STOICHIOMETRY OF MICROBIAL GROWTH AND

PRODUCT FORMATION

By:

Rianne Alipio

Sherry Banton

Danna Bitara

Glyza Villanea

Elajah Zaragoza

Elemental Balance

Elemental Balance• Can easily be written when the compositions of substrates, products, and cellular material are known

• Only biomass is produced without product formation

• One mole of biological materials is defined as the one mole of biological materials is defined as the amount containing 1 gram atom of carbon, such as CHαOβNδ

• Assumption: No extracellular products other than H2O and CO2 are produced

Eq(1)

CHmOn + aO2 + bNH3 cCHαOβNδ + d H2O + e CO2 Substrate Nitrogen source Dry biomass

where

CHmOn : 1mole of carbohydrate cCHαOβN δ : 1 mole of cellular material

Elemental BalancesEq(2)

C : 1= c + e H: m + 3b = c + 2dαO: n + 2a = c + d + 2eβN: b = c δ

Respiratory Quotient (RQ)

Eq(3)

RQ = e/a

We have five equations and five unknowns (a, b, c, d, e). With a measured value of RQ, these equations can be solved to determine the stoichiometric coefficients.

DEGREE OF REDUCTION

DEGREE OF REDUCTION

0The degree of reduction, g, for organic compounds may be defined as the number of equivalents of available electrons per gram atom C.

0 the concept of degree of reduction has been developed and used for proton–electron balances in bioreactions

0The degrees of reduction for some key elements are C = 4, H = 1, N = -3, O = -2, P = 5, and S = 6.

Degrees of reduction for various organic compounds

Examples of how to calculate the degree of reduction for substrates.

Methane (CH4): 1(4) + 4(1) = 8g = 8/1 = 8

Glucose (C6H12O6): 6(4) + 12(1) + 6(-2) = 24g = 24/6 = 4

Ethanol (C2H5OH): 2(4) + 6(1) + 1(-2) = 12g = 12/2 = 6

Consider the aerobic production of a single extracellular product.

CHmOn + Oα 2 + bNH3 cCHαOβN δ+ dCHxOy Nz + eH2O + fCO2

substrate biomass product

The degrees of reduction of substrate, biomass, and product are

Ƴs = 4 + m - 2n

Ƴb = 4 + - 2 - 3α β δ

Ƴp = 4 + x -2y – 3z

0NOTE that for CO2, H2O, and NH3 the degree of reduction is zero.

 

Consider the aerobic reaction of a single extracellular product

CHmOn + a O2 + bNH3 c CHαOβN𝝳 + d CHxOyNz + e H2O + f CO2

Substrate biomass product

This equation can lead to 0 Elemental balance on C, H, O and N0 Available electron balance0 Energy balance0 Total mass balance

0Carbon Balance Eq. 7.100Nitrogen Balance:Eq. 7.110Available-Electron BalanceEq. 7.12

(on a molar basis) (on a molar basis)

0An energy balance for an aerobic growth Eq. 7.13 heat evolved per equivalent of available electrons transferred to Oxygen 26.95 kcal/g of available electrons transferred to Oxygen

0Equation 7.12 can be written as Eq. 7.14a

Eq. 7.14b fraction of available electrons in the organic substrate that is transferred to oygen fraction of available electrons that is incorporated to biomass fraction of available electrons that is incorporated to extracellular products 

EXAMPLE

Assume that experimental measurements for a certain organism have shown that cells can convert two-thirds (wt/wt) of the substrate carbon (alkane or glucose) to biomass.

Theoretical Prediction of Yield Coefficients

Theoretical Prediction of Yield Coefficients

For Aerobic Fermentations

• The growth yield per available electron in oxygen molecules is approximately 3.14 gdw/electron when ammonia is used as the nitrogen sources.

• The number of available electron per oxygen molecule is four.

•When the number of oxygen molecules per mole of substrate is known, the growth yield coefficient Yx/s can easily be calculated

The ATP yield (Y x/ATP )

Anaerobic Fermentation

• approximately 10.5 gdw cells/ mol ATP

Aerobic Fermentation

• this yield varies between 6-29.

When the energy yield of metabolic pathway is known, the growth yield Yx/s can be calculated using the following equation

Y x/s = Y x/ATP N