The rate of an enzyme catalysed reaction was measured with increasing substrate concentrations in the presence of 0 µM, 2 μM or5 μM of a non-reactive substrate analogue capable of binding to the active site. The following double reciprocal Lineweaver-Burkplot was produced from the data obtained. Match each of the following terms with the correct label (A-G) on the plotPrompts1/[5](B)(A)(c)(D)(E)(G)(F)AnswersSelect match2Select matchReaction with 2 μM of the substrateanalogue3-1/KMLast saved 10:26:12Questions Filter (16)Q SearchSelect matchKFirst Last >|公告MEDEThi10BE OF £ & 2 2 2 211.YouHaQuqubasheWOpe

Therefore:"1/S" = (F)"Reaction with 2 µM of the substrate analogue" = (B)"-1/Km" = (G)This makes…

Answered: The rate of an enzyme catalysed…

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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The purification of cytochrome C begins with 1) yeast homogenization using a bead beater in the presence of BME (a reducing agent) and a protease inhibitor from approximately 900 grams of Baker’s yeast. Then, 2) insoluble cell contents were removed by centrifugation at 4,000 x g for approximately ten minutes. The ruptured cells (lysate) after centrifugation had a total volume of 0 mL and a 1.0 mL aliquot was set aside for further analysis. The following data was obtained from the 1.0 mL aliquot to quantify the protein amount and purity: The absorbance at 410 nm of the aliquot was 0.460 (1 cm pathlength). The absorbance at 595 nm from a 1.0 mL Bradford Assay solution that was diluted by 250-fold from the aliquot was 0.681 (1 cm pathlength). Using the information given, Calculate the total protein amount in mg from the absorbance at 595 nm. Calculate the cytochrome C amount in mg from the absorbance at 410 nm using Beer’s Law.
The kcat and KM for chymotrypsin-catalysed cleavage of a synthetic substrate, S, were determined to be 60 s-1 and 0.5 mM, respectively, using steady-state kinetics. The concentration of enzyme in the assay was 5 x 10-8 M. ii) Draw a graph on the axes shown in Figure 1 below showing how the initial rate of the reaction varies with [S], making use of the values for KM and Vmax. Label the axes with appropriate titles and units.
The enzyme lysozyme kills certain bacteria by attacking a sugar called N-acetylglucosamine (NAG) in their cell walls. At an enzyme concentration of 2 × 10-6 M, the maximum rate for substrate (NAG) reaction, found at high substrate concentration, is 1 × 10-6 mol L¯'s1. The rate is reduced by a factor of 2 when the substrate concentration is reduced to 6 x 10-6 M. Determine the Michaelis-Menten constants Km and k, for lysozyme.
An enzyme catalyzes the reaction M ßàN. The enzyme is present at a concentration of 1 nM, and the Vmax is 2 M s-1. The Km for substrate M is 4 μM. a)Calculate kcat. b)What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an α’ of 2.0?
Assume that the experiments performed in the absence of inhibitors were conducted by adding 5 μL of a 2 mg/mL enzyme stock solution to an assay mixture with a total volume of 1 mL. Take into account that XYZase is a monomeric enzyme with a molecular mass of 45,000 Daltons. Hint: To calculate the ???? in units of per second (s−1), you must first determine the ???? in micromoles per second (μmol/sec). Please explain step by step
A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).
Could you help with Part A and B of the question attached
a) Determine kcat (in units of sec-1) for a particular enzyme, given the following information: Vo = 144 mmol/min; [S] = 2 mM; Km = 0.5 mM; Enzyme Molecular weight = 40,000 mg/mmole; 8 mg of enzyme used in assay generating this data. b) In general, explain how the total enzyme concentration affects turnover number and Vmax?
A Lineweaver-Burk plot generates a line with the following formula: y = 0.3x + 0.4. What is the Km of this enzyme?
200 ml of a 2% protein solution are available, containing an enzyme to be purified. Half of the sample is subjected to method A, consisting of fractionated precipitations, and 5 ml of final solution are obtained, with a protein concentration equal to 5 mg / ml and enzymatic activity equal to 2000 U / ml. The other half is subjected to method B, consisting of ion exchange chromatography, and a final solution of 10 ml is obtained, with a protein richness equal to 10 mg / ml, and with an enzymatic activity equal to 2000 U / ml. You want to know: a) Which of the methods has provided the purest enzyme. b) By which of the methods has the greatest amount of protein been obtained.
Shown below is a proposed mechanism for the cleavage of sialic acid by the viral enzyme neuraminidase. The kcat for the wild-type enzyme at pH =6.15, 37 °C is 26.8 s-1.(a) Describe the roles of the following amino acids in the catalytic mechanism: Glu117, Tyr409, and Asp149. List all of the following that apply:general acid/base catalysis (GABC), covalent catalysis, electrostaticstabilization of transition state.(b) Based on the information shown in the scheme, would you expect mutation of Glu 117 to Ala to have a greater effect on KM or kcat?(c) For the R374N mutant at pH = 6.15, 37 °C, kcat is 0.020 s-1, and KMis relatively unaffected. Based on this result, it seems that R374 is morecritical for catalysis than for substrate binding. Explain how R374 stabilizesthe reaction transition state more than the substrate (i.e., what feature of this reaction would explain tighter binding to the transition state vs. substrate?).
Following is a Dixon plot for PNPP inhibition by inorganic phosphates. What answer choice shows the correct combination of the mode of inhibition and Ki of the EI complex? - Mixed inhibition, Ki= -21.2 uM - Mixed inhibition, Ki= 1/16.5 uM - Competitive inhibition, Ki= 16.5 uM - Competitive inhibition , Ki= -1/16.5 uM - None of the above

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