Download LS1a Problem Set #2

Name:
TF Name
LS1a Fall 2014
Problem Set #5
Due Monday 11/17 at 6 pm in your TF’s drop box on the 2nd floor of the Science Center
I. Basic Concept Questions
1. (16 points) The Her2 receptor is a receptor tyrosine kinase, similar to the EGF receptor,
whose activation leads to cell growth and proliferation. In many breast cancers, Her2
receptor mutations have been isolated.
a. (4 points) Ligand binding causes the Her2 receptor to dimerize. How does dimerization
of receptor tyrosine kinases help transmit a signal from outside the cell to inside the
cell?
Receptor dimerization activates the receptor by bringing the intracellular kinase
domains of the receptors close enough together to “trans” autophosphorylate one
another’s activation loops.
b. (4 points) A mutation has been found in the transmembrane domain of the Her2
receptor in which a Val is changed to a Gln. This mutation causes the receptor to
dimerize in the absence of ligand. What effect would this mutation have upon the
activation state of the receptor in the absence of ligand? Briefly explain.
The receptor will be always activate, even in the absence of the ligand. The role of the
ligand is to activate the receptor by causing two receptor subunits to dimerize. If the
mutant subunits are already dimerized, then the receptor is already active, regardless
of the presence of ligand.
c. (4 points) Scientists can genetically modify the Her2 receptor to eliminate the entire
intracellular domain of receptor. If this modified receptor were the only Her2 receptor
expressed in a cell, will it be able to transmit a signal inside the cell upon ligand binding?
Why or why not?
No. The intracellular domain has the kinase domain, which is required for intracellular
signaling. If it is absent, the kinase activity would be absent as well, such that no
downstream signaling could occur.
d. (4 points) If a dimer were to form containing one normal, full-length subunit of the
receptor and one subunit of the receptor described in part (c), would this complex be
able to activate downstream targets? Briefly explain your reasoning.
No. Receptor activation occurs when one receptor monomer phosphorylates the
other. If this heterodimer forms, the monomer with the functional kinase domain will
have nothing to phosphorylate, while the monomer missing the kinase domain will be
unable to phosphorylate the other monomer. Transactivating phosphorylation
between the two receptor intracellular kinase domains is necessary to activate
downstream targets.
2. (16 points) While our bodies can most readily derive chemical energy from the sugar
monomer glucose, our bodies store glucose in large sugar polymers called glycogen. The
majority of the glycogen in our bodies is found in our liver and muscles, and it must be
broken down into glucose monomers before our cells can make use of it. Glycogen
phosphorylase is an enzyme that helps catalyze the breakdown of glycogen into glucose.
a. (4 points) Glycogen phosphorylase is activated by a serine/threonine kinase called
phosphorylase kinase. Briefly explain why cells must make use of two types of kinases:
those that phosphorylate serine and threonine side chains; and those that
phosphorylate tyrosine side chains?
Serine and threonine are similar enough in shape (differing only in threonine’s
additional methyl group attached to the β-carbon), that they can both fit into the
same active site of a kinase as long as the active site has a pocket that accepts
threonine’s additional methyl group.
Tyrosine has an aromatic phenyl group, a much larger/longer and differently shaped
side chain, such that the cell requires a different kinase with a larger active site to
accept it.
It is also acceptable to argue that the reactivity or nucleophilicity of a tyrosine side
chain is different from that of a threonine or serine side chain, such that a different
type of active site is necessary.
b. (4 points) Briefly explain why phosphorylating an amino acid side chain in a protein can
change its function.
Phosphorylation of amino acids causes a significant change in the chemical properties
of the side chains, converting a small, polar alcohol to a larger, charged phosphate
group, changing both the size and the charge of the side chain. These changes in the
properties of side chains can alter the structure and chemical properties of proteins.
c. (4 points) Glycogen phosphorylase can be turned off by a phosphatase, an enzyme that
catalyzes the dephosphorylation of a molecule. The phosphatase deactivates glycogen
phosphorylase by removing the same phosphoryl group from the protein that
phosphorylase kinase had attached in order to activate it. Below is a diagram of the
mechanism of a serine/threonine phosphatase.
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What catalytic role does the aspartate play to help catalyze the reaction? What is the
name of the amino acid that is being dephosphorylated?
Aspartate makes the water molecule oxygen more nucleophilic to attack the
phosphorous atom. The amino acid being dephosphorylated is serine.
d. (4 points) Draw the transition state associated with this reaction.
3. (4 points) Epidermal growth factor (“EGF”) is an extracellular signal that stimulates cell
growth and differentiation. You are studying two populations of cells to better understand
how the EGF signal is relayed within a cell. Both populations of cells express the EGF
receptor, but only one population of cells is treated with EGF.
a. (2 points) In which population of cells (EGF-treated or untreated) would you expect to
find MAPK in the nucleus?
MAPK would be found in the nucleus of EGF-stimulated cells.
b. (2 points) Why is it important for MAPK to be able to change its subcellular localization
upon activation by MAPKK?
It is important for MAPK to change its subcellular localization upon activation because
entering the nucleus when activated allows MAPK to phosphorylate transcription
factors.
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II. Applied Concept Questions
4. (28 points) Similar to the EGF receptor, the insulin receptor (“IR”) is a receptor tyrosine
kinase with an extracellular ligand-binding domain and an intracellular tyrosine kinase
domain. Whereas the EGF receptor (“EGFR”) is activated by binding EGF, the IR is activated
when it binds insulin.
a. (4 points) The insulin receptor exists on the cell surface as a pre-formed dimer (i.e., both
insulin receptor subunits are always bound to each other on the cell surface). A single
insulin molecule binds at the interface of the pre-formed insulin receptor dimer. The
binding of insulin by the insulin receptor triggers a conformational change that activates
the cytoplasmic kinase domains of the IR. Briefly describe how the binding of EGF to
EGFR differs from the binding of insulin to the IR. How does the number of EGF
molecules needed to bind and activate an EGFR dimer differ from the number of insulin
molecules needed to bind and activate an IR dimer?
EGF binds monomeric EGFR subunits. Two EGF molecules need bind an EGFR dimer in
order to activate it; only one insulin molecule needs to bind an IR dimer in order to
activate it.
b. (4 points) Shown below are two representations of the structure of insulin (green)
binding the insulin receptor dimer. One subunit of the IR is cyan, the other is magenta.
Stick representation
Space filling representation
In the images shown above, what is the predominant type of intermolecular interaction
between the side chains at the interface between insulin and the two IR subunits? What
is the thermodynamic driving force that stabilizes the interaction between insulin and
the insulin receptor?
The protein interface is dominated by phenylalanines and other hydrophobic amino
acid side chains, such that the predominant intermolecular interaction at the interface
should be van der Waals interactions.
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Given the amount of hydrophobic surface area present on insulin and the receptor
that gets buried when the ligand binds the receptor, the hydrophobic effect is the
thermodynamic driving force of this binding reaction.
c. (4 points) The binding of insulin to the IR causes a conformational change in the two
cytosolic kinase domains, allowing autophosphorylation to occur as one kinase domain
phosphorylates the other kinase domain and vice versa. Shown immediately below is
the structure of the inactive conformation of the insulin receptor kinase domain (“IRK”)
in blue. On the top of the next page is the structure of the active conformation of the
IRK in green. In the structure of the active conformation, the peptide substrate (pink),
ATP (cyan), and the magnesium ions (gray) occupy the catalytic cleft. For both
structures, the activation loop is colored beige-ish khaki.
On the bottom of the next page is another image of the inactive kinase, but this time
with the position and orientation of the peptide substrate from the structure of the
active conformation superimposed onto the inactive conformation.
INACTIVE conformation
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ACTIVE conformation
Superimposed figure:
superimposes the
position and orientation
of the peptide substrate
from the structure of
the ACTIVE
conformation onto the
structure of the
INACTIVE conformation
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Compare and contrast relevant components of the inactive and active conformation
structures to briefly explain why the activation loop has to move for the kinase to
become active.
In the inactive conformation, the activation loop sits where the peptide substrate
would sit in the active conformation, occluding the catalytic cleft.
[Not required, but cool: In particular, Y1162 is held by a hydrogen bond to D1132
where the phosphorylation site substrate tyrosine is held in the active conformation.]
d. (4 points) The IRK has three tyrosines on its activation loop that can be phosphorylated,
as shown on the structure on the right. Briefly explain why Y1163 in particular must be
phosphorylated in order for the kinase to adopt the active conformation.
Y1163 is the tyrosine that must be phosphorylated in order for the kinase to adopt the
active conformation because the phosphoryl group of pY1163 forms a salt bridge that
stabilizes the activation loop away from the catalytic cleft in the active orientation of
the active loop, allowing substrate peptides to bind.
e. (4 points) Mutating Y1163 from a tyrosine to a glutamate (“Y1163E”) results in an IRK
that is always active and does not require phosphorylation to be turned on. Mutating
Y1163 to alanine (“Y1163A”) prevents the enzyme from ever being catalytically active.
Briefly explain why the Y1163E and Y1163A mutations cause these changes to the
catalytic activity of IRK.
A Y1163E mutation will mimic the negative charge of pY1163 that is necessary to
move the activation loop out of the catalytic cleft.
A Y1163A mutation will prevent the activation loop from ever acquiring the negative
charge necessary to form the salt bridge that stabilizes the active conformation of the
activation loop, thereby preventing the enzyme from becoming active.
f.
(4 points) Shown below is a close-up on the active site of the IRK. The enzymatic
residues are shown as green, the substrate tyrosine in the phosphorylation site is
magenta, and ATP is colored cyan.
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Mutating Asp1132 to asparagine does not affect the binding affinity of the kinase for the
substrate, but it does prevent the enzyme from being catalytically active. Briefly explain
why an IRK D1132N mutant is catalytically nonfunctional.
D1132 is required to deprotonate the substrate hydroxyl group to make it a better
nucleophile so that it can attack the gamma phosphorus atom of the ATP molecule.
Asparagine cannot deprotonate the hydroxyl group of tyrosine, preventing the
enzyme from being catalytically active.
g. (4 points) Shown below is a close-up of glycine-rich loop that helps stabilize the
phosphate groups of ATP within the kinase active site.
Briefly explain how glycine can favorably interact with the phosphate groups of ATP.
Which atoms of glycine form what types of intermolecular interactions with the
phosphate groups? Why must the glycine residues that interact with the phosphate
groups be in a loop and not involved in secondary structure?
Glycine can make favorable interactions with ATP by orienting its backbone amine
groups towards the negatively-charged oxygens of the phosphate groups to form
hydrogen bonds/ion-dipole interactions. Since these interactions involve the hydrogen
bond donors that would otherwise stabilize secondary structure, glycines involved in
such interactions must be in a loop.
5. (16 points) Many signaling pathways use scaffold proteins to co-localize the kinases involved
in a kinase cascade. Scaffold proteins coordinate signaling cascades by physically assembling
the components of a signaling pathway.
a. (4 points) Despite their role in facilitating the transmission of a signal, scaffold proteins
were first identified as inhibitors of signaling pathways. Scaffold proteins were first
thought of as inhibiting signaling pathways because experiments that increased their
cellular concentration to levels higher than normal decreased MAPK signaling output. If
you were to engineer a prokaryotic promoter to enable more transcription of a gene
than what normally occurs in the cell, list two parameters you would change about the
nature of the promoter and how you would change them to artificially increase
transcription of a gene of interest.

Idealize the -10 and -35 sequences
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



Add a positive regulator DNA sequence and protein (such as CAP)
Remove a negative regulator and operator sequence (such as the repressor)
[Insert the gene into an actively transcribed region of the genome]
I’m sure there are others…
b. (4 points) Briefly explain why dramatically increasing the cellular concentration of
scaffold proteins would inhibit the transmission of a MAPK cascade.
By increasing the scaffold concentration but keeping the concentrations of the
proteins to which they bind constant, it is likely to create scaffolds that are bound by
some, but not all of the necessary kinases required for signaling (for example: one
scaffold molecule will be bound to a MAPKKK and a MAPK, while another scaffold may
be bound to a MAPKK and nothing else, etc).
Kinases are sequestered by scaffold proteins that are not fully occupied by kinases,
and as a result, they are unable to participate in the signaling cascade.
As the scaffolds each have multiple binding sites, you can treat them each individually
for their specific protein. I.e., when over expressing a scaffold, you swamp the cell
with binding sites for a MAPKKK, MAPKK, and MAPK. There will be very little free
kinase, each will likely be bound a scaffold, but if the scaffolds are over-expressed,
then each scaffold may have an incomplete set of MAPKs.
Another possibility is that the scaffold sequesters the final kinase (e.g., MAPK) and
prevents it interacting with its downstream target.
Scaffold proteins also prevent exponential signal amplification, and therefore
increasing the concentration of scaffolds lowers the amount of signal produced.
c. (4 points) Two separate MAPK kinase cascades in yeast are used to respond to mating
pheromones and osmotic stress signals to produce mating and stress responses,
respectively. Each MAPK kinase cascade uses a different scaffold that recruits a unique
set of kinases. The diagram below depicts how one scaffold (a “mating” scaffold)
specifically binds a “mating” specific MAPKKK (“m”MAPKKK), a mating specific MAPKK
(“mMAPKK”), and a mating-specific MAPK (“mMAPK”). Similarly, a distinct scaffold (a
“stress” scaffold) specifically binds stress-responsive kinases (“sMAPKKK,” “sMAPKK,”
and “sMAPK”).
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It is thought that one function of a scaffold protein is to physically assemble the
components of a signaling pathway to provide a specificity of information flow and
prevent crosstalk between distinct MAPK cascades in the cell. To test this hypothesis,
scientists engineered three chimeric mutant scaffold proteins, which are diagrammed
below. Mutant scaffold 1 contains includes binding sites for mMAPKKK, mMAPKK, and
sMAPK. Mutant scaffold 2 contains includes binding sites for mMAPKKK, sMAPKK, and
sMAPK. Mutant scaffold 3 contains includes binding sites for sMAPKKK, sMAPKK, and
mMAPK. [Note: these kinases are capable of phosphorylating more than one substrate,
such that mMAPKKK can phosphorylate both mMAPKK and sMAPKK.]
When yeast cells had their mating scaffold replaced with mutant scaffold 1, they
responded to the presence of pheromones by initiating a stress response. Briefly explain
why.
The MAPKKK bound by a scaffold determines which signals the MAPK kinase cascade
will receive as input and the MAPK bound by a scaffold determines which response
the cascade will generate.
The MAPKKK which mutant scaffold 1 binds to is mMAPKKK, which will be activated
by an upstream component of the mating pheromone pathway. The MAPK the
scaffold binds to is sMAPK, which will activate downstream components of the
(osmotic) stress response. By physically linking mMAPKKK to sMAPK (via mMAPKK),
this mutant scaffold causes a mating signal to be transformed into a stress response.
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d. (4 points) Identify the input signal to which mutant scaffolds 2 and 3 will respond
(mating pheromones or stress) and the output response that will be generated (a
mating response or a stress response).
Mutant scaffold 2 will respond to mating pheromones and respond by initiating a
stress response. Mutant scaffold 3 will respond to (osmotic) stress and respond by
initiating a mating response.
6. (8 points) Consider a generic MAPK cascade, in which MAPKKK phosphorylates MAPKK
which phosphorylates MAPK which phosphorylates a transcription factor.
a. (4 points) Predict whether or not you would expect to see downstream transcriptional
activity if a MAPKKK was mutated to be constantly “on” (“active”) but the active site of
MAPKK was mutated to be nonfunctional. Briefly explain why.
I would not expect to see downstream transcriptional activity since MAPKK is
downstream of MAPKKK. If MAPKKK is constantly “on,” then it can keep activating
MAPKK, but if MAPKK cannot activate the downstream MAPK, the signal will not be
propagated downstream.
b. (4 points) Nuclear localization signal sequences can be stronger or weaker depending on
how closely they resemble a consensus sequence. MAPK has a nuclear localization
sequence (“NLS”) that differs enough from the consensus that MAPK is cytosolic unless
it is phosphorylated, but once phosphorylated it is imported into the nucleus by a cargo
receptor protein. Does phosphorylation increase or decrease the Kd of a cargo receptor
protein for the MAPK NLS?
Decreases. When hypophosphorylated, the cargo receptor has a low affinity for the
NLS. Once MAPK is phosphorylated by MAPKK, the cargo receptor has a higher affinity
(lower Kd) for the MAPK NLS.
7. (7 points) Shown below is a diagram describing how a cyclin:cyclin-dependent kinase
complex (“CyclinD1:Cdk5”) regulates DNA replication. The diagram shows whether a protein
inhibits (“
”) or activates (“”) the protein downstream of it.
a. (3 points) Indicate whether each of the following will promote or inhibit DNA
replication:
i.
E2F is inactivated
Inhibit
ii. pRB is inactivated
Promote
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iii. Cyclin D1:Cdk4 is inactivated
Inhibit
b. (4 points) Two other proteins, Akt and GSK-3, act upstream of the pathway shown
above. When Akt is inactivated, DNA replication is inhibited. When GSK-3 is inactivated,
DNA replication is increased. Inactivating both Akt and GSK-3 increases DNA replication.
Determine whether Akt acts upstream or downstream of GSK-3 and fill in the blank
boxes above using the “
” and “” symbols to show how Akt and GSK-3 regulate
downstream proteins.
8. (5 points) Cells from a certain type of lung cancer contain a chromosomal rearrangement
that results in an in-frame fusion between a portion of the EML4 gene and a portion of the
ALK gene to create a fusion protein, as shown below.
H2N
ALK
EML4
COOH
The ALK portion of the fusion protein is derived from the intracellular segment of a receptor
tyrosine kinase called ALK. This segment contains a fully functional tyrosine kinase domain
that must be phosphorylated to become catalytically active. The EML4 portion of the fusion
protein does not have enzymatic activity, but it contains an amphipathic alpha helix with
one face of the helix displaying polar side chains and the other face displaying non-polar
side chains.
In cells that express the EML4-ALK fusion protein, ALK kinase activity is turned on all the
time. However, cells that only express the ALK kinase domain in isolation do not show
continuous kinase activity. How might the presence of the EML4 segment of the fusion
protein promote continuous activation of the ALK kinase? Relate your answer to the
chemical characteristics of the EML4 structure.
The EML4 segment will allow for increasing the effective concentration of ALK kinase
domains, either by associating with another EML4 segment in order to sequester the
exposed non-polar side chains from the surrounding water OR by localizing to a
membrane.
The dimerization/membrane localization likely enables the autophosphorylation of
neighboring ALK kinase domains, which leads to activation of their kinase activity.
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