Reactions 1: Y=0.2x + 0.8 Reactions 3: Y=0.2x + 0.5 1/Vmax = 0.8 1/Vmax = 0.5 Vmax= 1/0.8 mM/min Vmax= 1/0.5 mM/min Vmax = 1.25 mM/min Vmax = 2 mM/min Km = 0.2 x 1.25 Km = 0.25 Km = 0.2 x 2 Km = 0.4
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Match with the given reaction
A. The reaction with a non-competitive inhibitor
B. The reaction without inhibitor
C. The reaction with competitive inhibitor
D. The reaction with the fastest reaction
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- You have measured the following data for enzyme X. Substrate Vo concentration (micromolar (mM) Isec) 0.05 10 0.1 60 0.5 175 1 250 350 490 10 492 20 494 50 499 100 498 200 498Question: Calculate the Rate Ratio of Giardia among hikers that filter water compare to hikers that don't filter water. 1.00 0.23 0.33 0.11 E+ = filter E-= no filter D+ = Giardia D-= no Giardia Patient Number E+1 E+ 2 E+ 3 E+4 E-5 E-6 E-7 E-8 1 X 2 3 Day X 4 X X 5Answer the questions based on the data table. [s] mM 0.333 0.40 0.50 0.666 1.0 2.0 V (M/s) Uninhibited 1.65 x 107 1.86 x 107 2.13 x 107 2.49 x 107 2.99 x 107 3.72 x 107 V (M/s) x 107 Inhibitor A 1.05 x 107 1.21 x 107 1.43 x 107 1.74 x 107 2.22 x 107 3.08 x 107 Determine the Km and Vmax of the enzyme. Determine the type of inhibition imposed by inhibitor A. Determine the type of inhibition imposed by inhibitor B. V (M/s) x 107 Inhibitor B 0.794 x 107 0.893 x 107 1.02 x 107 1.19 x 107 1.43 x 107 1.79 x 107
- Would it be more difficult to meet a 70 ppb one hour averaged standard or 70 ppb 8-hr averaged standard and why?The biexponential equation for a two compartment model drug is C1,t = 50.0e-1.64t + 20.0e-0.147t. The drug is administered as a rapid IV bolus at t = 0. After 2.00 hours how many mg/l will the drug concentration in the central compartment be.11. Calculate KM and Vmax from the following data: [S] (µM) 0.1 0.2 0.4 0.8 1.6 vo (mM.s¯¹) 0.34 0.53 0.74 0.91 1.04
- 1. Calculate the MAWQ for an electronic balance with a linearity of 0.001 g and an acceptable error of 5% or 0.05.When 100 mg of a drug is administered as an IV bolus dose, the plasma concentrations (ug/mL) observed over time (hr) can be described by the following equation: C = 3.6e-0.105t C= 3-6e AUC 36 6 -0.105t 5. Calculate AUC(0-0) 0-105 = 34.28 6. What is the volume of distribution? Dose 100 VD= Co 3-6 7. Calculate the clearance of this drug. CI = KXVD CI = 0•105 x 27-8 %3D %3D Cl=2.9 8. What IV dose would be required to produce an initial plasma concentration of 10 ug/mL?Using the Michaelis-Mentan graph pictured (graphing initial velocity and PNPP concentration), identify Vmax and Km as well as possible. Then, using the Lineweaver-Burk graph (graphing 1/Vo and 1 / PNPP concentration), label the reciprocals of Vmax and Km on the graph and use them to calculate Km and Vmax.
- Determine the % inaccuracy by calculating the percent error of the micropipette at the set volume of 100 µL A. Standard deviation B. Covieffecrnt of varaiant C. % error (precision) Percent error = mean volume delivered -expected volume x 100 Expected volume RUN Weight Measurement (in g) 1 0.0985 2 0.0987 3 0.0985 4 0.0982 5 0.0984 6 0.0984 7 0.0981 8 0.0983 9 0.0982 10 0.0988 Total 0.9841 A. Standard deviation B. Covieffecrnt of varaiant C. % error (precision)What is the HVL of tissue for 1.25 MeV gamma-rays. How many half-layers of tissue are in the human forearm (assume an average thickness of 10 cm of tissue). What is the reduction in intensity as a 1.25 MeV gamma-ray passes through 10 cm of tissue? Assume the density of tissue is 1 g/cm2.Time (min) 0 15 30 45 60 75 Solution A Weight (8) 4.205 4.032 3.787 3.648 3.550 3.500 Change (%) Solution B Weight (8) 3.611 -4.11% 3.606 -6.07% 3.622 -3.67% 3.662 -2.69% 3.706 1.41% 3.732 Change (%) Solution с Weight (8) 4.257 -0.138% 4.213 0.444% 4.219 4.209 Change (%) 4.210 -1.03% Solution D Weight (8) 4.728 4.513 0.142% 4.350 1.10% 0.237% 4.254 1.20% 0.024% 4.068 0.675% 4.209 -0.024% 4.140 Change (%) -4.55% -3.61% -2.21% -4.37% 1.77% Graph your results below, plotting percent weight change versus time for all 4 unknown solutions on one set of axes, labeled clearly with units. Include a title for your graph.