ASSIGNMENT                        ED 73.03 BIOPROCESS TECHNOLOGY                                             BY           ...
Solution:MW of acetic acid = 60MW of Biomass (methane bacteria), CH1.4O0.40N0.20 = 22.6Degree of reduction for substrate i...
Or, 4b + 2e + 4f = 4 ........... (iv)Solving (ii) and (iv) we get: 16b –2e = 0.36 ......... (v)Solving eq (iii) and eq (v)...
Molecular weight of glucose                               =180Degree of reduction of substrate (ᵞ )                       ...
The biomass yield is only about 42% of the maximum possible biomass yield.Part (b)i) Methanol + O2 + ammonia              ...
SOLUTION:                             no OC6H12O6 + NH3                    2          = CH1.8O0.5N0.2 + CO2   +   H2O + C2...
Or, f = 1.5Now, Yield of ethanol from glucose (Yps)      = f*MW of product/MW of substance                                ...
4.11 Detecting unknown ProductsYeast growing in continuous culture produce 0.37g of biomass per g glucoseconsumed; about 0...
= 2.65 g mol of biomass/ g mol of substrateWe knowwγs - 4 a = cγB +fjγPOr, (6)(4) – (4)(0.69) = (2.65)(4.16) +fjγPOr, 24- ...
4.13 Oxygen demand for production of recombinant proteinProduction of recombinant protein by a genetically-engineered stra...
Degree of reduction of product (γP) = (4)(1) +(1)(1.55) +(-2)(0.31) +(-3)(0.25) = 4.18          δ=0.2, m= 0.25        c( M...
Oxygen demand is given bya = ¼(wγs – cγB – fjγp)  = ¼(6 x4 – 3.51 x 3.9 – 0)  = ¼ (24- 13.69)  = ¼(10.11)  = 2.53Hence 2.5...
MW of product i.e., acetic acid = (12)(2) +(1)(4) +(16) (2) = 60Degree of reduction of product (γP) = ½{(4)(2) +(1)(4) +(-...
Assignment bioprocess
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  1. 1. ASSIGNMENT ED 73.03 BIOPROCESS TECHNOLOGY BY YAKINDRA PRASAD TIMILSENA (ID 111332)Q.N. 4.7 Klebsiella aerogenes is produced from glycerol in aerobic culture withammonia as nitrogen source. The biomass contains 8% ash, 0.40g biomass isproduced for each g of glycerol consumed and no major metabolic products areformed. What is the oxygen requirement for this culture in mass terms?Solution:MW of glycerol = 92MW of Biomass (Klebsiella aerogenes), CH1.73O0.43N0.24 = 23.97/0.92 = 26.1Degree of reduction for substrate i.e. glycerol (ᵞ ) = 4.7 sDegree of reduction for biomass (ᵞ ) = 4x1+1x1.73-2x0.43-3x0.24 = 4.15 BNo. of carbon atoms in glycerol, C3H8O3 (w) = 3Yield of biomass (Yxs) = 0.40 gg-1 MW substrateNow, c = Yxs = 0.4x92/26.1 = 1.41 g mol biomass/g mol substrate MWcellsAnd oxygen requirement (a) = ᵞ)= B = 2.1Therefore, Oxygen demand is 2.1 g mol of O2 per mole of substrate consumed.Converting into mass form = 2.1*16/92 = 0.37 g oxygen per g substrate.Q.N. 4.8 Anaerobic digestion of volatile acids by methane bacteria is representedby the equation CH3COOH + NH3 → Biomass + CO2 + H2O + CH4The composition of methane bacteria is approximated by the empirical formulaCH1.4O0.40N0.20. For each kg acetic acid consumed 0.67 kg CO2 is evolved. Howdoes the yield of methane under these conditions compare with the maximumpossible yield?
  2. 2. Solution:MW of acetic acid = 60MW of Biomass (methane bacteria), CH1.4O0.40N0.20 = 22.6Degree of reduction for substrate i.e. acetic acid (ᵞ ) = 4.0 sDegree of reduction for biomass (ᵞ ) = 4x1+1x1.4-2x0.40-3x0.20 = 4.0 BNo. of carbon atoms in acetic acid, CH3COOH (w) = 2Degree of freedom of product (ᵞ ) = 8 pNo. of carbon atoms in the product i.e. methane (j) = 1Since 0.67 g CO2 is produced from each g acetic acid, 0.67gCO2 1m olCO2 60gaceticacidd= = 0.91 g moles of CO2 per g mol of substrate 1gAceticaci d 44gCO2 1m olacetica cidNow, elemental balance when a product is also formed:N balance: z + bi = cᵟ + fmOr, 0 + b*1 = c*0.20 + f*0Or, c = b/0.2Or, c = 5b .................... (i)Carbon balance: w = c + 0.91 + fjOr, 2 = 5b + 0.91 + f*1 (b = 5c from eq (i))Or, 5b + f = 2-0.91Or, 5b + f = 1.09 .................... (ii)O balance: y + 2a + bh = cβ + 2d +e + flOr, 2 + 0 + b*0 = c*0.40 + 2*0.91 + e + f*0Or, 2 = 0.40c + 1.82 + eOr, 0.40c + e = 0.18Or, 0.4*5b + e = 0.18 (from eq (i))Or, 2b + e = 0.18 ........................ (iii)H balance: x + bg = cα + 2e + fkOr, 4 + b*3 = c*1.4 + 2e + f*4Or, 4 + 3b = 1.4c + 2e + 4fFrom, eq (i) 4 + 3b = 1.4*5b + 2e +4f
  3. 3. Or, 4b + 2e + 4f = 4 ........... (iv)Solving (ii) and (iv) we get: 16b –2e = 0.36 ......... (v)Solving eq (iii) and eq (v) we get: b = 0.036From (i): c = 5* 0.036 = 0.18From (iii): e = 0.18 – 2* 0.036 Or, e = 0.108From (ii): f = 1.09 – 5*0.036 Or, f = 0.91The product yield is 0.91 g mol per g mol of the substrateAnd, fmax = w*(ᵞ )/j*(ᵞ ) s p = 2*4/1*8 =1Comparing the value of f and fmax we can say that f is only 91% of fmax.Question 4.9a) Cellulomonus bacteria used as a single cell protein for human or animal foodare produced from glucose under anaerobic condition. All carbon in the substrate isconverted into biomass; ammonia is used as nitrogen source. The molecularformula for biomass is CH1.56O0.54N0.16; the cells also contain 5% ash. How doesthe yield of biomass from substrate in mass and molar terms compare with themaximum possible biomass yield? b) Another system for manufacture of SCP is Methylophilus methylotrophus. Thisorganism is produced aerobically from methanol with ammonia as nitrogen source.The molecular formula for the biomass is CH1.68O0.36N0.22; the cells contain 6% ash.1) How does the maximum yield of biomass compare with the above? What is themain reason for the difference?2) If the actual yield of biomass from methanol is 42% the thermodynamicmaximum, what is the oxygen demand?Solution:a) Glucose + ammonia CH1.56 O 0.54 N0.16 + CO2 + H2 O (Biomass)
  4. 4. Molecular weight of glucose =180Degree of reduction of substrate (ᵞ ) s = 4.0No. of carbon atoms in the substrate (w) = 6 (for glucose)Molecular weight of biomass with ash = 24.44/0.95 = 25.73Degree of reduction of biomass (ᵞ )B =4Part (a)By elemental balance we have to find c,C balance: w = c +dOr, 6 = c + dOr, c = 6 – d ....................... (i)N balance: z + bi = cᵞOr, 0 + b*1 = c*0.16Or, b = 0.16c .......................... (ii)From (i) and (ii) b = 0.16 * (6-d)Or, b = 0.96 – 0.16d .................... (iii)O balance: y + 2a + bh = cβ + 2d +eOr, 6 + 0 + b*0 = c*0.54 + 2*d + eOr, 6 = 0.54c + 2d + e .................... (iv)H balance: x + bg = cα + 2eOr, 12 + b*3 = c*4 + 2*1.56Or, 12 + 3b = 4c + 3.12 ....................... (v)From (ii) and (v) we have: 12 + 3*0.16c = 4c + 3.12Or, c = 2.52 g mol of biomass/ g mol of glucose MW cellsNow, Biomass Yield (Yxs ) = c MWsubstrat e = 2.52 *25.73/180 = 0.36 g g-1Now, cmax= w*(ᵞ )/ (ᵞ ) = 6*4/4 = 6 g mol of biomass / g mol of glucose s BConverting to mass basis,Yxs, max = [(cmax)* (Molecular weight of cells)/ (Molecular weight of substrate)] = 6*25.73/180 = 0.86 gg-1
  5. 5. The biomass yield is only about 42% of the maximum possible biomass yield.Part (b)i) Methanol + O2 + ammonia CH1.68 O0.36 N0.22 + CO2 + H2 O (Substrate) (nitrogen source) (Biomass)Here,Molecular weight of methanol (CH4O) = 32Degree of reduction (ᵞ ) s =6No. of carbon atoms in the substrate (w) = 1 (for methanol)Molecular weight of biomass including ash = 22.52 / 0.94 = 23.95Degree of reduction of biomass (ᵞ )B = 4.3By relation, Maximum yield of biomass (cmax) =1.4ii) Actual yield of biomass from methanol (Yxs) = 0.42* thermodynamic maxm = 0.42*(cmax/w) = 0.42* (1.4/1) = 0.588 gg-1Then c = Yxs *(molecular weight of substrate / molecular weight of biomass) c = 0.588* 32 / 23.95 = 0.79Oxygen demand (a) = ¼ (w ᵞ – c ᵞ ) s B = ¼ (1*6 - 0.79*4.3) = 0.65 moles of O2/moles of substrate.QUESTION: 4.10Both Saccharomyces cerevisiae yeast and Zymomonas mobilis bacteria produceethanol from glucose under anaerobic conditions without external electronacceptors. The biomass yield from glucose is 0.11gg -1 for yeast and 0.05gg-1 for Z.mobilis. In both cases the nitrogen source is NH3. Both cell compositions arerepresented by formula CH1.8O 0.5N0.2. a. What is the yield of ethanol from glucose in both cases ? b. How do the yields calculated in (a) compare with the thermodynamic maximum?
  6. 6. SOLUTION: no OC6H12O6 + NH3 2 = CH1.8O0.5N0.2 + CO2 + H2O + C2H6O(Substrate) (Biomass) (Product)Molecular weight of glucose = 180Molecular weight of product (ethanol) = 46Molecular weight of biomass (yeast) = 24.6Molecular weight of (Z. Mobilis) = 24.6Degree of reduction (ᵞ ) for glucose = 4.0 sDegree of reduction (ᵞ for ethanol = 6.0 p)Degree of reduction (ᵞ ) for ethanol = 6.0 Bw for glucose = 6.0w for ethanol = 2Yield of biomass (Yxs) for yeast = 0.11gg-1Therefore, c = Yxs * Molecular weight of glucose / Molecular weight of Cells = 0.11 * 180 / 24.6 = 0.8 g mol biomass / g mol of glucoseBy elemental balance in the case of product formedN balance: z + bi = cᵟ + fmOr, 0 + b*1 = 0.8*0.20 + f*0Or, b = 0.16Carbon balance: w = c + d + fjOr, 6 = 0.8 + d + f*2Or, d + 2f = 5.2 .................... (i)O balance: y + 2a + bh = cβ + 2d +e + flOr, 6 = 0.8*0.50 + 2*d + e + f*1Or, 2d + e + f = 5.6 ................ (ii)From (ii) and (iii) we have 3f – e = 4.8Or, e = 3f - 4.8H balance: x + bg = cα + 2e + fkOr, 12 + 0.16*3 = 0.8*4 + 2*e + f*6Replacing e by 3f - 4.8 we have 2*(3f - 4.8) + 6f = 8.32Or, 6f - 9.6 + 6f = 8.32
  7. 7. Or, f = 1.5Now, Yield of ethanol from glucose (Yps) = f*MW of product/MW of substance = 1.5*46/180 = 0.38 gg-1Now,Yield of biomass (Yxs) for z. mobilis = 0.05gg-1Therefore, c = Yxs X Molecular weight of glucose / Molecular weight of cells = 0.05 X 180 / 24.6 = 0.37 g mol of biomass / g mol of glucoseBy elemental balance in the case of product formedN balance: z + bi = cᵟ + fmOr, 0 + b*1 = 0.37*0.20 + f*0Or, b = 0.074Carbon balance: w = c + d + fjOr, 6 = 0.8 + d + f*2Or, d + 2f = 5.2 .................... (i)O balance: y + 2a + bh = cβ + 2d +e + flOr, 6 = 0.8*0.50 + 2*d + e + f*1Or, 2d + e + f = 5.6 ................ (ii)From (ii) and (iii) we have 3f – e = 4.8Or, e = 3f - 4.8H balance: x + bg = cα + 2e + fkOr, 12 + 0.074*3 = 0.8*4 + 2*e + f*6Replacing e by 3f - 4.8 we have 2*(3f - 4.8) + 6f = 8.32Or, 6f - 9.6 + 6f = 9.022Or, f = 1.55Now, Yield of ethanol from glucose (Yps) = f*MW of product/MW of substance = 1.55*46/180 = 0.39 gg-1The yield of ethanol from Z. mobilis is greater than that from S. Cerevisae by 0.39-0.38= 0.01gg-1
  8. 8. 4.11 Detecting unknown ProductsYeast growing in continuous culture produce 0.37g of biomass per g glucoseconsumed; about 0.88g O2 is consumed per g cells formed. The nitrogen source isammonia, and the biomass composition is CH1.79O0.56N0.17. Are other products alsosynthesized?Given:Glucose + Ammonia + Oxygen Biomass + Carbon dioxide +Water +ProductC6H12O6 + aO2 +b NH3 cCH1.79O0.56N0.17 +dCO2+eH2O + fCjHkOlNm0.88g O2 is consumed per g cells formed.The biomass yield from the substrate (YXS) = 0.37g biomass/g glucoseProducts = ?Solution:Molecular weight of glucose = 180Molecular weight of biomass = (12)(1) + (1)(1.79) +(16)(0.56) +(14)(0.17) = 25.13Degree of reduction of glucose (γS) = (4)(1) + (1)(12) + (-2)(6) = 4Degree of reduction of biomass(γB) =(4)(1)+(1)(1.79)+(-2)(0.56)+(-3)(0.17) = 4.16w for glucose = 6a = 0.88g of O2/g cells = 0.88 g of O2 * Mol. Wt of biomass Mol. Wt. of oxygen = 0.88 g of O2 * 25.13 = 0.69 g mole of O2/g mole of biomass 32 c( MWofcells)Y XS MWofsubstrate Y XS ( MWofsubstrate) 0.37 180c ( MWofcells) 25.13
  9. 9. = 2.65 g mol of biomass/ g mol of substrateWe knowwγs - 4 a = cγB +fjγPOr, (6)(4) – (4)(0.69) = (2.65)(4.16) +fjγPOr, 24- 2.76 = 11.024 + fjγPOr, 21.24 = 11.024 + fjγPOr, fjγp = 10.216 g g-1Hence other product is also synthesized beside increase in the biomass when glucose isconsumed.4.12 Medium formulationPseudomonas 5401 is to be used for production of single-cell protein for animalfeed. The substrate is fuel oil. The composition of Pseudomonas 5401 isCH1.83O0.55N0.25. If the final cell concentration is 25g l-1, what minimumconcentration of (NH4)2SO4 must be provided in the medium if (NH4)2SO4 is thesole nitrogen source?Solution:Fuel oil + a O2 + b (NH4)2SO4 → c CH1.83O0.55N0.25 + d CO2 + e H2O(Substrate) (Biomass)MW of Substrate =MW of Biomass (CH1.83O0.55N0.25) = 26.13Final Concentration of biomass = 25g l-1
  10. 10. 4.13 Oxygen demand for production of recombinant proteinProduction of recombinant protein by a genetically-engineered strain ofEscherichia coli is proportional to cell growth. Ammonia is used as nitrogensource for aerobic respiration of glucose. The recombinant protein has an overallformula CH1.55O0.31N0.25. The yield of biomass from glucose is measured at 0.48gg-1; the yield of recombinant protein from glucose is about 20% that for cells. (a) How much ammonia is required? (b) What is the oxygen demand? (c) If the biomass yield remains at 0.48 g g -1, how much difference are the ammonia and oxygen requirements for wild-type E. coli unable to synthesis recombinant protein?Solution:Given:Glucose + Ammonia + Oxygen Biomass + Carbon dioxide +Water +ProductC6H12O6 + aO2 + b NH3 cCHαOβNδ + dCO2 + eH2O + fCH1.55O0.31N0.25……(a)The biomass yield from the substrate (YXS) = 0.48g biomass/g glucoseYield of recombinant protein (f) = 20% of that of biomass (i) Ammonia required = ? (ii) Oxygen demand = ? (iii) If no product is formed, ammonia and oxygen required = ?For ammonia, i= 1For glucose, w= 6 j =1Molecular weight of glucose = 180Molecular weight of biomass = 24.6 (for E. coli)Degree of reduction of glucose (γS) = (4)(1) + (1)(12) + (-2)(6) = 4General molecular formula for E. coli = CH1.5O0.5N0.2Degree of reduction for biomass (γB) = (4)(1) +(1)(1.5) +(-2)(0.5) +(-3)(0.2) = 3.9Molecular formula of recombinant protein product = CH1.55O0.31N0.25
  11. 11. Degree of reduction of product (γP) = (4)(1) +(1)(1.55) +(-2)(0.31) +(-3)(0.25) = 4.18 δ=0.2, m= 0.25 c( MWofbiom as )sY XS MWofsubstrate YXS ( MWofsubstrate) 0.48 180c 3.51 ( MWofbiom as ) s 24.6f =20% of c = 0.2(3.51) = 0.702The general formula for product stoichiometry isCwHxOyNz + aO2 +b HgOhNi cCHαOβNδ +dCO2+eH2O + fCjHkOlNm…… (b)Comparing equation (a) and (b), balancing for nitrogen, we get bi = cδ + fmor, b(1) = (3.51)(0.2) +(0.702)(0.25)or, b = 0.8775Hence 0.8775 g of ammonia is required per g of glucose consumed.Oxygen demand is given by the equationa = ¼(wγs – cγB – fjγp) = ¼(6 x 4 – 3.51x 3.9 – 0.702 x 1 x 4.18) =1/4(24 - 13.69 – 2.93) =1/4(7.38) = 1.845Case IIProduct = 0Then, balancing for nitrogen bi = cδ b(1) = (3.51)(0.2) b = 0.702Hence when wild –type E. coli is used, 0.702g of ammonia is required per g of glucose used.
  12. 12. Oxygen demand is given bya = ¼(wγs – cγB – fjγp) = ¼(6 x4 – 3.51 x 3.9 – 0) = ¼ (24- 13.69) = ¼(10.11) = 2.53Hence 2.53 g of oxygen is required per g of glucose consumed if wild type E. coli is used.4.14 Effect of growth on oxygen demandThe chemical reaction equation for conversion of ethanol (C2H6O) to acetic acid (C2H4O2) is: C2H6O + O2 C2H4O2 + H2OAcetic acid is produced from ethanol during growth of Acetobacter aceti, which has thecomposition CH1.8O0.5N0.2 . Biomass yield from substrate is 0.14g g-1. Ammonia is used asnitrogen source. How does growth in this culture affect oxygen demand for acetic acidproduction?Solution:C2H6O +aO2 +bNH3 cCH1.8O0.5N0.2 + dCO2 + eH2O + fC2H4O2Ethanol acetic acidYield of product from substrate (YPS) = 0.92g g-1Yield of biomass from substrate (YxS) = 0.14 g g-1Molecular formula of Acetobacter acetii = CH1.8O0.5N0.2Oxygen demand = ?Here,MW of biomass = (12)(1) +(1)(1.8) +(16)(0.5) +(14)(0.2) = 24.6Degree of reduction of biomass (γB) = (4)(1) +(1)(1.8) +(-2)(0.5) +(-3)(0.2) = 4.2MW of substrate i.e. ethanol = (12)(2) +(1)(6) +(16)(1) = 46Degree of reduction of substrate (γS) =1/2{(4)(2) + (1)(6) + (-2)(1)} = 12/2 = 6
  13. 13. MW of product i.e., acetic acid = (12)(2) +(1)(4) +(16) (2) = 60Degree of reduction of product (γP) = ½{(4)(2) +(1)(4) +(-2)(2)} = 4 For ethanol, w = 2 For acetic acid, j = 2We know, c( MWbiom ass)YXS MWsubstrate c(24.6)0.14 46 0.14 4.6c 0.26 46 f ( MWproduct)YPS MWsubstrate f (60)0.92 46 0.92 46f 0.71 60Oxygen demand is given by the equationa = ¼(wγs – cγB – fjγp) =1/4(2 x 6 – 0.26 x 4.2 – 0.71x 2 x 4) =1/4(12 – 1.092 – 5.68) = ¼(5.228) =1.307Hence oxygen demand is 1.307 g per g of substrate consumed.

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