1The 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
Correct Answer: The strength of binding between a single antigenic determinant and a single antibody binding site
Explanation:
Affinity describes the strength of the interaction between one epitope and one antigen-binding site of the antibody.
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2The 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
Correct Answer: The overall strength of binding from multiple antigen-antibody interactions
Explanation:
Avidity is the cumulative binding strength of multiple antibody-antigen bonds, reflecting the combined effect of all interactions.
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3Which 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
Correct Answer: IgM
Explanation:
IgM is a pentamer with up to 10 binding sites, giving it high avidity despite low affinity of individual sites.
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4Cross-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
Correct Answer: A different antigen that shares similar epitopes
Explanation:
Cross-reactivity happens when antibodies recognize structurally similar or identical epitopes on different antigens.
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5Which 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
Correct Answer: ABO blood group antibodies reacting with bacterial antigens
Explanation:
Antibodies against ABO antigens can cross-react with similar carbohydrate structures found on intestinal bacteria.
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6A 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
Correct Answer: A soluble antigen
Explanation:
Precipitation reactions occur when antibodies bind soluble antigens to form an insoluble complex that settles out of solution.
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7In 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
Correct Answer: Zone of equivalence
Explanation:
Maximum precipitation occurs at the equivalence zone where antigen and antibody concentrations are optimally balanced for large lattice formation.
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8Which 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
Correct Answer: Ouchterlony double diffusion
Explanation:
Ouchterlony double immunodiffusion allows antigen and antibody to diffuse in agar gel, forming precipitin lines at equivalence.
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9Agglutination 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
Correct Answer: Particulate antigens
Explanation:
Agglutination occurs when antibodies cross-link particulate antigens such as cells or beads, causing visible clumping.
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10The clumping of red blood cells by antibodies is specifically called what?
agglutination
Easy
A.Hemagglutination
B.Precipitation
C.Opsonization
D.Neutralization
Correct Answer: Hemagglutination
Explanation:
Hemagglutination is the agglutination of red blood cells, widely used in blood typing and viral assays.
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11The 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
Correct Answer: Excess antibody
Explanation:
In the prozone effect, too much antibody prevents proper lattice formation, causing a false-negative result at high antibody concentrations.
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12In 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
Correct Answer: A radioactive isotope
Explanation:
RIA uses a radioisotope-labeled antigen or antibody, allowing detection of binding by measuring radioactivity.
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13Radioimmunoassay 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
Correct Answer: High sensitivity for measuring low concentrations
Explanation:
RIA can detect very small quantities of hormones, drugs, and other substances due to its high sensitivity.
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14In 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
Correct Answer: An enzyme
Explanation:
ELISA uses an enzyme-linked antibody that converts a substrate into a detectable colored product.
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15In 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
Correct Answer: Two antibodies
Explanation:
A sandwich ELISA uses a capture antibody and a detection antibody, trapping the antigen between them.
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16A 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
Correct Answer: A color change from the substrate
Explanation:
The enzyme acts on its substrate to produce a colored product, and the intensity of color indicates the amount of bound antigen or antibody.
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17Immunofluorescence 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
Correct Answer: A fluorescent dye
Explanation:
Immunofluorescence uses fluorochrome-labeled antibodies that emit light when viewed under a fluorescence microscope.
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18Which 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
Correct Answer: Fluorescein isothiocyanate (FITC)
Explanation:
FITC is a widely used fluorochrome that emits green fluorescence when excited by appropriate wavelength light.
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19In 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
Correct Answer: The primary antibody specific for the antigen
Explanation:
Direct immunofluorescence uses a labeled primary antibody that binds the target antigen in a single step.
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20The 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
Correct Answer: Specific proteins
Explanation:
Western blotting separates proteins by electrophoresis and detects specific ones using labeled antibodies.
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21Antibody 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
Correct Answer: Antibody B forms a more stable complex because higher means greater affinity
Explanation:
The affinity constant directly measures the strength of a single antigen-antibody bond. A higher ( vs ) indicates tighter binding and a more stable complex.
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22Why 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
Correct Answer: Its pentameric structure provides up to 10 binding sites for multivalent interactions
Explanation:
Avidity is the cumulative strength of multiple binding interactions. IgM's pentameric form offers up to 10 antigen-binding sites, so even low-affinity individual sites combine to yield high overall avidity.
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23A 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
Correct Answer: Cross reactivity due to shared or similar epitopes between bacteria and host tissue
Explanation:
Cross reactivity occurs when an antibody raised against one antigen recognizes a structurally similar epitope on another. Streptococcal antigens share epitopes with cardiac tissue, a mechanism underlying rheumatic heart disease.
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24In 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
Correct Answer: Antigens X and Y share one or more common epitopes
Explanation:
A positive signal with a related antigen indicates cross reactivity, which arises from shared or structurally similar epitopes recognized by the same antibody population.
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25In a precipitation reaction, maximum precipitate forms at the zone of equivalence. What happens in the zone of antibody excess?
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
Correct Answer: Little precipitate forms because excess antibody keeps complexes small and soluble
Explanation:
Precipitation requires large cross-linked lattices formed at equivalence. In antibody excess (prozone), too many antibodies bind individual antigens, preventing lattice formation, so little precipitate forms.
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26In 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
Correct Answer: Ring diameter (squared) is proportional to antigen concentration
Explanation:
In single radial immunodiffusion, antigen diffuses into antibody-containing gel until equivalence. The square of the ring diameter is proportional to the antigen concentration, allowing quantitation.
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27In the Ouchterlony double diffusion test, two adjacent antigens form precipitin lines that fuse smoothly into a continuous arc. This indicates the antigens are:
A pattern of complete fusion (a smooth continuous arc) indicates a reaction of identity, meaning the two antigens share the same epitopes recognized by the antibody.
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28The 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
Correct Answer: Excess antibody prevents proper cross-linking and lattice formation
Explanation:
In the prozone, high antibody concentration saturates antigen sites individually, preventing the cross-linking needed for visible agglutination. Diluting the serum restores agglutination.
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29Why 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
Correct Answer: These IgG antibodies bind RBCs but cannot cross-link them, so anti-Ig is added to cause agglutination
Explanation:
Small IgG anti-Rh antibodies coat RBCs but cannot bridge them across the electrostatic repulsion. Adding anti-human globulin (Coombs reagent) cross-links the bound antibodies, producing visible agglutination.
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30In 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
Correct Answer: To make soluble antigens detectable through visible agglutination of the carrier cells
Explanation:
Soluble antigens produce precipitation, not agglutination. Coating them onto red cells (or latex beads) gives a particulate carrier so antibody binding causes visible agglutination.
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31In 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
Correct Answer: Signal decreases as unlabeled antigen increases because it competes with labeled antigen
Explanation:
In competitive RIA, labeled and unlabeled antigen compete for limited antibody sites. More unlabeled (sample) antigen displaces labeled antigen, so bound radioactivity decreases as sample concentration rises.
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32Which 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
Correct Answer: Its extremely high sensitivity due to detection of radioisotope labels
Explanation:
RIA can detect picogram to nanogram quantities because radioisotope detection is highly sensitive, making it ideal for low-abundance analytes like hormones.
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33In 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
Correct Answer: The antigen must have at least two distinct epitopes for capture and detection antibodies
Explanation:
A sandwich ELISA requires the antigen to bind both the immobilized capture antibody and the labeled detection antibody simultaneously, so it must possess at least two accessible, distinct epitopes.
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34In 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
Correct Answer: The patient's primary antibody bound to the coated antigen
Explanation:
In an indirect ELISA, antigen is coated on the plate, patient antibody binds it, and an enzyme-labeled anti-species (secondary) antibody then binds the patient's antibody to generate the signal.
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35If 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
Correct Answer: The detection antibody or enzyme conjugate failed to bind or was washed away
Explanation:
No color in an expected-positive well points to a failure in the detection step, commonly the enzyme-conjugated antibody not binding or being removed during washing, so no substrate conversion occurs.
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36In 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
Correct Answer: The secondary antibody targets the constant region common to primary antibodies of one species
Explanation:
The labeled secondary antibody is anti-immunoglobulin directed at the conserved constant (Fc) region of primaries from a given species, so one labeled reagent works with many unlabeled primaries and also amplifies signal.
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37A 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
Correct Answer: Use two primary antibodies from different species and secondaries with distinct fluorochromes
Explanation:
Multiplex immunofluorescence requires distinguishable signals. Using primaries from different host species with species-specific secondaries carrying spectrally distinct fluorochromes prevents cross-detection and allows separate visualization.
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38In 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
Correct Answer: Molecular weight, since SDS confers uniform negative charge
Explanation:
SDS coats proteins with uniform negative charge and denatures them, so migration through the gel depends essentially on molecular weight (size), which is the basis of SDS-PAGE separation.
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39During 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
Correct Answer: To occupy unbound sites on the membrane and reduce nonspecific antibody binding
Explanation:
Blocking saturates the remaining protein-binding sites on the membrane so that antibodies bind only their specific targets, minimizing background and false-positive signals.
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40In 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
Correct Answer: Chromogenic (e.g., DAB) staining is stable and viewed with a standard light microscope
Explanation:
IHC typically uses enzyme-linked antibodies producing a stable colored precipitate viewable under an ordinary light microscope alongside routine histology, unlike fluorescence which fades and needs a fluorescence microscope.
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41The 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.
Correct Answer:
Explanation:
From , the fractional occupancy is . When , . This is the operational definition of affinity: the free ligand concentration giving half-maximal binding equals .
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42Two 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
Correct Answer: Antibody Y binds multivalently with favorable geometry, so the bonus effect of cooperative binding outweighs X's per-site affinity
Explanation:
Avidity is the cumulative strength of multivalent binding and can greatly exceed the sum of individual affinities due to the avidity (bonus) effect. Favorable epitope spacing and antibody geometry let Y engage multiple sites simultaneously, so its functional avidity surpasses X despite lower intrinsic per-site affinity.
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43An 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
Correct Answer: B shares epitopes structurally similar but not identical to A, giving genuine cross-reactivity with reduced affinity
Explanation:
Cross-reactivity arises when a shared or structurally similar epitope is recognized. Lower affinity for B ( 5-fold reduced) and partial titer reflect imperfect complementarity to a related-but-nonidentical epitope, the hallmark of genuine cross-reactivity rather than assay error or nonspecific binding.
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44In 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
Correct Answer: Environmental microbial antigens resemble A and B carbohydrate epitopes, priming antibodies that cross-react with the corresponding blood group substances
Explanation:
Gut flora and environmental microbes bear carbohydrate structures mimicking A and B antigens. Individuals produce antibodies against the blood group antigen they lack; these are predominantly IgM and arise from cross-reactive priming by microbial epitopes, not from transfusion or intrinsic germline expression.
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45In 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
Correct Answer: Antigen excess forms small soluble complexes because each antibody bridges too few antigen molecules to build a lattice
Explanation:
Lattice formation requires balanced multivalent cross-linking. In antigen excess, antibody valences are saturated by separate antigen molecules, preventing extensive cross-linking. The resulting small soluble complexes do not precipitate, so measured precipitate falls in the antigen-excess zone.
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46In 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
Correct Answer: The two antigens are non-identical and share no common epitopes (reaction of non-identity)
Explanation:
Crossing lines indicate non-identity: each antigen–antibody system diffuses and precipitates independently, so the lines pass through one another. Fusion (a continuous arc) indicates identity, while a spur indicates partial identity.
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47In 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
Correct Answer: The square of the ring diameter is linearly proportional to antigen concentration
Explanation:
In the Mancini (endpoint) method, antigen diffuses radially until equivalence is reached at the ring boundary. The area of the ring (proportional to the square of the diameter, ) is directly proportional to antigen concentration, providing the calibration relationship.
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48A 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?
The prozone (antibody-excess) phenomenon causes false negatives at low serum dilutions: excess antibody saturates individual antigenic sites without cross-bridging particles. Dilution restores the optimal antibody-to-antigen ratio, allowing lattice formation and visible agglutination.
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49The 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
Correct Answer: They are IgG whose small span cannot bridge the electrostatic gap between red cells, so a secondary anti-IgG is needed to cross-link
Explanation:
Anti-Rh antibodies are typically IgG. The zeta potential keeps red cells apart at a distance greater than the reach of a small IgG bridging two cells, so agglutination does not occur directly. Anti-human globulin (Coombs reagent) cross-links the bound IgG molecules, producing visible agglutination.
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50In 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
Correct Answer: Decreases, because unlabeled analyte displaces labeled tracer from the antibody
Explanation:
Competitive RIA uses limited antibody and fixed tracer. Rising unlabeled analyte competes for the same sites, reducing the fraction of labeled analyte bound. Thus the bound radioactive signal is inversely related to analyte concentration—the basis of the competitive standard curve.
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51A 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
Correct Answer: Use antibody of higher affinity and reduce the amount of labeled tracer used
Explanation:
Sensitivity in competitive assays improves with higher-affinity antibody (so small analyte amounts effectively compete) and with lower tracer/antibody concentrations (so the system is more responsive to small changes). Excess tracer or low affinity blunts the competition and raises the detection limit.
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52In 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
Correct Answer: Excess analyte independently saturates both capture and detection antibodies before they can form a sandwich
Explanation:
The high-dose hook effect occurs when analyte is so abundant that it separately occupies capture and detection antibodies rather than being bridged between them. Fewer complete sandwiches form, so signal drops despite very high analyte, potentially causing dangerous underestimation.
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53A 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
Correct Answer: Competitive ELISA, because a small hapten cannot simultaneously bind two antibodies needed for a sandwich
Explanation:
Sandwich assays require two non-overlapping epitopes for capture and detection antibodies. Small haptens present only one epitope and cannot bridge two antibodies, so a competitive format—where hapten competes with labeled hapten for antibody—is the correct choice.
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54In 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
Correct Answer: Add a more effective blocking agent and include detergent (e.g., Tween-20) in wash buffers
Explanation:
Nonspecific adsorption to the plate causes high background. Effective blocking of unoccupied sites plus detergent in washes disrupts weak, nonspecific interactions while leaving high-affinity specific binding intact. Increasing conjugate or development time would raise both background and specific signal.
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55In 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
Correct Answer: Multiple labeled secondary antibodies bind each primary antibody, amplifying the fluorescent signal
Explanation:
Signal amplification in indirect IF arises because several labeled secondary antibodies can bind to the multiple epitopes of each primary antibody. This multiplies fluorophores per antigen site, increasing sensitivity relative to a single-label direct method.
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56When 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
Correct Answer: Use narrower emission filters and single-stained controls to set spectral compensation
Explanation:
Spectral overlap (bleed-through) is minimized by selecting appropriately narrow emission filters and by using single-stained controls to quantify and compensate for crosstalk. Simply increasing laser power worsens overlap, and photobleaching destroys real signal.
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57In 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
Correct Answer: The native protein is a disulfide-linked homodimer that is separated into monomers by reducing agent
Explanation:
Reducing agents (e.g., β-mercaptoethanol or DTT) break interchain disulfide bonds. A band at ~100 kDa without reductant collapsing to 50 kDa with reductant indicates a disulfide-linked homodimer dissociating into its monomeric subunits.
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58A 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
Correct Answer: Repeat with fresh sample containing protease inhibitors and compare band patterns
Explanation:
If the extra bands are degradation products, adding protease inhibitors to a freshly prepared lysate should reduce or eliminate them. Persistent extra bands under these conditions point instead to antibody cross-reactivity with other proteins, distinguishing the two causes.
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59In 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
Correct Answer: Heat-induced epitope retrieval to reverse formalin-induced cross-linking that masks the epitope
Explanation:
Formalin creates methylene cross-links that mask epitopes. Heat-induced epitope retrieval (antigen retrieval) partially reverses these cross-links, exposing the epitope so the antibody can bind. Extending incubation or adding fixation would not unmask a cross-linked epitope.
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60In 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
Correct Answer: Quench endogenous peroxidase with hydrogen peroxide before adding detection reagents
Explanation:
Endogenous peroxidase activity (e.g., in RBCs and granulocytes) reacts with the HRP substrate and produces false signal even without primary antibody. Pretreating with hydrogen peroxide quenches this endogenous activity, eliminating the background while preserving specific HRP-based detection.
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