web.pdx.edu/~rueterj/bi445/notes/ultrastructure.htm

Ultrastructure of a cyanobacterium

Size, volume, mass and carbon content

The following is an example for the composition of an cyanobacterial cell.

The cell has a diameter of 1 um and a length of 3 um (cylindrical)

The total volume of the cell is um^3 cell^-1

The total mass g cell^-1

The dry mass g cell^-1

carbon content moles C cell^-1

moles of carbon per cell = 21 fmol C cell^-1 ( 21 x 10^-12 moles C per cell)

The composition by component

functional component % of mass mol C cell^-1
Structure 20% 4
Photosythetic stuff 35% 7
Enzymatic and biosynthetic 40% 8
Reserve 5% 1

 

Ultrastructure

Make a slice through this that is 0.5 um x .1 um x .01 um (this equals 5 x 10-4 um^3 or 2 x 10^-4 of the total cell volume)

Each of these slices should have the following components

  component % of cell number of "objects" size of each "object"
S lipopolysacc 2 1 layer about 32 nm
murein layer 3 1 layer
lipid in plasmalemma membrane 10 1 layer 6 to 7 nm
DNA 5 portion of the strand 100 nm
 
P lipid in thylakoid membranes 15 4 to 7 layers 6 to 7 nm each
protein in the membrane 15 at least 150

PSI is 9 x 7 nm
PSII is 12 x 7 nm
Coupling factor 3 X 6 nm
phycobilisomes 8 x 29 x 36

soluble Calvin cycle enzymes 4 at least 50 many are 40,000 daltons or greater ( 10 nm)
 
E RNA in ribosome, tRNA and mRNA 20 at least 6 ribosomes
at least 80 tRNA
70S ribosomes 21 x 29 nm
Protein as soluble enzymes 20 at least 200 MR = 40,000 daltons
10 nm
 
R glycogen 5 ?  

 

If you layed this out on your desk at 1,000,000 time magnification, the 0.5 um would be 0.5 meters (about 19 inches) and the 0.2 um would then be about 8 inches. Imagine two 81/2 by 11 sheets of paper taped together with a 3 " overlap.