The document is Stephen Corvini's answers to questions on a molecular cell biology test. It contains his responses to 4 questions. For question 1, he summarizes the role of integral membrane proteins in red blood cells. For question 2, he discusses N-linked and O-linked oligosaccharides and their roles in protein glycosylation and transport to lysosomes. For question 3, he explains receptor tyrosine kinases and their signaling mechanisms. For question 4, he defines the functions of topoisomerase I and II.

1. Stephen Corvini
Take Home Test
Molecular Cell Biology
Dr. Bettica
November 25, 2011
Question #1:
The human red blood cell is a great example for studying protein-membrane function
interactions because these are simple cells in which the membrane is a crucial component. Red blood
cells act primarily as means of transportation within the body and their integral membrane proteins
work the regulate gas, ion and other substance transport between the internal and external
environments of the cell. Spectrin maintains the structural integrity of the cell membrane of the
erythrocyte. It forms a movable mesh network that allows for lateral movement of protein components
within the inter-membrane layer. Band 4.1, adducin, and tropomysin contribute to the determination of
actin filament length. Band 3 appears to acts as a channel protein within the membrane. Band 4.1 works
with ankyrin in order to form structural complexes within the membrane as well.
In regard to constraint of lateral movement within the membrane, Band 3 is constrained and
this appears to be regulated by ankyrin and protein 4.1. These proteins are attached to the mesh
network of 100nm spectrin filaments within the membrane. Spectrin is flexible and as a result allows for
the lateral movement of molecules. Band 3 is a component of junctional complexes in the membrane
and as a result does express some physical constraints in its overall lateral mobility.
If a red blood cell were deficient in spectrin and was lacking proper spectrin as a result, there
would be a highly negative impact on the lateral movement of Band 3. This protein relies on spectrin
and the flexible mesh network that it maintains in order to achieve a proper degree of lateral mobility.
Band 3 is mostly constrained by spectrin and if this protein were not sufficiently bound to Band 3 it is
possible there would be an uncontrolled degree of lateral mobility which would negatively effect the
molecular transport function of the membrane as well. Without proper spectrin binding there would be
uncontrolled movement of Band 3. Spectrin holds Band 4.1 and ankyrin in place around Band 3 and this
would be absent with improper spectrin binding. This would result in the improper constraint of Band 3.
Corvini 1

2. There are many methods which are used to isolate extrinsic and intrinsic proteins from the cell
membrane. Detergents are chemical substances that are used in order to isolate specific proteins. Ionic
detergents such as sodium dodecyl sulfate (SDS) and nonionic detergents, such as octylglucoside are
used to treat various protein types. All detergents remain monomeric in low concentrations and do not
readily effect the proteins that are present within the cell environment. It is only when they are present
in strong/critical amounts that they will form globular molecules known as micelles and work to break
down the membrane environment so that various proteins can be purified.
Ionic and nonionic detergents have various advantages and disadvantages. Nonionic detergents
can readily isolate functional protein molecules. This allows them to be studied. Ionic usually render
most proteins inactive and must be removed in order to reinstate functionality of these molecules.
Conversely, ionic detergents can solubilize the most hydrophobic proteins. Ionic detergent isolated
proteins cannot be used for studies because they have lost their functionality. Nonionic detergents allow
for collection of proteins that can be used for studies, but these detergents are not necessarily capable
of solubilizing all membrane types.
Question 1A:
In order to characterize the integral protein composition I would choose a proper detergent. For
purposes of this scheme, sodium dodecyl sulfate (SDS) would be effective. This would solubilize the
membrane portion surrounding the lysosome. I would then utilize a mixture of phospholipids and
incorporate them with the detergents allowing for a phospholipid molecule to be created that would
hopefully capture the desired protein. The detergent SDS would be removed; thus allowing the integral
protein of the lysosomal membrane to be successfully purified.
Question 1B:
I would analyze the purified protein through SDS polyacrylamide-gel electrophoresis. This would
expose the hydrophobic core of the protein. Through comparison with the hydrophobic cores of other
isolated proteins from the lysosomal membrane I could confirm it was a member of that particular
membrane type. This would help rule out its possibility for being a contaminant from other parts of the
cell.
Corvini 2

3. Question 2:
An N-linked oligosaccharide is one that is added to the NH2 group of an asparagines amino acid
within a protein. These are added within the lumen of the Endoplasmic Reticulum. They don’t act on
cytosol located proteins. This molecule is important for the process of protein glycosylation. The O-
linked oligosaccharides are added to the hydroxyl group which forms a side chain on the amino acid
serine, hydroxylysine, or threonine. This type of protein glycosylation occurs within the lumen of the
Golgi Apparatus. In general the N-linked are longer than the O-linked.
The enzyme oligosacchyral transferase catalyzes the transfer of oligosaccharides to the target
functional group with which the sugars most commonly bind. Prior to this transfer a precursor
oligosaccharide is modified before attaching to the components of the proper receptor protein. They are
prepared for synthesis when they are added to a carrier lipid dolichol molecule. Glucose, mannose, and
N-acetylglucosamine represent the different categories of N-linked oligosaccharides.
Once essential lysosomal enzymes are recognized by the mannose 6-phosphate receptor they
must be transported to the lysosome. The hydrolases bind to the MP6 receptor and receive a clathrin
coat. This forms a receptor dependent vesicle through the process of endocytosis. This vesicle is then
ejected from the trans-terminal end of the Golgi Apparatus. They bind to the lysosome where the acidic
pH dissociates the vesicle and delivers the hydrolase successfully. Phosphates are removed from the
hydrolase and the enzyme is incorporated into the lysosome.
Modifications of N-linked oligosaccharides occur within the lumen of the Endoplasmic
Reticulum. This is where they are added as well. Transfer of N-linked sugars happens with help of
oligosacchyral transferase on the luminal side of the Endoplasmic Reticulum membrane. O-linked sugars
are added and modified within the lumen of Golgi Apparatus. They undergo similar modifications and
additions as do the N-linked sugars, but are less common. Only about 10% are usually O-linked sugars.
Corvini 3

4. Question 2A:
I-cell disease patients exhibit a series of molecular symptoms. Normally, most hydrolytic
enzymes are missing from lysosomes, undigested substrates accumulate and this forms inclusions which
are very large within cells. This disease is due to a single gene defect and as a result can be considered
hereditary in nature. Hydrolases are missing from lysosomes in the blood as well. This is because they
are secreted rather than properly transported to their lysosome receiving cells.
Question 2B:
LDL proteins work to bind to specific LDL receptor proteins at a specific binding site. The
mannose 6-phosphate is contained in the area where the binding of multiple LDLs begins to form the
membrane of the transport vesicle known as the clathrin coat. Areas known as clathrin pits exist as this
structure forms and play a direct role in its construction. Adaptor proteins enclose the components of
the coat of the vesicle. The vesicle will contain M6P receptors and the desired enzymes. Endocytosis
ejects the vesicle which is released in coated form. It is uncoated and becomes a vesicle. It then fuses
with the endosome which separate hydrolases and LDL components. Hydrolases are then sent to
awaiting lysosomes and LDL components returned to cell membrane in a recycling endosome which
binds to them back into the original cell membrane.
Question 2C:
The mannose 6-phosphate molecules are normal because they are not directly effected by the
pH. Within the cell environment pH is maintained by an ATP driven-pump. The more basic pH may not
be high enough to successfully dissolve the membrane of the transport vesicle. Also, it is possible that
the LDL receptors are binding to the endosome, but not being successfully ingested. The binding of the
LDL may be enough to signal the release of the M6P molecules. Also, the pH may still be high enough to
degrade the vesicle membrane enough to allow M6P release. It is most likely that the LDL receptors will
not be properly recycled as they are in the normal pathway due to the nature of the mutant LDL cell
components.
Corvini 4

5. Question 3:
RTKs directly phosphorylate specific tyrosines on themselves and on a small set of intracellular
signaling proteins. They have two domains which usually consist of a cysteine rich domain connected to
a tyrosine-kinase domain. These regions are connected by a disulfide-bridge that passes through the
plasma membrane. There are various growth factor receptors that are linked to the tyrosine-kinase
domains. These molecules can also have immunoglobulin-type-III-like domains as well.
RTKs function in cell-signaling. Ligands, predominantly various extracellular proteins bind to
RTKs and transducer a variety of different signal processes throughout the inner environment of the cell.
Ephrins are among these ligands that bind to EpH receptors. Binding causes a shift in the conformational
shape of the transmembrane alpha-helix region specifically. This allows transmembrane proteins to
transmit signals by activating the respective kinase region.
RTKs work with extracellular binding proteins and receptor proteins which are located internally
within the cell. Ligand binding causes receptors to become dimerized which allows for the activation of
the kinase region. This results in the successful phosphorylation of tyrosines. The activation of these
molecules is vital to cell signal transduction. Once activated intracellular signaling proteins, particularly
those with SH2 domains such as Insulin and platelet derived growth factors work with RTKs.
The Src Homology 2 domain (SH2) allow for binding of various ligands to kinase regions. SH2
domains contain phosphotyrosine binding domains which allow for proteins that have these regions to
bind to RTKs and transmit intracellular signals.
The Ras protein acts as a signal center for a variety of signals. They contain monomeric GTPases
and serve to relay signals from cell surface receptors.
In the mechanism where epidermal growth factor (EGF) binds to its RTK receptor, Ras serves as
the in-between “switch” that triggers for emission of the signal by this mechanism. These nonionic
GTPases work to bind and relay signals from their respective RTKs. Essentially, they act as the middle
man between the initiating and terminal factors of a signal-transduction pathway.
Corvini 5

6. The Src Homology 3 domain (SH3) specifically binds to proline rich amino acid sequences. It is a
type of interaction protein and is very similar to that of SH2, except it possesses its own unique function.
The Rho family of Ras proteins serve to regulate the progression of the cell cycle. They function
specifically on various cytoskeletal elements which may be important during the cell cycle. They play a
central role in gene transcription and membrane growth. These molecules are most likely involved in the
issuing of signals that trigger progression past specific checkpoints of the cell cycle as well. It is involved
in the PI-3 kinase Akt pathway by acting as the primary signal transducer for this pathway which results
in production of molecular components need to prevent the occurrence of apoptosis, or cell death.
When considering oncogenic cells within the body there is an error that exists within the
functionality of RTK binding that leads to continuous proliferation. Mutant forms of these genes do not
encode for the normal amount of binding and as a result there is a difference in the signal cascade that
is issued into the intracellular cytosol environment. This overrides important checkpoints within the
cell’s growth cycle and causes for continuous signaling.
In the mechanism for transcription steroids, EGF, cAMP, and interferons play individually
important roles. Primarily, cAMP serves to activate kinases through the energy transferring process of
phosphorylation. Steroids often bind to the active kinase regions of RTKs and take part in the
transduction of intracellular signal cascades. EGF is needed to stimulate cell growth, production, and
signaling. Interferons increase resistance of cells to viral infection and also propogate greater
macrophage proliferation. All of these components work together to ensure the factors for initiation of
transcription are generated while working to preserve a safe environment in which these molecular
processes can occur.
Question 4:
Topsoisomerase I produces a transient single strand break within the DNA backbone. By
breaking the phosphodiester bonds it allows two sections of the DNA helix to rotate freely and swivel
around the cut area as a point of reference. This serves to relieve tension that may exist within the
Corvini 6

7. structural make up of the DNA helix. Topsoisomerase II creates a double strand break within the double
helix. This molecule works on areas where two strands of DNA have overlapped and become tangled in a
sense. It is necessary to untangle these strands because this situation causes a great amount of steric
strain on the backbone of the DNA. By cleaving the double helix at this region the strain caused by
supercoiling is greatly reduced. Topsoisomerase II is located predominantly in proliferating cells. Also,
topsoisomerase II is not part of special structure. It is a functioning enzyme that works in tandem with
the DNA double helix.
Question 4A:
The average number of base pairs per turn of the DNA double helix would be ten. The value of W would
be -65.
L=T+W
L–T=W
500 – 565 = W
W = - 65
Question 4B:
Topsoisomerase I would need to nick the original supercoiled molecule five times in order to reduce the
linking number as described (to a value of – 60).
L=T+W
L = 565 + (-60)
L = 505
Corvini 7