Question

A batch culture of E. coli was initiated with a medium containing 10 g/L of glucose, 2 g/L of ammonia, and 0.1 g/L of cells

A batch culture of E. coli was initiated with a medium containing 10 g/L of glucose, 2 g/L of ammonia, and 0.1 g/L of cells. At the end of culture, 0.1 g/L of glucose remained in the medium. A total of 5.1 g/L of E. coli were recovered. The elemental composition of the cells was determined to be (by mass) 53% C, 7.3% H, 12.0% N, and 19.0% O, with the remaining balance being ashes. a) Calculate the yield (by mass) of biomass based on glucose and nitrogen. b) Write down the stoichiometric equation for biomass formation from glucose and ammonia. c) Assume all the carbon that was not incorporated into biomass was used for energy generation and was completely oxidized to carbon dioxide. Also, assume that each mole of glucose produces 38 moles of ATP when completely oxidized to carbon dioxide. Calculate the biomass yield coefficient based on ATP.

Answer 1

Part (a)

Yield of Biomass, Y = grams of biomass produces/ grams of substrate utilised

Y= Total E.coli mass recovered/ Mass of glucose + ammonia + cells utilised

Y= 5.1 / ( (10-0.1) + 2 + 0.1

= 0.425

Part (b)

C6H12O6 (9.9g) + NH3(2g) ------> CHNO (5g)

For glucose

mol weight = 12x6 + 11x12 + 16x6 = 300

mole of glucose= 9.9/300 = 0.033

For NH3,

Mol weight = 14+ 1x3 = 17

mole of NH3 = 2/17 = 0.177

For Cell

mol weight = 12 +1 + 14 = 27

mole of cell for product =(91.3%x5)/27 = 0.169

So the Stoichiometric Equation will be

0.033 C6H12O6 (9.9g) + 0.177 NH3(2g)  ------> 0.169 CHNO (5g)

Part (c)

C6H12O6 (0.1g) ------> CO2 + ATP

Mole of remaining glucose substrate = 0.1/300 0.0003

1 mole of glucose forms 38 ATP then,

0.0003 mole of glucose will form, 38x0.0003= 0.0114 ATP

Molecular weight of ATP = 507 then,

Mass of ATP will be

No. of moles of ATPx Mol weight of ATP = 0.0114x507 = 5.7

Biomass Yield Coefficient = Mass of Glucose/ Mass of ATP

= 5.7/0.1 = 57