Presentation is loading. Please wait.

Presentation is loading. Please wait.

Lecture 6: Product Formation Stoichiometry

Similar presentations


Presentation on theme: "Lecture 6: Product Formation Stoichiometry"— Presentation transcript:

1 Lecture 6: Product Formation Stoichiometry
Dr. AKM Shafiqul Islam 05/03/08

2 Product Formation Stoichiometry
Product formation are classified into four classes Main product appear as a result of primary energy metabolism. Example : ethanol production during anaerobic growth of yeast Main product arises indirectly from energy metabolism. Example: citric acid formation during aerobic mold formation Secondary metabolite. Example: penicillin production in aerated mold culture Biotransformation. The product is obtained from substrate through one or more reactions catalyzed by enzyme in the cells. Example: steroid hydroxylation

3 Class 1: product appears is relatively constant proportions as the cell mass accumulates and substrate consumed if the substrate is glucose which contain 6 carbon, the number of glucose molecule used is a’/6. if the product contain np carbon per molecule, f’/np product molecules are formed. thus molar yield YP/S is given by

4 Case 2: the simple stoichiometry does not apply
Case 2: the simple stoichiometry does not apply. The product formation is not proportional to substrate utilization or cell mall increase.

5 Class 3 and 4: The stoichiometric depend on particular substrate or products involved. Product formation is completely uncoupled with cell growth. Example, anaerobic fermentation of glucose to ethanol. The best possible case is utilization of all substrate for product formation alone. Assuming cell growth is negligible

6 2/3 of the substrate carbon are used for product formation.
Yield factor can be write as

7 The upper bound for product yield can determine by assuming that no cells or other products accumulate. For example, for penicillin synthesis The maximum yield of penicillin is 1.2 g penicillin/(g glucose). But the actual value is much more lower (0.1 g/g), indicating useful remaining potential for organism and process improvement. Yield much lower than upper bound value indicate that significant substrate utilization supports growth, maintenance or synthesis of other products

8 Metabolic Energy Stoichiometry: Heat Generation and Yield Factor Estimation
Cellular heat production is the result of energy and growth metabolism. It is proportional relationship between heat generated and energy substrate utilized. If Ys is grams of cell mass produced per gram substrate consumed and DHs and DHc are heat of combustion of substrate and of cell mass material, we can write yield factor

9 This arises from two energy balance pathways
the heat generation DHs per gram of substrate completely oxidized minus YsDHc, the heat obtained by combustion of cell grown from same amount of substrate, will approximate the heat generation per gram of substrate consumed in the fermentation which produce the cell, H2O and CO2.

10

11 Example: The heat of combustion of pseudomonas fluorescens growing in glucose medium. First a reaction of cell combustion is written using measured cell composition, assuming combustion products are H2O, CO2 and N2. Assuming a heat of combustion of 104 kcal/mole O2, the heat released by combustion of bacteria is

12 If cell dry matter contain 10% ash, then the heat of combustion is

13 A yeast (CH1.66N0.13O0.40) is growing aerobically on arabinose (C5H10O5) and ammonium hydroxide (NH4OH) with a respiratory quotient of 1.4. Write the stoichiometric equation Determine the yield coefficient, YX/S, for this system in grams biomass/gram substrate What is the maximum biomass concentration that can be achieved in batch growth for this system if the initial substrate concentration is 10 g/L and the initial biomass concentration is 0.5 g/L

14 aC5H10O5 + bO2 + cNH4OH → CH1.66N0.13O dCO2 + eH2O Equations for coefficients: C atom balance: 5a = 1 + d (1) H atom balance: 10a + 5c = e (2) O atom balance: 5a + 2b + c = d + e (3) N atom balance: c = (4) Respiratory quotient: RQ = 1.4 = d/b (5)

15 2×(3) - (2) b = (5) d = (1) a = (3) e =

16 Stoichiometry of the equation
0.2823C5H10O O NH4OH→ CH1.66N0.13O CO H2O


Download ppt "Lecture 6: Product Formation Stoichiometry"

Similar presentations


Ads by Google