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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- At what substrate concentration would an enzyme with a kcat of 25.0 s-1 and a KM of 3.5 mM operate at 25% of its maximal rate? How many reactions would the enzyme catalyze in 45 seconds when it is fully saturated with substate, assuming the enzyme has one active site?1. Can you describe how electrostatic and steric considerations may lead to preferential stabilization of the transition state at an enzyme active site? 2. What factors are involved in “transition-state complementarity”?1. The concentration of substrate X is high. What happens to the rate of the enzyme-catalyzed reaction if the concentration of substrate X is reduced? Explain. 2. An enzyme has an optimum pH of 7.2. What is most likely to happen to the activity of the enzyme if the pH drops to 6.2? Explain
- 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?A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction The researchers begin to characterize the enzyme. (a) In the first experiment, with [Et] at 4 nM, they find that the Vmax is 1.6 uM/s. Based on this experiment, what is the kcat for happyase*? (Include appropriate units.) (b) In another experiment, with [Et] at 1 nM and [HAPPY] at 30 uM, the researchers find that V0 = 300 nM/s. What is the measured Km of happyase* for its substrate HAPPY? (Include appropriate units.) (c) Further research shows that the purified happyase* used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase* preparation and the two experiments repeated, the measured Vmax in (a) is increased to 4.8 uM/s, and the measured Km in (b) is now 15 uM. Based on this information, can you figure out what type of inhibitor is ANGER? (Use table 6.9 at the end of the…An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.
- a. What is the Vmax of this enzyme WITHOUT inhibitor? Please show your work. b. What is the Km of this enzyme WITHOUT inhibitor? Please show your work. c. The specificity constant of enzyme X is 8 x 10^7 /(M * seconds) What is the kcat of enzyme X WITHOUT inhibitor? Please show your work d. What was the concentration of enzyme used for measuring the kinetics of enzyme X WITHOUT inhibitor? Please show your work3.18: 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.1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the inhibitor icreaces, what can be said about the mode of inhibition and why? 1.3) calculate the turnover number for an enzyme, assuming Vmax is 0.5M.sec-1 and the concentration of the enzyme used is 0.002M . why is it usefull to know this? 1.4) discuss the mechanism of the bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues. make use of relavant graphs and diagrams to explain your answer.
- In a bisubstrate reaction, a small amount of the fi rst product P is isotopicallylabeled (P*) and added to the enzyme and the fi rst substrate A. No B or Q is present. Will A (= P—X) become isotopically labeled (A*) if the reaction follows a Ping Pong mechanism?An enzyme catalyzes a reaction in which substrate A is cleaved into two products, P and Q. In the catalytic mechanism, the enzyme converts A to an covalently-bound reaction intermediate X and product P, P then desorbs from the enzyme, and in a second chemical step, the enzyme converts the intermediate X in the EX complex to the final product Q (in EQ), which then desorbs from the enzyme E. You discover two inhibitors of this enzyme, I and J. I is a competitive inhibitor of the substrate A, and has nearly double the molecular weight of J. On the other hand, J is a mixed inhibitor of enzyme E, and its inhibitory effect on Km / Vmax (the slope effect from the double reciprocal plot) is greater than that of 1 / Vmax (the intercept effect in a double reciprocal plot). That is Kis < Kii . At low pH, the conversion of EX to EQ is greatly slowed, kcat is decreased, and the intercept effect of inhibitor J is elevated, that is, the value of Kii is diminished. When a high, fixed concentration…A 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.