Chapter 1
Life
1. (a) The number of cells in 10 L of saturated culture is :
10 L × 103 mL/L × 1010 cells/mL = 1014 cells.
2n = 1014 where n is the number of doublings.
n = 14/log 2 = 46.5
Since each doubling takes 20 min = 1/3 hour,
Time to reach a saturated culture is 46.5/3 = 15.5 hours.
(b) Volume of an E. coli cell = πρ2η = π × (1 × 10–6 m/2)2 × 2 × 10–6 m
= 1.57 × 10–18 m3
Volume of 1 km3 = (103 m)3 = 109 m3
Number of E. coli in 1 km3 = 109 m3/1.57 × 10–18 m3
= 6.37 × 1026 E. coli
,2 Chapter 1. Life
2n = 6.37 × 1026
n = log (6.37 × 1026)/log 2 = 89 doublings
Time to reach 1 km3 volume = 89/3 = 29.7 hours
2. See Figures 1-2 and 1-5. An animal cell possessing mitochondria, peroxisomes,
and cilia, in addition to a nucleus, has four lines of descent.
3. (a) For an E. coli cell:
Surface area of cylinder = 2πrh + 2πr 2
= 2π (1 × 10–6 m/2) × 2 × 10–6 m + 2π (1 × 10–6 m/2)2
= 7.85 × 10–12 m2
Volume = πr 2h = π (1 × 10–6 m/2)2 × 2 × 10–6 m = 1.57 × 10–18 m3
Surface-to-volume ratio = 7.85 × 10–12 m2/1.57 × 10–18 m3 = 5 × 106 m–1
For a eukaryotic cell:
Surface area = 4πr2
= 3 × 2/20 × 10–6 m = 3.0 × 105 m–1
Thus, the ratio of these two surface-to-volume ratios is
E. coli/eukaryotic = 5 × 106 m–1/3.0 × 105 m–1 = 17
Since cells must take in all nutrients through their surfaces, the E. coli cell can
absorb nutrients 17 times faster per unit volume. Thus, an E. coli cell can have a
17 times greater metabolism per unit volume than the eukaryotic cell, all else
being equal.
, Chapter 1. Life 3
(b) A single microvillus adds the volume
π × (0.1 × 10–6 m/2)2 × 1 × 10–6 m = 7.85 × 10–21 m3
and the surface area
π × 0.1 × 10–6 m × 1 × 10–6 m = 3.14 × 10–13 m2
to the brush border cell (the top of the cylinder is not added surface area since
the cell has this surface area without the microvilli). The area on the eukaryotic
cell that is covered with microvilli is
0.20 × 4π × (20 × 10–6/2)2 = 2.5 × 10–10 m2
There is one microvillus per (0.2 × 10–6 m)2 = 4 × 10–14 m2
Number of microvilli = 2.5 × 10–10 m2/4 × 10–14 m2 = 6250
= 4.19 × 10–15 m3 + 4.91 × 10–17 m3 = 4.23 × 10–15 m3
Area of the brush border cell
= 4π × (20 × 10–6 m/2)2 + 6250 × 3.14 × 10–13 m2
= 1.26 × 10–9 m2 + 1.96 × 10–9 m2 = 3.16 × 10–9 m2
Surface-to-volume ratio of cell with microvilli
= 3.16 × 10–9 m2/4.23 × 10–15 m3
= 7.47 × 105 m–1
Thus, the microvilli have increased the surface-to-volume ratio of the brush
border cell by a factor of 7.47 × 105/3.0 × 105 = 2.49.
4. (a) Volume of E. coli cell = π (1 × 10–6 m/2) × 2 × 10–6 m × 103 L·m–3 = 1.57 × 10–15 L
Number of moles of the protein in an E. coli
= 2 molecules/6.02 × 1023 molecules·mol–1 = 3.32 × 10–24 mol
Concentration of the protein = 3.32 × 10–24 mol /1.57 × 10–15 L = 2.11 × 10–9M
, 4 Chapter 1. Life
(b) 1 m M concentration contains 6.02 × 1023 × 10–3 molecules·L–1
= 6.02 × 1020 molecules·L–1
Number of molecules in an E. coli cell
= 1.57 × 10–15 L × 6.02 × 1020 molecules·L–1
= 9.45 × 105 molecules of glucose
5. (a) Volume of E. coli cell = π × (1 × 10–6/2)2 × 2 × 10–6 m3 = 1.57 × 10–18 m3
Volume of DNA
= π × (20 Å × 10–10 m·Å–1/2)2 × 1.6 mm × 10–3 m/mm = 5.03 × 10–21 m3
Fraction of volume of an E. coli cell occupied by DNA
= 5.03 × 10–21 m3/1.57 × 10–18 = 3.20 × 10–3
Volum e of hum an cell = 4 π × (20 x 10 /2) m 3 = 4.19 × 10
–6 3 –15
m3
(b) 3 ×
Volume of human DNA
= 700 × volume of E. coli DNA = 700 × 5.03 × 10–21 m3
= 3.52 × 10–18 m3
Fraction of volume of human cell occupied by DNA
= 3.52 × 10–18 m3/4.19 × 10–15 m3 = 8.40 × 10–4
6. Since it is likely that any life forms on the planet will be microscopic, they will
have to be detected by chemical means. Such life forms may not have
macromolecules that resemble those of terrestrial life forms but they will most
probably have some sort of a carbon based metabolism. Thus, if the life forms
were supplied with radioactively labeled nutrients, their ability to incorporate
the label into new compounds would indicate their existence. [Such experiments
were carried out on Mars by the Viking landers in 1976. The results were
negative. See Horowitz, H.N., The Search for Life on Mars, Sci. Am. 237(5): 52-61
(1977).]
7. With the earth so cold and dark, those eukaryotic species unable to withstand the
cold, particularly those from the tropics, would rapidly die out. As time passed,
surviving plants would be unable to photosynthesize so that they and the other
Life
1. (a) The number of cells in 10 L of saturated culture is :
10 L × 103 mL/L × 1010 cells/mL = 1014 cells.
2n = 1014 where n is the number of doublings.
n = 14/log 2 = 46.5
Since each doubling takes 20 min = 1/3 hour,
Time to reach a saturated culture is 46.5/3 = 15.5 hours.
(b) Volume of an E. coli cell = πρ2η = π × (1 × 10–6 m/2)2 × 2 × 10–6 m
= 1.57 × 10–18 m3
Volume of 1 km3 = (103 m)3 = 109 m3
Number of E. coli in 1 km3 = 109 m3/1.57 × 10–18 m3
= 6.37 × 1026 E. coli
,2 Chapter 1. Life
2n = 6.37 × 1026
n = log (6.37 × 1026)/log 2 = 89 doublings
Time to reach 1 km3 volume = 89/3 = 29.7 hours
2. See Figures 1-2 and 1-5. An animal cell possessing mitochondria, peroxisomes,
and cilia, in addition to a nucleus, has four lines of descent.
3. (a) For an E. coli cell:
Surface area of cylinder = 2πrh + 2πr 2
= 2π (1 × 10–6 m/2) × 2 × 10–6 m + 2π (1 × 10–6 m/2)2
= 7.85 × 10–12 m2
Volume = πr 2h = π (1 × 10–6 m/2)2 × 2 × 10–6 m = 1.57 × 10–18 m3
Surface-to-volume ratio = 7.85 × 10–12 m2/1.57 × 10–18 m3 = 5 × 106 m–1
For a eukaryotic cell:
Surface area = 4πr2
= 3 × 2/20 × 10–6 m = 3.0 × 105 m–1
Thus, the ratio of these two surface-to-volume ratios is
E. coli/eukaryotic = 5 × 106 m–1/3.0 × 105 m–1 = 17
Since cells must take in all nutrients through their surfaces, the E. coli cell can
absorb nutrients 17 times faster per unit volume. Thus, an E. coli cell can have a
17 times greater metabolism per unit volume than the eukaryotic cell, all else
being equal.
, Chapter 1. Life 3
(b) A single microvillus adds the volume
π × (0.1 × 10–6 m/2)2 × 1 × 10–6 m = 7.85 × 10–21 m3
and the surface area
π × 0.1 × 10–6 m × 1 × 10–6 m = 3.14 × 10–13 m2
to the brush border cell (the top of the cylinder is not added surface area since
the cell has this surface area without the microvilli). The area on the eukaryotic
cell that is covered with microvilli is
0.20 × 4π × (20 × 10–6/2)2 = 2.5 × 10–10 m2
There is one microvillus per (0.2 × 10–6 m)2 = 4 × 10–14 m2
Number of microvilli = 2.5 × 10–10 m2/4 × 10–14 m2 = 6250
= 4.19 × 10–15 m3 + 4.91 × 10–17 m3 = 4.23 × 10–15 m3
Area of the brush border cell
= 4π × (20 × 10–6 m/2)2 + 6250 × 3.14 × 10–13 m2
= 1.26 × 10–9 m2 + 1.96 × 10–9 m2 = 3.16 × 10–9 m2
Surface-to-volume ratio of cell with microvilli
= 3.16 × 10–9 m2/4.23 × 10–15 m3
= 7.47 × 105 m–1
Thus, the microvilli have increased the surface-to-volume ratio of the brush
border cell by a factor of 7.47 × 105/3.0 × 105 = 2.49.
4. (a) Volume of E. coli cell = π (1 × 10–6 m/2) × 2 × 10–6 m × 103 L·m–3 = 1.57 × 10–15 L
Number of moles of the protein in an E. coli
= 2 molecules/6.02 × 1023 molecules·mol–1 = 3.32 × 10–24 mol
Concentration of the protein = 3.32 × 10–24 mol /1.57 × 10–15 L = 2.11 × 10–9M
, 4 Chapter 1. Life
(b) 1 m M concentration contains 6.02 × 1023 × 10–3 molecules·L–1
= 6.02 × 1020 molecules·L–1
Number of molecules in an E. coli cell
= 1.57 × 10–15 L × 6.02 × 1020 molecules·L–1
= 9.45 × 105 molecules of glucose
5. (a) Volume of E. coli cell = π × (1 × 10–6/2)2 × 2 × 10–6 m3 = 1.57 × 10–18 m3
Volume of DNA
= π × (20 Å × 10–10 m·Å–1/2)2 × 1.6 mm × 10–3 m/mm = 5.03 × 10–21 m3
Fraction of volume of an E. coli cell occupied by DNA
= 5.03 × 10–21 m3/1.57 × 10–18 = 3.20 × 10–3
Volum e of hum an cell = 4 π × (20 x 10 /2) m 3 = 4.19 × 10
–6 3 –15
m3
(b) 3 ×
Volume of human DNA
= 700 × volume of E. coli DNA = 700 × 5.03 × 10–21 m3
= 3.52 × 10–18 m3
Fraction of volume of human cell occupied by DNA
= 3.52 × 10–18 m3/4.19 × 10–15 m3 = 8.40 × 10–4
6. Since it is likely that any life forms on the planet will be microscopic, they will
have to be detected by chemical means. Such life forms may not have
macromolecules that resemble those of terrestrial life forms but they will most
probably have some sort of a carbon based metabolism. Thus, if the life forms
were supplied with radioactively labeled nutrients, their ability to incorporate
the label into new compounds would indicate their existence. [Such experiments
were carried out on Mars by the Viking landers in 1976. The results were
negative. See Horowitz, H.N., The Search for Life on Mars, Sci. Am. 237(5): 52-61
(1977).]
7. With the earth so cold and dark, those eukaryotic species unable to withstand the
cold, particularly those from the tropics, would rapidly die out. As time passed,
surviving plants would be unable to photosynthesize so that they and the other
