Physics Of Cell HW1

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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 6µm2 and a volume of 1µm3. Also, express this volume in femtoliters. Make a corresponding estimate of the mass of such a bacterium. (b) Roughly 2–3 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 world’s 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 µm in length and 0.5 µm 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 µm⇥ 2 µm ⇡ 6 µm2. (2.28)

Its volume is:

Vcell = ⇡ ⇥ ✓ 1

2 µm

◆2 ⇥ (2µm) ⇡ 1µm3 (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�18 m3 ⇡ 10�15 kg = 1 pg.

(b) The fact that each bacterium has a mass of 1 pg implies that 2�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 µm in diameter and has a spherical shape with the same density as that of water. Let’s assume that the mass of a “typical” human body is roughly 80 kg. Further, let’s 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 Vcell⇢H2O

= 1 3 ⇥ 80 kg

4/3⇡ ⇥ (5⇥ 10�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�18 m3 ⇥ 1028 = 1010 m3, which is equal to 10 km3.

HW 1

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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 radius r = 25 nm and side length s = 100 nm. Its surface area is then ⇡rs+⇡r2 ⇡ 1·104 nm2, and the number of CA proteins making up the capsid is then roughly 104

4⇡ ⇡ 800 CA proteins. This result can also be obtained by multiplying the ready-made estimate of 3500 total GAG proteins by the ratio of the surface areas

GAG proteins total· surface area capsid surface area virion

= 3500· 10 4 nm2

4⇡ · 60 nm2 ⇡ 800 CA proteins in capsid, (2.49)

where the virion radius used is 60 nm instead of 65 nm because the outer 5 nm of the virion shell are taken up by a lipid bilayer.

2.9 Areas and volumes of organelles

(a) Calculate the average volume and surface area of mitochondria in yeast based on the confocal microscopy image of Figure 2.18(C). (b) Estimate the area of the endoplasmic reticulum when it is in reticular form using a model for its structure of interpenetrating cylinders of diameter d ⇡ 10 nm separated by a distance a ⇡ 60 nm, as shown in Figure 2.25.

(a) The mitochondria in this yeast are shaped like a cylinder with a diameter of 400 nm approximately (which could be the resolution limit of the microscope). The total extension of this cylinder is about 20 µm. This results in a total mitochondrial volume of ⇡(0.2 µm)2 ⇥ 20 µm ⇡ 2.5 µm3. The total area is 2⇡ ⇥ 0.2 µm⇥ 20 µm ⇡ 25 µm2.

Mitochondria are thus just a small fraction of the total yeast volume, which is around 500 µm3. (b) Each “cross”, the unit that gets repeated in this structure, can be approxi- mated by two cylinders of length a and diameter d. Therefore, its surface area is 2⇥ ⇡ ⇥ 10 nm ⇥ 60 nm ⇡ 4000 nm2. Now, each one of this units occupies a volume a3 ⇡ 0.22⇥ 10�3 µm3.

We assume that a fibroblast has a height of approximately 1 µm and, based on figure 2.15, we approximate the area of the fibroblast were the ER is present to be 25 µm2, around one fourth of the total area of the field of view. Therefore, in this volume of 25 µm3 we can fit 25 µm3/(0.22⇥10�3 µm3) ⇡ 105 such units. This in turn corresponds to a total surface area of 105 ⇥ 4000 nm2 = 400 µm2.

 

 

Chapter 3

When: Stopwatches at Many Scales

3.1 Growth and the logistic equation

In the chapter, we described the logistic equation as a simple toy model for constrained growth of populations. In this problem, the goal is to work out the dynamics in more detail. (a) Rewrite the equation in dimensionless form and explain what units this means time is measured in. (b) Find the value of N at which the growth rate is maximized. (c) Find the maximum growth rate. (d) Use these results to make a one-dimensional phase portrait like that shown in Figure 3.10.

(a) Recall that the logistic equation can be written as

dN

dt = rN(1� N

K ). (3.16)

If we now define ⌧ = rt, this implies that

d

dt =

d

d⌧

d⌧

dt (3.17)

Now we can rewrite the original logistic equation as

dN

d⌧ = N(1� N

K ), (3.18)

with time measured in units of 1/r. (b) By di↵erentiating the right side with respect to N , we have

d

dN [N(1� N

K )] = (1� N

K )� N

K = 1� 2N

K = 0 (3.19)

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proteins by 50% (we assume ideal situation). However, in the case of degrada- tion, the amount of removed protein will depend on its concentration and it can vary from protein to protein.

3.7 The sugar budget in minimal medium

In rapidly dividing bacteria, the cell can divide in times as short as 1200 s. Make a careful estimate of the number of sugars (glucose) needed to provide the carbon for constructing the macromolecules of the cell during one cell cycle of a bacterium. Use this result to work out the number of carbon atoms that need to be taken into the cell each second to sustain this growth rate.

The number of sugars used was already worked out in problem 2.5. In this solution, we consider the implications for fast growing cells. How does all of this sugar get into the bacterium. If we consider fast growth, each cell cycle is approximately 20 minutes (or 1200 seconds), so the intake is:

109 sugars

1200 seconds = 8⇥ 105 sugar molecules/sec

There are approximately 1000 transmembrane proteins that transport sugar in the membrane of an E. coli, so the rate at which each protein must work at is:

8⇥ 105 sugars sec · 1000 proteins ⇡ 800

sugar molecules

transmembrane protein · sec

3.8 Metabolic rates

Assume that 1 kg of bacteria burn oxygen at a rate of 0.006mol/s. This oxygen enters the bacterium by di↵usion through its surface at a rate given by � = 4⇡DRc0, where D = 2µm2/ms is the di↵usion constant for oxygen in water, c0 = 0.2mol/m

3 is the oxygen concentration, and R is the radius of the typical bacterium, which we assume to be spherical. (a) Show that the amount of oxygen that di↵uses into the bacterium is greater than the amount used by the bacterium in metabolism. For simplicity, assume that the bacterium is a sphere. (b) What conditions does (a) impose on the radius R for the bacterial cell? Compare it with the size of E. coli.

(a) To estimate the amount of oxygen di↵using into a bacterium we assume that all the oxygen arriving at the cell membrane is absorbed. Furthermore, assuming that the bacterium is a sphere of radius R = 1µm (the typical size of an E.coli cell), the rate of oxygen entering the cell is

� = 4⇡DRc0 = 3⇥ 109 1/s . (3.68)

Given that bacteria burn oxygen at a rate of r = 0.2mole/kg s and that the