How to calculate the percent inhibition of xanthine oxidase enzyme inhibition with a blank absorbance value lower than the sample absorbance? Can the formula (Asample-A blank)/Asample ) x 100% be used? Thank you if you are willing to explain in detail Thank you.
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- The accompanying figure shows three Lineweaver–Burk plots for enzymereactions that have been carried out in the presence, or absence, of aninhibitor. Indicate what type of inhibition is predicted based on eachLineweaver–Burk plot. For each plot indicate which line corresponds to thereaction without inhibitor and which line corresponds to the reaction withinhibitor present.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 dilutionVmax [S] Km + [S] Vo = Eadie-Hofstee plot Lineweaver-Burk (L-B) plot v=Vm-Km [S] Km 1 Vm Vm [S] 1 The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For uncompetitive inhibition The Michaelis-Menten equation becomes Vo-Vmax(S)/(Km+a'[S)) Put the Minchaelis-Menten equation for uncompetitive inhibition in the Lineweaver-Burk format. The slope of this plot will be
- Please fully explain (use illustrate where appropriate) the Modes of Enzyme Catalysis exemplified by the serine protease: Chymotrypsin. In your answer discuss employing the illustration whenever possible: the overall reaction mechanism, stability of the reaction transition state, proximity and orientation effects, acid-base catalysis, and covalent catalysis.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?columns) and absence (second column = control) of enzyme inhibitor. Both inhibitors were added in each reaction at a concentration of 2 mM. The enzyme concentration was similar in all and was approximately 0.001 Им. Calculate both Vmax and KM for the control using Lineweaver-Burk curve. b. Provide the type of inhibition for both? Find, KI, for the inhibitor binding to the enzyme, for experiments (2) and (3). d. Calculate the reaction Kcat for the Control in experiment (1). Draw a velocity versus [S] showing Michaelis-Menten curve for the Control. Clearly show Vmax and Ky for the enzyme a. C. e. [S] (mM) (1) V. {(umol/(ml.s)] (2) V. [[umol/(ml.s)]| 4.4 (3) V. umol/{ml.s] 6.6 11.4 76 14.6 8.6 26.6 16.4 29 8 17.8 24.6 28.2 16 45.8 24 60 40.8
- 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?Glucosidase I catalyzes hydrolysis of specific glucosidase I is a synthetic trisaccharide, glucose-al-2- glucose-al-3-glucose-a-O(CH₂) #COOCH3. Kinetic measurements oligosaccharides containing glucose. obtained using this trisaccharide as substrate in the deoxynorjirimycin at concentrations of 50 μM (), 100 μM absence (x-x) and presence of the inhibitor 1- A) were used to prepare the (-), and 200 μM (4 Lineweaver-Burk plot below: b) Page 3 12) 7. a) V/V (nmol/hr)-1 1.S 1.0- 0.5 1/Trisaccharide (mM)-! Estimate the values for Vmax and KM for the trisaccharide substrate in the absence of the inhibitor. 0.0 -1.0 0.0 One substrate for 1.0 2.0 Determine whether inhibition by 1-deoxynorjirimycin is competitive, non-competitive or neither.The table below shows the purification of liver lactate dehydrogenase. Purification Table of liver LDH Step Crude Extract Ammonium Sulfate DEAE-Sephadex 5'AMP-Sepharose Volume (mL) 200 100 112 40 LDH activity (Units/mL) (mg/mL) [Protein] 47.0 50.00 90.0 63.00 51.0 1.50 137.0/ 0.4 Calculate the LDH yield for the ammonium sulfate step. Provide answer to one decimal only. Show Transcribed Text this is biochemistry and the topic is protein purification could you pleaae show me the full workings even the workings for the comversions
- 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 answerBased on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Calculate the Km of the reaction represented by Line (B).Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. Calculate the catalytic efficiency of the reaction represented by Line (A).