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Antigen-antibody interactions and testing

11 ASCP MLS practice questions on Antigen-antibody interactions and testing, part of Immunology. The first 11 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

  1. Question 1 · Hard

    A 41-year-old man with severe aplastic anemia received equine antithymocyte globulin 9 days ago. He now has fever, an itchy rash, joint pain and enlarged lymph nodes, and his blood film shows plasmacytoid reactive lymphocytes. Infectious mononucleosis (IM) is suspected, and the laboratory performs a Davidsohn differential absorption tube test for heterophile antibodies that agglutinate sheep red cells. Unabsorbed serum: agglutination to a titer of 320. Serum absorbed with guinea pig kidney: no agglutination at any dilution. Serum absorbed with beef red cells: no agglutination at any dilution. Positive and negative control sera react as expected. These results show:

    1. AForssman antibodies
    2. BInfectious mononucleosis heterophile antibodies
    3. CSerum sickness heterophile antibodies
    4. DA mix of IM and Forssman antibodies
    Show answer

    Answer: C · Serum sickness heterophile antibodies

    Heterophile antibodies are told apart by which absorbing antigen removes them. Forssman antibodies are removed by guinea pig kidney but not by beef red cells, IM antibodies are removed by beef red cells but not by guinea pig kidney, and serum sickness antibodies are removed by both. Here both absorptions removed all agglutination, which fits serum sickness 1 to 2 weeks after a horse-derived antithymocyte globulin. IM antibodies, alone or in a mixture, would still agglutinate sheep cells after guinea pig kidney absorption. If mononucleosis is still suspected, EBV-specific antibodies can be tested.

    • A: Forssman antibodies are not removed by beef red cells, so that tube would still agglutinate.
    • B: IM antibodies are not removed by guinea pig kidney, so that tube would still agglutinate.
    • D: Any IM antibody in a mixture would survive guinea pig kidney absorption.
  2. Question 2 · Medium

    The precipitin curve shown was made by adding increasing amounts of antigen to a fixed amount of antibody. At point 3, the supernatant would contain:

    Figure: MedExams
    Curve of precipitate formed against increasing antigen added to a fixed amount of antibody, rising to a single peak and then falling. Point 1 is on the rising limb, point 2 at the peak and point 3 on the falling limb.
    1. AEqual free antigen and antibody
    2. BNeither free antigen nor antibody
    3. CFree antigen and soluble complexes
    4. DFree antibody, unbound to antigen
    Show answer

    Answer: C · Free antigen and soluble complexes

    Point 1, on the rising limb, is the zone of antibody excess (prozone), where free antibody stays in the supernatant. Point 2, at the peak, is equivalence: antigen and antibody build the largest lattice and neither is left free. Point 3, on the falling limb, is antigen excess (postzone): each antibody binds separate antigen molecules, the complexes stay small and soluble, and free antigen remains. Antigen excess causes false-negative precipitation tests and the hook effect in immunoassays, and diluting the sample corrects it.

    • A: Free antigen and free antibody would combine; they are not left free together.
    • B: Neither is left free only at equivalence, the peak (point 2).
    • D: Free antibody is found on the rising limb, in antibody excess (point 1).
  3. Question 3 · Medium

    In end-point (Mancini) radial immunodiffusion, read after diffusion is complete, the antigen concentration is directly proportional to:

    1. AThe ring diameter itself
    2. BThe square of the ring diameter
    3. CThe time taken for the ring to appear
    4. DThe square root of the ring diameter
    Show answer

    Answer: B · The square of the ring diameter

    In radial immunodiffusion, antigen diffuses from a well into agar containing antibody and forms a ring of precipitate. In the Mancini end-point method, read when the rings stop growing (about 48 to 72 hours), the antigen concentration is proportional to the diameter squared. In the Fahey-McKelvey kinetic method, read at a fixed earlier time, the diameter is proportional to the logarithm of the concentration. Standards of known concentration are run on every plate to draw the curve.

    • A: Concentration follows the area of the ring, so diameter alone is not linear with it.
    • C: End-point plates are read after ring growth stops, not timed.
    • D: This inverts the relationship; it is the diameter that grows with the square root of concentration.
  4. Question 4 · Easy

    The overall binding strength of a multivalent antibody, combining all of its binding sites, is called:

    1. ASpecificity
    2. BAvidity
    3. CAffinity
    4. DCross-reactivity
    Show answer

    Answer: B · Avidity

    Affinity is the strength of one binding site for one epitope. Avidity is the total strength when several sites bind at once, so a pentameric IgM with low-affinity sites can still bind tightly to a cell surface. High avidity makes complexes stable and reactions easier to see. IgG avidity testing helps separate recent from past infection, because avidity rises over the months after infection.

    • A: Specificity is the ability to tell one epitope from others.
    • C: Affinity is the strength of a single binding site.
    • D: Cross-reactivity is binding to a similar epitope on a different antigen.
  5. Question 5 · Hard

    Twofold dilutions of a patient's serum in an agglutination test give: undiluted 0; 1:2 1+; 1:4 3+; 1:8 4+; 1:16 4+; 1:32 3+; 1:64 2+; 1:128 1+; 1:256 0. What titer should be reported?

    1. A128
    2. B256
    3. C8
    4. D2
    Show answer

    Answer: A · 128

    The titer is the reciprocal of the highest dilution that still shows agglutination, here the 1:128 tube, so the titer is 128. The negative undiluted tube and weak 1:2 tube are a prozone: excess antibody coats each particle separately and blocks lattice formation. The strongest reactions, at 1:8 and 1:16, are where antigen and antibody are closest to equivalence. Had only undiluted serum been tested, this strongly positive patient would have been reported as negative.

    • B: The 1:256 tube shows no agglutination, so it is beyond the end point.
    • C: The 1:8 tube is only where the strongest reaction begins.
    • D: The 1:2 tube is the first positive tube, still inside the prozone, not the end point.
  6. Question 6 · Medium

    In Ouchterlony double diffusion, the lines formed by two neighboring antigen wells cross each other completely. This shows:

    1. AIdentity: the same antigen
    2. BAntibody excess in the center well
    3. CPartial identity: some shared epitopes
    4. DNon-identity: no shared epitopes

    Answer and explanation in LearnFree account

  7. Question 7 · Medium

    In a latex test for cryptococcal antigen, latex particles coated with antibody clump when the patient's CSF contains the antigen. This test format is:

    1. AReverse passive agglutination
    2. BAgglutination inhibition
    3. CPassive agglutination
    4. DDirect agglutination

    Answer and explanation in LearnFree account

  8. Question 8 · Easy

    Which kind of bonds hold an epitope in an antibody's binding site?

    1. ANoncovalent and reversible
    2. BCovalent but easily reversed
    3. CDisulfide bridges made on contact
    4. DCovalent and irreversible

    Answer and explanation in LearnFree account

  9. Question 9 · Medium

    In an agglutination inhibition test for a drug in urine, urine is mixed with anti-drug antibody before drug-coated latex particles are added. A positive result (drug present) is seen as:

    1. AA precipitin ring around the well
    2. BHemolysis of the indicator cells
    3. CAgglutination of the particles
    4. DNo agglutination of the particles

    Answer and explanation in LearnFree account

  10. Question 10 · Easy

    The heaviest precipitate forms when antigen and antibody are present in:

    1. AAny ratio, given enough time
    2. BAntibody excess (prozone)
    3. CAntigen excess (postzone)
    4. DOptimal proportions (zone of equivalence)

    Answer and explanation in LearnFree account

  11. Question 11 · Hard

    A strongly positive sample gives a negative agglutination result, which becomes positive when the sample is diluted. What is this called?

    1. AThe prozone phenomenon (antibody excess)
    2. BThe zone of equivalence
    3. CCross-reactivity
    4. DThe postzone phenomenon (antigen excess)

    Answer and explanation in LearnFree account

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