Microbiology question, please help. A biological wastewater treatment process co

Question

Microbiology question, please help.

A biological wastewater treatment process contains an organism with a dry weight of 7.35xl0-13 g/cell. The process effluent contains 20 mg cells (dry weight)/L and has a total density of 1.02 g/cm3. Assume that the organic fraction of the biomass is described by C5H7O2N. Compute the following quantities. Cell count (# of microorganisms/mL) in effluent Mass of protein (mg) in a liter of the effluent Organic nitrogen concentration (mg/L) in the effluent Effluent organic carbon concentration (mg/L) attributable to cells

Explanation / Answer

a.     20x10-3 g/L / 7.35x10-13 g/cell = 2.72 x 1010 cells/L

b. By convention the proteins are about 1% of the dry weight of microorganism.

20 mg/L x 1/100= 0.2 mg proteins/L

Note for c) and d):

C5H7O2N is a general “chemical” formula for microorganisms (a simplification).

You may calculate a “molar mass”

12x5 + 7x1 + 16x2 + 14 = 113 g/mol

And percent composition by elements:

C    100 x 60/113 = 53 %

N    100 x 14/113 = 12.4 %

O     100 x 32/113 = 28.3 %

In fact, e.g. for E. coli, these values are C 50 %, N 14 %, O 20 %, H 8 %, other 8%.

c. 20 mg/L x12.4/100 = 2.48 mg N/L

d. 20 mg/L x53/100 = 10.6 mg C/L

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