Sample pH vs Initial Velocity (AA450/60seconds)a bo160.00140.0012o.0010.00080.00060.00040 0002pH Samplewhat does it mean? Does it agree with what we know about enzymekinetics and why or why nol? If there were anomalies, what were they and what are some possible reasons for thisoccurring? Was an optimal pH identifiable? Why or why not?Initial Velocity (AA450/60seconds)

Enzymes are biological molecules that are capable of catalyzing a reaction by binding a substrate to…

Answered: Sample pH vs Initial Velocity (AA450/60…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Similar questions

At what substrate concentration would an enzyme with a Km of 0.005M operate at one quarter of its Vmax? What would be the Km for an enzyme if it is operating at 90% of its Vmax when the substrate concentration is 0.01M
5.50 1/V, min/umol 5.00 4.50 4.00 y = 0.9474x + 2.6649 y = 0.9997x + 2.032 0.00 1.00 2.00 2.50 3.00 1/[S], uM -1 Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in the presence of two concentrations of inhibitor, what would be the Vmax for the uninhibited enzyme? (bottom graph) Equation is given. Choose the one best answer. 3.50 3.00 2.50 2.00
For an enzyme kinetics experiment, a student prepared a reaction mixture by mixing 450 microliters of 0.75mM PNPP with 4.25ml of 0.2M Tris-HCl buffer. When he is ready to measure the absorbance, he added 0.3ml of Alkaline Phosphatase to the mixture and mixed thoroughly. What is the substrate concentration at the beginning of the reaction in mM ?
Describe reaction between AP and PNPP and how is used to determine the kinetic parameters of AP. i. İndicate the value for the activity of AP(which is 3.08) What does it mean? Is your enzyme active or not? What does it say about the purity of your enzyme? ii. How the purification process affects the activity?
The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 104 M. At a substrate concentration of 0.20 M, v = 43 μmol/min. Calculate the rate of reaction when the substrate concentration is tenfold lower, 0.020M. 43 μmol min rate of reaction → Vmax Км substrate concentration →
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 work
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (µM). 1.3 - Inhibitor + Inhibitor 2.50 0.62 2.6 4.00 1.42 6.5 6.30 2.65 13.0 7.60 3.12 26.0 9.00 3.58 What is the Vmax of the inhibited enzyme reaction?
An enzyme contains an active site aspartic acid with a pKa = 5.0, whichacts as a general acid catalyst. On the accompanying template, draw thecurve of enzyme activity (reaction rate) versus pH for the enzyme (assumethat the protein is stably folded between pH 2–12 and that the active siteAsp is the only ionizable residue involved in catalysis). Briefly explain theshape of your curve.
Why is it important to know the Km and Vmax values of an enzyme?
You 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 μM
BIOCHEMISTRY QUESTION The initial rate (VO) data as a function of substrate concentration [S] for an enzyme (Enzyme1) that obeys Michaelis-Menten kinetics are shown in the table below. The total enzyme concentration is 2.5 nM. [S](μm) 16.0 8.0 4.0 2.0 1.0 v (μMsec ¹) 48.5 32.7 19.8 11.1 5.88 (a) Calculate KM for the enzyme, using the estimated slope (slope = (y2-y1)/(x2-x1)) from a Lineweaver-Burk plot. Show your work and don't forget the units! (b) When the enzyme concentration is 2.5 nM, a Lineweaver-Burk plot of this data gives a line with a y- intercept of 0.0106 μ M-1 sec. Calculate kcat for the enzyme. (Show your work and don't forget to include units). (c) A second enzyme (Enzyme2) also obeys Michaelis-Menten kinetics. Its kcat and KM are 800 sec-1 and 5 µM, respectively. Which is the more efficient enzyme? Briefly explain d) Is Enzyme 1 diffusion limited? Explain.
Answer #2

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