Complement: surfaces, amplification and protection
Follow three complement starts through surface amplification, host protection, immune clearance and clinical pathway tests, with accessible mechanism diagrams and cases.
Complement is not an on/off switch for bacteria: it is a surface-sensitive enzyme network. A C3 tag may land on a human cell or a microbe, but what happens next depends on recognition and local protection. This full-free lesson follows that decision from first contact to clinical interpretation. [1][3]
Recognize the target and the starting signal
Classical initiation begins when C1q recognizes suitable IgM or clustered IgG bound to an antigen; free antibody is not equivalent to antibody arranged on a target. Lectin initiation uses mannose-binding lectin or related recognition molecules and associated proteases to recognize microbial carbohydrate patterns without antibody. Both cleave C4 and C2 and produce the same C3 convertase. The alternative pathway supplies a background surveillance route: spontaneous hydrolysis of intact C3 produces fluid-phase C3(H2O), which can associate with factor B and factor D. The hydrolyzed molecule is not identical to surface-attached C3b. Initial C3 cleavage can deposit C3b onto a nearby surface; previously generated C3b from any start can also feed alternative amplification. [1][3][7]
A target bearing C3b does not automatically receive the same response everywhere. Whether regulators recognize surrounding host glycans determines if the initial deposit persists. Pathogens can sometimes mimic host signatures, so never define self versus microbe by one universal sugar alone. [3]
Predict: bound IgM recruits C1; which two starts will use C4?
Classical and lectin initiation both consume C4; the alternative route does not require C4.
Start three routes, converge on C3
Compare two C4-dependent triggers with fluid tickover: C3(H2O)Bb cleaves C3, creating the first deposited C3b that can then form surface C3bBb. [1][9]
C1s in the classical route and lectin-associated proteases in the lectin route cleave C4 and C2, assembling C4b2a on a surface. A 2019 nomenclature paper proposes C4b2b for the same enzyme because it renames the larger catalytic C2 fragment; conventional C4b2a remains widely used. [9] Fluid C3(H2O)Bb produced during tickover precedes surface deposition; a deposited C3b seed instead recruits B and D to make surface C3bBb. [7] Each C3 convertase cleaves C3 into soluble C3a and surface-reactive C3b. Thus shared downstream products do not identify which initiating route fired. [1]
Adding a further C3b to classical or lectin C4b2a produces C4b2a3b; adding C3b to alternative C3bBb yields C3bBbC3b. These are C5-convertase assemblies, which cleave C5 into C5a and C5b. The notation matters less than the extra surface C3b that changes substrate selection. [1]
Predict: if C1 is unavailable but lectin recognition works, can C4 still be cleaved?
Yes. Lectin-associated proteases can cleave C4 without C1.
Amplify a seeded surface without confusing tickover
Follow the feedback: an existing tag recruits B and D to generate more tags. [3]
A surface-bound C3b binds factor B. Factor D cleaves bound B, releasing Ba and leaving Bb in the C3bBb enzyme. Properdin stabilizes this short-lived alternative convertase; it is not the protease that cuts B. C3bBb generates fresh C3b, some of which attaches nearby and recruits further B. This positive feedback also amplifies C3b seeded by classical or lectin initiation. A defect in B or D interrupts this amplification, whereas loss of properdin weakens convertase stability and increases susceptibility to meningococcal infection. [1][3][7]
Factor I cleavage converts C3b to iC3b when it has a cofactor. [7] iC3b cannot bind factor B to regenerate the alternative convertase, yet remains an opsonin for CR3 on phagocytes. Further processing produces C3d, which binds CR2/CD21 on B cells and can strengthen antigen-dependent B-cell responses when the fragment travels with the antigen. A mouse immunization experiment with antigen linked to C3d supports this principle, not a universal magnitude for human B cells. [14] CR1/CD35 recognizes C3b and C4b on phagocytes and erythrocytes and helps carry immune complexes toward removal; it also aids factor I cofactor activity. These receptor functions explain why a tag that no longer amplifies can still shape clearance and immune responses. A 1988 erythrocyte experiment measured CR1-dependent immune-complex binding and transfer between erythrocytes, not macrophage delivery or renal outcomes. Broader clearance physiology includes transfer toward phagocytic cells in liver and spleen; a transport deficit may contribute to tissue deposition without the 1988 experiment proving that clinical outcome. [7][8][15] C3 itself can have other nonproteolytic roles; do not claim that every C3 function requires C3b as an intermediate. [1][3]
Predict: after factor I cuts C3b to iC3b, does phagocyte recognition disappear?
No. iC3b remains recognizable by CR3 despite losing convertase-forming activity.
Protect host surfaces locally
Compare a host recognition and cleavage branch against possible amplification when surface regulation is poor. [3]
Factor H is a soluble decay accelerator and factor-I cofactor, whereas factor I is the C3b-cleaving protease. Factor H recognizes both deposited C3b and features enriched on many host surfaces, including sialic acid and glycosaminoglycans. This two-part recognition brings its regulatory domains to a vulnerable surface: factor H helps displace Bb from alternative convertase and acts as a cofactor for factor I. Factor I performs the actual proteolytic cleavage of C3b. Loss of factor H recognition can therefore harm kidney endothelium even when circulating factor H antigen is measurable. The experimental structural study supports combined C3b/C3d-region and host-glycan contacts, not an absolute rule that microbes lack sialic acid. [3]
Membrane cofactor protein CD46 supplies a local factor I cofactor on nucleated cells. CD55, also called decay-accelerating factor, dissociates C3 and C5 convertases; it does not cut C3b. CD59 prevents C9 polymerization in the terminal pore. This division of work distinguishes protecting the upstream amplification step from guarding the final membrane. A deficient regulator permits attack on host tissue; deficiency of an effector instead reduces antimicrobial function. [1][3]
Predict: if factor H still binds C3b but cannot recognize host glycans, where does regulation become unreliable?
Host surfaces lose efficient local factor H recruitment; factor I cannot substitute for missing local cofactor docking.
Separate tagging, inflammation and pores
C3b and iC3b decorate targets for ingestion by different complement receptors. C3a and especially C5a signal inflammation; C5a powerfully attracts and activates neutrophils. C5b instead recruits C6, C7, C8 and C9 to form C5b-9, a pore in a lipid membrane, not a hole drilled into a bacterial cell wall. Gram-negative meningococci are especially vulnerable when this terminal defense is missing. A patient with recurrent meningococcal disease may have a terminal defect, an alternative pathway defect, or pharmacologic blockade; identify the actual branch with functional tests rather than infection history alone. [1][2]
Early classical proteins also help handle immune complexes and dying cells. C3b/C4b deposition and erythrocyte CR1 transport complexes toward phagocytic clearance. When this clearance fails, immune complexes can deposit in tissues and give a lupus-like phenotype. Erythrocyte-to-erythrocyte CR1 transfer was measured directly, whereas delivery to phagocytes and renal effects require additional physiological and clinical evidence. [8][15] This is distinct from the immune-complex consumption that occurs during active lupus: similar low complement results can arise by different mechanisms. [1]
Predict: if C5 cleavage is blocked, which inflammatory output is directly lost along with the pore precursor?
C5a generation is lost along with C5b generation.
Use pathway logic in patients
CH50 asks whether the classical-through-terminal route functions; AH50 probes alternative-through-terminal function. Absent CH50 with preserved AH50 points toward an early classical component; absent AH50 with preserved CH50 suggests an alternative component or regulator. Both low can reflect C3 or terminal failure, consumption, an inhibitor drug, or poor specimen handling. Component and functional measurements, collection conditions, clinical history and repeat testing resolve the branch. A normal C3 concentration does not demonstrate a healthy regulator on an endothelial surface. C9 defects can preserve measurable CH50 despite clinically important loss of C9: a primary family report measured only slightly reduced CH50 with complete C9 deficiency. A borderline or retained screen requires component-specific confirmation when suspicion persists. [10] For functional testing, follow the receiving laboratory's handling rules: ARUP specifies timely serum separation and frozen transport, with distinct details for CH50 and AH50. An overnight unseparated sample is not a reliable basis for inherited deficiency. [11][12][1]
In a person receiving C5 inhibition, both functional pathways can read suppressed. CDC recommends MenACWY and MenB for complement-inhibitor recipients, yet vaccination leaves substantial residual meningococcal risk; clinicians may consider antibiotic prophylaxis, and symptoms require immediate assessment even after vaccination or prophylaxis. This is specialist prevention, not a guarantee. Factor H-associated thrombotic microangiopathy, renal C3 deposition and acquired GPI-anchor loss are separate clinical stories discussed in the companion deficiency lesson. [1][2][5]
Do not conflate kidney syndromes: a thrombotic microangiopathy produces fragmented red cells, thrombocytopenia and organ injury, prompting prompt collection of indicated ADAMTS13 plasma and STEC stool specimens while assessing other causes. Do not delay urgent empiric treatment for suspected TTP while awaiting ADAMTS13 results. C3 glomerulopathy instead involves dominant glomerular C3 deposits; C3 nephritic factors are heterogeneous acquired antibodies: some prolong C3bBb activity or oppose decay, but their effects on C5 conversion differ. Purified-IgG transfer can establish a regulatory effect in an assay, not by itself that an antibody caused a patient's renal disease. [13] Neither diagnosis follows from one low C3 result. A normal C3 also cannot exclude surface-selective complement-mediated microangiopathy. [5][6]
C1 inhibitor has a second job beyond restraining classical C1: it restrains contact-system kallikrein, limiting bradykinin generation. Deep recurrent swelling without typical itchy wheals can point toward bradykinin, but clinical pattern alone is insufficient. Compare C1 inhibitor antigen, functional activity and C4; C4 alone cannot rule it out. Airway symptoms require urgent assessment and treatment. [4]
Predict: both CH50 and AH50 fall after C5 blockade. Does that establish inherited C3 deficiency?
No. Both assays depend on terminal function, which a C5 inhibitor suppresses.
Practice with clinical evidence
Case 1
Show answer and explanations for case 1
A. A shared C3 defect (Why this does not fit)
C3 loss could impair opsonization and clearance. But a shared C3 failure would compromise AH50 as well as CH50. Preserved AH50 keeps isolated early classical failure ahead of shared-pathway loss.
B. An early classical defect (Best answer)
Preserved AH50 localizes the functional gap before the shared cascade, within classical initiation. Early classical activity helps tag complexes for removal, explaining the glomerular deposits. CH50 alone cannot identify C1q, C4 or C2 specifically; test components next.
C. An alternative factor D defect (Why this does not fit)
Factor D loss can predispose to bacterial infection. Its alternative-convertase role predicts a low AH50 rather than an isolated CH50 loss. Use the unaffected pathway to localize the missing initiation branch.
D. A shared terminal defect (Why this does not fit)
Terminal loss can produce invasive bacterial infections. Both functional screens require terminal assembly, whereas AH50 is intact here. A shared terminal defect cannot explain a selective classical assay failure.
Takeaway: CH50 alone cannot identify C1q, C4 or C2 specifically; test components next.
C1s can initiate antibody-dependent C4 cleavage. The preserved CH50 and missing Bb on C3bB place the gap in the alternative branch. C1s does not perform the B cleavage in this assay.
B. Properdin stabilization of C3bBb (Why this does not fit)
Properdin prolongs the life of an already formed alternative convertase. Here bound B has not become Bb before the rescue. A failure to form Bb precedes a possible stabilization defect.
C. Factor D cleavage of C3b-bound B (Best answer)
Bound B without Bb places the block at alternative proconvertase processing. Adding D alone restores Bb while the classical pathway remains functional. Factor D cuts C3b-bound B to build C3bBb.
D. Factor I processing of C3b (Why this does not fit)
Factor I inactivates C3b with a cofactor and can curb amplification. The missing product is Bb, which reappears on D addition rather than I addition. Substrate processing into Bb is not C3b inactivation.
Takeaway: Factor D cuts C3b-bound B to build C3bBb.
A. X fails a later terminal assembly step; Y fails C5 activation (Why this does not fit)
It can preserve C5a production despite absent pores. Y, not X, generates C5a and fails the supplied C5b-6 bypass.
B. Both samples fail C5 activation (Why this does not fit)
X lacks C5a and responds to a downstream terminal complex. Y already produces active C5a and does not respond to the same bypass.
C. Both samples fail a later terminal assembly step (Why this does not fit)
It fits preserved C5a and absent response to the C5b-6 bypass. X recovers permeability when C5b-6 is supplied despite absent endogenous C5a.
D. X fails C5 activation; Y fails a later terminal assembly step (Best answer)
Absent C5a with rescue by supplied C5b-6 places its demonstrated failure at C5 activation or formation of that complex. Preserved C5a with failure of the C5b-6 bypass places its demonstrated failure later in terminal assembly. They make a general resistance of the bacterial target an inadequate explanation.
Takeaway: Use an upstream product and a downstream bypass together to localize a reaction failure.
A. A C2-dependent early increment is preserved; B-dependent later amplification is deficient (Best answer)
The antibody-associated increment persists without factor B but requires C2-dependent activity. Later accumulation changes specifically with factor B supplementation and blockade. It identifies the measured antibody-associated increment without claiming that all background C3b has that origin.
B. C2-dependent initiation is absent; B-dependent amplification is preserved (Why this does not fit)
It would remove the measured early antibody-associated C3b increment. The patient sample needs factor B supplementation to restore the later gain.
C. Only C1 recognition is deficient; stronger factor B activity repairs recognition (Why this does not fit)
It acts upstream of the C4/C2 convertase. Antibody-associated C2-dependent C3 cleavage already occurs before factor B restores later gain.
D. Only terminal pore assembly is deficient; factor B directly replaces the missing pore protein (Why this does not fit)
It changes membrane attack downstream of C5 activation. Factor B restores C3b accumulation rather than supplying a terminal pore component.
Takeaway: Measure the initiating increment separately from subsequent amplification.
A. C4b deposition falls on both bead types (Why this does not fit)
Both starts ultimately use cleavage of C4. The reagent leaves supplied C4 cleavage intact and targets the Fc-binding site of P. Q provides the independently demonstrated mannose recognition route.
B. C4b deposition falls on dense-IgG beads but persists on mannose beads (Best answer)
Productive initiation on the dense antibody surface requires P recruitment. It prevents the dense antibody surface from recruiting the working proteolytic system. The independently binding Q route and the shared cleavage reaction remain available.
C. C4b deposition persists on dense-IgG beads but falls on mannose beads (Why this does not fit)
Its ability to cleave supplied substrate is preserved. The measured P recruitment needed to position that system is blocked. Q, not the Fc-binding protein P, binds the mannose surface.
D. C4b deposition persists unchanged on both bead types (Why this does not fit)
It uses the distinct Q binding interaction. Loss of P previously abolished that response and P add-back restored it.
Takeaway: Infer the affected recognition route, then use the independent trigger as a control.
A. The sugar-dependent increments persist; assembly on the independently pretagged surface is lost (Why this does not fit)
The reagent removes the recognition contact required to initiate it. C3b and the alternative assembly factors are added independently.
B. Both the sugar-dependent increments and independently seeded assembly are lost (Why this does not fit)
An inhibitor acting directly on C3 conversion could affect more than one initiating route. Its measured action is recognition docking and the downstream convertase control remains functional.
C. The sugar-dependent increments decline; convertase assembly on the independently pretagged surface remains available (Best answer)
It identifies a recognition route distinct from the antibody-triggered classical start. It removes docking needed for the sugar-dependent initiation increment. That chamber receives C3b and factors B and D independently of sugar docking.
D. Both responses persist because classical initiation is intact (Why this does not fit)
The classical route can still generate its early C4b response. The sugar target depends on a different recognition complex whose docking is blocked.
Takeaway: Compare trigger-dependent increments with an independently supplied downstream substrate.
A. Surface C3bBb amplification on particles (Why this does not fit)
Surface C3bBb could magnify later particle deposition. The B-associated C3-cleaving signal precedes particles and surface attachment. Compartment and timing exclude particle-bound amplification as its first source.
B. Factor-I production of fluid iC3b (Why this does not fit)
Factor I can process C3b to iC3b with factor H. Removing H increases rather than produces a cleaving enzyme before particles arrive. iC3b cannot assemble a B-dependent C3 convertase.
C. C5-convertase generation on the vessel (Why this does not fit)
A C5 convertase can form after a surface C3 convertase gains C3b. The measured early enzyme cleaves C3 in cell-free fluid. C5 substrate specificity and surface dependence make this a later event.
D. Fluid C3(H2O)Bb tickover before deposition (Best answer)
C3 hydrolysis can expose a B-binding site in fluid before any surface receives C3b. Factor H removal raises that early soluble activity, and particles acquire C3b only later. C3(H2O)Bb tickover and deposited C3bBb are distinct compartments.
Takeaway: C3(H2O)Bb tickover and deposited C3bBb are distinct compartments.
A. IgG stabilizes C3bBb without demonstrated terminal enhancement (Best answer)
Normal serum acquires decay resistance and sustained C3 cleavage after IgG transfer. The independently assayed C5 convertase does not increase C5 output. The experiment supports transferable C3-convertase regulation, not proof that the antibody caused the renal disease.
B. IgG accelerates decay of assembled C3bBb (Why this does not fit)
Decay acceleration would shorten an assembled enzyme lifetime. IgG transfer instead leaves more active enzyme after the decay challenge. Persistent C3 cleavage contradicts accelerated decay.
C. IgG stabilizes C3bBb and necessarily enhances C5 conversion (Why this does not fit)
Transferred IgG prolongs an assembled C3-cleaving enzyme. The separately assembled C5 convertase does not gain C5 cleavage or lysis in this assay. Stabilizing a C3 convertase does not prove enhancement of the terminal pathway.
D. IgG blocks new factor B cleavage before C3bBb forms (Why this does not fit)
Factor B processing is required before C3bBb can assemble. The experiment starts with an assembled enzyme whose survival increases after transfer. A formation block cannot account for prolonged life of that enzyme.
Takeaway: An IgG-transfer effect on C3 convertase does not prove the kidney diagnosis.
A. Both chambers recover terminal lysis because the free inhibitor was washed away (Why this does not fit)
It removes that soluble blocking reagent from the washed preparation. The experiment also removes soluble complement and adds none back.
B. The phagocyte chamber can recover uptake; washing alone does not restore cell-free complement lysis (Best answer)
The measured surface C3 fragments remain available as receptor ligands. Healthy phagocytes can recognize them through complement receptors. It receives no replacement soluble complement machinery for pore assembly.
C. Neither chamber can use the retained tags because all complement functions require soluble C5 (Why this does not fit)
Membrane attack requires terminal complement assembly. Phagocyte receptors can recognize deposited opsonic C3 fragments without a new pore reaction.
D. The cell-free chamber recovers lysis while the phagocyte chamber cannot recognize C3 fragments (Why this does not fit)
The healthy phagocytes have functional complement receptors. It has no added soluble complement and no phagocytes to use the retained tags.
Takeaway: A retained opsonic tag can support uptake after soluble components are removed.
Factor D is essential to form Bb in the convertase. The enzyme forms before it decays and is stabilized by properdin alone. Normal formation shifts attention from catalysis to lifetime.
B. C3 deficiency (Why this does not fit)
C3 deficiency could disrupt both functional routes. Observed C3bBb assembly and preserved CH50 argue against a shared C3 shortage. Use formation and paired functional assays before assigning central deficiency.
C. Properdin deficiency (Best answer)
Normal assembly demonstrates that B and D can form the enzyme. Rapid loss followed by specific properdin rescue identifies faulty stabilization. Confirm endogenous properdin concentration or activity before labeling an inherited deficiency.
D. C1q deficiency (Why this does not fit)
C1q initiates the classical route to CH50. That route is normal while the B-dependent enzyme has shortened survival. Isolated AH50 loss with rescued lifetime is not C1q loss.
Takeaway: Confirm endogenous properdin concentration or activity before labeling an inherited deficiency.
A. Ingestion declines while factor B recruitment is restored (Why this does not fit)
It changes phagocyte recognition of the retained tag. The intervention does not reverse factor I processing of the target coat.
B. Ingestion persists while factor B recruitment is restored (Why this does not fit)
The supplied experiment identifies CR3-dependent ingestion. It blocks that receptor while leaving the processed ligand unable to recruit factor B.
C. Ingestion persists while factor B recruitment remains deficient (Why this does not fit)
The processed target remains unable to recruit factor B. Its ingestion depends on the CR3 receptor that is selectively blocked.
D. Ingestion declines while factor B recruitment remains deficient (Best answer)
It no longer supports the alternative convertase assembly function of intact C3b. CR3 blockade interrupts phagocyte recognition of the processed opsonic ligand. Blocking a receptor does not convert the processed ligand back into intact C3b.
Takeaway: A processed tag can lose amplification while retaining a separately blockable uptake function.
A. Equal antigen occupancy with the same fragment amount on separate particles (Best answer)
They argue against lost antigen binding or generalized antigen-receptor signaling failure. The experiment has not separated physical linkage from the total amount of fragment supplied. It should change linkage while preserving antigen occupancy and fragment amount.
B. More untagged antigen without matching antigen-receptor occupancy (Why this does not fit)
A larger antigen stimulus can alter receptor occupancy and signaling. It changes antigen stimulation rather than holding it constant while changing presentation.
C. Linked antigen with less total fragment and unchanged receptor blockade (Why this does not fit)
It could test dependence on fragment amount under those conditions. It changes dose and keeps the physical linkage unresolved.
D. A lower direct receptor cross-linking dose without complement fragment (Why this does not fit)
The antigen receptor can still signal despite coreceptor blockade. Whether linked and separate fragment presentation differ at matched occupancy and amount.
Takeaway: Isolate the untested presentation variable instead of predicting a result that was never measured.
A. Extra C3 improves binding; restoring CR1 does not address carriage (Why this does not fit)
It could support formation of receptor ligands during a tagging reaction. The complex coats are already matched and the binding difference follows CR1 availability.
B. Restoring CR1 improves binding; extra C3 alone does not replace the missing receptor sites (Best answer)
It holds the ligand coat constant before the erythrocytes are introduced. Erythrocyte CR1 availability differs between the matched cells. Restoring receptor sites addresses the carriage limitation rather than adding more soluble component.
C. Both interventions prove that macrophage delivery and renal outcomes normalize (Why this does not fit)
It measures erythrocyte binding of already coated complexes. That assay does not measure macrophage delivery or kidney outcomes.
D. Neither intervention can change binding because the complement coat is fixed (Why this does not fit)
No, receptor availability can still determine binding to that coat. Cells with restored CR1 bind the equally coated complexes normally.
Takeaway: A controlled ligand coat lets a receptor deficit be tested independently of complement production.
A. Factor I alone restores local processing without recruitment; tethering cannot improve processing (Why this does not fit)
The patient protein retains its measured regulatory functions when it can encounter the substrate. Too little regulator is recruited to the standardized host-like surface. The experiment identifies missing local regulator recruitment, not deficient factor I activity.
B. Tethering restores cleavage even when factor I is omitted; local convertase decay remains unchanged (Why this does not fit)
It restores local availability of the regulator. Proteolytic C3b cleavage still requires factor I. Its convertase-decay activity can act once it is recruited.
C. Tethering restores factor-I-supported processing; omitting factor I prevents cleavage but not factor-H-mediated decay (Best answer)
Its measured cofactor activity is intact and the tether restores local recruitment. Factor H supplies cofactor support but factor I performs proteolysis. Factor H can still accelerate convertase decay without cleaving C3b.
D. Tethering cannot restore local regulation even with factor I; omitting factor I removes only decay activity (Why this does not fit)
A defective or inaccessible regulatory region could prevent local regulation. The patient regulatory activities are measured as intact and remain accessible after tethering. Factor I performs cleavage rather than the regulator's decay-accelerating action.
Takeaway: Local recruitment, convertase decay and proteolysis are separate requirements.
A. Add properdin to prolong assembled C3bBb (Why this does not fit)
Properdin can prolong a formed alternative convertase. The convertases already decay under factor H, while uncleaved C3b persists afterward. Stabilizing the enzyme would not solve missing cofactor-supported proteolysis.
B. Add CD55 to dissociate assembled C3bBb (Why this does not fit)
CD55 can release catalytic fragments from assembled convertases. Bb release is already rapid in the paired chamber but deposited C3b is not cut. More decay acceleration cannot substitute for conversion of residual C3b.
C. Add more factor H to displace Bb (Why this does not fit)
Factor H has a role in decay and serves as a cleavage cofactor. It already binds C3b and displaces Bb in the split assay. Extra bound cofactor cannot supply the missing protease when factor I alone corrects processing.
D. Supply active factor I with bound factor H (Best answer)
Prompt Bb release demonstrates an intact decay response. Persistent C3b after decay isolates the separate cofactor-dependent cleavage response. Added factor I acts with bound H to convert that residual tag into iC3b.
Takeaway: Factor H-supported decay and factor I proteolysis are distinct interventions.
Normal soluble cleavage shows factor H and factor I work in the shared serum. Only patient cells fail after washing, while introducing a membrane cofactor restores processing. CD46 supplies a local factor-I cofactor on nucleated cells.
B. Factor H soluble cofactor function (Why this does not fit)
Factor H acts in plasma and on selected surfaces. Patient and control cells share serum with intact soluble cleavage, yet only one cell population fails. Cell-restricted complementation is stronger evidence than antigen quantity.
C. CD55 convertase decay function (Why this does not fit)
CD55 shortens the life of surface convertases. The readout is factor I cleavage of already deposited C3b after washing, not enzyme lifetime. Decay acceleration does not supply factor-I cofactor activity.
D. CD59 terminal-pore restraint (Why this does not fit)
CD59 guards the terminal membrane complex. The measured difference precedes terminal components and is corrected by a C3b-processing cofactor. Pore inhibition cannot explain deficient C3b proteolysis.
Takeaway: CD46 supplies a local factor-I cofactor on nucleated cells.
A. Factor I cleavage of C3b (Why this does not fit)
iC3b generation relies on factor I and its cofactors. That process remains normal before rescue although C3bBb lasts too long. Retained processing points toward deficient decay rather than missing protease.
B. CD55-mediated convertase decay (Best answer)
Prolonged upstream convertase survival accounts for excess newly deposited C3b. Blinded GPI-regulator rescue shortens that lifetime while direct terminal resistance remains intact. CD55 accelerates decay of C3 and C5 convertases.
C. CD59-mediated C9 inhibition (Why this does not fit)
CD59 limits terminal C9 pore formation. Resistance to direct C5b-8/C9 challenge is preserved before rescue. Do not infer terminal protection loss from an upstream C3b rise.
D. CD46-mediated factor I cofactor activity (Why this does not fit)
CD46 helps factor I process C3b on nucleated cells. Normal iC3b generation makes a missing cofactor insufficient to explain the prolonged C3bBb lifetime. Separate cleavage of C3b from disassembly of a convertase.
Takeaway: CD55 accelerates decay of C3 and C5 convertases.
A. CD55 acceleration of convertase decay (Why this does not fit)
CD55 is a GPI-anchored upstream regulator. The terminal challenge bypasses convertases whose decay is already normal. A rescue of C9 incorporation is not evidence of restored CD55 decay.
B. Factor H control of C3b feedback (Why this does not fit)
Factor H can restrain surface C3b amplification. C3b is low before and unchanged after rescue, whereas direct terminal lysis falls. An upstream soluble regulator is not the discriminating terminal intervention.
C. CD59 restraint of terminal C9 assembly (Best answer)
Direct C5b-8/C9 challenge localizes vulnerability to pore completion. The blinded GPI protein reduces C9 incorporation while leaving upstream C3b unchanged. CD59 specifically protects host membranes against terminal C9 assembly.
D. CD46 support of C3b proteolysis (Why this does not fit)
CD46 aids factor I cleavage of surface C3b. Neither a C3b-processing deficit nor increased C3b appears in the paired assays. A terminal-only rescue cannot be assigned to C3b cofactor activity.
Takeaway: CD59 specifically protects host membranes against terminal C9 assembly.
A. C3 convertase cleavage of C3 (Why this does not fit)
C3 convertases are necessary for C3b opsonization. Both C3b and processed iC3b remain measurable on the target. Preserved upstream tagging argues against C3 cleavage blockade.
B. Factor I cleavage of C3b (Why this does not fit)
Factor I converts C3b to iC3b. Both fragments persist while two downstream C5-dependent outputs disappear. Inactivation of an opsonin does not simultaneously remove C5a and C5b.
C. C5a receptor signaling alone (Why this does not fit)
C5a signaling can explain reduced neutrophil recruitment. It cannot by itself eliminate C5b-9-mediated bacterial lysis. Paired inflammatory and terminal losses localize before C5 splits.
D. C5 convertase cleavage of C5 (Best answer)
Retained C3 tagging locates the intervention after C3 cleavage. Loss of neutrophil recruitment and terminal lysis tracks loss of both C5a and C5b despite retained C5 antigen. Inhibited C5 cleavage removes both branches of its output.
Takeaway: Inhibited C5 cleavage removes both branches of its output.
Dye entry indicates a permeability defect before lysis. Unchanged wall polymer and pore-blocker rescue localize it to the lipid bilayer. C5b-9 makes a membrane pore rather than digesting a wall.
B. The peptidoglycan polymer (Why this does not fit)
A weakened peptidoglycan wall could also cause lysis. The wall label is retained while a pore blocker prevents early dye entry. Test polymer degradation separately from membrane permeability.
C. The outer capsule polysaccharide (Why this does not fit)
A capsule can alter complement access to bacteria. The purified terminal mixture changes dye entry without an observed capsule-cleaving reaction. Upstream evasion differs from the direct site of a terminal pore.
D. The C3b coat on the bacterium (Why this does not fit)
C3b coats microbes and can recruit complement proteins. This assay supplies C5b through C9 directly and measures permeability without changing the coat. A tag is not the membrane structure perforated by the MAC.
Takeaway: C5b-9 makes a membrane pore rather than digesting a wall.
A. At alternative convertase assembly on C3b (Why this does not fit)
Alternative assembly can contribute to AH50. A selective alternative defect would not account for the newly absent CH50 plus reduced C5a after an intact classical start. The bypass result points to a shared step after C3 tagging.
B. At the shared generation of C5b from C5 (Best answer)
Both functional screens require terminal lysis. Preserved C3b with reduced C5a places the new block between C3 tagging and C5-derived outputs. Direct C5b-6 bypass restores downstream lysis, consistent with impaired production of C5b.
C. At terminal C9 polymerization after C5b-8 (Why this does not fit)
Loss of C9 polymerization would prevent lysis after a terminal bypass. C5b-6 plus C7 through C9 restores lysis in the treated sample. The supplied terminal machinery therefore works under these assay conditions.
D. At classical C4 cleavage before C3 (Why this does not fit)
Early C4 cleavage can affect CH50. The alternative screen also becomes absent while C3b deposition remains measurable. A classical-only initiation lesion does not explain both screens and a terminal bypass rescue.
Takeaway: A C5b bypass after preserved C3 tagging locates the new shared functional block.
A. Persistent C5 deficiency (Why this does not fit)
C5 deficiency can depress both pathway screens. Normal results on a correctly handled repeat contradict fixed terminal absence. Repeat sensitive functional assays before labeling an inherited defect.
B. Constitutional C3 deficiency (Why this does not fit)
C3 deficiency can also affect both screens. The abnormality disappears when collection and transport follow test instructions. A genuine fixed C3 absence should not normalize with specimen handling.
C. Preanalytical functional loss (Best answer)
The first specimen violated timely separation and frozen transport requirements. Both screens normalize when collection and handling change. Suspect a preanalytical effect rather than a persistent pathway defect.
D. Fixed alternative consumption (Why this does not fit)
Consumption can lower complement function during active disease. Only the handling condition changes while repeat screens normalize. A single compromised specimen cannot establish ongoing in vivo consumption.
Takeaway: Suspect a preanalytical effect rather than a persistent pathway defect.
A. Inherited C2 deficiency (Why this does not fit)
C2 defects may impair immune-complex clearance. A fixed C2 absence would not explain normal interflare function plus concurrent C3 and C4 decline. Track changes in multiple components rather than using one phenotype.
B. Inherited factor D deficiency (Why this does not fit)
Factor D loss can reduce alternative function. The flare also lowers C4 and previously both screens were normal. A constitutive alternative lesion does not explain episodic multi-protein loss.
C. Isolated C9 deficiency (Why this does not fit)
C9 loss may alter terminal hemolytic results. The C3/C4 concentrations change with disease activity while interflare screens recover. Terminal deficiency is not an explanation for transient upstream consumption.
D. Acquired multi-component consumption (Best answer)
The biopsy and flare provide a setting for complement activation by immune complexes. C3, C4 and both functions fall together after normal interflare results. An acquired consumption pattern is more likely than a fixed isolated deficiency.
Takeaway: An acquired consumption pattern is more likely than a fixed isolated deficiency.
A. Residual hemolysis despite marked C9 dysfunction (Best answer)
Detectable CH50 initially suggests some terminal activity survives. Selective C9 add-back boosts lysis in patient serum and in a matching C9-depleted control. C9 dysfunction can coexist with residual hemolysis; confirm the component directly.
B. Incomplete C8 production with isolated C9 rescue (Why this does not fit)
C8 loss can impair terminal lysis. The patient responds to C9 rather than C8 add-back in the matched component assay. An add-back result identifies the limiting component better than residual CH50.
C. Preserved terminal function excluding component loss (Why this does not fit)
Measurable CH50 may seem reassuring. Selective rescue by purified C9 shows that the screen misses a functional limitation. Residual activity never by itself excludes every terminal component defect.
D. Upstream C3 cleavage failure corrected by C9 (Why this does not fit)
C3 cleavage precedes terminal pore assembly. Adding terminal C9 cannot restore a truly absent upstream C3 substrate. A downstream component rescue localizes the limiting step downstream.
Takeaway: C9 dysfunction can coexist with residual hemolysis; confirm the component directly.
A. Repeat AH50 before clinical assessment (Why this does not fit)
A functional screen can document drug effect. It cannot exclude meningococcal disease in a symptomatic treated patient. Testing must not postpone evaluation of possible meningococcemia.
B. Arrange prompt emergency evaluation for meningococcal disease (Best answer)
C5 inhibition maintains susceptibility even after both recommended vaccines. Fever with headache and petechiae is a concerning new syndrome requiring immediate assessment and rapid treatment. Prevention lowers neither the urgency nor the need to evaluate symptoms.
C. Presume another infection and schedule next-day follow-up (Why this does not fit)
Many infections cause fever and headache. Petechiae during complement inhibition warrant immediate consideration of meningococcal disease rather than next-day follow-up. A plausible alternative infection does not safely lower triage urgency.
D. Use vaccine history alone to defer assessment (Why this does not fit)
Completing MenACWY and MenB is appropriate prevention. CDC still describes substantial breakthrough risk during complement inhibition. Vaccination history must not override new meningococcal symptoms.
Takeaway: Prevention lowers neither the urgency nor the need to evaluate symptoms.
