The kinetics data shown in this graph are from the enzyme lactate dehydrognase, which catalyzes the redox reaction lactate + NAD* <-> pyruvate + NADH.Calculate the Km and Vmax values with correct units0.040.031/vo(min uM-) 0.02Recall:0.01Vo = Vmax[S]/(Km + [S])1/vo = (K/Vmax)(1/[S]) + 1/Vmax-0.02-0.010.010.020.030.040.050.060.071/[NAD*] (µM)

The graph that is peesnted is LB plot or Line weaverburk plot.This is a double reciprocal plot where…

Answered: The kinetics data shown in this graph…

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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82.21
(b) You are investigating the effects of several agents on the activity of alcohol dehydrogenase. The enzyme activity data are shown in the table below. Construct a [substrate] vs. activity plot and a double-reciprocal plot for this enzyme. Be sure to label all axes. Determine the Vmax and KM for AD from the graphs in each type of plot. AD activity (nM/min) AD activity + agent A (nM/min) AD activity + agent B (nM/min) [Alcohol] (nM) 0.1 14 2 0.5 50 7 8. 1.0 65 10 30 2.0 72 12 45 4.0 80 14 62 8.0 85 15 75 32.0 90 16 90
Calculate AG for this reaction under the following conditions: 37°C, pH 7, [Pyruvate] = [CO₂] = 4.0 mM, [OAA] = 2.0 mM, [ATP] = 3.5 mM, [P;] = 5.0 mM, and [ADP] = 1.8 mM. Use 2 signficant figures. AG= i kJ.mol-1 Under these conditions, the reaction is (Tolerance is +/- 2%)
If 287.9 umol of enzyme X has a Vmax = 47.8 mmol/sec, what is the value of kcat %3D sec-1? Please report answer with 1 decimal place. Please do not report units. Your Answer: Answer units MacBook Air 888 F5 F4 F3 F2 %23 %24 %24
You have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. ?maxVmax(μmol min−1) ?MKM(mM) Wild‑type 100 10 Mutant 1 0.1 Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1k−1 is much larger than ?2,k2, which of the following statements are correct? The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. ?0=V0=
Your current Vmax (4x10-3 A/S) from the Michaelis-Menten plot has the units of A/s and was determined using 0.1 mg of ADH in a total volume of 1.25 mL. Use the Beer-Lambert equation followed by simple conversion to express your estimated Vmax in (mol )/s per 1 mg ADH. The molar absorptivity of NADH at 340 nm is 6220 L/(mol*cm).
Refer to the figure shown here, and determine the value of E for the overall oxidation/reduction reaction (refer to the book/lecture slides if you need help with the overall reaction). 121/202 + 2H+ + 2e → H₂0 NAD + H* + 2e → NADH O-1.136 volts O 0.496 volts O+1.136 volts voltsm -0.496 volts EU (volts) +0.816 - 0.320
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.
(1) Enzy Matic, a Biochemistry student is studying the mode of inhibition of maleate to fumarase enzyme. Fumarase is the enzyme in the Kreb's cycle that converts fumarate to malate. He obtained the following kinetics data from her experiment: Velocity (mM/min) Inhibited 1.38 [S], mM Uninhibited 1.75 1.94 2.17 3.00 5.50 2.26 1.67 2.85 3.55 2.13 2.97 10.5 4.39 3.83
Intramitochondrial ATP concentrations are about 5 mM, and phosphate concentration is about 10 mM. Consider that ADP is five times more abundant than AMP. a. Calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. b. Calculate ∆G' for ATP hydrolysis under these conditions (∆G0' for ATP hydrolysis is -32.2 kJ/mol) The energy charge is defined as ( [ATP] + 1/2 [ADP] ) / ( [ATP] + [ADP] + [AMP] )
82.16
The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: Vo (micromol/min) Substrate added (mM) 200 1.0 400 2.0 800 4.0 1600 8.0 3200 16.0 The enzyme concentration used in all reactions is 106 M. Which of the following is true: O A) The Km for the reaction must be less than 1 mM ○ B) It is possible to determine Km for the reaction, but not Vmax or kcat OC) It is possible to determine only kcat, not Vmax or Km D) It is not possible to determine Vmax for the reaction OE) The ratio Vmax/Km for the reaction cannot be determined

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