1. You are studying the enzyme catalyzed reaction below, and you find the KM is 3.3x104 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 k1
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- The enzyme lysozyme kills certain bacteria by attacking a sugar called N-acetylglucosamine (NAG) in their cell walls. At an enzyme concentration of 2 × 10-6 M, the maximum rate for substrate (NAG) reaction, found at high substrate concentration, is 1 × 10-6 mol L¯'s1. The rate is reduced by a factor of 2 when the substrate concentration is reduced to 6 x 10-6 M. Determine the Michaelis-Menten constants Km and k, for lysozyme.1). The initial velocity of an enzyme-catalyzed reaction has been determined at several substrate concentrations with and without an inhibitor I. The total enzyme concentration [E]ot present in each experiment is 1.20 x 10 M. [1] = 5.00 x10 M. (Note: v initial rate or velocity.) [S], M 1.25E-3 2.50E-3 5.00E-3 10.0E-3 v, M/min, no inhibitor 0.480E-3 0.660E-3 0.810E-3 0.920E-3 v, M/min, 5.00x10 M 9.43E-5 1.41E-4 1.70E-4 2.10E-4 b). What kind of inhibition occurred. c). Calculate the value of Ki = a). Use the Lineweaver-Burk plot to determine the values of Vmax and KM at [1] = 0.00 mM, and the values of Vmax and KM at 0.500 mM of I. it in 'How many molecules ofA research group discovers a new enzyme they decide to name happyase. This enzyme has a Michaelian behavior and catalyzes the chemical reaction below: HAPPY ↔ SAD The researchers begin to characterize the enzyme. 1-In the first experiment, the researchers use a total enzyme concentration ([Et]) of 4 nM. They find that the happyase has a Vmax of 1.6 µs-1 in these conditions. Calculate the k2 of happyase based on this finding. Show your work and include appropriate units. 2-In another experiment, with [Et]=1 nM and [HAPPY]=30 µM, the researchers find that V0 = 300 nM.s-1. Calculate the Km of happyase for its substrate HAPPY based on this finding. Show your work and include appropriate units. 3-Further research shows that the purified happyase enzyme used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. After the careful removal of ANGER, the measured Vmax is increased to 4.8 µM.s-1, and the calculated Km becomes equal to 15 µM. Compare…
- The initial rates of an enzyme-catalyzed reaction have been determined for five different [S] (see table below). [S]o (mol.L-¹) 1.0 x 10 1.5 x 10-4 2.0 x 10-4 -4 5.0 x 10 7.5 x 10-4 -1 Vo (μmol.L-¹ min-¹) 28 35 42 63 75 1- Determine graphically the value of Km (first with the Michaelis-Menten representation and then with the Lineweaver-Burk representation, then compare by reporting on the Michaelis-Menten representation the values found by the Lineweaver-Burk representation) 2- Calculate the concentration of a competitive inhibitor 1, with a K₁ value of 2.4.104 M, whose action would quadruple the apparent value of Km.You are working on an enzyme that obeys standard Michaelis-Menten kinetics. Based on the following reaction expression, what is the Km value for this enzyme? E+SESE + P . . . k₁ = 880.8 M-¹5-1 k.₁ = 42.18 S-1 k₂ = 56.29 S-1A synthetic substrate, the para-nitrophenylacetate (PNPA), is used to monitor enzyme activity of protein P. The product of the hydrolysis reaction absorbs at 410 nm with a molar extinction coefficient of 4 000 M-¹.cm-¹. 1- Write the reaction catalyzed by the protease P using the pNPA substrate. 2- The enzyme extract is too concentrated and a 1/300 dilution is needed for enzyme tests. Considering that you would need at least 600 µL of diluted enzyme extract for activity tests, indicate which volume of buffer and enzyme extract you must use for the dilution.
- The 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-1A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.Enzyme A catalyzes the reaction S → P and has a KM of 50 μM and a Vmax of 100 nM s–1. EnzymeB catalyzes the reaction S → Q and has a KM of 5 mM and a Vmax of 120 nM s–1. When 100 μM ofS is added to a mixture containing equal amounts of enzymes A and B, which reaction product (Por Q) will be more abundant after 1 minute of reaction?
- An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s3. (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]-¹You find in the literature that an enzyme purified very close to homogeneity has a specific activity of 12,300 U/mg. What is likely to be the largest source of error in the quoted value? U (unit) typically is defined as the amount of enzyme that is required to consume 1.00 umoles of substrate per minute under some given set of conditions. Units are typically determined spectrophotometrically by measuring the rate of formation of product.