Answer to Question #108649 in Biology for amir

Question

In the activated sludge process, we "activate" the sludge by accumulating a
high concentration of biomass through sludge settling and recycle. You analyze
the biomass in an activated sludge process and find that its total biomass
concentration is 3,000 mg/1 as volatile suspended solids, which is the same
as organic dry weight. Your bacteria are well represented by cocci having a
diameter of 0.8 J,lm, 80 percent water content, and 90 percent organic matter
in the dry weight. If the total number of metabolically active bacteria is
4 . 1013/1, what fraction of the total biomass is active bacteria?

Answer 1

The steps required for the caclulation of the total biomass are as following:

• find a volume of a single bacteria cell: V_cocci = (π × d³) / 6. As a diameter of a single cocci equals 0.8 μm, V_cocci = (π × 0.8³) / 6 = 0.27 μm³.
• find a total mass of a single bacteria cell: m_cocci = V_cocci × d_cocci. As the mean density of a bacteria cell equlas d = 1.1 g/cm³, m_cocci = 0.27 μm³ × 1.1 g/cm³ = 2.7 × 10^-13 cm³ × 1.1 g/cm³ = 2.97 × 10^-13 g.
• find a mass of a dry weight of a single cell: m_{cocci_dry} = (100% - 80%) × m_cocci / 100% = 20% × 2.97 × 10^-13 g / 100% = 5.94 × 10^-14 g.
• find a biomass in a dry weight of a single bacteria cell: m_{cocci_biomass} = 90% × 5.94 × 10^-14 g / 100% = 5.346 × 10^-14 g.
• find the total biomass in metabolically active bacteria: m_{total_biomass} = number of bacteria × m_{cocci_biomass} = 4 × 10¹³ × 5.346 × 10^-14 g = 2.14 g

Answer: 2.14 g