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1. A ligand binds more tightly to the folded state (N) of a protein than to the unfolded state (U). Show that the ligand stabilizes the protein and calculate by how much (DDGfold = ?)
2. An enzyme E binds a substrate S and a cofactor C. The equilibrium dissociation constant Kd,S of the enzyme-substrate complex ES is 1 μM, for EC it is 10 μM. When the cofactor C is present, Kd,s’ is decreased to 0.1 μM. What is the value for the dissociation constant Kd,C’ of the enzyme-cofactor complexing the presence of substrate S? Calculate the interaction energy DDGint for cofactor and substrate binding.
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Proteins are polymers of amino acids which are folded in three dimensional structure by both covalent and non-covalent interactions. Ligands are molecules that influence the function of protein (proteins such as an enzyme, receptor, etc). Thus, ligand produces a signal by binding to the target proteins. Ligands can bind strongly to only folded protein and the unfolded protein will not be able to recognize the ligand. Ligand also stabilizes the protein.
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- Which of the following statements about inhibition is true? a. Allosteric inhibitors and allosteric activators are competitive for a given enzyme. b. If an inhibitor binds the active site, it is considered noncompetitive. c. If an inhibitor binds to a site other than the active site, this competitive inhibition. d. A noncompetitive inhibitor is believed to change the shape of the enzyme, making its active site inoperable. e. Competitive inhibition is usually not reversible.3. MUTPase from ASFV (aDUT) and from swine (listed as sDUT) were studied in the absence and present of the DUTP substrate. Using the melting temperature data provided below, how does adding the substrate affect enzyme stability? Explain your choice in 25 words or less. TemperatureL°CT protein Tm by thermal denaturation (°C) aDUT 83.1 ± 0.2 SDUT 61.8 + 0.2 aDUT-DUTP-Mg SDUT-dUTP-Mg 84.5 + 0.1 62.7 + 0.2 a. The substrate makes aDUT less stable and SDUT more stable b. The substrate makes aDUT more stable and $DUT less stable c. The substrate makes both ADUT and SDUT less stable d. The substrate makes both aDUT and SDUT more stable sh (United States) D. Focus rch 7:15 80°F ENG 7/10/2 DELL F3 F43.18: An enzyme E binds a substrate S and a cofactor C. The equilibrium dissociation constantKd,S of the enzyme-substrate complex ES is 1 μM, for EC it is 10 μM. When the cofactor Cis present, K’d,S is decreased to 0.1 μM. What is the value for the dissociation constant K’d,C of the enzyme-cofactor complex in the presence of substrate S? Calculate the interactionenergy ΔΔGint for cofactor and substrate binding.
- 1. For enzymatic reaction, a mechanism was proposed by Michaelis and Menten as follows: ES k, and k,' ES à E and P k,. E + S a. Use steady state assumption, derive expression for the reaction rate. Where E is concentration of enzyme, S substrate, ES complex of E and S, E = E, – ES. (If you have difficulty in doing it, please consult lecture note) b. Assume K = 0.038 mol.L' at 25 °C, when the substrate concentration is 0.156 Mol.L', the rate of the reaction is 1.21 m mol/L.s. The maximum rate of conversion reaction is reached at high substrate concentrations. Calculate the maximum rate of this enzyme catalyzed reaction.1. You are studying the enzyme catalyzed reaction below, and you find the KM is 3.3x10-4 M,. You also find that k1 is 4.3x106 M-'s-1. What is the dissociation constant (KD) for the enzyme/substrate complex? k1 k2 E+S ES E+P k.12. Cofactors, coenzymes, and prosthetic groups are important non-protein substances that are required for enzyme function. This is a table of cofactors/coenzymes/prosthetic group we've discussed. Draw the functional portion of the cofactor/pro and explain the functional role in each reaction. a. Cofactor Name thiamine pyrophosphate (TPP) oxidized Lipoamide/lipoic acid/lipoyl-lysine Coenzyme A (CoASH) flavin adenine dinucleotide (FAD) reduced nicotinamide adenine dinucleotide (NADH + H+) oxidized nicotinamide adenine dinucleotide (NADP+) biotin Reaction Pyruvate -> acetaldehyde Pyruvate -> Acetyl CoA aKetoglutarate -> Succinyl-CoA Succinate -> Fumarate Pyruvate -> lactate 6-Phosphogluconate -> Ribulose 5- phosphate Bicarbonate + pyruvate -> oxaloacetate Structure of only the cofactor Functional or catalytic role
- 3. (a) The beakers below represent different conditions for measuring the initial velocity (vo) of an enzyme-catalyzed reaction under the steady-state approximation. The gray "donut-shaped" struc- tures represent the enzyme. The blue, filled circles represent free substrate molecules. The red cir- cles represent substrate molecules bound in the active site of the enzyme forming the Michaelis (ES) complex. Indicate in the diagram of the double reciprocal plot which kinetic parameters or variables each of the three "beaker conditions" represent either alone or in combination with another. A B 455 C V₁™¹ [So]-¹2. The initlal velocity for an enzyme-catalyzed reaction has been determined at a number of different substrate concentrations. Data are given below: [S (mM) Va ( umoles product/min) 5 22 10 39 20 65 50 102 100 120 200 135 500 147 Use the data above to create a Michaelis-Menten plot in Excel, plotting ye vs [S]. Include a copy of your plot, with axes labeled appropriately. Estimate Vmax and KM from this graph. 3. From the data given in problem #2 above, determine the value of the dissociation constant (Ko) for the ES complex. What is the value of the affinity or association constant (KA)?1. Consider the three-dimensional model of the tertiary structure of an enzyme below. Amino acids involved in binding are shaded blue, and amino acids involved in catalysis are shaded red. A. Suppose research has shown that amino acid 82 in the red shaded region is lysine, an amino acid with a positively-charged side chain. This lysine is critical for catalysis. Other studies have found that amino acids 12 and 62 in the blue region are both phenylalanine, an amino acid with a nonpolar side chain, and are critical for substrate binding. These amino acids are relatively close in the active site but are separated by 20-70 amino acids in the primary structure. Using what you know about protein structure, explain how amino acids separated in the primary structure can come close together in the active site. B. Use this information and figure 4.2 in your book to answer the following questions: Do you think changing amino acid 82, lysine, an amino acid with a positively-charged side…
- 1. Consider an enzyme-catalyzed reaction giving the following results at a fixed enzyme concentration of 1 x 10-6 M in the presence or absence of 5 mM or 10 mM concentrations of an inhibitor I [S] (mol/L) 0.0010 0.0025 0.0050 0.0100 0.0500 0.1000 a) Control 0.00167 0.00333 0.00500 0.00667 0.00909 0.00952 v (m/s) [I] = 5 mM 0.00090 0.00200 - 0.00333 0.00500 0.00833 0.00909 [I] = 10 mM 0.00062 0.00142 0.00250 0.00400 0.00769 0.00869 Estimate Vmax and KM for all cases. What type inhibitor is [I]? Estimate K₁ and/or Kı' depending on the type inhibitor. Estimate Kcat. Estimate Kcat/KM.An enzyme is present at a concentration of 1 nM and has a Vmax of 2 µM s-'. The Km for its primary substrate is 4 µM. Calculate kcat- kcat Calculate the apparent Vmax and apparent Km of this enzyme in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 2. Assume that the enzyme concentration remains at 1 nM. apparent Vmax µM s-1 apparent Km = µM7., For the enzyme reaction mechanism with an inhibitor that produces product from both ES and EIS, E+S1→ ES -2➜ E+ P E+1+3 → El El +S4→ EIS -5→ El + P with reaction rate constants of k₁ and k-1 (and so on) for each of the reactions: A. Write the enzyme balance B. Write each of the equations that result from applying the quasi steady state assumption. C. Write each of the equations that result from applying the rapid equilibrium assumption. D. Based on the mechanism, what type of inhibitor is this? (competitive, uncompetitive or non-competitive) Why?