UMEÅ UNIVERSITY
Department of Molecular Biology
Per Holmfeldt
NAME:………………………………………………
Identity number: ……………………………………………
CODE
Examination
5MO021 / 3MB002 Cell Biology, V09
Date:
Time:
Place:
Allowed aids
February 20, 2009
09.00 - 15.00
Sal E, Humanisthuset
English-Swedish dictionary
Translation pen
A handwritten A4 “Fusklapp”
The exam contains two parts, I) multiple choice questions and II) short answer questions. Each part
starts with instructions on how to answer the questions.
READ INSTRUCTIONS CAREFULLY BEFORE YOU BEGIN THE EXAM.
Questions: 47
Maximum point:
Grade
49
Pass with distinction / VG (80%)
39
Passed / G (60%)
29
Please!
Sign your name and identification number on the first page. Enter your code on all
of the following pages.
Show your ID (legitimation) when you hand in your exam.
Results:
EXAM:
GRADE
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Part I: Multiple Choice questions,
- Circle the correct answer
- Mark only one answer per question
- If you change your answer, be sure that all previous marks are erased completely. Incomplete
erasures may be read as multiple answers
- Each correctly answered question will be awarded 0.5p
Q1: The diameter of most animal cells ranges from
A) 1.0 to 10 µm.
B) 0.01 to 0.1 µm.
C) 10 to 100 µm.
D) 100 to 1000 µm.
E) 0.1 to 1.0 µm.
Q2: Which of the following is a feature found in BOTH prokaryotic and eukaryotic cells?
A) extensive array of intermediate filaments
B) multiple linear chromosomes in each cell
C) ribosomes that accomplish protein synthesis
D) mitotic spindles that partition chromosomes into the daughter cells
E) All of the above are found in both types of cells.
Q3: The image below is a picture of a single-celled organism named Euglena. The specimen
was viewed with a(n)
A) Light microscope
B) Fluorescence microscope
C) Transmission electron microscope
D) Scanning electron microscope
E) None of the above
2.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q4. Knowing what you know about phospholipid bilayers; if you wanted to build your own
membrane protein, what would your amino acids need to have in order to stay in the
membrane?
A negatively charged groups
B) hydrophilic groups
C) positively charged groups
D) hydrophobic groups
E) glycosylations
Q5: What make parts of the endoplasmic reticulum rough? The presence of
A) budding vesicles
B) proteins in the membrane
C) ribosomes
D) glycosylations
E) intermediate filaments
Q6: The proteins that make up the electron transport chain in animals are located
A) on the outer mitochondria membrane
B) on the inner mitochondria membrane
C) in the mitochondria matrix
D) in the cytoplasm
E) in the nucleus
Q7: CFTR is a human plasma membrane glycoprotein. If a cell is currently synthesizing
CFTR, in what areas of the cell will these proteins be found?
A) Plasma membrane, Golgi apparatus and smooth endoplasmic reticulum
B) Plasma membrane, Golgi apparatus and rough endoplasmic reticulum
C) Plasma membrane, lysosomes and rough endoplasmic reticulum
D) Plasma membrane, vesicles, Golgi apparatus and rough endoplasmic reticulum
E) Plasma membrane
Q8: A researcher made an interesting observation about a protein that was translocated into the
rough endoplasmic reticulum (ER) during its synthesis and eventually ending up in the plasma
membrane. Significantly, the protein in the plasma membrane was found to be slightly larger
than the cognate protein in the ER. The protein was probably changed in the
A) Golgi apparatus
B) smooth endoplasmic reticulum
C) mitochondrion
D) nucleus
E) lysosome
3.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q9: Clathrin, a member of the coating protein family, is responsible for:
A) Bending the membrane into a pit
B) Directing the vesicles to the proper compartment
C) Pinching off of the vesicle from the donor compartment
D) Fusing the vesicle to the target compartment
E) The transport of the vesicle along microtubules
Q10: Lysosomes can be expected to be present in large numbers in cells which
A) secrete peptide hormones
B) are migrating
C) are actively dividing
D) carry out phagocytosis
E) replicate DNA
Q11: A cell makes both a signaling molecule and the receptor for that signaling molecule in
which mode of signaling?
A) Paracrine
B) Endocrine
C) Autocrine
D) Neuronal
E) Contact dependent
Q12: Which of the second messengers listed below remains bound to the plasma membrane?
A) Diacylglycerol
B) Ca2+
C) IP3
D) cAMP
E) cGMP
Q13: Ca2+/calmodulin-dependent protein kinase
A) can be activated by nuclear receptors
B) binds four calcium ions for full activation
C) can remain active long after the original stimulus that led to activation has
disappeared
D) is an integral membrane protein
E) is activated by calcium released from the Golgi apparatus
4.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q14: Which of the signaling receptors are/is generally activated by dimerization induced by
binding to two sites on their ligand?
A) Gated ion channels
B) G protein-coupled receptors
C) Receptor tyrosine kinases
D) Steroid hormone receptors
E) All of the above
Q15: Place the following steps in the correct order in the mammalian Ras signalling pathway.
1. Activation of Ras
2. Activation of a tyrosine kinase transmembrane receptor
3. Activation of MAP kinase
A) 3Æ1Æ2
B) 2Æ3Æ1
C) 2Æ1Æ3
D) 1Æ2Æ3
E) 1Æ3Æ2
Q16: What is the key event that leads to association of a protein with a proteosome?
A) phosphorylation
B) acetylation
C) glycosylation
D) ubiquitination
E) methylation
Q17: Which of the following would INHIBIT the onset of mitosis?
A) binding of M Cyclin to Cdk
B) phosphorylation of Cdk by Wee1
C) phosphorylation of Wee1 by Cdk
D) dephosphorylation of Cdk by Cdc25
E) None of the above
Q18: The stage of mitosis when chromosomes condense to form rod-shaped structures visible
under the microscope is called:
A) interphase
B) prophase
C) metaphase
D) anaphase
5.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
E) telophase
Q19: When ATP binds to the head of myosin II it promotes:
A) binding of myosin to a new actin subunit
B) pivoting of the myosin head and generation of movement
C) release of actin
D) formation of myosin filaments
E) none of the above
Q20: Which of the following processes during animal cell division is not mediated by
microtubules?
A) Movement of the chromosomes to the poles of the cell
B) Movement of the chromosomes to the metaphase plate
C) Contraction of the cleavage furrow
D) Positioning of the cleavage furrow
E) Separation of the centrosomes
Q21: You have a test-tube that contains purified monomeric actin protein. You incubate the
actin at 37oC in a buffer that includes ATP. After 30 minutes you find that some of the actin
has polymerized. Now you perform the same experiment, except that you ALSO add each of
the proteins listed below. Which of these proteins cause a DECREASE in the amount of actin
polymerization that occurs?
A) Fimbrin
B) Thymosin
C) Profilin
D) Formin
E) Tropomyosin
Q22: The following modification occurs on nuclear lamins, which causes nuclear lamina to
disassemble and nuclear envelope to rupture during mitotic entry
A) Glycosylation
B) Ubiquitination
C) Phosphorylation
D) GTP binding
E) Cyclin binding
6.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q23: A given haploid S. cerevisiae (baker’s yeast) strain has a temperature-sensitive mutation
that inactivates its hexokinase enzyme at the high (restrictive) temperature. (Hexokinase
catalyzes the first reaction in the glycolysis pathway.) A culture of this mutant is grown on
glucose/minimal salts medium (glucose is the only energy source under these conditions) and
then shifted to the restrictive temperature for a couple of hours. Which of the alternatives
below describes the cell cycle stage of most of the cells in the culture?
A) G1 phase
B) S phase
C) G2 phase
D) M phase
E) A mix of cells in G1, S, G2 and M
Q24: Which of the following is NOT a change experienced by a typical cell committed to
apoptosis?
A) Loss of mitochondrial membrane functions
B) Cytoskeleton collapses
C) DNA breaks into fragments
D) Cell swells and ultimately bursts
E) Nuclear envelope disassembles
Q25: Dye injected into an epithelial cell might be able to enter an adjacent cell through a
A) tight junction
B) microtubule
C) desmosome
D) adherence junction
E) gap junction
Q26: Glycosaminoglycan polysaccharides are found on serine and threonine residues in
A) proteoglycans
B) collagens
C) laminins
D) elastins
E) integrins
Q27: What is acquired during the progression of ALL tumors?
a) loss of the RB gene
B) mutations
7.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
C) the ability to synthesize mitogens
D) activation of tumor suppressors
E) loss of the p53 gene
8.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Part II: Short answer questions,
- Depending on the question at hand, the required answer may range from a couple of words to
several sentences. Use the allocated space as an indication on how much information you
should provide to receive full credit.
- Graphic can be used to illustrate matters but must always be accompanied by explanatory
text
- Answers can be written in either Swedish or English.
Q28: You are exploring a rather inhospitable planet which has “seas” filled with a hydrophobic
liquid. Surprisingly, there are living organisms in these seas whose cytoplasm is hydrophobic
to a similar degree. These organisms have membranes made primarily of phospholipids
arranged in a bilayer. What is the most probable orientation of these phospholipids? Motivate
your answer. (1p)
The amphiphatic phospholipids are probably orientated in a manner exposing the
hydrophobic tails towards the hydrophobic solutions and hiding the hydrophilic heads
against the same solutions (0.5p). This orientation is the most energetically favourable
(0.5p)
Q29: How is it possible for some molecules to be at equilibrium across a biological membrane
and yet not be at the same concentration on both sides of the membrane? (2p)
The equilibrium distribution of a molecule across a membrane depends on concentration
and membrane potential (the electrochemical gradient) (1p). A charged molecule will
respond to both components of the electrochemical gradient and will distribute
accordingly (1p).
Points were also rewarded for “sequestration” like binding which lowers the
concentration of a particular molecule specie without actually lowering the physical
presence of that specie.
9.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q30: You clone a sheep by somatic cell nuclear transfer. In doing so, you take the nucleus
from a mammary gland cell from “Sheep A” and insert it into the cytoplasm of an enucleated
egg taken from “Sheep B.” You then implant the newly-nucleated egg into a pseudopregnant
“Sheep C” You name the resulting newborn clone “Sheep D”. “Sheep D” is born with a genetic
disease that “Sheep B” has. Explain how this is possible, and what may have caused the
disease. (2p)
The disease is showing cytoplasmic inheritance, because the cytoplasm of the oocyte used
to make the clone was made by Sheep B (1p), thus a mitochondrial mutation most likely
caused the disorder (1p).
Q31: The ER signal sequence is removed from proteins once they are delivered into the ER.
Why is the nuclear localization signal (NLS) NOT removed from nuclear proteins once they
are delivered into the nucleus? (1p)
Nuclear proteins need to be re-imported into the nucleus after completion of mitosis (1p)
Points were also rewarded for need for nuclear-cytoplasmic shuttling and disastrous
effect of cutting a non-linear NLS
Q32: You are studying patients with a genetic disorder that affects function of the lysosome.
You have discovered that in these patients, a mutation in a certain gene is causing a lysosomal
hydrolase called “degradase” to be missing from their lysosomes. Otherwise their lysosomes
are normal, and contain normal levels of all the other lysosomal hydrolases. You also find that
there are high levels of “degradase” in the bloodstream of these patients. Suggest an
explanation for the disease, i.e. what protein is mutated, and what is the consequence of the
mutation? Be as specific as possible. (2p)
The enzyme “degradase” is mutated (1p). Possible in a manner so it cannot be recognized
by the Golgi localized enzyme that adds the phosphorylation on specific mannoses,
facilitating their lysosomal targeting (1p).
It is unlikely that the enzyme that catalyze the phosphorylation was mutated, since no
other hydrolases are affected, but points was awarded here also, since this mechanism
can not completely be “eliminated”.
10.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q33: Many transcription factors shuttle into and out of the nucleus. One such factor that has
been widely studied is STAT5, a protein of 90 kD. In unstimulated erythroid cells STAT5 is
present exclusively in the cytosol. After binding of erythropoietin (Epo) to the Epo receptor,
the protein tyrosine kinase JAK2 becomes activated. JAK2 then phosphorylates STAT5 on a
tyrosine near its C-terminus. STAT5 then moves into the nucleus where it activates expression
of a set of genes. After removal of Epo, the phosphate group on the specific tyrosine residue of
nuclear STAT5 is removed and the (non-phosphorylated) STAT5 moves immediately (i.e.
within 5 min.) from the nucleus back into the cytosol.
A) Assume you can add segments of amino acids or even sequences encoding entire proteins to
the C-terminus of STAT5 and still have a fully functional STAT5 protein. How could you
monitor experimentally the movement of STAT5 into and out of the nucleus in response to Epo
addition or removal in living cells? (1p)
The best method for following protein localization in living cells is by making a
fluorescent protein fusion. One could fuse the coding region for GFP onto either end of
the STAT5 gene, creating fusion protein (1p).
B) You treat erythroid cells with Epo for 15 min. and observe that about half of the STAT5
proteins are phosphorylated on tyrosine while half is not. You treat these cells with a nonionic
detergent which dissolves all cell membranes, including the nuclear membrane. Then you
incubate the cell lysate at 4°C for an hour with small plastic beads coupled with antibodies to
the importin protein. You then recover the beads by centrifugation and thereafter analyze the
bound proteins. You find STAT5 in the mixture of bound proteins. Would you expect the
bound STAT5 to be tyrosine phosphorylated or unphosphorylated, or a mixture thereof?
Motivate your answer. (1p)
The bound STAT5 would be phosphorylated. Importin is responsible for nuclear import,
which only occurs when STAT5 is phosphorylated. (1p)
C) You take the same detergent lysate of Epo-treated cells as in Part B, and incubate it at 4°C
for an hour with small plastic beads coupled with antibodies to the Ran protein. You then
recover the beads by centrifugation and analyze the bound proteins. Again, you find STAT5 in
this mixture of bound proteins. Would you expect the bound STAT5 to be tyrosine
phosphorylated, unphosphorylated, or a mixture thereof? Motivate your answer. (1p)
The bound STAT5 will be unphosphorylated. Ran only binds cargo during nuclear
export, which only occurs when STAT5 is unphosphorylated. (1p)
11.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
D) You have made a mutant of STAT5 in which amino acids 18 – 23 are replaced by alanines.
When expressed (by recombinant DNA in erythroid cells) this mutant, called STAT5-6A, is
present exclusively in the cytosol as long as Epo is not added. Epo addition causes the protein
to move normally into the nucleus (and activate transcription normally). However, after Epo
removal, the STAT5-6A derivative becomes dephosphorylated normally but remains in the
nucleus. Provide an explanation for the function of amino acids 18 – 23 in STAT5. (1p)
These 6 amino acids most probably contains the nuclear export sequence of STAT5 (1p)
Q34: Confused and overwhelmed by the number of events involved in different signaling
pathways, your classmate declares that signaling cascades are unnecessarily overcomplicated.
Give two reasons why the complexity of signaling pathways might be beneficial to cells. (2p)
The complexity of signaling cascades are advantageous for the following reasons:
-- A cascade amplifies small signals, so that a low concentration of the signal can
stimulate an amplified response. (1p)
-- Having multiple steps in a cascade means that there are multiple points/opportunities
for regulation and many opportunities for integration of other signals. (1p)
Q35: Umeå Derived Growth Factor (UDGF) is a recently identified mitogen that activates a
Receptor Tyrosine Kinase. Urinary bladder cells do not express the receptor for UDGF, and
when these cells are grown in culture, their proliferative rate is unaffected by addition of
UDGF to their growth medium. However, you have cloned the gene for the UDGF receptor,
and have successfully transfected the cDNA for the UDGF receptor into urinary bladder cells.
You establish that the UDGF receptor is expressed on the cell surface of the transfected
bladder cells. Surprisingly, the proliferation of these cells remains unchanged when UDGF is
added to the culture medium. Propose an explanation for this negative result. (1p)
The cell most likely does not express the downstream components of the UDGF signalling
pathway (1p)
Other more creative solutions have also been rewarded some points……….
12.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q36: You are studying the effects of various mutations in the β-adrenergic receptor (a Gprotein coupled receptor), which, when bound to epinephrine, mediates an increase in
intracellular levels of cAMP (as compared to the low basal levels of cAMP in cells not exposed
to epinephrine). How would the mutations listed in A) – C) affect cAMP concentration?
Indicate whether the mutations will cause levels of cAMP to rise normally in response to
epinephrine, will cause constitutively high levels of cAMP, or will cause constitutively low
levels of cAMP. Motivate your answers.
A) A mutation in the Gsα subunit that inhibits nucleotide exchange (0.5p)
This will cause constitutively low levels of cAMP. Gsα will remain bound to GDP all the
time (because the GDP cannot be exchanged for GTP), which will prevent Gsα’s
dissociation from Gβγ. Thus adenylyl cyclase will never get activated. (0.5p)
B) A mutation in the Gβ subunit that increases its affinity for the Gsα subunit in its GTP bound
state (0.5p)
This will cause constitutively low levels of cAMP. Gsα will be unable to dissociate from
Gβγ, so Gsα will not be able to leave the trimeric complex and bind and stimulate
adenylyl cyclase. (0.5p)
C) A mutation in Gsα that prevents GTP hydrolysis (0.5p)
This will cause constitutively high levels of cAMP. Gsα will remain bound to GTP all the
time. This GTP-bound Gsα will remain bound to adenylyl cyclase, continuously
stimulating cAMP production.
Q37: Cofilin preferentially binds to the older parts of actin filaments and promotes their
disassembly. How does cofilin distinguish between old and new parts of actin filaments? (2p)
G-Actin binds to ATP which will be hydrolyzed upon polymerization. However, as long
as the polymerization rate exceed the hydrolysis rate new F-actin will contain ATP and
old F-actin ADP (1p). Cofilin binds preferably F-actin containing ADP (1p).
13.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q38: There are no known motor proteins for intermediate filaments. Based on what you know
about filament assembly and motor proteins; provide an explanation for this observation. (2p)
Molecular motors need structurally polar filaments to be able to translocate to a specific
end, thus creating polarized transport (1p). Intermediate filaments are structurally nonpolar and arranged randomly throughout the cytosol and can thereby not be used for
directed transport (1p)
Note: Observe that the plus and minus ends of Actin filaments and microtubules has
NOTHING to do with electric charge!
Q39: Kinesin carries vesicles over long distances along microtubule tracks in the cell. Are the
two motor domains of a kinesin molecule essential to accomplish this task, or could a oneheaded motor protein function just as well? Motivate your answer. (1p)
Two heads are better than one. A two headed kinesin motor can move a vesicle for long
distances along a microtubule track because it holds on with one ‘hand’, while it releases
and binds with the other. A one-headed motor would lose its way each time it released the
microtubule to take a step. (1p)
Q40: G1 cyclin activates Cyclin Dependent Kinase 4/6 (CDK4/6), and the formed complex
phosphorylates the Retinoblastoma protein (Rb). Indole-3-carbinol (I3C), a compound found in
vegetables such as Brussels sprouts, inhibits CDK4/6 kinase activity. Knowing the cell cycle as
you do, would you expect I3C to promote, suppress or have no net effect on the proliferation of
tumor cells? Motivate your answer. (2p)
I3C suppresses the proliferation if the tumor cells have acquired self-sufficency in
mitogen signaling through mutations upstream of Rb phosphorylations. (1p)
I3C will have no effect on the proliferation if the tumor cells have acquired self-sufficency
in mitogen signalling through mutations downstream or at the level of the
14.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
phosphorylation of Rb. (1p)
If you scored two points here; VERY WELL DONE!
Q41: Overexpression of Cdc6 leads to re-replication in fission yeast so that cells end up with
more DNA than in normal cells. You have identified a Cdc6 temperature-sensitive mutant that
gives a very similar phenotype (i.e., re-replication of DNA) at the non-permissive temperature.
However, you find that this mutant has increased Cdc6 protein levels with no change in mRNA
levels. You map the mutation to a specific lysine residue that is changed to alanine in the
mutant Cdc6 protein sequence. Provide an explanation for how this lysine mutant leads to
increased Cdc6 protein levels and DNA re-replication. (2p)
Cdc6 facilitate re-replication by recruiting and possibly activating DNA helicases at the
replication origins (0.5p) Cdc6 levels are decreased through proteosomal degradation of
Cdc6 ubiquitinated (0.5p) on specific lysine residues. However this lysine can not be an
ubiquitin target (or a phosphorylation target either) since the degradation obviously
works at the permissive temperature. The mutation must create a conformational
change that prohibit phosphorylation or ubiquitination (1p)
Q42: Drug companies screen for so called small molecule inhibitors (i.e. different chemical
compounds) that interfere with specific biological processes. How will cell cycle progression
be affected when a cell is treated with a small molecule inhibitor that binds to kinetochores and
specifically blocks their ability to interact with microtubules? Motivate your answer. (2p)
The cells will be blocked in prometaphase (you get credit for metaphase also) since
unattached kinetochores will activate the spindle assembly checkpoint (1p). This is
accomplished by inhibiting the complex formation between cdc20 and APC/C, which
together forms the ubiqutin ligase that promote entry into Anaphase. (1p)
15.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q43: p53 is thought to act as a sensor of cellular stress. Cellular levels of the p53 protein are
known to rise in response to DNA damage and hyper-proliferation. Increased levels of p53 can
lead to cell cycle arrest or apoptosis. The p53 gene is frequently mutated in human cancers.
However, a subset of tumors possesses two wild type copies of the p53 gene, expresses normal
level of p53 mRNA, but no p53 protein can be detected. Karyotyping analysis of some of the
tumors reveals over-amplification of a locus that is unlinked from the p53 locus. Based on what
you have learned about p53 during this course, propose a model to explain the above
observations. (1p)
Degradation of p53 is facilitated by the ubiqutin ligase Mdm2. If Mdm2 levels would be
increased, through gene amplification, p53 levels would possibly be kept low even though
p53 stabilizing signals would be present. (1p)
Q44: You administrate stable short double-stranded RNA with a sequence complementary to
the mRNA coding for Caspase-3 (a protein with a half life of ~6h) into the cytosol of primary
fibroblasts.
A) How will the fibroblasts respond if you simultaneously radiate them with a heavy dose of
ionizing radiation? Motivate your answer. (1p)
Cells will die through caspase-3 dependent apoptosis, since caspase-3 depletion through
RNAi has not had time to occur. (1p)
B) How will the fibroblasts respond if you radiate them with a heavy dose of ionizing radiation
three days after the administration of the short double-stranded RNA? Motivate your answer.
(1p)
Cells will not die since Caspase-3 is depleted and apoptosis is dependent on caspase-3
(0,5p). But cell proliferation will be inhibited upon DNA damage through a p53
dependent mechanism (0,5p).
16.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
Q45: The blood-brain barrier is a highly selective filter that prevents many molecules from
entering the brain. Its effectiveness depends on tight junctions between neighboring endothelial
cells in the capillaries of the brain. Can steroid hormones be expected to cross the blood-brain
barrier? Motivate your answer. (1p)
Yes, because they are hydrophobic molecule and can pass through the cell membranes of
the endothelial cells of the bloodbrain barrier. (1p)
Q46: Increased expression of Myc in human cancer cells may be caused by chromosomal
translocation or through gene amplification. In your studies of the oncogenic properties of the
Myc protein, a plasmid carrying the myc-gene under the control of a strong promoter is
transfected into primary murine fibroblasts. You are surprised to find that rather than showing
increased proliferation, the Myc over-expressing cells undergo apoptosis.
A) You decide to transfect the Myc expression plasmid into primary fibroblasts isolated from a
Rb knockout mouse. Do you think the cells will undergo apoptosis or will become
hyperproliferative? Motivate your answer. (1p)
They will die through apoptosis. myc will activate p53 through Arf. p53 will transcribe
pro-apoptotic proteins facilitating apoptotic entry (1p)
B) You decide to transfect the Myc expression plasmid into primary fibroblasts isolated from a
p53 knockout mouse. Do you think the cells will undergo apoptosis or will become
hyperproliferative? Motivate your answer. (1p)
They will proliferate since the arf/p53 pathway is disabled and an apoptotic response is
impossible and myc will only promote cell proliferation. (1p)
Q47: A single amino acid change in Ras eliminates its ability to hydrolyze GTP, even in the
presence of a GTPase-activated protein (GAP). Roughly 30% of human cancers have this
change in Ras. You have identified a small molecule that prevents SH2 domains from
recognizing phosphorylated tyrosines. Would you expect this molecule to be effective in the
treatment of cancers that express this common, mutant form of Ras? Motivate your answer.
(2p)
17.
UMEÅ UNIVERSITY
Department of Molecular Biology
February 20, 2009, 09.00-15.00
EXAMINATION
5MO021 / 3MB002 Cell Biology, V09
(maximum points shown within parenthesis)
…………………………………….
Results
SH2 domains are used at the receptor tyrosine kinase level of RTK/Ras/MAPK signalling
(1p)
The small molecule would not be effective in treating cancers with mutationally activated
Ras proteins. Activated Ras signals independently of the receptor tyrosine kinase, thus,
preventing receptors from dimerizing would have no effect on activity of the mutant Ras.
(1p)
Other answers have been somewhat rewarded if they have been rational (even though
incorrect technically).
18.