# Big and Small Numbers in Biology

##### Stage: 4 Challenge Level:

QUESTION 1

Size of smallest virus: 20 nm

Size of largest bacterium: 0.75 mm (This bacterium is called Thiomargarita namibiensis and is so large that it is visible by the naked eye).

Reasoning

We assume that the virus is generally spherical, and that the bacterium is cylindrical, with diameter 0.1 mm.

Therefore, the volume of the smallest virus is $$V_1 = \frac{4}{3} \pi r^3 \approx 3.35 \cdot 10^4 \textrm{nm}^3$$

The volume of the largest bacterium is on the other hand $$V_2 = \pi r^2 h \approx 0.0236 \textrm{mm}^3 = 2.36 \cdot 10^{16} \textrm{nm}^3$$

So, in principle, we could fit

$$N = \frac{V_2}{V_1} = 7.04 \cdot 10^{11} \textrm{ viruses}$$

inside a Thiomargarita namibiensis.

QUESTION 2

A bacterium which is 200mm long is probably one of the longest bacteria known (since most species have length between 0.5 and 5 microns).

QUESTION 3

Average mass of a bacterium: $9.5\cdot 10^{-13} \textrm{g}$

Average volume of a bacterium: $0.6 \mu \textrm{m}^3$

Average density of soil: $1.602 \textrm{g} / \textrm{cm}^3$

Reasoning

2500 million bacteria weigh $m = 2500\cdot 10^6 \cdot 9.5\cdot 10^{-13}\textrm{g} = 2.375 \cdot 10^{-3} \textrm{g}$.

So, the percentage of bacteria contained in 1g of soil is

$$2.375 \cdot 10^{-3} = 0.2375 \textrm{%}$$

Now, 1g of soil has volume $\frac{1}{1.602} = 0.624 \textrm{cm}^3$. So, if the 2500 million bacteria are evenly spread in the soil, they each get a volume

$$V = \frac{0.624}{2500 \cdot 10^6} = 2.496 \cdot 10^{-10} \textrm{cm}^3 = 2.496 \cdot 10^2 \mu \textrm{m}^3$$

around them.

If we consider this volume around them as a sphere, the sphere's radius will be given by

$$r = \sqrt[3]{\frac{3V}{4\pi}}$$

and so, substituting for V as found above, we obtain $r = 3.906 \mu \textrm{m}$ as the distance between two neighbouring bacteria.

QUESTION 4

Recall that the surface area of a sphere is given by $S = 4\pi r^2$. Then, the surface area of the Earth is $$S = 4\pi \cdot (6\cdot10^{6})^2 \textrm{m}^2 \approx \ 4.53 \cdot 10^{14} \textrm {m}^2$$

Hence, if the humans were evenly spread out on it, each would have an area of

$$A = \frac{4.53 \cdot 10^{14} \textrm {m}^2}{6,673,031,923} \approx 6.78 \cdot 10^{4} \textrm{m}^2$$

around them.

QUESTION 5

Now, if we only restrict to the land, we have only 30% of the area found above. Hence, each human would have an area of

$$A = 0.3\cdot (6.78 \cdot 10^{4} \textrm{m}^2) = 2.03 \cdot 10^{4} \textrm{m}^2$$

around them.

QUESTION 6

A reasonable way to measure if humans or bacteria are more densely packed, is to evaluate how many bacteria "fit" into one's bacterium "personal space" and compare that to the corresponding fact for humans.

We found that a bacterium has a space $V_1 = 2.496 \cdot 10^2 \mu \textrm{m}$ around it, whereas its average volume is $V_2 = 0.6 \mu \textrm{m}^3$. Hence, the number of bacteria fitting inside is

$$N = \frac{V_1}{V_2} = 416$$

Now, we will perform a similar calculation for humans, noting that each needs approximately $0.5 \textrm{2}$ around them when standing. Then, the number of human's fitting inside one's "personal space" is

$$N = \frac{2.03 \cdot 10^4}{0.5} = 40600$$

Hence, the number of humans fitting inside the personal space is much higher than the corresponding fact for bacteria, we conclude that we humans are much less packed than bacteria in fertile soil.

QUESTION 7

Total fertile (arable and agricultural) land surface area: 62.6 million km $^2$

Reasoning

In total, we see that the volume of fertile soil is $$V = 62.6 \cdot 10^6 \cdot 10^6 \textrm{m}^2 \cdot 10^{-2} \textrm{m} = 62.6 \cdot 10^{10} \textrm{m}^3 = 62.6 \cdot 10^{19} \textrm{mm}^3$$

So, if we assume that we can find one bacterium in every $2.496 \cdot 10^2 \mu \textrm{m}$, we estimate that the total number of baceria in the fertile soil of the Earth is around

$$N = \frac{62.6 \cdot 10^{28}}{2.496 \cdot 10^2} = 2.5 \cdot 10^{27}$$

QUESTION 8

In general, we would expect the depth of fertile land to be much more than 10cm, since we would hope that this extends at least as much as a plant's roots do. This reasoning is correct, and indeed fertile soil does extend to the depth of the plants' roots (which can be up to several meters). However, most of the bacteria are contained in the topsoil, which is on average the first 10 cm below the ground. So our estimation about the depth is actually quite good.

QUESTION 9

Average blood content in an adult male: 5.2 litres

Reasoning

Since there are 5.2 litres = 5.2$\times$10$^6$ mm$^3$ of blood in the human body, we calculate that in the average adult male there are about

$$300,000 \times 5.2 \times 10^6 = 1.52 \times 10^{12} \textrm{ platelets}$$

QUESTION 10

Diameter of an erythrocyte or a lymphocyte: 7 $\mu \textrm{m}$ (they have roughly the same size)

Reasoning

First of all we need to make an assumption about the other dimensions of the sample. Given that a microscope slide is 7.5 cm long and 2.5 cm wide, it is fairly reasonable to assume that a circular drop of blood in the centre of the slide has radius about 1cm.

Therefore, the drop occupies a volume of

$$V = \pi r^2h=\pi \times 1\textrm{cm}^2 \times 2 \mu \textrm {m} = 2\pi \times 10^{-1} \textrm {mm}^3 \approx 0.63 \textrm{mm}^3$$

So, we expect that in this small sample there are about 2.5 - 4 million erythrocytes and 600 - 2000 lymphocytes (since we are given the density per cubic millimetre).

Each erythrocyte or lymphocyte should occupy an area of $$A_1 = \pi r^2 \approx 38.5 \mu \textrm{m}^2 = 38.5 \times 10^{-8} \textrm{cm}^2$$.

Thus, the about 3 million cells, should occupy an area of $A = 3 \times 10^6 \times A_1 = 1.15 \textrm{cm}^2$.

The area of the drop is, however relatively larger, at $A = \pi r^2 \approx 3.14 \textrm{cm}^2$, so the overlaps shouldn't be too many, as there is enough space around each cell.