1. pH Effects a. In the enzyme mechanism of lysozyme, two acidic amino acid residues, Asp52 and Glu35, are critical for catalytic activity. If we assume normal side chain pKa = 4.07), what proportion of values for Asp (pKar = 3.90) and for Glu (pKaR enzyme molecules will have both Asp52 and Glu35 in the correct ionization state at pH 5.0 (the pH optimum for lysozyme)? b. Are the traditional pKa values likely to be correct within the protein? What pKa changes might be present within lysozyme?
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- 5. You discover a new cysteine protease similar to papain. Cysteine proteases are proteolytic enzymes that utilize a cysteine residue in the active site for the nucleophilic attack of a peptide bond, while a second residue acts as a base for proton abstraction in the reaction (acid base catalysis). In this novel enzyme the pKa of the cysteine residue is significantly lowered by the active site environment to pKa=4. a. Knowing that the protease shows highest activity between pH 4-6, what could be the identity of the base residue acting as a in the reaction? Explain and write the expected mechanism for the reaction. SH он. The optimal conditions for salivary lysozyme (hydrolyzing glycoproteins of bacterial wall) are 37 C - temperature and pH is 5.2. Explain the decrease in this enzyme activity if the temperature will rise up to 60 °C and pH will be changed to 8.0. To answer the question: a) draw the graph of the velocity dependency on temperature and pH; b) calculate the relative enzyme activity if 10 mg of lysozyme catalyzes the formation of 5 uM of the product per 2 minutes. Concidor NH3: 5.A7. Consider what can be concluded about the catalytic site (substrate binding site) of phosphatase and the binding site of NaF from the type of inhibition shown by NaF for phosphatase and illustrate it schematically. The inhibition typre is pure noncompetitive inhibition.
- What is the catalytic efficiency of Catalase ? Table. The values of KM and kcat for some Enzymes and Substrates Enzyme Carbonic anhydrase Substrate CO2 HCO3 KM (M) 1.2 x 10-2 2.6 x 10-2 Kcat (s-1) 1.0 x 106 4.0 x 105 Catalase H2O2 2.5 x 10-2 1.0 x 107 Urease Urea 2.5 x 10-2 4.0 x 105 O A. 4 x 108 M-s-1 O B. 4 x 108 M-1.s-1 OC25x 10-9 M-s1 D. 2.5 x 102 M-1.s-1 OE 1.0 x 107 s1The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1RuBP carboxylaseis by no means an idesl enzyme. Describe some of the problems with its active site and its substrate specificity. If we compare the amino acid sequences of this enzyme from many different species, they are almost identical. What is the significance of this uniformity?
- . The allosterically regulated enzyme ATCase binds aspartic acid as a substrate and acylates the a-amino group. Succinate acts as a competitive inhibitor of ATCase because it binds the active site but can't be acylated. The dependence of vo on [aspartic acid] for ATCase is shown in panel (a) of the accompanying figure. Panel (b) shows the effect of increasing [succinate] on v, when [Asp] is held at a low concentration (see thick vertical arrow in panel (a)). Note that in panel (b), vo is not zero when [succinate] =0 (see thin horizontal arrow). Explain the shape of the curve in panel (b). Why does v, increase initially, before decreasing at higher [succinate]? Co0- COO CH2 CH, HC -NH, CH, COO COO Asp Succinate [Asp) [Succinate] [Asp] in experiment b (a) (b)Studies at diff erent pH’s show that an enzyme has two catalytically important residues whose pKs are ∼4 and ∼10. Chemical modifi cation experiments indicate that a Glu and a Lys residue are essential for activity. Match the residues to their pKs and explain whether they are likely to act as acid or base catalysts.The diagram below shows the substrate binding cleft for a protease, providing the substrate structure, and indicating the residues (using one-letter code) that line the four specificity pockets. F 1 M The protease is known to cleave the amide linkage between W and E residues for substrates containing the WEFD sequence. 1 O Match the specificity pocket (identified by red number) to the residue that occupies it. 3 R Match the specificity pocket (identified by red number) to the residue that occupies it. 4 W 2 1. D 2. E 3. F 4. R 5. W 6. Y
- Chymotrypsin has the highest affinity for which of the following substrates: Table. The values of KM and kcat for some Enzymes and Substrates Enzyme Chymotrypsin Ки (М) 4.4 x 10-1 8.8 x 10-2 6.6 x 104 Kcat (S-1) 5.1 x 10-2 1.7 x 10-1 1.9 x 102 Substrate N-acetylglycine ethyl ester N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester Catalase H2O2 2.5 x 10-2 1.0 x 107 Urease Urea 2.5 x 10-2 4.0 x 105 OA. N-acetylglycine ethyl ester OB. N-acetylvaline ethyl ester OC. N-acetyltyrosine ethyl ester D. UreaThe KMof the enzyme for the substrate adenosine is 3 × 10ꟷ5M. The product inosine acts as an inhibitor of the reaction, with an inhibition constant (KI, the dissociation constant for enzyme-inhibitor binding) of 3 × 10ꟷ4M. However, a transition state analog,Inhibits the reaction with KIof 1.5 × 10ꟷ13M. Explain why 1,6-dihydroinosine serves as a better inhibitor of adenosine deaminase than inosine. Elaborate on your answe6F. What conformational state is stabilized by y in ATP synthase? Why might achieving this state require energy input from the PMF?