Exercise 8 - Exercise 8: Analysis 1. C1 x V1 = C2 x V2 C1 x...

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Exercise 8: Analysis 1. C1 x V1 = C2 x V2 C1 x 10μL = 0.322 mg / mL x .02 mL C1 = .644 mg / mL The concentration of protein in the LE 1:20 dilution is: 0.644 mg/mL. 2. C1 x V1 = C2 x V2 C1 x 50μL = 0.644 mg / mL x 1mL C1 = 12.88 mg / mL The concentration of protein in the Original Liver Crude Extract is: 12.88 mg/mL. 3. C1 x V1 = C2 x V2 C1 x 4μL = .324 mg / mL x .02 mL C1 = .162 mg / mL The concentration of protein in the Original Tripe Crude Extract is: 0.162 mg / mL 4. Total protein in the Original Liver Crude Extract = (Protein in Original Liver Crude Extract) (Total volume of the Original Liver Crude Extract) Total protein = (12.88 mg / mL)(4 mL) = 51.52 mg or .05152 g 5. Relative amount of protein in the Original Liver Crude Extract = (Total amount of protein in the extract) / (Amount of wet weight of the tissue extracted) Relative amount = (.05152 g) / (4.0 g)
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This note was uploaded on 02/22/2010 for the course BIO 50100 taught by Professor Brand during the Fall '09 term at University of Texas at Austin.

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