Scl 3401
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This document outlines a group assignment for an introduction to statistics and data analysis course. Students will analyze data on 143 fish contaminated by toxic discharges from a chemical plant. The data includes the species, length, weight, and DDT level of each fish. Students must calculate measures of central tendency and variability for DDT levels and weights. They must also construct graphical displays like stem-and-leaf plots, dot plots, boxplots, relative frequency distributions, and histograms to analyze and compare the data.
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2.3 A feeling for the numbers: microbes as the unseen majority
(a) Use Figure 2.1 to justify the assumption that a typical bacterial cell (that
is, E. coli) has a surface area of 6袖m2 and a volume of 1袖m3. Also, express
this volume in femtoliters. Make a corresponding estimate of the mass of such
a bacterium.
(b) Roughly 23 kg of bacteria are harbored in your large intestine. Make an
estimate of the total number of bacteria inhabiting your intestine. Estimate
the total number of human cells in your body and compare the two figures.
(c) The claim is made (see Whitman et al., 1998) that in the top 200m of
the worlds oceans, there are roughly 1028 prokaryotes. Work out the total
volume taken up by these cells in m3 and km3. Compute their mean spacing.
How many such cells are there per milliliter of ocean water?
(a) E. coli has (roughly) the shape of a cylinder that is 2 袖m in length and 0.5 袖m
in radius. For those that are so inclined, the bacterium can alternatively be
treated as a spherocylinder, though the results will not change in any interesting
way. Using these numbers we calculate the area of an E. coli to be:
Acell = 1 袖m 2 袖m 6 袖m2. (2.28)
Its volume is:
Vcell =
1
2
袖m
2
(2袖m) 1袖m3 (2.29)
= 1 fL. (2.30)
If we assume that the density of a bacterium is the same as that of water, the
mass of one bacterium is 103kg/m3 10鐃18 m3 10鐃15 kg = 1 pg.
(b) The fact that each bacterium has a mass of 1 pg implies that 2鐃3 kg worth
of bacteria in the intestines of one person amounts to 2 3 1015 bacteria.
Assume that the size of a typical human cell is roughly 10 袖m in diameter and
has a spherical shape with the same density as that of water. Lets assume that
the mass of a typical human body is roughly 80 kg. Further, lets assume
that thirty percent of the human mass corresponds to cells. On the basis of
these assumptions, we find that the number of the cells in a human body is
approximately
Mhuman
VcellH2O
=
1
3 80 kg
4/3 (5 10鐃6 m)3 1000 kg/m3
' 5 1013. (2.31)
By this estimate, the number of bacterial cells outnumbers the number of human
cells by more than a factor of ten.
(c) Using (a) we can estimate the volume of 1028 prokaryotes to be about
10鐃18 m3 1028 = 1010 m3, which is equal to 10 km3.
HW 1
27
section of this cylinder with the same cross-sectional area. The schematic of
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can be used up to form the capsid.
From the micrographs, the capsid can be approximated as a cone with base
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2.3 A feeling for the numbers: microbes as the unseen majority
(a) Use Figure 2.1 to justify the assumption that a typical bacterial cell (that
is, E. coli) has a surface area of 6袖m2 and a volume of 1袖m3. Also, express
this volume in femtoliters. Make a corresponding estimate of the mass of such
a bacterium.
(b) Roughly 23 kg of bacteria are harbored in your large intestine. Make an
estimate of the total number of bacteria inhabiting your intestine. Estimate
the total number of human cells in your body and compare the two figures.
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the worlds oceans, there are roughly 1028 prokaryotes. Work out the total
volume taken up by these cells in m3 and km3. Compute their mean spacing.
How many such cells are there per milliliter of ocean water?
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in radius. For those that are so inclined, the bacterium can alternatively be
treated as a spherocylinder, though the results will not change in any interesting
way. Using these numbers we calculate the area of an E. coli to be:
Acell = 1 袖m 2 袖m 6 袖m2. (2.28)
Its volume is:
Vcell =
1
2
袖m
2
(2袖m) 1袖m3 (2.29)
= 1 fL. (2.30)
If we assume that the density of a bacterium is the same as that of water, the
mass of one bacterium is 103kg/m3 10鐃18 m3 10鐃15 kg = 1 pg.
(b) The fact that each bacterium has a mass of 1 pg implies that 2鐃3 kg worth
of bacteria in the intestines of one person amounts to 2 3 1015 bacteria.
Assume that the size of a typical human cell is roughly 10 袖m in diameter and
has a spherical shape with the same density as that of water. Lets assume that
the mass of a typical human body is roughly 80 kg. Further, lets assume
that thirty percent of the human mass corresponds to cells. On the basis of
these assumptions, we find that the number of the cells in a human body is
approximately
Mhuman
VcellH2O
=
1
3 80 kg
4/3 (5 10鐃6 m)3 1000 kg/m3
' 5 1013. (2.31)
By this estimate, the number of bacterial cells outnumbers the number of human
cells by more than a factor of ten.
(c) Using (a) we can estimate the volume of 1028 prokaryotes to be about
10鐃18 m3 1028 = 1010 m3, which is equal to 10 km3.
HW 1
27
section of this cylinder with the same cross-sectional area. The schematic of
the mature virion in fig. 2.31(C) shows that the CA proteins come together to
form the capsid with inward pointing spokes, in the same way that the GAG
polyproteins form the initial outer shell of the virion. This means that CA and
GAG have the same surface areas per protein. Because the surface area of the
capsid is less than that of the virion and because each CA protein is cleaved o
a GAG polyprotein, this fact immediately implies that not all the CA proteins
can be used up to form the capsid.
From the micrographs, the capsid can be approximated as a cone with base
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