Confluence cells added onto 100-wells plate, 2 ml trypsin added onto cells and 6 ml media added to dilute trypsin, 1 ml was used for counting, and average number of Cells counted was 24(DF 2) find out the the number of cells/ml and total number of cells in 7 ml? How many cells you need for 100 wells, assuming Number of cells/well is 10 to the power 4 cells?
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- Use correct sig figs The concentration of a purified monoclonal antibody was measured using UV280 nm. The sample was diluted (200 μL of purified antibody in 800 μL buffer) prior to analysis using a spectrophotometer. Calculate the concentration of antibody in the purified fraction if the Abs=0.95 of the diluted antibody. The molar absorptivity is known to be 191,411.6 M-1cm-1 and the molecular weight is 150 kDa. The pathlength for the cuvette is 1 cm.KCI for injection contains 40 mEq in 20 mL. An order requires 15 mEq, how many mL are needed?5. a) Thin layer chromatography (TLC) and high performance liquid chromatography (HPLC) can be used asexamples to explain difference between adsorption and partition. For each of these chromatographic techniques,identify:+ the stationary and mobile phase+ how the mobile phase movesb) How is the Rf value for a spot on a TLC plate calculated? What can the Rf value be used for? In order toachieve the best separation of several components of different polarity in a mixture spotted on a TLC plate,what guideline should one follow regarding the choice of a developing solvent? (i.e. polar, non-polar, etc.)
- Time point (min) Absorbance of culture at 660nm Approximate cell concentration Approximate # cells in 1mL extract 0 0.298 1.49 x 108 cells/mL 1.49 x 108 cells 10 0.316 1.58 x 108 cells/mL 1.58 x 108 cells 20 0.374 1.87 x 108 cells/mL 1.87 x 108 cells 30 0.429 2.145 x 108 cells/mL 2.145 x 108 cells 40 0.512 2.56 x 108 cells/mL 2.56 x 108 cells 50 0.544 2.72 x 108 cells/mL 2.72 x 108 cells 60 0.607 3.035 x 108 cells/mL 3.035 x 108 cells a. Using these data, prepare a growth curve of this strain ofEscherichia coli (E. coli).b. Estimate the doubling time for this strain of E. Coli. Clearly showhow you estimated this value from the empirical data presented.From this standard curve and chart below, does the separation of molecules in the mixture appear successful from the gel filtration? Is there a clearlydefined separation between molecules? Explain your conclusions. Parameters required for calculation of coefficient (Kd) for unknown protein Volume eluted (mL) Which variable does this volume represent in the equation for Kd? Fraction with maximal DNP-Aspartate detected 36 Vt Fraction with maximal Protein detected 24 Ve Fraction with maximal Blue dextran detected 6 VoWIDAL TEST Principle: The test depends on the ability of antibody in the patient’s serum to agglutinate the stained bacterial antigens. When this occurs, the aggregates become clearly visible to the naked eye. Task: Illustrate(draw) how the antigens aggregate in the Widal Test
- In precipitin ring testing: What is the purpose of incubating 0.9% saline over equine albumin antiserum in a second tube?Indole test what the result shows, what the result means, thank you!For the serial dilution, your stock solution must have a concentration of 3.5 mg/mL. How much diluent must be added to the 5.3 mg/mL red cell to prepare the stock solution? Show pertinent solution/s. What are the initial concentrations used for tubes 3, 5, and 6? Show pertinent solutions. What are the dilutions of the last positive tube and first negative tube respectively? Show computation.
- how did you get 104 as the dilution factor? I got 10-2 as the dilution factor for the CNFL plateIn an electrochemical cell, Q = 0.00025 and K = 0.70. What can you conclude about Ecell and E°cell? Ecell is positive and E°cell is positive Ecell is negative and E°cell is positive Ecell is negative and E°cell is negative Ecell is positive and E°cell is negativeApple puree was analyzed for petulin by HPLC-MS-MS after SPE clean-up. The procedure was 10.0g of puree + 10μl of a 10μg/ml solution of isotopically labeled petulin as internal standard were treated with 10.0ml of pectinase and acetic acid, centrifuged, and filtered. Four ml of the filtrate was passed through a SPE cartridge. The petulin was eluted with 2.0ml of ethyl acetate. The sample was evaporated to dryness and the residue dissolved in 1.0ml of the mobile phase. The analyte signal was 127 and the internal standard 197. Calculate the concentration of petulin in the sample in μg/g (RRF=1).