g&g_ch_1_p_3-4

g&g_ch_1_p_3-4 - G&G Ch. 1 Problem 3 (pg. 29-30)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 3 (pg. 29-30) 3a. Convert everything to m first. Length of E. coli cell : 2μm = 2x10 -6 m Pinhead diameter : 0.5mm = 5x10 -4 m How many cells fit across the diameter? Simple division: 5x10 -4 m 2x10 -6 m = 250 cells 3b. OK, we want volume (L). Change everything to cm…remember, 1cm 3 = 1mL Length of E. coli cell : 2μm = 2x10 -4 cm Diameter of E. coli cell : 0.8μm = 8x10 -5 cm Volume of a cylinder : V = πr 2 h V = π(4x10 -5 cm) 2 (2x10 -4 cm) = 1x10 -12 cm 3 = 1x10 -15 L 3c. Alright, now change everything to m…makes life a lot easier. Length of E. coli cell : 2μm = 2x10 -6 m Diameter of an E. coli cell : 0.8μm = 8x10 -7 m Surface area of a closed cylinder : SA=2πrh + 2 πr 2 SA=2π(4x10 -7 m)(2x10 -6 m) + 2π(4x10 -7 m) 2 = 6.03x10 -12 m 2 6.03x10 6 m -1 Surface to volume ratio? Ratio = SA V = 6.03x10 -12 m 2 1x10 -18 m 3 = r = 4x10 -5 cm r = 4x10 -7 m 1x10 -12 mL =
Background image of page 1

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

View Full DocumentRight Arrow Icon
Problem 3 (cont’d) 3d. First convert everything to mg and mL. Volume of
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/16/2008 for the course BIS 102 taught by Professor Hilt during the Summer '08 term at UC Davis.

Page1 / 4

g&g_ch_1_p_3-4 - G&G Ch. 1 Problem 3 (pg. 29-30)...

This preview shows document pages 1 - 3. Sign up to view the full document.

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