Unformatted text preview: 7.06 Spring 2004 PS 3 Key 1 of 7 Problem Set 3 2004
1) You are a geneticist investigating the signal transduction pathway that allows yeast
cells to sense and respond to mating factors. It has been known for some time that
mating type a cells secrete a factor while mating type alpha cells secrete alpha factor.
Mating type a cells respond to alpha factor by forming a mating projection known as a
“shmoo.” Conversely, mating type alpha cells respond to a factor by also forming a
shmoo. Assembly of the shmoo structure allows yeast cells of opposite mating type to
find one another and enhances fusion of two haploid cells. This structure can be easily
visualized by a conventional light microscope. a Alpha factor
a shmoo a factor
a a a. How would you isolate mutants that are unable to sense or respond to the mating
You can start with a culture of mating type a or alpha cells. Mutagenize the cells by
chemicals or irradiation. Add a factor if the cells were mating type alpha or alpha
factor if they were mating type a. Examine the cells under a light microscope and
look for ones that fail to project a shmoo.
b. How would you isolate genes responsible for the observed defect?
You would obtain a cDNA library from a WT strain and transform it into your
collection of mutants. The plasmid that rescues the ability to shmoo should contain
a WT copy of the gene that is defective in the mutant.
c. Your competitors recently published a paper that defines the genetic network
responsible for the yeasts’ mating response. Preliminary evidence suggests that the
mating type factors bind to a G-protein coupled receptor that is linked to a signaling
cascade that might resemble a MAP kinase-like pathway. The genes known to be
involved are shm1, shm2, shm3, shm4 and shm5 (shm for shmoo). The authors failed to
order these genes into a pathway. In order to salvage your scientific career, you perform
some epistasis analysis to determine the order of function of these genes.
Using a set of mutants in the mating response you produced the following data:
i. A constitutively active shm4 mutant is unable to rescue the phenotype (no shmoo) of a
shm3 deletion. 7.06 Spring 2004 PS 3 Key 2 of 7 ii. The expression of any other constitutively active gene in the pathway cannot rescue
the phenotype of shm5.
iii. The phenotype of a shm3 deletion cannot be rescued by expression of constitutively
active forms of shm1 or shm2.
iv. A constitutively active shm2 cannot rescue the phenotype of non-phosphorylatable
v. A constitutively active shm4 rescues the phenotype of a non-phosphorylatable shm1.
Based on this data, propose an organization of these genes in a linear pathway.
shm2 ‡ shm1 ‡ shm4 ‡ shm3 ‡ shm5 2) Answer the following questions pertaining to MAP kinase signaling and Ras.
A) What are some general similarities and differences between G-protein coupled
receptor signaling and MAP kinase signaling?
Both pathways can lead to activation of MAP kinase. Both involve GTPase
switch proteins (Ga and Ras) that are on when bound by GTP and off when
bound by GDP. The Ras GTPase needs a GEF (Sos) and a GAP to function
properly, whereas the Ga subunit does not. Ras also does not bind the hormone
receptor directly, in contrast to the G-protein, which is activated by receptor
binding. G-protein coupled receptor signaling leads to rapid amplification of
signal through the production of a 2nd messenger molecule, the MAP kinase
pathway does not.
B) What is the importance of the GTPase activating protein (GAP) in the MAP
kinase pathway? How does a mutation that prevents GAP binding to Ras promote
The GAP attenuates Ras signaling by promoting GTP hydrolysis. Without the
GAP, the Ras GTPase is not very active and will remain GTP bound for an
extended period of time. Activated Ras leads to transcription of genes that
promote cell division, thus hyperactive Ras will lead to uncontrolled cell growth
and potentially cancer.
C) Why is an inhibitor of farnesyl transferase a potent anti-cancer drug?
An inhibitor of farnesyl transferase will prevent Ras from attaching to the
membrane (Ras is anchored in the membrane by a farnesyl group), and Ras 7.06 Spring 2004 PS 3 Key 3 of 7 cannot be activated without stable association with the plasma membrane.
Because many tumors depend on activated Ras for growth stimulation, they are
very sensitive to disruptions in the MAP kinase pathway and can be selectively
killed with this drug.
D) Why is the fly eye such a powerful experimental system in which to study
receptor tyrosine kinase signaling?
Drosophila is a genetically tractable organism, and the sevenless RTK pathway that
functions within the compound eye is not necessary for fly survival. Thus one can
perform a genetic screen for mutants in this pathway, and the isolated mutant flies
will be completely viable. By looking for enhancers and suppressors of the
sevenless mutation, investigators identified many key components of the RTK
E) In yeast and higher eukaryotes, multiple MAP kinase pathways exist within a cell.
They are activated by different signals and lead to different cellular responses, yet
share many components. How is it that these pathways can use the same
components yet achieve different responses within the same cell?
Large scaffold proteins exist that stabilize the interaction of specific downstream
effector proteins. These scaffold proteins allow specific kinases to interact with
components of one signaling pathway but not another, establishing specific signaling
modules within the cell. 3. (i) You have a human cell line that doesn’t respond to EGF. Your advisor suggests that
you do a functional complementation assay to identify the gene that is mutated. Explain
how you would do this, making sure to specify the source of the genetic material in your
library, if you choose to use one. In your assay you need a way to monitor for cells that
now respond to EGF, which of the following would be most useful for doing this and
a) an antibody to phosphorylated histone H1 protein, a known substrate for MAPK
b) a version of the above cell line that has the GFP cDNA under the control of the serum
response element (SRE)
You would make a cDNA library from a WILDTYPE human cell line i.e. one that
responds to EGF. You would transfect this into your MUTANT cell line and assay
for rescue of the phenotype - it now responds to EGF. A stable cell line expressing
GFP under control of the SRE would be the preferred way to assay for
complementation a s r escue c ould b e e asily v isualized w ith a f luorescence
microscope. If you chose to use the antibody to phosphorylated histone H1 you
would have to make replicates of all your transfected cells as you would need to lyse
the cells and run a western to assay for presence of phosphorylated histone H1. 7.06 Spring 2004 PS 3 Key 4 of 7 (ii) The cDNA you identify from your assay encodes the EGFR. For each of the
following EGFR mutants explain how and why levels of phosphorylated MAPK would
differ, if at all, from cells expressing wildtype EGFR, in the absence and presence of
a) a mutant EGFR that lacks the cytosolic tyrosine kinase domain
b) a mutant EGFR that is expressed 1000-fold more than wildtype such that the
membrane is saturated with EGFR
c) a mutant EGFR whose tyrosine residues in the kinase domain have been mutated to
a) absence of EGF: same as wildtype as the pathway should be off
presence of EGF: same response as in the absence of EGF since no signaling can
occur since the tyrosine kinase domain is missing
b) absence of EGF: levels of phospho-MAPK should be close to what you would see
in the wildtype when EGF is present. This is because there are so many receptors on
the cell surface that they will dimerise and signal even in the absence of hormone
presence of EGF: levels of phospho-MAPK should be the same or higher than
wildtype when EGF is present since there is constant signaling through the pathway.
If the levels of EGF are not high enough to saturate all receptors in the wildtype
cells then you would expect increased levels of phospho-MAPK in the mutant.
c) absence of EGF: same as wildtype as the pathway should be off
presence of EGF: same response as in the absence of EGF since no signaling can
occur since the tyrosine kinase domain can't be phosphorylated
(iii) You have a kinase dead version of EGFR (i.e. it is full-length protein but it can't
phosphorylate another EGFR monomer) tagged with a His tag to permit purification.
You also have a wild-type EGFR that is tagged with a “Myc” epitope to allow
purification. You would like to investigate whether EGFR molecules auto-phosphorylate
themselves (intra-molecular phosphorylation) or trans-phosphorylate one another (intermolecular phosphorylation).
You mix the two versions of EGFR together and add
radioactive ATP. Which version would isolate to test whether EGFR molecules transphosphorylate or autophosphorylate each other (the status of EGFR phosphorylation can
be known by detecting the incorporation of radioactive phosphate on the protein)?
You would isolate the kinase dead version and determine if it is phosphorylated. If
autophosphorylation were the mechanism, WT EFGR should not be able to 7.06 Spring 2004 PS 3 Key 5 of 7 phosphorylate any kinase dead EGFR. However, if it were transphosphorylation,
WT EGFR can phosphorylate kinase dead and other WT EGFR molecules.
If you purify WT EGFR using the Myc epitope (e.g. by antibody purification) and find
that it is phosphorylated, does it help you differentiate between trans – vs. autophosphorylation? Why or why not?
No, it does not. Because the EGFR molecules are WT, their phosphorylation can be
due to an intramolecular, autophosphorylation mechanism or an intermolecular
If you purify the kinase dead EGFR using the His tag and find that it is phosphorylated,
does it help you differentiate between the possibilities? Why or why not?
Yes, because the only way kinase dead EGFR molecules can become phosphorylated
is if the WT kinase active molecules phosphorylated them. Therefore, EGFR
molecules can phosphorylate each other in trans.
Imagine that you have a cell-line that expresses equivalent amounts of wildtype EGFR
and a mutant EGFR that is constitutively active. How would this affect signaling in the
presence and absence of EGF?
In the absence or presence of EGF levels of phospho-MAPK will be equivalent to the
level seen in wildtype cells in the presence of EGF. 4. You are studying cell/cell signaling in Wnt deficient cell line. In this cell line, Wnt is
required to induce the Wnt pathway, which leads to expression of hedgehog (Hh). Once
Hh is expressed, it is secreted by the cells inducing the Hh signaling pathway. Induction
of the Hh pathway leads to cell proliferation in this cell line; Without Wnt, the cells don’t
die, but they do not undergo cell division. You make several mutants of proteins in Wnt
and Hh pathway components and overexpress them (50 times more than wild type) in the
cell line. You measure cell proliferation before and after Wnt treatment for each of the
mutants. For the following scenarios determine whether the cells would proliferate
without Wnt and whether addition of Wnt would induce cell proliferation. Explain your
A) You overexpress a B-catenin mutant that has a serine to alanine mutation that
prevents its phosphorylation by GSK3.
Answer: This mutation would cause constitutive expression of Hh independent of
Wnt so the cells would always be dividing. 7.06 Spring 2004 PS 3 Key 6 of 7 B) You overexpress a Dsh mutant that cannot inhibit GSK3.
Answer: Without Wnt, cells would not proliferate. Addition of Wnt would not
induce cell proliferation b/c the mutant Dsh cannot keep GSK3 from
phosphorylating _B-Catenin so _B-Catenin cannot induce Hh expression.
C) You overexpress a GSK3 mutant that is constitutively active and a _B-Catenin mutant
that cannot be phosphorylated.
Answer: If _-Catenin cannot be phosphorylated, then it would accumulate in the
cytosol and subsequently translocate to the nucleus and active Hh. This would
occur independent of Wnt, so cells would always be proliferating. It does not matter
if GSK3 is always on b/c the _B-Catenin is unresponsive to phosphorylation. D) You overexpress a constitutively active form of Ptc.
Answer: Without Wnt there would be no cell proliferation. Addition of Wnt would
not induce cell proliferation b/c a constitutively active form of Ptc would not relieve
its inhibition of Smo even in the presence of Hh so no proliferation would occur.
E) You overexpress a mutant of Smo that is insensitive to Ptc inhibition.
Answer: Cells would proliferate independent of Wnt signal. If Ptc cannot inhibit
Smo, then Smo would be constitutively active and always inducing cell proliferation.
This mutant would proliferate in the absence of Hh, thus it would not need Wnt to
induce Hh expression.
5) Cell junctions stabilize interactions and promote local communication between
2+ A) Fill in the following table for each type of cell junction: B) Specify how Ca affects
each of these cell junctions.
Tight junction Adherens
junction Cell-cell or cellECM interaction?
Cell-cell Cell-cell Extracellular
Occludin & claudin
from adjacent cells E-cadherins from
adjacent cells Intracellular
Membrane occludin Impermeable
- ZO-1/2/3 - actin
Rigid cellMembrane Ecell
cadherin - / attachment
catenins œ actin and 7.06 Spring 2004 Desmosomes Gap junction PS 3 Key Cell-cell Cell-cell 7 of 7 desmoglein and
Connexin hemichannels from myosin belt
- desmoplakin - Strong cell
cell keratin filaments attachment
of ions &
cytosols adjacent cells Hemidesmosomes Cell-ECM Cellular integrins ECM laminin Focal adhesion Cell-ECM Cellular integrin –
ECM fibronectin Membrane integrins
- plectin - keratin
- vinculin - actin
attachment 2+ Tight junctions are disrupted in the absence of Ca in the medium (unknown
Adherens junctions and desmosomes are disrupted in the absence of Ca in the
medium since these structures contain members of the cadherin family and these
are Ca - dependent.
The channels of gap junctions are closed in the presence of high intracellular Ca
As far as we know, hemi-desmosomes and focal adhesion are not affected by Ca .
2+ 2+ 2+ 2+ ...
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- Spring '11