Download 7.06 Problem Set #7, Spring 2005

7.06 Problem Set #7, Spring 2005
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a. You are a grad student studying genetic alterations that lead to tumorigenesis.
While plating primary cell cultures from a healthy person and a tumor from a cancer
patient, you mix up the samples. A colleague suggests that you use the differences in
growth properties between the two samples to distinguish them. Describe how the
growth properties of cancer cells differ from those of healthy primary cells.
Unlike normal primary cell cultures, cancer cells will grow indefinitely in culture. Normal
primary cultures will stop dividing after several passages. (A passage is when cells are replated at low density to allow cells to grow and divide up to high cell density again.) Cancer
cells also lack contact inhibition, and will grow on top of each other once the cells have
divided enough times to form a confluent monolayer, unlike normal cells. In addition,
cancer cells will grow in soft agar, unlike normal cells. Finally, cancer cells will often grow
in media that does not contain any growth factors, while normal primary cell cultures have
a strict dependence on growth factors.
b. When studying the cancer cells from part a), you discover that there is high Ras
activity in the cells, but you can find no mutations in the Ras gene. You then start
investigating different Receptor Tyrosine Kinases. You find by immunofluorescence,
using a monoclonal antibody against the kinase domain, that one specific RTK is now
located throughout the cytosol instead of in the plasma membrane. You then perform
Western blot analysis on protein extracts from these cancer cells, and find that the RTK
is much larger than normal. What kind of mutation in the RTK is indicated by these
results, and how does that specific type of mutation lead to cancer?
The chromosome containing the gene that encodes the RTK protein must have undergone a
translocation with another chromosome, so that a chimeric gene was formed. A chimeric
gene is a gene that is a fusion between two different normally occurring genes that are
usually physically separated. In this case, a chimeric gene must have been formed between
a gene that encodes a cytoplasmic protein and the gene that encodes the RTK. This
chimeric gene, when transcribed and translated, now produces a chimeric protein whose N
terminus is from a random cytoplasmic protein, and whose C terminus is from the kinase
domain of the RTK. The protein fused to the kinase domain probably is a dimer (or some
sort of oligomer), so that the kinase domains of the RTK are now constitutively dimerized,
and therefore constitutively active, due to the constitutive dimerization of the N terminus of
the chimeric protein. Constitutively active RTK would lead to constitutive activity of Ras,
and thus the promotion of growth and division when it is inappropriate for the cells to
grow and divide.
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c. In addition to differences in growth properties, cancer cells also have morphological
differences, such as a disorganized cytoskeleton. The Rho family of small GTPases is
involved in cytoskeleton regulation, and the pathways that Rho proteins are involved in
are implicated in cancer development. Their regulation of the cytoskeleton is thought to
be important in cell migration during cell division and invasion of other tissues. In an
experiment performed a decade ago, it was found that transfecting NIH 3T3 cells with a
construct that expressed a specific mutant form of RhoA could transform these cells.
Based on this experiment, would you predict that RhoA is a proto-oncogene or a tumor
suppressor gene? Would you predict that the RhoA mutation is a gain-of-function
mutation or a loss-of-function mutation? Explain your answers.
RhoA is a proto-oncogene, and this mutation must be a gain-of-function mutation. Mouse
cells and human cells contain most of the same genes, and very important human genes
such as RhoA almost always have highly conserved homologs in the mouse. When you
transform this mutant version of RhoA into mouse cells (which are RhoA+/+), you see the
phenotype of transformation. This means that the mutant form of RhoA is causing a
dominant phenotype. Most dominant phenotypes are caused by gain-of-function
mutations. If the RhoA mutation had been loss of fuction mutation, then the transfection
of such a mutated gene into RhoA+/+ cells would have made no difference (as long as the
RhoA mutation was not a dominant negative mutation!). Gain-of-function mutations occur
in proto-oncogenes, and thereby cause cancer. Thus the wild-type version of the RhoA
gene is a proto-oncogene.
d. In class, we learned about the experiment by the Weinberg lab, in which an
oncogenic version of Ras was discovered in cells from a human bladder carcinoma.
This was done by transfecting human carcinoma DNA into NIH 3T3 cells (which is a cell
line that can grow indefinitely) and looking for the formation of foci (clumps of cells that
grow on top of each other).
When oncogenic Ras is transfected into Rat Embryonic Fibroblast (REF) cells (a
wild-type primary cell culture), it could only partially transform the cells. However, if a
hyper-active mutant version of the myc gene is introduced into the REF cells along with
the oncogenic version of Ras, the cells are fully transformed. Why did the REF cells
require another oncogene to become transformed, while NIH 3T3 cells did not?
Unlike the REF cells, the NIH 3T3 cells are fibroblast cells that already contain mutations
(such as the deletion of the genes encoding the cell cycle inhibitor p16) that allow indefinite
division. Transformation requires multiple mutations to bring about the loss of regulation
over cell proliferation.
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2. You are an oncologist working at Children’s Hospital in Boston.
Two patients come
to you with retinoblastoma. Based on their descriptions below, discuss whether the
disease is familial or sporadic, and why:
a. The patient is a 15-year-old boy with a retinoblastoma in one eye.
This patient has sporadic retinoblastoma. The tumor has appeared in only one eye, which
suggests that a very rare event has occurred. This event is the loss or mutation of both
copies of the Rb gene. Each somatic cell in this patient’s body is homozygous for the WT
allele of Rb, except for the cells in the tumor, which are Rb-/-. Since the likelihood of
mutating or losing both copies is extremely low, it has taken fifteen years for this boy to
acquire a mutation in each of the two Rb genes in a single cell.
b. The patient is a 3-year-old girl with a retinoblastoma in both eyes.
This patient has familial retinoblastoma. Usually, the chances of losing or mutating both
copies of the Rb gene are very low. This patient has developed two tumors at a very young
age, indicating that she had at least two cells, one in each eye, which accumulated
mutations in both copies of the Rb gene. This indicates that this patient had a
predisposition to cancer, in the form of an inherited mutant allele of Rb. Each somatic cell
in this patient’s body is heterozygous for the Rb gene (Rb+/-), except the cells of the tumor,
which are Rb-/-. In each eye, at least one cell acquired the second mutation in Rb
necessary to result in a tumor. There is a greater probability for this one mutational event
per cell to occur than for two mutational events per cell to occur, as discussed in part a) of
this question.
c. You are talking with a colleague at the hospital about familial inheritance of a
predisposition to cancer. Your colleague says that patients with the familial form of
retinoblastoma must have had a parent with retinoblastoma. Why is this not always the
case?
Many children born with a predisposition to retinoblastoma do have a parent that
also had the disease. The parent has a mutated copy of Rb in every somatic cell in his/her
body. Therefore, 50% of the germ cells (sperm or egg) in this parent contain the WT Rb
allele, and 50% of the germ cells contain the mutant allele of Rb. Thus either copy maybe
passed on to the child, who has a 50% chance of being predisposed to retinoblastoma.
Alternatively, a parent may be homozygous for the WT allele of Rb, but through a DNA
repair or replication error during gametogenesis, one of the copies of Rb became lost or
mutated in a gamete or gamete progenitor cell. If a child is produced from this gamete,
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then the child will inherit this mutant allele of Rb and thus have one mutant copy of Rb in
every cell in the child’s body. Thus he or she will be predisposed to retinoblastoma, even
though the parent was not predisposed to retinoblastoma.
d. The normal function of the Rb gene (that is mutated in retinoblastomas) is to regulate
cell cycle progression through control of a family of transcription factors called E2F.
The general pathway Rb is involved in is shown below.
p16
--|
Cyclin D/Cdk4
--|
Rb
--|
E2F 
upregulation of S phase genes
Which of the genes in this pathway are potential tumor suppressor genes and why?
The p16 gene and the Rb gene are the two tumor suppressor genes shown in this pathway.
The WT function of tumor suppressor genes is to negatively regulate cell growth, or “apply
brakes” to cell cycle progression in response to the proper signals. It is upon loss of both
copies of this gene that diploid cells can begin to undergo unregulated cell proliferation,
which may lead to cancer. The normal functions of both p16 and Rb are to inhibit cell
cycle progression, and thus both are potentially tumor suppressors. To fit the definition of
a tumor suppressor, a gene must not only have the characteristic just described, but also
must be found mutated in cancers. In fact, both p16 and Rb are found mutated or deleted
in human cancers, and are thus tumor suppressors.
e. To better understand retinoblastoma, you remove cells from a retinoblastoma and
culture them. You find these cells contain the same exact loss of function point
mutation in both copies of Rb. Describe two realistic ways that this may have come
about during tumorigenesis.
It is very unlikely that both copies of a gene will spontaneously and independently
acquire the same point mutation. Both mitotic recombination and non-disjunction are
examples of events are examples that can lead to loss of heterozygosity (LOH) in the Rb
gene such that both copies of Rb contain the same point mutation.
These cells may have acquired two identical point mutations through mitotic
recombination. In this case, one copy of the gene acquired a point mutation. The cell may
have undergone DNA damage in the region of the WT copy of Rb, and then the cellular
DNA repair machinery repaired the break using the homologous chromosome as a
template for DNA synthesis. Because the homologous chromosome contains a point
mutation in Rb, this mutation will be copied onto the other chromosome. Now both copies
of the Rb gene would have the same point mutation, and there will be no functional copy of
Rb.
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These cells may also have acquired two identical point mutations through nondisjunction (NDJ) during cell division. In NDJ, a mistake is made during the metaphaseanaphase transition that allows both sister chromatids to segregate to the same cell. The
cell that received one copy of the chromosome that contains Rb will die, but the other cell
will contain three copies of this chromosome. This partially triploid cell will randomly lose
one of the three chromosomes during the next round of cell division. If the inappropriately
segregated pair of chromosomes contains a mutation in Rb (and both will have the same
mutation because these chromosomes are sister chromosomes) and the chromosome lost
during cell division is the one that contains the WT copy of Rb, the daughter cell will
contain two non-functional copies of Rb with the same point mutation.
3.
In class, we discussed how the SV40 virus causes the transformation of monkey
kidney cells into cancerous cells. SV40 is not the only type of virus that can lead to
cancer in an infected individual. This problem focuses on several others.
a. Rous Sarcoma Virus (RSV) is a retrovirus that induces tumors within a few days in
birds. However, most other oncogenic retroviruses induce cancer after a longer period
of time (months or years). Why is it that some cancer-inducing retroviruses are slowacting, while others like RSV are fast-acting?
RSV’s RNA genome encodes v-src, a mutated version of the host proto-oncogene csrc that causes the dominant phenotype of inducing cell transformation. Slow-acting
retroviruses do not contain mutated oncogenes in their genomes. These slow-acting viruses
lead to cancer by eventually integrating into host DNA in the vicinity of a proto-oncogene
and activating its expression. The long terminal repeats flanking the retroviral DNA can
act as strong enhancers or promoters to genes near which it has incorporated. Insertional
proto-oncogene activation is the more common mechanism by which retroviruses cause
cancer.
Another reason that RSV acts much faster than other cancer-causing retroviruses is
that the tumor formation that results from infection by slow-acting retroviruses often
requires the accumulation of additional mutations in the infected cell. In contrast, v-src is
sufficient to induce tumor formation quickly.
b. Human papillomaviruses (HPVs) are a family of DNA viruses that cause genital
warts. HPVs have been predominantly associated with cervical cancer. One of the
proteins encoded by the HPV virus is E5. E5 is a short transmembrane protein that
forms a dimer or trimer. E5 can form a stable complex with endogenous host PDGF
receptor. Explain how this could lead to cell transformation.
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E5 causes constitutive activation of the PDGF receptors even in the absence of the
hormone. This is because activation of the PDGF receptor normally occurs because of the
dimerization of PDGF receptors that occurs upon interaction with its ligand PDGF. E5
causes dimerization of PDGF receptor when cells are infected with HPV (as opposed to
when cells receive PDGF signal from other cells in the body). This inappropriate activation
of the PDGF receptor then leads to constitutive growth signals being sent through the cell
to the nucleus.
c. HPV also encodes proteins E6 and E7. It has been shown that adding E6 and E7
to normal cells is sufficient to transform them and induce uncontrolled mitosis. Initially,
it was not clear how E6 and E7 accomplished, this but recently it has been
demonstrated that E6 binds and inhibits p53, while E7 is known to bind and inhibit Rb.
Why do these findings explain how HPV causes warts (which are benign tumors), and
how HPV infection increases one’s risk of developing cancer?
p53 and Rb are both tumor suppressors. Loss of Rb leads to premature S phase, because
Rb normally inhibits E2F from activating transcription of S phase genes. Inactivation of
Rb leads to constitutive release of active E2F transcription factors. p53 induces
transcription of several key inhibitors of cell cycle as well as inducers of apoptosis that
must operate when DNA damage is detected. Since benign tumors result from cells that
can divide more frequently than they are supposed to, inhibitors of p53 and Rb lead to the
formation of benign tumors. Since cancer is a disease caused by progressive accumulation
of mutations that all permit uncontrolled cell division, loss of the function of the p53 and
Rb proteins greatly increases the chances that a tumor will develop. This is because fewer
mutations are required to lead to cancer in a cell that has already lost function in two
important cell cycle regulators.
d. In class, we discussed how it is not always obvious why certain mutations
predispose people to certain types of cancer. For instance, it is not clear why a loss of
BRCA1 function leads specifically to breast cancer.
Spleen focus-forming virus (SFFV) induces erythroleukemia in mice. It has been
shown that the SFFV envelope protein gp55 can bind and activate Epo receptors.
Explain why it makes sense that SFFV infection leads to the development of
erythroleukemia, but not other kinds of cancer.
Proliferation of erythroid progenitor cells and development of mature red blood cells
requires Epo and corresponding Epo receptor. If a viral protein can stimulate the Epo
receptor, it would lead to inappropriate and continuous stimulation of erythroid
progenitors and thus lead to cancer. Given that the Epo receptor is only expressed and
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utilized in blood progenitor cell differentiation and development, a virus that produces a
protein that activates the Epo receptor can only cause transformation of blood cells.
4. You decide to study the well-known involvement of mutations in DNA mismatch
repair machinery in the development some types of colon cancer. You have at your
disposal a primary culture of normal colon cells as well as a cell line of colon cancer
cells deficient in DNA mismatch repair. When you treat the cells with TGF-β and then
analyze whole-cell extracts, you see the following results on a Western Blot:
Normal colon
cell line
TGF-β
--
+
Colon cancer
cell line
--
+
PAI-1—
p15—
a. Explain how a defect in mismatch repair could lead to the observed experimental
finding. Also briefly explain how your result provides a mechanistic explanation for
enhanced tumorigenicity in the colon cancer cell line.
The formation of cancerous cells from wild-type cells requires the accumulation of several
mutagenic events in the cell. Both loss-of-function mutations in tumor suppressor genes as
well as increased/ unregulated expression or activation of proto-oncogenes are generally
required for a cell to become transformed. Such events occur through translocations, point
mutations, amplifications, insertions, or deletions of genetic material. Normally, DNA
repair mechanisms are extremely efficient at picking up and fixing such assaults on the
integrity of the genetic material. Defects in DNA repair lead to a global increase in
mutations in all genes in the cell. This means that there is an increased risk that tumor
suppressor genes and proto-oncogenes can acquire mutations in cells that lack the proper
repair mechanisms.
One tumor suppressor gene, the gene that encodes the TGF-beta receptor, is frequently
mutated during normal DNA replication due to “slippage” of DNA polymerase over a
sequence of 10 adenines in a row that occurs in this gene. Slippage of the DNA polymerase
results in a TGF-beta receptor gene that has either 9 or 11 adenines in a row, instead of 10.
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This frameshift mutation must be repaired by the mismatch repair machinery. Otherwise
a frameshift in the coding sequence of the TGF-beta receptor gene persists, leading
expression of non-functional mutant receptor.
Wild-type TGF- beta receptor expression is necessary for sensing and responding to the
anti-proliferative effects of TGF- beta. The signaling cascade downstream of TGF-beta
leads to changes in gene expression that include induction of the cell cycle inhibitor p15 and
induction of the extracellular matrix protease inhibitor PAI-1. p15 halts the cell cycle, and
PAI-1 inhibits the degradation of ECM proteins. Degradation of ECM proteins allows cells
to uproot and travel through the body, which is what leads to metastasis (spreading) of
tumors.
b. You are asked to suggest how to treat the patient whose colon tumor was used to
grow the colon cancer cell line from part a). Would it be beneficial to treat this person
with a drug that specifically inhibited the activity of Smad proteins?
It would not be beneficial to treat this patient with an inhibitor of the activity of Smad
proteins. Smad proteins are normally activated by the binding of TGF-beta to its receptor.
Smad proteins send the signal to the nucleus that TGF-beta has been sensed, and genes
such as PAI-1 and p15 should be transcribed. This person has a deficiency in TGF-beta
signaling that is, in part, causing her cancer. She may benefit from somehow receiving
wild-type TGF-beta receptor or by having her Smad proteins activated even though her
TGF-beta receptor is not functional. Treating a patient for a disorder caused by the lossof-function of an important protein is often much more complicated than treating a patient
for a disorder caused by the gain-of-function or hyper-activity of a protein (which can often
be specifically inhibited using drugs).
c. UV-irradiation causes DNA damage that is repaired by the nucleotide excision repair
machinery. The efficiency of the nucleotide excision repair machinery can be quantified
in human cultured skin fibroblast cells by a Host Cell Reactivation assay (HCR). In this
assay, one takes plasmids containing a gene that encodes luciferase, a reporter
enzyme that emits luminescence. One then UV-irradiates the plasmids to damage
them. One then transfects these UV-irradiated reporter plasmids into cells, waits 48
hours, and measures how much luminescence the cell can emit. The process of
repairing these plasmids such that they can encode functional luciferase and emit
luminescence is known as Host Cell Reactivation (HCR). The extent of HCR after 48
hours is a measure of the efficiency of nucleotide excision repair.
You obtain the following HCR values at 48 hours after transfection of the damaged
reporter plasmids into fibroblast skin cells that were isolated from eleven randomly
selected patients at a nearby hospital.
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Patient #
Patient’s age
1
2
3
4
5
6
7
8
9
10
11
14
18
24
33
43
55
67
67
78
81
94
HCR--% of saturated
luciferase expression at 48
hours after transfection
52
48
60
2
58
40
64
38
46
34
54
The incidence of skin cancer in humans is known to increase with age. Why might this
be? Is the increased risk of skin cancer due to a gradual decrease in the ability of skin
cells to do nucleotide excision repair?
There is no obvious correlation between the age of the patient and the HCR value after 48
hours. Thus, the increased incidence of skin cancer in older patients cannot be explained by
reduced excision repair efficiency. Rather, this increased incidence is probably due to the
accumulation of unrepaired mutations over the years that eventually translates into the
transformation of skin cells.
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For your interest, a Pearson’s correlation statistic shows that the correlation coefficient is
not statistically different from 0, with a 95% CI of -0.5776 to 0.6215. The p value for this
correlation is 0.9202.
d. The 33 year old patient from part c) has virtually no nucleotide excision repair
capacity, as evidenced by the HCR value of 2 in the table above. What hereditary
defect would you expect to find in this patient? What types of DNA damage would this
patient by particularly susceptible to?
You would predict that this patient has Xeroderma Pigmentosum, a hereditary disease
characterized by a deficiency in one or more of the genes involved in nucleotide excision
repair. Without this DNA repair mechanism, these patients are vulnerable to mutations
that cause bulges or other irregularities in the shape of the double helix of DNA. In
particular, UV-irradiation that causes pyrimidine dimers between adjacent pyrimidine
nucleotides is especially harmful in these patients and leads to early-onset skin cancers,
including melanomas, basal cell carcinomas, and squamous cell carcinomas.
NOTE: Question #2 from this year’s Problem Set #3 is highly relevant for
this unit. We suggest that you redo this question at this time.
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