Cell Communication
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AP Biology › Cell Communication
A neuron releases neurotransmitter Y into a synaptic cleft. Postsynaptic Cell P responds, but only when Y is released locally at the synapse. When Y is injected into the bloodstream at the same concentration, Cell P does not respond. Which of the following best explains why synaptic release is effective but bloodstream injection is not?
Synaptic release requires Y to enter the nucleus, which cannot occur from the blood
Bloodstream injection converts Y into a steroid that cannot bind membrane receptors
Synaptic release creates a high local concentration near receptors before Y is diluted systemically
Bloodstream injection prevents vesicle fusion, so Y cannot be released at synapses
Synaptic signaling is long-distance endocrine signaling, so blood delivery is unnecessary
Explanation
This question assesses understanding of cell communication via signal transduction pathways. The correct answer is A because synaptic release delivers neurotransmitter Y at a high local concentration directly to receptors on Cell P, enabling effective binding before dilution, as opposed to bloodstream injection where Y is diluted systemically and fails to reach threshold levels. Evidence from the stimulus shows that Cell P responds only to local synaptic release, not to the same concentration in the blood, highlighting the importance of proximity and concentration gradients in synaptic signaling. This mechanism ensures specific, rapid communication between neurons without affecting distant cells. A tempting distractor is B, which suggests bloodstream injection converts Y into a steroid, but this is wrong due to the misconception that delivery method alters molecular structure, when the issue is concentration and localization. A transferable strategy is to consider signal concentration and delivery mode when analyzing differences between local and systemic signaling.
In a plant root, Cell M releases signal Z into the apoplast. Cells within 2–3 cell diameters respond, but cells farther away do not. When the same amount of Z is injected directly into the xylem, distant cells begin responding. Which of the following best explains the change in responding cells?
Injection into xylem causes Z to become hydrophobic, allowing it to cross membranes freely
Distant cells respond because xylem injection increases Z gene expression in those cells
Injection into xylem enables long-distance transport of Z, expanding the range of exposure
Apoplastic signaling is endocrine signaling, while xylem signaling is synaptic signaling
Local signaling requires receptors, but xylem signaling does not require receptors
Explanation
This question assesses understanding of cell communication via signal transduction pathways. The correct answer is A because injecting Z into the xylem allows for long-distance transport through the plant's vascular system, enabling it to reach and activate receptors in distant cells that were previously unaffected by local apoplastic diffusion. Evidence from the stimulus shows that normally only nearby cells (within 2–3 cell diameters) respond to apoplastic Z, but xylem injection expands the response to distant cells, indicating a shift from local to systemic signaling. This highlights how transport pathways influence signal range in plants. A tempting distractor is B, which suggests xylem injection makes Z hydrophobic, but this is wrong due to the misconception that transport alters signal properties, when the key is the distribution pathway. A transferable strategy is to consider transport mechanisms like xylem when evaluating changes in signaling range.
A signaling molecule S is released by cells and normally acts locally because it is rapidly degraded by an extracellular enzyme. A mutant tissue lacks the degrading enzyme, and cells farther from the source now respond to S. Which of the following best explains the expanded signaling range in the mutant?
Reduced extracellular degradation allows S to persist longer and diffuse farther before binding receptors
Loss of the enzyme converts local signaling into synaptic signaling by creating axons
Reduced degradation decreases receptor specificity, allowing any cell surface protein to bind S
Reduced degradation prevents secretion of S, so only distant cells receive the signal
Loss of the enzyme forces S to enter the nucleus of distant cells to be activated
Explanation
This question assesses understanding of cell communication via signal transduction pathways. The correct answer is A because lacking the degrading enzyme allows S to persist longer in the extracellular space, enabling it to diffuse farther and bind receptors on distant cells that were previously unaffected. Evidence from the stimulus shows that normally S acts locally due to rapid degradation, but in the mutant, distant cells respond, indicating expanded signal stability and range. This demonstrates how degradation regulates signaling distance. A tempting distractor is B, which suggests loss of the enzyme forces S into the nucleus, but this is wrong due to the misconception that extracellular enzymes affect intracellular signal processing. A transferable strategy is to assess signal stability and degradation when signaling range changes in mutants.
Cells in a dish release signal S. When S is neutralized by a specific antibody added to the medium, the same cells show reduced response compared with untreated controls. The cells are known to express the receptor for S on their own surfaces. Which of the following best explains the signaling mode involved?
Which of the following best explains the signaling mode suggested by these observations?
Autocrine signaling, because cells respond to a signal they themselves secrete
Quorum sensing, because antibodies increase signal concentration at high density
Juxtacrine signaling, because antibodies remove plasmodesmata between the cells
Synaptic signaling, because antibodies block neurotransmitter release from vesicles
Endocrine signaling, because antibodies can circulate only in the bloodstream
Explanation
This question assesses understanding of cell communication via signal transduction pathways. The signaling mode is autocrine, as cells release signal S and express its receptor, responding to their own secretion, and neutralizing S with antibody reduces this self-response. Antibody addition to the medium decreases response in the same cells, indicating self-signaling. Cells express the receptor, supporting autocrine loop disruption. A tempting distractor is choice D, suggesting juxtacrine via plasmodesmata removal, but this reflects the misconception that antibodies affect cell contacts, whereas they target diffusible signals. To approach similar questions, use neutralization effects to distinguish self-signaling from other modes.
A hormone J is released into blood and reaches many tissues. In tissue U, J produces a response only when cells are pretreated with a drug that increases membrane permeability to ions. Without the drug, J still binds to its receptor but no response occurs. Which of the following best explains the drug’s effect on signaling?
Which of the following best explains why increasing ion permeability enables J’s response?
The drug increases blood flow, causing more J to be produced by the gland
The drug transports J into the nucleus where all receptors are located
The response likely depends on ion movement across the membrane as part of signal transduction
The drug converts J into a different hormone that binds a new receptor
The drug makes receptors unnecessary by allowing J to diffuse directly into ribosomes
Explanation
This question assesses understanding of cell communication via signal transduction pathways. The response likely depends on ion movement across the membrane as part of signal transduction, so the drug enables this by increasing permeability, allowing J to trigger response after binding its receptor. Without the drug, J binds but no response occurs, indicating ion flux is required for transduction. Pretreatment with the drug restores the pathway in tissue U. A tempting distractor is choice E, suggesting the drug bypasses receptors for direct diffusion to ribosomes, but this reflects the misconception that signals act without receptors, ignoring binding evidence. To approach similar questions, identify downstream transduction requirements like ion involvement from pharmacological evidence.
Two ligands, L1 and L2, are added separately to the same cell type. L1 binds receptor R1 and L2 binds receptor R2. When an antibody blocks only R1, the cell still responds to L2 but not L1. Which of the following best explains the specificity of the responses?
Which of the following best explains why blocking R1 affects only L1 signaling?
Ligand-receptor binding is specific, so blocking R1 prevents only L1 from activating its receptor
Antibodies block ligands directly, so L2 is unaffected because it is not a protein
R1 and R2 are identical receptors, so blocking one should block both responses equally
Blocking R1 increases diffusion of L2, allowing L2 to activate R2 more strongly
Blocking R1 converts R2 into an inactive receptor by removing its transmembrane domain
Explanation
This question assesses understanding of cell communication via signal transduction pathways. Ligand-receptor binding is specific, so blocking R1 with an antibody prevents only L1 from activating its receptor, while L2 still binds R2 and triggers response. L1 and L2 act separately on distinct receptors, and the antibody targets only R1, preserving L2 signaling. This demonstrates receptor specificity in transduction pathways. A tempting distractor is choice E, suggesting R1 and R2 are identical so blocking one affects both, but this reflects the misconception that all receptors are nonspecific, ignoring evidence of selective blocking. To approach similar questions, use selective inhibitor effects to map ligand-receptor specificity.
In a culture dish, neuron A releases neurotransmitter X into a synaptic cleft. Muscle cell B contracts only when it expresses receptor R on its membrane. When an R-blocking drug is added, X is still released but B does not contract; a nearby cell lacking R shows no response in either condition. Which of the following best explains the specificity of this cell-to-cell communication?
Cells lacking receptor R still detect neurotransmitter X because all membranes allow X to diffuse inside.
Neurotransmitter X changes its structure in the cleft to match receptors on any nearby cell membrane.
Muscle cell B responds because receptor R binds neurotransmitter X, initiating signaling only in cells with R.
The R-blocking drug prevents neuron A from releasing neurotransmitter X into the synaptic cleft.
Muscle cell contraction occurs because neurotransmitter X directly catalyzes ATP hydrolysis on the cell surface.
Explanation
This question assesses understanding of cell communication via signal transduction pathways, focusing on the specificity of ligand-receptor interactions. The specificity is evident because muscle cell B contracts only when expressing receptor R, and a nearby cell without R shows no response, indicating that signaling requires receptor binding. When the R-blocking drug is added, neurotransmitter X is released but B does not contract, confirming that the response depends on X binding to R to initiate the pathway. Choice B accurately explains that the response occurs only in cells with R, highlighting how receptors confer specificity in cell-to-cell communication. A tempting distractor is choice D, which is incorrect because cells without R do not respond, stemming from the misconception that signals can freely enter any cell without specific receptors. For similar problems, identify how experimental manipulations like blockers reveal the role of receptors in selective signaling.
A receptor R is normally found on the plasma membrane and binds ligand L outside the cell. A mutation deletes R’s transmembrane domain, and the mutant R is secreted into the extracellular fluid. Cells with only mutant R show no response to L, even though L binds mutant R in solution. Which of the following best explains the loss of signaling?
Which of the following best explains why signaling fails with the secreted receptor?
Mutant R causes L to diffuse faster, preventing any binding interactions
Secreted R cannot relay ligand binding across the membrane to initiate a cellular response
Deleting a transmembrane domain converts L into a different ligand type
L cannot bind receptors unless receptors are located in the nucleus
Secreted R increases L concentration inside the cell by active transport
Explanation
This question assesses understanding of cell communication via signal transduction pathways. The secreted mutant receptor R cannot relay ligand L binding across the membrane to initiate transduction, as it lacks the transmembrane domain needed to anchor and transmit the signal intracellularly. Normally, R binds L outside and signals inside via its transmembrane domain, but mutation causes secretion, so binding occurs in solution without transduction. Cells with only mutant R show no response despite binding, confirming the domain's role. A tempting distractor is choice B, claiming secreted R increases L inside by transport, but this reflects the misconception that soluble receptors act as carriers, whereas they fail to transduce signals. To approach similar questions, analyze receptor structure modifications and their impact on signal relay across membranes.
A secreted ligand Z triggers a response only when presented in pulses every 5 minutes; constant exposure to the same average concentration produces little response. Binding assays show that receptors bind Z in both conditions. Which of the following best explains why pulsing changes signaling effectiveness?
Which of the following best explains why pulses of Z produce a stronger response than constant Z?
Pulses convert Z into a steroid, enabling intracellular receptor binding
Constant Z prevents diffusion, so Z cannot reach the receptor-binding site
Pulses increase Z solubility, allowing it to cross the membrane without receptors
Constant Z increases cell wall thickness, blocking receptor access to the ligand
Constant Z likely causes receptor desensitization or internalization, reducing signaling over time
Explanation
This question assesses understanding of cell communication via signal transduction pathways. Constant exposure to ligand Z likely causes receptor desensitization or internalization, reducing signaling over time, while pulses allow recovery and stronger responses despite the same average concentration. Binding occurs in both conditions, but pulsing maintains effectiveness by preventing adaptation. This explains why constant Z produces little response compared to pulses. A tempting distractor is choice B, claiming pulses increase solubility, but this reflects the misconception that delivery mode alters molecular properties, whereas adaptation affects receptor responsiveness. To approach similar questions, consider how signal presentation timing influences receptor dynamics and adaptation.
A scientist compares two signaling molecules: peptide P and steroid S. P causes a response only when added outside intact cells, while S causes a response even when added to cell-free cytosolic extracts containing receptor proteins. Which of the following best explains the difference in how P and S are detected?
Which of the following best explains the most likely receptor locations for P and S?
Both P and S require plasmodesmata for receptor binding
P is detected by intracellular receptors, while S is detected by membrane receptors
Both P and S bind directly to phospholipids, so receptors are unnecessary
P is detected by membrane receptors, while S is detected by intracellular receptors
Both P and S are detected only by receptors embedded in the cell wall
Explanation
This question assesses understanding of cell communication via signal transduction pathways. Peptide P is detected by membrane receptors, while steroid S is detected by intracellular receptors, as P works only outside intact cells, but S activates cytosolic extracts without membranes. P, being hydrophilic, cannot cross membranes, requiring surface receptors for transduction. S, being lipophilic, enters cells and binds cytosolic receptors directly. A tempting distractor is choice A, reversing the receptor locations, but this reflects the misconception that peptides use intracellular receptors, ignoring their inability to cross membranes. To approach similar questions, compare molecule properties and experimental contexts to determine receptor types.