Question

In the top 170 m of the world's oceans, there are about 1028 prokaryotes. *The shape of the cells are not specified*

1. Estimate the total volume taken up by these cells in m3.

2. Estimate the total volume taken up by these cells in km3.

3. What is the mass of these cells in kg?

4. Compute their mean spacing

Answer 1

Answer

1. There are 1028 prokaryotes in the top 170m of the worlds ocean. The volume occupied by a single prokaryote is 10^-18 m³ in general.

Therefore, the total volume V = n*v, where n is the number of cells and v is the volume occupied by a prokaryote.

                                           V = n*v

                                              = 1028*10^-18 m³

2. The total volume taken up by these cells in Km³

                                             1 m³ = 1000,000,000 Km³

                                                     = 10⁹ Km³

Therefore, total volume = 1028*10^-18*10⁹ = 1028*10^-9 km³

3. The mass is the product of density and volume.

               Mass = density * volume

Here, volume is the volume of the ocean where the prokaryotes are grown.

In general, worlds ocean area = 361.1*10⁶ Km²

                                                = 361.1*10⁶*10⁶

                                                = 361*10¹² m²

                                                = 361*10¹¹m³

Volume for 170m of the worlds ocean = 170*361*10¹¹m³ = 6.2*10¹⁵ m³

From this the density can be calculated as, D = n/volume of the ocean considered

                                                                        = 1028*6.2*10¹⁵

                                                                       = 3.91*10³⁵ cells / m³

Therefore, mass = density / volume

                           = 3.91*10³⁵ / 1028*10^-9

                           = 3.054*10⁴⁴ cells = 3.054*10⁴⁴*10³ kg = 3.054*10⁴⁷ kg

4. Mean spacing is the number of cells in the given volume.

                           = 1028 / 6.2*10¹⁵ = 16.58*10^-16 cells / m³