Biochemistry Quiz: G Protein Coupled Receptors Second Messengers

What this quiz covers

This quiz focuses on G Protein Coupled Receptors Second Messengers, giving you a quick way to practice the rules, question types, and explanations that matter most for Biochemistry.

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. GTP serves as the primary phosphate donor for the phosphorylation of downstream kinases like PKA.
2. The hydrolysis of GTP provides the necessary energy for the Gα subunit to bind to its effector enzyme.
3. GTP binding induces a conformational change in Gα, promoting its dissociation from the Gβγ dimer. (correct answer)
4. GTP acts as a substrate, along with ATP, for adenylyl cyclase to synthesize a novel second messenger.

Explanation: GTP does not act as a phosphate donor for kinases (that's ATP) or as a substrate for adenylyl cyclase. Its role is allosteric. The binding of GTP (exchanged for GDP) into the nucleotide-binding pocket of the Gα subunit causes a significant conformational change. This new conformation has low affinity for the Gβγ dimer and the receptor, causing the Gα-GTP monomer to dissociate and become free to interact with its effector protein.

Question 2

1. Generation of a small, diffusible second messenger from a membrane-bound precursor.
2. Activation of a membrane-associated effector enzyme by a G-protein α-subunit.
3. Phosphorylation of specific serine and threonine residues on downstream target proteins.
4. Mobilization of Ca²⁺ ions from the endoplasmic reticulum via an IP₃-gated channel. (correct answer)

Explanation: The Gαq pathway activates phospholipase C, which generates IP₃. IP₃ binds to receptors on the endoplasmic reticulum, causing the release of stored Ca²⁺. This is a hallmark of the Gαq pathway. The other options are features of both pathways: Gαs produces cAMP and Gαq produces IP₃/DAG (A); Gαs activates adenylyl cyclase and Gαq activates PLC (B); PKA (from Gαs) and PKC (from Gαq) are both Ser/Thr kinases (C).

Question 3

1. An increased rate of glycogenolysis due to constitutive activation of the PKA signaling cascade. (correct answer)
2. A decreased rate of gluconeogenesis due to feedback inhibition of key enzymes by high levels of cAMP.
3. A shift towards fatty acid synthesis due to the dephosphorylation of acetyl-CoA carboxylase.
4. A complete cessation of glycolysis as PKA phosphorylates and inhibits phosphofructokinase-1.

Explanation: By inhibiting the GTPase activity of Gαs, cholera toxin locks it in an active state. This leads to constitutive activation of adenylyl cyclase, high levels of cAMP, and persistent activation of Protein Kinase A (PKA). In hepatocytes, PKA phosphorylates and activates phosphorylase kinase, which in turn activates glycogen phosphorylase, stimulating glycogen breakdown (glycogenolysis).

Question 4

1. Adenylyl cyclase will be inhibited, but the K⁺ channel will not be activated.
2. The K⁺ channel will be activated, but adenylyl cyclase activity will remain at basal levels.
3. Neither adenylyl cyclase inhibition nor K⁺ channel activation will occur. (correct answer)
4. Both pathways will be constitutively active because the toxin uncouples the G-protein from the receptor.

Explanation: Pertussis toxin prevents the Gαi protein from interacting with the activated receptor, thereby blocking the GDP-for-GTP exchange. This means the Gαi protein can never become activated. Since activation is required to release both Gαi-GTP and the Gβγ dimer, neither of the downstream signaling events (inhibition of adenylyl cyclase or activation of the K⁺ channel) can occur.

Question 5

1. cAMP levels will decrease with the hormone alone and remain at basal levels with the toxin pre-treatment. (correct answer)
2. cAMP levels will decrease with the hormone alone and increase above basal levels with the toxin pre-treatment.
3. cAMP levels will increase with the hormone alone and remain at basal levels with the toxin pre-treatment.
4. cAMP levels will not change in either condition as Gαi does not affect cAMP production.

Explanation: The hormone activates a Gαi-coupled receptor. Gαi inhibits adenylyl cyclase, so adding the hormone alone will cause cAMP levels to decrease below the basal level. Pertussis toxin prevents the activation of Gαi by uncoupling it from the receptor. Therefore, in cells pre-treated with the toxin, adding the hormone will have no effect, and cAMP levels will remain at their basal production level.

Question 6

1. An attenuated and shortened signal due to the inability of the G-protein to hydrolyze GTP.
2. A complete lack of signaling, as the receptor cannot be properly activated without GRK.
3. An amplified and prolonged signal due to the failure of arrestin to bind and promote desensitization. (correct answer)
4. The G-protein will be unable to dissociate from the receptor, trapping it in an inactive state.

Explanation: GRKs phosphorylate activated GPCRs, which creates a binding site for arrestin. Arrestin binding sterically hinders G-protein coupling and promotes receptor endocytosis, a process called homologous desensitization. If GRK is non-functional, the receptor will not be phosphorylated, arrestin will not bind, and the receptor will remain active on the cell surface for longer, leading to a prolonged and amplified signal.

Question 7

1. AC activity will be maximally stimulated because the Gαs pathway is dominant over the Gαi pathway.
2. AC activity will be near the basal, unstimulated level due to opposing regulatory inputs. (correct answer)
3. AC activity will be completely inhibited as Gαi binding is typically much stronger than Gαs binding.
4. AC activity will oscillate as the Gαs and Gαi subunits alternately bind and dissociate from the enzyme.

Explanation: Gαs activates adenylyl cyclase, while Gαi inhibits it. When both pathways are maximally stimulated, the stimulatory effect of Gαs-GTP and the inhibitory effect of Gαi-GTP on adenylyl cyclase will counteract each other. Assuming comparable expression levels and coupling efficiencies, the net result will be an activity level close to the basal state.

Question 8

1. A rapid increase in cytosolic Ca²⁺ levels, followed by activation of Protein Kinase C (PKC).
2. Activation of phospholipase C (PLC) via a positive feedback mechanism involving PKC.
3. Activation of Protein Kinase C (PKC) without a significant release of Ca²⁺ from the ER. (correct answer)
4. Generation of inositol trisphosphate (IP₃) due to the allosteric activation of PLC.

Explanation: In the Gαq pathway, PLC generates both DAG and IP₃. DAG recruits PKC to the membrane, and Ca²⁺ (released by IP₃) is required for its full activation. Phorbol esters mimic DAG, so they will recruit and partially activate PKC. However, because they do not cause the generation of IP₃, there will be no release of Ca²⁺ from the endoplasmic reticulum (ER). Therefore, PKC is activated without the typical IP₃-mediated calcium signal.

Question 9

1. The Gαs subunit will be unable to bind to the activated receptor, preventing initiation of the signal cascade.
2. The Gβγ subunit will remain permanently associated with the Gαs subunit, blocking effector enzyme activation.
3. Adenylyl cyclase will be persistently activated because the Gαs subunit is locked in its active conformation. (correct answer)
4. The signal will terminate normally via receptor desensitization, which is independent of GTP hydrolysis.

Explanation: GTP binding to Gαs causes its activation and dissociation from Gβγ. The intrinsic GTPase activity of Gαs hydrolyzes GTP to GDP, which terminates the signal. Because GTPγS cannot be hydrolyzed, the Gαs subunit remains in the active, GTP-bound state, leading to persistent activation of its effector, adenylyl cyclase. Receptor desensitization is a separate, slower mechanism and cannot compensate for the permanently active G-protein.

Question 10

1. The dissociation of the Gαq subunit from the Gβγ dimer upon receptor activation.
2. The hydrolysis of phosphatidylinositol 4,5-bisphosphate (PIP₂). (correct answer)
3. The intrinsic GTPase activity of the Gαq subunit that terminates the signal.
4. The release of Ca²⁺ from intracellular stores, which is mediated by protein kinase C.

Explanation: The function of activated Gαq-GTP is to bind to and activate phospholipase C (PLC). PLC's enzymatic activity is the hydrolysis of PIP₂ into the second messengers IP₃ and DAG. If Gαq cannot bind to PLC, PLC will not be activated, and this hydrolysis step will be directly blocked. Ca²⁺ release is a downstream effect of IP₃, and PKC is downstream of DAG, so these are also impaired but less directly.

Question 11

1. PKA activity would decline more slowly because the degradation of cAMP to AMP would be hindered. (correct answer)
2. PKA activity would decline more rapidly because the inhibitor triggers a feedback loop that deactivates PKA.
3. PKA activity would remain permanently elevated as cAMP can no longer be broken down.
4. PKA activity would be unaffected, as its deactivation depends on phosphatases, not cAMP levels.

Explanation: PKA is activated by cAMP. Signal termination requires the removal of cAMP, which is accomplished by phosphodiesterases (PDEs) hydrolyzing cAMP to AMP. A PDE inhibitor blocks this degradation pathway. Therefore, after the stimulus is removed and adenylyl cyclase activity returns to basal levels, cAMP levels will decrease much more slowly, prolonging the activation state of PKA.

Question 12

1. The pathway will be constitutively active at the level of PKA, regardless of cAMP concentration. (correct answer)
2. The pathway will be completely inactive because PKA cannot be formed properly.
3. Adenylyl cyclase will be inhibited through a potent negative feedback mechanism.
4. cAMP levels will be extremely high due to a compensatory increase in synthesis.

Explanation: Normally, the regulatory subunits of PKA bind to and inhibit the catalytic subunits in the absence of cAMP. cAMP binding causes the regulatory subunits to dissociate, releasing the active catalytic subunits. If the regulatory subunit cannot bind to the catalytic subunit at all, the catalytic subunit will be constitutively active, constantly phosphorylating its targets, independent of the upstream signal (ligand, G-protein, or cAMP levels).

Question 13

1. Phosphorylation of the specific cellular protein substrates of Protein Kinase C.
2. The recruitment of Protein Kinase C from the cytosol to the plasma membrane.
3. Activation of gene transcription for genes whose expression is dependent on PKC activity.
4. Generation of diacylglycerol (DAG) at the plasma membrane by phospholipase C. (correct answer)

Explanation: The signaling cascade proceeds as follows: GPCR → Gαq → PLC → PIP₂ cleavage into DAG and IP₃. DAG and Ca²⁺ (released by IP₃) cooperate to activate PKC. The inhibitor blocks PKC's kinase activity, which is the final step. All upstream events, including the generation of DAG by PLC, will still occur normally. PKC recruitment to the membrane (B) also occurs upon DAG generation, but the question asks about a process, and DAG generation is the most fundamental upstream process that will occur.

Question 14

1. The response will be abolished because the ligand can no longer bind to the altered receptor.
2. The response will switch from cAMP production to the generation of IP₃ and DAG. (correct answer)
3. The signal duration will be significantly shortened due to faster GTP hydrolysis by Gαq.
4. The cellular response will be unchanged as the Gβγ subunit is identical in both pathways.

Explanation: The intracellular loops of a GPCR determine its coupling specificity to different G-proteins. By changing the loop, the receptor has been rewired. Instead of activating Gαs, which stimulates adenylyl cyclase to produce cAMP, it now activates Gαq. Gαq stimulates phospholipase C, which cleaves PIP₂ to produce IP₃ and DAG. Thus, the primary second messenger output is completely changed.

Question 15

1. Both glycogen synthesis and glycogenolysis will be maximally activated, creating a futile cycle.
2. Glycogen synthase will be dephosphorylated and active, while glycogen phosphorylase will be inactive.
3. Glycogen synthesis will be inhibited, and glycogenolysis will be stimulated. (correct answer)
4. The metabolic state will be unchanged because insulin signaling can override the effects of the mutation.

Explanation: Constitutively active Gαs leads to chronically high levels of cAMP and active PKA. PKA has two key effects on glycogen metabolism: 1) It phosphorylates and inactivates glycogen synthase, halting glycogen synthesis. 2) It phosphorylates and activates phosphorylase kinase, which in turn phosphorylates and activates glycogen phosphorylase, stimulating glycogenolysis. The net effect is a strong shift away from storage and towards glucose release.

Question 16

1. PLC-β preferentially produces DAG over IP$_3$ due to substrate availability limitations, resulting in sustained DAG production with minimal IP$_3$ formation
2. IP$_3$ is rapidly metabolized by IP$_3$ 3-kinase and IP$_3$ 5-phosphatase, while DAG is more slowly metabolized by DAG lipase and DAG kinase (correct answer)
3. The GPCR undergoes rapid desensitization that selectively affects IP$_3$ production while maintaining DAG synthesis through alternative phospholipase activation
4. IP$_3$ and DAG are produced in equimolar amounts initially, but IP$_3$ is sequestered by binding proteins while DAG remains free in the membrane

Explanation: When PLC-β cleaves PIP₂, it produces equimolar amounts of IP₃ and DAG. The differential time courses reflect their different metabolic fates. IP₃ is rapidly metabolized by IP₃ 3-kinase and 5-phosphatase, leading to its quick clearance. DAG persists longer because it's metabolized more slowly by DAG lipase and DAG kinase. Choice A is incorrect because PLC-β produces equimolar amounts of both products. Choice C is incorrect because receptor desensitization would affect both products equally since they come from the same enzymatic reaction. Choice D is incorrect because there are no significant IP₃-binding proteins that would sequester it.

Question 17

1. PKC-mediated receptor phosphorylation causes desensitization, since PKC is activated by DAG produced in this pathway (correct answer)
2. Multiple GRK isoforms are present, and blocking only one isoform is insufficient to prevent receptor desensitization
3. PLC-β phosphorylation reduces enzyme activity, requiring kinases other than GRKs that are sensitive to staurosporine
4. Constitutive receptor internalization occurs independent of phosphorylation, blocked by staurosporine effects on endocytic machinery

Explanation: In Gα_q-coupled receptor signaling, PLC-β produces both IP₃ and DAG. DAG activates PKC, which can phosphorylate the same GPCR, leading to desensitization (heterologous desensitization). Since staurosporine blocks all kinases including PKC, it prevents this desensitization. The specific GRK inhibitor doesn't prevent desensitization because GRKs are not the primary kinases involved in this pathway's desensitization. Choice B is unlikely since specific GRK inhibitors typically affect multiple isoforms. Choice C is incorrect because PLC-β phosphorylation typically enhances rather than reduces its activity. Choice D is incorrect because the mechanism described involves kinase activity, not just internalization.

Question 18

1. The compound requires both direct binding to adenylyl cyclase and Gα$_s$ for full activation because adenylyl cyclase exists in different conformational states that respond optimally to combined stimulation (correct answer)
2. Gα$_s$ depletion indirectly affects the compound's binding affinity for adenylyl cyclase through allosteric mechanisms that reduce the enzyme's sensitivity to direct activators
3. The compound activates adenylyl cyclase independently of Gα$_s$, but Gα$_s$ depletion reduces total adenylyl cyclase protein levels through decreased transcription of the enzyme
4. The compound's effect depends on baseline adenylyl cyclase activity, which is partially maintained by constitutive low-level Gα$_s$ activation that is eliminated by siRNA treatment

Explanation: Adenylyl cyclase can be activated by direct binding of activators, but Gα_s binding induces conformational changes that enhance the enzyme's activity. The compound can activate the enzyme alone (hence activity remains when Gα_s is depleted), but maximal activation requires both the compound and Gα_s working together. This is similar to how some adenylyl cyclase isoforms show synergistic activation. Choice B is incorrect because the compound still works without Gα_s, indicating binding isn't affected. Choice C is incorrect because the effect is seen acutely, not through transcriptional changes. Choice D is incorrect because constitutive Gα_s activity would be minimal and wouldn't account for such a large difference.

Question 19

1. Enhanced contraction because thapsigargin increases cytosolic Ca$^{2+}$ levels, which synergizes with the agonist-induced Ca$^{2+}$ release to produce stronger muscle activation
2. Normal contraction because the agonist can still activate PLC-β and produce IP$_3$, and thapsigargin only affects the Ca$^{2+}$-ATPase without interfering with IP$_3$ receptor function
3. No effect on contraction because smooth muscle contraction depends primarily on extracellular Ca$^{2+}$ influx rather than intracellular Ca$^{2+}$ release mechanisms
4. Reduced contraction because thapsigargin depletes the ER Ca$^{2+}$ stores that would normally be released in response to IP$_3$ produced by agonist stimulation (correct answer)

Explanation: When you encounter questions about G-protein coupled receptor signaling and calcium homeostasis, focus on tracing the complete pathway from receptor activation to the final cellular response. The α$_1$-adrenergic pathway works through a clear sequence: receptor activation → Gα$_q$ → PLC-β → IP$_3$ production → IP$_3$ binds to ER receptors → Ca$^{2+}$ release from ER stores → muscle contraction. Thapsigargin disrupts this pathway by inhibiting the ER Ca$^{2+}$-ATPase pump, which normally maintains high calcium concentrations in the ER. When this pump is blocked, calcium leaks out and the ER stores become depleted. Therefore, when the α$_1$-agonist stimulates IP$_3$ production, there's little to no calcium available for release, resulting in reduced contraction. Option A incorrectly assumes thapsigargin enhances the response. While thapsigargin does initially release calcium as it depletes stores, this transient effect doesn't synergize with subsequent agonist treatment. Option B misses the critical point that even though IP$_3$ is still produced, depleted calcium stores mean IP$_3$ has nothing to release. Option C incorrectly dismisses the importance of intracellular calcium stores in smooth muscle contraction—while extracellular calcium influx does contribute, ER calcium release is a major component of smooth muscle activation. Remember this principle: disrupting any step in a signaling cascade affects the final response, even if upstream components remain functional. Always trace the complete pathway to predict experimental outcomes.

Question 20

1. Synergistically increase cAMP levels above those achieved by the full agonist alone because both ligands can bind simultaneously to different sites on the same receptor
2. Have no additional effect on cAMP levels because the receptor is already maximally occupied and activated by the full agonist at high concentrations
3. Decrease cAMP levels below those produced by the full agonist alone because the partial agonist competes for binding but produces submaximal receptor activation (correct answer)
4. Completely block cAMP production because partial agonists function as competitive antagonists in the presence of full agonists at saturating concentrations

Explanation: A partial agonist competes with the full agonist for the same binding site but produces lower maximal activation even when all receptors are occupied. In the presence of high concentrations of full agonist, the partial agonist will compete for binding and replace some of the full agonist molecules, but each receptor bound by partial agonist will be less active than when bound by full agonist. This results in a net decrease in cAMP levels. Choice A is incorrect because they compete for the same site. Choice B is incorrect because the partial agonist can still bind and replace full agonist. Choice D is incorrect because partial agonists retain some intrinsic activity.