degree of reduction
DESCRIPTION
Bioprocess Engineering(Shuler and Kargi)TRANSCRIPT
STOICHIOMETRY OF MICROBIAL GROWTH AND
PRODUCT FORMATION
By:
Rianne Alipio
Sherry Banton
Danna Bitara
Glyza Villanea
Elajah Zaragoza
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
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.
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
(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
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.