RESULTS Table 1. Absorbance values of BSA standards. ● ● Test Tube No. BSA (μL) DDW (μL) 100 80 60 40 j Protein Concentration (mg/mL) 1 0 2 20 3 40 4 60 5 80 6 200 Please make graphs of the results and show your computations your reports Please show the slope and y-intercept; the equation used for the concentration of protein in the extract sample 20 A562 0.2 0.4 0.6 0.8 2 A562 A562 Duplicate AVERAGE ??? 0.099 0.451 ??? ??? 0.691 ??? 0.978 1.142 ??? 2.557 ??? 0.087 0.421 0.673 0.959 1.164 2.451
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- 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.50 mg/50 ml kinetin stock is available You need 2.5 mg/l kn for Ms media calculate amount of kn is required? Explain the role of surface sterilising chemicals in aseptic culture initiation and function of sucrose in MS medium?Example of a Protein Purification Scheme: Purification of the Enzyme Xanthine Dehydrogenase from a Fungus Volume Total Total Specific Percent Fraction (mL) Protein (mg) Activity Activity Recovery 1. Crude extract 2. Salt precipitate 3. Ion-exchange chromatography |4. Molecular-sieve chromatography 5. Immunoaffinity chromatography 3,800 22,800 2,460 0.108 100 165 2,800 1,190 0.425 48 65 100 720 7.2 29 40 14.5 23 1.8 275 152.108 11 Calculate the specific activity of step#4. Note that percent recovery=% Yield.
- Separation of Amino Acids by Thin Layer Chromatography Lab Questions 1. Describe in detail Thin layer chromatographic experiment. Example: the theory behind it, how youwould prepare the materials to spot on the plate with different mobile amino acids and unknown andhow TLC Plate is developed and the reasoning behind which solvent/ solvent mixture should be used,along how to correctly identify of the unknown. 2. Calculate the Rf value if a solute travelled 5 cm from the base spot and the solvent front is 10 cmfrom the origin? 3. In a TLC experiment using a 70:30 mixture of Petroleum ether and ethyl acetate, a student noted thedevelopment of spots in the origin, what can you suggest about this observation?10 If3 ml of culture is diluted by adding 9 ml of water and the absorbancy reading of the diluted culture is 0.082, what is the actual absorbancy?Now prepare a 500 ml media having ¼ strength MS with 5mM BAP, 5.5 mg/l Kn and coconut water 10%. Stocks are MS Macro 10X, micro 20X, FeEDTA 10X, Organic 100X, BAP and Kn 5gm/100ml. If you need anything else then you add accordingly. BAP has a MW of 225.3. Sucrose and Myo-inositol is not mentioned in the question. These you need to add. Consistency of the medium is also mentioned.
- Large proteins travel/migrate faster in both gel filtration chromatography and SDS- PAGE. O True O FalseSodium Dodecyl Sulfate-Polyacrylamide Gel Electrophoresis (SDS-PAGE)1. What are the functions of the sample buffer and sample reducing agent? Why do the samples need to be heated before you load them on the gel?2. What determines the current in your gel? What could cause the current in your gel unit to be lower than expected? Is there anything that could cause it to be higher than expected?3. Why do you need to wash the gel before staining it? Why use warm water?2Cemical analysis of a patient’s urine using a reagent strip gave negative protein and negative glucose results. However, using a separate microalbumin reagent strip, the result was positive. Earlier in the day, the urine controls gave the expected results with both types of reagent strips. 1. Is there a discrepancy in the patient’s test results obtained with the two strips? 2.Should the urine chemistry controls be rerun? 3.What is the most likely explanation for the results?
- REMOVAL OF WATER FROM MICROBIAL CELLS USING FREEZE DRYING METHOD. Before starting the freeze-drying process, the weight of the test tube, Wi which is 15.33g and test tube with microbial cell pellets, Ww which is 34.94g were weighed using an electronic balance. Plus, final weight of the beaker with test tube and cells is 70.35g. Then, the average amount of water removal and average cell yield were calculated by using the formulas given which were 9.81g and 0.5001. As a conclusion, there was 9.81g of water had been removed by using the technique freeze-drying for 24 hours by using a freeze dryer. Question: 1.From this statement, please explain what process actually has happened during this freeze-drying method. 2.Make a clear conclusion that can made from this experiment.(explain briefly)What's the amount of protein if the measurement at A280 = 1.120, dilution factor is 100 and total volume of extract is 45100ml of LB media with 25 μg/ml of Amp and 100 μg/ml of Kan final concentration. You have 100ml of LB provided and Amp and Kan stocks at 100 mg/ml and 50 mg/ml provided. Determine how much of each antibiotic stock solution you need to add to 100ml of LB to reach desired antibiotic concentration.