The enzyme adenosine kinase (AdoK) is critical for psilocybin biosynthesis. Adokis a kinase that catalyzes the phosphorylation of adenosine (Ado) toAdenosine-5'-monophosphate (AMP) using ATP. You obtain the following results of an enzymeanalysis of Adok from the Mushroom P. cubensis:Total enzyme concentration (M) Substrate KM (M)AdoATPMa3535keat (sec ¹)1.4 x 10³1.4 x 10³33.511.6(a) Is the AdoKk enzyme a more efficient enzyme for ATP or Ado substrate? Explain(b) Does the enzyme approach the diffusion limit for either substrate? Explain(c) What is the Lineweaver Burk plot equation for the Adok enzyme for the substrate Ado([S] = [Ado])? (Write the Lineweaver-Burk equation)

Catalytic Efficiency Catalytic efficiency is a measure of how well an enzyme converts the substrate…

Answered: The enzyme adenosine kinase (AdoK) is…

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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Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM) 0.20 per minute 0.50 per minute 1.0 per minute 2.0 per minute 5.0 per minute
The following Michaelis-Menten plot shows the response of phosphofructokinase-1 (PFK-1), an enzyme in the glycolytic pathway, to the presence of ATP and AMP. Based on the data below, which of the following statements apply? Select all that apply. No inhibitors (low (ATP)) 1 mM ATP + 0.1 mM AMP 1 mM ATP 1.0 2.0 (Fructose-6-phosphate) mM Fig. : Regulation of PFK activity AMP is an allosteric modifier that enhances the binding of the substrate fructose 6-phosphate а. O b. AMP is an allosteric modifier that reduces the binding of the substrate fructose 6-phosphate ATP is an allosteric modifier that reduces the binding of the substrate fructose 6-phosphate O c. O d. ATP is an allosteric modifier that enhances the binding of the substrate fructose 6-phosphate Phosphofructokinase activity
The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.
(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
The KM for the reaction of chymotrypsin with Substrate A is 8.8 x 10-4 M, while the KM for the reaction of chymotrypsin with Substrate B is 8.7 x 10-3 M. Which of the following statements are likely true? Chymotrypsin has a higher apparent affinity for Substrate A. The Vmax would be higher in the presence of Substrate A. The kcat would be higher with Substrate A. The V max would be higher in the presence of Substrate B. Two of the above are true.
Table 1 shows the kinetic data that have been obtained for glucoamylase from Aspergillus niger at different temperatures in the production of glucose from maltodextrin. The rates of reaction measured are using an enzymes concentration of 0.003 μmolml-1. Outline and describe the reaction scheme of the production of glucose from maltodextrin by using glucoamylase.
In an enzymatic reaction, the equations that correspond with and without inhibition according to linewear-burk are calculated, these are with inh: 3.5x + 6 = y without inh: 3.5 + 10 = y. calculate alpha factor
Studies were carried out to investigate some kinetic properties of alkaline phosphatase (ALP) extracted from Lepus townsendii liver. Incubation of ALP extract with 4- nitrophenylphosphate (4-NPP) formed the basis for determination of enzyme activity. Spectrophotometric method was used to assay the enzyme activity, and the kinetic constants- maximum enzyme velocity (Vmax) and Michealis-Menten constant (Km) were evaluated. Figure 2 shows the Lineweaver-Burk plot of ALP assay in the presence of inhibitor NaH;PO4 (0) and the control (+) analysis. [I]: concentration of inhibitor = 0.67 mM. 1/V, uM*min x 104 40- 30 20 y-Intercept 10 x-intercept -1/K. (1+ [1]/K)/ 12 1/[S)x 102 (M) Figure2 (i) Estimate Km and Vmax in the absence and presence of the inhibitor. (ii) What is the type of inhibition illustrated by the inhibitor? Explain.
The change in the amount of product B in a reaction which is catalyzed by Enzyme A is given in the table below. Determine the Vmax and Km of the enzyme in the absence and presence of C. Is C an inhibitor or an activator of enzyme A? Determine the type of inhibition or activation. Rate of formation of Product B in the presence of 25 mg/mL C (mM/min) Substrate Rate of formation of (mM) Product B (mM/min) 0.5 21.5 14.1 1 30.2 22.2 1.5 34.9 27.0 2.5 39.8 33.1 3.5 42.0 36.3
Table 1 shows the kinetic data that have been obtained for glucoamylase from Aspergillus niger at different temperatures in the production of glucose from maltodextrin. The rates of reaction measured are using an enzymes concentration of 0.003 μmolml-¹. Table 1: Glucoamylase Activity at different temperatures Maltodextrin Concentration (μmol/L) 0.5 1 1.5 2.2 5.5 11 Initial Reaction Velocity (μmol/L.s) T = 40°C 2.1 3.5 4.9 5.6 589 8.3 9.3 T = 60°C 5.2 9.7 13 16 23 31
A marine microorganism contains an enzyme that hydrolyzes glucose-6-sulfate. The activity test is based on the measurement of the rate of glucose formation. The enzyme in a cell-free extract has a KM of 6.7 x 10-9 M and a Vmax of 300 nM min-1 . Galactose-6-sulfate is a competitive inhibitor of the enzyme. To a concentration of glucose-6-sulfate 2 x 10-5 M and galactose-6-sulfate 10-5 M, the speed initial is 1.5 nM min-1 . Calculate the Ki for galactose-6-sulfate.
One mg of enzyme has a Km of 5 x 10-5 M and a Vmax of 700 µmoles/liter/sec. The observed velocity in a cell extract containing one mg of enzyme and 2 x 10-4 M substrate was found to be 200 µmoles/liter/sec. This observation indicated that an inhibitor was present in this cell extract. True or False

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