If 10 µM enzyme was used to obtain the data in the summary plot below, V-max = uM sec-1 and Km = µM for this enzyme. (Enter numeric values to the nearest integer; Do NOT write unts.) 200 150 100 50 0 10 20 30 40 50 S-init (µM) V-init (µM sec-l)
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- The streptomycin in 500 g of a broth was determined by addition of 1.25 mg of the pure antibiotic containing 14C. The specific activity of this preparation was found to be 240 cpm/mg for a 30-min count. From the mixture, 0.112 mg of purified streptomycin was isolated which produced 675 counts in 60.0 min. Calculate the concentration in parts per million streptomycin in the sample.PBL 1: An enzyme with a Km = 1 X 102 M was assayed to convert the substrate with initial co ncentration of 3 X 10$M. Find the time required to convert 70% of substrate when Vmax = 2.167 M/min. Km In So 1 - (So – S) t = Vmax S' Vmax S = So - xS,From a kinetics experiment, the Vmax was determined to be 325µM*min^-1. for the kinetic assay, 0.1mL of a 0.25mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 75,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec^-1) for the enzyme
- You are trying to come up with a drug to inhibit the activity of an enzyme thought to have a role in liver disease. In the laboratory the enzyme was shown to have a Km of 1.0 x 10-6 M and Vmax of 0.1 micromoles/min.mg measured at room temperature. You developed an uncompetitive inhibitor. In the presence of 5.0 x 10-5 M inhibitor, the apparent Vmax was determined to be 0.02 micromoles/min.mg. What is the Ki of the inhibitor?The catalytic activity of an insect aminopeptidase was investigatedusing an artificial peptide substrate. The Vmax was 4.0 × 10-7 M/s and the KM was 1.4 × 10-4 M. The enzyme concentration used in the assay was 1.0 × 10-7 M. What is the value of kcat? 2.86 x 10-3 M 0.25 M/s 4.0 s-1 2.86 x 104 s/M 0.25 s-1Find the apparent Km and apparent Vmax for the enzyme reaction containing the inhibitor and identify the type of inhibitor The increment on the X axis is 25uM The increment on the Y axis is 8 sec uM We VISI O appvmax-0.05microM/sec; appKm=0.02microM: mixed O appvmax-0.05microM/sec; appKm=0.02microM; uncompetitive O appvmax-0.075microM/sec, appKm-0.013microM; competitive O appvmax-0.05microM/sec; appKm-0.013microM; uncompetitive O appvmax 75microM/sec; appKm-13.3microM; uncompetitive
- analyte concentration(C)(mg/ml) injection volume (ul) elution time (time) peak DAD signal(mAU) caffeine 1 1 4.67 302.85 aspartame 5 1 7.53 15.83 benzoic acid 1 1 8.14 89.98 saccharin 1 1 1.91 84.86 mixture(add everything above with 1:1:1:1 ratio) 1 4.47 69.58 How to get the concentration of the mixture in this case?what is the concentration of the unknown sample using serial diluitions: 0.05M 1.7472nm, 0.025M 0.8569nm, 0.0125M 0.4167nm, 0.00625M 0.1958nm the epsilon value is 0.0101 and l is 1 cmGiven the Ksp for Ag2CrO4 is equal to 1.20 x 10-12, calculate the percent error in your experimentally determined value for average Ksp (this will be an extremely large percentage – think about why).
- You are studying the absorbance properties of the COVID-19 virus. You notice at (4.9500x10^-1) absorbance, the molar absorptivity of the G variant is (4.7887x10^5) M1 cm in hexane at (5.3700x10^2) nm. The concentration was determined to be (3.60x10^-4) M. What is the pathlength of the cuvette in nanometers (nm)? Please do not include units in the answer and present your answer in scientific notation. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 AnswerChemistry An aquifer contaminated with petroleum is found to have the following component concentrations at a particular site: benzene158 ppm toluene124 ppm ethylbenzene91 ppm xylene45 ppm n-heptadecane161 ppm pristane 84 ppm Provide an estimate for the age of the spill at this site using (a) BTEX ratio and (b) nC17:Pr ratio. Show your calculations and use units throughout. Give proper s.f. for the answer.The image below shows a Lineweaver-Be plot for an Enzyme catalysed reaction. Units for the Y-axis is (s/mmol) and X-axis is (1/mM). Use the data on the plot to calculate Vmax and Km for this reaction. -0.2 0.20 0.15 0.10 0.05 0.00 0.0 ENZYME CATALYSED REACTION 0.2 0.4 O Vmax = -0.089 mmol/s; Km = 0.0145 mM O Vmax = 0.0145 s/mmol; Km = -0.089 1/mM Vmax = 68.97 mmol/s; Km = 11.24 mM O Vmax = 11.24 mM; Km = 68.97 mmol/s 0.6 y = 0.1624x + 0.0145 R² = 0.9967 0.8 1.0 1.2