Presentation on theme: "1. The crystal structure for the enzyme you are studying has recently been published. The structure reveals a Glu45 reside in the active site and you propose."— Presentation transcript:
1. The crystal structure for the enzyme you are studying has recently been published. The structure reveals a Glu45 reside in the active site and you propose it represents a substrate binding residue. Sequence comparison among 5 orthologs of the enzyme shows Glu 45 is absolutely conserved. Kinetic studies of an E45A point mutants results in a fold lower K m for substrate and a fold lower V max relative to wild type enzyme. Interpret these results with respect to the likely role of Glu45 in the catalytic cycle of the enzyme. The lower V max is consistent with the mutation resulting in a decreased k cat, which is what one would expect if Glu45 is a catalytic group in the active site. Therefore, the observed effect on V max is consistent with the hypothesis. The effect of the mutation on K m suggest one of two possibilities: This mutation could result in an increased binding affinity as evidenced by the decrease in K m. More likely, the enzyme follows steady state kinetics for which K m =k 2 /k 1, for which a decrease in k 2 (I.e., k cat ) would result in a lower K m as observed. The data presented do not allow one to distinguish between the two alternative interpretations.
2. For the enzyme in question 1 you next prepare a E45D point mutant and repeat the kinetic studies. These studies show a K m that is statistically indistinguishable from wild type enzyme but a V max that is 10-fold lower. Interpret these new results with respect to the effect of the E45D point mutant. The 10-fold lower Vmax suggests that mutation of Glu 45 to Asp is not a conservative mutation. Since the side chain of Asp is one methylene shorter than Glu, the results suggests a decrease in V max due to a correspondingly greater distance between the carboxlate side chain and intermediate in the reaction cycle with which it must interact.
3. Contrast the effect on pK a for an aspartate versus a lysine residue within a hydrophobic microenvironment of an enzyme active site. A hydrophobic microenvironment will result in a localized lower dielectric constant, which favors uncharged species. Therefore, the carboxylate side chain of aspartate will appear to have a higher pK a (favoring the uncharged protonated form over a broader pH range) and the lysine will exhibit a lowered pK a (favoring an uncharged unprotonated form over a broader pH range).
4. AlphaCP1 is a polyC/U binding protein that binds to polyC/U-containing 3' UTR regions of certain mRNA species to stabilize the message and suppress translation. Phosphorylation of CP1 by S6 protein kinase is thought to result in reduced binding to the mRNA and release of the protein from the polyC/U region, allowing translation initiation. You have identified Ser 395 as a putative phosphorylation site for S6 protein kinase. To test this hypothesis, you generate an CP1S395A point mutant and find that it binds polyC/U but fails to undergo S6 protein kinase- mediated phosphorylation. You are feeling pretty good about the data until your advisor voices a concern that the result is not conclusive since the mutation may have partially altered the structure of CP1 in a way that allows mRNA binding but prevents binding of S6 kinase. Propose an experiment in sufficient detail that resolves this question. The question is whether S6 kinase can distinguish wild type from mutant α CP1. One way would be to express recombinant wild type and mutant α CP1. (If done in parallel this would also obviate potential differences due to differences in the preparation methods.) One would then measure the initial rate of S6 kinase-catalyzed phosphorylation of wild type α CP1 in order to determine V max and K m. One would then repeat the experiment at different constant concentrations of point mutant (which cannot be phosphorylated) and observe whether the mutant is a competitive inhibitor. A secondary plot of K m,apparent versus [mutant] o would allow determination of the K I, which should equal the K m for wild type protein if the kinase is unable to distinguish between the two proteins.
1/v o 1/[ CP1] o -Mutant Increasing concentrations of mutant 1/K m 1/K m,app K m,app [Mutant] o KmKm Slope= K m /Ki
6. Defend the statement that general acid catalysis in an enzyme catalyzed reaction is a type of proximity effect. Proximity effects in enzyme catalysis arise by the precise orientation of amino acid side chains within the active site. In the case of general acid catalysis, this precise orientation results in the weak acid group being positioned to optimally donate the proton within the catalytic cycle. Therefore, general acid catalysis during an enzyme reaction is a special case of a proximity effect.
7. In 25 words or less thoroughly but succinctly define entropy-enthalpy compensation. The unfavorable transition state entropy of activation ( S ‡ ) is compensated by the favorable enthalpy of binding ( H) of the reactants in the active site.
8. A student in your lab is using a 125 I-labeled peptide hormone to study its binding to a membrane receptor. When the radiolabeled hormone was purchased the product sheet listed a specific radioactivity of 53 milliCuries/mmole. If the half life for 125 I is 60 days, what is the specific radioactivity of the probe after 45 days? Show your work. k = 0.693/60 days = days -1 Specific activity after 45 days = 53 mCi/mmolee days-145 days = 53 mCi/mmole0.59 = 32 mCi/mole
9. You are studying the role of mRNA binding proteins in translational regulation of gene expression. You have found in a genetic screen that the ubiquitin conjugating enzyme E2 epf interacts with the RNA binding protein AUF1. To prove this interaction, you express E2 epf as an N-terminal GST fusion protein and demonstrate in a glutathione-Sepharose pull down assay that 100 nM concentration of the fusion is able to bind AUF1 (see Lane 2 of attached figure). As a control, 100 nM GST alone fails to bind AUF1 (see Lane 1 of attached figure). [In this variation on an IP assay, the glutathione binding domain Glutathione S-Transferase (GST) is genetically fused to the amino terminus of E2 epf. The resulting chimeric protein is incubated with a putative ligand, after which glutathione-linked beads are added to precipitate your GST fusion protein and any bound ligand. The presence of bound ligand can be tested by Western blot, as displayed in the figure which has been stained with anti-AUF1 antibody.] The literature reports that AUF1 can bind another RNA binding protein CP1; to confirm this, you incubate the same concentration of AUF1 with 100 nM GST- CP1 (Lane 3 of figure). Out of curiosity, the last experiment is repeated but you also add 100 nM E2 epf to the mixture of GST- CP1 and AUF1 (Lane 4 of figure). 1234
Provide a succinct thermodynamic argument supporting the hypothesis that E2 epf binds CP1. For extra credit, is there an alternative hypothesis also consistent with the observations? Because the density of the AUF1 band is greater in lane 4, the Western blot suggests that a combination of CP1 and E2 epf bind AUF1 with greater affinity than either alone. Because E2 epf and CP1 each bind AUF1, one can construct the following model for the process: AUF1 + E2 epf AUF1E2 epf + CP1 + CP1AUF1 + E2 epf CP1AUF1E2 epf K -1 K -2 In the scheme, the equilibrium constants represent K dissociation constants. Because the bands of bound AUF1 in lanes 2 and 3 are nearly equal, K -1 K -2 ; however, because the combination of E2 epf and CP1 bind AUF1 with greater affinity, K -1 K -1, requiring K -2 K -2.
The model requires E2 epf and CP1 to form a complex which binds AUF1 tighter than either alone. The tighter binding can be most easily envisioned as involving additive binding surfaces, as shown schematically: E2 epf AUF1 CP1 There is one other model that is fits the available data that does not require additive binding: AUF1 can bind E2 epf and aCP1 at different sites that result in a conformation change that makes the other component bind tighter. This model would follow the same scheme as above but not require interaction of E2 epf and aCP1: E2 epf AUF1 CP1
10.The plasma membrane of human cells contains a ubiquitously distributed family of tissue-specific receptor complexes involved in regulating cell adhesion, growth, and differentiation. These complexes contain a member of the tetraspanin superfamily of transmembrane proteins that is non-covalently associated with one or more membrane receptors or glycoproteins. The tetraspanins also bind an additional 14 kDa subunit termed Leu-13. The complex present on B lymphocytes (a type of immune cell circulating in the blood) is reported to be composed of the tetraspanin CD81 (25 kDa) which is associated with the transmembrane receptor CD19 (91 kDa) and the 110 kDa transmembrane glycoprotein CD21 (see illustration). You have recently found that a protein kinase, PI 3-kinase, binds to the cytoplasmic side of CD19.
In normal freshly isolated B lymphocytes and in cultured Daudi cells (an established leukemic B lymphocyte cell line used in tissue culturing) Leu-13 is absent from this complex or present at low levels. However, addition of pure recombinant Leu-13 to either fresh B lymphocytes or Daudi cells reversibly reconstitutes the complex. Reconstitution of the complex is accompanied by complete inhibition of cell division. In this experiment, cell division is quantitated by measuring the incorporation of the radioactive nucleotide 3 H- thymidine into DNA during the 2 hours following addition of recombinant Leu- 13 to the cultures. You compare this data to an experiment conducted under identical conditions but with freshly isolated normal human B lymphocytes. Data for both experiments are summarized in Table 1 for experiments conducted at 37 o C. What is the quantitative difference, if any, in the binding of Leu-13 to CD81 present on normal B lymphocytes versus Daudi cells?