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Unformatted text preview: 7.06 Spring 2004 PS 2 KEY 1 of 7 7.06 Problem Set 2 Answer Key
1) You have started working in lab that investigates signaling pathways downstream of TGFß
receptors. One of the senior graduate students has been working on a screen for novel TGFß’s
and has one that she thinks might interest you—she calls it TGFß57. She’s already cloned the
gene so you can make pure TGFß57. You decide to take up the challenge of characterizing this
new growth factor.
a) You would first like to find some high affinity receptors that bind your novel growth
factor. Design an experiment to identify the cell-surface TGFß57 receptors.
Answer: There is no one right way to do this, however, a common method is to radiolabel
the purified growth factor with iodine-125 (125I) and culture the labeled growth factor with
cultured cells (you would probably use the same fibroblast cell line that’s been used in
earlier experiments with TGFß57). Then you’d treat the culture with a chemical crosslinker to covalently bind the growth factor to any receptors it bound. You can then use
standard purification methods to purify the labeled receptors from cell membranes (nonionic detergents to separate the receptors from the cell membrane and chromatography or
electrophoresis to separate your receptor from other proteins).
You would now like to investigate the binding properties of TGFß57.
b) You decide to look at the maximal physiological response of TGFß57R to TGFß57.
Based on the graph below determine the Kd (in relative concentration) of TGFß57 binding, the
percentage of receptors occupied at 50% of maximal physiological response, and the percentage
of maximal physiological response when half the receptors are occupied. What does this tell you
about the percentage of bound receptors needed to achieve maximal response?
*** NOTE: y-axis label: Fraction of maximum cellular response Answer: Kd is about 1.2. At half-maximal physiological response, only about 26% of the
receptors are bound. When 50% of receptors are bound, about 77% of the maximal
physiological response is elicited. This tells you that cells do not nee to have 100% of the
receptors bound by ligand in order to achieve a maximal response. 7.06 Spring 2004 PS 2 KEY 2 of 7 c) You would now like to determine the binding affinity of TGFß57 for the TGFß57
receptor that you have been working with. Design an experiment to measure the Kd of TGFß57
binding high- affinity receptors in a cultures fibroblast cell line. Explain how you would deal
with TGFß57 binding nonspecifically to low-affinity receptors.
Answer: You’d want to design some kind of binding assay to measure receptor binding.
What is commonly done is to take a sample of cells and incubate them with increasing
concentrations of 125I-labeled TGFß57. After incubation, you separate the cells from
unbound TGFß57 (usually by centrifugation) and the amount of radioactivity bound to
them is measured. This gives you the total binding of TGFß57 to receptors. In order to
determine specific binding to high-affinity receptors you have to measure nonspecific
binding. This is determined by repeating the binding assay in the presence of a 100-fold
excess of unlabeled TGFß57, which saturates all the specific high-affinity sites. In this case
all the labeled V binds to nonspecific sites. To determine specific binding you simple
calculate the difference between total binding and nonspecific binding.
d) Using the specific binding curve below, determine the Kd of TGFß57 for high-affinity
receptors. Also, find the total number of receptors per cell.
*** NOTE: y-axis label: [I125] TGFb-57 Bound (molecules per cell) Answer: The Kd is determined by finding the concentration of TGFß57 at half-maximal
binding. Kd= 10nM. The total number of receptors per cell is 19,000.
e) You’d now like to determine the sensitivity of these cells to TGFß57. You know that the Kd
for binding TGFß57 to its receptor is about 1X10-8 and you know there are a total of 19,000
receptors per cell. You have previously determined that only 13% of the 19,000 receptors must
be bound to obtain the maximal cellular response. Determine the TGFß57 concentration needed
to induce the maximal response. What would be the concentration of TGFß57 required to obtain
maximal response if there were only 5000 receptors per cell, what is the fold difference 7.06 Spring 2004 PS 2 KEY 3 of 7 compared to the normal concerntration you just calculated, and would this make the cell more or
less sensitive to TGFß57? Why might a cell want to regulate the number of receptors to a given
Answer: To calculate the concentration of TGFß57 to induce maximal response, you use
the equation [L] = Kd/(RT/[RL])-1. [RL] is the number of TGFß57-occupied receptors
needed to induce maximal response—which is 13% of 19,000. Plugging all the values in to
the equation gives you [TGFß57] = 1X10-9M. If the total number of receptors was reduced
to 5000, the [RL] would still be 2470, then [TGFß57] = 5X10-9M—a 5-fold increase in the Kd
of TGFß57. This decrease makes the cell less sensitive to TGFß57 because there needs to be
a higher concentration of TGFß-57 released in order to achieve the maximal response. A
cell might want to regulate the number of receptors in order to direct physiological or even
f) If TGFß57 binding to its receptor had occurred with a low affinity (a Kd higher than 1X-7M),
you couldn’t have measured the Kd by performing the binding assay from above. How could you
have determined the Kd if this had been the case?
Answer: You could perform a competition assay with another ligand that binds to the
same receptor with high affinity. In this type of assay, increasing amounts of an unlabeled,
low-affinity ligand are added to a cell sample with a constant amount of the radio-labeled
high-affinity ligand. Binding of unlabeled competitor blocks binding of the radioactive
ligand to the receptor. Consequently the concentration of competitor required to inhibit
binding of half the radioactive ligand approximates the Kd value for binding if the
competitor to the receptor.
2. A colleague in your lab discovers that the binding of TGFb57 to its receptor turns on
expression of a gene that inhibits cell growth, SloGro. SloGro is frequently mutated in cancers,
and you suspect that components involved in the TGFb57 signaling pathway might also be
relevant to cancer research. You decide to set up a cell culture screen to identify proteins
involved in TGFß57 signaling. Your screen is a success, and you isolate a number of cell lines
that have mutations in the TGRb57 signaling pathway.
A) Explain how the following mutations would affect TGFb57 signaling. Assume that the basic
components of the b57 pathway are identical to the canonical TGFb signaling pathway.
i)The MH1 domain of Smad 3 is missing.
The MH1 domain contains a NLS sequence. Thus Smad 3 will not localize to the nucleus
even in the presence of TGFb57 and will not activate transcription of SloGro.
ii)R1 has a truncated cytoplasmic tail and cannot be phosphorylated.
The R1 kinase is activated by phosphorylation of its cytoplasmic tail by RII. A truncated
version of R1 that cannot be phosphorylated will be unable to phosphorylate Smad3, and
will not activate the signaling cascade. 7.06 Spring 2004 PS 2 KEY 4 of 7 iii) Smad 7 is constitutively active.
Smad 7 blocks the ability of R1 to phosphorylate Smad 3, which prevents it from entering
the nucleus and activating transcription.
B) You also isolate a version of Smad3 that cannot bind DNA, and find that this mutant version
exerts a dominant negative effect on the TGFß57 pathway. Give one possible mechanistic
explanation for how mutant Smad3 could act dominant negatively on the pathway.
One possible scenario is as follows: Smad 3 can’t bind DNA, and therefore the complex
between mutant Smad3 and Smad4 can’t activate transcription of target genes. Because
the complex that binds DNA and activates transcription involves two Smad 3 proteins, the
mutant version gets incorporated into complexes with wild type Smad 3, effectively
inactivating complexes with wild type protein as well.
3) Epineprhine increases the concentration of glucose 1-phosphate in wild type liver cells.
Explain how the following mutations would affect the glucose 1-phosphate concentration within
an epinephrine stimulated liver cell.
i) The cell contains a PKA with a nulceotide binding site that is unable to bind cAMP.
cAMP binding liberates the active version of PKA from its psuedosubstrate. A PKA that is
unable to bind cAMP will therefore not be activated upon epinephrine stimulation and
glucose 1-phosphate concentration will not increase.
ii) The cell contains an unphosphorylatable version of Inhibitor of Phosphoprotein Phosphatase.
Inhibitor of Phosphoprotein phosphatase will not inactivate phosphoprotein phosphatase in
its unphosphorylated form. Thus phosphoprotein phosphatase will be active even in the
presence of active PKA, and glucose 1-phosophate concentration will not increase.
iii) The cell contains a mutant version of Gsa that cannot hydrolyze GTP .
Gsa that cannot hydrolyze GTP will remain active once turned on. This will constitutively
activate adenylyl cylase and all the downstream targets of PKA, which will lead to
increased glucose 1-phosphate concentration. 4)
a) Given the following values, calculate the electrochemical equilibrium potential of each ion
(F=23, 062cal/(mol V), R = 1.987 cal/K mol, T = 293K).
K+out = 3 mM Na+out = 117 mM Cl-out = 120 mM 7.06 Spring 2004 K+in = 90 mM PS 2 KEY Na+in = 30 mM 5 of 7 Cl-in = 4 mM E = (RT/zF) ln [ionout/ionin]
EK = - 85.9 mV
ENa = 34.4 mV
ECl = - 85.9 mV
b) What does the electrochemical equilibrium potential reflect/what is true at the electrochemical
At the electrochemical equilibrium potential, the concentration gradient for the ion is
balanced by the electrical gradient. Hence, there is not net flux of ions across the
membrane. That is, the rate (notice rate, not concentration) of movement of the ion in both
directions is equal.
c) If one reduces the magnitude of the K+ concentration gradient across the cell, one will
increase/decrease (circle one) the equilibrium potential. Why?
One would decrease the equilibrium potential. Remember that at the electrochemical
equilibrium potential, the concentration gradient is balanced by the electrical gradient. If
we decrease the magnitude of the concentration, the electrical gradient will also necessarily
decrease. Another way to think about it, is as follows: by increasing extracellular K+, we
are decreasing the flow of K+ out of the cell because the concentration gradient that makes
this process energetically favourable has decreased. Assuming the Na+ concentration
remains the same, its flow in the cell remains the same. So, we have decreased the flow
(current) of K+ out of the cell while not changing the flow of Na+ into the cell. As a result,
the cell should depolarize.
d) Assume that the resting membrane of a cell is determined primarily by the equilibrium
potential of potassium. If you insert an electrode into this cell and clamp the membrane voltage
to +20 mV, would K+ be moving into or out of the cell through the potassium channels (K+out = 3
mM, K+in = 90 mM)?
From above, the equilibrium potential of potassium is –87.2 mV. The ratio of K+ (out/in) is
1/30. At +20 mV, the ratio of K out/in would be:
E = 0.02V, E = 0.059log (out/in), (out/in) = 10(E/0.059) = 891.3.
Since the present K+ concentration ratio of 1/30 is less than that required at +20 mV, the
potassium will leave the cell until the concentration outside is nearly 3 orders of magnitude
greater than the one on the inside. 7.06 Spring 2004 PS 2 KEY 6 of 7 e) You isolate cells that have only K+ channels. You also have a drug that causes this channel to
open. With a voltage clamp apparatus, you find, as expected, that in the presence of the drug the
membrane potential is dependent on the concentration of extracellular K+ used in the experiment.
At one concentration of K+ of 5 mM, you find that the membrane potential is -80.5 mV. From
this information, calculate the intracellular K+ concentration.
E = - 0.0805 V, E = 0.059 log (Kout/Kin)
Kout/Kin = 10(E/0.059) = 0.043
Kin = Kout / 0.043 = 115.7 mM
5) You are doing a summer internship in a lab that studies a recently discovered receptor called
RKR. It is a G-protein coupled receptor that is expressed in kidney cells and found to be mutated
in patients with a kidney disease. You sequence the receptor in 100 patients and find that the
mutations fall into two classes, A and B. Unlike wildtype RKR, both mutants fail to increase
cellular levels of cAMP.
(i)The ligand for the RKR receptor is called MC. How could you find out if the mutant receptors
are expressed on the surface of kidney cells?
First you would transfect cDNAs for wt and mutant RKRs into cultured human kidney
cells. These receptors would have been tagged e.g. using GFP, and then you would incubate
the ligand with the cells and then look under the microscope to see whether the receptor is
localized at the membrane and you would examine the relative levels of expression between
mutant and wild type.
(ii)You find that mutant A is localized to the membrane at approximately the same levels as
wildtype but mutant B has very low levels of RKR receptor on the surface. Mutation A lies in the
conserved C3 loop, which is known to be important for binding and activation of the Ga subunit.
How could you determine whether Ga becomes activated in cells expressing mutant A?
One way to do this would be to use the FRET system mentioned in class. You would tether
a YFP tag to the Gb subunit and a CFP tag to the Ga subunit. In the absence of ligand both
subunits are close together and so when CFP is excited at 440nm FRET occurs between
CFP and YFP and so light of 527nm is emitted. When ligand is bound to wildtype receptor
the Ga subunit binds GTP and dissociates from the Gbg subunits. Now when CFP is
excited, light emits at 490nm since no FRET can occur.
(iii) While trying to understand the basis for mutant B you look at receptor localization in the
presence and absence of ligand using the GFP tagged versions of the receptors from part (i).
What do you expect to see in the wildtype cells after extended exposure (30mins) to ligand?
You would expect to see the receptor localized at the plasma membrane when ligand is not
present. After 30 mins of ligand exposure you would expect to see some of the GFP 7.06 Spring 2004 PS 2 KEY 7 of 7 fluorescence in the cytoplasm in a punctate pattern, since upon extended interaction with
the ligand the receptors are internalized into endosomes using b-arrestin.
In the cells expressing mutant B the receptor is found to localize predominantly to the cytoplasm
in the presence or absence of ligand. How could you explain this? How could you demonstrate
b–arrestin could be constitutively binding to the mutant RKR receptor in the presence and
absence of ligand, resulting in the receptor being internalized and hence there is insufficient
receptor on the surface to activate the required signaling pathways. To demonstrate this
you could express a version of b-arrestin that is tagged with YFP and look for
colocalization with mutant RKR-GFP in the endosomes in the absence of the MC ligand.
(iv) The RKR receptor has a conserved motif in the cytoplasmic portion of the protein that
contains Ser-Ser-Thr. You want to show that if you prevent the constitutive b-arrestin mediated
endocytosis of the mutant B version of the RKR receptor that it will then be capable of signaling
Your advisor suggests that you mutate all three residues to alanine in the mutant receptor? Why
does he choose these three residues? Why does he suggest mutating them to alanine?
After extended exposure to ligand, GPCRs become phosphorylated and then b-arrestin can
bind to the phosphorylated residues and this results in endocytosis of the receptor and
termination of signaling. Serine, threonine and tyrosine are all residues that can be
phosphorylated and hence if they are converted to another residue then you won’t get
internalization of the receptor.
He chose alanine because it is small and so probably won’t affect the secondary structure of
the protein and also it is used in mutagenesis studies because it allows you to determine the
functional importance of the side chain of an amino acid – which is where phosphorylation
occurs on serine and threonine. ...
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