Biochemistry Quiz: Rtks And Mapk Signaling

What this quiz covers

This quiz focuses on Rtks And Mapk Signaling, 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. The signal duration will be significantly shortened because Ras rapidly hydrolyzes GTP to GDP without GAP assistance.
2. The total amount of phosphorylated ERK will be reduced due to an inability to efficiently activate Ras via its GEF.
3. The signal will be prolonged because the intrinsic GTPase activity of Ras is slow, keeping it in the active GTP-bound state for longer. (correct answer)
4. The initial activation of Raf by Ras will be prevented because GAP is a required cofactor for the Ras-Raf interaction.

Explanation: The correct answer is C. Ras has a slow intrinsic GTPase activity that hydrolyzes GTP to GDP, turning itself off. GTPase-Activating Proteins (GAPs) dramatically accelerate this process, acting as the primary 'off switch'. A loss-of-function mutation in Ras-GAP means this 'off switch' is broken. Therefore, upon activation by a GEF, Ras will remain in its active, GTP-bound state for a much longer period, leading to a prolonged downstream signal. Choice A is incorrect because GAPs accelerate, not inhibit, GTP hydrolysis. Choice B is incorrect because Ras activation by GEFs is unaffected; the inactivation step is what is impaired. Choice D is incorrect because GAP's role is in deactivation, not in mediating the interaction with the downstream effector Raf.

Question 2

1. Both Raf and ERK will be in an unphosphorylated and inactive state due to pathway blockage.
2. Raf will be phosphorylated and active, but ERK will be in an unphosphorylated and inactive state. (correct answer)
3. Both MEK and ERK will be phosphorylated and active, as the inhibitor is likely ineffective.
4. ERK will be phosphorylated and active, but Raf will be in an unphosphorylated and inactive state.

Explanation: The correct answer is B. The MAPK cascade proceeds in the order Ras → Raf → MEK → ERK, where each component activates the next via phosphorylation (except for Ras activating Raf). An inhibitor of MEK's kinase activity will block the step where MEK phosphorylates ERK. Therefore, the signal will propagate normally up to MEK. Ras will activate Raf, and Raf will phosphorylate and activate MEK. However, since MEK's enzymatic activity is inhibited, it cannot phosphorylate ERK. As a result, Raf will be active (and phosphorylated), but ERK will remain inactive (and unphosphorylated). Choice A is incorrect because Raf is upstream of the block and will be activated. Choice C is incorrect because the inhibitor prevents MEK from phosphorylating ERK. Choice D incorrectly inverts the pathway's order.

Question 3

1. The therapy will be ineffective because the oncogenic signal originates from a point downstream of the drug's target. (correct answer)
2. The therapy will be very effective because inhibiting the RTK prevents Ras activation, which is required for B-Raf function.
3. The therapy will be effective because the RTK inhibitor will also allosterically inhibit the mutant B-Raf protein.
4. The therapy's effectiveness depends on cellular levels of MEK, which can be activated by alternative pathways.

Explanation: When analyzing cancer drug therapies, you need to map out the signal transduction pathway and identify where the mutation occurs relative to where the drug acts. This determines whether blocking upstream signals can affect downstream oncogenic activity. The MAPK pathway flows: RTK → Ras → B-Raf → MEK → ERK. In this patient, the V600E mutation makes B-Raf constitutively active, meaning it no longer requires upstream activation signals to function. The drug targets the RTK at the very beginning of this cascade. Since the oncogenic signal originates from the mutated B-Raf protein itself, blocking the RTK upstream cannot shut down the aberrant signaling. The mutant B-Raf will continue phosphorylating MEK regardless of RTK status, making the therapy ineffective. This makes choice A correct. Choice B incorrectly assumes that B-Raf still requires Ras activation. While normal B-Raf does need Ras, the V600E mutation bypasses this requirement entirely. Choice C suggests allosteric inhibition between the RTK inhibitor and B-Raf, but these proteins don't interact directly—they're separated by multiple pathway steps. Choice D focuses on MEK levels and alternative pathways, but the primary issue isn't pathway redundancy; it's that the oncogenic signal originates downstream of the drug target. Remember this principle: when evaluating targeted cancer therapies, always trace where the oncogenic mutation sits in the pathway relative to the drug's target. Drugs cannot effectively treat cancers caused by mutations downstream of their target proteins.

Question 4

1. No signaling will occur because neither mutant protein alone is fully functional to propagate the signal.
2. Signaling will occur, but only through Mutant A, as Mutant B's inactive kinase domain prevents its participation.
3. Signaling will be constitutively active because the formation of heterodimers bypasses the need for ligand binding.
4. Signaling will occur because the kinase domain of Mutant A will phosphorylate the tyrosine residues on Mutant B. (correct answer)

Explanation: When analyzing receptor tyrosine kinase (RTK) signaling, remember that successful signal transduction requires both kinase activity to phosphorylate tyrosines AND available tyrosine residues to serve as phosphorylation targets. The key insight here is that these functions can occur in trans between different receptor molecules within a dimer. Upon ligand binding, Mutant A and Mutant B form heterodimers. Mutant A contributes a functional kinase domain, while Mutant B provides intact cytoplasmic tyrosine residues. The active kinase domain of Mutant A can phosphorylate the tyrosine residues on Mutant B in trans, creating phosphotyrosine docking sites for downstream signaling proteins like SH2 domain-containing adapters. This complementation allows the signaling pathway to proceed normally. Option A incorrectly assumes that each individual receptor must be fully functional, missing the concept of trans-phosphorylation between dimeric partners. Option B fails to recognize that Mutant B participates crucially by providing phosphorylation substrates, even though its kinase domain is inactive. Option C misunderstands the system entirely—ligand binding is still required for dimerization and kinase activation, and there's nothing constitutive about this arrangement. The correct answer is D because trans-phosphorylation between receptor subunits is a fundamental mechanism in RTK signaling that allows functional complementation. Study tip: For RTK questions, always consider whether the essential components (ligand binding, kinase activity, and phosphorylatable tyrosines) can be distributed across multiple receptor molecules in a complex. Trans-phosphorylation is a recurring theme in receptor biology.

Question 5

1. It will be constitutively active because the negative charges of the glutamate residues mimic phosphorylation. (correct answer)
2. It will be completely inactive because glutamate cannot be phosphorylated by the Raf kinase.
3. It will be active but will only be able to phosphorylate the threonine residue on ERK, altering its specificity.
4. It will act as a dominant negative, binding to Raf but preventing the phosphorylation of wild-type MEK.

Explanation: When you encounter questions about protein phosphorylation and kinase regulation, focus on how phosphate groups alter protein structure and activity through electrostatic interactions. Phosphorylation typically adds negative charges that can induce conformational changes leading to activation or deactivation. MEK requires phosphorylation on two serine residues by Raf for activation. The key insight here is understanding phosphomimetic mutations – amino acid substitutions that mimic the electrostatic effects of phosphorylation. Glutamate residues carry negative charges at physiological pH, similar to phosphate groups. When the two serine residues are replaced with glutamate, these negative charges simulate the phosphorylated state, likely keeping MEK in its active conformation without requiring Raf phosphorylation. This makes the mutant MEK constitutively active, supporting answer A. Answer B is incorrect because the question isn't about whether glutamate can be phosphorylated – it's about whether glutamate can functionally replace phosphoserine. Answer C misunderstands the mutation's effect; replacing serine residues in MEK's activation sites with glutamate wouldn't alter MEK's dual-specificity toward ERK's threonine and tyrosine residues. Answer D is wrong because this mutant MEK wouldn't bind Raf (it doesn't need Raf activation) or act as a dominant negative. Remember that phosphomimetic mutations using acidic amino acids (glutamate, aspartate) are common experimental tools to study constitutively active proteins. When you see serine/threonine→glutamate substitutions in kinase regulation questions, think "mimicking phosphorylation" rather than "preventing phosphorylation."

Question 6

1. The mutated RTK cannot bind ATP, which prevents autophosphorylation and subsequent downstream signaling events.
2. The absence of phosphotyrosine residues prevents the recruitment of adaptor proteins containing SH2 domains, such as Grb2. (correct answer)
3. The phenylalanine residues mimic a phosphorylated state, which leads to constitutive inhibition of the kinase domain itself.
4. The RTK cannot be dephosphorylated by protein phosphatases, which traps the receptor in an inactive conformation.

Explanation: The correct answer is B. RTK activation requires autophosphorylation on specific tyrosine residues in the cytoplasmic tail. These phosphotyrosine sites act as docking points for proteins with SH2 (Src Homology 2) or PTB domains, such as the adaptor protein Grb2. Mutating these tyrosines to phenylalanine, which cannot be phosphorylated, eliminates these docking sites. Even though the kinase is active and can dimerize, it cannot recruit the necessary downstream components to propagate the signal. Choice A is incorrect because the stem states the kinase domain is functional, implying it can bind ATP. Choice C is incorrect as phenylalanine does not mimic phosphotyrosine; acidic residues like aspartate or glutamate are used for that purpose. Choice D is incorrect because the problem is a failure to initiate the signal via phosphorylation, not a failure to terminate it via dephosphorylation.

Question 7

1. The Receptor Tyrosine Kinase (RTK), as it cannot recruit Grb2 when thiophosphorylated.
2. Ras, because it is directly thiophosphorylated by the RTK in this pathway.
3. MEK (MAPKK), as it requires ATP hydrolysis to phosphorylate its substrate, ERK. (correct answer)
4. ERK (MAPK), as it would be the final kinase in the cascade to be thiophosphorylated.

Explanation: The correct answer is C. This question requires analyzing the flow of the kinase cascade. 1) The RTK will use γ-S-ATP to thiophosphorylate itself. 2) This will recruit Grb2/Sos and activate Ras (which uses GTP, not ATP). 3) Active Ras will activate Raf. 4) Raf, a kinase, will use γ-S-ATP to thiophosphorylate its substrate, MEK. Now, MEK is thiophosphorylated. 5) For MEK to act as a kinase and phosphorylate ERK, it needs to bind and hydrolyze ATP. Since only the non-hydrolyzable analog is available, MEK cannot transfer a phosphate to ERK. Therefore, MEK becomes thiophosphorylated but cannot activate its target. Choice A is incorrect because thiophosphorylated RTKs can still recruit SH2-domain proteins. Choice B is incorrect because Ras is a G-protein activated by GTP binding, not phosphorylation. Choice D is incorrect because ERK will not become thiophosphorylated at all, as its activator (MEK) is stalled.

Question 8

1. The mutation locks Ras in a GDP-bound state, which has been shown to constitutively activate the Raf kinase.
2. The G12V mutation enhances the binding affinity of Ras for its GEF, Sos, leading to a massive increase in GTP loading.
3. The mutation prevents Ras from interacting with the cell membrane, causing it to activate aberrant cytosolic pathways.
4. By being trapped in the GTP-bound state, Ras continuously activates its downstream effectors, regardless of upstream signals. (correct answer)

Explanation: The correct answer is D. Ras functions as a molecular switch, being 'on' when bound to GTP and 'off' when bound to GDP. The G12V mutation specifically cripples the GTPase activity, which is the 'off' switch. By being unable to hydrolyze GTP to GDP, the protein becomes trapped in the active, GTP-bound conformation. This leads to persistent, unregulated stimulation of downstream effectors like the Raf kinase, driving cell proliferation even in the absence of growth factors. Choice A is incorrect because the GDP-bound state is the inactive form. Choice B is incorrect because the primary defect is in GTP hydrolysis (the 'off' switch), not GTP loading (the 'on' switch). Choice C is incorrect because Ras must be localized to the membrane to function correctly.

Question 9

1. The initial rate of Ras activation will be significantly slower in the mutant cell line due to improper Sos folding.
2. The peak level of ERK phosphorylation will be lower in the mutant cell line due to rapid pathway desensitization.
3. The duration and overall intensity of MAPK pathway activation will be increased in the mutant cell line. (correct answer)
4. The mutant Sos will be unable to bind to the Grb2 adaptor protein, preventing any activation of the pathway.

Explanation: The correct answer is C. The described mechanism is a negative feedback loop: the pathway's output (active ERK) inhibits an early step (Sos activity). In a wild-type cell, this feedback helps to attenuate the signal and turn it off after an appropriate duration. In the mutant cell line, this feedback mechanism is broken because Sos cannot be inhibited by ERK. As a result, Sos will remain active for longer, leading to prolonged activation of Ras and the entire downstream MAPK cascade. This results in a signal that is stronger (higher integrated intensity) and more sustained. Choice A is incorrect because the initial activation would be normal. Choice B is incorrect because the lack of negative feedback would lead to a higher, not lower, signal. Choice D is incorrect as the stem specifies the mutant is otherwise fully functional.

Question 10

1. The RTK will fail to dimerize upon ligand binding, as phosphatase activity is required for this conformational change.
2. The autophosphorylation of the RTK and phosphorylation of downstream targets will be more robust and prolonged. (correct answer)
3. Ras will be unable to exchange GDP for GTP, as this process is directly catalyzed by a tyrosine phosphatase.
4. The kinase activity of Raf and MEK will be directly inhibited, as they are sensitive to high levels of vanadate.

Explanation: The correct answer is B. Protein kinases add phosphate groups to activate signaling components, while protein phosphatases remove them to terminate the signal. Protein tyrosine phosphatases (PTPs) specifically remove phosphates from tyrosine residues. By inhibiting PTPs, sodium orthovanadate removes the 'off switch' for tyrosine phosphorylation. This means that once the RTK is autophosphorylated upon ligand binding, it will remain phosphorylated (and thus active) for a longer period. This leads to a more sustained and stronger signal throughout the pathway. Choice A is incorrect; dimerization is ligand-induced. Choice C is incorrect; Ras activation is mediated by a GEF, not a phosphatase. Choice D is incorrect because vanadate inhibits phosphatases, which would indirectly lead to more kinase activity, not less.

Question 11

1. Normal MAPK activation occurs because Grb2 can still bind SOS and activate Ras through alternative membrane recruitment mechanisms that bypass RTK interaction requirements.
2. Delayed but eventually normal MAPK activation occurs because other adaptor proteins can compensate for Grb2 function by recruiting SOS to activated RTKs through redundant binding mechanisms.
3. Enhanced MAPK activation occurs because the mutant Grb2 cannot bind RTKs but retains constitutive SOS binding, leading to continuous Ras activation independent of growth factor signals.
4. Severely impaired MAPK activation occurs because Grb2 cannot be recruited to activated RTKs, preventing the formation of RTK-Grb2-SOS complexes necessary for Ras activation at the membrane. (correct answer)

Explanation: When approaching MAPK pathway questions, focus on the sequential steps required for signal transduction from the membrane to the nucleus. The RTK-Grb2-SOS complex formation is absolutely critical for proper Ras activation. The mutant Grb2 creates a fatal break in the signaling chain. While it can still bind SOS, it cannot attach to phosphorylated RTK tyrosine residues. This means that when growth factors bind to RTKs and cause receptor autophosphorylation, the Grb2-SOS complex cannot be recruited to the membrane where Ras is located. Since Ras must be activated at the membrane (where it's anchored), SOS needs to be brought there via the RTK-Grb2 bridge. Without this recruitment mechanism, SOS remains cytosolic and cannot exchange GDP for GTP on Ras, severely blocking MAPK activation. Option A is wrong because alternative membrane recruitment mechanisms for Grb2 don't exist when its RTK-binding domain is defective. Option B incorrectly assumes sufficient compensatory mechanisms exist - while other adaptors like Shc exist, they don't fully compensate for Grb2 loss. Option C misunderstands the mutation's effect; the Grb2-SOS interaction isn't constitutively active, and without membrane recruitment, this binding is actually detrimental since it sequesters SOS away from Ras. Remember that signal transduction pathways require proper subcellular localization. A functional domain mutation doesn't just reduce efficiency - it can completely break essential protein-protein interactions that enable spatial organization of signaling complexes.

Question 12

1. Protein synthesis inhibition prevents DUSP degradation by blocking proteasome component production, leading to enhanced phosphatase activity that paradoxically maintains ERK activation through complex feedback mechanisms.
2. The protein synthesis inhibitor blocks production of negative regulatory proteins including DUSPs, preventing the normal desensitization response and maintaining sustained ERK phosphorylation levels. (correct answer)
3. Inhibiting protein synthesis reduces cellular ATP levels, which decreases kinase activities upstream of ERK while simultaneously reducing phosphatase function, resulting in net maintenance of ERK phosphorylation.
4. The protein synthesis inhibitor enhances growth factor receptor stability by preventing synthesis of receptor-degrading enzymes, leading to sustained upstream signaling and continued ERK activation.

Explanation: The data shows that prolonged stimulation normally induces DUSP expression, which dephosphorylates and inactivates ERK (negative feedback). When protein synthesis is blocked, DUSPs cannot be produced, preventing this negative feedback and allowing ERK to remain active. This is a classic example of how cells use induced phosphatase expression to terminate prolonged MAPK signaling. Option A is incorrect because protein synthesis inhibition would block DUSP production, not degradation. Option C is wrong because the effect is specifically about blocking DUSP synthesis, not general metabolic effects. Option D doesn't explain the DUSP observation and focuses on receptor stability rather than pathway regulation.

Question 13

1. The phosphatase inhibitor prevents receptor dimerization, leading to sustained RTK activation and enhanced MAPK signaling through reduced negative feedback regulation.
2. The phosphatase inhibitor blocks receptor internalization and degradation, maintaining surface receptor levels and prolonging the duration of MAPK pathway activation.
3. The phosphatase inhibitor prevents dephosphorylation of activated signaling intermediates, sustaining phosphorylation cascades and extending MAPK pathway responses. (correct answer)
4. The phosphatase inhibitor enhances growth factor binding affinity to the receptor, increasing initial signal strength and amplifying downstream MAPK activation.

Explanation: Tyrosine phosphatases normally terminate RTK and MAPK signaling by dephosphorylating activated kinases and adaptor proteins in the pathway. Inhibiting these phosphatases would prevent the normal 'off' signal, leading to sustained phosphorylation of pathway components and prolonged MAPK activation. Option A is incorrect because phosphatases don't prevent dimerization - they act on phosphorylated residues. Option B is wrong because phosphatases don't directly control receptor internalization. Option D is incorrect because phosphatases don't affect ligand-receptor binding affinity.

Question 14

1. Increased Raf expression overcomes MEK inhibition by activating alternative downstream pathways that bypass the blocked MEK-ERK signaling cascade entirely.
2. Elevated Raf levels lead to stronger MEK activation signals that can outcompete the inhibitor for MEK binding sites, restoring pathway function through mass action effects. (correct answer)
3. Higher Raf expression activates parallel kinase cascades that converge on ERK through MEK-independent mechanisms, maintaining cellular proliferation signals despite MEK blockade.
4. Increased Raf expression enhances feedback inhibition loops that normally downregulate the pathway, creating compensatory activation of growth-promoting signals through receptor upregulation.

Explanation: In kinase inhibitor resistance, increased expression of upstream kinases can overcome competitive inhibition through mass action - higher Raf levels produce more activated Raf molecules that can phosphorylate and activate more MEK molecules, some of which may escape inhibitor binding and restore downstream ERK activation. Option A is incorrect because MEK specifically activates ERK, and there are no major alternative pathways that bypass this step. Option C is wrong because ERK is primarily activated by MEK - there are no significant MEK-independent pathways to ERK. Option D is incorrect because increased Raf wouldn't enhance feedback inhibition, and this mechanism doesn't explain resistance to MEK inhibition.

Question 15

1. RTK-A will dominate the response because its higher ligand affinity ensures greater receptor occupancy, leading to stronger MAPK activation despite weaker individual kinase activity.
2. RTK-B will dominate the response because its stronger kinase activity compensates for lower receptor occupancy, producing more robust MAPK pathway activation overall.
3. Both receptors will contribute equally to MAPK activation because the high affinity-weak kinase combination of RTK-A balances the low affinity-strong kinase combination of RTK-B.
4. The responses will be additive but RTK-A will contribute more because at 10 nM ligand concentration, RTK-A achieves higher fractional occupancy than RTK-B. (correct answer)

Explanation: At 10 nM ligand concentrations: RTK-A (Kd=1 nM) will have ~91% occupancy (10/(10+1)), while RTK-B (Kd=50 nM) will have ~17% occupancy (10/(10+50)). Even though RTK-B has stronger kinase activity per receptor, the much higher occupancy of RTK-A means more total activated receptors, likely resulting in RTK-A contributing more to the overall response. The responses would be additive since both activate the same pathway. Option A is incorrect because it ignores the kinase activity differences. Option B underestimates the importance of receptor occupancy. Option C is wrong because the combinations don't perfectly balance at these concentrations.

Question 16

1. Initial strong ERK activation followed by rapid decline as PKA activation increases, then sustained low-level ERK activity reflecting the inhibited but not completely blocked MAPK pathway. (correct answer)
2. Delayed ERK activation as the growth factor signal must first overcome PKA-mediated Raf inhibition, followed by normal sustained ERK activity once pathway equilibrium is established.
3. Oscillating ERK activity as PKA and growth factor signals compete for control of Raf activity, creating cyclical activation and inhibition of the MAPK pathway.
4. Enhanced ERK activation compared to growth factor alone because PKA-mediated Raf phosphorylation creates a priming event that sensitizes the MAPK pathway to subsequent stimulation.

Explanation: Growth factor stimulation would initially activate RTK-Raf-MEK-ERK signaling. As forskolin simultaneously activates adenylyl cyclase, rising cAMP levels activate PKA, which then phosphorylates and inhibits Raf. This creates a pattern of initial ERK activation that declines as PKA activity increases, but some ERK activity would remain because PKA inhibition is typically partial, not complete. Option B is incorrect because both signals start simultaneously. Option C is wrong because there's no mechanism described for oscillations. Option D is incorrect because PKA phosphorylation inhibits rather than primes Raf activity.

Question 17

1. The mutation would cause normal initial MAPK activation but dramatically prolonged ERK phosphorylation because mutant Ras remains in the active GTP-bound state much longer than normal. (correct answer)
2. The mutation would prevent MAPK pathway activation because reduced GTPase activity impairs Ras's ability to cycle between GDP and GTP-bound states necessary for signal transduction.
3. The mutation would result in enhanced initial MAPK activation because reduced GTPase activity allows more efficient GTP binding and stronger interaction with downstream effector proteins.
4. The mutation would have minimal effects on MAPK signaling because GAPs (GTPase-activating proteins) can compensate for reduced intrinsic GTPase activity by providing alternative Ras inactivation mechanisms.

Explanation: When you encounter questions about Ras mutations in cancer, focus on understanding the normal Ras cycle and how disruptions affect signal duration rather than signal initiation. Ras proteins function as molecular switches that cycle between active (GTP-bound) and inactive (GDP-bound) states. Upon growth factor stimulation, Ras exchanges GDP for GTP, becoming active and recruiting downstream effectors like RAF to initiate the MAPK cascade. The signal is terminated when Ras hydrolyzes GTP back to GDP through its intrinsic GTPase activity. This mutation reduces GTPase activity by 90% while preserving GTP binding and effector interactions. This means Ras can still become activated normally when growth factors stimulate GTP binding, and it can still interact with downstream proteins like RAF. However, once activated, the mutant Ras remains "stuck" in the GTP-bound state much longer because it cannot efficiently hydrolyze GTP to GDP. This leads to normal initial MAPK activation followed by dramatically prolonged ERK phosphorylation, making answer A correct. Answer B is wrong because the mutation doesn't prevent the GDP→GTP exchange needed for activation. Answer C is incorrect because the mutation doesn't enhance GTP binding or effector interactions—these remain normal. Answer D is flawed because while GAPs do assist Ras inactivation, they work by enhancing the intrinsic GTPase activity, which is severely compromised in this mutant. Study tip: Remember that oncogenic Ras mutations typically cause prolonged signaling rather than preventing or enhancing signal initiation—it's about getting stuck "on," not turning on stronger.

Question 18

1. Normal MAPK activation occurs because receptor dimerization alone is sufficient to recruit and activate downstream signaling proteins through conformational changes.
2. Partial MAPK activation occurs because the truncated receptor can still recruit adaptor proteins through its remaining cytoplasmic tail, bypassing kinase activity requirements.
3. No MAPK activation occurs because autophosphorylation of receptor tyrosine residues is essential for creating docking sites for downstream signaling proteins. (correct answer)
4. Enhanced MAPK activation occurs because the absence of kinase domain prevents negative feedback phosphorylation, leading to sustained pathway stimulation.

Explanation: RTK signaling requires autophosphorylation of specific tyrosine residues in the receptor's cytoplasmic tail to create docking sites for adaptor proteins like Grb2 and SOS. Without the kinase domain, autophosphorylation cannot occur, preventing recruitment of downstream signaling components and blocking MAPK pathway activation. Option A is wrong because dimerization alone is insufficient - phosphorylation is required. Option B is incorrect because adaptor protein binding specifically requires phosphotyrosine residues. Option D is wrong because without kinase activity, there would be no initial signal to sustain.