7.06_2004_PS1key - 7.06 Spring 2004 PS 1 key 1 of 11 7.06...

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Unformatted text preview: 7.06 Spring 2004 PS 1 key 1 of 11 7.06 Problem Set 1 1. You would like to purify a protein(s) that is responsible for some activity (call it X) in the cell. Fortunately, you have at your disposal a very simple colorimetric assay to monitor the activity you are interested in. a) List at least 3 possible techniques you could use to purify the activity out of a crude cell lysate and the basis by which these methods separate molecules of interest (note: you do not know the identity of the activity producing molecule, so antibodies cannot be used). Method 1) Rate-zonal Centrifugation 2) Ion-exchange Chromatography 3) Gel Filtration Chromatography Property Mass Charge Mass How do you know you have isolated the activity of interest? You need to realize that after fractionation, each sample/fraction must be tested for the activity X by using the “colorimetric assay.” b) You use the colorimetric assay to measure the enzyme’s activity in the crude lysate and found it to be 5000 units. When you examine your fractions, however, none seem to contain any measurable activity. How could this have happened? The enzyme is a multi-subunit complex that requires all subunits for activity – you may have disrupted the complex during purification (i.e. you purified the subunits away from one another). Alternatively, the protein may have precipitated/denatured under your purification conditions. c) As previously mentioned, your lysate contained 5000 units of activity. But, unlike the situation outlined in (b), what if one fraction contains 50, 000 units of activity. Under what circumstances could this happen (i.e. how could you have 10x more activity in your purified fraction than in your total lysate)? The enzyme’s activity in the crude lysate may be restrained by an inhibitor. During purification, you may have separated the two, unleashing the full activity of the enzyme. d) You decide to use Gel Filtration to purify your activity. What property are you exploiting by using gel filtration? How would you actually measure this property? Mass. You must run proteins of known molecular weight through the column and determine in which fractions they elute. You can estimate the size of your protein by comparing the fraction(s) it eluted into with the fractions that the molecular weight standards eluted into. 7.06 Spring 2004 PS 1 key 2 of 11 e) You perform the gel filtration experiment and measure the activity using your nifty colorimetric assay. The following results are obtained (the elution position of molecular weight standards is superimposed on the profile). Elution Profile 10 kDa 30 kDa 50 kDa 70 kDa 140 100 kDa 120 Activity (U) 100 80 60 40 20 0 0 10 20 30 40 50 60 70 80 90 Fraction Number What is the estimated molecular mass of the molecules in the sample with the highest activity? Fraction 50 has the highest activity and contains molecules of approximately 50 kDa. 100 7.06 Spring 2004 PS 1 key 3 of 11 f) However, after performing SDS-PAGE on fractions containing the highest levels of activity and staining the gel with Coomassie Blue (a dye that reveals proteins), you observe the following: Fractions 49 MW Standards 50 51 52 100 80 70 50 30 10 How could you reconcile the difference between the molecular mass obtained by gel filtration with the one determined by SDS-PAGE? A single band appears in each fraction with a molecular mass of ~25 kDa. However, the gel filtration column suggests that the protein, under native, non-denaturing, conditions has a mass of 50 kDa. Therefore, it is likely that the activity we isolated in fractions 49-51 results from a dimeric complex between two subunits of mass 25 kDa. Note, the two subunits can be identical polypeptides (i.e. our protein is a homodimer) or different polypeptides that coincidentally have the same molecular mass (i.e. a heterodimer). What if, instead of the gel shown in (f), you obtained the following pattern? Fractions MW Standards 49 50 51 52 100 80 70 50 30 10 What would you conclude about the protein’s organization given this gel? Two distinct bands appear in each fraction with molecular mass of >10 (~15) kDa and >30 (~35) kDa. Therefore, it is likely that the activity we isolated in fractions 49-51 results from a heterodimeric complex of two subunits of mass 35 and 15 kDa. In the gel filtration column, the protein is purified under 7.06 Spring 2004 PS 1 key 4 of 11 native, non-denaturing, conditions and therefore exhibits the molecular mass of the 50 (35 +15) kDa complex. g) After successfully purifying the protein, you intend to determine its identity. List two methods that you could use to identify the protein of interest if you are given a test tube of purified enzyme. 1) Edman Degradation 2) Mass-Spec 2. You have decided to do a UROP and end up working in a lab that studies transport proteins. This doesn’t concern you, however, since you learned all about transport proteins in 7.06. Your graduate student advisor is attempting to create an in vitro system involving cellular Ca2+ flux in order to study Ca2+ triggered intracellular signaling. However, your advisor is busy with other problems and doesn’t have time to choose the best Ca2+ pump for the system. He thus asks you to find the best Ca2+ pump for his system. (a) Being the expert that you are, you know the first thing to do is to collect different kinds of Ca2+ pumps to test. List two methods of enriching for Ca2+ pumps in the membrane. Answer: One approach is to extract and purify a specific transport protein, then reincorporate into a liposome. A second approach is to express the gene encoding the transport protein at high levels in a cell type that doesn’t normally express the protein. (b) Now that you’ve obtained the initial rates of Ca2+ intake by the pumps, you can determine the Vmax and Km of each Ca2+ pump. What does Vmax and Km tell you about a given Ca2+ pump? Answer: The Vmax gives you a measure of the maximal rate of the pump, in other words it tells how well it would transport Ca2+ if every pump was in the process of moving an ion of Ca2+ at every possible moment. The Km tells you how well the pump binds its substrate, in this case Ca2+. The higher the affinity of a transport protein for Ca2+ (in this case), the more specific that transporter is for transporting Ca2+. (c) Using the initial rate data below, calculate the Vmax and Km for each Ca2+ pump. 7.06 Spring 2004 80 PS 1 key 5 of 11 Ca2+ pump A Ca2+ pump B Ca2+ pump C Initial rate of Ca2+ export into liposome, Vo 70 Ca2+ pump D 60 50 40 30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 12 2+ External [Ca ] uM Answer: For pumpA, Vmax=75 and Km=2.2mM. PumpB, Vmax= 37.5 and Km= 3.5mM. PumpC, Vmax= 15 and Km= 4.5mM. PumpD, Vmax=2 and Km=6mM (d) Based on this data, which Ca2+pump would you recommend to your graduate student and why? Also, why would some do some Ca2+ pumps have higher Kms than others? Answer: Based on this data, the best pump is Ca2+ pumpA. This pump has the highest maximal rate and is the most specific for Ca2+. Your initial isolation experiment didn’t test for Ca2+ specific pumps, just for transport proteins that were capable of transporting Ca2+. Therefore the primary function of some of these pumps might not be Ca2+ transport. However, because of their structure they are still capable of transporting Ca2+ when its true substrate is absent, or when the concentration of Ca2+ is high enough to out-compete the true substrate. (e) Ca2+ pumps fall under the category of ATP-powered pumps. Explain how ATPpowered pumps work. Also explain how ion channels work, what the three types of transporters are and how each one works. Answer: ATP-pumps use the energy of ATP hydrolysis to transport ions or small molecules against their electrochemical gradient. Ion channels allow specific ions (determined mostly by the structure of the channel) to flow down their electrochemical gradient (no extra energy is required). The three types of transporters are: uniporters, which utilize conformational changes to transport specific molecules down their concentration gradient; symporters and antiporters drive the transport of one molecule against its concentration gradient by transporting a second molecule down its concentration gradient. Symporters 7.06 Spring 2004 PS 1 key 6 of 11 transport both molecules to the same side of a membrane, while antiporters move the two molecules to opposite sides of the membrane. 3. (a) It is known that capsaicin, the active ingredient in hot and spicy foods, increases the permeability of sensory neuron plasma membranes to cations. You speculate that there is a specific ion channel within these neurons that binds capsaicin and is responsible for this increased permeability. In your lab, you have the ability to visualize the calcium uptake of cells using fluorescence microscopy and a calcium sensitive dye. How would you identify the capsaicin receptor? Make a cDNA library from the sensory neurons, transfect pools of clones (individual cDNAs in plasmids) into cells that do not have a capsaicin dependent ion channel, and look for a specific pool that increases calcium uptake in cells. Then iteratively subdivide the pool until a single clone is identified and sequence the gene. (b) You successfully clone the capsaicin receptor and decide to study its properties as an ion-channel using patch clamping. (i) The resting membrane potential of an oocyte not treated with capsaicin is –65 milliVolts. The sodium concentration of the patch electrode is 90 mM, and you have measured the intracellular sodium concentration to be 8mM. What is the ∆G for the inward movement of sodium ions at 37˚C? Is the process thermodynamically favored? ∆Gc = RT * ln (Nain/Naout) ∆Gm = zFE ∆G = ∆Gc + ∆Gm cal/(K*mol) R = 1.987 F = 2.3 10^4 cal/(V*mol) ∆G = zFE + RT ln (Nain/Naout) ∆G = (1) * (2.3E4 cal/V*mol) * (-0.065 V) + (1.987 cal/(K*mol) * (310 K) * (ln 8/90) ∆G = -1.495 + -1.49 ∆G = -2.98 Because ∆G is negative the process is thermodynamically favored. (ii) When performing a single channel patch clamp of an oocyte expressing the receptor, you add 1µM capsaicin to the media and measure a current of 3pA. How many sodium ions does the capsaicin receptor let into the cell per second? Ampere = Coulomb/Second (3 * 10^-12 C/s) * (1 mole/ 96500 C) * (6 * 10^23 molecules/mole) = 1.86 * 10^7 molecules of sodium per second 7.06 Spring 2004 PS 1 key 7 of 11 (iii) You find the transport of cations by the receptor to be independent of ATP. What type of transport mechanism does the receptor use? Facilitated diffusion. The capsaicin receptor is a channel protein that allows cations to move down their concentration gradient. This process is thermodynamically favored and thus ATP-independent. (iv) How does the type of transport utilized by the capsaicin receptor differ from the transport mechansim of the Na+/K+ ATPase? The Na+/K+ ATPase uses active transport. It moved two potassium ions into the cell as three sodium ions are moved out. Both ions are transported against their concentration gradient, thus in order for the process to occur, the pump must couple their transport to the hydrolysis of the high energy phosphodiester bond of ATP. 4. While studying families that have a severe skin blistering disease you find linkage to a gene you name epidermo-1. The genomic structure is shown below, exons are depicted as boxes, introns as straight lines: 1 2 3 4 Sequencing of affected individuals reveals a deletion of exon 3. You decide to try to develop a mouse model of the disease. (a) Using only a computer how would you determine whether mice have a homologous gene? and whether it has been shown to be expressed in the skin of mice? Since the evolutionary distance between mice and humans is only seventy five million years you could do a blast search against the NCBI nr database or against the Ensembl DNA database using the sequence of one of the human exons as a probe – using intronic sequence is not recommended since there is less selective pressure on intronic sequence than on exons and so it usually diverges at a faster rate. To find out if epidermo-1 is expressed in the skin of mice you would take part of the mRNA sequence (i.e. the exons) and blast it against the EST database and if there are any hits then you can check what tissue/developmental time point the ESTs were cloned from. (b) You decide to recreate in mice the mutation found in the affected individuals. Based on the human disease phenotype your supervisor thinks that Epidermo-1 may be involved 7.06 Spring 2004 PS 1 key 8 of 11 in forming cell-cell junctions and therefore may be essential during development of the mouse. How would you modify your gene targeting strategy so as to ensure the targeted mice were not embryonic lethal? LoxP sites are shown in yellow, the neomycin resistance gene is shown in green. 1 2 2 3 NeoR 4 3 Using the loxP sites allows for a conditional knockout of exon 3. Once the targeted mice have been generated they need to be crossed to a mouse strain that expresses Cre in the skin. (c) The mutant mice you generate develop a skin blistering disease. To try and investigate the function of Epidermo-1 you take skin sections from both wild type and mutant mice and using an antibody against the N-terminus you carry out some immunostaining. In wild-type skin Epidermo-1 is found to co-localize with desmocollin and desmoglein, two proteins found in desmosomes. Since a portion of exon 3 is composed of a stretch of 22 hydrophobic amino acids that can form an a -helix, where might you expect to find the mutant Epidermo-1 protein? Why? You might expect to find the mutant protein in the cytoplasm since it is missing the part of the protein that is likely to be the transmembrane portion. The interior of the plasma membrane is hydrophobic and therefore amino acids that are in contact with the fatty acyl chains of the bilayer need to have hydrophobic side-chains. Stretches of 20-25 hydrophobic amino acids can form an a -helix that traverses the plasma membrane. This is the principle secondary structure that the membranespanning portion of a protein takes, the other being the b-barrel – a single b-sheet is not sufficient to satisfy all of the H-bonds in the polypeptide backbone. Both structures maximize H-bonding between peptides that are in the bilayer, which is an energetically favorable outcome since there is no water present in the interior of the membrane. (d) You attempt to culture keratinocytes from these mutant mice and find that, unlike wild type keratinocytes, they don’t form a compact monolayer and detach from the cell 7.06 Spring 2004 PS 1 key 9 of 11 culture dish very easily. In an attempt to demonstrate that Epidermo-1 is involved in cellcell interactions at desmosomes you treat wild type keratinocytes with EGTA, a calcium chelator, and find that they detach from the dish just like the mutant cultures. Why does this occur? Desmosomes, adherens junctions, and tight junctions all need high concentrations of calcium to stay intact. EGTA is a calcium chelator and so removing the calcium from the culture media prevents the formation of these junctions and so the cells lose their adhesive properties and detach. 5. Below are five of the ways by which proteins can associate with the plasma membrane: (i) as a single-pass transmembrane protein (ii) as a multi-pass transmembrane protein (iii) via a covalent fatty acid chain (iv) via an oligosaccharide to phosphatidylinositol (v) as a peripheral protein via electrostatic interactions or a lipid binding domain (a) Give an example of each and detail the conditions by which you could purify each of these proteins from the cell membrane. Single-pass transmembrane protein: e.g. glycophorin A on erythrocytes or EGFR (epidermal growth factor receptor). Multi-pass transmembrane protein: e.g. bacteriorhodopsin, G-protein coupled receptors such as b-adrenergic receptor. Both single and multi-pass transmembrane receptors can be purified by adding detergent to cells to disrupt the membrane. Ionic detergents, such as SDS, will denature the protein and so prevent any functional assays being carried out. Non-ionic detergents, such as Triton X-100, can be used to purify these proteins in a non-denatured form. Covalent fatty acid chain: e.g. v-Src is attached to the membrane via a myristoyl group linked to its N-terminal glycine. Proteins can also be palmitoylated allowing them to attach to the membrane. They can be purified by treating cells with detergents. Oligosaccharide linkage to phosphatidylinositol: e.g. Thy-1 is attached via a GPI (glycosylphosphatidylinositiol) linkage. Such proteins are usually only found on the extracellular side of the membrane. It can be purified by treating with a PI-specific phospholipase C. Peripheral protein: e.g. Phospholipase A2. It can be extracted by breaking up the electrostatic interactions using high salt concentrations or a high pH. Some peripheral membrane proteins need calcium ions to form electrostatic interactions 7.06 Spring 2004 PS 1 key 10 of 11 with the phospholipid head groups and so these proteins can be released by adding a calcium chelator. (b) Why is it that polypeptides that enter the bilayer tend to pass completely through it rather than changing direction midway? Changing direction midway would involve making a bend in the polypeptide backbone, which is not energetically favored as it decreases the number of hydrogen bonds formed. Therefore, from a thermodynamic viewpoint it is much better if a protein passes through the entire bilayer. (c) You have cloned a multipass transmembrane receptor and are trying to determine whether the N-terminus is on the cytoplasmic or the extracellular side of the bilayer. How would the following information help you determine this: (i) position of glycosylated residues (ii) a series of monoclonal antibodies whose epitopes have been mapped (iii) a recombinant form of the protein that has been tagged at the N-terminus (i) Glycosylated residues are always found on the extracellular portion of a transmembrane protein – in a future lecture on protein secretion and modification we will discover why this is. (ii) If you have a monoclonal antibody you could incubate cells that express the receptor with this antibody, wash away any unbound antibody, cross-link bound antibody and then add a secondary antibody coupled to a fluorescent molecule. Antibodies can’t diffuse through the bilayer and so the only way in which you could get fluorescence is if the epitope that the antibody is recognizing is exposed on the cell surface. Since you have a series of monoclonals whose epitope targets are known you can easily determine which parts of the protein are on the extracellular side of the membrane. (iii) If you engineer the protein to have an epitope tag e.g. HA (hemagglutinin) at the N-terminus and express the recombinant protein in cultured cells, then all you have to do is incubate the cells with an antibody against the HA tag – if it binds to the cell surface then the N-terminus (along with the HA tag) must be on the extracellular side of the membrane. You find that the protein has four cysteine residues, two on the extracellular side, two intracellularly. Which are more likely to form disulfide bonds and why? The cysteine residues that are found on the extracellular side are the only ones that could form disulfide bonds because the cytoplasm of a cell is a reducing environment and for a disulfide bond to form the sulfhydrl group of cysteine needs to oxidize. 7.06 Spring 2004 PS 1 key 11 of 11 ...
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This note was uploaded on 01/23/2012 for the course LSM lsm1301 taught by Professor Seow during the Spring '11 term at National University of Singapore.

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