AP Biology Quiz: Signal Transduction Pathways

What this quiz covers

This quiz focuses on Signal Transduction Pathways, 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. Mutating receptor R so H cannot bind its extracellular domain (correct answer)
2. Increasing intracellular ATP concentration to provide more substrate
3. Adding a competitive inhibitor that binds the active site of enzyme E
4. Deleting the catalytic subunits of PKA while leaving cAMP levels unchanged
5. Overexpressing PDE so cAMP is rapidly hydrolyzed even with inhibitor present

Explanation: This question assesses the skill of analyzing signal transduction pathways by identifying changes that abolish the impact of a modulator on downstream enzyme activity. The pathway begins with hormone H binding receptor R to activate G protein Gα, stimulating adenylyl cyclase to produce cAMP from ATP, which activates PKA to phosphorylate and enhance enzyme E activity, while PDE hydrolyzes cAMP to limit the signal. The PDE inhibitor prolongs cAMP elevation only when H is present, leading to a larger and sustained increase in E activity, but has minimal effect without H due to low basal cAMP production. Mutating receptor R to prevent H binding blocks pathway initiation, eliminating the inhibitor's ability to enhance E activity since no cAMP is generated in response to H. A tempting distractor is choice E, which suggests overexpressing PDE to rapidly hydrolyze cAMP despite the inhibitor, but this overlooks that the inhibitor's effect depends on upstream activation, not just PDE levels, reflecting a misconception about signal dependency. A transferable strategy is to dissect how inhibitors affect pathway dynamics and test upstream blocks to confirm dependency on initial signal transduction steps.

Question 2

1. Receptor internalization normally reduces signaling by removing active receptors from the membrane (correct answer)
2. Receptor internalization is required to activate adenylyl cyclase at the membrane
3. Blocking internalization increases cytosolic ATP, which directly activates PKA
4. Blocking internalization prevents cAMP synthesis but extends PKA activity by feedback
5. Receptor internalization increases ligand concentration by importing L into the cytosol

Explanation: This question assesses the skill of analyzing a signal transduction pathway. Receptor internalization normally reduces signaling by removing active receptors from the membrane, so blocking it prolongs PKA activity as receptors continue activating the pathway despite unchanged binding. In the pathway, L activates R to increase cAMP and PKA, with normal transient response, but blocking internalization extends it. This supports desensitization via internalization. Choice B is tempting but wrong because it suggests internalization activates cyclase, which is a misconception as the prolonged signal indicates internalization terminates signaling. A transferable strategy is to manipulate endocytosis to study desensitization mechanisms in GPCR pathways.

Question 3

1. Increase PI3K activity to raise PIP$_3$ levels despite phosphatase conversion (correct answer)
2. Inhibit Ad phosphorylation so PI3K binds more strongly to the receptor fragment
3. Remove membrane lipids so Akt remains cytosolic and is phosphorylated more efficiently
4. Block ATP binding to Akt so it stays phosphorylated and continues phosphorylating X
5. Add extra PIP$_2$ so the phosphatase has more substrate and produces more PIP$_3$

Explanation: This question assesses the skill of analyzing signal transduction pathways by countering phosphatase effects in PI3K-Akt signaling. The lipid phosphatase converts PIP3 to PIP2, reducing PIP3, Akt membrane recruitment, phosphorylation, and X phosphorylation. Increasing PI3K activity raises PIP3 levels, opposing the phosphatase and restoring Akt activation and X phosphorylation. This boosts production to overcome degradation, maintaining the lipid signal. A tempting distractor is choice E, adding PIP2 to increase PIP3 via phosphatase, but phosphatases degrade, not produce PIP3, confusing reaction direction. For balancing enzymes, enhance the opposing activity to shift equilibrium toward the desired product.

Question 4

1. Express a PKA catalytic subunit variant that is active without cAMP binding (correct answer)
2. Increase phosphodiesterase activity further to stabilize cAMP by faster turnover
3. Decrease receptor number so AC is activated more strongly per receptor
4. Block ion flux through Ch so phosphorylation remains detectable for longer
5. Add a ligand antagonist to prevent receptor activation and reduce channel dephosphorylation

Explanation: This question assesses the skill of analyzing signal transduction pathways by selecting ways to sustain phosphorylation despite rapid cAMP breakdown. Rapid cAMP breakdown by phosphodiesterase activation causes quick loss of PKA activity and channel Ch dephosphorylation, as Reg inhibits PKA when cAMP falls. Expressing a PKA catalytic subunit active without cAMP bypasses cAMP dependence, maintaining phosphorylation even with low cAMP. This variant escapes Reg inhibition, sustaining the signal. A tempting distractor is choice B, increasing phosphodiesterase further, but this would worsen cAMP loss, not sustain it, confusing turnover with stabilization. To maintain signals, introduce components insensitive to the degrading factor.

Question 5

1. Overexpressing GAP to accelerate GTP hydrolysis and shorten G activity duration
2. Increasing GEF activity to promote more frequent exchange of GDP for GTP on G
3. Inhibiting effector enzyme E so it cannot be activated by G-GTP (correct answer)
4. Adding more ligand so receptor activation can override constitutive G activity
5. Reducing cytosolic GDP concentration to favor spontaneous GTP binding to G

Explanation: This question assesses the skill of analyzing a signal transduction pathway. The pathway activates GTPase G via receptor-stimulated GEF to promote GDP-GTP exchange, with GAP accelerating hydrolysis for inactivation and G-GTP activating effector E. Choice C restores regulation in mutant G cells by inhibiting E, preventing its activation despite constitutive G-GTP from impaired hydrolysis, thus controlling E activity without ligand. This counters the high basal E activity in mutants lacking GTPase function. A tempting distractor is A, overexpressing GAP, but this misconception assumes GAP can accelerate hydrolysis in a mutant that cannot hydrolyze GTP, whereas it fails to inactivate the mutant. A transferable strategy is to target downstream effectors when upstream regulators are constitutively active in signaling mutations.

Question 6

1. G
   remains active longer because it cannot hydrolyze GTP, sustaining adenylyl cyclase activation (correct answer)
2. The toxin prevents ligand from dissociating, so receptors stay occupied and increase cAMP permanently
3. Adenylyl cyclase becomes a ligand-gated channel that stays open after covalent modification
4. cAMP remains elevated because phosphodiesterase converts cAMP into ATP more slowly
5. cAMP elevation persists because the receptor moves into the cytosol where it produces more ligand

Explanation: This question assesses the skill of analyzing a signal transduction pathway. The toxin prevents GTP hydrolysis by the Gα subunit, which normally deactivates the G protein after activation, leading to prolonged activation of adenylyl cyclase and sustained cAMP elevation even after the ligand is removed. In the pathway, ligand binding to the GPCR causes Gα to exchange GDP for GTP and activate adenylyl cyclase, but without hydrolysis, Gα remains in the active GTP-bound state. This explains the prolonged cAMP signal as the pathway fails to terminate properly. Choice B is tempting but wrong because it assumes the toxin affects ligand dissociation from the receptor, which is a misconception about the toxin's target being the G protein rather than the receptor-ligand interaction. A transferable strategy is to map out the sequence of activation and deactivation steps in G protein-coupled pathways to identify where disruptions occur.

Question 7

1. Inhibiting PP increases kinase C phosphorylation, extending kinase C activity toward enzyme Y (correct answer)
2. Inhibiting PP prevents ligand X from binding receptor R, increasing kinase A activation
3. Inhibiting PP decreases kinase B phosphorylation, causing kinase C to activate enzyme Y faster
4. Inhibiting PP directly phosphorylates enzyme Y, bypassing kinases A, B, and C
5. Inhibiting PP reduces ATP availability, slowing dephosphorylation and increasing Y activity

Explanation: This question assesses the skill of analyzing a signal transduction pathway. The pathway activates kinase A upon ligand X binding receptor R, leading to sequential phosphorylation of kinases B and C, which then activates enzyme Y, with phosphatase PP normally dephosphorylating C to reduce its activity. Choice A accounts for increased Y activity in PP-inhibited cells because inhibiting PP sustains kinase C phosphorylation, extending its activity and enhancing Y phosphorylation despite normal A activation. This results in a larger Y response, as seen in the experiment. A tempting distractor is C, suggesting inhibited PP decreases B phosphorylation, but this misconception confuses dephosphorylation targets, as PP acts on C to prolong, not accelerate, downstream activation. A transferable strategy is to assess how phosphatase inhibition affects phosphorylation persistence in multi-step kinase cascades.

Question 8

1. Activation of PKC by $\mathrm{Ca^{2+}}$ binding, preventing phosphorylation of transporter T (correct answer)
2. Autophosphorylation of receptor R, preventing recruitment of PLC$\gamma$
3. Synthesis of IP$_3$ from PIP$_2$, preventing second messenger formation
4. Binding of growth factor G to receptor R at the cell surface
5. Insertion of transporter T into the membrane by vesicle trafficking

Explanation: This question assesses the skill of analyzing a signal transduction pathway. The Ca²⁺ chelator directly blocks activation of PKC by preventing Ca²⁺ binding, which is required alongside DAG for PKC to phosphorylate transporter T and increase ion transport. In the pathway, G activates R autophosphorylation, PLCγ recruitment, IP₃ production, and Ca²⁺ release, but without free Ca²⁺, PKC cannot activate despite IP₃ rising. This step is disrupted downstream of IP₃ but upstream of transport. Choice C is tempting but wrong because it describes IP₃ synthesis as preventing second messenger formation, which is a misconception reversing the role of PLCγ in generating messengers. A transferable strategy is to identify cofactor requirements like Ca²⁺ in kinase activation to predict blocks in RTK pathways.

Question 9

1. Increase PDE activity to accelerate cAMP breakdown after PKA activation (correct answer)
2. Mutate receptor R to prevent ligand L binding at the extracellular surface
3. Block ATP production so adenylyl cyclase cannot generate any second messenger
4. Remove channel C from the membrane so phosphorylation has no cellular purpose
5. Increase cytosolic ribosomes to enhance synthesis of PKA catalytic subunits

Explanation: This question assesses the skill of analyzing a signal transduction pathway. Increasing PDE activity would accelerate cAMP breakdown, reducing the duration of PKA activation and thus shortening the Ca²⁺ influx response without affecting initial receptor binding by ligand L. In the pathway, L activates the G protein and adenylyl cyclase to produce cAMP, which activates PKA to phosphorylate channel C, but faster cAMP hydrolysis by PDE would limit this. The PDE inhibitor experiment shows that slowing cAMP breakdown prolongs the response, supporting that enhancing PDE would have the opposite effect. Choice B is tempting but wrong because mutating receptor R to prevent binding would block the initial response entirely, which is a misconception about targeting downstream steps to modulate duration without preventing initiation. A transferable strategy is to identify negative regulators like degradative enzymes in pathways and consider how altering their activity affects signal timing.

Question 10

1. Ligand binding to the extracellular domain of the receptor channel
2. Opening of the receptor channel to permit Na$^+$ influx into the cytosol (correct answer)
3. Kinase K phosphorylation of protein V to increase V activity
4. Second messenger M activation of kinase K in the cytosol
5. Na$^+$-sensitive enzyme N synthesis to increase intracellular enzyme concentration

Explanation: This question assesses the skill of analyzing a signal transduction pathway. The pathway involves ligand binding opening a receptor channel for Na⁺ influx, activating enzyme N to produce messenger M, which stimulates kinase K to phosphorylate and activate protein V. Choice B is most immediately prevented by removing extracellular Na⁺, as ligand binding occurs but without Na⁺ influx, N remains inactive, blocking M production and downstream V activation. This is evidenced by unchanged M levels and V activity in the absence of Na⁺. A tempting distractor is A, ligand binding to the receptor, but this misconception assumes binding requires Na⁺, whereas the results show binding persists without influx. A transferable strategy is to pinpoint the first disrupted step in ion-dependent pathways by examining immediate downstream outputs.

Question 11

1. Increase activity of the phosphatase that dephosphorylates S after signaling begins (correct answer)
2. Decrease receptor affinity for L so fewer receptors are occupied at the start
3. Inhibit the kinase upstream of S so phosphorylation cannot be initiated
4. Add more ligand L so receptor activation persists and S remains phosphorylated longer
5. Block ATP synthesis so phosphorylation reactions proceed longer using stored phosphate

Explanation: This question assesses the skill of analyzing signal transduction pathways by shortening signal duration without affecting peaks in kinase cascades. Phosphatase inhibitor increases peak and duration of S phosphorylation, suggesting increasing phosphatase activity would accelerate dephosphorylation, shortening duration while leaving initial peak intact. This enhances termination without altering activation. No binding changes confirm post-activation regulation. A tempting distractor is choice C, inhibiting upstream kinase, but this eliminates the peak, not just shortens duration, confusing initiation with termination. To modulate duration, target regulators of signal decay like phosphatases.

Question 12

1. Increase phosphatase activity further to maintain M3 in a dephosphorylated state
2. Inhibit M2 so M3 is not phosphorylated and cannot phosphorylate X
3. Use an M3 variant lacking the phosphatase recognition site, reducing dephosphorylation (correct answer)
4. Remove pheromone P so the receptor cannot activate the G protein pathway
5. Decrease ATP availability so phosphorylation reactions occur at higher rates

Explanation: This question tests the ability to analyze components and interventions in a signal transduction pathway. The pathway involves pheromone P activating a GPCR and MAP kinase cascade where M1 activates M2, M2 activates M3, and M3 phosphorylates X, with a phosphatase limiting M3 activity by dephosphorylation. Using an M3 variant lacking the phosphatase recognition site would reduce dephosphorylation, prolonging M3 activation and increasing phospho-X despite phosphatase overexpression. This directly protects M3 from excessive negative regulation. A tempting distractor is choice A, increasing phosphatase activity, but this misinterprets the goal by assuming more dephosphorylation helps, when it would further decrease M3 activity and phospho-X. A transferable strategy is to alter enzyme recognition sites to mitigate overactive negative regulators in phosphorylation cascades.

Question 13

1. Activation of Gαq by the GPCR, preventing PLC from cleaving membrane PIP2
2. Binding of IP3 to its ER receptor, preventing IP3 formation in the cytosol
3. Formation of the Ca2+-calmodulin complex needed to activate Cmk (correct answer)
4. Phosphorylation of PLC by Cmk, preventing IP3 accumulation after Z addition
5. Diffusion of ligand Z through the membrane, preventing it from reaching the GPCR binding site

Explanation: This question analyzes the PLC-IP3-Ca2+ signal transduction pathway and the effect of Ca2+ chelation. The pathway flows: ligand Z → GPCR → Gαq → PLC → IP3 → ER Ca2+ release → Ca2+ binds calmodulin → Ca2+-calmodulin complex activates Cmk → Cmk phosphorylates enzyme V. The Ca2+ chelator binds free cytosolic Ca2+, preventing it from binding to calmodulin, which means the Ca2+-calmodulin complex cannot form to activate Cmk (choice C). Since IP3 levels rise normally, the chelator doesn't affect upstream steps. Students often think chelators affect Ca2+ release (choice B), but they actually sequester Ca2+ after it's already in the cytosol. When analyzing Ca2+ signaling, distinguish between Ca2+ release and Ca2+-dependent downstream events.

Question 14

1. Reduced cAMP because unphosphorylated GPCRs cannot activate Gs
2. More sustained cAMP because arrestin binding is reduced, prolonging G protein activation (correct answer)
3. Decreased Ca2+ influx because PKA requires GPCR phosphorylation to become active
4. No change because arrestin blocks epinephrine binding rather than G protein coupling
5. Lower enzyme complex activity because new channels must be inserted into the membrane

Explanation: This question analyzes how preventing receptor desensitization affects signal duration. Normally, activated GPCRs are phosphorylated by specific kinases (like GRKs), allowing arrestin to bind and block further G protein activation, thus terminating signaling. The kinase inhibitor prevents GPCR phosphorylation, so arrestin cannot bind to terminate Gs activation. This allows the epinephrine-bound receptor to continue activating Gs proteins, producing more sustained cAMP levels and prolonged PKA activity compared to normal desensitization. Choice A incorrectly suggests unphosphorylated GPCRs cannot activate G proteins, confusing the role of phosphorylation in desensitization versus initial activation. When analyzing receptor regulation, distinguish between mechanisms for initial activation (ligand binding) and subsequent inactivation (phosphorylation and arrestin binding).

Question 15

1. Increase phosphatase D activity so P is dephosphorylated efficiently without recruitment by U (correct answer)
2. Inhibit kinase K so P remains phosphorylated longer after ligand removal
3. Increase extracellular chloride to enhance channel opening persistence
4. Mutate receptor R to bind ligand L more tightly to shorten signaling
5. Increase transcription of U to overcome the binding mutation by mass action

Explanation: This question assesses the skill of analyzing signal transduction pathways by determining modifications that can alter signal termination in a mutant context. The normal pathway features ligand L binding receptor R to activate kinase K, which phosphorylates protein P, opening a chloride channel to change membrane potential, while protein U recruits phosphatase D to dephosphorylate P and terminate the signal. In the mutation preventing U from binding P, phosphorylated P persists, leading to prolonged channel opening after L removal because dephosphorylation is impaired. Increasing phosphatase D activity allows efficient dephosphorylation of P even without U-mediated recruitment, reversing the persistence by enabling faster signal shutoff in mutant cells. A tempting distractor is choice E, which proposes increasing U transcription to overcome the mutation by mass action, but this misunderstands that the binding defect prevents U from interacting with P regardless of quantity. A transferable strategy is to trace how mutations disrupt feedback or termination steps in pathways and evaluate changes that compensate by enhancing alternative regulatory mechanisms.

Question 16

1. Inhibit guanylyl cyclase to reduce cGMP synthesis and favor channel closure (correct answer)
2. Increase cGMP production to compensate for reduced PDE activity
3. Block rhodopsin activation so transducin is not activated by light
4. Open $\mathrm{Na^+}$ channels directly to increase hyperpolarization amplitude
5. Increase ATP hydrolysis to speed GTP binding to transducin

Explanation: This question assesses the skill of analyzing a signal transduction pathway. Inhibiting guanylyl cyclase would reduce cGMP synthesis, favoring channel closure and increasing hyperpolarization to compensate for reduced PDE activity in the mutant. In the visual transduction pathway, light activates rhodopsin to transducin to PDE, decreasing cGMP to close Na⁺ channels, but lower PDE means less cGMP drop and smaller hyperpolarization. Reducing cGMP production enhances the effect. Choice B is tempting but wrong because increasing cGMP would worsen the mutant's reduced response, which is a misconception about balancing synthesis and degradation. A transferable strategy is to adjust opposing enzymes like cyclases and diesterases to modulate cyclic nucleotide levels in sensory pathways.

Question 17

1. Loss of PKA catalytic activity so cAMP cannot drive phosphorylation of downstream targets (correct answer)
2. Reduced ligand affinity of R so receptor occupancy cannot reach threshold
3. Inactive adenylyl cyclase so cAMP cannot increase after receptor activation
4. Overactive phosphodiesterase so cAMP is eliminated before binding PKA
5. Failure of X to be synthesized so phosphorylation cannot be observed

Explanation: This question assesses the skill of analyzing a signal transduction pathway. The defect is loss of PKA catalytic activity, as normal L binding and cAMP increase occur, but X phosphorylation is reduced, and a cAMP analog fails to restore it, indicating the issue is downstream of cAMP. In the pathway, L activates R, Gα, adenylyl cyclase, and cAMP to stimulate PKA for X phosphorylation. The analog bypassing cyclase but not restoring function points to PKA itself. Choice C is tempting but wrong because inactive adenylyl cyclase would prevent cAMP increase, which is a misconception since cAMP rises normally in the mutant. A transferable strategy is to use analogs or activators to test pathway components and localize defects in second messenger systems.

Question 18

1. Phosphatases normally remove phosphate from P, and inhibiting them prolongs the phosphorylated state (correct answer)
2. Phosphatases normally add phosphate to P, and inhibiting them increases phosphorylation
3. Phosphatase inhibition prevents receptor R from binding L, increasing free ligand concentration
4. Phosphatase inhibition blocks ATP synthesis, causing kinases to work more efficiently
5. Phosphatase inhibition increases membrane permeability, allowing L to enter the cytosol

Explanation: This question assesses the skill of analyzing a signal transduction pathway. Phosphatases normally remove phosphate from P, and inhibiting them prolongs the phosphorylated state, increasing magnitude and duration as dephosphorylation is blocked while kinase activity continues. In the pathway, L activates R to phosphorylate P rapidly, and phosphatase inhibition enhances this without affecting binding. Kinase inhibitor blocking confirms phosphorylation dependence. Choice B is tempting but wrong because it reverses phosphatases' role to adding phosphate, which is a misconception of their dephosphorylating function. A transferable strategy is to use enzyme inhibitors to dissect the balance between kinases and phosphatases in regulating signal strength.

Question 19

1. The scaffold increases signaling efficiency by colocalizing kinases to promote sequential phosphorylation (correct answer)
2. The scaffold converts F into a second messenger that directly phosphorylates T
3. The scaffold prevents F from dissociating from R, increasing receptor occupancy
4. The scaffold transports phosphorylated T into the nucleus to sustain the response
5. The scaffold increases ATP synthesis, supplying energy for phosphorylation reactions

Explanation: This question assesses the skill of analyzing a signal transduction pathway. The scaffold increases signaling efficiency by colocalizing M1, M2, and M3 near the receptor, promoting rapid sequential phosphorylation in the MAPK cascade despite normal receptor and G protein activation. In the yeast pathway, F activates R and G protein to the cascade, phosphorylating T, but without the scaffold, efficiency drops, leading to lower and slower T phosphorylation. This explains the scaffold's role in organization. Choice D is tempting but wrong because it suggests the scaffold transports T to the nucleus, which is a misconception as the effect is on phosphorylation rate, not localization. A transferable strategy is to consider protein complexes and scaffolds when analyzing speed and efficiency in multi-step kinase pathways.

Question 20

1. S2 phosphorylation depends on Ca$^{2+}$ signaling and does not require cAMP production (correct answer)
2. Adenylyl cyclase inhibition increases IP$_3$ production, compensating for loss of cAMP
3. Adenylyl cyclase is downstream of CaK, so inhibiting it selectively activates pathway 2
4. The inhibitor blocks ligand binding, but S2 phosphorylation persists due to receptor recycling
5. S1 phosphorylation occurs in the nucleus, while S2 phosphorylation occurs only at the membrane

Explanation: This question assesses the skill of analyzing signal transduction pathways by explaining inhibitor specificity in parallel signaling branches. The adenylyl cyclase inhibitor blocks S1 phosphorylation via pathway 1 (cAMP-PKA) but not S2 via pathway 2 (IP3-Ca2+-CaK), showing S2 depends on Ca2+ signaling independently of cAMP. This specificity arises because the pathways diverge after receptor activation, with the inhibitor targeting only the cAMP branch. The unchanged S2 confirms parallel, non-overlapping transduction. A tempting distractor is choice B, suggesting inhibition increases IP3, but there's no evidence of compensation, misunderstanding independence versus crosstalk. To dissect parallel paths, use selective inhibitors and observe which outputs are affected.