4) An enzyme catalyzes the reaction XY. The KM for the substrate X is 4 µM, and the Keat is 20 min-1. a) In an experiment, [X] = 6 mM, and vo = 480 nM min. What was [E] total used in the reaction? b) In another experiment, [E]total is 0.5 μM, and the measured vo is 5 μM min1. What was [X] used in the experiment? c) Calculate the catalytic efficiency of this hypothetical enzyme. Based on what you have seen in the textbook and lecture, how does this hypothetical enzyme compare to the best real enzymes?
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- 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).Enzyme X has a molecular weight of 48,000. It converts substrate Z into product Y. Z absorbs at 340 nm, and Y absorbs at 480 nm. A.) At what wavelength would you measure the change in absorbance to assay for enzyme X? Would the absorbance increase or decrease over time? B.)If Vmax = 60 μmol/min and you used 400 μL of a 0.1 mg/mL solution of enzyme, what is the turnover number?If Vmax for a reaction is 10 μM ⋅ s-1 and the KM is 0.5 μM, what is the reaction velocity when the substrate concentration is 2 μM?
- 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?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.If a data from an enzyme experiment is plotted as a Lineweaver - Burk plot, and the Vmax is 0.02 sec/mol, and x-intercept is -2.5 mM-1, then what is the KM value?
- 11) Using the data above, construct Lineweaver-Burk plots. First, graph the inverse of the reaction velocity (rate of reaction) data in column II versus the inverse of the methanol concentration in column I. Then, on the same graph, plot the inverse of the reaction velocity (rate of reaction) data in column III versus the inverse of the methanol concentration in column IThe KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 × 10−2 M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 × 10−4 M. (a) Which substrate has the higher apparent affi nity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax?For an enzyme kinetics experiment, a student prepared a reaction mixture by mixing 450 microliters of 0.75mM PNPP with 4.25ml of 0.2M Tris-HCl buffer. When he is ready to measure the absorbance, he added 0.3ml of Alkaline Phosphatase to the mixture and mixed thoroughly. What is the substrate concentration at the beginning of the reaction in mM ?
- As a reference model, a 7 L baffled stirred-tank reactor with a 3:1 (H:D) aspect was employed to scale down to a 7 mL tiny bioreactor.Dt = 0.16 m; Di = 0.07 m; N = 100 rpm are the dimensions of the huge tank.1)Using geometric similarities, determine the dimensions of the micro bioreactor (Dt, Di, and H).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.In 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