Exercise 6 Analysis

# Exercise 6 Analysis - Exercise 6 Analysis Giselle Zornberg...

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Exercise 6 Analysis Giselle Zornberg October 18, 2007 1. mg Chl/mL = {(.246) x (0.020)} + {(0.445) x (0.0080)} mg Chl/mL = 0.00848 2. Solve for C 1 if V 1 = 0.1 mL, C 2 = 0.00848, and V 2 = 10mL C 1 V 1 = C 2 V 2 C 1 = C 2 V 2 /V 1 C 1 = (10mL)(.00848)/(0.1mL) C 1 = 0.848 mg Chl/mL in the homogenizer containing isolated spinach chloroplasts 3. a. mg Chl a /mL = {(.445)(0.013)} – {(0.246)(.0027)} mg Chl a /mL = 0.0051208 b. mg Chl b /mL = {(.246)(.023)} – {(.445)(.0047)} mg Chl b /mL = 0.0035665 4. mg Chl b /mL = {mg total Chl/mL} – {mg Chl a /mL} mg Chl b /mL = 0.00848 - 0.0051208 mg Chl b /mL = 0.0033592 This result is the same as the result calculated in 3b to three significant figures. The differences in value may be due to approximations used by the empirical formulas when calculating chloroplast concentrations. In order to keep results accurate, calculations should be done using data with three significant figures. 5. [mg Chl

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Exercise 6 Analysis - Exercise 6 Analysis Giselle Zornberg...

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