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HW_6-25-10

# HW_6-25-10 - G&G Ch 1 Problem 3(pg 29-30 3a This question...

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G&G Ch. 1 Problem 3 (pg. 29-30) 3a. This question is designed to get you to marvel at the miniscule size of the cell. 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 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 =

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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 5 mg/mol)(1x10 -6 mol/mL) Number of molecules: (1x10 -6 mol/mL)(1x10 -12 mL)(6.023x10 23 molecules/mol) = 602,300 molecules 3e. It would help here to know that 1 Da = 1.66x10 -24 g (I use Da for Daltons instead of D). Convert everything to mg and mL. MW of one protein : 40kDa = 4x10 4 (1.66x10 -24 g) = 6.64x10 -17 mg Volume of E. coli cell (from 3b) : 1x10 -15 L = 1x10 -12 mL Concentration of one molecule in one cell: = 6.64x10 -17 mg 1x10 -12 mL 6.64x10 -5 mg/mL = 1 molecule 6.023x10 23 molecules/mol 1.66x10 -24 mol = 1.66x10 -24 mol 1x10 -15 L 1.66x10 -9 M = 1.7nM In mg/mL?
G&G Ch. 1 (cont’d) Problem 3 (cont’d) 3f. Convert everything to a common unit...I suggest m. Volume of E. coli cell (from 3b) : 1x10 -15 L = 1x10 -18 m 3 Volume of a sphere : V = = 0.063 (or 6.3%) Fraction of cell volume occupied by ribosomes: (15000)(4.2x10 -24 m 3 ) 3g. Molecular genetics…yummy. MW of E. coli DNA : 3x10 9 Da = 4π(1x10 -8 m) 3 = 4.2x10 -24 m 3 (1x10 -18 m 3 ) = (# ribosomes)(ribosome vol.) (cell vol.) 4πr 3 3 3 MW of average bp : 660Da Length imparted by average bp : 0.34nm = 3.4x10 -7 mm Total length of E. coli DNA: 3x10 9 Da 660Da/bp ( ) (3.4x10 -7 mm/bp) = 1.6mm Number of bp in E. coli genome: 3x10 9 Da 660Da/bp ( ) = 4.6x10 6 bp Hypothetical max. number of proteins: 4.6x10 6 bp (3bp/a.a.)(360a.a./protein) 4300 proteins = Radius of a ribosome : 10nm = 1x10 -8 m

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G&G Ch. 1 (cont’d) Problem 4 (pg. 30) 4a. Now we get even smaller. Again, convert to cm for later conversion to L. Length of mitochondrion : 1.5μm = 1.5x10 -4 cm 4b. Oxaloacetate…extraordinarily fascinating.
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