Download Problem Set #1 - Answers 2/11/2002 7.06 Spring `02 ANSWER

Problem Set #1 - Answers
2/11/2002
7.06 Spring ‘02
Question 1
A glycoprotein with a molecular weight of 83 kDa was isolated from detergent extracts of
erythrocyte membranes. This protein was found to be a single peptide chain of 780 amino
acids. The amino acid sequence of this protein was determined and found to have 3 αhelical segments of hydrophobic amino acids at positions 151-175, 336-360, and 570-594.
The only tyrosine residues in the protein were found at positions 122, 437, 523, 703 and
757. In an effort to determine the orientation of this protein in the membrane, you conduct
the following experiments:
a. You treat the intact erythrocytes with the enzyme lactoperoxidase and the isotope 125I.
This enzyme will radioactively label tyrosine residues. You then wash the cells, extract
the protein from the membrane and perform peptide mapping experiments. Through
these experiments you determine that the only tyrosines that were labeled are at
positions 703 and 757.
b. You determine the position of the carbohydrate unit on the protein and find that it is
covalently linked to the asparagine at position 302.
Based on the data provided above, use the lipid bilayer diagram below to answer the
following questions concerning the orientation of your protein in the membrane:
i.
Draw the polypeptide chain and indicate the N-terminus & C-terminus of the protein
with respect to the intracellular and extracellar sides.
ii. Label the position of the α-helical segments and include the amino acid numbers to
indicate the orientation of each α-helix.
iii. Label the location and the amino acid number for each of the tyrosines on the
peptide chain. Be sure to indicate which of the tyrosines have been labeled with 125I.
iv. Label the position of the carbohydrate (asparagine) molecule with respect to the
peptide.
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Problem Set #1 - Answers
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v. Explain briefly why some, but not all of the tyrosines are labeled with
7.06 Spring ‘02
125
I?
Only those tyrosines that are on the outside of the cell are accessible by the
lactoperoxidase enzyme since the enzyme cannot pass through the cell membrane.
c. To better understand the structure of this protein, you treat another batch of
erythrocytes with a cocktail of different proteases. You then wash away the proteases,
disrupt and extract the membrane proteins using a detergent. You run the extract using
SDS PAGE and then perform an immunoblot (also called a Western blot) of the gel using
polyclonal antibodies you have prepared against the isolated protein. You find that your
antibodies recognize 2 different peptide fragments that are approximately 180 & 260
amino acids long. The amino acid sequences of these fragments are different from one
another.
vi. In the box below, show the position of the two bands from the western blot:
vii. Next to your two bands, list the molecular weight of each peptide and label which
segments of the protein, using amino acid numbers, are represented by each of the
bands (show your calculations).
To estimate the molecular weight of a polypeptide chain, multiply the number of
amino acids in the chain by 110 daltons (a weighted average of the amino acid
molecular weights):
1) 336 – 594 = 258 amino acids X 110 Daltons = 28,380 Daltons = ~28 kDa.
2) 1 – 175 = 175 amino acids X 110 Daltons = 19,250 Daltons =~19 kDa.
(Note: this only provides a rough estimation of weight.)
Question 2
After doing some further experiments, you learn that your membrane protein is an ion
channel. A number of cell signaling mechanisms open Ca2+ channels in the membrane
allowing a transient increase in Ca2+ concentration which activates certain Ca2+ responsive
proteins in the cytosol. Answer the following questions based on a cytosol Ca2+
concentration of 10-4 mM and an extracellular Ca2+ concentration of 1 mM:
a. If a cell has a membrane potential of -0.05 V at 37 oC, what is the Ca2+ equilibrium
potential:
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Problem Set #1 - Answers
VCa 2+ =
2+
out
2+
in
Ca
RT
Ln
zF
Ca
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cal 

2
 ( 310 K )
1 mM
mol i K 

Ln −4
=
=
cal  10 mM

4
2  2.3 × 10

mol iV 

( 0.0135V )( 9.21)
= 0.123 V
b. Following the increase in Ca2+ that occurs upon signaling, the cytosolic Ca2+ level is soon
returned to normal by a membrane Ca2+ ATPase that uses the energy of ATP hydrolysis
to pump the Ca2+ out of the cell. A Ca2+ ATPase present in the membrane of the cell was
found to pump out 1 mole of Ca2+ for 1 mole of ATP hydrolyzed. Assume that the ∆G for
ATP hydrolysis is -12 kcal/mole. Calculate the overall change in free energy that occurs
per mole of Ca2+ pumped from the cell by the Ca2+ ATPase as it seeks to maintain the
cytosolic Ca2+ level at its normal concentration of 10-4 mM:
∆GCa2+ = RT Ln
2+
Caout
cal 
cal 
kcal
1 mM


− zFE =  2
− ( 2 )  2.3 × 104
 ( 310 K ) Ln −4
 ( −0.05 V ) = 8
2+
Cain
mol iV 
mol
10 mM
 mol i K 

∆GCa 2+ + ∆GATP = ∆Goverall
8
kcal
kcal
kcal
− 12
= −4
mol
mol
mol
(The overall ∆G is negative, and therefore favorable)
Question 3
a. GLUT1 is a glucose transporter found in many cells. The Km of GLUT1 for D-glucose is
1.5mM, and the Vmax for GLUT1 transport of glucose is 500 µmoles transported per
milliliter of cells.
v, umoles D-glucose/mL cells
i.
Plot velocity of glucose transport vs. extracellular glucose concentration.
500
i. f acilitated transport
450
iii. mutant w ith Km=7.5mM
400
350
300
iv. Mutant w ith 1/2 # of transporters
on the cell surf ace
(V max=250umoles/mL)
250
200
150
ii. passive transport
100
50
0
0
20
40
60
80
100
[D-glucos e ], m M
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Problem Set #1 - Answers
ii.
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Is GLUT1 transport of glucose an example of facilitated or passive transport across
the plasma membrane? Draw what the other kind of transport would look like on
your graph.
GLUT1 transport of glucose is FACILITATED transport.
iii. How would this graph change if the cells expressed only a mutant version of GLUT1
with a higher Km for glucose?
iv. How would this graph change if the cells expressed wild type GLUT1, but only had
half as many transporters on their surfaces?
b. What is the velocity of glucose transport mediated by the GLUT1 transporter under
physiological conditions when the extracellular concentration of D-glucose is 5mM?
v=
Vmax
K
1+ m
C
v=
500µ moles / mLcells 500 ×10−6
=
moles / mLcells = 380µ moles / mLcells
1.5 ×10−3 M
1.3
1+
5 × 10−3 M
c. If D-mannose has a Km of 20mM but the same Vmax as D-glucose, what extracellular
concentration of D-mannose would be needed to reach the same velocity of transport as
you determined in part b?
v=
V max
Km
1+
C
C=
vK m
(380µ moles / mLcells)(20mM )
C=
= 63mM
(500µ moles / mLcells − 380µ moles / mLcells )
Vmax − v
d. There is a high free energy barrier to transporting glucose across the plasma membrane
in the absence of GLUT1 and other transporter molecules. (Why?) Transporters lower
the energetic barrier and facilitate transport across the membrane in both directions. If
transporters are not unidirectional, what causes glucose to be transported from outside
to inside cells instead of vice-versa?
Glucose transport across the membrane is unfavorable because the hydrophobicity
of the interior of the plasma membrane makes it nearly impermeable to watersoluble polar molecules such as glucose. There are no polar groups to solvate
glucose in the interior of the bilayer.
The concentration gradient of glucose (higher [glucose] outside the cell, lower
[glucose] inside the cell) causes it to be transported from outside to inside.
f.
When GLUT1 transporters are reconstituted in liposomes and then incubated with
glucose, the levels of glucose transport into and out of the liposomes reaches a state of
equilibrium. Why do erythrocytes in the body that absorb glucose not achieve
equilibrium between import and export of glucose? [Hint: What happens to glucose
once it enters these cells?]
When glucose is transported across the membrane of a liposome, it does not
undergo any chemical modifications and can easily be re-transported back out of
the liposome. In contrast, when glucose is transported into an erythrocyte, it is
rapidly phosphorylated and becomes glucose-6-phosphate. (This is the first step
in the glycolytic pathway.) Glucose-6-phosphate cannot be transported out of the
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Problem Set #1 - Answers
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cell by GLUT1. As all glucose transported into the cell is immediately
phosphorylated, there is effectively no increase in the intracellular [glucose].
Therefore the glucose gradient is maintained, the rate of glucose entry into the
cell does not slow down, and, since there is no regular glucose inside the cell to be
exported, it does not reach equilibrium between import and export.
Question 4
Cells have several ways of coupling energetically non-favorable processes to energetically
favorable processes in order to drive them forward. What drives the following processes
forward in cells?
a. glucose transport by GLUT1
Glucose gradient across the cell membrane drives to transport into the cell.
b. diffusion of EtOH across the plasma membrane
EtOH gradient across the cell membrane allows diffusion into the cell.
(This is an example of passive transport.)
c. export of Ca++ from cells by the Ca++ pump
Hydrolysis of ATP to ADP & 2Pi produces lots of energy and drives Ca++
transport against its gradient.
d. glucose transport by the Na+/glucose symporter
Glucose transport against its gradient is coupled to the favorable transport of
Na+ down its gradient. (Both the 2 Na+ ions and the glucose molecule travel in
the same direction across the membrane.)
e. Ca+ and Na+ by the Na+/Ca++ antiporter
The energy from the transport of three Na+ ions down the Na+ gradient is
harnessed to drive one Ca+ ion out of the cell against its gradient. (Na+ and
Ca+ ions move in opposite directions across the membrane.)
Question 5
Define the following terms:
a. Nucleus
The nucleus is the subcellular membrane-enclosed organelle that encloses the
chromosomes and separates them from the rest of the cytosol in eukaryotic
cells.
b. SDS PAGE
SDS-PAGE (Sodium Dodecyl Sulfate PolyAcrylamide Gel Electrophoresis) is a
method for separating polypeptides (proteins) by size. SDS is an anionic
detergent that denatures proteins and coats them with a negative charge.
When these proteins are subject to an electric field, they migrate through a
polyacrylamide matrix toward the + electrode. Proteins are separated by size
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as it takes longer for bigger polypeptides to snake through the gel matrix than
it does for shorter polypeptides.
c. Monoclonal & Polyclonal antibodies
Antibodies are proteins that interact with particular sites (epitopes) on
antigens (often proteins). Monoclonal antibodies (one clone) are typically
generated from mice and recognize one single epitope within a protein.
Monoclonal antibodies are produced by hybridoma cell lines, fusions of B-cell
lymphoma cells and cells from an immunized mouse’s spleen. These
hybridoma lines secrete antibodies into the culture media as they grow.
Polyclonal antibodies can be generated in a variety of animals (rabbits, rats,
goats…) and are harvested by bleeding the animals and purifying the
antibodies from the serum by affinity purification. Polyclonal antibodies (many
clones) recognize multiple epitopes within a protein.
d. Epitope
An epitope is the short peptide sequence within a protein or other antigen that
is recognized and bound by an antibody.
e. Amphipathic
An amphipathic molecule has both regions which prefer to interact with water
(hydrophilic) and regions which prefer to be secluded from water
(hydrophobic). A phospholipid is a good example.
f.
pH (formula)
pH is a measure of acidity. Lower pH = more acidic (more H+).
pH = − log[ H + ]
g. Conformational Change
A conformational change is an alteration in the secondary or tertiary structure
of a protein that occurs when the protein is modified (phosphorylated, etc.),
bound to another protein, or bound to an enzymatic substrate (in enzymes, this
is called “induced fit”). Conformational changes involve movement of the
protein substructures/secondary elements relative to one another, but no
cleavage of the polypeptide.
h. Detergent
Detergents are amphipathic molecules. Ionic detergents such as sodium
deoxycholate and sodium dodecylsulfate (SDS) contain charged groups and
nonionic detergents such as Triton X-100 lack charged groups. (See pp83-4.)
Detergents disrupt membranes by intercalating into phospholipids bilayers and
solubilizing lipids and proteins. Detergents can be used to denature proteins.
Proteins normally hide their hydrophobic residues from water within their
interiors. Detergents have hydrophobic regions that interact with the normally
hidden hydrophobic side chains and hydrophilic regions that interact with
water; the interactions between strong detergents and hydrophobic regions of
proteins decrease the energy of solvation of these regions and cause proteins
to unfold.
i.
Conditional Mutants
Conditional mutants contain genes whose gene products can carryout the wild
type functions under certain conditions (often 25°C) but whose gene products
become unstable and non-functional under other conditions (high or low
temperature).
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Problem Set #1 - Answers
j.
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Complementation
In genetics, complementation is the restoration of a wild type function. If two
mutants with the same mutant phenotype complement one another, then their
mutations are likely in two different genes. Thus mutant alleles isolated in
genetic screens can be placed into complementation groups which usually
correspond to the # of different genes represented.
k. Integral & Peripheral Membrane Proteins
An integral membrane protein is inserted through the plasma membrane (via
hydrophobic alpha helices, beta barrels, etc.) and cannot be dissociated from
the membrane except under harsh conditions (high salt or detergent). A
peripheral membrane protein can associate with the membrane by using a
hydrophobic tail, usually via a myristolated residue or other fatty acid
modification of an amino acid, or via association with an integral membrane
protein or the polar head groups of membrane phospholipids. Peripheral
membrane proteins are more easily disrupted from the membrane and are
typically soluble in water while integral membranes that have been disrupted
from the membrane are typically not soluble in water and have to be
reconstituted in micelles or liposomes.
l.
Transfection
Transfection is a method for introducing DNA into tissue culture cells.
Transfection of a plasmid can be used to induce expression of any gene of
interest, mutants forms of genes (inc. dominant negative versions), or tagged
versions of genes whose protein products are used for fluorescent microscopy
(GFP-tagged) or biochemistry (HA-, myc-tagged, etc.).
m. Transformation
Transformations are used to transfer DNA into bacterial or yeast cells. This
method can be utilized to introduce plasmids into bacteria for amplification or
protein expression and to generate yeast strains that have integrated plasmids
for gene knock-outs, protein tagging, protein truncation, etc., or yeast strains
that carry CEN/ARS plasmids that are self-replicating and carried through cell
division as minichromosomes.
n. Chromosome Non-Disjunction
Chromosome non-disjunction occurs when a pair of replicated sister
chromatids both move to the same pole of the mitotic spindle during anaphase.
This results in one too many chromosomes in one daughter cell and one too
few in the other daughter.
o. Haploid and Diploid
Haploid cells have only one copy of each chromosome and, hence, only one
allele of each gene. Diploids have two copies of each chromosome, and
therefore two alleles of each gene. In yeast, a haploid cell has 1C DNA content,
while a diploid has 2C. C=the normal complement of chromosomes. Humans
are diploid, but gametes (egg, sperm) are haploid.
p. Aneuploid
An aneuploid cell has an inappropriate number of chromosomes. (Too few or
too many.) Aneuploid cells may have several copies of one or more
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chromosomes and only one of another. They may also have translocations or
duplications of chromosome segments.
q. Allele
An allele is one version of a gene sequence. The wild type allele is the allele
that normally functions in a cell. Mutant alleles encode non-functional (or
partially functional) gene products, and conditional alleles only behave as
mutants under inducible conditions (high temperature, etc.). Often there is
more than one wild type allele within a population. This is called allelic
variation and each site of variation within the genetic code that does not
adversely affect function is called a polymorphism.
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