical of GPCRs and rapid cAMP increase. Basic signaling principles assign intracellular loops to effector coupling. A tempting distractor is B, which is incorrect because it mislocates binding to cytosolic parts, a misconception of GPCR topology. In signal transduction questions, use predicted structures to assign functions to receptor regions.
A researcher studies a signaling pathway in animal cells where ligand M causes a rapid opening of a specific ion channel in the plasma membrane, increasing cytosolic Na$^+$. The ligand binds to receptor T on the cell surface. In cells expressing a receptor T variant lacking most of its cytosolic tail, ligand binding still occurs, but the Na$^+$ increase is greatly reduced. Direct application of a channel-opening drug restores Na$^+$ influx in both cell types. Which of the following best explains the function of receptor T’s cytosolic tail in early signal transduction?
The cytosolic tail binds Na$^+$ and transports it across the membrane after ligand binding.
The cytosolic tail enables the receptor to enter the nucleus to open ion channels from inside.
The cytosolic tail interacts with intracellular proteins that promote channel opening after receptor activation.
The cytosolic tail converts Na$^+$ into a second messenger that activates the channel.
The cytosolic tail is required for ligand M synthesis and secretion from the signaling cell.
Explanation
This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because the receptor variant lacking the cytosolic tail binds ligand M but shows reduced Na⁺ increase, indicating the tail is essential for transducing the signal to open channels, while the channel-opening drug restores function, confirming channels are present but not activated. This suggests the cytosolic tail interacts with intracellular proteins, such as G proteins or adapters, to promote channel opening via second messengers. Basic signaling principles show that receptor cytosolic domains are crucial for coupling to effectors in pathways like those involving ion channels. A tempting distractor is A, which is wrong due to the misconception that receptors directly transport ions, ignoring that the response involves separate channels and transduction steps. For signal transduction questions, analyze receptor domain functions by comparing wild-type and mutant behaviors to determine their role in intracellular signaling interactions.
A scientist adds ligand G to cells and observes rapid receptor dimerization at the plasma membrane using fluorescence. Shortly after dimerization, receptor phosphorylation increases. When a drug prevents dimerization, ligand binding still occurs but phosphorylation is greatly reduced. Which of the following best explains why dimerization is important early in this signaling pathway?
Dimerization is required for ligand G to be synthesized and secreted into the extracellular space.
Dimerization allows G to diffuse through the receptor into the cytosol as the first response.
Dimerization brings cytosolic kinase domains close enough to phosphorylate each other or nearby sites after ligand binding.
Dimerization converts the receptor into a transcription factor that immediately binds DNA.
Dimerization prevents phosphorylation by blocking access of ATP to the receptor.
Explanation
This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is A because preventing dimerization reduces phosphorylation despite G binding, indicating dimerization positions kinase domains for autophosphorylation. Fluorescence shows rapid dimerization leading to phosphorylation, aligning with RTK principles. The drug's effect highlights dimerization's importance. A tempting distractor is D, which wrongly claims dimerization blocks phosphorylation, based on the misconception that aggregation inhibits, though it enables. When analyzing signal transduction questions, observe structural changes like dimerization to link them to enzymatic activation.
Two cell types express the same membrane receptor for ligand Q. After Q addition, Cell type 1 shows a rapid increase in cGMP, while Cell type 2 shows no change in cGMP. Radiolabeled Q binds equally to both cell types. A biochemical assay shows that only Cell type 1 contains a membrane-associated guanylyl cyclase that can be activated by the receptor. Which of the following best explains the difference in early intracellular response?
Cell type 1 increases cGMP because Q is transported into the cytosol and acts as an enzyme.
Cell type 2 must express a different receptor because equal ligand binding cannot occur otherwise.
Cell type 1 produces cGMP because Q binding directly opens nuclear pores for cGMP entry.
Cell type 2 fails to respond because Q is degraded in the extracellular fluid after binding.
Cell type 2 lacks the downstream effector needed to convert receptor activation into cGMP production.
Explanation
This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is A because Cell type 1 shows cGMP increase with receptor-associated guanylyl cyclase, while Cell type 2 lacks this effector, explaining the differential response despite equal Q binding. In basic signaling principles, receptors like those for nitric oxide or peptides can directly activate guanylyl cyclase to produce cGMP as a second messenger. The evidence of equal binding but response only in cells with the cyclase supports that the difference lies in downstream transduction components. A tempting distractor is B, which is wrong because it assumes different receptors are needed for equal binding, reflecting the misconception that binding affinity determines response without considering intracellular effectors. A transferable strategy for signal transduction questions is to compare cell types by examining shared receptors versus differing downstream components.