In rapidly dividing bacteria the cell can divide in times as

In rapidly dividing bacteria, the cell can divide in times as short as 1200s. 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.

Solution

About half of the dry mass of a bacterium generally consist carbon. That is 0.1 x 10^-12g. in which, half value is weight of cell wall and the other half value is cytoplasm related carbon atoms. Therefore, a bacterium generally has 0.05 x 10^-12 g

In a dividing bacterium, a cell wall needs double of a single cell wall. 1 mol of glucose has 180 g weight [C6H12O6; 72+12+96 =180] which made of 6 x 10²³ molecules.

Therefore, 180 g glucose has 6 x 10²³ molecules; 0.05 x 10^-12 g glucose have

= [6 x 10²³ / 180] x 0.05 x 10^-12

= 0.0017 x 10¹¹ = 17 x 10⁷ glucose molecules.