Maximum velocity of an enzyme-catalyzed reaction dolgomsH 7 T (8a. Is determined from the slope of the line of product formationzouza11 901 ml,0,.201over time for a particular substrate concentration. vx011(eb. Is determined from reaction rate late in the reaction dc. Is the rate where addition of more substrate does not cause a rateincrease.SO 121 91 to ydinitis of zomit 001 vainittsd. Changes with increases in substrate concentration(IIWhich of the following is not an assumption in Michaelis Mentenskinetics?91 21 noito926 v basylsts noitsbyri riguard(SEa. [ES]~ constantb. [E] >> [S] 00 to fu291 6 26 9022it ni[*] 92691T (EIc. [E] << [S]gnul art ni a.. Hq o bot Hq edt 1swol ind. [S] + [ES] [Sbaid atidinni slusslom 398-E 900Tantern anibnid yd 29tie gnibnid

For a one-substrate enzyme-catalyzed reaction, the Michaelis-Menton equation shows the quantitative…

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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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when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.
126 nmol of enzyme catalyzes the conversion of 2.8 ugram of substrate (molar mass- 350 g/mol) in t second at 37C at Vman what is the turnover number kcat for the enzyme in units of s R Show all your work to recenve credit. For the toolbar, press ALTF10 (P) or ALTFNF10 (Mac). %3B .... BIUS Paragraph Arial 14px
The following data were obtained in a study of an enzyme that is known to follow the Michaelis-Menten kinetics. Approximately, what is the value of Km tor this enzyme? Substrate added, umol/L Vo uM/sec 48 97 175 187 192 197 2 16 100 1000 198 OA 1 uM O8.2 pM OC 4 uM. OD. 8 uM OE 16 IM OF 1000 uM OG 97 uM/sec OH 175 UM/sec
1. A gluscose trasporter in teh liver cell binds with Km of 40mM. Calculate what fraction of vmax the Vo operating when blood glucose is 10mM 2. Draw a Michaelis-Menten kinetic curve and identify Vmax and Km. Indicate on the curve where the reaction is first order, where the reaction is zero order, where the reaction operates at half-saturation, and the range where most enzymes operate Be very clear about your labels.
41 The following data describe an enzyme-catalyzed reaction (hydrolysis of carbobenzoxyglycyl-L-tryptophan) Plot these results using the Lineweaver-Burk method, and determine values for KM and Vmax Velocity (mM.sec-) 0 024 0 036 0 053 0 060 0 061 0 062 Substrate Concentration (mM) 25 50 10 0 15 0 200 25 0 42 If the KM of an enzyme for its substrate remains constant as the concentration of the inhibitor increases, what can be said about the mode of inhibition and why? 43 Calculate the turnover number for an enzyme, assumıng Vmax IS 05 M sec1 and the concentration of the enzyme used is 0 002 M Why is it useful to know this? 44 Dıscuss the mechanism of the Bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues Make use of relevant graphs and diagrams to explain your answer
An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min−110 µmol⋅min−1 when the substrate concentration is five times greater than the ?M.KM. What is the ?maxVmax of this enzyme?
An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.
1). The initial velocity of an enzyme-catalyzed reaction has been determined at several substrate concentrations with and without an inhibitor I. The total enzyme concentration [E]ot present in each experiment is 1.20 x 10 M. [1] = 5.00 x10 M. (Note: v initial rate or velocity.) [S], M 1.25E-3 2.50E-3 5.00E-3 10.0E-3 v, M/min, no inhibitor 0.480E-3 0.660E-3 0.810E-3 0.920E-3 v, M/min, 5.00x10 M 9.43E-5 1.41E-4 1.70E-4 2.10E-4 b). What kind of inhibition occurred. c). Calculate the value of Ki = a). Use the Lineweaver-Burk plot to determine the values of Vmax and KM at [1] = 0.00 mM, and the values of Vmax and KM at 0.500 mM of I. it in 'How many molecules of
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
enzyme-inhibitor complex requires 450 kJ.mol-1 to dissociate and that it displays kinetics somehow similar to non-competitive inhibition, does this is inhibitor good to use to inhibit toxanthine oxidase in the case of hyperuricemia and gout?
The Lineweaver - Burk plot (Figure 1) shows an enzyme-catalyzed reaction in the absence and presence of 0.1µM inhibitor (ketoconazole). O Estimate Vmax and Km in the absence and presence of the inhibito.. (ii) Determine the type of inhibition shown by the inhibitor. Explain. 0.1- 0.08 0.06 - With inhibitor 0.04 0.02 Without inhibitor 0.4 -0.2 -0.02 0.2 0.4 0.6 0.8 1.0 1/{S\(&M-1) 1/vo (pmol-11 min)
1/x10exp-7 From the following enzyme-catalyzed kinetic data, estimate K, and V 'm 25 20 15 10 1.2 0.4 0.6 0.8 0.2 1/Sx10exp-3 10

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