Exercise 6 Analysis

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online