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g&g_ch_1_p_3-4

# g&g_ch_1_p_3-4 - G&G Ch 1 Problem 3(pg 29-30 3a Convert...

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G&G Ch. 1 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 =

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G&G Ch. 1 (cont’d) Problem 3 (cont’d) 3d. First convert everything to mg and mL. Volume of E. coli cell (from 3b) : 1x10 -15 L = 1x10 -12 mL MW of glucose (C 6 H 12 O 6 ): 180g/mol = 1.8x10 5 mg/mol = 0.18mg/mL Concentration of glucose : 1mM = 1x10 -6 mol/mL Convert the concentration: (1.8x10
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g&g_ch_1_p_3-4 - G&G Ch 1 Problem 3(pg 29-30 3a Convert...

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