Demonstration of antigen antibody interaction by ELISA method
Easy
A.Enzyme-Linked Immunoselective Antibody
B.Electrophoretic Linked Immuno Serum Assay
C.Enzyme-Linked Immunosorbent Assay
D.Enzyme-Labeled Immunosorbent Antigen
Correct Answer: Enzyme-Linked Immunosorbent Assay
Explanation:
ELISA stands for Enzyme-Linked Immunosorbent Assay, a technique that detects antigen–antibody interactions using an enzyme-linked reagent.
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2ELISA is primarily used to detect the interaction between which two molecules?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Lipid and protein
B.Antigen and antibody
C.DNA and RNA
D.Enzyme and substrate only
Correct Answer: Antigen and antibody
Explanation:
ELISA is an immunological assay based on the specific binding between an antigen and its corresponding antibody.
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3Which type of plate is most commonly used to perform an ELISA?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Glass slide
B.Petri dish
C.Agar plate
D.96-well microtiter plate
Correct Answer: 96-well microtiter plate
Explanation:
ELISA is routinely carried out in a 96-well microtiter plate, which allows many samples to be tested at once.
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4The enzyme in an ELISA reaction produces a detectable signal by acting on a specific:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Buffer salt
B.Antibody
C.Substrate
D.Antigen
Correct Answer: Substrate
Explanation:
The enzyme attached to the detection reagent converts a substrate into a colored (or luminescent) product, giving the measurable signal.
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5Which enzyme is commonly conjugated to antibodies in ELISA?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Restriction endonuclease
B.Horseradish peroxidase (HRP)
C.Reverse transcriptase
D.DNA polymerase
Correct Answer: Horseradish peroxidase (HRP)
Explanation:
Horseradish peroxidase (HRP) and alkaline phosphatase are the enzymes most frequently used as labels in ELISA.
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6In a typical ELISA, a positive result is usually indicated by:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.A drop in temperature
B.Release of gas bubbles
C.A color change in the well
D.Formation of a precipitate ring
Correct Answer: A color change in the well
Explanation:
When the antigen–antibody–enzyme complex acts on the substrate, a color develops, indicating a positive result.
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7Which instrument is used to measure the color intensity of the final ELISA product?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Centrifuge
B.Autoclave
C.ELISA reader (spectrophotometer)
D.pH meter
Correct Answer: ELISA reader (spectrophotometer)
Explanation:
An ELISA plate reader measures absorbance (optical density) of each well, which is proportional to the amount of analyte.
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8The step of adding a protein solution to block unoccupied sites on the plate is called:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Elution
B.Blocking
C.Coating
D.Washing
Correct Answer: Blocking
Explanation:
Blocking uses proteins such as BSA or milk to cover empty binding sites, preventing non-specific binding and reducing background.
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9Why is a washing step included between ELISA stages?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.To change the plate color
B.To remove unbound reagents
C.To increase temperature
D.To add more antigen
Correct Answer: To remove unbound reagents
Explanation:
Washing removes unbound antibodies, antigens, and enzyme conjugates, ensuring the signal reflects only specific binding.
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10Which ELISA format uses two antibodies that bind to different sites of the same antigen?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Direct ELISA
B.Competitive ELISA
C.Sandwich ELISA
D.Dot ELISA
Correct Answer: Sandwich ELISA
Explanation:
In sandwich ELISA, the antigen is captured between a coating (capture) antibody and a detection antibody.
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11In an indirect ELISA, the enzyme label is attached to the:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Antigen
B.Substrate
C.Secondary antibody
D.Primary antibody
Correct Answer: Secondary antibody
Explanation:
In indirect ELISA, an unlabeled primary antibody binds the antigen, and an enzyme-linked secondary antibody detects the primary antibody.
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12The molecule that is coated onto the plate surface in a direct ELISA is usually the:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Antigen
B.Enzyme
C.Substrate
D.Stop solution
Correct Answer: Antigen
Explanation:
In direct ELISA, the antigen is immobilized on the plate and then detected by an enzyme-labeled antibody.
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13The specificity of ELISA is mainly due to:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Wash buffer composition
B.Plate material
C.Random enzyme activity
D.Specific antigen–antibody binding
Correct Answer: Specific antigen–antibody binding
Explanation:
ELISA relies on the highly specific recognition between an antibody and its target antigen.
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14Which of the following is a common application of ELISA?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Amplifying DNA fragments
B.Separating proteins by size
C.Detecting antibodies against a pathogen
D.Sequencing genes
Correct Answer: Detecting antibodies against a pathogen
Explanation:
ELISA is widely used in disease diagnosis, such as detecting antibodies or antigens for infections like HIV.
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15In competitive ELISA, the sample antigen competes with labeled antigen for binding to the:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Antibody
B.Plate surface only
C.Substrate
D.Enzyme
Correct Answer: Antibody
Explanation:
In competitive ELISA, unlabeled sample antigen and labeled antigen compete for a limited number of antibody binding sites.
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16What is the purpose of the stop solution in an ELISA?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.To halt the enzyme–substrate reaction
B.To wash the wells
C.To coat the plate
D.To dilute the sample
Correct Answer: To halt the enzyme–substrate reaction
Explanation:
The stop solution ends the color-forming reaction at a fixed time, allowing accurate and reproducible absorbance readings.
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17The optical density (OD) measured in an ELISA is generally proportional to the:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Number of wells used
B.Temperature of the room
C.Volume of wash buffer
D.Amount of analyte present
Correct Answer: Amount of analyte present
Explanation:
In most ELISA formats, higher analyte concentration produces more colored product and thus a higher OD value.
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18Which of these is NOT a standard component of an ELISA test?
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Microtiter plate
B.Substrate
C.Antibody
D.Thermal cycler
Correct Answer: Thermal cycler
Explanation:
A thermal cycler is used in PCR, not ELISA. ELISA uses antibodies, substrate, and a microtiter plate.
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19In ELISA, the term immunosorbent refers to the:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Washing buffer
B.Free enzyme in solution
C.Antigen or antibody bound to a solid surface
D.Colored end product
Correct Answer: Antigen or antibody bound to a solid surface
Explanation:
"Immunosorbent" indicates that an immune reactant (antigen or antibody) is adsorbed/immobilized onto the solid plate surface.
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20A major advantage of ELISA over many other immunoassays is that it:
Demonstration of antigen antibody interaction by ELISA method
Easy
A.Works only on DNA
B.Avoids the use of radioactive labels
C.Cannot be quantified
D.Requires no antibodies
Correct Answer: Avoids the use of radioactive labels
Explanation:
ELISA uses enzyme labels instead of radioisotopes, making it safer, sensitive, and suitable for routine laboratory use.
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21In a sandwich ELISA, a researcher coats the plate with a capture antibody, adds the sample antigen, then adds a detection antibody conjugated to an enzyme. What is the primary requirement for the antigen being detected using this format?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.The antigen must be denatured before coating
B.The antigen must be a small hapten
C.The antigen must possess at least two distinct epitopes
D.The antigen must be enzyme-labeled
Correct Answer: The antigen must possess at least two distinct epitopes
Explanation:
In a sandwich ELISA, the capture antibody binds one epitope and the detection antibody binds a second, distinct epitope. Therefore the antigen must have at least two accessible epitopes, which is why the method is unsuitable for small monovalent haptens.
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22A student observes a strong color signal in the negative control wells of an indirect ELISA. Which of the following is the most likely cause?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.The plate was washed too many times
B.Inadequate blocking leading to non-specific binding
C.The primary antibody concentration was too low
D.Too little substrate was added
Correct Answer: Inadequate blocking leading to non-specific binding
Explanation:
High background in negative controls usually indicates that unbound sites on the well were not properly blocked, allowing antibodies or conjugates to bind non-specifically. Effective blocking with proteins like BSA or casein prevents this.
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23Which enzyme-substrate pair is commonly used in ELISA and produces a yellow color upon reaction?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Glucose oxidase with ABTS
B.Alkaline phosphatase with pNPP
C.Horseradish peroxidase with luminol
D.Beta-galactosidase with X-gal
Correct Answer: Alkaline phosphatase with pNPP
Explanation:
Alkaline phosphatase hydrolyzes p-nitrophenyl phosphate (pNPP) to produce a soluble yellow product measured at 405 nm. HRP with luminol gives chemiluminescence, and X-gal gives a blue product.
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24In an indirect ELISA used to detect anti-HIV antibodies, what is directly immobilized on the microtiter plate surface?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Patient's serum antibody
B.Purified viral antigen
C.Chromogenic substrate
D.Enzyme-labeled antibody
Correct Answer: Purified viral antigen
Explanation:
In an indirect ELISA for antibody detection, the specific antigen is coated onto the plate. Patient antibodies then bind the antigen and are detected by an enzyme-conjugated secondary anti-human antibody.
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25A competitive ELISA shows an inverse relationship between analyte concentration and signal. This means that:
Demonstration of antigen antibody interaction by ELISA method
Medium
A.High analyte gives a low color signal
B.Color develops only in the absence of enzyme
C.High analyte gives a high color signal
D.Signal is independent of analyte concentration
Correct Answer: High analyte gives a low color signal
Explanation:
In competitive ELISA, sample analyte competes with labeled antigen for antibody binding sites. More analyte means less labeled antigen binds, producing less color. Hence signal is inversely proportional to analyte concentration.
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26Why are wash steps critical between each stage of an ELISA protocol?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.To remove unbound reagents and reduce background
B.To dilute the substrate for reaction
C.To increase enzyme activity
D.To denature the bound antigen
Correct Answer: To remove unbound reagents and reduce background
Explanation:
Washing removes unbound antibodies, antigens, and conjugates that would otherwise generate non-specific signal. Proper washing improves the signal-to-noise ratio and assay specificity.
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27In an ELISA, the optical density (OD) of a test sample is and the cut-off value is . How should this result be interpreted?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Positive, since OD exceeds the cut-off
B.Invalid, since OD must equal the cut-off
C.Negative, since OD exceeds the cut-off
D.Positive only if OD is below cut-off
Correct Answer: Positive, since OD exceeds the cut-off
Explanation:
In ELISA, a sample is considered positive when its OD is greater than the established cut-off value. Since , the sample is positive.
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28Which component provides the specificity of antigen detection in a direct ELISA?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.The stop solution
B.The chromogenic substrate
C.The enzyme-conjugated primary antibody
D.The blocking buffer protein
Correct Answer: The enzyme-conjugated primary antibody
Explanation:
In a direct ELISA, the enzyme-labeled primary antibody binds the target antigen specifically. This antigen-antibody recognition determines the assay's specificity; the substrate merely reports enzyme activity.
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29The addition of a stop solution (e.g., dilute ) to an HRP-TMB ELISA reaction serves to:
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Halt the enzymatic reaction and stabilize the color
B.Wash away unbound conjugate
C.Start the enzymatic reaction
D.Block free binding sites
Correct Answer: Halt the enzymatic reaction and stabilize the color
Explanation:
Stop solution terminates the enzyme reaction at a fixed time, converting the blue TMB product to a stable yellow color read at 450 nm. This ensures reproducible, time-controlled OD measurements.
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30A sandwich ELISA is generally more sensitive than a direct ELISA primarily because:
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Signal amplification occurs through multiple antibody layers
B.The antigen is directly labeled with enzyme
C.Fewer wash steps are required
D.No blocking step is needed
Correct Answer: Signal amplification occurs through multiple antibody layers
Explanation:
The sandwich format uses a capture antibody plus a labeled detection antibody, and often a secondary conjugate, allowing signal amplification and higher sensitivity compared to a single-antibody direct ELISA.
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31In an indirect ELISA, the secondary antibody is described as anti-species. What does this mean?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.It is raised against immunoglobulins of the host species of the primary antibody
B.It is raised against the coated antigen
C.It is raised against the blocking protein
D.It is raised against the enzyme substrate
Correct Answer: It is raised against immunoglobulins of the host species of the primary antibody
Explanation:
The secondary antibody recognizes the constant regions of antibodies from the species that produced the primary antibody (e.g., anti-rabbit IgG). This allows one labeled secondary to detect many different primary antibodies.
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32During ELISA optimization, a technician finds that increasing the primary antibody beyond a certain concentration no longer increases signal. This plateau indicates:
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Degradation of the substrate
B.Loss of enzyme activity
C.Incomplete blocking of the plate
D.Saturation of available antigen binding sites
Correct Answer: Saturation of available antigen binding sites
Explanation:
Once all accessible antigen epitopes are occupied, adding more antibody cannot increase binding, so the signal plateaus. This reflects saturation kinetics of the antigen-antibody interaction.
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33Which of the following best explains why a standard curve is prepared in a quantitative ELISA?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.To confirm plate coating uniformity only
B.To measure the pH of each well
C.To determine the wash buffer volume
D.To relate measured OD values to known antigen concentrations
Correct Answer: To relate measured OD values to known antigen concentrations
Explanation:
A standard curve plots OD against known concentrations of standards. Unknown sample concentrations are then interpolated from their OD values, enabling accurate quantification.
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34A false-negative ELISA result for antibody detection could occur if:
Demonstration of antigen antibody interaction by ELISA method
Medium
A.The blocking step was too long
B.The plate was incubated at too low a temperature during coating
C.The substrate was too concentrated
D.The sample was collected during the early window period before seroconversion
Correct Answer: The sample was collected during the early window period before seroconversion
Explanation:
During the window period, antibody levels are below the detection threshold, producing a false-negative result even in an infected individual. This is a key limitation of antibody-based ELISA.
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35Which property makes microtiter plates made of polystyrene suitable for ELISA?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.They chemically digest antigens
B.They passively adsorb proteins onto their surface
C.They actively transport antibodies across membranes
D.They emit fluorescence spontaneously
Correct Answer: They passively adsorb proteins onto their surface
Explanation:
Polystyrene wells bind proteins through hydrophobic and electrostatic interactions, immobilizing antigens or antibodies. This passive adsorption is the basis for capturing reagents in ELISA.
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36If the enzyme conjugate in an ELISA is accidentally omitted, what result is expected?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Color only in negative controls
B.No color development in any well
C.Uniform strong color in all wells
D.Immediate color even without substrate
Correct Answer: No color development in any well
Explanation:
The enzyme catalyzes the substrate reaction that produces color. Without the enzyme conjugate, no substrate conversion occurs, so no color develops regardless of antigen-antibody binding.
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37In a competitive ELISA for a small hapten, why is the competitive format preferred over a sandwich format?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Small haptens produce color on their own
B.Small haptens bind irreversibly to plastic
C.Small haptens have only one epitope and cannot bind two antibodies simultaneously
D.Small haptens cannot be labeled with enzymes
Correct Answer: Small haptens have only one epitope and cannot bind two antibodies simultaneously
Explanation:
Sandwich ELISA needs two epitopes for capture and detection antibodies. Small haptens typically present a single epitope, so a competitive format, which requires only one binding site, is used instead.
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38The wavelength at which OD is measured in an ELISA depends primarily on:
Demonstration of antigen antibody interaction by ELISA method
Medium
A.The color of the substrate reaction product
B.The volume of blocking buffer
C.The size of the antigen
D.The number of wash steps
Correct Answer: The color of the substrate reaction product
Explanation:
Each chromogenic product absorbs maximally at a characteristic wavelength (e.g., TMB stopped product at 450 nm, pNPP at 405 nm). The reader is set to this absorbance maximum for accurate quantification.
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39A technician runs an ELISA and gets weak signal in all wells including positive controls. Which cause is most consistent with this observation?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.The washes were skipped entirely
B.The enzyme conjugate had lost activity
C.Blocking buffer was omitted
D.Too much antigen was coated
Correct Answer: The enzyme conjugate had lost activity
Explanation:
Uniformly weak signal, including in positive controls, points to a failure in signal generation, such as a degraded or inactive enzyme conjugate. Omitting blocking or washes would instead raise background.
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40Which statement correctly distinguishes direct from indirect ELISA?
Demonstration of antigen antibody interaction by ELISA method
Medium
A.Direct uses a labeled primary antibody, indirect uses a labeled secondary antibody
B.Direct detects antibodies, indirect detects antigens only
C.Direct uses two antibodies, indirect uses only one
D.Direct requires no coating step, indirect requires two coatings
Correct Answer: Direct uses a labeled primary antibody, indirect uses a labeled secondary antibody
Explanation:
In direct ELISA the primary antibody itself is enzyme-labeled. In indirect ELISA an unlabeled primary antibody is detected by an enzyme-labeled secondary antibody, adding an amplification step.
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41In a sandwich ELISA, a researcher observes strong signal in the negative control wells that contain only buffer and no antigen. The capture antibody, detection antibody, and enzyme-conjugate were all added normally. Which explanation most likely accounts for this false-positive background?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.The primary antibody has too high an affinity for the target antigen
B.The antigen concentration in the sample was far above the linear range of the assay
C.The detection antibody is directly binding to inadequately blocked plastic surfaces and the capture antibody
D.The substrate was added before the stop solution, shortening the incubation
Correct Answer: The detection antibody is directly binding to inadequately blocked plastic surfaces and the capture antibody
Explanation:
In sandwich ELISA, negative controls lacking antigen should give no signal. Signal without antigen indicates non-specific binding of the detection/conjugate antibody to unblocked plastic or cross-reactivity with the capture antibody, requiring improved blocking or antibody pairs.
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42A competitive ELISA is used to quantify a small hapten. In this format, the relationship between analyte concentration and measured absorbance is:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.Inversely proportional — higher analyte gives lower signal
B.Sigmoidal and always increasing with analyte
C.Independent — signal is fixed regardless of analyte
D.Directly proportional — higher analyte gives higher signal
Correct Answer: Inversely proportional — higher analyte gives lower signal
Explanation:
In competitive ELISA, sample analyte competes with a labeled/immobilized competitor for limited antibody. More analyte displaces more labeled competitor, so signal decreases as analyte concentration rises.
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43The 'hook effect' (prozone) in a sandwich ELISA produces which counterintuitive result at very high antigen concentrations?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.A perfectly linear signal extending to infinite antigen levels
B.Complete loss of capture antibody from the well surface
C.A falsely high signal due to increased enzyme turnover
D.A falsely low signal because excess antigen saturates capture and detection antibodies separately
Correct Answer: A falsely low signal because excess antigen saturates capture and detection antibodies separately
Explanation:
At very high antigen levels, antigen individually saturates both capture and detection antibodies before sandwich formation, preventing bridging. This yields a paradoxically low signal, so high samples must be diluted and re-tested.
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44A standard curve for a direct ELISA gives absorbance values that plateau at high antigen concentrations. Which statement about interpreting an unknown sample reading in this plateau region is correct?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.The plateau indicates the enzyme has been denatured
B.Plateau readings should be multiplied by the blank absorbance
C.The plateau values can be used directly as they are the most accurate
D.The sample must be diluted and re-assayed because readings in the plateau underestimate true concentration
Correct Answer: The sample must be diluted and re-assayed because readings in the plateau underestimate true concentration
Explanation:
The plateau reflects antibody/binding-site saturation, so the assay is no longer sensitive to concentration changes. Samples reading in this region must be diluted into the linear working range for accurate quantification.
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45Why is TMB (3,3',5,5'-tetramethylbenzidine) commonly stopped with sulfuric acid () before reading at 450 nm rather than reading the blue product at 650 nm?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.The acid-stopped yellow product is more stable and gives higher, more reproducible absorbance at 450 nm
B.Acid destroys the HRP enzyme, which is required for accurate reading
C.450 nm is the only wavelength ELISA readers can detect
D.The blue product cannot be measured by any spectrophotometer
Correct Answer: The acid-stopped yellow product is more stable and gives higher, more reproducible absorbance at 450 nm
Explanation:
Acid stops the HRP reaction and converts the blue radical cation to a stable yellow diimine with strong absorbance at 450 nm. This endpoint reading is more reproducible than kinetic reading of the transient blue product at 650 nm.
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46A four-parameter logistic (4PL) model is fitted to an ELISA standard curve. What advantage does the 4PL provide over a simple linear fit?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.It forces all data through the origin for better accuracy
B.It converts absorbance directly into enzyme units without standards
C.It accurately models the sigmoidal curve including the low- and high-dose asymptotes
D.It eliminates the need for replicate wells
Correct Answer: It accurately models the sigmoidal curve including the low- and high-dose asymptotes
Explanation:
ELISA dose-response data are inherently sigmoidal. The 4PL model captures the upper and lower plateaus, the slope, and the inflection point, giving accurate interpolation across the full curve, unlike a linear fit valid only in a narrow midrange.
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47In an indirect ELISA for detecting anti-viral antibodies in serum, samples from previously vaccinated but uninfected individuals give positive signals. This is best described as:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.Loss of assay sensitivity due to over-blocking
B.A hook effect at low antibody concentrations
C.Cross-reactivity leading to reduced clinical specificity of the assay
D.Failure of the secondary antibody conjugate
Correct Answer: Cross-reactivity leading to reduced clinical specificity of the assay
Explanation:
Vaccine-induced antibodies binding the coated antigen produce true immunological signal but not from infection, lowering the assay's clinical specificity. Distinguishing vaccination from infection requires targeting antigens absent from the vaccine.
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48A lab measures the following triplicate OD values for one well set: 0.42, 0.44, and 0.98. Before calculating concentration, the analyst should:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.Average all three values because ELISA data are always reliable
B.Discard the two lowest values and keep 0.98
C.Flag 0.98 as a likely outlier (e.g., a bubble or pipetting error) and investigate before averaging
D.Multiply the three values together to reduce noise
Correct Answer: Flag 0.98 as a likely outlier (e.g., a bubble or pipetting error) and investigate before averaging
Explanation:
The value 0.98 deviates greatly from the tight 0.42/0.44 pair, giving a very high coefficient of variation. Such an outlier commonly arises from bubbles or pipetting error and should be flagged and investigated rather than blindly averaged.
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49Which sequence correctly orders the reagent additions in a standard sandwich ELISA?
Demonstration of antigen antibody interaction by ELISA method
Hard
Sandwich ELISA immobilizes capture antibody first, blocks unbound sites, captures antigen, then adds detection antibody and enzyme conjugate to form the sandwich, followed by substrate, stop solution, and reading. Order is essential to build the immune complex correctly.
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50An ELISA has a limit of detection (LOD) defined as the mean blank signal plus . If the mean blank OD is and the blank standard deviation is , the LOD in OD units is:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
LOD OD units. Signals above this threshold are considered distinguishable from background noise.
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51During ELISA optimization, checkerboard titration is performed. Its primary purpose is to:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.Determine the optimal combination of coating antigen and antibody dilutions that maximizes signal-to-noise
B.Measure the enzyme kinetics of the substrate reaction
C.Calibrate the spectrophotometer at multiple wavelengths
D.Establish the storage stability of the microplate over time
Correct Answer: Determine the optimal combination of coating antigen and antibody dilutions that maximizes signal-to-noise
Explanation:
Checkerboard titration cross-titrates two variables (e.g., coating and antibody dilutions) in a grid to identify the combination giving the best specific signal relative to background, which is central to ELISA optimization.
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52A researcher switches from a direct ELISA to an indirect ELISA using the same primary antibody. What is the main expected consequence?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.Increased sensitivity due to signal amplification from multiple labeled secondary antibodies binding one primary
B.Reduced sensitivity because fewer enzyme molecules are involved
C.The primary antibody no longer needs to bind antigen
D.Complete loss of specificity for the antigen
Correct Answer: Increased sensitivity due to signal amplification from multiple labeled secondary antibodies binding one primary
Explanation:
Indirect ELISA uses an enzyme-labeled secondary antibody; several secondaries can bind each primary, amplifying signal and increasing sensitivity compared with directly labeling the primary, at the cost of an extra step and possible background.
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53Two anti-target antibodies are being evaluated as a capture/detection pair for a sandwich ELISA. Which property is most critical for them to function together?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.They must both be polyclonal antibodies
B.They must recognize non-overlapping (distinct) epitopes on the antigen
C.They must have identical affinity constants
D.They must be raised in the same host species
Correct Answer: They must recognize non-overlapping (distinct) epitopes on the antigen
Explanation:
A sandwich requires both antibodies to bind the same antigen simultaneously. If they share overlapping epitopes they compete and cannot bridge; recognizing distinct epitopes is the essential requirement for a functional pair.
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54A serum sample gives an OD of . From a 4PL standard curve, this corresponds to . The sample was diluted before assay. The concentration in the original undiluted serum is:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The interpolated value reflects the diluted sample, so multiply by the dilution factor: in the original serum.
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55Why does inadequate washing between ELISA steps typically increase background rather than decrease specific signal?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.It permanently denatures the coated antigen
B.Unbound enzyme conjugate remains in the wells and generates non-specific colour development
C.It removes the capture antibody from the well surface
D.It reduces the pH of the substrate buffer below optimal
Correct Answer: Unbound enzyme conjugate remains in the wells and generates non-specific colour development
Explanation:
Washing removes unbound conjugate and reagents. Insufficient washing leaves residual enzyme-labeled antibody that converts substrate everywhere, raising background across all wells including negatives and reducing signal-to-noise.
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56In assay validation, an ELISA shows 95% sensitivity and 80% specificity. In a population where disease prevalence is only 2%, what is the main practical concern with a positive result?
Demonstration of antigen antibody interaction by ELISA method
Hard
A.Sensitivity must be recalculated as 80%
B.Low positive predictive value — most positives will be false positives due to low prevalence
C.The negative predictive value will be near zero
D.The assay cannot detect true positives at all
Correct Answer: Low positive predictive value — most positives will be false positives due to low prevalence
Explanation:
With low prevalence and imperfect specificity, false positives from the large healthy population outnumber true positives, so the positive predictive value is low. Confirmatory testing is needed before acting on a positive.
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57Biotin-streptavidin systems are frequently incorporated into ELISA detection. The main rationale is that:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.The extremely high-affinity biotin–streptavidin interaction and multivalency amplify and stabilize the signal
B.The system removes the need for any enzyme in the assay
C.Streptavidin directly cleaves the substrate to produce colour
D.Biotin blocks all non-specific binding sites on the plate
Correct Answer: The extremely high-affinity biotin–streptavidin interaction and multivalency amplify and stabilize the signal
Explanation:
The biotin–streptavidin bond has one of the highest known non-covalent affinities (), and streptavidin's four biotin sites allow amplification, giving robust, stable signal enhancement in ELISA detection.
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58An ELISA plate reader corrects readings using a reference wavelength (e.g., 570 nm subtracted from 450 nm). The purpose of this dual-wavelength correction is to:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.Increase the enzymatic reaction rate
B.Convert absorbance to fluorescence units
C.Extend the linear range of the standard curve to infinity
D.Compensate for optical imperfections such as scratches, fingerprints, or bubbles in the plate
Correct Answer: Compensate for optical imperfections such as scratches, fingerprints, or bubbles in the plate
Explanation:
The reference wavelength is chosen where the chromogen does not absorb, so subtracting it removes non-specific optical interference (scratches, dust, bubbles) that affects both wavelengths equally, improving reading accuracy.
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59A researcher wants to detect a conformational epitope that is destroyed by drying antigen onto the plate. The most appropriate ELISA design modification is:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.Increase the coating buffer pH to 12 to preserve structure
B.Use a capture (sandwich) format so the antigen is held in near-native conformation by an antibody
C.Use a direct ELISA with prolonged plate drying
D.Coat at 65 °C to speed adsorption
Correct Answer: Use a capture (sandwich) format so the antigen is held in near-native conformation by an antibody
Explanation:
Direct adsorption can denature conformational epitopes. A capture antibody holds the antigen in solution-phase-like, near-native conformation, preserving conformational epitopes for detection—making the sandwich format the appropriate choice.
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60The intra-assay coefficient of variation (CV) of an ELISA is calculated from replicate wells as . If replicate ODs have mean and standard deviation , the CV is:
Demonstration of antigen antibody interaction by ELISA method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
. Intra-assay CVs are typically expected to be below about 10%, so this result is at the acceptable borderline.
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