SM(2001) T MEXEGiven, Iso s ri Glucose + Pi→ Glucose-6-P + H₂O K'eq = 3.9 x 10-3 M-1 utoe MATP + H₂O → ADP + PibsK'eq = 2 x 105 Mrink ats1014 Glucose +DADetermine K 'eq for a coupled reactionenimenteGlucose + ATP →q8)otsdGlucose-6-P + ADP 100K'eq?Ollom\xA12800.0=8).noitoso stericeoria-lf)undili

Consider a reaction ; A B. Its biochemical equilibrium constant (K'eq 1) will be;Now consider a…

Answered: Given, isas ri Glucose + Pi→…

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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2. The two diagrams to the right il- lustrate plots of steady-state ki- 5FA 0.8 netic studies to characterize the in- nH = 3.5 0.6- teraction of heart muscle phos- phofructokinase-1 with a non-phy- siological, synthetic substrate fruc- tose-6-sulfate. Because the kcat is 0.4- 0.2- smaller than that for the natural 10 μΜ 20 μΜ 48 μΜ substrate, higher enzyme concen- trations could be used. The results show the influence of increasing 0.2 0.4 concentrations of ATP on the initial -0.6- > velocity of the enzyme catalyzed reaction in the presence of no -0.8 4 12 20 28 36 44 52 60 68 76 84 92 1.2 2.0 2.4 AMP (•), 10 µM AMP (•), 20 µM AMP (-), and 48 µM AMP (). [ΑΤPΙ (μM) log[ATP] (µM) (а) how ATP interacts with the enzyme in the case of no AMP (•). Write the reaction in words catalyzed by the enzyme for the alternative substrate, describe (b) with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme. Explain the physical significance of the displacement of the…
1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…
The activity of the enzyme β-galactosidase produced bywild-type cells grown in media supplemented with different carbon sources is measured. In relative units, thefollowing levels of activity are found:Glucose Lactose Lactose + glucose0 100 1Predict the relative levels of β-galactosidase activity incells grown under similar conditions when the cells arelacI−, lacIS, lacO+, and crp−.
Lysozyme is a catalytic enzyme that cleaves NAG-NAM units in the bacte- rial cell wall, and is found in external secretions of the body. Why does this
What is the Keq of Reaction #3? Show your work & clearly label your answer. Is Reaction #3 favorable under mammalian cellular conditions when the concentration of dGTP is 0.2 M, dCMP is 20 mM, dGMP-dCMP is 7 mM, and PPi is 10 mM? Use △G to support your answer.
Begining with 1 M concentrations of each reactant and product at pH=7 and 25.0 degrees C, calculate the K'eq (to one decimal point) of the reaction Pyruvate + NADH+H+ <=> Lactate + NAD+.Note the temperature of this reaction will not affect the standard reducton potential delta E'o in the table 13-7b. please provide a comprehensive explanation with each step taken.
While furthering your studies of iocane powder you determine that a ceular enzyme, iocase will degrade it. Under normal conditions, iocase shows km 0f 1.85 uM and a kcat of 76. The drug sildenafil was found to have no effect on the km of iocanse yet the kcat was found to drop 58 minutes^-1, when the drug was added and no other changes were made in the reaction conditions. What can be inferred from this data? (remember kcat = vmax/et) Sildenafil acts as an uncompetitive inhibitor of iocanse Sildenafil acts as a noncompetitive inhibitor of iocanse The effect of sildenafil could be countered by adding more iocane to the reaction The y-intercept of the sildenafil inhibited iocanse would appear to be lower on a Lineweaver burke plot as compared to uninhibited iocanase The slope of a Lineweaver burke plot of the inhibited versus uninhibited iocanase would remain unchanged
Que:- An Michaels - Menten Kenetirs eneyme obeying the following parameters K2= 1•4 x10%M's" , K,= 2:5x1o°5' gave - 1 K3 = l-5 x10s ca) So this enzyme ctalytically pufet ? Explain your heasoning . (b) What is Vmaz , ik 1-4 nmol/me and saturabing eveyrme are usecd ? substrate
-Inhibitor +Inhibitor [S] (mM) V0&νβσπ;(μmol/sec). νο&νβσπ: &ν βσπ (μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Vmax of this enzyme WITH iinhibitor?
Can you explain the answe?
A solution of [U 14C] glucose-1-phosphate (specific activity = 16,000 cpm/mmole) was incubated with glycogen and glycogen phosphorylase, an enzyme which adds glucose units on to glycogen. Radioactivity was incorporated into the glycogen primer at a rate = 2550 cpm/min. The rate of the enzymatic reaction in units of mmole glucose incorporated per minute is: (a) 0.016 mmol/min (b) 0.57 mmol/min (c) 0.16 mmol/min (d) 5.7 mmol/min
Calculcate Kcat for PNP substrate for both enzyme concentrations. enzyme volume: 20 ul Bovine Intensince Alkaline phosphatase molecular weight: 140,000 Bovine intenstine Alkaline phosphatase activity: 300 units/ml and 14 units/mg extinction coefficient PNP: 18.5 abs (mM-1 cm-1) Vmax: 0.332 moles/sec a) enzyme 1 concentration: undiluted b) enzyme 2 concentration: 1:1 dilution

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Given, isas ri Glucose + Pi→ Glucose-6-P ATP + H₂O → ADP + Pi Determine K 'eq for a coupled reaction Tot 24 h Glucose + ATP 18) (OUT) I MEX + H₂O K'eq = 3.9 x 10-3M-1uto beriasen K'eq=2 x 105 Marbl 93A otsdge Glucose-6-P + ADP K'eq?