KM is determined by measuring the reaction velocity of two enzymes (X andY) at different concentrations. The curves for X and Y were sigmoid andhyperbolic, respectively. How are these graphs different, and why?

Enzymes are usually protein molecules which catalyzes a biochemical reaction by decreasing its…

Answered: KM is determined by measuring the…

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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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
At body temperature (37°C), k of an enzyme-catalyzed reaction is 2.3X10¹⁴ times greater than k of the uncatalyzed reaction. Assuming that the frequency factor A is the same for both reactions, by how much does the enzyme lower the Eₐ?
The total concentration of enzyme in a reaction, [E], is made up of the concentration of enzyme bound to substrate, [ES], and the concentration of enzyme still free in solution, [Ef]. Similarly, the total amount of substrate is made up of [Sf] and [ES]. We can assume that the concentration of enzyme is much less than that of the substrate, [E] << [S]. Assuming the steady state condition and the relationships between [E], [Ef] and [ES], and similar ones for S, given in lectures, derive an expression for the saturation factor, , in terms of [S] and . (Note that [E] and [S] denote the total amounts of enzyme and substrate added to the reaction, respectively. You may assume that [S]>>[E].)
Draw Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and Vmax values for the inhibited and uninhibited rxns? Is the inhibitor competitive or noncompetitive? [S] (M) 10,000 1 x 104 2.000-5 x 104 666673 1.5 x 10-3 400- 2.5 x 10-³ 2005x103 V, No Inhibitor (Arbitrary units) 0.026 38,461 0.092 10.87 37100 0.136 7.35 Wak 0.150 -67 0.165 6.06 L slope: 0033 yint=5057 полесте um.x=0.198 km 6.5x 10-4 0.010 0.040 25 10.086 11.63 0.120 8.33 0.142 7.04 2500 V, Inhibitor Present (Arbitrary Units) 100 ↓ $op.= .009 yint: 5.21 900x12 km: 1.73 y/o conf. V10-3 321000
You design an enzyme assay and choose a substrate concentration equal to the Km of the enzymatic reaction. You measure the rate of the reaction at this substrate concentration to be 75.0 μmol/min. Calculate the Vmax of this reaction in μmol/min. Express your answer with one decimal place. Your previous attempt suffered from low signal-to-noise ratio when you used a substrate concentration equal to the Km. So you decide to increase the substrate concentration to 0.45 mM. You remeasure the rate of the reaction at this new substrate concentration and find it to be 135.0 μmol/min. Using the same Vmax value found above, calculate the Km value of this enzymatic reaction. Express your answer in μM with one decimal place. Considering the previous two questions, what would be the rate of the reaction at a substrate concentration of 0.01 mM? Express your answer in μmol/min with one decimal place.
Compound A is the substrate for two enzymes, El and E2, their reaction rates, r1 and r2,Determine the Km and rmax for both enzymes, with respect to the concentration of A. Which set of data is more likely to be for El and which for E2, and why? Concentration of 0.2 0.6 1.2 3 4 5 6 8 12 15 A (mM) Reaction rate (r.) (mmol/L'min) 3.33 4.29 4.62 4.76 4.84 4.88 4.9 4.92 4.94 4.95 4.96 4.97 Reaction rate (r,) (mmol/L*min) 0.09 0.23 0.38 0.5 0.6 0.67 0.71 0.75 0.8 0.82 0.86 0.80
what is the Vmax and Km of the enzyme in the reaction asap please.
From a series of flasks with a constant concentration of enzyme the following initial velocities weretaken, they were obtained as a function of the concentration of the substrate.a) Calculate the KM and Vmax kinetic parameters of the three forms (Lineweaver-Burk, Eadie-Hofstee, Dixon).b) Analyze which are the atypical data that cause a low correlation, which can be eliminated and explain youranswer.
Consider two enzymes A and B, which are not related. However, the two enzymes coincidentally share the same value for Vmax. Using this information, can you derive conclusions how the km values for enzymes A and B compare to each other?
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?
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.
Calculate 1/[S] and 1/V to complete the table. Use this data to draw a Lineweaver-Burke plot, with lines for ‘no inhibitor’ and ‘with inhibitor’. Be sure to label your axes and lines. What kind of inhibitor is it? Estimate Vmax and Km for the uninhibited reaction.

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KM is determined by measuring the reaction velocity of two enzymes (X and Y) at different concentrations. The curves for X and Y were sigmoid and hyperbolic, respectively. How are these graphs different, and why?