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- 5.5 Explain the effect of each type of inhibitor on the apparent kinetic parameters: (1) Inhibitor binds only free enzyme does Km increase, decrease, not change? does Vmax increase, decrease, not change? (2) Inhibitor binds only ES complex does Km increase, decrease, not change? does Vmax increase, decrease, not change? (3)Inhibitor binds E and ES equally does Km increase, decrease, not change? does Vmax increase, decrease, not change?MATHEMATICAL The hydrolysis of a phenylalanine-containing peptide is catalyzed by -chymotrypsin with the following results. Calculate KM and Vmax for the reaction. PeptideConcentration(M)Velocity(Mmin1)2.51045.010410.010415.01042.21065.81065.91067.11062.13 Calculate the reaction enthalpy, entropy, and free energy, AH°,, ASº,, and AG°,, for the amidation of glutamate to glutamine (1) at standard temperature, (2) at 37°C, and (3) at 75°C (a) taking into account the temperature dependence of AH and AS, and (b) assuming that AH and AS are temperature-independent. What can you conclude about the importance of Cp? What does the result mean for organisms that live at moderate temperature (37°C; mesophilic organisms) and thermophilic organisms that live at 75°C? Glutamine: AH°; = -826 kJ mol², Cp,m = 184 J mol1 K1, Sm = 195 J mol1 K1 f%3= Glutamate: AH°; = -1010 kJ mol1, Cp,m = 175 J mol1 K1, Sm = 188 J mol1 K1 H2O: AH°¡ = -290 kJ mol1, Cp,m = 75 J mol1 K1, Sm = 70 J mol1 K1 NH,*: AH°; = -133 kJ mol1, Cp,m = 80 J mol1 K1, Sm = 113 J mol1 K1 %3D
- At 37 °C, the serine protease subtilisin has kcat = 50 s-1 and KM = 1.4 × 10-4 M. It is proposed that the N155 side chain contributes ahydrogen bond to the oxyanion hole of subtilisin. J. A. Wells and colleagues reported (1986, Phil. Trans. R. Soc. Lond. A 317:415–423) the following kinetic parameters for the N155T mutant of subtilisin: kcat = 0.02 s-1 and KM = 2 × 10-4 M.(a) Subtilisin is used in some laundry detergents to help remove protein-type stains. What unusual kind of stability does this suggest for subtilisin?(b) Subtilisin does have a problem in that it becomes inactivated by oxidation of a methionine close to the active site. Suggest a way to make abetter subtilisin.(c) Is the effect of the N155T mutation what you would expect for a residuethat makes up part of the oxyanion hole? How do the reported values ofkcat and KM support your answer?(d) Assuming that the T155 side chain cannot H-bond to the oxyanion intermediate, by how much (in kJ/mol) does N155 appear to stabilize…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. Urea4.14 The standard Gibbs energy for the hydrolysis of ATP to ADP is -31 kJ mol-. What is the Gibbs energy of reaction in an environment at 37°C in which the ATP, ADP, and P; concentrations are all (a) 1.0 mmol dm and (b) 1.0 µmol dm?
- There are parts A-C for this picture included. A) What type of enzyme is Malate Dehydrogenase? choices: Hydrolase, Isomerase, Ligase, Oxideoreductase, Transferase, or Translocase B) Which of the following statements are true in biochemical standard conditions? There can be more than 1. Choices: The reaction is spontaneous since ∆G°' is positive The reaction is spontaneous since ∆G°' is negative The reaction is not spontaneous since ∆G°' is positive The reaction is not spontaneous since ∆G°' is positive The equilibrium favors products since K is greater than 1 The equilibrium favors reactants since K is greater than 1 The equilibrium favors products since K is less than 1 The equilibrium favors reactants since K is less than 1 The reaction is always at equilibrium C) If the concentration of Oxaloacetate is 10^7 times lower than the concentration of Malate D.,Is the reaction Spontanuous? Choices: No, because RTInQ is very positive Yes, because RTlnQ is very…Calculate the net charge of the following peptide sequence DAVIRSAXSUEY at pH 7 а. -1 b. -2 с. -3 d. 0 е. 1Stearidonic acid (C18H28O2) is an unsaturated fatty acid obtained from oils isolated from hemp and blackcurrant (see also Problem 10.11). a. What fatty acid is formed when stearidonic acid is hydrogenated with excess Hạ and a Pd catalyst? b. What fatty acids are formed when stearidonic acid is hydrogenated with one equivalent of Hz and a Pd catalyst? c. Draw the structure of a possible product formed when stearidonic acid is hydrogenated with one equivalent of H2 and a Pd catalyst, and one double OH stearidonic acid bond is isomerized to a trans isomer. d. How do the melting points of the following fatty acids compare: stearidonic acid; one of the products formed in part (b); the product drawn in part (c)?
- Structures given of the amino acids alanine (Ala), methionine (Met) and threonine (Thr) yOH H2N CO2H H2N CO2H H2N `CO2H Q3 (a) How could the sequence of Ala-Met-Thr be distinguished from that of Thr-Ala-Met by tandem ESI- MS? Explain in detail. (b) Draw the structures of products from a trypsin- catalysed reaction of Ala-Lys-Ser.Nonspecific elution of affinity bonded macromolecules is used in affinity chromatography explain why?(b) The values of kinetic parameters for a variety of synthetic ester and peptide substrates ofa- chymotrypsin are compared for the reaction scheme below where ES is the Michaelis complex, ES' E+S Substrate K₁ N-Ac-Trp–OC₂H5 N-Ac-Phe-OC2H5 N-Ac-Leu-OC2H5 N-Ac-Phe-CONH2 N-Ac-Tyr-p-nitro-anilide ES K₂ 3.5 13.0 3.2 1.7 K-₁ is the acylenzyme, P₁ is the first product to be released, P2 is the second product, k2 is the acylation rate constant, k3 is the deacylation rate constant, and kcat = k2 k3/( k2 + k³). In the table below, N- Ac = N-acetyl; -CONH₂ = carboxamide; -OC₂H5 = ethyl ester; p-nitroanilide = −NH-C6H4-NO2 simulating a peptide group. ES' K2 (S-¹) K3 (S-1) 0.84 2.2 0.19 P₁ K3 0.073 H → E+ P2 Kcat (S-1) 0.82 1.9 0.18 0.070 0.038 KM (MM) 0.08 1.3 4.2 24.0 0.35 1 Kcat/ KM (mM-¹ s¯¹) 10.3 1.5 0.04 0.003 0.11 The value of kcat for N-Ac-Phe-OC2H5 is two-fold greater than that for the L-tryptophanyl analog and more than 10-fold greater than the value of kcat for the ester substrate…