Download e. None of the above

MCB 4211 Practice Questions
Circle the letter of the BEST answer from the alternatives listed. (Read all of the
alternatives.)
1. What is the portion of an antibody molecule that crystallizes when dialyzed against an
appropriate dialysis buffer?
a. The allotype.
b. The Fc region.
c. The F(ab) region.
d. The isotype.
e. The light chains.
2. Pepsin digests antibodies to produce fragments that can still bind antigen. How would
you determine if the fragment can bind to one or two epitopes?
a. By its ability to interfere with intact antibody’s ability to precipitate the antigen
to which both can bind.
b. By its ability to initiate complement mediated cytotoxicity.
c. By its ability to initiate opsonization.
d. By its ability to initiate a fluorescent reaction
e. By electrostatic xerography.
3. Adjuvants work to non‐specifically augment the immune response to simultaneously
injected antigen. Which of these mechanisms do adjuvants use to produce this
augmentation?
a. Adjuvants enhance antibody association constants.
b. Adjuvants augment complement protein concentrations.
c. Adjuvants emulsified with antigen result in slow release of antigen to the
immune system.
d. Adjuvants decrease sex hormone concentrations.
e. Adjuvants augment the levels of carrier proteins in the serum.
4. Predict the effect of the surgical removal of the thymus at birth.
a. No immune function will develop.
b. The humoral response to all antigens will be eliminated.
c. Cytotoxic T cells and regulatory T cells will be absent, and the animal will have
a diminished T Dependent humoral immune response.
d. The thymus, as the bursal equivalent in mammals, will not be there to enable B
cell differentiation.
e. There will be no effective thyroid development, and goiter will result from the
lack of thyroxin.
5. Inbreeding, in mice as in humans, results in deleterious effects to the offspring. The
formal definition of an inbred mouse strain is one that has resulted from brother x sister
mating for more than 20 generations. Why is a mouse strain that has been through this
process NOT completely genetically homogeneous?
a. There is a statistical probability that some loci will be selected as heterozygotes
through 20 successive generations of breeding pairs.
b. It is possible that a balanced lethal condition will be maintained as a
heterozygote.
c. Random mutation contributes to increases in heterozygosity.
d. The two sexes of the strain have different genetic elements on the sex
chromosomes.
e. All of the above.
6. All of the following are secondary lymphoid tissues except:
a. Spleen
b. Bursa of fabricius
c. Mucosal associated lymphoid tissues like the Peyer’s patches
d. Lymph nodes found at the base of the jaw
e. Cells that can migrate into the lymphoid tissue in order to activate other cells.
7. Granulocytes fall into three major groups and several subgroups. Which is untrue of
these cells?
a. They share a common hematopoietic progenitor.
b. They differentiate in the bone marrow.
c. They must pass through the Bursal equivalent to mature.
d. They provide non‐specific immune functions.
e. They have a different 90-degree (side scatter) light scatter profile from
lymphocytes as measured in the flow cytometer.
8. What is the purpose of the thymidine and hypoxanthine in the selective media (“HAT
media”) that Kohler and Milstein used to isolate hybridomas?
a. To prevent bacterial growth.
b. To prevent the fused cell (hybridoma) from growing too rapidly.
c. To facilitate the lysing of sheep RBCs after the fusion step.
d. To block the de novo pathway of nucleotide synthesis.
e. To provide raw material for the salvage pathway of nucleotide synthesis.
9. An antibody association constant (K) can be determined by placing a known amount of
antibody in dialysis tubing in a bath of radiolabeled antigen. What must be true of this
system to calculate the constant?
a. The antibody and antigen concentrations must not be equimolar
b. The dialysis membrane must be impermeable to the antigen
c. The antibody must be pentameric
d. The system must reach equlibrium before the amount of radioactivity in each
compartment is measured
e. All of the above
10. Prior exposure to the cowpox (vaccinia) virus increase one's chances of survival
during subsequent exposure to smallpox because
a. The cowpox has an adjuvant effect.
b. The immune system has been non‐specifically stimulated and is ready for
anything.
c. Antigen similarity between pathogens induces cross‐reactive immunity.
d. A general anti‐viral immunity has been induced
e. All of the above
11. An animal with severe combined immunodeficiency (SCID) will be unable to
recognize xenogeneic grafts because
a. The molecular structures characterizing the cell surfaces of the graft are
identical to its own.
b. The vascularization process in these mutants is defective.
c. Macrophage antigen processing is defective in these mutants.
d. Development of both cellular and humoral immunity in SCID animals is
unsuccessful.
e. None of the above
12. Congenic lines of mice differ from their partner strains by
a. Single differential loci.
b. Differential loci surrounded by relatively large pieces of linked genetic
material.
c. Fifty percent of their genetic material.
d. Twenty five percent of their genetic material.
e. Fifty percent of their genetic material.
13. Antibodies are usually bivalent, but can have larger numbers of antigen combining
sites. Which isotype can have ten antigen combining sites in one secreted
immunoglobulin molecule?
a. IgG
b. IgM
c. IgY
d. IgA
e. IgF
14. Pepsin digests antibodies to produce fragments that can still bind antigen. If these
fragments are then treated with heat and a reducing agent like 2‐mercaptoethanol,
describe the effect on antigen binding
a. The fragments will become monovalent
b. The fragments will become cross‐reactive with antigens that share some
structural homology
c. There will be no effect on antigen binding
d. This treatment “humanizes” the antibody so that it is no longer antigenically
distinct from other human antibodies
e. None of the above
15. The gradual release of antigen from an emulsification with mineral oil is called the
“depot effect”, and results in an enhanced immune response to that antigen. What term is
used for the mineral oil in this context?
a. The mineral oil acts as a carrier protein.
b. The mineral oil serves as a hapten for the antigen.
c. The mineral oil is an adjuvant.
d. The mineral oil blocks suppressive elements of the signal transduction cascade
and so is called a signal antagonist.
e. The mineral oil has no effect on immunity.
16. Secondary immune tissues are distinguished from primary immune tissues because
a. Surgical removal of secondary tissues does not eliminate entire lineages of
hematopoietic cells.
b. The originating embryonic tissues that produce primary and secondary tissues
are different.
c. Non‐self antigens are not presented to primary immune tissues, while
secondary immune tissues are exposed to these antigens
d. All of the above (a‐c) are true
e. There are no differences between primary and secondary immune tissue
capacities.
17. The ELISA can be configured to detect the presence of serum antibodies that will
bind to an immobilized antigen by using a secondary antibody‐enzyme conjugate
followed by the addition of a chromogenic (color‐producting) substrate. What kind of
secondary antibody‐enzyme conjugate would be the WORST choice as a way to detect
antigen binding by a specific serum antibody?
a. an anti‐idiotypic alkaline phosphatase conjugate
b. an anti‐isotype‐β‐galactosidase conjugate
c. an anti‐allotype‐alkaline phosphatase conjugate
d. an anti‐light chain‐alkaline phosphatase conjugate
e. an anti‐Fc region‐alkaline phosphatase conjugate
18. Removing the Bursa of Fabricius from a chicken will result in:
a. The inability to process antigen
b. Immediate death
c. An augmented immune response
d. The inability to produce antigen‐specific antibody
e. The ability to phagocytose bacterial pathogens
19. The disulfide bridges found in an immunoglobulin molecule connect:
a. Only heavy chain to heavy chain
b. Only light chain to light chain
c. The two variable regions
d. Heavy chain to heavy chain, light chain to heavy chain, and also interconnect
the ends of the immunoglobulin domains on each chain
e. Two antibodies
20. The major forms (isotypes) of mammalian antibodies are
a. IgG, IgD, IgY
b. IgG, IgM, IgA, IgX, IgY
c. Kappa and lambda
d. IgG, IgD, IgA, IgM, and IgE
e. Fab (I), Fab (II), Fc, F(ab’)2, and FcReceptor
21. Totipotent stem cells differ from pleuripotent hematopoietic stem cells because:
a. Totipotent stem cells originate in the mother and are not found in the
developing fetus.
b. Totipotent stem cells can mature into non-hematopoietic cells and the
hematopoietic stem cells cannot.
c. Totipotent stem cells produce teratocarcinoma cells and the hematopoietic stem
cells produce hemocytes.
d. Totipotent stem cells are found in the Bursal equivalent and hematopoietic stem
cells are found in the thymus.
e. Neither cell type can mature into lymphocytes.
22. Macrophages and T lymphocytes
a. Both originate in the spleen
b. Both originate in the Bursa of Fabricius
c. Share a common stem cell progenitor
d. Are both phagocytes
e. None of the above.
23. Haptens are defined as molecular structures
a. That can induce an inflammation
b. That stimulate cell division
c. That bind to the Fc receptor
d. That can be bound by antibody, but cannot stimulate antibody formation on
their own.
e. That have various isoforms
24. Radial immunodiffusion produces a precipitin line in the matrix because
a. Complement activation produces residual insoluble proteins
b. Antibodies chemically react with the surrounding agarose matrix and produce a
change in the agarose transparency
c. The formation of immune complexes can result in large molecular aggregates
that become insoluble
d. None of the above
e. All of the above
In the experimental results shown in figure below, three haptens (listed in the column on
the left side of the chart) were individually crosslinked to albumin (immunizing carrier
protein), and then used as an immunogen to produce three different immune sera. These
same three haptens (listed across the top of the table) were also each coupled to a
different carrier protein (keyhole limpet hemocyanin, KLH), and used as the antigen to be
tested with each immune serum in a solution immunoprecipitation reaction. The
results indicated by "0" refers to an absence of immunoprecipitate; "+" and “+” refer to
slight precipitation results, and "+++" or “+++” refers to the largest immunoprecipitation
results. The reaction in the first parentheses in each of the boxes labeled “A, B, C, D, E,
F, G, H, or I” is the result of a mixture of 0.05% concentration of antigen added to the
antisera. The second parentheses in each of the boxes in the table reflects the size of the
precipitin reaction when 0.01% concentration of antigen was added to the antiserum.
Note: the data has been taken from Table 23 of Landsteiner, K. The Specificity of
Serological Reactions; Harvard University Press: Cambridge, Massachusetts, 1945, p
156. The next questions (27- 29) refer to the chart shown in figure 2 below.
25. Which precipitation reactions are examples of a cross-reaction between the antibody
and a noncognate antigen?
a. A, E, I
b. A, B, C
c. A, D, G
d. C, D, G, H
e. none of the above
26. If the carrier protein used in the precipitation reactions had been albumin instead of
KLH, what would the pattern of precipitations have been?
a. Every box would have had “+++”
b. All boxes would have had “0”
c. There would have been no change in the pattern from what is shown in figure 2.
d. None of the above.
e. The only precipitations would have been seen in A, E, and I
27.The results of this study show that
a. Antibody recognition of molecular features can be extremely precise.
b. Unconjugated tartaric acid is an exellent immunogen.
c. Ionizing side chains can alter the antigenic character of a molecule.
d. The molecular weight of an antigen determines its immunogenicity.
e. BSA is an excellent carrier protein
28. Adjuvants work in several different ways to augment immune capacity. These ways
include
a. Non-specific stimulation of phagocytosis
b. Non-specific mitogenic stimulation of lymphocyte proliferation
c. Gradual and sustained release of the immunogen
d. All of the above
e. None of the above.
29. In the antibody structure depicted above, identify the location of the carboxy terminus
of the heavy chain
a. Indicated by the arrow starting at A
b. Indicated by the arrow starting at B
c. Indicated by the arrow starting at C
d. Indicated by the arrow starting at D
e. Indicated by the arrow starting at E
30. In the antibody structure depicted above, identify the location of the VL (variable
domain of the light chain
a. Indicated by the arrow starting at A
b. Indicated by the arrow starting at B
c. Indicated by the arrow starting at C
d. Indicated by the arrow starting at D
e. Indicated by the arrow starting at E
31. In the antibody structure depicted above, the location of the heavy chain carboxy
termini in an F(ab’)2 fragment (e.g. where pepsin would cleave) is
a. Indicated by the arrow starting at A
b. Indicated by the arrow starting at B
c. Indicated by the arrow starting at C
d. Indicated by the arrow starting at D
e. Indicated by the arrow starting at E
32. The multiple “–S-S-“ structures in the antibody structure shown in figure 3
a. Indicate the number of joining chains that hold the antibody structure together’
b. Represent the locations of the sugars that are part of the carbohydrates
associated with antibody molecules
c. Are a graphic representation of pi bonds that hold the antibody to its cognate
antigen
d. Are serine amino acids along the linear amino acid sequence of the heavy and
light chains
e. Are interchain and intrachain disulfide bridges
33. In Kohler and Milstein's paper describing the first production of a hybridoma cell line
secreting antibody with a predetermined specificity (in this case, anti-sRBC antibody),
they use a modified radial diffusion assay to determine the isotype of the antibody.
Remember, they allowed the radial diffusion to occur, and then overlaid with sheep
erythrocytes mixed with agarose, followed by application of guinea pig serum as a source
of complement. The result of the experiment was
a. The immunoprecipitated anti-sRBC diffused into the sRBC cell layer and lysed
the cells.
b. The immunoprecipitated anti-sRBC could not reach the sRBC layer to activate
the complement cascade
c. The anti-isotype antibodies were able to identify the anti-sRBC isotype without
the need for the overlying sRBC
d. All anti-sRBC antibodies are of the IgG isotype
e. None of the above
34. The ELISA assay of antibody-antigen interactions employs
a. A radioactive probe
b. Enzyme digestion of the antibody to produce Fab and Fc fragments
c. Bovine serum albumin (BSA) as a way to enhance the specificity of the
antibody-antigen reaction
d. All of the above
e. An antibody-enzyme conjugate to detect specific immune complex formation
35. The light chains and heavy chains are joined by
a. Disulfide bond
b. Ionic bond
c. Hydrogen bond
d. B and C
e. A, B, and C
SHORT ANSWER QUESTIONS
36. Describe why an antibody cross-reaction might occur, and give an example.
37. Why does an HGPRT- mutant die in HAT media?
38. In which quadrant of the dot plot shown below would a flow cytometer plot the cells
that are neither apoptotic nor infected with virus (upper left, lower left, upper right, or
lower right)? What tag might have been used to detect each of these phenotypes?