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Name: TF Name LS1a Fall 2014 Problem Set #6 Due Wednesday 12/10 at 6 pm in your TF’s drop box on the 2nd floor of the Science Center I. Basic Concept Questions 1. (10 points) Shown below is part of a peptide that can be hydrolyzed by a protease. a. (2 points) The P1 pocket of the protease binds to the tryptophan in the peptide, and the P2 pocket binds to the serine. Circle the peptide bond that will be hydrolyzed by the enzyme. b. (4 points) The protease that cleaves this substrate shows a strong preference for substrates that contain nonpolar, cyclic amino acid side chains such as tryptophan or phenylalanine at position R1. The P1 pocket binds Trp more tightly than Phe. In terms of entropy and enthalpy, why would you expect a binding pocket that can accommodate tryptophan to bind tryptophan more favorably than phenylalanine? In terms of entropy, tryptophan is larger and contains more hydrophobic surface area; as a result, more ordered water would be released when it binds to the enzyme, leading to a more favorable (i.e., a larger or more positive) ΔS of binding. In terms of enthalpy, the larger size of tryptophan provides it with greater surface area over which van der Waals interactions can form compared to phenylalanine. In addition, a pocket that binds tryptophan may accept a hydrogen bond from tryptophan’s side chain –NH group. This hydrogen bond acceptor could only make a dipole:induced dipole interaction with phenylalanine, which would also be enthalpically unfavorable. c. (4 points) The P2 pocket of the protease contains a residue that forms a hydrogen bond to the serine in the substrate. If the serine in the substrate is mutated to alanine, would you expect Kd for binding of the protease to the substrate to increase or decrease? Briefly explain your answer. The P2 pocket normally forms a hydrogen bond with serine that it cannot with alanine. The mutation therefore breaks a favorable bond between the P2 pocket without replacing it, making the binding of the substrate less favorable. As a result, the Kd for that binding interaction would increase with the mutation. 2. (16 points) The diagram below shows the binding interactions between a drug and six amino acid side chains of an enzyme that binds to the drug. The numbers indicate the locations of the specified amino acids within the enzyme’s protein sequence. a. (10 points) In the table below, name the amino acid given its number in the protein sequence and indicate the strongest type of intermolecular interaction that can take place between the amino acid side chain and the drug (e.g., van der Waals, ionic, etc…). Amino acid # Name of amino acid Strongest type of interaction possible with drug 57 Aspartate ionic 84 Tryptophan van der Waals (or induced-dipole:induced dipole) [ion:induced-dipole is also acceptable] 199 Glutamate ion:dipole 231 Threonine hydrogen bond 2 330 Phenylalanine van der Waals (or induced-dipole:induced dipole) [ion:induced-dipole is also acceptable] b. (6 points) The tight binding that is observed between this drug and the enzyme is enhanced by the rigid nature of the drug as well as its ability to displace water molecules from the space that is occupied by the drug when it binds to the enzyme. Briefly discuss in your own words how each of these factors contributes favorably to the free energy of binding between the enzyme and the drug. The rigidity of the drug helps to minimize the decrease in its entropy that occurs when it binds to the enzyme (i.e., ΔSD is made less negative). Because this decrease in entropy is unfavorable, the free energy of binding is made more favorable (more negative) when the drug is pre-rigidified such that is has less entropy to lose upon binding. The displacement of ordered water molecules that were previously bound to the enzyme’s drug-binding pocket, releasing them into bulk solvent upon drug binding, increases the entropy of the water molecules (i.e., ΔSW > 0). 3. (14 points) The reaction coordinate diagram below compares the energy landscape of an uncatalyzed reaction with that of the same reaction catalyzed by an enzyme. a. (2 points) What is the free energy change associated with binding of the enzyme to the substrate? 3 1 kJ/mol (either positive or negative is ok, we don’t care about the sign) b. (6 points) What is the difference between ΔG‡catalyzed and ΔG‡uncatalyzed? Show the calculations you used to arrive at your answer. 1 kJ/mol ΔG‡uncatalyzed = 10 – 6 = 4 kJ/mol ΔG‡catalyzed = 8 – 5 = 3 kJ/mol ΔG‡uncatalyzed - ΔG‡catalyzed = 4 – 3 = 1 kJ/mol (again, either positive or negative value for energy is fine) c. (6 points) A mutant version of the enzyme is discovered that generates product at a much slower rate than the wild-type (“un-mutated”) protein. A reaction coordinate diagram of the uncatalyzed reaction and the reaction catalyzed by the mutant protein is shown below. Why does the mutant protein yield product at a slower rate than the wild-type protein? The energy landscape of the reaction catalyzed by the mutant protein shows much tighter binding affinity for the substrate. The affinity of the enzyme for the substrate is so great that the enzyme no longer lowers the energy of the transition state relative to the energy of the E+S complex (ΔG‡ is now 5 kJ/mol for the mutantcatalyzed reaction). 4 The mutant protein generates product much more slowly because it raises the energy of activation for the reaction, slowing down the reaction rate. The mutant protein raises the energy of activation of the reaction relative the wild-type enzyme by stabilizing the E+S complex more so than the wild-type enzyme does. 5 II. Applied Concept Questions 4. (20 points) Similar to aspartyl proteases, cysteine proteases cleave peptides using acid/base catalysis. The enzyme first cleaves the peptide bond, releasing the aminoterminal peptide fragment in a two-step process shown below as steps (a) and (b). The carboxy-terminal peptide fragment remains attached to the cysteine of the enzyme, resulting in an “acyl-enzyme” intermediate as labeled below. In the final step (not shown), the histidine positions a water molecule to attack the carbonyl of the acyl-enzyme intermediate to release the carboxy-terminal peptide and regenerate the cysteine of the active site. a. (2 points) Is the sulfur of cysteine acting as a nucleophile or an electrophile in step (a)? Sulfur acts as a nucleophile that attacks the carbonyl carbon. b. (6 points) What role does histidine perform in catalyzing step (a)? What role does histidine perform in catalyzing step (b)? In step (a), it is acting as a catalytic base by abstracting a proton off of cysteine to make it a better nucleophile. In step (b), it donates its proton, acting as a catalytic acid. 6 c. (6 points) Draw the transition state for step (a) of this reaction. d. (6 points) As shown below, backbone amide hydrogens help to orient the substrate appropriately in the active site. In addition to orienting the substrate, what other role do the two backbone amide hydrogens perform to help catalyze this reaction? They stabilize growing negative charge on the carboxyl oxygen that occurs specifically in the transition state, thereby catalyzing the reaction by stabilizing the transition state. The hydrogen bonds help make the carbonyl carbon more electrophilic (acid catalysis) by placing partially-positive hydrogens near the oxygen, stabilizing a greater negative charge on the oxygen, pulling electron density towards the oxygen an away from the electrophilic carbon. 7 5. (14 points) Morphine is a drug that binds to the opioid receptor and is given to patients suffering from chronic pain. a. (8 points) The binding pocket of the opioid receptor bound to morphine is shown below (the morphine molecule is in bold). For the two analogs shown below, would the Kd of binding to the same opiod receptor be higher or lower than the Kd of morphine binding to the receptor? Briefly explain whether the change in Kd is principally due to a change in Hbinding or Sbinding, and why. i. (4 points) Analog 1: Analog 1 would have a larger Kd than morphine because of a change in Hbinding. When morphine is free (not bound by the pocket), its positively-charged -NH+- group makes an ion:dipole interaction with water. When the pocket is free, its D147 also makes an ion:dipole interaction with water. When morphine binds the pocket, D147 and the -NH+- form an ion:ion bond, and the released water molecules form a hydrogen bond between them. The binding of this region of morphine to D147 therefore replaces two ion:dipole bonds 8 with an ionic bond and a hydrogen bond. The Hbinding is roughly zero as the energetic sum of two ion:dipole bonds is roughly equivalent to the energetic sum an ionic bond plus a hydrogen bond. Analog 1 replaces the positively-charged -NH+- in morphine with a neutral -CH- group, which would not interact with water since it is nonpolar. The binding of analog 1 therefore replaces an ion:dipole bond that the free pocket could form with water with an ion:induced dipole bond when the pocket binds this region of the drug. Replacing the positively charged nitrogen with a neutral carbon therefore makes Hbinding for binding analog 1 positive by replacing, making the binding of analog 1 less favorable than the binding of morphine. [One could also argue for full credit that making the drug more hydrophobic by replacing the -NH+- with a -CH- group allows for more water to be released upon analog 1 binding, such that SW is even more positive for Analog 1 than for morphine.] ii. (4 points) Analog 2: Analog 2 would have a larger Kd than morphine because of a change in Sbinding. Analog 2 lacks a bridging carbon-carbon bond that morphine has, and is therefore less rigid than morphine. Analog 2 would therefore suffer a greater loss of entropy upon binding compared to morphine, making the binding reaction less favorable. b. (6 points) An opioid receptor mutant that replaces the aspartate at position 147 with a leucine was found. Which drug (morphine, analog 1, or analog 2) would have the smallest Kd of binding with this mutant opioid receptor? Briefly explain your answer. Analog 1 would have a smaller Kd than morphine or analog 2 for binding to this mutant opiod receptor. For morphine and analog 2, the binding reaction would require that the positively-charged nitrogen lose a ion:permanent-dipole interaction with water when the drug is free in order to be make an ion:induceddipole interaction with leucine, such that Hbinding > 0. Analog 1 would not have this increase in Hbinding. Additionally, analog 1 substitutes a carbon for this positively-charged nitrogen, making it more hydrophobic. When analog 1 is free in solution, a solvation shell of water would form around this hydrophobic region. When analog 1 binds the mutant opiod receptor, the previously-ordered water molecules are released to the bulk solvent, such that SW > 0. 9 6. (16 points) Adenosine hydrolase is an enzyme that hydrolyzes adenosine into ribose and adenine. Shown below on the left is a diagram of the structure of an inhibitor bound to the active site of the enzyme. The inhibitor binds much more tightly to the enzyme than the substrate, adenosine (shown on the right), due to a critical binding interaction between the inhibitor and glutamate 166, an amino acid in the active site of the enzyme. The interaction between E166 and the inhibitor is indicated in the diagram. a. (10 points) Shown below are two proposed mechanisms for adenosine hydrolysis (note that mechanism 1 shows only the first step of a multi-step process). Draw the transition state associated with each mechanism. Indicate partial charges where appropriate. 10 b. (6 points) Given the importance of the binding interaction between E166 and the inhibitor, which of the mechanisms shown above, 1 or 2, is more likely to be the mechanism catalyzed by this enzyme? Briefly explain your answer. The structure of the inhibitor and the critical interaction that it forms with the negative glutamate suggests that mechanism 1 is the mechanism that leads to adenosine hydrolysis. Mechanism 1 involves an accumulation of positive charge on the oxygen of the ribose ring (the position equivalent to the position of the positive nitrogen in the inhibitor) during both the transition state and in the intermediate. The interaction with glutamate lowers the energy of the transition state in order to make the reaction proceed more quickly. Mechanism 2 does not lead to an accumulation of positive charge on this oxygen, so if that mechanism took place, the interaction with glutamate would be unnecessary towards stabilizing the transition state, and therefore would be unlikely to bind an inhibitor (since inhibitors bind active sites tightly by mimicking the transition state). 7. (10 points) The HIV Tat protein is required for high-level expression of viral genes in infected host cells. a. (2 points) Why is Tat necessary for HIV virulence? Tat is required to recruit the kinase that phosphorylates RNA polymerase. This phosphorylation is needed for efficient (i.e., processive) transcription of the viral genome. b. (4 points) Shown below is a diagram of the secondary structure of the TAR sequence. The short nucleotide sequence shown below on the right significantly reduces Tat-mediated enhancement of viral gene expression when injected in excess into the nuclei of infected cells. However, the sequence does not bind to HIV Tat by itself. 11 5’-CUGGCUAACUAGGGA-3’ 5’– Propose an explanation that accounts for the observed results. The sequence is the same as the region highlighted in red, and is therefore complementary to the portion of the hairpin sequence highlighted in blue. By adding an excessive amount of the 15-base RNA polynucleotide strand shown to the right, the 15-base polynucleotide would base pair with the blue region of TAR, preventing formation of the hairpin structure. Because Tat recognizes and binds to the hairpin structure, it is unable to do so in the presence of the short nucleotide sequence. c. (4 points) A mutant Tat protein is found that shows the same level of weak transcription as is observed in the absence of Tat. However, the mutant Tat protein still binds to TAR with the same affinity as the wild-type (“non-mutant”) Tat protein. Briefly explain what could account for this. Most likely, the mutant Tat is no longer able to recruit the protein kinase that is required to phosphorylate RNA polymerase. Failure to recruit the protein kinase results in transcription that is not processive. 12