AP Biology Quiz: Introduction To Signal Transduction

What this quiz covers

This quiz focuses on Introduction To Signal Transduction, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Ligand G is transported into the cytosol, where it directly synthesizes cGMP from GTP.
2. Ligand G activates a receptor that promotes GTP binding to a G protein, enabling activation of guanylyl cyclase. (correct answer)
3. Ligand G increases transcription of guanylyl cyclase, leading to rapid accumulation of cGMP.
4. Ligand G blocks phosphodiesterase export, causing cGMP to accumulate outside the cell and diffuse inward.
5. Ligand G binds the receptor and is converted into cGMP on the extracellular surface of the membrane.

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.

Question 2

1. The cytosolic tail is required to couple ligand binding to activation of intracellular signaling proteins (correct answer)
2. Deleting the cytosolic tail prevents Ligand L from binding to the extracellular domain
3. The cytosolic tail normally blocks ligand binding, so deletion should increase phosphorylation
4. Tyrosine phosphorylation requires ligand entry into the nucleus, which is blocked by mutation
5. Phosphorylation fails because the mutation reduces ATP production in mitochondria

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.

Question 3

1. Molecule X diffuses through the membrane and directly activates adenylyl cyclase in the cytosol.
2. Molecule X binds receptor R, triggering a conformational change that activates a G protein leading to cAMP production. (correct answer)
3. Molecule X binds receptor R and is transported into the cell by receptor-mediated endocytosis to raise cAMP.
4. Molecule X increases transcription of the adenylyl cyclase gene, causing higher cAMP within seconds.
5. Molecule X binds receptor R to supply ATP for cAMP synthesis in the extracellular fluid.

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.

Question 4

1. L must be converted into a peptide outside the cell before any Ca$^{2+}$ response can occur.
2. R binding to L initiates an intracellular interaction that triggers Ca$^{2+}$ release from internal stores. (correct answer)
3. L binds a cell-surface receptor that directly pumps Ca$^{2+}$ into the cytosol.
4. R mutation increases membrane permeability, preventing L from entering the cytosol.
5. L binding to R causes immediate synthesis of new Ca$^{2+}$ channels in the plasma membrane.

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.

Question 5

1. Z binding activates a G protein whose  subunit inhibits adenylyl cyclase, reducing cAMP production. (correct answer)
2. Z binding blocks cAMP diffusion out of the cell by closing membrane pores.
3. Z binding directly degrades cAMP by acting as a phosphodiesterase enzyme.
4. Z binding causes receptor endocytosis, which immediately consumes cAMP as ATP is hydrolyzed.
5. Z binding increases transcription of a cAMP inhibitor, lowering cAMP within seconds.

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.

Question 6

1. Phosphatases remove phosphate groups from signaling proteins, opposing kinase-driven activation states (correct answer)
2. Phosphatases are required to add phosphate groups to receptors to initiate ligand binding
3. Phosphatases degrade the ligand outside the cell, reducing receptor occupancy over time
4. Phosphatases transport phosphate into the cytosol, enabling phosphorylation of Protein S
5. Phosphatases increase signaling by synthesizing ATP from ADP at the plasma membrane

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.

Question 7

1. The cytosolic loops interact with and activate a G protein, linking receptor binding to cAMP production (correct answer)
2. The cytosolic loops form the ligand-binding pocket, so mutation should prevent ligand binding
3. The cytosolic loops transport cAMP into the extracellular space, so mutation increases intracellular cAMP
4. The cytosolic loops are required for ribosome binding, enabling rapid synthesis of cAMP enzymes
5. 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.

Question 8

1. The cytosolic tail binds Na$^+$ and transports it across the membrane after ligand binding.
2. The cytosolic tail interacts with intracellular proteins that promote channel opening after receptor activation. (correct answer)
3. The cytosolic tail is required for ligand M synthesis and secretion from the signaling cell.
4. The cytosolic tail converts Na$^+$ into a second messenger that activates the channel.
5. The cytosolic tail enables the receptor to enter the nucleus to open ion channels from inside.

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.

Question 9

1. Dimerization brings cytosolic kinase domains close enough to phosphorylate each other or nearby sites after ligand binding. (correct answer)
2. Dimerization is required for ligand G to be synthesized and secreted into the extracellular space.
3. Dimerization allows G to diffuse through the receptor into the cytosol as the first response.
4. Dimerization prevents phosphorylation by blocking access of ATP to the receptor.
5. Dimerization converts the receptor into a transcription factor that immediately binds DNA.

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.

Question 10

1. Cell type 2 lacks the downstream effector needed to convert receptor activation into cGMP production. (correct answer)
2. Cell type 2 must express a different receptor because equal ligand binding cannot occur otherwise.
3. Cell type 1 increases cGMP because Q is transported into the cytosol and acts as an enzyme.
4. Cell type 1 produces cGMP because Q binding directly opens nuclear pores for cGMP entry.
5. Cell type 2 fails to respond because Q is degraded in the extracellular fluid after binding.

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.

Question 11

1. ATP provides phosphate groups used by kinases to phosphorylate target proteins after receptor activation. (correct answer)
2. ATP is the ligand that binds the receptor, so removing ATP blocks reception.
3. ATP is required to form the receptor's extracellular domain through rapid protein synthesis.
4. ATP prevents ligand N from binding by competing for the same binding pocket.
5. ATP is converted into ligand N at the membrane, so analogs eliminate ligand availability.

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 replacing ATP with a nonphosphorylatable analog prevents protein phosphorylation despite N binding, indicating ATP supplies phosphate for kinase-mediated transduction within 10 seconds. This matches principles where kinases use ATP to phosphorylate targets. The analog's effect isolates ATP's role. A tempting distractor is B, which wrongly makes ATP the ligand, based on the misconception that energy molecules bind receptors, though binding is unchanged. When analyzing signal transduction questions, substitute analogs to confirm molecular requirements in steps like phosphorylation.

Question 12

1. Ligand Z binds an intracellular receptor after diffusing through the plasma membrane (correct answer)
2. Ligand Z must bind to a cell wall receptor to be transported into the cytosol
3. Ligand Z is too large to cross the membrane and therefore requires endocytosis for reception
4. Ligand Z binds a membrane receptor that directly converts ATP into steroid second messengers
5. Ligand Z initiates signaling by increasing transcription of membrane receptors during reception

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on the reception step for hydrophobic ligands. The correct answer is A because Ligand Z is a hydrophobic steroid-like molecule that can diffuse through the plasma membrane to bind an intracellular receptor, as shown by the receptor-ligand complex moving to the nucleus in Cell Type 1, enabling direct genomic responses. In Cell Type 2, anchoring the receptor to the membrane prevents this intracellular interaction, resulting in no early second messenger changes, which supports that reception occurs cytosolically rather than at the surface. Basic signaling principles state that nonpolar ligands like steroids bypass membrane receptors and act intracellularly, often as transcription factors. A tempting distractor is B, which is wrong because it assumes a cell wall receptor is needed for transport, a misconception applying plant cell wall properties inappropriately to general animal-like cells. For signal transduction questions, distinguish between hydrophilic and hydrophobic ligands to determine the location of reception and subsequent pathway steps.

Question 13

1. Epinephrine enters the cytosol and directly activates adenylyl cyclase by binding its active site
2. Epinephrine binds a plasma membrane receptor that activates a G protein, stimulating adenylyl cyclase to make cAMP (correct answer)
3. Epinephrine binds DNA-associated proteins, rapidly increasing cAMP by altering transcription of cyclase genes
4. Epinephrine dissolves in the lipid bilayer, and cAMP rises due to increased membrane fluidity and ion leakage
5. Epinephrine is converted to cAMP at the cell surface, and cAMP then diffuses through the membrane into the cytosol

Explanation: This question tests your ability to analyze signal transduction by interpreting experimental evidence about epinephrine signaling. The membrane-impermeable analog producing the same effect proves the receptor must be on the cell surface, while the lack of effect in cell-free cytosol shows epinephrine cannot directly activate intracellular components. The blocking compound that prevents cAMP elevation confirms a specific plasma membrane protein (receptor) is required, and the rapid 10-second response indicates a G protein-coupled receptor activating adenylyl cyclase to produce cAMP as a second messenger. Choice A incorrectly suggests epinephrine enters the cell, contradicting the membrane-impermeable analog data. When analyzing signal transduction experiments, use multiple pieces of evidence to trace the pathway: membrane permeability tests reveal receptor location, timing indicates direct vs. transcriptional effects, and blocking studies identify required components.

Question 14

1. Early signaling occurs at the plasma membrane upon ligand–receptor interaction, independent of receptor internalization (correct answer)
2. Receptor internalization is required to allow ligand M to enter the cytosol and directly activate the kinase
3. Endocytosis generates kinase activity by mechanically stretching the cytoskeleton, which acts as the receptor
4. Ligand M activates the kinase by serving as a phosphate donor that covalently modifies the kinase active site
5. Kinase activation occurs because receptor internalization increases ATP synthesis, raising kinase activity nonspecifically

Explanation: This question requires analyzing signal transduction timing to distinguish early signaling from receptor trafficking. The kinase activation at 15 seconds occurring even when endocytosis is blocked proves early signaling happens at the plasma membrane immediately upon ligand-receptor binding, independent of the slower internalization process at 2 minutes. This demonstrates that receptor activation and initial signal transduction precede receptor endocytosis, which serves other functions like signal termination or trafficking. Choice B incorrectly claims internalization is required for ligand entry and kinase activation, contradicting the evidence that blocking endocytosis doesn't prevent the early response. To analyze signaling dynamics, distinguish immediate membrane-initiated events (seconds) from slower processes like internalization (minutes) or transcription (hours).

Question 15

1. Ligand binding to the receptor is required to initiate receptor activation that leads to downstream phosphorylation (correct answer)
2. Ligand binding is not required because phosphorylation occurs through random collisions of cytosolic proteins
3. Ligand binding triggers phosphorylation by increasing mitochondrial ATP production over long time scales
4. Ligand binding causes phosphorylation because the ligand is itself a kinase that enters the cytosol
5. Ligand binding fails because receptors require a cell wall to orient properly in the membrane

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on the reception step linking to transduction. The correct answer is A because ligand binding activates the receptor, initiating the cascade leading to phosphorylation, as shown by the requirement for the receptor, rapid 15-second response, and failure with a mutated binding site. This is supported by basic signaling principles where reception triggers downstream events. The absence of effect without receptor confirms specificity. A tempting distractor is B, which is incorrect because it suggests random activation, a misconception ignoring ligand-receptor dependency. In signal transduction questions, evaluate receptor mutations to determine necessity of binding for pathway initiation.

Question 16

1. Ligand binding to the extracellular domain is prevented because the kinase domain determines ligand specificity.
2. The receptor cannot catalyze phosphorylation needed to create docking sites for adaptor proteins. (correct answer)
3. The mutation forces L to bind intracellular receptors instead of membrane receptors.
4. The receptor becomes constitutively phosphorylated because kinase domains inhibit phosphorylation.
5. Adaptor proteins bind directly to L, so receptor phosphorylation is not relevant to early signaling.

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is B because removing the kinase domain prevents receptor phosphorylation despite normal L binding, indicating the domain is essential for catalyzing phosphorylation that creates docking sites for adaptor proteins. This aligns with RTK signaling principles where ligand binding induces autophosphorylation, enabling adaptor binding within 30 seconds. The rapid timeframe and specific loss of phosphorylation confirm the kinase domain's role in early transduction. A tempting distractor is A, which incorrectly ties the kinase to ligand specificity, stemming from the misconception that intracellular domains control extracellular binding, though binding is unchanged. When analyzing signal transduction questions, use mutation effects on phosphorylation and binding to pinpoint roles in reception versus transduction.

Question 17

1. Opening of plasma membrane Ca$^{2+}$ channels allows extracellular Ca$^{2+}$ influx to raise cytosolic Ca$^{2+}$
2. Generation of IP$_3$ leads to Ca$^{2+}$ release from the endoplasmic reticulum into the cytosol (correct answer)
3. Ligand enters mitochondria and releases Ca$^{2+}$ by disrupting oxidative phosphorylation membranes
4. Ca$^{2+}$ increases because the receptor directly synthesizes Ca$^{2+}$ from ATP in the cytosol
5. Ca$^{2+}$ increases because the cell requires more Ca$^{2+}$ for contraction after ligand exposure

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on the transduction step releasing second messengers. The correct answer is B because receptor activation leads to IP₃ production, which binds ER receptors to release stored Ca²⁺, as shown by the response persisting without extracellular Ca²⁺ but being blocked by IP₃ receptor inhibitors. This is consistent with basic signaling principles involving PLC-IP₃ pathways for internal Ca²⁺ mobilization in muscle cells. The rapid nature supports a second messenger cascade. A tempting distractor is A, which is wrong because it relies on extracellular influx, a misconception contradicted by the chelator experiment. For signal transduction questions, differentiate between internal release and external influx by examining dependency on extracellular ions.

Question 18

1. cAMP functions inside cells as a second messenger, and a membrane-impermeable analog cannot reach cytosolic targets. (correct answer)
2. cAMP analogs must bind the extracellular domain of the receptor to activate adenylyl cyclase.
3. Extracellular cAMP analog prevents ligand E from being secreted, eliminating receptor binding.
4. Extracellular cAMP analog is converted into ligand E by the receptor, so signaling should be identical.
5. cAMP acts only in the nucleus, so extracellular analog should enter through nuclear pores.

Explanation: This question assesses the ability to analyze signal transduction pathways, particularly the intracellular role of second messengers like cAMP. The correct answer is A because the membrane-impermeable cAMP analog added extracellularly fails to elicit a response, indicating it cannot access intracellular targets, unlike the cAMP produced inside by ligand E activation of adenylyl cyclase. Basic signaling principles highlight that second messengers like cAMP act cytosolically to activate proteins such as protein kinase A, and the analog's impermeability prevents this, as supported by the lack of response despite mimicking cAMP's structure. The stimulus shows ligand E increases cAMP rapidly, confirming its intracellular generation and function. A tempting distractor is B, which is incorrect because it assumes cAMP analogs act as ligands on extracellular receptors, a misconception confusing first and second messengers. A transferable strategy for signal transduction questions is to distinguish between extracellular ligands and intracellular messengers by considering membrane permeability and site of action.

Question 19

1. Ligand binding activates a G protein that synthesizes Na$^+$ as a second messenger in the cytosol
2. Ligand binding directly gates an ion channel, and Na$^+$ influx changes membrane potential (correct answer)
3. Ligand binding induces transcription of Na$^+$ transporters, increasing Na$^+$ influx over hours
4. Ligand binding causes Na$^+$ to diffuse through the lipid bilayer without protein involvement
5. Membrane potential fails because impermeant cations prevent ligand-receptor recognition

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on the reception and early response involving ion channels. The correct answer is B because the receptor functions as a ligand-gated ion channel, allowing Na⁺ influx upon binding, which changes membrane potential, as indicated by the millisecond-scale response and the protein's transmembrane helices typical of channels. Replacing extracellular Na⁺ with an impermeant cation prevents potential change despite binding, confirming Na⁺ influx is the direct early response. Basic signaling principles show that ligand-gated channels provide rapid electrical responses via ion flow without second messengers. A tempting distractor is A, which is wrong because it misattributes Na⁺ as a synthesized second messenger, a misconception confusing ion channels with G protein pathways. For signal transduction questions, examine response timing and receptor structure to differentiate direct gating from cascade mechanisms.

Question 20

1. Ligand A activates a pathway that stimulates adenylyl cyclase, whereas ligand B activates PLC leading to Ca$^{2+}$ release. (correct answer)
2. Ligand A enters the nucleus to generate cAMP, whereas ligand B diffuses into mitochondria to release Ca$^{2+}$.
3. Ligand A binds directly to adenylyl cyclase, whereas ligand B binds directly to Ca$^{2+}$ ions in the cytosol.
4. Ligand A and ligand B both activate the same receptor, but different temperatures change the second messenger produced.
5. Ligand A increases membrane cholesterol to create cAMP, whereas ligand B decreases cholesterol to create Ca$^{2+}$.

Explanation: This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is A because inhibiting adenylyl cyclase blocks only ligand A's cAMP response, while inhibiting PLC blocks only ligand B's Ca²⁺ response, indicating distinct transduction pathways where A activates adenylyl cyclase via one receptor type and B activates PLC leading to IP₃ and Ca²⁺ release via another. Both ligands not crossing the membrane confirms receptor-mediated mechanisms. Basic signaling principles differentiate GPCR pathways, such as Gs stimulating cAMP and Gq activating PLC for Ca²⁺ mobilization. A tempting distractor is D, which is wrong due to the misconception that the same receptor can switch messengers based on external factors like temperature, disregarding evidence for ligand-specific receptors and pathways. For signal transduction questions, compare pathways by using selective inhibitors to identify unique enzymes and second messengers activated by different ligands.