A fatty acid hydroxylase enzyme has a Km for palmitic acid of 9.97 micromolar and a turnover rate of 1212.3 micromole substrate per minute per micromole enzyme. If 63 nanomolar enzyme is mixed. with an unknown concentration of substrate and the reaction rate was 53.69 micromolar substrate per minute, how much substrate was in the reaction to start with? Report your answer in micromolar to the nearest 0.1 uM.

A fatty acid hydroxylase enzyme has a Km for palmitic acid of 9.97 micromolar and a turnover rate of 1212.3 micromole substrate per minute per micromole enzyme. If 63 nanomolar enzyme is mixed. with an unknown concentration of substrate and the reaction rate was 53.69 micromolar substrate per minute, how much substrate was in the reaction to start with? Report your answer in micromolar to the nearest 0.1 uM.

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter21: Biochemistry

Section: Chapter Questions

Problem 28QAP: . Describe the lock-and-key model for enzymes. Why are the shapes of the enzyme and its substrate...

See similar textbooks

Similar questions

. Describe the lock-and-key model for enzymes. Why are the shapes of the enzyme and its substrate important in this model? What does it mean to say that an enzyme is inhibited by a particular molecule? What happens if this inhibition is irreversible? Can you think of a situation in which it might be advantageous to be able to inhibit an enzyme?
Compare the lock-and-key and induced-fit models for substrate binding to an enzyme.
Compare the activation energy in uncatalyzed reactions and in enzyme-catalyzed reactions.
Describe the difference between graphs showing temperature versus reaction rate for an enzyme-catalyzed reaction and an uncatalyzed reaction.
8 The optimal temperature for the action of lactate dehydrogenase is 36°C. It is irreversibly inactivated at 85°C, but a yeast containing this enzyme can survive for months at —10°C. Explain how this can happen.
The figure below shows the dependence of the enzyme's rate, v (in uM/min), as a function of substrate concentration, S (in mM). Also shown is the dependence of the rate in the presence of an inhibitor, present at a concentration of 0.2 mM. Based on this information, the Vmax of the enzyme is approximately: 1/v 1- 0.5 1/[S] O A. 0.25 pM/min O B. 0.5 pM/min O C. 1.0 uM/min D. 2 pM/min O E. 4 µM/min
Can you please help explain how to approach these two problems of calculating the turnover number and average time?
:The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics Vo (mmol/min) S (mmol/L) 0.8 3 4 217 325 433 488 1,000 :The Km for this enzyme is approximately 647 mM 4 O mM 1 O mM 1,000 Ⓒ mM 3 O mm 6 O
Determine the kcat of the reaction given [E] = 2.5E-7 and the enzyme has a mass of kDa. (Km = 9.86E-6 micromol; Vmax 140.85 micromol/minute) HO.HO [S] (M) 1/[S] vo (μmol/min) 1/vo OHO (M) (μmol/min) 2.50E-06 28 4.00E+05 3.57E-02 4.00E-06 40 2.50E+05 2.50E-02 1.00E-05 70 1.00E+05 1.43E-02 2.00E-05 95 5.00E+04 1.05E-02 4.00E-05 112 2.50E+04 8.93E-03 1.00E-04 128 1.00E+04 7.81E-03 2.00E-03 139 5.00E+02 7.19E-03 1.00E-02 140 1.00E+02 7.14E-03 y = 7.0 × 10-8 x + 0.0071
. Relation between Reaction Velocity and Substrate Concentration: MichaelisMenten Equation: At what substrate concentration would an enzyme with a kcat of 30.0s−1 and a Km of 0.0050 M operate at one-quarter of its maximum rate?
None
What mutant of Dihydroorotase affected the most it's turnover rate? Estimate its impact on the enzyme's turnover rate. Table 3: Kinetic Parameters for Mutants of Dihydroorotase kcat (s-1) Km (mM) kca/ K„ (M¯' s-') wild type D250A 160 +8 <0.01% 12 +1 <0.01 1.7 + 0.2 1.5 (0.2) × 10$ nd 6.3 (0.3) x 103 nd nd 4.3 (0.1) x 10! nd 1.7 (0.1) x 104 ndº D250E D250H 1.9 + 0.3 nd ndº 0.51 +0.1 nd 0.9 + 0.1 D250N <0.01 0.022 + 0.002 <0.01 15 +1 D250S R20Q R20K