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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View Full DocumentThus 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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 Spring '08
 Makemson
 Biochemistry, pH

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