Chemistry Exam #2 Answer Key -- November 4, 2008 There are 6 pages

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1 ame: Answer Key 1 Chemistry Exam #2 Answer Key -- ovember 4, 2008 There are 6 pages 1. (10 pts) The proteins found in silk fibers are formed from stacked antiparallel β-sheets in which long stretches are composed of alternating glycine and alanine residues. This results in all the alanine sidechains being on the same side of the sheet. Draw a detailed structure of a portion of a silk antiparallel β-sheet containing at least 6 amino acids in each strand and 3 strands wide. Be sure to indicate hydrogen bonds with dotted lines. The stereochemistry at the alpha carbon should also be clearly shown. 2. (10 pts) What are the two major experimental methods used for determining the detailed three dimensional structure of proteins? Compare and contrast these two methods, clearly indicating their strengths and weaknesses. Both MR and Crystallography require relatively large amounts (mg's) of protein. In both techniques, some parts of the structure may not be well defined. Both techniques may have difficulty with integral membrane proteins. MR Determined in solution, the "normal" condition of proteins But must be very soluble Protons are observed Can be used for following dynamic processes (e.g., folding) Can be used to determine pka of groups on the protein Limited to relatively small proteins (< 50 kd) Proteins may need to be labeled by growing on 15 and C13 enriched media if > 15 kd Typical RMSD of backbone atoms ~ 1 Å Reported as an ensemble of structures that fit the constraints X-ray Crystallography Must be able to grow well-ordered crystals Protons are normally not seen, -bonding patterns must be inferred. Crystal contacts may perturb orientation of flexible surface features Can be used for large or small proteins xygen, nitrogen, and carbon atoms can't normally be distinguished, must be inferred eed to solve "phase problem", e.g., by making heavy metal derivative Typical RMSD of backbone atoms ~0.1 Å Reported as a single "average" structure

2 ame: Answer Key 2 3. (25 pts) Assume that you are an experienced student in a lab and that you are demonstrating a purification for citrate synthase (CS) to a new student. Briefly explain to the new student what is happening in the various steps and answer the questions at the end of each section. a) After some initial steps to isolate the mitochondria (which contain the citrate synthase), you lyse the mitochondria. To this lysate you add ammonium sulfate. You centrifuge the solution, remove the supernatant, and discard the pellet. To the supernatant you add more ammonium sulfate. nce again you centrifuge the sample, but this time you keep the pellet. Why is ammonium sulfate often added during the early stages of protein purifications? What is the rationale for the two-step addition of ammonium sulfate? As the concentration of ammonium sulfate increases the solubility of proteins decreases and they start to precipitate out of solution. Different proteins will precipitate at different ammonium sulfate concentrations. So we can purify our protein by first adding ammonium sulfate to a concentration below the concentration needed to precipitate out our protein, but high enough that other contaminant proteins precipitate. We can then remove the supernatant, which still contains our protein, and then add more ammonium sulfate to precipitate it out. This time we take the precipitate (since it contains our protein), but leave behind the supernatant which now contains other contaminant proteins. b) You solubilize the ammonium sulfate pellet and dialyze it overnight against large volumes of a buffer solution at p 7.2. What does the dialysis do? Dialysis is used to remove small molecules from the protein solution. (Small molecules can go through the pores of the dialysis membrane, but proteins cannot.) In this case the dialysis removes the ammonium sulfate and replaces it with the desired buffeer. c) You run the dialyzed solution over a gel filtration column. Following the protocol, you collect the first protein fraction that exits from the column, and discard the rest of the fractions that elute from the column later. What property of a molecule determines how it elutes in gel filtration chromatography? What does the fact that you collect the first protein fraction tell you about the protein? What would be a simple method to monitor for the presence of protein in the eluted fractions? Gel filtration separates molecules on the basis of size (and to some extent on shape). Small molecules must travel through the pores of the gel, and thus take longer to elute than large molecules. The fact that the CS is in the first protein fraction indicates that it is one of the largest protein molecules in the sample. The simplest method for monitoring for the presence of protein in the eluted fractions would be to measure the Absorption at 280 nm. d) The pi of CS is 5.6. The next step in the purification involves ion exchange chromatography. Would you use an anion exchange resin or a cation exchange resin? Briefly explain. As part of your explanation indicate what functional group is on the ion-exchange residue you chose. ow would you elute your protein from the column? At p 7.2 a protein with a pi of 5.6 will be negatively charged. Therefore you would want to use an anion exchange resin. A typical anion exchange resin contains positively charged ammonium groups. These positively charged groups will attract the negative charge on the protein and cause it to be retained on the column. The typical method for eluting the protein is to increase the ionic strength (for example by increasing acl concentration). Another method would be to lower the p so that the protein becomes positively charged.

3 ame: Answer Key 3 e) The final, purified protein is analyzed by SDS-PAGE. The SDS-PAGE shows a single band with a MW of 42,000±2000. owever from the gel filtration column you estimate a MW of 240,000±16,000. Describe briefly how SDS-PAGE works (start by describing what SDS-PAGE stands for). Then explain the difference in MW determined by SDS-PAGE and gel filtration. What can you infer about the structure of the CS from this data? SDS-PAGE -- Sodium Dodecyl Sulfate - PolyAcrylamide Gel Electrophoresis In this technique, proteins are exposed to high concentrations of the detergent sodium dodecyl sulfate. This detergent causes the proteins to denature and assume a rodlike shape. The large number of detergent molecules bound to the protein result in a net negative charge. The protein sample is applied to a polyacrylamide gel containing an electrolyte and an electric potential is applied to the gel. The protein molecules (coated with SDS) move through the gel toward the positive end of the electric field at a rate that is related to the logarithm of their molecular masses. The important distinction between SDS-PAGE and regular gel filtration is that the proteins are denatured in SDS-PAGE and in normal gel filtration they are in their native conformation. This means that if a protein is composed of multiple polypeptide chains, these chains will show up as separate bands on SDS-PAGE (assuming they are not cross-linked by disulfide bonds). In normal gel-filtration a protein composed of several chains will run together. Thus, in the case of CS, we can infer that the protein exists as a hexamer of 40,000 D subunits. f) You later decide to make a mutant form of CS in which several Ser residues on the surface are changed to Lys. Will the pi of the mutant protein be higher, lower, or about the same as the native enzyme? igher g) What experimental procedure(s) could be used to confirm that the mutant CS protein that you isolated has the desired amino acid sequence. Digest with trypsin and use mass spectrometry to determine the sequence of the resulting peptides. 4. (5 pts) ow many amino acids are there in each turn of an alpha helix? 3.6 The following sequence is an alpha helix from flavodoxin. Circle the residue whose side-chain is oriented in the most similar direction to the isoleucine sidechain. Asp-Asp-Arg-Ile-Lys-Ser-Trp-Val-Ala-Glu-Leu-Lys-Ser-Glu

4 ame: Answer Key 4 5. (18 pts) In an article in the European Journal of Biochemistry entitled "Ligand-binding properties and structural characterization of a novel rat odorant-binding protein variant" (Eur. J. Biochem. 2000, 267, 3079) the binding of 1-aminoanthracene (1-AMA) to the odorant-binding protein (BP) isolated from rat nasal mucus was studied. (ote: I am snot making this up!) A shorthand for the binding of AMA to BP is shown below, along with the fluorescence as a function of 1-AMA concentration. AMA + BP AMA BP a) Using the definition of, derive an expression for the fluorescence as a function of 1-AMA concentration. (int: This is similar to the derivation for the binding of 2 to myoglobin.) In your expression use to represent the dissociation constant and F max to represent the maximum fluorescence (i.e., the fluorescence when protein is completely saturated with 1-AMA.) ote that the observed fluorescence will be equal to the fraction of protein with 1-AMA bound times the maximum fluorescence. BP AMA [BP][AMA] = [BP AMA] BP + AMA [BP AMA] = [BP][AMA] (Eqn. 2) Fraction bound = [BP AMA] = [BP] + [BP AMA] Substituting in Eqn. 2 [BP][AMA] [BP] + [BP][AMA] = [BP] [BP] [AMA] ( + [AMA]) Fraction bound = [AMA] + [AMA] Fluorescence = F max [AMA] + [AMA] b) From the data shown in the figure estimate. (Briefly explain how you obtained this estimate.) = 1 µm. When [AMA] =, Fluorescence = 0.5 F max By inspection of the plot, it looks like F max is approximately 120. When Fluorescence = 60, [AMA] ~ 1 µm. c) Assume you find another molecule and determine that it has a of 3 mm for binding to the rat odorant-binding protein. Does this new molecule bind more tightly or less tightly than 1-AMA? Less Tightly

5 ame: Answer Key 5 6. (6 pts) Briefly compare and contrast myoglobin and hemoglobin in terms of their structure, oxygen binding characteristics, and physiological function. Myoglobin is a monomer (1 polypeptide chain), while hemoglobin is a tetramer (α 2 β 2 ). Although only 18% of the residues in the polypeptide chain in myoglobin are identical to those in the α or β chains of hemoglobin, the secondary and tertiary structures of these polypeptides are nearly identical, consisting of mostly α helices. The polypeptide chain of myoglobin and each of the polypeptide chains of hemoglobin contain one heme group that binds oxygen. The binding of oxygen to hemoglobin results in a conformational change that changes the relative orientations of the α and β subunits. Myoglobin binds oxygen with a "normal" hyperbolic binding curve and a p50 of 2.8 torr. Each tetrameric emoglobin can bind 4 oxygen atoms. The binding is not hyperbolic, but rather sigmoidal, reflecting cooperative interaction between the binding sites in a hemoglobin tetramer. That is, binding of oxygen at one site increases the affinity for binding at the other sites. The p50 for hemoglobin binding to oxygen in vivo is around 26 torr. The binding affinity in hemoglobin is affected by p (Bohr effect), the binding of BPG, and the binding to C 2. Myoglobin is an intracellular protein in vertebrate muscle. Its physiological role is to facilitate oxygen diffusion in muscle. emoglobin is the major protein in red blood cells and its role is to transport oxygen from the lungs (or gills) to other tissues in the body (e.g., the muscles). Since hemoglobin binds oxygen more weakly than myoglobin, oxygen is transferred from hemoglobin to myoglobin.

6 ame: Answer Key 6 7. (12 pts) Draw as precisely as possible a plot showing the oxygen binding curve (Y 2 versus p 2 ) for hemoglobin under normal conditions. When you go to higher altitudes your body rapidly adapts by increasing the concentration of BPG in red blood cells. Use a dotted line to show the oxygen binding curve in the presence of the increased BPG and show graphically how this helps compensate for the lower external oxygen pressure. (Assume the venous p 2 is 30 Torr, normal arterial p 2 is 100 torr, and at high altitude the arterial 2 drops to 60 torr.) Briefly explain the mechanism by which BPG changes hemoglobin's affinity for oxygen. n the graph above you can see that the decreased 2 affinity seen in the +BPG curve above leads to a larger fraction of 2 being delivered than would occur if the BPG concentration were not increased. BPG binds tightly to the T state hemoglobin but not the R state. Thus, increasing the amount of BPG shifts the equilibrium from the R state (high oxygen affinity) to the T state (low oxygen affinity) form, resulting in lower oxygen affinity.

7 ame: Answer Key 7 8. (7 pts) Describe a general pathway by which proteins fold into their 3-dimensional structure. (Include general time frames for these steps.) From pages in your text. 1. < 5 ms: Rapid formation of local segments of secondary structure (α helices and β sheets). This forms a "molten globule" containing a lot of the secondary structure, but little of its tertiary structure ms: The secondary structure becomes stabilized and tertiary structure begins to form. During this stage, the nativelike elements are thought to take the form of subdomains that are not yet properly docked to form domains s: The protein undergoes a series of complex motions in which it attains its relatively stable internal side chain packing and hydrogen bonding while it expels the remaining water molecules from its hydrophobic core. 4. In multidomain and multisubunit proteins, the respective units then assemble in a similar manner, with a few slight conformational adjustments required to produce the protein's native tertiary or quaternary structure. 9. (7 pts) What are prions, why were they controversial when first proposed, list two diseases that are caused by prions, and describe how prions cause disease? See pages in your text. Prions are infections protein particles that lack nucleic acids and cause disease. They were controversial when first proposed since it was thought that all infectious diseases required nucleic acids in order to be propagated. Diseases caused by prions include: Scrapie, Mad-cow disease (aka bovine spongiform encephalopathy, BSE), kuru, Creutzfeldt-Jakob disease (CJD). Prions are proteins that induce abnormal folding of a normal cellular protein (normally in the brain). The abnormally folded proteins catalyze the formation of more abnormally folded proteins (i.e., the process is autocatalytic). The abnormally folded proteins aggregate into amyloid fibrils that then disrupt brain function.