Question

BIOCHEMISTRY QUESTION

E. Coli cells have up to 70,000 ribosomes per cell. Each ribosome has one molecule of 16S rRNA.

a) Calculate the maximal concentration of 16S rRNA in a cell (~2x1 micrometer, assume a shape of a cylinder).

b) If it was possible to extract 100% of all 16S rRNA molecules from one E. Coli cell into a 10microliter solution, what would be the concentration of 16S rRNA in this solution?

Answer 1

(a)

Given that each ribosome contains one molecule of 16s rRNA and the total no. of ribosomes per cell ~70,000.

Volume of cell (dimentions of 2*1 microns) is to be computed, given that it is to be approximated as a cylinder.

Volume of cylinder = π r² h where r is the radius and h is the height. Here r is 0.5 micron (half of width) and h is 2 micron.

Volume is therefore 3.14*(0.25)*2 = 1.57 cubic microns = 1.57 * 10^-18 m³ (approx.)

Maximal concentration = no. of molecules of 16s rRNA per cell/ volume of the cell

no. of moles of 16s rRNA per cell = 70,000/6.02 x 10²³ = 1.16 * 10^-19 (6.02 x 10²³ is the Avogadro no.)

Con. = no. of moles of 16s rRNA/ volume of cell (in L) = 1.16 * 10^-19/ 1.57 * 10^-18 = 0.116/1.57 = 0.073 M = 73 mM

(b) now all the molecules which were dispersed in 1.57 * 10^-18 L of cytoplasm is now diluted to 10 uL solution. We know that M1*V1=M2*V2 where M1 is the molarity of the substance in V1 volume and M2 is the molarity of the substance in V2 volume.

Since we know M1, V1 (molarity and volume which we computed earlier) and the new volume V2 (10 uL), we can calculate the new molarity of 16s rRNA in the solution.

73 mM* 1.57 * 10^-18 = X * 10* 10^-6

X = (73 x 1.57*10^-18)/10^-5 mM = 1.15 * 10^-13 mM = 0.115 femtomolar (fM).

I hope this helps :)