EXERCISE 6 ANALYSIS

# EXERCISE 6 ANALYSIS - Bio 205L EXERCISE 6 ANALYSIS 1 Chlorophyll concentration in 80 acetone solution mg Chl mL =(A645)x(0.020(A663)x(0.0080

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Bio 205L EXERCISE 6 ANALYSIS 1. Chlorophyll concentration in 80% acetone solution mg Chl / mL = {(A 645 )x(0.020)} + {(A 663 )x(0.0080)} = (0.360 abs x 0.020) + (0.640 abs x 0.0080) = (0.0072) + (0.00512) = 0.01232 mg Chl / mL 2. Chlorophyll concentration in homogenizer containing isolated spinach chloroplasts C1 x V1 = C2 x V2 C1 x 0.1 mL = 0.01232 mg ChL/mL x 10mL C1 = 1.232 mg Chl / mL 3. (a) Mg Chl a / mL = {(A 663 ) x (.013)} - {(A 645 ) x (.0027)} = (0.640 x .013) – (0.360 x .0027) = (0.00832) – (0.000972) = 0.007348 mg Chl a / mL (b) Mg Chl b / mL = {(A 645 ) x (.023)} - {(A 663 ) x (.0047)} = (0.369 x .023) – (0.640 x .0047) = (0.008487) – (0.003008) = 0.005479 mg Chl b / mL 4. Mg Chl b/ mL = {mg total Chl / mL} – {mg Chl a/mL} = (0.01232 mg Chl / mL) – (0.007348 mg Chl a / mL) = 0.004972 mg Chl b / mL The two values for mg Chl b / mL only differ by only 0.000507 mg Chl b / mL. This was probably the result of experimental error in measuring the wavelengths of the chlorophyll. 5.

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## This note was uploaded on 06/15/2008 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 6 ANALYSIS - Bio 205L EXERCISE 6 ANALYSIS 1 Chlorophyll concentration in 80 acetone solution mg Chl mL =(A645)x(0.020(A663)x(0.0080

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