Molecular cell take home 1


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Take home exam which was part II of our first exam.

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Molecular cell take home 1

  1. 1. Stephen CorviniTake Home TestMolecular Cell BiologyDr. BetticaNovember 25, 2011Question #1: The human red blood cell is a great example for studying protein-membrane functioninteractions because these are simple cells in which the membrane is a crucial component. Red bloodcells act primarily as means of transportation within the body and their integral membrane proteinswork the regulate gas, ion and other substance transport between the internal and externalenvironments of the cell. Spectrin maintains the structural integrity of the cell membrane of theerythrocyte. It forms a movable mesh network that allows for lateral movement of protein componentswithin the inter-membrane layer. Band 4.1, adducin, and tropomysin contribute to the determination ofactin filament length. Band 3 appears to acts as a channel protein within the membrane. Band 4.1 workswith 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 andthis appears to be regulated by ankyrin and protein 4.1. These proteins are attached to the meshnetwork of 100nm spectrin filaments within the membrane. Spectrin is flexible and as a result allows forthe lateral movement of molecules. Band 3 is a component of junctional complexes in the membraneand 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, therewould be a highly negative impact on the lateral movement of Band 3. This protein relies on spectrinand 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 ispossible there would be an uncontrolled degree of lateral mobility which would negatively effect themolecular transport function of the membrane as well. Without proper spectrin binding there would beuncontrolled movement of Band 3. Spectrin holds Band 4.1 and ankyrin in place around Band 3 and thiswould be absent with improper spectrin binding. This would result in the improper constraint of Band 3. Corvini 1
  2. 2. There are many methods which are used to isolate extrinsic and intrinsic proteins from the cellmembrane. Detergents are chemical substances that are used in order to isolate specific proteins. Ionicdetergents such as sodium dodecyl sulfate (SDS) and nonionic detergents, such as octylglucoside areused to treat various protein types. All detergents remain monomeric in low concentrations and do notreadily effect the proteins that are present within the cell environment. It is only when they are presentin strong/critical amounts that they will form globular molecules known as micelles and work to breakdown the membrane environment so that various proteins can be purified. Ionic and nonionic detergents have various advantages and disadvantages. Nonionic detergentscan readily isolate functional protein molecules. This allows them to be studied. Ionic usually rendermost 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 isolatedproteins cannot be used for studies because they have lost their functionality. Nonionic detergents allowfor collection of proteins that can be used for studies, but these detergents are not necessarily capableof solubilizing all membrane types.Question 1A: In order to characterize the integral protein composition I would choose a proper detergent. Forpurposes of this scheme, sodium dodecyl sulfate (SDS) would be effective. This would solubilize themembrane portion surrounding the lysosome. I would then utilize a mixture of phospholipids andincorporate them with the detergents allowing for a phospholipid molecule to be created that wouldhopefully capture the desired protein. The detergent SDS would be removed; thus allowing the integralprotein of the lysosomal membrane to be successfully purified.Question 1B: I would analyze the purified protein through SDS polyacrylamide-gel electrophoresis. This wouldexpose the hydrophobic core of the protein. Through comparison with the hydrophobic cores of otherisolated proteins from the lysosomal membrane I could confirm it was a member of that particularmembrane type. This would help rule out its possibility for being a contaminant from other parts of thecell. Corvini 2
  3. 3. Question 2: An N-linked oligosaccharide is one that is added to the NH2 group of an asparagines amino acidwithin a protein. These are added within the lumen of the Endoplasmic Reticulum. They don’t act oncytosol 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 acidserine, hydroxylysine, or threonine. This type of protein glycosylation occurs within the lumen of theGolgi Apparatus. In general the N-linked are longer than the O-linked. The enzyme oligosacchyral transferase catalyzes the transfer of oligosaccharides to the targetfunctional group with which the sugars most commonly bind. Prior to this transfer a precursoroligosaccharide is modified before attaching to the components of the proper receptor protein. They areprepared for synthesis when they are added to a carrier lipid dolichol molecule. Glucose, mannose, andN-acetylglucosamine represent the different categories of N-linked oligosaccharides. Once essential lysosomal enzymes are recognized by the mannose 6-phosphate receptor theymust be transported to the lysosome. The hydrolases bind to the MP6 receptor and receive a clathrincoat. This forms a receptor dependent vesicle through the process of endocytosis. This vesicle is thenejected from the trans-terminal end of the Golgi Apparatus. They bind to the lysosome where the acidicpH dissociates the vesicle and delivers the hydrolase successfully. Phosphates are removed from thehydrolase and the enzyme is incorporated into the lysosome. Modifications of N-linked oligosaccharides occur within the lumen of the EndoplasmicReticulum. This is where they are added as well. Transfer of N-linked sugars happens with help ofoligosacchyral transferase on the luminal side of the Endoplasmic Reticulum membrane. O-linked sugarsare added and modified within the lumen of Golgi Apparatus. They undergo similar modifications andadditions as do the N-linked sugars, but are less common. Only about 10% are usually O-linked sugars. Corvini 3
  4. 4. Question 2A: I-cell disease patients exhibit a series of molecular symptoms. Normally, most hydrolyticenzymes are missing from lysosomes, undigested substrates accumulate and this forms inclusions whichare very large within cells. This disease is due to a single gene defect and as a result can be consideredhereditary in nature. Hydrolases are missing from lysosomes in the blood as well. This is because theyare 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. Themannose 6-phosphate is contained in the area where the binding of multiple LDLs begins to form themembrane of the transport vesicle known as the clathrin coat. Areas known as clathrin pits exist as thisstructure forms and play a direct role in its construction. Adaptor proteins enclose the components ofthe coat of the vesicle. The vesicle will contain M6P receptors and the desired enzymes. Endocytosisejects the vesicle which is released in coated form. It is uncoated and becomes a vesicle. It then fuseswith the endosome which separate hydrolases and LDL components. Hydrolases are then sent toawaiting lysosomes and LDL components returned to cell membrane in a recycling endosome whichbinds to them back into the original cell membrane.Question 2C: The mannose 6-phosphate molecules are normal because they are not directly effected by thepH. Within the cell environment pH is maintained by an ATP driven-pump. The more basic pH may notbe high enough to successfully dissolve the membrane of the transport vesicle. Also, it is possible thatthe LDL receptors are binding to the endosome, but not being successfully ingested. The binding of theLDL may be enough to signal the release of the M6P molecules. Also, the pH may still be high enough todegrade the vesicle membrane enough to allow M6P release. It is most likely that the LDL receptors willnot be properly recycled as they are in the normal pathway due to the nature of the mutant LDL cellcomponents. Corvini 4
  5. 5. Question 3: RTKs directly phosphorylate specific tyrosines on themselves and on a small set of intracellularsignaling proteins. They have two domains which usually consist of a cysteine rich domain connected toa tyrosine-kinase domain. These regions are connected by a disulfide-bridge that passes through theplasma membrane. There are various growth factor receptors that are linked to the tyrosine-kinasedomains. These molecules can also have immunoglobulin-type-III-like domains as well. RTKs function in cell-signaling. Ligands, predominantly various extracellular proteins bind toRTKs 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 conformationalshape of the transmembrane alpha-helix region specifically. This allows transmembrane proteins totransmit signals by activating the respective kinase region. RTKs work with extracellular binding proteins and receptor proteins which are located internallywithin the cell. Ligand binding causes receptors to become dimerized which allows for the activation ofthe kinase region. This results in the successful phosphorylation of tyrosines. The activation of thesemolecules is vital to cell signal transduction. Once activated intracellular signaling proteins, particularlythose 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. SH2domains contain phosphotyrosine binding domains which allow for proteins that have these regions tobind to RTKs and transmit intracellular signals. The Ras protein acts as a signal center for a variety of signals. They contain monomeric GTPasesand serve to relay signals from cell surface receptors. In the mechanism where epidermal growth factor (EGF) binds to its RTK receptor, Ras serves asthe in-between “switch” that triggers for emission of the signal by this mechanism. These nonionicGTPases work to bind and relay signals from their respective RTKs. Essentially, they act as the middleman between the initiating and terminal factors of a signal-transduction pathway. Corvini 5
  6. 6. The Src Homology 3 domain (SH3) specifically binds to proline rich amino acid sequences. It is atype 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 functionspecifically on various cytoskeletal elements which may be important during the cell cycle. They play acentral role in gene transcription and membrane growth. These molecules are most likely involved in theissuing of signals that trigger progression past specific checkpoints of the cell cycle as well. It is involvedin the PI-3 kinase Akt pathway by acting as the primary signal transducer for this pathway which resultsin 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 thefunctionality of RTK binding that leads to continuous proliferation. Mutant forms of these genes do notencode for the normal amount of binding and as a result there is a difference in the signal cascade thatis issued into the intracellular cytosol environment. This overrides important checkpoints within thecell’s growth cycle and causes for continuous signaling. In the mechanism for transcription steroids, EGF, cAMP, and interferons play individuallyimportant roles. Primarily, cAMP serves to activate kinases through the energy transferring process ofphosphorylation. Steroids often bind to the active kinase regions of RTKs and take part in thetransduction of intracellular signal cascades. EGF is needed to stimulate cell growth, production, andsignaling. Interferons increase resistance of cells to viral infection and also propogate greatermacrophage proliferation. All of these components work together to ensure the factors for initiation oftranscription are generated while working to preserve a safe environment in which these molecularprocesses can occur.Question 4: Topsoisomerase I produces a transient single strand break within the DNA backbone. Bybreaking the phosphodiester bonds it allows two sections of the DNA helix to rotate freely and swivelaround the cut area as a point of reference. This serves to relieve tension that may exist within the Corvini 6
  7. 7. structural make up of the DNA helix. Topsoisomerase II creates a double strand break within the doublehelix. This molecule works on areas where two strands of DNA have overlapped and become tangled in asense. It is necessary to untangle these strands because this situation causes a great amount of stericstrain on the backbone of the DNA. By cleaving the double helix at this region the strain caused bysupercoiling 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 withthe 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 wouldbe -65.L=T+WL–T=W500 – 565 = WW = - 65Question 4B:Topsoisomerase I would need to nick the original supercoiled molecule five times in order to reduce thelinking number as described (to a value of – 60).L=T+WL = 565 + (-60)L = 505 Corvini 7