What type of inhibition is observed from the shift of the Lineweaver-Burke plot ahown inthe graph below where the solid Ine represents the uninhibited enzymatic reaction while thebroken line represents the inhibited enzymatic reaction?A. Irreveraible inhíbítionB. Noncompetitive inhíbitionC. Competitive inhíbitionD. Uncompetitive inhibitionPotassium cyanide ia a polson which combines with cytochrome A3 to prevent bindingof oxygen to the enzyme without altering the Km of the reaction with respect to reducedcytochrome c. Which type of inhibition does this represent?A. Irreveraible inhibitionc. Competitive inhibitionD. Uncompetitive inhibitionB. Noncompetitiveinhibition_10.Which of the following enzyme classes catalyze reactions in which two molecules becomedissociated from each other?A. Kinaseв. Нуdrolasec. IsomeraseD. LigaseWhich of the following enzyme classes catalyze reactions in which two molecules becomecovalently linked to each other?A. KinaseB. Hydrolase1.C. IsomeraseD. Ligase

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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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in cells, an enzyme catalyzes the reaction ab → a + b. it was isolated, however, as an enzyme that carries out the opposite reaction a + b → ab. explain the paradox.
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 substrateA is cleaved into two products, P and Q. In the catalytic mechanism, the enzyme converts A to a 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 Kiiis diminished. When a high, fixed concentration of…
Which of the following is true under the following conditions: an enzyme displaying Michaelis-Menten kinetics where the enzyme concentration is 10 nM, the substrate concentration is 45 mM, and the Km is 50 µM? a) The enzyme has low catalytic efficiency for the substrate. b)The rate of catalysis is near half-maximal velocity. c)The enzymatic reaction is near maximal velocity. d)Halving the substrate concentration has little effect on the catalytic rate. e) There is not enough information provided.
Which of the following best explains why enzyme catalysis is affected by a change in pH? A. Change in pH alters ionization states of serine in the active site involved in nucleophilic catalysis B. The ionization states of his, asp and glu involved in acid/base catalysis are altered with change in pH C. Change in pH alters ionization states of contact amino acids in the active site D. All enzymes have optimum pH
(a) Give the substrate for each of the following enzymes: (i) urease (ii) fructose oxidase (b) What is the difference between competitive and noncompetitive inhibition? (c) In feedback control, what type of regulator slows down the catalytic activity of the reaction series? Why aren't intermediate products in a reaction sequence used in feedback control?
The important amino acid residues in the active site of enzyme X were found to be aspartic acid and lysine. In an experiment, the pH of the buffer solution in which enzyme X was dissolved was changed from pH 7 to pH 5. What would be the expected effect of this change on the enzyme’s catalytic activity? A. decrease B. increase C. unchanged D. cannot be determined
The reaction: A + B <-> P + Q is catalysed by enzyme E. Draw a Cleland diagram for a Non-sequential (Ping-Pong) mechanism for this reaction.
It's a three part question based on the chart provided asking: a) Which of these enzymes has the weakest binding of substrate? (I chose fumarase since it has lowest Km but that was wrong so I don't understand) b) Which enzyme has the fastest conversion of ES? (I would assume this means highest Kcat value, so should be catalase) c) Which enzyme is most closely catalytically perfect? (I'm assuming this means highest Kcat/Km ratio, so I was thinking Crotonase, although I don't know what classifies an enzymes as 'catalytically perfect', please explain)
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…
An enzyme catalyzes the reaction M ßàN. The enzyme is present at a concentration of 1 nM, and the Vmax is 2 M s-1. The Km for substrate M is 4 μM. a)Calculate kcat. b)What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an α’ of 2.0?
(a) What are the Vmax and Km (including units) for Tigerzyme? b) What are the Vmax and Km (including units) for Tigerzyme in the presenceof the inhibitor Tar Heel? (c) What are the Vmax and Km (including units) for Tigerzyme in the presenceof the inhibitor Blue Devil?

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