For the following calculations, please provide your answers in whole numbers (no decimals). A research group has discovered a new enzyme they denote XC-95, which catalyzes the conversion of its substrate, propanol, to propionic acid, the product. The researchers begin to characterize the enzyme. In the first experiment, with [E]=4x10-5 mM, they find that Vmax=4 µM s-1. Based on this experiment, the Kcat for XC-95 in units of s-1 is A/ . In their second experiment, with [propanol]=0.01 mM, the researchers find that v₁-2000 nM s-1. The measured KM of XC-95 for propanol in units of µM is A/ Further researc showed that the purified XC-95 used in the first two experiments was actually contaminated with reversible inhibitor of the enzyme, butanol. When butanol is removed from the purified enzyme a the first two experiments are repeated, Vmax is found to be 4 μM s-1, and the measured KM becomes 10 μM. Based on this new data, the mechanism of inhibition by butanol is BEST described by which of the following: Competitive Uncompetitive Mixed Non-competitive A/ and the value of a is A .

For the following calculations, please provide your answers in whole numbers (no decimals). A research group has discovered a new enzyme they denote XC-95, which catalyzes the conversion of its substrate, propanol, to propionic acid, the product. The researchers begin to characterize the enzyme. In the first experiment, with [E]=4x10-5 mM, they find that Vmax=4 µM s-1. Based on this experiment, the Kcat for XC-95 in units of s-1 is A/ . In their second experiment, with [propanol]=0.01 mM, the researchers find that v₁-2000 nM s-1. The measured KM of XC-95 for propanol in units of µM is A/ Further researc showed that the purified XC-95 used in the first two experiments was actually contaminated with reversible inhibitor of the enzyme, butanol. When butanol is removed from the purified enzyme a the first two experiments are repeated, Vmax is found to be 4 μM s-1, and the measured KM becomes 10 μM. Based on this new data, the mechanism of inhibition by butanol is BEST described by which of the following: Competitive Uncompetitive Mixed Non-competitive A/ and the value of a is A .

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

See similar textbooks

Similar questions

A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).
You plot a reaction progress curve and find the equation of the line to be y = -0.0961x + 0.7142. Calculate the U, U/μL and U/mL of lysozyme activity in this fraction. The lysozyme activity is done identical to what is indicated in the lab manual.
From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
In your acid phosphatase enzyme kinetics lab, you constructed a Lineweaver-Burk plot. Lefs assume that you graphicaly obtain 1/Vmax - 9.33 (umoliey and -1Km--0.012 M. Which of the following is correct? A) Vmax = 9.33 (no units) B) Km = 0.012 (no units) C) Vmax = (1/Vru" = 0.107 umolis D) Km =-14-1K = 83.3 pM E) Both C) and D) are correct
Please note the following Lineweaver-Burk plot for the enzyme Virbraniumase reacting with a substrate: -0.4 . 1/V 5 4 3 2 -0.2 0 0.2 y = 5.2781x + 1.3338 R² = 0.9967 0.4 0.6 0.8 1/[S] 1 Based on the information provided, what is the Vmax for this reaction?
If a data from an enzyme experiment is plotted as a Lineweaver - Burk plot, and the Vmax is 0.02 sec/mol, and x-intercept is -2.5 mM-1, then what is the KM value?
An enzyme-catalyzed reaction is allowed to proceed for 3.4 minutes under steady-state conditions (with large excess of substrate). If the enzyme concentration is 7.9 nM, and the concentration of product formed is 6.3 µM, what is the kcat of this enzyme? Please give your answer to two significant figures in s1.
An enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.
Compounds I, II, and III are in the following biochemical pathway: precursor compound I enzyme A enzyme B compound II ► compound III enzyme Mutation a inactivates enzyme A, mutation b inactivates enzyme B, and mutation c inactivates enzyme C. Mutants, each having one of these defects, were tested on minimal medium to which compound I, II, or III was added. Fill in the results expected of these tests by placing a plus sign (+) for growth or a minus sign (-) for no growth in the table below. Minimal medium to which is added Strain with mutation Compound I Compound II Compound III
Proline racemase catalyzes the conversion between L-proline and D-proline. The Km and kcat for this reaction are 0.15 M and 550/sec respectively. If the enzyme concentration is 1.45 X 10-5 mmole/ml what is the Vmax of this reaction?
During the early stages of an enzyme purification protocol, when cells have been lysed but cytosolic components have not been separated, the reaction velocity-versus-substrate concentration is sigmoidal. As you continue to purify the enzyme, the curve shifts to the right. Explain your results. This is an allosteric enzyme and you must use a Lineweaver-Burk plot to determine KM and Vmax correctly. This is an enzyme that displays Michaelis-Menten kinetics and you purify away an inhibitor. This is an allosteric enzyme and during purification you purify away an activator. This is an allosteric enzyme displaying a double-displacement mechanism and during purification you purify away one of the substrates: This is an enzyme that displays Michaelis-Menten kinetics, and you must use a Lineweaver-Burk plot to determine KM and Vmax correctly.
Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some resistant bacteria. The mass of this enzyme in Staphylococcus aureus is 29.6 kd. The amount of penicillin hydrolyzed in 1 minute in 10 ml solution containing 10–9 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change during the assay. [Penicillin] µM Amount hydrolyzed (nanomoles) 1 0.11 3 0.25 5 0.34 10 0.45 30 0.58 50 0.61 a) Plot V0 versus [S] and 1/V0 1/[S]. Does penicillinase appear to obey Michaelis-Menten kinetics? b) What is the value of KM? c) What is the value of Vmax? d) What is the turnover number of penicillinase under these conditions?