hanges in delta-S for the folding of a protein is positive and provides the most significant contribution to thermodynamic driving force for folding. True or False
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- A generalized enzyme active site is shaped like a hemisphere with a radius of 45Å. The active site holds the following amino acids in a homeostatic solution (pH = 7.38): -HAVARILKHAVARILKHAVARILK- Assuming the charge is distributed uniformly along the hemisphere, determine the force at which this active site acts upon a single ATP molecule at the center of the hemisphere.Regarding the physical condition (characteristics of the solution/environment) in which an enzyme finds itself: sometimes enzymes are most active at a certain level and still active but to a lesser degree at a level above or below the optimal level. Describe the likely reason for the enzyme’s decrement in function at levels other than the optimal level?True or False Immobilization improves the stability of the enzyme. EnaLne, has a half-life of 10 days in free solution, but under identical conditions of temperature, pH, and medium composition, the measured half-life of a packed column is 30 days. The enzyme is immobilized in a porous sphere 5 mm in diameter.
- Since KM is an intrinsic property of an enzyme, its value does not depend on the enzyme concentration. True or FalseThe process of protein decomposition by pepsin enzymes in stomach is assumed and modeled as a batch reactor. If the protein concentration over time is measured as follows: Write the relationship between time (t), protein concentration (CA). Assuming that the enzyme reaction follows Michaelis-Menten kinetics, write the required constant (Vmax, Km), and the initial concentration of the protein is CA0.Which of the following statements is right? O For a protein-ligand interaction P +LSP•L, 0 = 0.5 when [L] = Kd. all of the answers are correct Peptide bonds are not weak non-covalent interactions. For an enzymatic reaction that follows Michaelis-Menten kinetics, v/Vmax = 0.5, when [S] = KM. The initial rate of an enzyme catalyzed reaction approaches Vmax, as 1/[S] approaches 0.
- Many types of cells are able to use membrane-spanning transport proteins and a source of energy to drive the active transport of glucose across the cell membrane. Leť's consider a specific protein that is able to transport (from outside the cell into the cytoplasm) one molecule of glucose for each molecule of ATP the protein hydrolyzes. The details of this process are not important for this problem except that you can treat it like equilibrium process. The overall reaction and the associated AG at 298 K for this process are given below. Glc(out) + ATP + H20 = Glc(in) + ADP + P at T = 298 K, AG = -31.3 kJ mol-1 Additional Information: • P. is "inorganic phosphate" (i.e. PO4²- & related species). At equilibrium, [P] = 3 x 10-3 M. • The cell maintains a ratio of ATP to ADP of about 20 (i.e. [ATP]/[ADP] = 20). • The maximum ratio of [Glc(in)] to [Glc(out)] occurs when the system is at equilibrium. a) Write an expression for the equilibrium constant for this process. b) Your expression for…Which statements are false? Initial velocities of enzyme reactions are best obtained in the absence of product because it simplifies analysis. Initial velocities refer to the velocity of the reaction right after it is initiated. The velocity of the reaction as a function of measuring time are curved just like an isothermal binding curve because of substrate binding to the enzyme. Initial velocities correspond to the pre-steady state condition for free enzyme. Initial velocities can sometimes be measured by spectroscopy such as UV/Vis spectroscopy when monitoring the production of NADH from NAD+. The velocity of the reaction will eventually go to zero. The reaction will reach equilibrium because of the presence of the enzyme. It is always better to use substrate rather than product to measure enzyme kinetics.You will perform the protocol below for the calf intestinal alkaline phosphatase (CIP) provided. For each reaction, your final enzyme concentration should be 10 nM CIP. Note: Enzymes purchased are typically labelled with their “units of activity” (U), as this relates to how much enzyme is needed to catalyze a reaction. The 100 nM CIP provided has approximately 3 U/mL and was diluted 1 in 1,000 from a 500 U/mL purchased enzyme. 1) Create a table (similar to the one below) to help you determine and keep track of what to add to each of the cuvettes in which your reactions will be measured. The five different concentrations of PNPP should be: 25, 50, 100, 200, 300 μM. Each reaction will be in a final volume of 1 mL and contain 10 nM alkaline phosphatase. Concentrations of stock solutions: 1.0 mM PNPP, 100 nM calf intestinal phosphatase
- Which of these statements about enzyme-catalyzed reactions is false? The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction. The Michaelis-Menten constant Km equals the [S] at which V = 1/2 V, max: At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration. The rate of a reaction decreases steadily with time as substrate is depleted. If enough substrate is added, the normal V, of max a reaction can be attained even in the presence of a competitive inhibitor.The enzyme mutase which is important for the synthetic of tyrosine and phenylalanine in saccharomyces cerevisiae has been studied as an example of an allosteric enzyme. Tyrosine acts as a negative effector for this enzyme. What effects would you see on the action of the enzyme were you to increase the concentration of tyrosine? The chorismate mutase would shift to its R conformation The curve showing the kinetics or chorismate mutase would shift to the right The curve showing the kinetics of chorismate mutase would shift to the left The chorismate mutase would become saturated more rapidlyWhich of the following statements are true about the relationships of [S], KM, and Vmax? (Choose all that are true) As the [S] is increased, vo approaches the limiting value, Vmax KM = Vmax/2 The rate of product formed, vo, is at Vmax when [S] <<< KM KM and Vmax assist in finding the rate of the enzyme catalyzed reaction only if the reaction is considered irreversible.