AP Biology · Learn by Concept

AP Biology Help: Cell Communication

Review real example questions for Cell Communication in AP Biology.

Question 1 / 10

0 of 10 answered

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?

All questions

Question 1

Synaptic release creates a high local concentration near receptors before Y is diluted systemically (correct answer)
Bloodstream injection converts Y into a steroid that cannot bind membrane receptors
Synaptic release requires Y to enter the nucleus, which cannot occur from the blood
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.

Question 2

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 enables long-distance transport of Z, expanding the range of exposure (correct answer)
Injection into xylem causes Z to become hydrophobic, allowing it to cross membranes freely
Local signaling requires receptors, but xylem signaling does not require receptors
Apoplastic signaling is endocrine signaling, while xylem signaling is synaptic signaling
Distant cells respond because xylem injection increases Z gene expression in those cells

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.

Question 3

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 (correct answer)
Loss of the enzyme forces S to enter the nucleus of distant cells to be activated
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

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.

Question 4

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?

Autocrine signaling, because cells respond to a signal they themselves secrete (correct answer)
Synaptic signaling, because antibodies block neurotransmitter release from vesicles
Endocrine signaling, because antibodies can circulate only in the bloodstream
Juxtacrine signaling, because antibodies remove plasmodesmata between the cells
Quorum sensing, because antibodies increase signal concentration at high density

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.

Question 5

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?

Ligand-receptor binding is specific, so blocking R1 prevents only L1 from activating its receptor (correct answer)
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
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

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.

Question 6

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?

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. (correct answer)
The R-blocking drug prevents neuron A from releasing neurotransmitter X into the synaptic cleft.
Cells lacking receptor R still detect neurotransmitter X because all membranes allow X to diffuse inside.
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.

Question 7

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?

Constant Z likely causes receptor desensitization or internalization, reducing signaling over time (correct answer)
Pulses increase Z solubility, allowing it to cross the membrane without receptors
Constant Z prevents diffusion, so Z cannot reach the receptor-binding site
Pulses convert Z into a steroid, enabling intracellular receptor binding
Constant Z increases cell wall thickness, blocking receptor access to the ligand

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.

Question 8

A peptide ligand binds a receptor on immune cells. When extracellular pH is lowered, ligand binding decreases sharply, but receptor abundance on the membrane is unchanged. Which of the following best explains the reduced binding at low pH?

Low pH alters charge interactions and protein conformation, reducing ligand–receptor affinity (correct answer)
Low pH increases receptor gene transcription, temporarily diluting receptors across the membrane
Low pH causes the ligand to become nonpolar and diffuse through the membrane, avoiding receptors
Low pH blocks blood flow, preventing long-distance endocrine delivery of the peptide to cells
Low pH prevents vesicles from forming, so receptors cannot be inserted into the membrane

Explanation: This question assesses understanding of cell communication via signal transduction pathways. The correct answer is A because low extracellular pH alters charge interactions and protein conformation in the ligand or receptor, reducing their binding affinity without changing receptor abundance on the membrane. Evidence from the stimulus indicates that binding decreases sharply at low pH, but receptor levels remain unchanged, pointing to a direct effect on interaction dynamics. This is common in immune signaling where pH influences receptor-ligand pairs. A tempting distractor is C, which claims low pH makes the ligand nonpolar, but this is incorrect due to the misconception that pH changes molecular polarity, ignoring that peptides remain polar. A transferable strategy is to consider environmental factors like pH when binding affinity changes without alterations in receptor number.

Question 9

In a plant leaf, Cell A releases a small peptide signal into the extracellular fluid. Nearby Cell B responds within seconds only if it has membrane protein R. When Cell B is treated with a protease that removes extracellular protein domains, the peptide is still present outside the cell but Cell B no longer responds. A different nearby cell lacking R shows no response under any condition. Which of the following best explains the loss of Cell B's response after protease treatment?

The protease degraded the peptide signal before it could enter Cell B by diffusion
The protease removed the extracellular binding site required for peptide recognition by receptor R (correct answer)
The protease increased transcription of receptor R, preventing immediate signaling responses
The protease converted the peptide into a lipid-soluble signal that bypasses receptors
The protease blocked gap junction formation, preventing long-distance signaling between cells

Explanation: This question assesses understanding of cell communication via signal transduction pathways. The correct answer is B because the protease treatment removes extracellular protein domains, including the binding site on receptor R, preventing the peptide signal from binding and initiating a response in Cell B. Evidence from the stimulus shows that the peptide is still present outside the cell after treatment, but Cell B no longer responds, indicating the issue is with receptor functionality rather than signal availability. Additionally, cells lacking R show no response under any condition, confirming that R is essential for signal detection. A tempting distractor is A, which suggests the protease degraded the peptide signal, but this is incorrect due to the misconception that proteases act on free signals rather than membrane-bound receptors, ignoring that the peptide remains present. A transferable strategy is to distinguish between disruptions to ligands versus receptors when signaling fails despite ligand presence.

Question 10

A bacterial population uses quorum sensing: cells secrete small molecule Q that diffuses in the medium. At low cell density, Q remains low and cells do not respond. At high density, Q accumulates and binds receptor protein inside the bacteria, triggering a rapid physiological response. Which of the following best explains why response depends on density?

High density increases extracellular Q concentration so more Q enters cells and binds intracellular receptors (correct answer)
High density changes bacterial DNA sequence, creating new receptors that recognize Q
Low density prevents diffusion of Q, so Q cannot leave the secreting cells
High density makes Q bind directly to ribosomes, eliminating the need for receptor proteins
Low density causes Q to become a neurotransmitter that requires synapses to function

Explanation: This question assesses understanding of cell communication via signal transduction pathways. The correct answer is A because at high cell density, the extracellular concentration of Q increases, allowing more Q to enter cells and bind intracellular receptors, triggering the response, whereas at low density, Q remains too dilute. Evidence from the stimulus shows that Q diffuses in the medium and binds inside bacteria, with response occurring only at high density when Q accumulates. This is the basis of quorum sensing in bacterial populations. A tempting distractor is C, which suggests low density prevents Q diffusion, but this is incorrect due to the misconception that density affects diffusion physics, when it's actually about concentration buildup. A transferable strategy is to consider density-dependent concentration effects in population-level signaling like quorum sensing.