Homework For You



Fall2018_BIOL203Homework 11. You discover a small protein containing only 10 amino acids, called interleukin-87 (IL-87), that is involved in modulating the immune response. You are interested in purifying this small peptide so that you can study its function further. The following analyses (A-E) were performed on the purified peptide to determine its sequence and structure. A. Upon reaction with fluro-2,4-dinitrophenol (FDNB or Sanger reagent), you obtain DNP-Arg and DNP-Asp, leading you to believe your protein may not be pure after all. B. You then perform isoelectric focusing on IL-87 and observe only one band, as expected, at pH 6.25. IL-87 is then treated with dithiothreitol (DTT) and the protein is again subject to isoelectric focusing. Much to your relief, you obtain two bands, one at pH of 10 and one at pH of 2.85. Explain what you know so far about the protein (after steps A and B) and why you are relieved. C. Each peptide is then analyzed for its amino acid composition. One peptide, peptide A, contains a 1:1:1:1:1 molar ratio of Y, I, K, R, C. The other peptide, peptide B, has a molar ratio of 1:1:1:1:1 for C, D, I, N, E. D. Peptide A is digested with chymotrypsin and only one product results which contains (R, C, Y, I, K). However digestion with trypsin results in one 3-mer containing (C, I, K) and also free tyrosine and free arginine. In the first round of Edman degradation on the 3-mer, PTH-Ile is obtained.What is the sequence of peptide A? E. Peptide B is subject to Edman degradation. After the first two amino acids have been removed, the remaining fragment now has a neutral isoelectric point. The remaining 3-mer is reacted with FDNB and DNP-Asn is isolated. The disulfide bridge between peptide A and B is thought to occur between the C-terminal amino acid of peptide B and some residue in peptide A.What is the sequence of peptide B? Show how the isoelectric points given in B prove this sequence and structure are correct. 2. A. You are interested in studying resilin, a protein thought to be responsible for the ability of fleas to jump many times higher than their own length. This protein is purified and then analyzed as follows. Fractions are collected from a gel filtration column that is a run with a buffer containing relatively low salt concentration. You collect 8mg of protein from these pooled fractions and discover that the protein contains 0.124mg of isoleucine and 0.209mg of glutamate after complete hydrolysis of the protein. Isoleucine has a molecular weight of 131g/mol and glutamate of 147g/mol.What is the minimum molecular weight of resilin given these data? Show your calculations. B. A standard curve is produced from this same gel filtration column using proteins of known weight and measuring their relative elution volumes. The molecular weight of resilin is determined to be 34kDa by measurement of its relative elution volume and use of the standard curve.How many isoleucine and glutamate residues are present in 1 molecule of resilin? How do you know? C. Your labmate repeats this experiment but does not use the buffer you gave her. She incubates the protein instead in a buffer with higher salt concentration before running the experiment. Now the purified resilin elutes into two separate fractions, with two different relative elution volumes. This confuses you initially. You then perform N-terminal sequencing analysis by reacting the protein in each of these two separate fractions with FDNB and discover that the protein in the fraction with a lower elution volume, protein A, volume results in isolation of DNP-A. N-terminal analysis of the protein in the fraction with the higher elution volume, protein B, results in isolation of DNP-B. What might you now conclude about resilin? Identify the DNP derivatives. D. The first fraction, the one that eluted in less volume, contained ‘protein A’ and was found to be 5.24% isoleucine by weight after complete acid hydrolysis and contain no glutamate. The second fraction, the one that eluted in a greater volume, contained ‘protein B’ that was found to be 3.675% glutamate by weight after complete acid hydrolysis but contained no isoleucine.Explain what new information this gives you about resilin. In other words what is the minimum molecular weight of proteins A and B? How might these organize into a native resilin molecule? (Note: you still do not know the actual weights of proteins A and B) (10 points)E. To determine the actual size of the subunits that might be present in resilin, you subject the native protein (34kDa) to SDS-PAGE. The following standard curve was obtained using proteins of known molecular weight.You observe two different bands on the SDS-PAGE gel. The relative migration of one of the bands is 1.0859 and the relative migration of the other band is 1.044. The band with the higher relative migration is three times as dark as the one with the lower relative migration. Note that the results are identical when DTT is present in the buffer. Explain what you now know about the subunit composition of resilin in terms of proteins A and B and how the SDS-PAGE data can be interpreted to lead you to this result. Also explain which subunits contain glutamate, isoleucine, alanine and methionine. (10 points)F. Your labmate, sorry-not-sorry she did the gel filtration experiment in the high salt buffer now makes two suggestions. She suggests you react native resilin with FDNB again. She says you should get a 1:3 molar ratio of the following DNP derivatives, A and B. She claims this will confirm your results. Additionally she suggests that if no lysine, histidine, aspartate or arginine are present, subunit B should bind a cation exchanger at physiological pH and thus elute separately from subunit A. Is she right on both counts? (8 points)3. The amino acid sequence of several leucine zipper proteins are indicated in the figure below. This family of proteins are homologous at the protein sequence level. The leucine zipper motif is comprised of a single ?-helix. One portion of this ?-helix, termed the leucine zipper, homodimerizes with another molecule of an identical protein when acting as a transcription factor. The other major domain is termed the DNA-binding domain, and as its name suggests, is the part of the ?-helix that interacts with negatively charged DNA. Proteins with leucine zippers are often important transcription factors, or proteins that are involved in stimulating transcription of genetic material into mRNA. Highlighted residues are conserved among different proteins and species but many other positions have similar chemical character.A. A portion of the leucine zipper region of the transcription factor Jun is shown below. Give a reason for the periodicity of the bolded and unbolded residues. Also explain how this could explain the energetics of dimerization of the leucine zipper proteins in particular.Val-Lys-Thr-Leu-Lys-Ala-Gln-Asn-Ser-Glu-Leu-Ala-Ser-Thr-Ala-Asn-Met-Leu-Thr-Glu-Gln-Val-Ala-Gln-LeuB. A portion of the DNA binding domain of the transcription factor Jun is shown below:Arg-Lys-Arg-Met-Arg-Asn-Arg-Ile-Ala-Ala-Ser-Lys-Cys-Arg-Lys-Arg-LysIs the isoelectric point of this domain acidic, basic, or neutral and how do you know? Why is this the expected isoelectric point for a protein domain that ‘binds DNA’? Recall that nucleic acid is housed in the nucleus, a compartment that is around pH 7. C. How might the leucine zipper region “work” biologically? In other words, what drives dimerization?D. The rest of the DNA-binding domain contains residues which are important for DNA-sequence specific interactions with the protein. When these monomers interact with their specified DNA sequence, the two leucine zipper motifs on the adjacent monomers are brought near each other in space. How might these leucine zipper proteins function as transcription factors? How do they ‘work’? 4. The primary sequence of the protein resilin is depicted in Figure 1. Resilin is a ‘super elastic rubber’ found in the cuticle or exoskeleton of the regions of insects that subjected to repeated extension and retraction. Elasticity is the ability of a protein to resume its original shape after being stretched, bent, or otherwise distorted in some way. Regions of the wings of locusts or the hindlimbs of fleas contain resilin which is thought to be responsible for the ‘resiliency’ of these tissues to damage that could otherwise result from jumping multiple times higher than one’s body length or beating one’s wings at an extremely high rate. This biological rubber has an elasticity that allows resilin to be stretched up to 300% before the tissue breaks. Figure 1: Resilin - Exons 1, 2 and 3 each comprise separate functional protein domains. Exon 1 extends out of the cell into the extracellular space and exon 3 is intracellular. Exon 2 is a chitin binding domain, chitin being the major structural polysaccharide in the membrane that comprises the exoskeleton of insects. Its function is thus to anchor the extra- and intracellular subunits of resilin to the exoskeleton of insects. 975995100901500A. Exon 1, which extends outside of the cell membrane, contains 18 ‘elastic repeat units’ (depicted as yellow disks in Figure 1) which are 15-mer repeats of these amino acids: GGRPSDSYGAPGGGN. These segments are attached by short (2-5) amino acid segments of opposite polarity to the 15-mer repeat units. Additionally, at the N- and C- terminus of this exon are larger segments that also have opposite polarity to the 15-mer repeat. In other words, the blue and yellow regions of exon 1 in the diagram above have opposite polarity. Exon 1, when studied in isolation, takes an extended conformation in water. In other words, it does not take any secondary structure but retains a mostly disordered structure. Upon heating, or removal of water, individual exon 1 peptides begin to aggregate into stacks, as depicted below. Explain why the protein folds this way with and without water present. You should use your knowledge of amino acids to deduce the polarity of the 15-mer repeat and thus the polarity of the connecting regions. Consider glycine ambivalent in polarity – neither polar nor nonpolar.B. Exon 3, which is the intracellular domain, contains 11 ‘elastic repeat units’ (blue disks) which are 13-mer repeats of this amino acid sequence: GYSGGRPGGQDLG. These segments are also attached by amino acids of opposite polarity, but the connecting segments are larger than they were in exon 1 (5-10aa). Additionally, there is a large block of amino acids with opposite polarity to the 13-mer in the center exon 3, as indicated by the larger blue space in Figure 1. Exon 3 is able to form a micelle-like structure in water, as shown below. The teal blue disks are the 13-mer and the dark blue lines are the connecting regions (ignore the lines and letters on the outside). Micellar StructureHow does the primary sequence of this exon relate to its ability to form micelles?C. The micellar structures of exon 3 (shown above) are able to assemble into polymers when exon 3 is isolated from the other exons. The size and organization of the micellar polymers is more random in water, though the polymer does still form when water is present. Upon heating exon 3 micellar structures themselves become more organized and the polymer becomes more tightly organized as well. These polymer structures are shown below (each circle is one of the micellar structures on the left – ignore the red). Micellar polymerHow might this behavior of resilin, in vitro, give us a model for its ‘elasticity’ in vivo? Explain how the looser water-containing structure and the tighter water-excluding structure may give us a general mechanism for the elasticity of biological proteins. D. Now for the hard part – the biology. In actual cells, it is thought that the micellar form of exon 3 is not present. Instead, it remains relatively extended and unstructured like exon 1 does because of the cytosolic interaction and the association with the chitin binding domain which is in the membrane. Instead, it has been shown that exon 3 takes a reversible ?-turn conformation when physically stressed – for instance, stressed by the pressure that results from the pre-jump posture of the limb. Upon release of the mechanical stress during the jump, exon 3 resumes its disordered state. This unfolded-folded-unfolded transition is referred to as an ‘energy storage’ capacity. If you think about it, in order for the bug to jump, the energy used to contract the muscles in the pre-jump posture must be released as work (flying or jumping) rather than lost as heat. Explain, in terms of entropy, how exon 3 of resilin appears to confer the ability of the tissue to have ‘high energy storage capacity’ during the movement of jumping or flying as well as how exon 1 of resilin allows it to be ‘super-elastic.’ 5. ?82 lysine projects into the central cavity of the hemoglobin tetramer from the ?-chains. It has the highest pKa of any residue in that region after the quarternary structure is taken into account. This lysine residue is known to oppose the T-state when it is charged in part because it stabilizes the pKa of neighboring His143 so it does not participate in the Bohr effect when the lysine is present. Note that Lys82 and His143 are within range to interact by electrostatic, dipole or H-bonding interactions, whichever would be appropriate for residues of this type, and at the appropriate pH. A. It was discovered that mutation of this lysine to an aspartate results in reduced oxygen affinity. Note that the pKa of His143 decreases in the deoxygenated state compared to the oxygenated state without the lysine present. How does this possibly explain the lowered oxygen affinity? Think about the Bohr effect and what this usually means about the pKa of histidine residues in hemoglobin during state transitions. Do not consider 2,3-BPG at this point.B. It was also discovered that normal hemoglobin binds 2,3-BPG, a highly negative molecule, at this central site. Predict the effects on the oxygen binding curve of this LysAsp mutation in the presence of BPG. Also explain how BPG binding might explain the normal changes in response to BPG in the binding curve for wild type hemoglobin. Recall that Lys82, at physiological pH, opposes transition to the T-state when it is projecting into the central cavity and unbound to BPG. C. At 1M NaCl, it was discovered that the oxygen binding properties of both mutant and wild type hemoglobin are the same over a range of pH values and in the presence or absence of BPG. The presence of 1M NaCl with HbA, wild type, causes a rightward shift compared to HbA without salt. Give a suggestion as to how NaCl suppresses the effects of the LysAsp mutation. D. On the surface of red blood cells there is a passive antiporter than moves bicarbonate and chloride ions in opposite directions. Bicarbonate is produced from carbon dioxide and water by the activity of an enzyme called carbonic anhydrase. Chloride as well as several other inorganic ions have been shown to reduce oxygen affinity by hemoglobin when these anions are bound to the central cavity. Why is it important that this transporter is an antiporter? In what circumstances will bicarbonate move out of the red blood cell? By what mechanism will chloride reduce oxygen affinity? Why does influx of chloride in this circumstance make some biological sense? ................
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