missing part...(the concentrations are known ) of an experimental drug in physio
ID: 575578 • Letter: M
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missing part...(the concentrations are known ) of an experimental drug in physiological saline.... .
ete od SOCK Solutions (the concentrations are known you prepare will be used in assessing concent (0.07wv) to be used in developing a spectrophotometric assay for this drug. The standard curve your disposal a primary standard solution of the drug in sal w be used in assessing concentrations of the drug in samples having unknown concentrations. You have at this STD). The linear range of the assay for this drug a primary standard solution of the drug in saline having a concentration of 0.0300 g/mL (about 2.500 mL of e linear range of the assay for this drug is anticipated to be 0.120 mg/ml to approximately 3.00 mg/mL and u need to prepare SEVEN stocks by diluting the primary standard drug solution (STD) that fall within this range of concentrations so that you perform the assay using these stocks and develop a standard curve. Minimally, you will need to have (arter preparation of all of the stocks) enough of each solution you prepare to run five assay trials. For each assay, 0.125 mL of the appropriate sample is reacted with 0.875 ml of reagent to produce a colored product; the color of the reaction solution is proportional to the concentration of the drug in the solution. So, the absorbance values (dependent variable) obtained for these reaction mixtures by spectrophotometric analysis (at the appropriate waveleng or ngnt) when plotted versus concentration (independent variable) provides the required standard enter CONSIDERATIONS: So, let's think this through - in order to have enough of each standard solution to carry out five assay trials: 5 trials X 0.125 mL for each = 0.625 mL. So, we must have -0.750 ml of each stock solution left over after making all of the dilutions/stocks that are required. The preparation of -1.5 to 2 ml of each stock solution should be adequate for this purpose - that would provide sufficient volumes (0.75 - 1.25 mL) of each solution for making the dilutions (reserving the 750 UL of each stock solution for the 5 assays). The prepared stocks should have following concentrations (these fall within the linear range stated above): 0.125 mg/mL (stock G): 0.250 mg/ml (stock F): 0.500 mg/ml (stock E); 0.750 mg/mL stock D); 1.00 mg/mL (stock C): 1.50 mg/mL (stock B); and 3.00 mg/mL (stock A). Again, the stock concentrations span the linear range of the assay and vary effectively - which should result in a well-defined standard curve (plot of Absorbance vs Concentration (mg/mL). Please note that stocks C, E, F, and G vary geometrically (from CHE, EF, and FG the fold dilution is the same (2X)); and stocks A, B, and D vary geometrically, too (also by 2-fold). CALCULATIONS - Let's get started: From the 0.0300 g/mL STD drug solution - converting units = 30.0 mg / ml Now we have the same concentration units as stated for the stock solutions that will be prepared Prepare stock 'A' which must have a concentration of 3.00 mg/mL 30.0 mg/mL STD 73.00 mg/mL (desired concentration) = 10x dilution of the STD will make 'A' this means you need 1V of STD drug solution plus 9 V saline (diluent) (1V + 9V = 10x dilution) a o ano ml of A which = 2000 ut divided by 10X dilution = 200 uL of STD (1V): 1800 ul of saline (90)Explanation / Answer
Stock “A” has a concentration of 3.00 mg/mL of the drug and the desired concentration of the drug in stock “B” is 1.50 mg/mL.
The dilution factor is (3.00 mg/mL)/(1.50 mg/mL) = 2X; therefore, the stock “B” must contain 1V of stock “A” and 1V of the diluent to make a total dilution of 2X.
The total volume of stock “B” is 1.500 mL = (1.500 mL)*(1000 µL/1 mL) = 1500 µL.
Volume of stock “A” required = (1500 µL)/(2) = 750 µL and volume of diluent to be added = (1500 – 750)µL = 750 µL (ans).
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