Answers Problem Set 4 2005
Problem 1:
MAb1: MAb1 is specific for the A allotype of gamma 1. You conclude this because it will react
with all of the gamma 1 bearing cells in the A strains mouse, none in the B strain mouse, and half
in the AB mouse in which half of the lymphocytes would be expected to be of each allotype.
Mab2: MAb2 is specific for the B allotypes for the same reasons.
MAb3: Mab3 is specific for the gamma 1 isotype because it recognizes all of the gamma 1
bearing cells in all three strains.
The pattern expected for a BC mouse:
If you mix MAb1 and MAb2 the pattern you would expect for the AB mouse is:
Mabs 1 and 4 are of the same specificity.
Non-labeled Mab4 binds to the antigen on the surface and prevents the subsequent binding of
labeled Mab1.
If you add unlabeled MAb2 before adding the indicated labeled Mab the pattern would be:
The antibodies produced by a hybridoma are monoclonal and contain only one type of heavy and
light chain. Hence when you take them apart and reassemble they, they correctly reassemble the
same antibody that you had before. In contrast, the anti-DNP-BSA present in an antiserum is
Answers Problem Set 4 2005
heterogeneous with a very large number of different variable regions. Therefore the probability
of the correct light and heavy chain coming together to make a functional antibody is small.
Answers Problem Set 4 2005
Problem 2:
In the transgenic there are B cells expressing both human and murine  chains.
This is not what you would expect for a normal mouse in which expression of a
transgene with a membrane exon prevents expression if the endogeneous  gene.
However this observation may explain why these animals develop autoimmune
disease.
Mouse JH - rearrangement in Hy but not liver of Tg or non-Tg - rearrangement
correlated with expression
Human JH - gene was rearranged before used for producing transgenic so some
pattern in liver and Hy; functional absent from non-tg as expected
Yes - because have murine and human s.
 or







Problem 3:

#1
#2
#3
#4
#5
ORDER:
IgM 
IgA 
IgG 
IgE 
IgA 

In a normal mouse, the  light chain alleles are the first to rearrange. If there has been a
successful  light chain rearrangement then that B cell will express that functional  light chain
on the surface of its cell and that B cell will never rearrange its light chain alleles due to allelic
exclusion. If both attempts to rearrange the  alleles result in the production of non-functional 
light chains, then the B cell will attempt to rearrange its  light chain alleles. As a result, if a
mature B cell has both its  and  light chain alleles rearranged, then you know that it must be
expressing a  light chain on its surface. What is weird about the unique mouse is the fact that
out of 55 hybridomas that were looked at, all of them seem to be expressing  light chains!
Normally, you would expect to see at least some B cells that have functionally rearranged their 
light chains! Remember, at this point we have not seen the FACS data so we cannot assume that
allelic exclusion is not occurring!
Answers Problem Set 4 2005
By analyzing the FACS data, we can conclude the following:


There are B cells with IgMa on their surface, others with IgMa on their surface,
others with IgMb on their surface and others with IgMb on their surface. There is
nothing surprising about this.
What is surprising is that we see B cells with BOTH IgMa and IgMb heavy chains on
their surface! We also see B cells with BOTH  and  light chains on their surface!
This FACS data is not consistent with the interpretation of the Southern blot because we said that
all of the hybridomas we made seemed to be expressing the  light chain on their surface. We
argued that this was so because all of the  light chain rearrangements must have been nonfunctional. As a result, we assumed that there were no hybridomas (B cells) that were expressing
 light chains. The FACS data is not consistent with this interpretation because according to this
analysis, we do see cells that have the  light chain on their surface.
This unique mouse seems to be different from what we know about a normal mouse because
allelic exclusion doesn’t seem to be occurring in the B cells of this unique mouse. If allelic
exclusion were occurring, we shouldn’t see BOTH the IgMa and the IgMb heavy chains on the
surface of one cell. We also wouldn’t see BOTH the  and the  light chains on the surface of
one cell.
These are the results that you would expect because the supernatant should contain secreted
antibody. When you run an SDS-PAGE gel, you are breaking the disulfide bonds that hold the
heavy and the light chains together. As a result, when you run the gel, transfer the gel to a
membrane and conduct a western with a reagent that detects heavy and light chains, you see a
band that runs slower (top) and a band that runs faster (bottom). The top band represents the
heavy chains and the bottom band represents the light chains.
We determined above that there is no allelic exclusion occurring in the B cells in this unique
mouse. As a result, you could have one B cell expressing a functional IgMa heavy chain, a
functional IgMb heavy chain, 2 functional  light chains, and 2 functional  light chains
(assuming all rearrangements were functional). You could imagine that any 2 functional heavy
chains could pair with any 2 functional light chains. The result would be a B cell that has many
different antibodies on its surface that have different specificities! In other words, each of these
different antibodies would be specific for a different antigen. Remember, this is on the surface of
ONE B cell. If you immunized this unique mouse with the Dang protein and then made
hybridomas, you know that the hybridomas must be producing an antibody that recognizes the
Dang protein. After all, the only way that B cell would respond to the Dang protein is if it had an
antibody on its surface that recognized it. As a result, we know that all 5 of these hybridomas
must have membrane bound antibodies and secreted antibodies that recognize the Dang protein.
When we collect the supernatant and remove all of the antibodies that bind to the Dang protein
Answers Problem Set 4 2005
and discard them and then run what REMAINS on the gel, transfer the gel to a membrane, and do
a western, this is why we see what we see in each of the lanes:
Lane 2 Obviously, this particular B cell (and now hybridoma) must have produced only one
functional heavy chain and one functional light chain. The antibody that it produced was specific
only for Dang. When you remove all of the Dang-specific antibodies, nothing remains in the
supernatant and that is why you no longer detect heavy and light chains.
Lanes 1, 3, 4, and 5 These B cells must have produced more than one functional heavy
chain AND/OR more than one functional light chain. As a result, these B cells had many
different antibodies on their surface that had different specificities. Due to the fact that this B
cell responded to Dang, one of those antibodies must have been specific for Dang. When that B
cell became a plasma cell, it would not only be secreting antibodies specific for the Dang protein,
but would secrete all of the other antibodies that had different specificities as well. As a result,
even though you remove all of the Dang-specific antibodies, you will still have all of the other
antibodies with different specificities left behind! When you run this on a gel you still see heavy
and light chains! You see less protein then before because you have removed the antibodies that
were specific for the Dang protein.
The problem with the transgene is that it violates the 12/23 rule and hence VL cannot rearrange to
JL. It can be fixed by changing the spacing of one heptamer/nonamer.
For the human heavy chain transgene you fail to see surface expression therefore it probably
lacks the transmembrane region. Lacking the transmembrane region it fails to turn-off the
rearrangement of the murine mu ge
Problem 4
a. The B lymphocytes simultaneously express  and light chains. The normal mouse expresses
either  or . Expression of two different light chains could result in the expression of antibodies
of two specificities which might lead to unwanted autoreactivity.
b. You would expect to precipitated 50% from the normal mouse (half of the B cells express ;
see the FACS); You might precipitate up to 100% from the sick mouse since B cells expressing
both  and  might have mixed molecules with one  light chain.
c. Yes, there is a defect in signaling such that the “sick” mouse fails to send the appropriate
signal to stop L chain rearrangement when a functional antibody is expressed on the surface.
d.
Answers Problem Set 4 2005
Non-transgenic Mouse
murine
murine
human
NormalTransgenic Mouse
Weird Transgenic Mouse
human 
human 
e. There is allelic exclusion of the heavy chains so that heavy chain expressed on the surface can
send the correct signal.
f. Cells not expressing IgM - i.e. non-B cells (such as T cells or macrophages) or B cells that
have isotype switched. They cannot be cells with a  of a different allotype since there are only 2
alleles.
g.
Tran sgen ic Mouse with memb ra ne d ele te d mu
murin e
mu rin e
Tra nsge nic Mo usewi th co mple te hum an mu
hu man 
hu man 
h. The expression of the human mu with the membrane exon on the surface prevents
rearrangment of the murine mu. [For most transgenics and for all of the examples given in class
a rearranged heavy chain gene is used. This rearranged gene is expressed before endogenous
rearrangement can take place so you don’t see expression of the endogenous gene. However,
since it was not explicitly stated that a rearranged gene was used, credit was given for expression
of both endogenous and transgene as long as allelic exclusion was seen. Remember from the
results shown above the “sick” mouse shows allelic exclusion at the H chain locus even thought
it does not show allelic exclusion at the L chain locus]. The transgene lacking the membrane
exon would
not be expressed on the surface and would not inhibit rearrangement of the endogenous gene.
FACS analysis looks at molecules on the surface of the cell.