4. An enzyme-catalyzed reaction is studied in the absence and presence of an inhibitor andthe following data was obtained.[S] inmmoles/LVelocity in mmoles/L/min-1No inhibitorWith inhibitor1.251.720.981.672.041.172.502.631.475.003.331.9610.004.172.38Calculate thea. Km of the enzyme in the reaction without inhibitorb. Km' of the enzyme in the reaction with inhibitorC.Vmax of the uninhibited reactiond. Vmax' of the inhibited reaction1/V1/[S]What kind of inhibitor was added to the enzyme-catalyzed reaction in #4?

Enzymes are involved in the catalysis of various reactions. There are different classes of enzymes…

Answered: An enzyme-catalyzed reaction is studied…

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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Given the initial rate data for the reaction A + B → C, determine the rate expression for the reaction. [A], M [B], M A[C]/At (initial) M/s 0.0418 0.122 2.73 x 10-4 0.0836 0.122 10.9 × 10 4 0.0418 0.488 2.73 x 104 O A[C] At ○ A[C] ΔΙ = 0.439 Ms[A][B]² = 1.27 M²s¹ [A][B] ● A[C] = 6.53 × 10³ s¹ [A] At O A[C] = 0.156 M's¹[A]² At ○ A[C] At = 5.35 x 102 M2s[A][B]
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
The typical Michaelis-Menten equation mathematically describes the overall rate of the reaction as V (this is because biologists don't like math). What does V actually mean? (write the definition of V in differential equation form). V= d( )/dt Reaction rate Substrate concentration V max ·½V TURK
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?
Chymotrypsin has the highest affinity for which of the following substrates: Table. The values of KM and kcat for some Enzymes and Substrates Enzyme Chymotrypsin Ки (М) 4.4 x 10-1 8.8 x 10-2 6.6 x 104 Kcat (S-1) 5.1 x 10-2 1.7 x 10-1 1.9 x 102 Substrate N-acetylglycine ethyl ester N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester Catalase H2O2 2.5 x 10-2 1.0 x 107 Urease Urea 2.5 x 10-2 4.0 x 105 OA. N-acetylglycine ethyl ester OB. N-acetylvaline ethyl ester OC. N-acetyltyrosine ethyl ester D. Urea
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
The following data have been obtained for two different initial enzyme concentrations for an enzyme-catalyzed reaction. [E]=0.015 g/l) (g/l-min) [S] (g/l) v([E]=0.00875 g/l) (g/l-min) 1.14 20.0 0.67 0.87 10.0 0.51 0.70 6.7 0.41 0.59 5.0 0.34 0.50 4.0 0.29 0.44 3.3 0.39 2.9 0.35 2.5 a. Find K b. Find V for [E]=0.015 g/l. c. Find V for [E]=0.00875 g/l. d. Find k2. Use Eddie-Hostee plot to find your answer. Compare your result using Hanes- Wolf plot.

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