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- 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…You have performed protein purification on your new favorite enzyme using a protocol which involves the following steps/samples: crude extract, ammonium sulfate cut, ion exchange and gel filtration. You need to run 25ug of protein from your crude extract sample on an SDS-PAGE gel. You have determined that the protein concentration of your crude extract sample is 2.1mg/ml. Your total sample volume is 30ul. You have water for your diluent and 6x SDS-loading dye to prepare your sample. List the components of your prepared sample. Note: you will only have access to a P20 and a P200 to prepare this sample.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.
- 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.200 ml of a 2% protein solution containing an enzyme that you want to purify. Half of the sample is subjected to method A, consisting of fractionated precipitations and 5 ml of final solution are obtained, with a concentration protein 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, with protein richness equal to 10 mg / ml and with an activity enzymatic also equal to 2000 U / ml. You want to know: a) Which of the two methods has provided the purest enzyme. b) By which of the methods the greatest amount of enzyme has been obtained.A 0.5 M phosphate buffer at pH 6.5 and a 10 mM substrate stock solution are available for preparing enzyme assays. Each assay must have a total volume of 3 mL and a final phosphate concentration of 100 mM is desired. Using the substrate concentration of 3.33 mM and 100 µL of the enzyme solution, calculate the volume of buffer, volume of substrate stock solution, and volume of water that should be place in a cuvette for the assay.
- 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 stepA 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).Using a detailed scheme, propose a step-wise protocol to purify protein B by ion exchange chromatography (Explain your logic/choices).
- Under most in vitro assay conditions, the enzyme is used in catalytic amounts (10 to 10" M). Estimate the concentration of an enzyme in a living cell. Assume that (a) fresh tissue is 80% water and all of it is intracellular, (b) the total soluble protein in a cell represents 15% of the wet weight, (c) all the soluble proteins are enzymes, (d) the average molecular weight of a protein is 150,000, and (e) about 1000 different enzymes are present.A colleague has successfully purified the enzyme sphingomyelinase from bovine brain. She provides you with the following incomplete purification table: Specific Activity (nmol/min) (nmol/min/mg) (%) Purification Factor (-fold) Total Total Purification Step Activity Yield Protein (mg) 100 1 Whole homogenate 40-60% (NH4);SO4 CM-sepharose ConA-sepharose Gel filtration Isoelectric focusing 88,625 5,685 925 544 51,700 32,067 29,500 31,680 1,855 420 5 0.03 (A) (B) (C) Per your colleague's request, you agree to calculate: (A) the final specific activity, (B) the final % yield, and (C) the final fold-purification. - x x x x *xx x xThe purification continues with a cation exchange step in which the positively charged cytochrome C protein is separated from negatively charged DNA and other proteins. The cation exchange eluate (volume of solution collected) had a total volume of 42.0 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 diluted 5-fold was 0.474 (1 cm pathlength). The absorbance at 595 nm from a 1.0 mL Bradford Assay solution that was diluted by 100-fold from the aliquot was 0.195 (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.