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SBPM SSN Worksheet #1
Sept. 18, 2002
1.
Cyclin D/CDK4 is known to play an important role in regulating the G1  S checkpoint.
Using your knowledge of the retinoblastoma (Rb) gene, (a) explain this mechanism.
Hypophosphorylated Rb is
complexed with the E2F
transcription factor and inhibits the
transcription of genes required for
the S phase. During G1, levels of
Cyclin D/CDK4 build up, and
eventually, it phosphorylates the Rb
protein through its kinase activity.
The phosphorylated Rb can no
longer bind the transcription factor
E2F, allowing it to bind DNA and
cause transcription of genes critical
for progression through S phase.
(b) Rb is a well studied gene known to cause childhood tumors. Approximately 60% of
retinoblastomas are sporadic while the remaining are hereditary. Cytogeneticists have associated
a deletion of chromosome band 13q14 with retinoblastoma. In a patient with sporadic
retinoblastoma, would you expect to see tumors in one or both eyes? Why?
“Two hits” are required on the Rb gene for it to cause cancer. Without any functional Rb, E2F is free
to bind to DNA and continuously promote transcription of S phase genes. In a patient with a tumor in
only one eye, we assume that he has two normal Rb alleles, but he gained “two hits” due to somatic
mutation in one of his eye retinal cells. Therefore, this is a case of sporadic and not hereditary. We
would NOT expect to see the 13q14 deletion in his normal cells.
In patients with hereditary retinoblastoma, they carry a germline mutation on one of their Rb
alleles. Therefore, they only require “one hit” for the gene to become cancerous, and we expect that
a second mutation (somatic) will occur at least in some retinal cell (the product of the target cell
population 107 and the probability of a second mutation, 10-6, exceeds unity).
(c)
Is Rb a tumor suppressor gene or a protoncogene?
Rb is a tumor suppressor gene because its loss-of-function causes cancer. Also, the Rb gene
product‘s normal function is to inhibit the transcription of genes necessary for continuation through the
cell cycle; thus it inhibits cell division and can be thought of as a tumor suppressor for that reason.
2.
Why do X-linked dominant diseases show variable expression in affected females?
Barr Bodies. Since males only have one X chromosome, all X-linked disorders are expressed
whether they are dominant or recessive. But in females, who have two X's, one of the X's is randomly
inactivated in embryonic life in each cell. Because all of that cell's clonal descendants have the same
inactive X, only certain cells will express the mutated allele. The inactivated X is called a Barr body.
3.
Design a protein that will be a plasma membrane associated, ligand activated, cation
transporter.
Several sequences that will be membrane-spanning, forming a tube with hydrophobic residues facing
out, and hydrophilic residues facing in. Anionic residues surround the perimeter at some point along
the tube will select for cations and exclude anions. Moreover, there will be an extracellular sequence
that has affinity for the ligand, whose binding causes an opening of the “tube” portion. Since the
protein is membrane-bound, the mRNA will code for (a) signal sequence(s) which is/are cleaved off.
4.
Signaling through tyrosine receptor kinase (trk) molecules is an important way for cells to
receive extracellular cues to activate gene transcription. Answer the questions regarding the
following labels on the diagram of the RAS signal transduction pathway.
Label 1. After the growth factor binds, what two things happen? Receptor dimerization and
autophosphorylation
Label 2. This molecule contains what type of domain? SH2 (SRC Homology 2)
Label 3. What is this molecule called? SOS or GEF (Guanine – nucleotide Exchanging Factor)
Labels 4a/4b. This molecule (4a) is exchanged for (4b). GDP (4a) is exchanged for GTP (4b)
Label 5. This molecule (5) becomes active and phosphorylates downstream targets to activate the
MAP kinase pathway. RAF
How can mutation of the above RAS signal transduction pathway contribute to tumor formation?
Mutated Ras is unable to hydrolyze GTP, meaning that it is constitutively active. This causes the
target genes to be transcribed constitutively as well, leading to uncontrolled cell growth and
proliferation.
Hypothesize what would happen in the following scenarios:
a) Homozygous mutation causing deletion of the extracellular domain:
Without the steric interaction of an extracellular domain, the TRK molecule is able to get within
close enough proximity to cause autophosphorylation. Thus, signaling would be rampant, and
the cell would be able to replicate without the presence of extracellular growth factors. Thus,
the cell would become cancerous.
b) Heterozygous mutation causing deletion of the intracellular domain:
Without the intracellular domain, autophosphorylation cannot occur, and the cell will not be able
to respond to extracellular growth factors and will probably die. This mutation is dominant.
Mutant receptors can pair up with the wild type copy, and so ~75% of receptor dimers will have
either one or two nonfunctioning intracellular domains. The remaining ~25% will probably not
suffice to provide a sufficiently robust response to GFs and the cell will be at a major growth
disadvantage and will probably die off. N.B. It is not a rule that 25% of the normal amounts of
any gene product will automatically cause death or malfunction. In other situations, perhaps
with less vital signal molecules or in situations where the cell may have alternative routes of
signaling, the cell might not necessarily die.
In terms of structure and activation, how does a tyrosine receptor kinase (trk) differ from a G
Protein Coupled Receptor (GPCR)?
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GPCR is a seven-pass transmembrane receptor whereas trk’s only possess a single alpha helix
(both are oriented with the N-terminus in the extracellular space).
Binding of a ligand to GPCR causes activation, whereas it causes dimerization of trk’s which may
in turn cause activation.
Trk’s are activated via autophosphorylation and then recruitment of SH2-containing molecules.
GPCRs undergo a conformational change in response to ligand binding. This confirmational
change is sufficient to cause the exchange of GDP for GTP, directly activating the Gα subunit.
5.
Describe the changes in the O2 saturation curve of hemoglobin under the following
conditions:
(a) Increase in blood pH.
The curve shifts to the left. Recall the Bohr effect, from which we know that an increase in [H +]
decreases hemoglobin’s O2 affinity. Here, we have an increase in pH, which means that [H+] is lower,
so the affinity increases (i.e., P50 decreases).
(b) pCO2 is high.
The curve is shifted to the right. CO2 decreases hemoglobin’s O2 affinity (i.e., P50 increases).
(c) Let’s say there is a variant of hemoglobin with a greater affinity for 2,3-DPG. What then?
The variant will have a right-shifted curve. 2,3-DPG binding decreases hemoglobin’s O2 affinity (i.e.,
P50 increases).
(d) The patient has carbon monoxide (CO) poisoning. CO binds hemoglobin with an affinity 250
times that of O2. Describe how the O2 saturation curve looks different and state the implications
for O2 delivery to the tissues.
The curve levels off at a lower O2 saturation and is shifted to the left. Since the affinity of CO for
hemoglobin is high, it can compete effectively with O2 for binding sites even though its relative
concentration may be low. By taking up binding sites, CO reduces hemoglobin’s overall O2-binding
capacity. The remaining binding sites, however, have an increased affinity for O2 (pseudocooperative binding?) (i.e., P50 decreases). The consequence of both of these changes is reduced O 2
delivery to the tissues, because the hemoglobin is carrying less O 2 and it is less willing to give up the
O2 it is carrying.
6.
What mechanism could a drug use to inhibit the motility of intracellular bacteria that co-opt
the cytoskeleton for their movement? What might be some side effects?
These kinds of bacteria use Arp 2/3 complex, an actin nucleation protein, which causes a burst of
actin polymerization behind them resulting in the bacteria moving forward. Blocking Arp2/3 would
immobilize the bacteria. However, Arp2/3 is used at the leading edge of moving cells; inhibiting it
would result in impaired cell motility. This would inhibit the movement of phagocytes to the infected
area, which is not good.
7.
Name three distinct locations that a cell might deliver its proteins to and describe the
mechanisms that it would use to get them there.
Nucleus: NLS on finished polypeptide and importin function
Cytosol: mRNA simply gets transcribed by free ribosome. All proteins that do not have a sorting
signal remain in the cytosol by default.
ER lumen (pre-modifications that occur in golgi): polypeptide has signal sequence near amino
teminus which that causes SRP to bind, which stops translation and then allows it to continue on the
rER with continous translocation into the lumen. N.B. the SRP recognition site is cleaved from
polypeptide product under normal conditions. See also below for proteins that end up in ER.
ER membrane, PM and lysosome membrane (as integral proteins): same as above PLUS the
presence of at least one “start transfer signal” which is hydrophobic and is inserted into the
membrane as the protein is being synthesized and anchors it in the membrane of the ER. (The
presence of additional stop transfer and start transfer signals can generate multipass proteins.) After
translation, vesicle containing the integral membrane peptide must be exported to golgi apparatus (to
cis face) and then out of Golgi appartus (from trans face). N.B. carbohydrate tags are added to
polypeptides while they are in the gogli lumen which direct the golgi machinary to send them in
vesicles bound for different places (ER smooth or rough, to PM or to lysosome).
Out of cell, ER lumen or lysosome lumen; same as ER lumen PLUS protein must receive
modifications in the Golgi apparatus that allow it to be packaged in vesicles bound for the cell surface,
ER or lysosome respectively. Protein must NOT have start transfer sequences for membrane
integration.
8.
What is the purpose of the glucose-alanine shuttle? When would it be activated? Describe
the steps of the shuttle and where these steps take place.
The glucose-alanine shuttle eliminates nitrogenous waste products that are generated by the
oxidation of amino acids in muscle. The nitrogenous waste cannot be eliminated directly from the
muscle because the muscle does not have the ability to make urea.
Muscle protein is broken down to amino acids during periods of starvation. The carbon skeleton of
these amino acids can be used as fuel to satisfy local energy needs.
1)
2)
3)
1)
2)
3)
4)
The steps involved in the glucose-alanine shuttle are:
In muscle:
Proteins broken down into amino acids
Amino group transferred from amino acid to pyruvate (via transamination), generating alanine
Alanine exits muscle and is transported to liver via the glucose-alanine shuttle
In liver:
Amino group removed from alanine by liver, generating pyruvate
Amino group taken up by urea
Pyruvate converted to glucose via gluconeogenesis
Glucose transported back to muscle to restore pyruvate levels for next transamination reaction