6. An enzyme catalyzed reaction has a KM of 1 mM and a Vmax of 5 nM/s. What is the reaction velocity when the substrate concentration is (a) 0.25 mM (b) 1.5 mM, and (c) 10 mM?
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- A bacterial enzyme catalyzes the hydrolysis of maltose as shown in the reaction given below: Maltose + H2O -> 2 glucose If the reaction has a Km of 0.135 mM and a V max of 65 umol/min. What is the reaction velocity when the concentration of maltose is 1.0 mM? (Please take note of the units)A bacterial enzyme catalyzes the hydrolysis of maltose as shown in the reaction given below: Maltose + H2O -> 2 glucose If the reaction has a Km of 0.135 mM and a V max of 65 m mol/min. What is the reaction velocity when the concentration of maltose is 1.0 mM?Given an enzyme with a KM= 10 mM and Vmax = 100 mmol/min. If [S] = 100 mM, which of the following will be true? a. A 10 fold increase in Vmax would increase the velocity 10 fold b. A 10 fold decrease in KM would increase the velocity c. A 10 fold increase in Vmax would decrease velocity 20 fold d. Both A) and B) e. 15000 pM
- 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-sYou are working on an enzyme that obeys standard Michaelis-Menten kinetics. You have determined the Vmax to be 0.1 mol/sec and the Km to be 2.5 mM. What would the rate of the reaction be when the substrate concentration is 20 mM? 0.09 MS-1 O 0.133 Ms-1 O 0.18 Ms ¹ 9 Ms-1 O 0.018 Ms-1 0.2 MS-1If Vmax for a reaction is 10 μM ⋅ s-1 and the KM is 0.5 μM, what is the reaction velocity when the substrate concentration is 2 μM?
- An enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The Km for the substrate is 5.2 uM. What is the initial velocity (V0) at a substrate concentration of 15.2 uM? Report your answer to three significant figures in units of uM/s.The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)3. (а) 0.0050 M operate at one-quarter of its maximum rate? At what substrate concentration would an enzyme with a kcat of 30.0 s-1 and a Km of (b) trations [So]: ½ Km, 2 KM, and 10 KM. Determine the fraction of Vmax that would be obtained at the following substrate concen- (c) ( 1 (HIV-1) has been the object of innumerable studies to develop effective chemotherapeutic agents. It has been shown that p6* known as the late assembly protein is an inhibitor of HIV protease. An assay was developed using an artificial polypeptide substrate containing a p-nitrophenylalanine residue at the cleavage point that undergoes a small change in absorption at 295 nm upon bond hydrolysis that could be followed spectrophotometrically. The cleavage of the peptide bond is shown schematically on the right. Results of the assay are given in the table below. The protease of the human immunodeficiency virus- Lys NH2 Ala Nle Arg Ala Val-Nle-NH-CH- C-NH-Glu CH2 Lys NH2 Ala NO2 Nle H2OH NHIV-1 protease Ala Vo…
- Q is an analog of substrate A that binds to enzyme X and produces the following kinetics:[A] V0 (µmole/ml/min) [Q] = 0 [Q] = 0.5 µM [Q] = 2 µM1 µM 10 7 43 µM 20 16 1010 µM 35 31 2330 µM 43 41 3680 µM 47 46 43a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of dataon your graph) and determine the following: the KM and Vmax of the enzyme X in theabsence of Q, and the KMapp the Vmaxapp at each concentration of Q. b) What type of inhibition is this?c) Calculate the inhibition constant (Ki) for Compound QYou are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μMSuppose that the data shown in the margin are obtained for an enzyme -catalyzed reaction. [S] (mM) V (mmol/ml min) 0.1 3.33 0.2 5.0 0.5 7.14 0.8 8.00 1.0 8.33 2.0 9.09 a.) Determine the Km and Vmaxb.) Assuming htat the enzyme present in the system had a concentration of 10-6M, calculate the turn -over number