Ch1-3 ans

Ch1-3 ans - Chapter 1 Intro/Cells/Biomolecules 1 The size...

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Chapter 1 Intro/Cells/Biomolecules 1. The size of Cells and Their Components (Ans) a) The magnified cell would have a diameter of 50 x 10 4 μm = 500 mm, or 20 inches, which is about the diameter of a large pizza. b) The radius of a globular actin molecule is 3.6 nm/2 = 1.8 nm; the volume of the molecule, in cubic meters, is 4/3[3.14(1.8 x 10 9 m) 3 ] = 2.44 x 10 26 m 3 . The total number of actin molecules that could fit inside the cell is found by dividing the muscle cell volume by the actin molecule volume. Cell volume = 4/3[3.14(25 x 10 6 m) 3 ] = 6.5 x 10 14 m 3 . Thus the number of actin molecule in the hypothetical muscle cell is (6.5 x 10 14 m 3 ) (2.44 x 10 26 m 3 ) = 2.66 x 10 12 . 2.7 x 10 12 molecules This is approximately 2.7 trillion actin molecules. c) The radius of the mitochondrion is 1.5 μm/2 = 0.75 μm, therefore the volume of a spherical mitochondrion is 4/3[3.14(0.75 x 10 6 m) 3 ] = 1.77 x 10 18 m 3 . The number of mitochondrion in the hypothetical liver cell is (6.5 x 10 14 m 3 ) (1.77 x 10 18 m 3 ) = 36,723 . 37,000 mitochondria d) The volume of the eukaryotic cell is 6.5 x 10 14 m 3 as determined above, which is 6.5 x 10 8 cm 3 or 6.5 x 10 8 mL. One liter of a 1m M solution has (0.001 mol/1000mL) (6.02 x 10 23 molecules/mol) = 6.02 x 10 17 molecules/mL. The total number of glucose molecules is the product of the cell volume and glucose concentration: (6.5 x 10 8 mL)(6.02 x 10 17 molecules/mL) = 3.91 x 10 10 molecules . 3.9 x 10 10 molecules Thus there are approximately 39 billion glucose molecules in this cell. e) The concentration ratio of glucose/hexokinase is 0.001 M /0.00002 M , or 50/1, meaning that each enzyme molecule would have about 50 molecules of glucose available as substrate. 2. Components of E. coli (Ans) a) The volume of a single E. coli cell can be calculated from B r 2 h : B (0.4) 2 (2) = 3.14(4 x 10 5 cm) 2 (2 x 10 4 cm) = 1.0 x 10 12 cm) 3 = 1 x 10 15 L Density (g/L) multiplied by volume (L) gives the weight of a single cell: (1.1 x 10 3 g/L)(10 15 L) = 1.1 x 10 12 g . 1 pg b) The diameter of E. coli cell contained within the cell wall (i.e., excluding the thinkness of the cell wall) is 0.8 0.02 = 0.78 μm; the height is 2 0.02 = 1.98 μm. The volume of the E. coli cell that does not include the cell wall (using B r 2 h ) is 3.14[(7.8 x 10 5 cm) (2) cm] 2 (1.98 x 10 4 cm) = 9.5 x 10 13 cm 3

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Thus the E. coli volume within the cell wall occupies (9.5 x 10 13 ) (1 x 10 12 ) = 0.95 or 95% of the total volume. This means that the cell wall must account for 5% of the total volume for this bacterium. c) Ribosomal radius is 9 nm = 0.9 x 10 6 cm. Volume of a single ribosome is 4/3 B r 3 = 1.33(3.14)(0.9 x 10 6 cm) 3 = 3.0 x 10 18 cm 3 This volume, multiplied by the total number of ribosomes, gives the total volume occupied by ribosomes: 15,000(3.0 x 10 18 cm 3 ) = 4.6 x 10 14 cm 3 = 0.046 x 10 12 cm 3 Given a cell volume of 9.5 x 10
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Ch1-3 ans - Chapter 1 Intro/Cells/Biomolecules 1 The size...

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