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Dec 11, 2012 (4 years and 8 months ago)

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Midterm 1 MCB150 Fall 2006

Name________________________________________


1

MCB 150 Midterm Exam #1 (100 points total)

Please write your full name on each page of the exam!!

The exam consists of 9 questions (6 pages). Each has a different point count as indicated.

Please limit your answer to the space provided.

Questions are not
necessarily in order of difficulty, so please look over the whole test before spending alot
of time on one question.

Good Luck!!


1a. (5 pts) What are PRRs and PAMPs? Give two examples of each.

PRRs are Pattern recognition receptors; receptors that recogn
ize PAMPS

PAMPS are pathogen associated molecular patterns; ligands for PRRs

Examples:

PAMPs: LPS, dsRNA, CpG DNA, (specific ligands)



PRRs: TLRs. NODs, Scavenger, etc


1b. (4 pts) What does TLR9 recognize? Explain why its ligand is unique to microbes
v
ersus mammalian cells?

It recognizes unmethylated CpG DNA

Its ligand is unique because we methylate our CpG DNA and microbes do not


1c. (4 pts) Below is a figure from our discussion section paper showing the proliferation
of splenocytes from wild and kno
ck
-
out mice treated with the indicated ligands. Explain
why LPS treated splenocytes were included in this experiment.


They were included as a positive
control to show that the splenocytes
are not defective in their ability to
respond and that the lack o
f response
to ODN in the TLR9
-
/
-

cells is
specific




1d. (4 pts) Below is another figure from our discussion section paper showing the
survival of mice following injection with CpG oligodinucleotides (ODN). What are the
wild type mice dying from? Why
are the knockout mice surviving?

The wildtype mice are dying from septic shock from an overproduction of cytokines.
The TLR9
-
/
-

mice are not dying because they are are not producing cytokines in
response to CpG(since they lack the ability to respond to CpG

through TLR9)



Midterm 1 MCB150 Fall 2006

Name________________________________________


2

2. (6 pts) For each technique listed below, indicate which of the statements below apply.


Western blot



c (f) (b)


Flow cytometry


a f (b)


FACS


a g f (b)


Sandwich ELISA

b d


RIA




b (a)


Immunofluorescence microscopy e f (
a)

(letter is parenthesis means OK if they did or did not include.)


a)

Can be used to determine the number of cells in a cell suspension that express a particular antigen.

b)

Sensitive method used to determine the quantity of antigen in a complex sample.

c)

Can be

used to determine the molecular weight of an antigen.

d)

Requires 2 different antibodies to the same antigen.

e)

Provides anatomical/spatial information about location of antigen.

f)

Can be used to detect multiple different antigens simultaneously in the same samp
le.

g)

Can be used to physically isolate cells based on marker expression


3. (5 points) Below is a ribbon diagram of a light chain. Indicate on the diagram the
regions that correspond to the :

V domain

C domain

Intrachain disulfide bond (only 1 needed)

HV
loops

part of the light chain that corresponds to the V
-
J DNA rearrangement junction =
CDR3

















disulfide bond


C domain



V domain

CDR3

HV loops

Midterm 1 MCB150 Fall 2006

Name________________________________________


3

4. (4 pts) In order to generate antibodies in a rabbit against the hapten DNP, you need

both a
carrier

and an
adjuvant
. Briefly explain both terms and explain how each one
contributes to antibody production.

Carrier: a protein that is covalently coupled with hapten; hapten small, not a protein, so
has to be linked to protein in order to be
recognized by immune system.

Adjuvant: material that that is mixed with antigen to boost antibody production: slows
release of antigen and stimulates innate immunity (contains TLR agonists).


5. (8 pts) For each antibody isotype listed below, indicate whi
ch of the statements below
apply.
(letter is parenthesis means OK if they did or did not include.)

IgG a e


IgM c d (a)


IgD a f


IgE a e g


IgA b e

a)

Dimeric (2 antigen binding sites/molecule).

b)

Tetrameric (4 antigen binding sites/molecule).

c)

Dec
americ (10 antigen binding sites/molecule).

d)

First antibody secreted in an immune response.

e)

Production requires class switch recombination.

f)

Exists in only a transmembrane form.

g)

Binds to Fc receptors on mast cells.


6. (10 points total, 2 points each) What i
s (are) each of the following and
why are they
important
. (Two sentence limit for each):


Allelic exclusion

An individual B cell expresses only one HC locus (maternal or paternal) and one LC
locus. Ensures that antibody response can be regulated appropri
ately.


follicular dendritic cell

Cell in germinal center that displays antigen on its surface and helps to select high
affinity B cells during affinity maturation



5

component of the surrogate light chain, part of preBCR. Signals to developing B cell to
shut down HC rearrangement and induce LC rearrangement.


germinal center

site of B cell proliferation and affinity maturation during T dependent B cell response to
antigen.


defensins

small polypeptide present in skin and other tissues that kills bacteria.

Midterm 1 MCB150 Fall 2006

Name________________________________________


4

7a. (5 pts) Below is a simplified diagram of an unrearranged light chain locus with the 1
-
turn RSS shown in black and the 2
-
turn RSS in grey. Label the gene segm
ents and
indicate on the diagram where a rearrangement might take place. Sketch the rearranged
DNA locus below.











Other answers possible


7b. (6 pts) Sketch a
simplified

diagram (similar to the light chain locus diagram show
above) of an unre
arranged heavy chain locus, labeling the location of the 1
-
turn and 2
turn RSS and the gene segments. Indicate where rearrangements might take place and
sketch a rearranged DNA locus version of the heavy chain locus below the un
-
rearranged
DNA locus.









Other answers possible


7c. (5 pts) Briefly explain why receptor editing can occur efficiently with the light chain
locus, but not the heavy chain.


After rearrangement the D segments are deleted. The only remaining RSS are from V
and J segments, wh
ich cannot recombine with each other because of the 12/23 rule.











V segments

J segments

Constant Region

V segments

J segments

Constant
Region

D segments

Midterm 1 MCB150 Fall 2006

Name________________________________________


5

8. (8 pts) Below are listed several events that occur during B cell development. Fill in
the table below to indicate which statements apply to which events.





statemen
t


event

Directly
involves rag
proteins

Directly
effected by
AID mutation

Alters the DNA
encoding
antibody
proteins

Occurs after
antigen
encounter

Involves
alternative
mRNA
processing

Class switch to
IgG, IgA, IgE


*

*

*


Production of
secreted antibody




*

*

Kappa V
-
J joining


*


*



Somatic
hypermutation


*

*

*


IgD expression






*



9a. (5 pts) The left panel below shows an idealized flow cytometry plot of spleen cell
from wild type mice. In the plots to the right, fill in what results you w
ould expect for
spleen cells from a rag2 gene knockout mouse and an AID gene knockout mouse.













9b. (3 pts) For the AID knockout mouse, you decide to further investigate the defect by
examining the average affinity o
f antibodies (determining the equilibrium binding
constants) produced in mutant mice after repeated immunizations with chicken
ovalbumin (ova). Your first step is to generate a panel of monoclonal antibodies from
immunized mutant and wild type mice. Expl
ain why you need to include this step.

You need homogeneous preparations of individual antibodies in order to measure
equilibrium binding constants. These individual Ka measures can then be compared
between the knock out and wt mice. In a polyclonal prep
, one can only measure the
estimated affinity of all the different antibodies mixed together, hence you can not get a
sense of how the affinity has matured after repeated immunization.



9c. (2 pt) You notice that all of the monoclonals generated from the
knockout mice are
secreting IgM, whereas those from the
mutant (should be “wild type”)

are producing
predominatly IgG. Explain this observation.

AID is required for class switching, so mutants can only make IgM

IgG

IgG

IgG

Ig
Splenocytes from wild type mouse

Splenocytes from rag2
-

mouse

Splenocytes from AID
-

mouse

IgM

IgM

Midterm 1 MCB150 Fall 2006

Name________________________________________


6


9d. (6 pts) You decide to initially evaluat
e your monoclonal antibodies using assays that
rely on lattice formation. Under which conditions would you expect antibody
-
antigen
lattices to form:

a)

individual monoclonal antibodies with soluble, monomeric ovalbumin (ova).
no

b)

mixture of several different
monoclonals with soluble, monomeric ova.
yes

c)

individual monoclonal antibodies with cells displaying ovalbumin on their surface
as a transmembrane protein.
yes


9e. (4 pts) In order to determine the affinities of your monoclonal antibodies for
ovalbumin, yo
u decide to use the technique of Surface Plasmon Resonance using a
biosensor that detects changes in resonance angle as a read out of the degree of protein
binding to another protein immobized on a chip. You consider 2 different setups: (a)
antibody imm
obilized on chip with soluble antigen or (b) antigen immobilized on chip
with soluble antibody. Say which setup would be better for determining equilibrium
binding constants and briefly explain why.

Only (a) would work. In (b) the antigen becomes a multi
meric ligand when immobilized
on the chip. Since the antibody is dimeric, binding at one site will effect the other and K
cannot be determined. (1 pt for right answer and 3pts for correct explanation)




9f. (3 pts) After completing your affinity me
asurements, you generate the graph of
equilibrium dissociation constants (K
d
) shown below. Indicate on the graph which
samples are from mutant or wild type mice. Briefly explain your answer.











Wild type antib
odies will be higher affinity (lower Kd). Since AID mice cannot undergo
somatic hypermutation/affinity maturation, they will have lower affinity.


9g. (3 pts) You find that the antibodies from the mutant mice bind at least as tightly
(avidly) to cells dis
playing ovalbumin as those from wild type mice, in spite of the
affinity differences seen in the graph above. Explain this observation.

Avidity=valency
x
affinity. Mutant antibodies are mostly IgM=decavalent, Wt is
predominantly IgG=bivalent. The increas
ed valency of the interaction with the
multmeric ligand can compensate for the reduced affinity of these antibodies.

Mutant

or
wildtype?

Kd
d

10
-
9
M

10
-
7
M

10
-
8
M

Mutant or
wildtype?