Exercise 7 Analysis - Exercise 7 Analysis Giselle Zornberg...

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Exercise 7 Analysis Giselle Zornberg October 25, 2007 1. C 1 x V 1 = C 2 X V 2 C 1 = concentration of protein in the LE 1:20 dilution V 1 = 4µL = .004mL V 2 = 20µL = .02mL C 2 =0.2318 mg/mL C 1 (.004mL) = (.2318 mg/mL) (.02mL) C 1 = 1.159 mg/mL 2. C 1 x V 1 = C 2 X V 2 C 1 = concentration of protein in the Original Liver Crude Extract V 1 = 50µL = .05mL V 2 = 1000µL = 1mL C 2 =1.159 mg/mL C 1 (.05mL) = (1.159 mg/mL) (1mL) C 1 = 23.18 mg/mL 3. C 1 x V 1 = C 2 X V 2 C 1 = concentration of protein in the Original Tripe Crude Extract V 1 = 10µL = .01mL V 2 = 20µL = .02mL C 2 =0.1974 mg/mL C 1 (.01mL) = (0.1974 mg/mL) (.02mL) C 1 = 0.3948 mg/mL 4. Total protein in the Original Liver Crude Extract = (mg/mL of protein in the Original Liver Crude Extract)(Total volume of Original Liver Crude Extract) Total protein in Original Liver Crude Extract: (23.18 mg/mL)(6mL) Total protein in Original Liver Crude Extract: 139.08mg 5. Relative amount Original Liver Crude Extract = (total amount of protein in extract)/ (wet weight of tissue extracted) Total protein: 139.08mg Wet weight: 2.07g = 2070mg Relative amount of Original Liver Crude Extract: (139.08mg)/2070mg
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This lab report was uploaded on 03/20/2008 for the course BIO 205L taught by Professor Hanson during the Fall '07 term at University of Texas at Austin.

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Exercise 7 Analysis - Exercise 7 Analysis Giselle Zornberg...

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