Adenosine deaminase catalyzes the hydrolytic …

Adenosine deaminase catalyzes the hydrolytic … BMB/Ch 178 final Fall, 2015 Instructions Finals are due Friday, Dec. 11 by 6 pm. Please drop off the completed finals in a box outside of Shu-ou’s office (150 Braun). The final exam is 3 hours and you will have use of lecture notes and course material. You are not allowed to discuss the exam with others and all the work should be your own. You are not allowed to use online resources to aid in the completion of final. You cannot consult exams from previous years. You will be tested based on lecture notes and problem sets. This exam seeks to test that you have a firm and broad grasp of the concepts presented throughout the course, and the ability to make connections between experimental data and molecular mechanisms. Please be concise in your answers. In most cases the answer is fairly specific, and either you know it or you do not. You won’t be expected to describe minutiae, and writing long paragraphs in the hope that you will chance upon the right key words will not help you. Your are encouraged to use a word processor for the final but it is not a requirement. If you do type your exam on a computer you don’t have to count time for printing out. If you write your answers that answers by hand please make sure that they are legible! You won’t get credit for can’t be read. Where figures are required you can either draw them or make them using a data analysis software (such as excel). Please set a timer when you start and stop immediately when the time limit is reached. You can do work after the time limit is up but you cannot expect full credit for this. In such cases please note explicitly when the time is up. Please note that this is done purely on the honor system, which Caltech takes seriously. Problem 1. Transition state theory, intrinsic binding energy (18 points) 1. Plot the K and k/K values as a function of K. Don’t forget that free energy analysis mcatmi applies on the log, instead of linear, scale. Which parameter shows strong correlation with the binding affinity of the phosphoamidate inhibitor? (6 points) correlation of Km with Kicorrelation of kcat/Km with Ki32.5 y = -2.8842 + 1.0266x R= 0.94259 2.52 21.5)mm1.51/Kcat 10.5log K log (k0.50 0-0.5 -0.5-12.533.544.555.52.533.544.555.5 log Klog Kbindingbinding 2. Based on this information, what type of inhibitor is the phosphoamidates for thermolysin? List your reasons. (4 points) Phosphoamidates are transition state analogue inhibitors. The strong dependence of kcat/Km, rather than Km, on the substituent Y indicates that the structural changes contribute to transition state stabilization, rather than ground state binding. As the binding of these inhibitors is also strongly affected by substituent Y in the same trend, these inhibitors mimic the binding of the reaction transition state. 3. What principle of enzymatic catalysis do these data illustrate? (4 points) Binding interactions of enzyme with substrate groups not directly involved in chemistry are used to enhance the reaction rate rather than substrate binding, likely by inducing the enzyme into more active conformations 4. Based on the above information, what is the putative transition state structure of this reaction? (4 points) The transition state for thermolysin resembles the phosphoamidates, with tetrahedral geometry at the reactive carbon and/or accumulation of negative charges on the carbonyl oxygen Problem 2. Alkaline Phosphatase: catalytic strategies, kinetics 101 (44 points) Alkaline phosphatase (AP) catalyzes the hydrolysis of phosphate monoesters using an active site serine as a covalent nucleophile (Eq 2). (2) 1. Why is it advantageous for the enzyme to use serine 102, instead of an active site-bound water, as the nucleophile? (4 points) The chemical reactivity of serine hydroxyl is similar to that of water. However, it is easier to position the nucleophilic hydroxyl use a covalent bond, than using noncovalent binding interactions. 2. Figure 2 depicts the active site structure with an inhibitor, vanadate, bound. 2+a. What are the catalytic roles of each Zn ion? (4 points) Zn1: helps activate the nucleophiles (Ser-OH and water) by lowering their pK, thus a generating a high concentration of reactive Figure 2 Structure of the AP active site. hydroxide ions at neutral pH. Zn2: coordinates the leaving group oxygen to help stabilize developing negative charges in the transition state. b. What is the catalytic role of Arg166? (2 points) rg 166: positions the reactive phosphoryl group. A c. What kind of inhibitor would vanadate be (e.g., competitive, noncompetitive, or uncompetitive)? List your reasons. (4 points) Vanadate will be a competitive inhibitor, as it binds to and reacts with the free enzyme, not the E•S complex. d. Suppose that you have an assay to measure the hydrolysis of a substrate, p-nitrophenol phosphate (pNPP), by AP. What experiments can you do to test the mode of inhibition by vanadate? Draw the predicted results from your model (don’t worry about numbers, only the patterns are important) (6 points) To test this hypothesis, one can measure the reaction rate constants with and without vanadate. As a competitive inhibitor, vanadate will weaken substrate binding (i.e., increase K), decrease the second order m reaction: E + S , products (i.e., lower k cat /K), but will not affect the rate of reaction m from a fully formed E•S complex (i.e., k cat value is unchanged). A representative graph will look like this: 3. In traditional assays for the AP reaction, the reaction is monitored by absorbance changes. Multiple turnover reactions are followed and reasonable data can only be obtained with at least 0.5 µM pNPP substrate because of the low absorption coefficients of pNPP. However, one of the products of the reaction, P, binds to AP with K < 0.1 nM. id (a) Will this assay accurately measure the reaction rates? What is the problem? (4 points) The measurement using the spectroscopic assay is problematic because of serious product inhibition (the amount of P generated during the reaction quickly exceeds its K for AP, so id reaction is inhibited due to competitive binding of P). i (b) To measure reaction rate constant more accurately, a radioactive assay is developed in 32which P-labeled substrates, at concentrations < 0.1 nM, is used. The concentration of AP is varied and in excess to substrate. The following reaction time course can be observed (Fig. 3): What are the advantages of this assay over the pNPP based spectroscopic assays? (4 points) This radioactive assay is superior to the Figure 3 Time course of AP reaction with a spectroscopic assay because: radioactive assay. (i) It avoids product inhibition problems. By using substrate concentrations < 0.1 nM, the amount of Pgenerated during the reaction is far i below its K of 0.1 µM for AP and therefore will not significantly inhibit the reaction. d (ii) By using enzyme in excess over substrate, the reaction becomes pseudo-first-order and the full reaction time course can be followed, which allows more accurate determination of reaction rate constant and facile separation of the contribution of different reaction constituents. 4. AP is non-specific with respective to its leaving group. This allows a Brønsted relationship between reaction rate and leaving group ability to be constructed for a series of alkyl phosphates (Fig. 4): (a) Estimate the leaving group Brønsted slope of this reaction (4 points) The leaving group Brønsted slope is –0.85. (b) What does this result suggest about the nature of the transition state for the AP reaction? (4 points) Figure 4 Leaving group A large leave group Brønsted slope suggests that the dependence for reaction of AP. transition state is dissociative (metaphosphate-like), with extensive bond breaking to the leaving group. 5. Being one of the most ancient phosphomonoesterases, Alkaline phosphatase also harbors promiscuous activity towards phosphate diesters and sulfate esters. Suggest how this catalytic promiscuity could have contributed to the evolution of phosphodiesterases and sulfatases. (4 points) The catalytic promiscuity of alkaline phosphotase allows products of duplicated genes to harbor low levels of activity for hydrolyzing phosphodiesters and sulfate esters. These activities could be preserved in environments where these activities offer selective advantage. Natural selection could then have improved these promiscuous activities, ultimately leading to the evolution of efficient sulfatases and phosphodiesterases. 6. In solution, hydrolysis of phosphate diesters goes through a roughly synchronous transition state midway between the ‘dissociative’ and ‘associative’ extremes. Alkaline phosphatase could catalyze the phosphodiesterase reaction by forcing its transition state to resemble that of a monoesterase reaction, or alternatively by stabilizing the synchronous transition state for the diesterase reaction. Which possibility is more likely and why? (4 points) The transition state in solution represent the lowest possible reactive structure among all the possible transition state species. It would cost less binding energy from the enzyme to stabilize the solution transition state, than to distort the transition state to make it fit the enzyme active site structure. Therefore, it is more likely that alkaline phosphatase catalyzes the hydrolysis of phosphodiesters by using its active site to stabilize the solution transition state of this reaction. Problem 3. Cooperativity and allostery (18 points) Ire1 is a sensor of unfolded proteins in the endoplasmic reticulum. In response to unfolded proteins detected by its luminal domain, Ire1 oligomerizes, and activates its endonuclease activity in the cytoplasmic domain (Figure 5). Ire1’s RNase activity is also allosterically regulated by the binding of nucleotides (ATP or ADP) and by phosphorylation of its activation loop. Figure 5 Ire1 activity controlled by multiple allosteric factors. 1. Figure 6 shows the activation profiles for Ire1. What does this data suggest about the role of nucleotides in Ire1 activation? (2 points). Nucleotides shift Ire1 towards a more nuclease-active conformation, the same activating conformational change is also favored by oligomerization. In which nucleotide-bound state is Ire1 more active, ADP or Figure 6 Activation profile of ATP? (4 points) Ire1 with different ADP-bound Ire1 is more active. nucleotides. E denotes Ire1. 2. Explain why the cooperativity of Ire1 activation is reduced in the presence of ADP than ATP. (4 points) Cooperativity results from conformational changes of an enzyme towards more active structures upon binding cofactors or oligomerization. Ire1 is already close to the active form when it is bound with ADP, so further shifts in conformational equilibrium towards the active state induced by oligomerization do not provide additional rate advantage. Therefore cooperativity is higher in the less active ATP-bound enzyme than the more active, ADP-bound enzyme. 3. Phosphorylation of Ire1 also activates its RNase activity. In the simplest model, how would the cooperativity of Ire1 activation change if it is phosphorylated, and why? (4 points) Phosphorylation of Ire1’s activation loops further shifts the conformational equilibrium of the enzyme towards the activated state, and will further decrease the cooperativity brought upon by Ire1 oligomerization. 4. Substitution of the ,-phosphoryl oxygen of ADP (ADP,S) significantly reduces the activation 2+2+2+of Ire1, an effect that is rescued by using metal ions alternative to Mg, such as Mn or Cd (Fig. 7a). Oligomerization of Ire1 is affected in the same manner (Fig. 7b). What does this suggest about the mechanism of ADP coordination and Ire1 activation? (4 points) 2+The ,-phosphoryl oxygen of ADP coordinates a key Mg ion, which upon substitution by sulfur 2+coordinates better with Mn. This metal ion interaction is important for inducing the oligomerization of Ire1 and conformational changes that lead to its activation. a.b. Figure 7 Activation (a) and oligomerization (b) profile of Ire1 with different nucleotides and metal ions. OD measures the light scattering from large Ire1 oligomers. 600 Problem 4: Single Molecule Kinetics (20 points) The Tetrahymena group I ribozyme catalyzes the cleavage of an oligonucleotide substrate: CCCUCUA , CCCUCU + pA 5OH5 Binding of the oligonucleotide substrate to the enzyme occurs in two-steps, involving (i) base pairing between the substrate and the ribozyme to form a P1 duplex; and (ii) docking of the P1 duplex into the ribozyme active site. A model of the structures of the ribozyme in the docked and undocked states based on crosslinking data (substrate highlighted in blue) is shown below Single molecule fluorescence is used to study the docking of P1 in this ribozyme. To immobilize the ribozyme on a glass surface, a 3’-extension is added to the ribozyme that anneals with a DNA oligo, which is tethered to the glass surface via a biotin tag (Fig. 1). a) To study the docking of P1 by FRET, where will you place the fluorescent dyes? (2 points) Which pair of fluorescent probes will you use for this FRET study, and why? (2 points) A pair of Cy3 and Cy5 dyes (other FRET pairs with overlapping absorption and emission spectrum can also be used; GFP and YFP are not acceptable) can be used for FRET. One of them can be placed on the 3’-end of the substrate, the other at the 3’-end of the DNA tether that is close to the P1 duplex crosslinking site in the docked state (other locations that produce a different FRET value in the docked and the undocked states are acceptable). b) The following fluorescence time traces were observed for the wildtype all-ribose substrate (rS) and for a modified substrate, -3mS, in which docking of P1 is substantially destabilized by a methoxy substitution of a 2’-hydroxyl group. Which one of these time traces belongs to rS, and which one belongs to -3mS? (4 points) The all ribose substrate should stay predominantly in the docked (high FRET) state while the opposite is true for the -3mS substrate. Therefore, the left panel refers to the all ribose substrate, the right panel refers to the -3mS substrate c) The dwell times of molecules in the docked and undocked states are analyzed and their population distributions are shown below: Based on these data, estimate the rate constant for docking and undocking, and the equilibrium constant for docking of the all-ribose and the -3mS substrates. (8 points) All ribose substrate: estimated half-time in the undocked state ~ 0.5 sec -1 k (docking) = ln2/(0.5 sec) = 1.4 s estimated half-time in the docked state ~ 3 sec -1 (undocking) = ln2/(3 sec) = 0.23 sk K (docking) = k (docking) / k (undocking) = 6 –3mS substrate: estimated half-time in the undocked state ~ 0.6 sec -1k (docking) = ln2/(0.6 sec) = 1.2 s estimated half-time in the docked state ~ 0.06 sec -1 k (undocking) = ln2/(0.06 sec) = 12 s K (docking) = k (docking) / k (undocking) = 0.1 (4) Which kinetic parameter does the -3m substitution primarily affect? (2 points) Based on this, can you suggest why the -3m substitution disfavors docking of the P1 duplex? (2 points) The docking kinetics is similar between the all-ribose and -3mS substrates, while the undocking rate is 50-fold faster with the -3mS substrate. Therefore the –3 methoxy substitution disfavors P1 docking by destabilizing the docked state, presumably due to disruption of tertiary interactions with the 2’-hydroxyl and/or steric clashes introduced by the methoxy group