Components of E . coli E . coli cells are rod-shaped, about...

Components of $E .$ coli $E .$ coli cells are rod-shaped, about $2 \mu \mathrm{m}$ long and $0.8 \mu \mathrm{m}$ in diameter. The volume of a cylinder is $\pi r^{2} h$, where $h$ is the height of the cylinder. (a) If the average density of $E .$ coli (mostly water) is $1.1 \times 10^{3} \mathrm{g} / \mathrm{L},$ what is the mass of a single cell? (b) $E$. coli has a protective cell envelope $10 \mathrm{nm}$ thick. What percentage of the total volume of the bacterium does the cell envelope occupy? (c) $E$ coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical ribosomes (diameter 18 $\mathrm{nm}$ ), which carry out protein synthesis. What percentage of the cell volume do the ribosomes occupy?

Components of $E .$ coli $E .$ coli cells are rod-shaped, about $2 \mu \mathrm{m}$ long and $0.8 \mu \mathrm{m}$ in diameter. The volume of a cylinder is $\pi r^{2} h$, where $h$ is the height of the cylinder.
(a) If the average density of $E .$ coli (mostly water) is $1.1 \times 10^{3} \mathrm{g} / \mathrm{L},$ what is the mass of a single cell?
(b) $E$. coli has a protective cell envelope $10 \mathrm{nm}$ thick. What percentage of the total volume of the bacterium does the cell envelope occupy?
(c) $E$ coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical ribosomes (diameter 18 $\mathrm{nm}$ ), which carry out protein synthesis. What percentage of the cell volume do the ribosomes occupy?

Key Concepts

Spherical Volume Calculation

This concept uses the formula for the volume of a sphere, V = (4/3)?r³, to calculate the space occupied by spherical particles such as ribosomes. It is essential for determining the proportion of cellular volume taken up by spherical organelles or sub-structures and is key in many areas of biophysics and cell biology.

Shell Volume Calculation

This concept pertains to the calculation of the volume of a thin shell or layer surrounding a three?dimensional object, such as a cell envelope. It typically involves subtracting the volume of the inner core from the volume of the overall structure, a technique that is useful in quantifying the contribution of boundary layers to the overall volume.

Unit Conversion and Dimensional Analysis

This concept involves converting measurements across different units (for example, nanometers to micrometers or liters to cubic meters) and ensuring dimensional consistency in calculations. It is a critical technique in all scientific computations to ensure that quantities are compared appropriately and calculated accurately.

Cylindrical Volume Calculation

This concept involves using the mathematical formula for the volume of a cylinder, V = ?r²h, to model the shape of a cell and calculate its volume. It is fundamental in biological contexts for estimating the size of rod-shaped cells and is applicable across many fields that require spatial measurements of cylindrical objects.

Density and Mass Relationship

This concept connects the physical properties of density and volume to determine mass. By multiplying the density by the volume, one can calculate the total mass of an object, a principle that is widely used in physics and biology to relate an organism's composition (such as its water content) with its overall mass.

Transcript

00:01 So this question is dealing with, given the e.

00:08 Coli to figure out, first of all, volume of the e.

00:14 Coli, given that the diameter is 0 .8 micrometers, and the height or the way that i've got it drawn here is two micrometers.

00:24 And then also using the equation of volume equals pi r squared times h.

00:45 To another.

00:49 One is looking at the cell, it is having the dimensions in terms of that's, okay, that's going to be the h of the when we saw for the volume of a cylinder of v equals pi r squared times height.

01:10 So the height is two.

01:17 So half of them, so then take 0 .4 times, 0 .4, because that's r squared, times 2 times 3 .14, and that gives you of the cylinder that is the rod -shaped bacterial cell.

01:38 So when you do the mathematics of it, you get volume is equal to 10 .48.

01:49 Then...

01:50 10 .048.

01:52 10 .48 into the equation of density equals mass over volume.

02:01 Volume would be 10 .48.

02:04 And then you also have a number that they've given you for the density, the total density.

02:10 That is 1 .1 times 10 to the third power, which is exactly the same thing as 1100.

02:16 So you essentially then take 1100 density times 10 .48 volume to get the volume out of the fraction equal to mass.

02:28 So then therefore mass is 1100 times 10 .48 or 11 ,053.

02:38 That's how many grams are in a cell that has the dimension of two micrometers for its height, 0 .8 micrometers for its diameter.

02:50 And i had made the error of kind of repeatedly saying, but it's actually the volume of the cell is 10 .0 .4.

03:00 And the mass again of this kind of cell is going to be 11 ,055 grams.

03:08 It has a density of 1 .1 times 10 to the third power are due to the bacterial envelope, the outer thick outer covering on the outside after the cell wall and is the volume of the cylinder when it's at the point four loss and including 10, nanometers or .01.

04:02 And what i get is the volume of the whole thing, including the envelope, is 10 .4.

04:08 When i just only have a hundredth and get the volume of it, the actual envelope itself, find as the answer for the volume of the envelope as 0 .00628.

04:42 So then we're doing a comparison...

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