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- Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?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 answerWhat is the concentration of enzyme (in mM) needed to achieve a Vmax of 8.00 mM/s if the enzyme has a turnover number of 1.50s-1?
- Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?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 turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?
- What is the likely keq for an enzyme that has a deltaG =-18.3kj/mol?During the early stages of an enzyme purification protocol, when cells have been lysed but cytosolic components have not been separated, the reaction velocity-versus-substrate concentration is sigmoidal. As you continue to purify the enzyme, the curve shifts to the right. Explain your results. This is an allosteric enzyme and you must use a Lineweaver-Burk plot to determine KM and Vmax correctly. This is an enzyme that displays Michaelis-Menten kinetics and you purify away an inhibitor. This is an allosteric enzyme and during purification you purify away an activator. This is an allosteric enzyme displaying a double-displacement mechanism and during purification you purify away one of the substrates: This is an enzyme that displays Michaelis-Menten kinetics, and you must use a Lineweaver-Burk plot to determine KM and Vmax correctly.7.8 The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by turnover number=?max[E]t=?2(?max is often written as ?max)turnover number=Rmax[E]t=k2(Rmax is often written as Vmax) If the enzyme molecule has more than one active site, then [E]t[E]t is multiplied by the number of active sites to determine its effective concentration. Determine the value of the turnover number of the enzyme carbonic anhydrase, given that ?maxRmax for carbonic anhydrase equals 249 μmol⋅L−1⋅s−1249 μmol⋅L−1⋅s−1 and [E]t=2.16 nmol⋅L−1[E]t=2.16 nmol⋅L−1 . Carbonic anhydrase has a single active site. turnover number =
- From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.SAMSUNG NAL (1308 A 01-2021-FSCBIO124-Biology1-A X ive.google.com/file/d/1mawGqo2tZnOIFPc7w0Gpr1VyRRMpNInM/view QUESTION 2 The Michaelis-Menten and the Lineweaver-Burk plots can be used to determine the Km and Vmax of the enzyme-catalysed reaction. Explain the significance of Km value in enzyme catalysed reaction. By using succinate dehydrogenase as the enzyme, sketch a Lineweaver-Burk plot that shows an enzyme catalysed with and without competitive inhibitor. Explain what will happen if malonate is added to the reaction. Relate the Lineweaver-Burk plot to the effect of malonate on Km and Vmax values. Using the knowledge from cell biology, explain how the enzyme is being synthesis. rch 81°F A ENG F4In your acid phosphatase enzyme kinetics lab, you constructed a Lineweaver-Burk plot. Lefs assume that you graphicaly obtain 1/Vmax - 9.33 (umoliey and -1Km--0.012 M. Which of the following is correct? A) Vmax = 9.33 (no units) B) Km = 0.012 (no units) C) Vmax = (1/Vru" = 0.107 umolis D) Km =-14-1K = 83.3 pM E) Both C) and D) are correct