# Answer to Question #108649 in Biology for amir

Question #108649
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?
1
2020-04-20T09:59:58-0400

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

• find a volume of a single bacteria cell: Vcocci = (π × d3) / 6. As a diameter of a single cocci equals 0.8 μm, Vcocci = (π × 0.83) / 6 = 0.27 μm3.
• find a total mass of a single bacteria cell: mcocci = Vcocci × dcocci. As the mean density of a bacteria cell equlas d = 1.1 g/cm3, mcocci = 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: mcocci_dry = (100% - 80%) × mcocci / 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: mcocci_biomass = 90% × 5.94 × 10-14 g / 100% = 5.346 × 10-14 g.
• find the total biomass in metabolically active bacteria: mtotal_biomass = number of bacteria × mcocci_biomass = 4 × 1013 × 5.346 × 10-14 g = 2.14 g

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