03-390 Immunology Exam II - 2016 Name:______________________

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03-390 Immunology Exam II - 2016 Name:______________________
03-390 Immunology
Exam II - 2016
Instructions: This exam contains 15 questions on 6 pages for a total of 100 points. On questions with
choices/options all of your attempts will be graded and you will be awarded the highest grade.
1. (6 pts) Sketch the immunoglobulin component of a Bcell receptor. Indicate on your diagram: i) the different
polypeptide chains and their structural domains, ii) the
location of antigen binding, iii) an FAB fragment, iv)
representative disulfide bonds.
Diagram should show a Y shaped molecule with:
-two identical light chains, with two Ig folds, VL and CL.
-two identical heavy chains, with four (or five, IgM&IgE) domains, VH, CH1,CH2,CH3
-antigen binds at the amino terminal of both domains (ends of the Y)
-an Fab fragment is light chain + VH+CH1
-disulfide bonds are found within each Ig fold, between the light and heavy chain, and between the
heavy chains.
2. (4 pts) The following is a sequence alignment of two light chains from antibodies with different specificities. The
middle line indicates sequence identity (single letter amino acid) or similarity (+). Identify the following on the
sequences: i) hypervariable loop(s) by circling, ii) constant region with a box.
Antibody A
Antibody B
3. (5 pts) You produce an antibody that recognizes unique epitopes on the surface of cancer cells. You treat the
antibody with trypsin to produce Fv fragments and find that the preparation no longer binds to the epitope.
Why does the Fv fragment no longer bind to the antigen, and what could you do to restore binding?
The Fv fragment dissociates into VL and VH domains, neither of which can bind the antigen.
It would be necessary to make a scFV, by cloning the VL and VH DNA sequences and joining
them in frame with a short (15-20) residue flexible polar linker (Gly & Ser)
Points on page:______________
03-390 Immunology
Exam II - 2016
4. (6 pts) Please do one of the following choices:
Choice A: Why are IgG molecules good at physical blocking of pathogens
Choice B: Why are IgM more proficient at agglutination of pathogens.
Choice C: Why is IgM particularly good at activating complement while most forms of IgG are not?
Choice D: How do Fc receptors enhance pathogen destruction by either macrophages or NK cells?
Choice E: How do babies benefit from the immune system of their mothers?
Choice A: The smaller IgG (monomer versus IgM pentamer) can reach a higher density on the surface of
the pathogen.
Choice B: IgM are larger so they can span between pathogens, they also have 10 antigen binding sites, so
they could bind to two or more pathogens.
Choice C: Activation of complement (classical pathway) requires two Fc regions within 40-50 A of each
other. Since IgM is a pentamer with 5 Fc regions, it is easy to accomplish this.
Choice D: Antibodies bound to the pathogen will target the pathogen-antibody complex to the Fc
receptor on the macrophage, leading to efficient internalization of the pathogen by receptor mediated
endocytosis. In the case of NK cells, the interaction between the Fc and the Fc receptor will activate
the NK cell and cause the antibody bearing cell to be killed (only necessary to discuss macrophage or
NK cells).
Choice E: The mother’s IgG crosses the placenta, providing immunity in utero. Breast milk contains IgA
that provides protection of the mucosal membranes in the gut of the feeding infant.
5. (10 pts) Describe, or sketch, the major steps that occur in B-cell development, beginning from stem cells and
ending with plasma cells/BMEMORY cells. In your answer you should include:
Location of the events in the body.
Checkpoint(s) associated with that event, if applicable.
General nature of DNA rearrangements that occur at each event, you do not need to discuss the
detailed mechanism for these changes.
Key cell(s) involved in the process.
In bone marrow (3 pts)
 Stem cells rearrange HC and then LC genes to produce function V-exons, providing diversity
 Checkpoints are for a functional heavy chain and light chain
 Last check point is self-tolerance, if passed the B-cell goes to the blood.
In Lymph node (maturation) (2 pts)
 Interacts with follicular dendritic cells to complete maturation and 2 nd self-tolerance check point.
They then circulate and enter other lymph nodes/secondary lymphatic organs (e.g. spleen).
In Lymph node (activation) (5 pts)
 B-cell acquires antigen, presents on MHC II to Th cell. (1 pt)
 Signaling events occur leading to cytokine release by Th cell (1 pt)
 Activated B-cell undergoes class switching (1 ½ pts)
 Undergo affinity maturation to make higher affinity antibodies that will bind better to pathogen ( 1 ½
 Plasma cells and memory cells develop.
Points on page:______________
03-390 Immunology
Exam II - 2016
6. (6 pts) The following is a segment of DNA:
i) Correct the mistakes in the diagram. Briefly justify your answer.
ii) Does this DNA represent
D1 D2 D3
J1 J2 J3
the light or heavy chain?
2 2
1 1 1 1 1 1
iii) What is the minimum
number of chains that can be generated from this DNA?
i) The joining of segments requires that a one-turn be paired with a two turn. In this diagram V cannot join
to D, The “1” turn signals at the 3’ end of each V-segment should be “2” (3 pts)
ii) Heavy, since there are D segments. (1 ½ pt)
iii) 27, since any D can join with any J and any V can then join with DJ, 3 x 3 x 3. (1 ½ pt)
7. (6 pts) Select one of the following and briefly describe its contribution to the diversity of antibodies.
Choice A: p-bases
Choice B: n-bases
Choice C: junctional diversity (imprecise joining/crossover)
Choice D: CL and the CH1 interactions.
Choice A: p-bases are added during repair of the hairpin during joining when the top strand is cleaved by
Rag1/Rag2. Repair synthesis by DNA polymerase adds new bases.
Choice B: n-bases are added by terminal transferase during the joining of segments in the heavy chain.
They increase diversity because the bases added are random.
Choice C: When the 1 and 2 turn joining occurs the actual position of the joint is not precise. Cleavage of
the bottom strand by Rag1/Rag2 produces a 3’ overhand that is removed by exonucleases, the loss of
sequences also increases diversity.
Choice D: This protein-protein interaction allows the pairing of any heavy chain to any light chain,
increasing the number of antibodies since the total number of Ab is the # of HC x # of LC.
8. (5 pts) Please do one of the following choices:
Choice A: What happens to a B-cell if it is self-reactant to cell-surface antigens? How is the response different
for soluble self-antigens?
Choice B: How does co-expression of IgM and IgD occur on the surface of B-cells? Briefly describe the events at
the molecular scale (e.g. polyA addition, mRNA splicing)
Choice C: How is the production of soluble antibody different from the production of membrane bound
antibody? Briefly describe the events at the molecular scale (e.g. polyA addition, mRNA splicing)
Choice A:A B-cell that is self-reactant to membrane (cell surface) antigens will undergo receptor editing,
trying different light chains to become self-tolerant. In the case of soluble antigens the B-cell becomes
unresponsive or anergic.
Choice B: The initial mRNA transcript contains both IgM and IgD constant exons. The constant exons are
followed by two polyA addition sites. If the second polyA site is used a membrane bound IgM is produced.
If the fourth polyA site is used, the IgM constant exons are spliced out and a membrane bound IgD is
Choice C: At the end of each constant region there are exons to produce a soluble antibody and exons that
encode a membrane anchor. Each of these are followed by a polyA addition site. If the first is used, then
only the S exon remains and the antibody is soluble. If the second is used, the membrane exons are used
(S exon removed by alternative splicing).
Points on page:______________
03-390 Immunology
Exam II - 2016
9. (6 pts) What is allelic exclusion and why is it an important step in B-cell development.
Once an allele of the light or heavy chain is successful rearranged by VJ or VDJ joining,
rearrangement of the other allele is suppressed. This results in a single specificity (4 pts)
 This is efficient with regard to antibody production – all the antibodies produced by the plasma cell
will bind to the pathogen.
 It also greatly reduces the chance of producing a self-reactive B-cell
(it is only necessary to give one reason, 2 pts).
10. (8 pts) The activation of B-cells by TH cells requires a number of signaling events between the B and TH cells
that occur in sequential order.
i) Select any one of the signaling events in B-cell activation and briefly discuss the event you have chosen and
the prior event that caused the signaling event that you have chosen (6 pts).
ii) Why are there multiple signals events? What might happen if TH cells could be activated solely by MHC IIpeptide complexes (2 pts)?
i) Signaling events are (4 pts for correct pairing, 2 pts for sensible description.)
TH cell
MHC II + peptide → TCR + CD4
(specific recognition of MHC II + peptide by TCR, CD4 binds to MHC II)
← CD40L
(CD40L is elevated on T-cells after 1st signal (MHC-TCR)
→ CD28
(b7 is elevated on B-cell after receiving CD40-CD40L signal)
Cytokine Rec.
← IL2, IL4, IL5
(cytokines are released when b7-CD28 signal is received)
ii) The multiple signaling events ensure that the correct cells are being activated (2 pts) (Activation of a TH
may produce cytokines that have undesirable effects on nearby cells).
11. (10 pts) Please do one of the following choices:
Choice A: Class switching: i) How does it occur, ii) why is it an important attribute of the acquired immune
system, iii) what governs which class is produced?
Choice B: Why is affinity maturation (somatic hypermutation) an important event after class switching? Briefly
describe the process by which affinity maturation occurs.
Choice A:
i) homologous switch regions are found between the constant exons for the different types of
heavy chain. After DNA damage recombination occurs between the different switch
regions, deleting the ones between the VH exon and the one that is to be used. (This moves
the desired heavy chain constant exons adjacent to the VH exon, RNA splicing will join the
ii) Different classes of antibodies have different biological functions, optimizing the
iii) the cytokine environment of the activated B-cell.
Choice B:
 Class switching goes from IgM which has 10 Fv regions to IgG which has two, the reduction
in binding sites decreases the avidity – the IgG will not bind as well as the IgM, so it is
necessary to increase the affinity of the Fv region to restore the binding (5 pts).
 Random mutations are made in the V exons of both the heavy and light chain. Those B-cells
that produce higher affinity antibodies will capture more antigen, be more likely to be
activated by Th cells, and will come to dominate the B-cell population. (5 pts)
Points on page:______________
03-390 Immunology
Exam II - 2016
12. (6 pts) Please do one of the following choices:
Choice A: Briefly describe how B-cell regulation can be detrimental to the production of antibodies against
related pathogens.
Choice B: A B-cell response to an antigen does not result in class switching or the production of memory cells.
What are the properties of the antigen and why are only IgM antibodies produced by plasma cells?
Choice A: The cross reacting antibodies from the first pathogen can suppress the production of
antibodies to the second, related, pathogen by binding to the Fc receptor on the B-cell.
Choice B: The antigen has multiple epitopes that cause crosslinking of the B-cell receptor,
activating the B-cell without Th support. IgM is the default class, class switching would be
required to change the type of heavy chain.
13. (10 pts) Compare and contrast the peptide binding properties of a class I MHC and a class II MHC. How are
they similar? How do they differ? A well labeled diagram is a suitable answer.
In both cases the peptides bind in extended (beta) configuration to allow the formation of
hydrogen bond between the mainchain atoms of the peptide and the MHC. (3 ½ pts)
In addition, for both I and II, a few residues on the peptide have specific interactions with
the MHC (anchor residues) (3 ½ pts)
Class I MHC can only bind peptides 8-9 in length, class II can bind peptides of any length
(but the extent of interaction is the same, roughly 8-9 residues (3 pts).
14. (6 pts) The following shows the gene structure of the MHC region of
an organism. Briefly justify your answers for all parts.
i) How many different class I MHC molecules would be found on an
inbred individual (2 pt)?
One, since there is only one class I gene (A) and in an inbred
animal the same alleles would be found on both chromosomes.
C C E E E
iii) How many different class II MHC molecules would be found on an inbred APC (2 pts)?
One sub-group C, and two subgroup E. The C chains cannot pair with the E. Either alpha chain
of the sub-group E can pair with the single beta chain. The proteins produced from both
chromosomes would be the same.
ii) How many different class I MHC molecules would be found on an outbred individual (2 pts)?
Two, because the alleles would be different
Points on page:______________
03-390 Immunology
Exam II - 2016
15. (6 pts) Two different inbred strains of mice (A and A ) are infected with a virus. The Ak strain generally
survives the infection, while the Ab strain does not. Give one possible explanation for this observation.
Strain of haplotype b does not have MHC alleles that can present peptides from the virus, so
neither Tc or TH cells can be activated. The strain with haplotype k can present peptides and
can activate Tc (via class I) for an antiviral response and Th for an antibody response.
Points on page:______________
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