Unit 3: Antigen - antibody interactions - Practice Quiz

BTS511 — Immunology 60 Questions
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1 The affinity of an antibody refers to which of the following?

strength of antigen antibody interaction Easy
A. The rate at which antibodies are produced
B. The strength of binding between a single antigenic determinant and a single antibody binding site
C. The number of antigens an antibody can bind at once
D. The total strength of binding when multiple sites interact

2 The term avidity best describes which concept?

strength of antigen antibody interaction Easy
A. The specificity of an antibody for one antigen
B. The overall strength of binding from multiple antigen-antibody interactions
C. The binding strength of a single epitope to a single site
D. The speed of antigen clearance

3 Which antibody class generally has the highest avidity due to its multiple binding sites?

strength of antigen antibody interaction Easy
A. IgG
B. IgE
C. IgA
D. IgM

4 Cross-reactivity occurs when an antibody binds to which of the following?

cross reactivity Easy
A. Only the exact antigen that induced it
B. A completely unrelated protein with no shared structure
C. Its own light chain
D. A different antigen that shares similar epitopes

5 Which classic example demonstrates cross-reactivity in blood group serology?

cross reactivity Easy
A. Complement binding to red cells
B. Rh antibodies binding to platelets
C. IgE reacting with mast cells
D. ABO blood group antibodies reacting with bacterial antigens

6 A precipitation reaction typically involves an antibody reacting with which type of antigen?

precipitation Easy
A. An intracellular antigen only
B. A soluble antigen
C. A cell surface antigen only
D. A particulate antigen

7 In a precipitation curve, maximum precipitate forms in which zone?

precipitation Easy
A. Zone of antigen excess
B. Zone of equivalence
C. Zone of antibody excess
D. Zone of dilution

8 Which technique uses precipitation in a gel where both antigen and antibody diffuse toward each other?

precipitation Easy
A. ELISA
B. Ouchterlony double diffusion
C. Western blot
D. Flow cytometry

9 Agglutination reactions involve antibodies reacting with which type of antigen?

agglutination Easy
A. Free-floating peptides
B. Particulate antigens
C. Soluble antigens only
D. Nucleic acids

10 The clumping of red blood cells by antibodies is specifically called what?

agglutination Easy
A. Hemagglutination
B. Precipitation
C. Opsonization
D. Neutralization

11 The prozone phenomenon in agglutination is caused by which condition?

agglutination Easy
A. Excess antigen only
B. Low temperature
C. Excess antibody
D. Absence of antibody

12 In a radioimmunoassay (RIA), what is used as the label to detect antigen-antibody binding?

radioimmunoassay Easy
A. A radioactive isotope
B. A fluorescent dye
C. A colored latex bead
D. An enzyme

13 Radioimmunoassay is particularly valued for which of the following characteristics?

radioimmunoassay Easy
A. Requiring no specialized equipment
B. Producing a color change visible to the eye
C. High sensitivity for measuring low concentrations
D. Being free of any safety concerns

14 In an ELISA, the label attached to the antibody is typically which of the following?

enzyme linked immunosorbent assay Easy
A. A magnetic particle
B. A radioactive isotope
C. A heavy metal
D. An enzyme

15 In a sandwich ELISA, the antigen is captured between which two components?

enzyme linked immunosorbent assay Easy
A. Two substrates
B. Two antibodies
C. An enzyme and a substrate
D. Two antigens

16 A positive ELISA result is most commonly detected as which observable change?

enzyme linked immunosorbent assay Easy
A. Formation of a precipitin line
B. A color change from the substrate
C. Emission of radioactivity
D. Clumping of cells

17 Immunofluorescence uses antibodies labeled with what type of molecule?

immunofluorescence Easy
A. A radioactive isotope
B. A latex particle
C. An enzyme
D. A fluorescent dye

18 Which fluorescent dye is commonly used to label antibodies and emits green light?

immunofluorescence Easy
A. Fluorescein isothiocyanate (FITC)
B. Horseradish peroxidase
C. Alkaline phosphatase
D. Iodine-125

19 In direct immunofluorescence, the fluorochrome is attached to which antibody?

immunofluorescence Easy
A. The primary antibody specific for the antigen
B. A secondary anti-immunoglobulin antibody
C. The substrate
D. The antigen itself

20 The Western blot (immunoblot) technique is used to detect which type of molecule?

immunoblot Easy
A. Specific proteins
B. RNA molecules
C. DNA fragments
D. Carbohydrates only

21 Antibody A binds an epitope with an affinity constant , while Antibody B binds the same epitope with . What can be concluded?

strength of antigen antibody interaction Medium
A. Antibody A forms a more stable complex because lower means slower dissociation
B. Antibody A has higher valency, explaining its lower
C. Both antibodies bind with equal stability since affinity is independent of
D. Antibody B forms a more stable complex because higher means greater affinity

22 Why does IgM often show high avidity despite its individual binding sites having relatively low affinity?

strength of antigen antibody interaction Medium
A. It binds only monovalent antigens with high specificity
B. It undergoes somatic hypermutation more rapidly than IgG
C. Its heavy chains chemically modify epitopes to increase affinity
D. Its pentameric structure provides up to 10 binding sites for multivalent interactions

23 A patient recovering from a streptococcal infection develops antibodies that also react with heart tissue. This phenomenon is best explained by:

cross reactivity Medium
A. Complete loss of antibody specificity after infection
B. Increased antibody affinity toward all self-antigens
C. Conversion of IgG antibodies into autoantibodies by heat
D. Cross reactivity due to shared or similar epitopes between bacteria and host tissue

24 In an ELISA, an antiserum raised against Antigen X also gives a positive signal with Antigen Y. Which is the most likely explanation?

cross reactivity Medium
A. Antigen Y has a higher molecular weight than Antigen X
B. Antigens X and Y share one or more common epitopes
C. The secondary antibody is binding directly to Antigen Y
D. The antiserum has lost all specificity for Antigen X

25 In a precipitation reaction, maximum precipitate forms at the zone of equivalence. What happens in the zone of antibody excess?

precipitation Medium
A. Only monovalent complexes form, increasing precipitation
B. Little precipitate forms because excess antibody keeps complexes small and soluble
C. No antigen-antibody binding occurs at all
D. Maximum precipitate forms due to abundant antibody

26 In radial immunodiffusion, the diameter of the precipitin ring is related to antigen concentration in what way?

precipitation Medium
A. Ring diameter is independent of antigen concentration
B. Ring diameter is inversely proportional to antigen concentration
C. Ring diameter decreases as antigen concentration increases
D. Ring diameter (squared) is proportional to antigen concentration

27 In the Ouchterlony double diffusion test, two adjacent antigens form precipitin lines that fuse smoothly into a continuous arc. This indicates the antigens are:

precipitation Medium
A. Present in antibody excess only
B. Partially identical with a spur formation
C. Completely non-identical
D. Identical (immunologically indistinguishable)

28 The prozone phenomenon in an agglutination test results in a false-negative result because:

agglutination Medium
A. The antigen has degraded before testing
B. Excess antibody prevents proper cross-linking and lattice formation
C. There is too little antibody to bind the antigen
D. Complement inhibits agglutination at low dilutions

29 Why is the Coombs (antiglobulin) test needed to detect certain anti-Rh antibodies on red blood cells?

agglutination Medium
A. IgG antibodies naturally agglutinate RBCs without assistance
B. These antibodies are too large to bind red cells directly
C. The test removes antigens from the RBC surface before testing
D. These IgG antibodies bind RBCs but cannot cross-link them, so anti-Ig is added to cause agglutination

30 In passive (indirect) hemagglutination, soluble antigen is first adsorbed onto red blood cells. What is the purpose of this step?

agglutination Medium
A. To convert the antigen into an antibody
B. To make soluble antigens detectable through visible agglutination of the carrier cells
C. To increase the affinity of the antibody for the antigen
D. To prevent cross reactivity with other antigens

31 In a competitive radioimmunoassay (RIA), how does the measured radioactive signal relate to the amount of unlabeled antigen in the sample?

radioimmunoassay Medium
A. Signal decreases as unlabeled antigen increases because it competes with labeled antigen
B. Signal increases proportionally with unlabeled antigen
C. Signal is unaffected by the amount of unlabeled antigen
D. Signal increases only when labeled antigen is absent

32 Which property makes RIA particularly useful for measuring hormones present at very low concentrations?

radioimmunoassay Medium
A. Its use of colorimetric enzyme substrates
B. Its independence from antibody specificity
C. Its extremely high sensitivity due to detection of radioisotope labels
D. Its reliance on visible precipitation for readout

33 In a sandwich ELISA, the antigen is captured between two antibodies. What is a key requirement of the antigen for this format to work?

enzyme linked immunosorbent assay Medium
A. The antigen must be a small hapten with a single epitope
B. The antigen must be enzymatically active itself
C. The antigen must have at least two distinct epitopes for capture and detection antibodies
D. The antigen must be radioactively labeled beforehand

34 In an indirect ELISA measuring antibody in a patient's serum, what does the enzyme-conjugated secondary antibody bind to?

enzyme linked immunosorbent assay Medium
A. The patient's primary antibody bound to the coated antigen
B. The plastic well surface
C. The enzyme substrate before color develops
D. The coated antigen directly

35 If the substrate is added to an ELISA well but no color develops in a sample expected to be positive, which is the most likely technical cause?

enzyme linked immunosorbent assay Medium
A. The plate was incubated for too short a time only in positive wells
B. The antigen concentration was too high to detect
C. The substrate reacted too strongly with the enzyme
D. The detection antibody or enzyme conjugate failed to bind or was washed away

36 In indirect immunofluorescence, why can a single fluorochrome-labeled secondary antibody be used with many different primary antibodies?

immunofluorescence Medium
A. The secondary antibody binds antigens directly regardless of the primary
B. The secondary antibody targets the constant region common to primary antibodies of one species
C. Each primary antibody is itself labeled with the same fluorochrome
D. The secondary antibody recognizes the fluorochrome only

37 A researcher wants to simultaneously visualize two different proteins in the same cell using immunofluorescence. What is the essential requirement?

immunofluorescence Medium
A. Use only a direct method with one labeled antibody
B. Use the same fluorochrome for both proteins
C. Ensure both proteins share identical epitopes
D. Use two primary antibodies from different species and secondaries with distinct fluorochromes

38 In a Western blot, proteins are separated by SDS-PAGE before transfer to a membrane. What property primarily determines this separation?

immunoblot Medium
A. Native three-dimensional conformation
B. Isoelectric point of each protein
C. Antibody affinity for each protein
D. Molecular weight, since SDS confers uniform negative charge

39 During a Western blot, why is the membrane incubated with a blocking agent such as non-fat milk before adding the primary antibody?

immunoblot Medium
A. To increase the transfer efficiency of proteins
B. To label the target protein with a fluorescent tag
C. To denature the transferred proteins further
D. To occupy unbound sites on the membrane and reduce nonspecific antibody binding

40 In immunohistochemistry (IHC), what is the main advantage compared to immunofluorescence for routine diagnostic tissue examination?

immunohistochemistry Medium
A. It only works on live tissue samples
B. Chromogenic (e.g., DAB) staining is stable and viewed with a standard light microscope
C. It uses radioactive labels for higher sensitivity
D. It requires no antibodies for detection

41 The intrinsic affinity of an antibody for a monovalent hapten is described by the association constant . If at equilibrium the concentration of free hapten equals the reciprocal of , what fraction of antibody binding sites are occupied?

strength of antigen antibody interaction Hard
A.
B.
C.
D.

42 Two antibodies, X and Y, bind the same multivalent antigen. Antibody X has higher intrinsic affinity per site, but antibody Y shows higher functional avidity in cell-binding assays. Which explanation is most consistent with these observations?

strength of antigen antibody interaction Hard
A. Antibody Y binds multivalently with favorable geometry, so the bonus effect of cooperative binding outweighs X's per-site affinity
B. Antibody Y has a higher per site, which always increases avidity
C. Antibody X must be monomeric IgG while Y must be a Fab fragment
D. Avidity and affinity are identical quantities, so the data must be experimentally in error

43 An antiserum raised against antigen A reacts with antigen B at 20% of its titer against A. Scatchard-type analysis shows the anti-A antibodies bind B with a that is 5-fold lower than for A. Which interpretation is most accurate?

cross reactivity Hard
A. B is identical to A and the assay is malfunctioning
B. Cross-reactivity requires equal values, so this is not true cross-reactivity
C. B shares epitopes structurally similar but not identical to A, giving genuine cross-reactivity with reduced affinity
D. The antibodies are polyreactive and bind B nonspecifically with equal affinity

44 In the ABO blood group system, anti-A and anti-B isohemagglutinins arise without prior transfusion. What best explains their origin through cross-reactivity?

cross reactivity Hard
A. The antibodies are germline-encoded and require no antigenic stimulus at all
B. They are IgG antibodies produced only after subclinical hemolysis
C. Environmental microbial antigens resemble A and B carbohydrate epitopes, priming antibodies that cross-react with the corresponding blood group substances
D. Fetal exposure to maternal red cells directly induces them in all individuals

45 In a quantitative precipitin curve, increasing antigen beyond the equivalence zone causes the amount of precipitate to decrease. What is the mechanistic reason for this prozone-analogous 'postzone' behavior?

precipitation Hard
A. The equilibrium constant reverses sign in antigen excess
B. Antigen excess forms small soluble complexes because each antibody bridges too few antigen molecules to build a lattice
C. Antibody denatures irreversibly at high antigen concentrations
D. Excess antigen catalyzes proteolysis of the immune complexes

46 In double immunodiffusion (Ouchterlony), two adjacent antigen wells are tested against a central antiserum. The precipitin lines cross each other forming an X. What does this pattern indicate?

precipitation Hard
A. The two antigens are immunologically identical (reaction of identity)
B. The two antigens share some epitopes with a spur (partial identity)
C. The two antigens are non-identical and share no common epitopes (reaction of non-identity)
D. The antiserum contains no antibodies to either antigen

47 In radial immunodiffusion (Mancini method) at endpoint, the relationship between the diameter of the precipitin ring and antigen concentration is best described as:

precipitation Hard
A. The ring diameter is linearly proportional to antigen concentration
B. The square of the ring diameter is linearly proportional to antigen concentration
C. The logarithm of the diameter is proportional to the square of concentration
D. The ring diameter is inversely proportional to antigen concentration

48 A serum with a very high antibody titer gives a negative agglutination result at low dilution but strong agglutination at higher dilutions. What phenomenon explains the false negative at low dilution?

agglutination Hard
A. Prozone effect: antibody excess coats particles monovalently and prevents lattice cross-linking
B. The antibody is nonagglutinating IgG that only works when diluted
C. Complement consumption inhibits agglutination at low dilution
D. Postzone effect: antigen excess prevents precipitation

49 The antiglobulin (Coombs) test is required to detect certain anti-Rh antibodies in hemagglutination assays. Why do these antibodies fail to directly agglutinate red cells?

agglutination Hard
A. They lack antigen-binding sites and only fix complement
B. They are IgG whose small span cannot bridge the electrostatic gap between red cells, so a secondary anti-IgG is needed to cross-link
C. They bind only soluble antigen and cannot attach to cell surfaces
D. They are IgM that is too large to reach neighboring cells

50 In a competitive RIA, patient sample analyte competes with a fixed amount of radiolabeled analyte for limited antibody. As the concentration of unlabeled analyte in the sample increases, the bound radioactivity signal:

radioimmunoassay Hard
A. Increases, because more total analyte binds antibody
B. Decreases, because unlabeled analyte displaces labeled tracer from the antibody
C. First increases then decreases in a bell-shaped curve
D. Remains constant, because antibody is in excess

51 A researcher wishes to improve the lower limit of detection of a competitive RIA. Which change would most effectively lower the detection limit?

radioimmunoassay Hard
A. Increase the amount of labeled tracer to boost total counts
B. Increase incubation temperature to accelerate dissociation
C. Use antibody of lower affinity to widen the dynamic range
D. Use antibody of higher affinity and reduce the amount of labeled tracer used

52 In a sandwich ELISA at very high analyte concentrations the measured signal paradoxically falls, giving a hook effect. What is the underlying cause?

enzyme linked immunosorbent assay Hard
A. Excess analyte independently saturates both capture and detection antibodies before they can form a sandwich
B. Detection antibody dissociates faster at high analyte levels
C. High analyte denatures the capture antibody on the plate
D. The substrate is exhausted before the enzyme can act

53 A competitive ELISA and a sandwich ELISA are compared for measuring a small hapten (~400 Da). Which format is appropriate and why?

enzyme linked immunosorbent assay Hard
A. Sandwich ELISA, because it always gives higher sensitivity
B. Either format works equally well for haptens
C. Sandwich ELISA, because small molecules bind capture antibody more tightly
D. Competitive ELISA, because a small hapten cannot simultaneously bind two antibodies needed for a sandwich

54 In an indirect ELISA measuring anti-viral IgG, background signal is unacceptably high in negative controls. Which single change most specifically reduces nonspecific binding without affecting true signal?

enzyme linked immunosorbent assay Hard
A. Add a more effective blocking agent and include detergent (e.g., Tween-20) in wash buffers
B. Extend the substrate development time
C. Increase the concentration of the enzyme-conjugated secondary antibody
D. Reduce the number of wash steps to preserve bound antibody

55 In indirect immunofluorescence, a primary rabbit antibody is detected with a fluorophore-labeled anti-rabbit secondary. Compared with direct immunofluorescence, the indirect method gives stronger signal primarily because:

immunofluorescence Hard
A. The fluorophore is chemically brighter when conjugated to secondary antibodies
B. The primary antibody binds antigen with higher affinity when unlabeled
C. Secondary antibodies bind antigen directly, doubling the epitopes detected
D. Multiple labeled secondary antibodies bind each primary antibody, amplifying the fluorescent signal

56 When performing dual-color immunofluorescence with FITC (green) and a red fluorophore, bleed-through causes the FITC channel to show false red-region signal. Which correction is most appropriate?

immunofluorescence Hard
A. Use narrower emission filters and single-stained controls to set spectral compensation
B. Use a single filter cube for both fluorophores to simplify imaging
C. Photobleach the red fluorophore before imaging FITC
D. Increase excitation laser power on both channels equally

57 In a Western blot, a protein known to be 50 kDa migrates as a band at ~100 kDa under non-reducing conditions but at 50 kDa with reducing SDS-PAGE. The most likely explanation is:

immunoblot Hard
A. SDS binding is doubled without reducing agent, doubling apparent mass
B. The reducing agent proteolytically cleaves the protein in half
C. The native protein is a disulfide-linked homodimer that is separated into monomers by reducing agent
D. The protein is glycosylated only under non-reducing conditions

58 A Western blot shows the target band plus several unexpected higher- and lower-molecular-weight bands. Which action best distinguishes genuine cross-reactivity of the antibody from proteolytic degradation of the target?

immunoblot Hard
A. Repeat with fresh sample containing protease inhibitors and compare band patterns
B. Reduce transfer time to sharpen only the true band
C. Strip and reprobe with the same primary antibody
D. Increase antibody concentration to enhance all bands equally

59 In formalin-fixed paraffin-embedded tissue, an antibody that works on frozen sections gives no staining. Which pretreatment most directly addresses the likely cause?

immunohistochemistry Hard
A. Heat-induced epitope retrieval to reverse formalin-induced cross-linking that masks the epitope
B. Additional formalin fixation to stabilize the epitope
C. Longer primary antibody incubation at higher concentration
D. Switching to a fluorescent secondary antibody

60 In an IHC assay of a tissue rich in endogenous peroxidase using an HRP-based detection system, strong staining appears even in the no-primary-antibody control. The most appropriate corrective step is:

immunohistochemistry Hard
A. Increase the number of primary antibody layers
B. Use a higher concentration of the DAB chromogen
C. Skip the blocking step to reduce reagent interference
D. Quench endogenous peroxidase with hydrogen peroxide before adding detection reagents