Solution Key- 7.012 Problem Set 6 - CSAIL People

Solution Key- 7.012 Problem Set 6
Question 1
Following is the schematic of a signal transduction pathway that is activated by the binding of
Epidermal growth factor (EGF), produced by one cell type, to its specific membrane receptor on a
target cell. The major steps involved in this pathway are outlined below:
• EGF ligand binds to the EGF receptor.
• Ligand bound EGF receptors become active through phosphorylation and homodimerization.
• Active EGF receptor causes Ras to exchange its bound GDP for GTP and become active.
• Active Ras activates the kinase cascade (RAF, MEK and MAPK) through phosphorylation.
• This increases the expression of c-myc gene which results in cell proliferation.
EGF Receptor
MAPK
MAP
K
Cytosol
P
c-myc transcription
Cell proliferation
Plasma membrane
a) What would be the effect of each of the following treatments/mutations on cell proliferation?
Explain why you see this effect.
i. A mutation that results in the constitutive homodimerization of EGF receptor.
The homodimerization of EGF receptor is required for the activation of Ras-Raf –MEK-MAPK kinase cascade that
increases c-myc gene expression. Therefore a mutation that results in constitutive dimerization of the EGF
receptor will result in constitutive / uncontrolled cell proliferation irrespective of the presence or absence of EGF.
ii. Treatment of cells with a Raf specific phosphatase.
This treatment will inactivate the Raf kinase enzyme. As a result there will be no phosphorylation and activation
of MEK and MAPK. Hence there will be no c-myc gene expression and no cell proliferation even in the presence
of EGF.
iii. Treatment of cells with PD9809, an inhibitor of MAPK.
This treatment will inactivate the MAPK enzyme. As a result there will be no phosphorylation and activation of
MAPK, no c-myc gene expression and no cell proliferation even in the presence of EGF.
EGF
RAS-GTP
Nucleus MAPK
RAF
MEK
b) Consider the following cells that have mutations in different components of the EGF signal
transduction pathway.
• Mutant 1 (m1): Ras protein that continues to stay in its GDP bound form.
• Mutant 2 (m2): RAF protein that lacks its kinase domain.
• Mutant 3 (m3): MAPK that lacks a nuclear localization sequence.
• Mutant 4 (m4): EGF receptor that lacks its extracellular domain.
• Mutant 5 (m5): MAPK that is constitutively phosphorylated at its active site.
• Mutant 6 (m6): c-myc gene that has a constitutively active promoter.
P
P
P
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Question 1 continued
Complete the table for each of the following homozygous double mutants in the presence of EGF
relative to wild type (wt). Also explain the change in cell proliferation relative to wt cells in the
presence of EGF.
Mutations in
the cell
wt/wt Yes
m1/m1 and
m2/m2
m3/m3 and
m6/m6
m4/m4 and
m5/m5
c-myc
expressed
(Yes/No?)
Cell proliferation increased/decreased /unchanged? Explain.
No Decreased proliferation. Both Ras and Raf kinases are constitutively inactive and so
the cascade is never turned on to enhance c-myc gene expression that is required for
cell proliferation.
Yes Increased proliferation. Since c- myc gene has a constitutively active promoter it will
always be transcribed independent of the active/inactive state of MAPK, which results
in increased cell proliferation.
Yes Increased proliferation. MAPK is constitutively active in its phosphorylated form,
which causes increased c-myc expression that is required for cell proliferation.
c) Ligand mediated signaling events can be classified as endocrine, autocrine or paracrine. Of what is
EGF signaling pathway an example? Explain your choice.
The EGF ligand is secreted by one cell type and it mediates its effect by binding to the EGF receptors on another
cell. This is an example of paracrine signaling.
Question 2
You decide to engineer mammalian cell lines, each expressing a specific mutant variant of either the
EGF ligand or the EGF receptor (EGFR). Note: A cell line is a single type of cell which continuously grows in
culture.
Cell line-1 has a mutation that results in the deletion of only the signal sequence of EGF ligand.
Cell line-2 has a mutation that results in the deletion of only the transmembrane domain of EGFR.
Cell line-3 has a mutation that results in the deletion of both the signal sequence and transmembrane
domain of EGFR.
You incubate each of these mutant cell lines with fluorescent antibodies that specifically bind either to
EGF or the EGFR. You then observe these cell lines under the fluorescent microscope to study the
localization of EGF ligand or EGFR.
a) In cell line-1…….
i. Where do you expect to find the EGF ligand (cell membrane/cytosol/ cell culture medium/)? Explain
your choice.
A protein is secreted if and only if it has a signal sequence. In comparison a protein, in order to get localized to the
cell membrane, has to have a signal sequence and a hydrophobic /transmembrane domain sequence. Since the
signal sequence is deleted, the EGF ligand will be localized in the cytosol.
ii. For the EGF ligand…
• In the template DNA strand, where do you expect to see the base sequence that
corresponds to signal sequence of EGF (close to the 5’ end or the 3’ end)?
You would expect the signal sequence to be located near the 3’ end of the template DNA strand.
• In the mRNA transcript of EGF, where do you expect to see the base sequence that
corresponds to signal sequence of EGF (close to the 5’ end or the 3’ end)?
Once transcribed the signal sequence will be on the 5' end of the mRNA.
• In the EGF ligand, where do you expect to see the signal sequence (close to the N- or C-
terminus)?
During translation the signal sequence will be at the N-terminus of the translated protein.
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Question 2 continued
b) If cell line-2 is incubated with EGF ligand, do you expect these cells to proliferate? Answer as Yes/No
and explain your choice.
The EGF receptor variant in this cell line has the signal sequence but lacks the hydrophobic / transmembrane
domain sequence. Therefore this protein variant will be secreted by the cell in the surrounding medium. Thus the
EGF receptor will not be available at the cell membrane to bind to EGF ligand and activate Ras-Raf–MEK-
MAPK kinase pathway that is required for cell proliferation.
d) If cell line-3 is incubated with EGF ligand, do you expect these cells to proliferate? Answer as
Yes/No and explain your choice.
The EGF receptor variant in this cell line lacks both the signal sequence and the hydrophobic / transmembrane
domain sequence. Therefore this protein variant will remain localized in the cytosol. Thus the EGF receptor will
not be available at the cell membrane to bind to EGF ligand. Therefore the Ras-Raf –MEK-MAPK kinase pathway
will not be stimulated, there will be no transcription of c-myc gene and hence there will be no cell proliferation.
Question 3
a) What is membrane potential? Is membrane potential exclusively present in neurons?
All cells have an unequal distribution of charges across their plasma membrane and this results in the outside
being more positive compared to the inside. This unequal distribution of charges across the plasma membrane is
called membrane potential. This is a property of all cells and not exclusively the neurons.
b) How does a membrane potential relate to an action potential?
The resting membrane potential is an electrochemical potential of all cells, and can be exploited to do some forms
of cellular work. The action potential is the response within a neuron that can also be relayed to other cells. It is a
transient change of the membrane potential in a tiny patch of the membrane. This transient change is propagated
down the length of the neuronal axon and can result in signals from one neuron to another.
c) Shown below is a plot of axonal membrane potential as a function of time, measured at a single
position on the axon.
C
+50mV
-70mV
A
B
Time
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Question 3 continued
Complete the following table for each of the three time points indicated in the graph on page 3. Name
the phase of membrane potential that describes the time point, the channels/pumps that are involved
(Select from Na+/K+ ATPase pump, open K+ channels, voltage gated Na+ channels, voltage gated K+ channels,
voltage gated Ca+ channels, ligand gated channels) and explain how these channels/pumps maintain that
phase.
Time
point
A Resting
Phase All the ion channel/
Pump involved
Na+/K+ pump
& open K+ channels
B Depolarization Voltage gated Na+
Channels
C Repolarization Voltage gated
K+Channels
Explanation
Na+/K+ pump creates a difference in concentration for both
Na+ and K+ across the cell membrane. The relative
concentration of Na+ is high outside as compared to inside
and the relative concentration of K+ is low outside as
compared to inside the cell. The open K+ channel always
allows K+ to flow out of the cell. As K+ moves out of the cell,
down its concentration and electrical gradients, the inside of
the cell becomes negative with respect to the outside of the cell.
K+ moves out of the cell until the driving force of the
concentration gradient equals the driving force of the
electrical gradient, which for most cells sets the resting
membrane potential at -70 mV.
These channels, once activated, allow the Na+ ions to move
into the cells down its concentration and electrical gradient
making the inside more positive relative to outside thus
depolarizing the membrane.
These channels, once activated, allow the K+ ions to move out
of the cells making the inside more negative relative to outside
thus repolarizing the membrane.
d) The following schematic shows the tracings of action potentials recorded in the same type of
neurons isolated from several mouse mutants. Each mutant has a defect in only one of the channels or
pumps involved in action potential. You stimulate each neuron, record the electrical activity and plot
your data below. Using the data below, state the channel /pump that most likely has the defect in each
of the mutants and explain how this defect may account for the observed changes in the action
potential compared to the wild type neuron.
Wild type Mutant A
mV mV
770m
V70
mV
Time
Mutant B
770m
V70
mV
mV mV
Time
Time 770m
V70
mV
Time
Mutant C
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Question 3 continued
Mutants Which channel or pump is
most likely defective?
A Voltage gated K+ channel or voltage
gated Na+ channel.
B Na+K+ ATPase Pump or open K+
channel
C Voltage gated Na+ channel
How does this defect account for the observed
change in the action potential?
The cell is slow to repolarize, suggesting that either the
Voltage-gated K+ is slow to open or fails to open or the
Voltage-gated Na+ is slow to close.
Na+/K+ pump and / or K+ pump are less active than usual.
No action potential is initiated even though threshold is
reached. Voltage-gated Na+ channel fails to open.
e) A student shows you some new data, obtained from a newly isolated neuronal cell line and claims
that the amplitude of action potential varies with the amount of stimulation (1X, 5X and 25X). You are
quite skeptical about these data and think the student has made a mistake. Explain.
The student has made a mistake because the frequency of the action potential can increase with the stimulus but
the amplitude remains unchanged.
f) Indicate, on the graph below, how a normal neuron would react if the strength of stimulus is
increased from 1X to 5X or 25X.
mV mV mV
mV
Time Time Time
Stimulus IX 5X 25X
Question 4
Cystic fibrosis (CF), is an autosomal recessive disorder. This is caused by a defect in a gene called the
cystic fibrosis transmembrane conductance regulator (CFTR) gene located on chromosome 7 that makes
the CFTR protein. This protein functions as a chloride channel across the cell membrane, controls the
movement of water in tissues and maintains the fluidity of mucus and other secretions.
a) Below are three different options for a hypothetical linear stretch of amino acids present in the CFTR
protein channel.
Option 1: ……… - arg-cys-ala-his-trp-val-glu-lys-arg ……….
Option 2: ……… -leu-ala-gly-cys-phe-val-gly-trp-met……….
Option 3: ……… -leu-ser-gly-lys-phe-val-gly-tyr-met ………..
i. Which option most likely forms a stretch of CFTR receptor that spans/comes in contact with the
lipid bilayer? Explain why you selected this option.
Option #2 since this option contains a continuous stretch of non-polar/ hydrophobic amino acids.
ii. The CFTR protein has a regulatory binding site that allows its activation by phosphorylation.
Which of the above options can most likely be phosphorylated.
Only serine, tyrosine and threonine amino acids have -OH groups in their sidechains and can therefore be
phosphorylated. Since ser and tyr are present only in option #3 therefore this is the most likely option for
phosphorylation.
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Question 4 continued
b) CF trait is most common in people with European heritage that show the deletion of a codon
corresponding to phenylalanine amino acid at 508 th position in the CFTR protein. In general, the alleles
that correspond to an autosomal recessive trait may exist at a fairly high frequency in some regions of
the world. Provide two reasons to explain why the CF trait is so common in some regions.
It may so be that the allele responsible for the occurrence of this disease somehow provides resistance to some other
infection, which is otherwise fairly common in these regions. Hence many individuals in these regions are the
carriers for the disease but are phenotypically normal and yet have resistance to some other infection that is fairly
common in that region. This may also be owing to the bottle neck and founder effect where most people in the
region have a common ancestry.
c) Blood test is one of the diagnostic methods for CF. You have the blood samples from 5 newborns and
you want to PCR amplify and sequence the CFTR genes from these samples to look for mutations.
Based on what information would you design a primer pair that you could use to PCR amplify and
sequence the CFTR gene from these newborns?
The sequence of the wild type CFTR gene is known. You will use this information to design the primers to PCR
amplify and test the CFTR genes in these newborns. Or you can use the flanking regions of the CFTR gene to
design your primer pair.
You also analyze the A locus that is close to the CFTR gene. The only difference between the wild type
allele (A) and the allele associated with the CF ( A’) is that A’ allele lacks an internal EcoR1 restriction
site which is present in the A allele.
EcoR1
A allele
A’allele
The A locus from the DNA of five newborns is amplified using PCR. The PCR products are digested
with EcoR1 and the resulting fragments are separated by agarose gel electrophoresis as shown below.
1 2 3 4 5 Normal
b) Based on the PCR analysis of the A locus, identify the newborns as normal/ having CF / having CF
trait and write the genotypes at the A locus for individuals #1 through #5.
Samples Genotype at the A locus Normal/CF trait/ CF disease
1 A’A’ CF disease
2 AA Normal
3 AA Normal
4 AA’ CF trait
5 AA’ CF trait
Control AA
(-)
A’
A
(+)
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Question 4 continued
e) Individuals with CF often experience respiratory illness for which they are prescribed a wide
spectrum of antibiotics. Even after being treated with antibiotics these patients may ultimately require
lung transplant. Explain why.
Excessive and continuous treatment with antibiotics creates an environment for the bacterial cells to develop
resistance against antibiotics. This may ultimately lead to a point where lung transplant remains the only option
for the affected individual.
f) In 1985, the specific gene that causes cystic fibrosis was identified. Recently, laboratory tests of gene
therapy have been successful. Explain how gene therapy may be useful in treating these patients.
Gene therapy is the insertion of genes into an individual's cells and tissues to treat a disease such as CF in
which a deleterious mutant CFTR allele is replaced with a functional one. Gene therapy may be classified into the
following types. Germ line gene therapy; In the case of germ line gene therapy, germ cells, i.e., sperm or eggs,
are modified by the introduction of functional genes, which are ordinarily integrated into their genomes.
Therefore, the change due to therapy would be heritable and would be passed on to later generations. This new
approach, theoretically, should be highly effective in counteracting genetic disorders and hereditary diseases.
However, many jurisdictions prohibit this for application in human beings, at least for the present, for a variety of
technical and ethical reasons. Somatic gene therapy; In the case of somatic gene therapy, the therapeutic genes
are transferred into the somatic cells of a patient. Even if 10-15% of cells receive the wild type copy of the gene,
the amount of protein made may be sufficient enough for the affected individual to lead a normal life. With such a
therapy any modifications and their corresponding effects will be restricted to the individual patient only, and
will not be inherited by the patient's offspring.
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