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 with ammonia as nitrogen source. The biomass contains 8% ash, 0.40g biomass is produced for each g of glycerol consumed and no major metabolic products are formed. 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 (ᵞs) = 4.7Degree of reduction for biomass (ᵞB) = 4x1+1x1.73-2x0.43-3x0.24 = 4.15No. of carbon atoms in glycerol, C3H8O3 (w) = 3Yield of biomass (Yxs) = 0.40 gg-1

Now, c =Yxs ×MW substrate

MWcells = 0.4x92/26.1 = 1.41 g mol biomass/g mol substrate

And oxygen requirement (a) = ¼¿ᵞB) = ¼(3∗4.7−1.41∗4.15)= 2.1

Therefore, 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 represented by the equation CH3COOH + NH3 → Biomass + CO2 + H2O + CH4

The composition of methane bacteria is approximated by the empirical formula CH1.4O0.40N0.20. For each kg acetic acid consumed 0.67 kg CO2 is evolved. How does the yield of methane under these conditions compare with the maximum possible yield?

Solution:

MW of acetic acid = 60MW of Biomass (methane bacteria), CH1.4O0.40N0.20 = 22.6Degree of reduction for substrate i.e. acetic acid (ᵞs) = 4.0Degree of reduction for biomass (ᵞB) = 4x1+1x1.4-2x0.40-3x0.20 = 4.0No. of carbon atoms in acetic acid, CH3COOH (w) = 2Degree of freedom of product (ᵞp) = 8No. of carbon atoms in the product i.e. methane (j) = 1

Since 0.67 g CO2 is produced from each g acetic acid,

d =

0 .67 gCO21gAceticacid

×1molCO244 gCO2

×60 gaceticacid1molaceticacid = 0.91 g moles of CO2 per g mol of substrate

Now, elemental balance when a product is also formed:

N balance: z + bi = cᵟ + fm

Or, 0 + b*1 = c*0.20 + f*0Or, c = b/0.2 Or, 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 = 0.18 Or, 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 + 4f

From, eq (i) 4 + 3b = 1.4*5b + 2e +4f

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 = 0.91

The product yield is 0.91 g mol per g mol of the substrate

And, fmax = w*(ᵞs)/j*(ᵞp)= 2*4/1*8= 1

Comparing the value of f and fmax we can say that f is only 91% of fmax.

Question 4.9

a) Cellulomonus bacteria used as a single cell protein for human or animal food are produced from glucose under anaerobic condition. All carbon in the substrate is converted into biomass; ammonia is used as nitrogen source. The molecular formula for biomass is CH1.56O0.54N0.16; the cells also contain 5% ash. How does the yield of biomass from substrate in mass and molar terms compare with the maximum possible biomass yield? b) Another system for manufacture of SCP is Methylophilus methylotrophus. This organism 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 the main reason for the difference? 2) If the actual yield of biomass from methanol is 42% the thermodynamic maximum, what is the oxygen demand?

Solution:

a) Glucose + ammonia CH1.56 O 0.54 N0.16 + CO2 + H2O (Biomass)

Molecular weight of glucose =180 Degree of reduction of substrate (ᵞs) = 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 (ᵞB) = 4

Part (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 = cᵟOr, 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 + e Or, 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

Now, Biomass Yield (Yxs ) =

c×MW cellsMWsubstrate

= 2.52 *25.73/180 = 0.36 g g-1

Now, cmax= w*(ᵞs)/ (ᵞB) = 6*4/4 = 6 g mol of biomass / g mol of glucose Converting to mass basis,Yxs, max = [(cmax)* (Molecular weight of cells)/ (Molecular weight of substrate)]

= 6*25.73/180 = 0.86 gg-1

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 + H2O (Substrate) (nitrogen source) (Biomass)

Here,Molecular weight of methanol (CH4O) = 32Degree of reduction (ᵞs) = 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 (ᵞB) = 4.3By relation, Maximum yield of biomass (cmax) =1.4

ii) Actual yield of biomass from methanol (Yxs) = 0.42* thermodynamic maxm = 0.42*(cmax/w)= 0.42* (1.4/1) = 0.588 gg-1

Then c = Yxs *(molecular weight of substrate / molecular weight of biomass)

c = 0.588* 32 / 23.95 = 0.79

Oxygen demand (a) = ¼ (w ᵞs – c ᵞB) = ¼ (1*6 - 0.79*4.3) = 0.65 moles of O2/moles of substrate.

QUESTION: 4.10

Both Saccharomyces cerevisiae yeast and Zymomonas mobilis bacteria produce ethanol from glucose under anaerobic conditions without external electron acceptors. 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 are represented 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?

SOLUTION:

C6H12O6 + NH3 no O

2 = CH1.8O0.5N0.2 + CO2 + H2O + C2H6O

(Substrate) (Biomass) (Product)

Molecular weight of glucose = 180 Molecular weight of product (ethanol) = 46 Molecular weight of biomass (yeast) = 24.6Molecular weight of (Z. Mobilis) = 24.6Degree of reduction (ᵞs) for glucose = 4.0Degree of reduction (ᵞp) for ethanol = 6.0Degree of reduction (ᵞB) for ethanol = 6.0w for glucose = 6.0w for ethanol = 2

Yield of biomass (Yxs) for yeast = 0.11gg-1

Therefore, c = Yxs * Molecular weight of glucose / Molecular weight of Cells= 0.11 * 180 / 24.6 = 0.8 g mol biomass / g mol of glucose

By elemental balance in the case of product formed

N balance: z + bi = cᵟ + fm

Or, 0 + b*1 = 0.8*0.20 + f*0Or, b = 0.16

Carbon 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 = 4.8 Or, e = 3f - 4.8

H balance: x + bg = cα + 2e + fkOr, 12 + 0.16*3 = 0.8*4 + 2*e + f*6

Replacing e by 3f - 4.8 we have 2*(3f - 4.8) + 6f = 8.32

Or, 6f - 9.6 + 6f = 8.32Or, f = 1.5

Now, Yield of ethanol from glucose (Yps) = f*MW of product/MW of substance= 1.5*46/180 = 0.38 gg-1

Now,Yield of biomass (Yxs) for z. mobilis = 0.05gg-1

Therefore, 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 glucose

By elemental balance in the case of product formed

N balance: z + bi = cᵟ + fm

Or, 0 + b*1 = 0.37*0.20 + f*0Or, b = 0.074

Carbon 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 = 4.8 Or, e = 3f - 4.8

H balance: x + bg = cα + 2e + fkOr, 12 + 0.074*3 = 0.8*4 + 2*e + f*6

Replacing e by 3f - 4.8 we have 2*(3f - 4.8) + 6f = 8.32Or, 6f - 9.6 + 6f = 9.022Or, f = 1.55

Now, Yield of ethanol from glucose (Yps) = f*MW of product/MW of substance= 1.55*46/180 = 0.39 gg-1

The yield of ethanol from Z. mobilis is greater than that from S. Cerevisae by 0.39-0.38= 0.01gg-1

4.11 Detecting unknown Products

Yeast growing in continuous culture produce 0.37g of biomass per g glucose consumed; about 0.88g O2 is consumed per g cells formed. The nitrogen source is ammonia, and the biomass composition is CH1.79O0.56N0.17. Are other products also synthesized?

Given:

Glucose + Ammonia + Oxygen Biomass + Carbon dioxide +Water +Product

C6H12O6 + aO2 +b NH3 cCH1.79O0.56N0.17 +dCO2+eH2O + fCjHkOlNm

0.88g O2 is consumed per g cells formed.

The biomass yield from the substrate (YXS) = 0.37g biomass/g glucose

Products = ?

Solution:

Molecular weight of glucose = 180Molecular weight of biomass = (12)(1) + (1)(1.79) +(16)(0.56) +(14)(0.17) = 25.13

Degree 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 = 6

a = 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 biomass32

Y XS =c ( MWofcells )MWofsubstrate

c=Y XS ( MWofsubstrate )( MWofcells)

=0 . 37×18025 . 13

= 2.65 g mol of biomass/ g mol of substrate

We know

wγs - 4 a = cγB +fjγP

Or, (6)(4) – (4)(0.69) = (2.65)(4.16) +fjγP

Or, 24- 2.76 = 11.024 + fjγP

Or, 21.24 = 11.024 + fjγP

Or, fjγp = 10.216 g g-1

Hence other product is also synthesized beside increase in the biomass when glucose is consumed.

4.12 Medium formulation

Pseudomonas 5401 is to be used for production of single-cell protein for animal feed. The substrate is fuel oil. The composition of Pseudomonas 5401 is CH1.83O0.55N0.25. If the final cell concentration is 25g l-1, what minimum concentration of (NH4)2SO4 must be provided in the medium if (NH4)2SO4 is the sole 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

4.13 Oxygen demand for production of recombinant protein

Production of recombinant protein by a genetically-engineered strain of Escherichia coli is proportional to cell growth. Ammonia is used as nitrogen source for aerobic respiration of glucose. The recombinant protein has an overall formula CH1.55O0.31N0.25. The yield of biomass from glucose is measured at 0.48g g-

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 +Product

C6H12O6 + aO2 + b NH3 cCHαOβNδ + dCO2 + eH2O + fCH1.55O0.31N0.25……(a)

The biomass yield from the substrate (YXS) = 0.48g biomass/g glucose

Yield 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= 1

For glucose, w= 6

j =1

Molecular weight of glucose = 180

Molecular weight of biomass = 24.6 (for E. coli)

Degree of reduction of glucose (γS) = (4)(1) + (1)(12) + (-2)(6) = 4

General molecular formula for E. coli = CH1.5O0.5N0.2

Degree of reduction for biomass (γB) = (4)(1) +(1)(1.5) +(-2)(0.5) +(-3)(0.2) = 3.9

Molecular formula of recombinant protein product = CH1.55O0.31N0.25

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

Y XS =c ( MWofbiomass )MWofsubstrate

c=Y XS ( MWofsubstrate )( MWofbiomass )

=0 . 48×18024 . 6

=3. 51

f =20% of c = 0.2(3.51) = 0.702

The 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 = 0.8775

Hence 0.8775 g of ammonia is required per g of glucose consumed.

Oxygen demand is given by the equation a = 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) = 1.845

Case II

Product = 0

Then, balancing for nitrogen

bi = cδ

b(1) = (3.51)(0.2)

b = 0.702

Hence when wild –type E. coli is used, 0.702g of ammonia is required per g of glucose used.

Oxygen demand is given by

a = ¼(wγs – cγB – fjγp) = ¼(6 x4 – 3.51 x 3.9 – 0) = ¼ (24- 13.69) = ¼(10.11) = 2.53

Hence 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 demand

The chemical reaction equation for conversion of ethanol (C2H6O) to acetic acid (C2H4O2) is:

C2H6O + O2 C2H4O2 + H2O

Acetic acid is produced from ethanol during growth of Acetobacter aceti, which has the composition CH1.8O0.5N0.2 . Biomass yield from substrate is 0.14g g-1. Ammonia is used as nitrogen source. How does growth in this culture affect oxygen demand for acetic acid production?

Solution:

C2H6O +aO2 +bNH3 cCH1.8O0.5N0.2 + dCO2 + eH2O + fC2H4O2

Ethanol acetic acid

Yield of product from substrate (YPS) = 0.92g g-1

Yield of biomass from substrate (YxS) = 0.14 g g-1

Molecular formula of Acetobacter acetii = CH1.8O0.5N0.2

Oxygen demand = ?

Here,

MW of biomass = (12)(1) +(1)(1.8) +(16)(0.5) +(14)(0.2) = 24.6

Degree of reduction of biomass (γB) = (4)(1) +(1)(1.8) +(-2)(0.5) +(-3)(0.2) = 4.2

MW of substrate i.e. ethanol = (12)(2) +(1)(6) +(16)(1) = 46

Degree of reduction of substrate (γS) =1/2{(4)(2) + (1)(6) + (-2)(1)} = 12/2 = 6

MW of product i.e., acetic acid = (12)(2) +(1)(4) +(16) (2) = 60

Degree of reduction of product (γP) = ½{(4)(2) +(1)(4) +(-2)(2)} = 4

For ethanol, w = 2

For acetic acid, j = 2

We know,

Y XS=c ( MWbiomass)¿ MWsubstrate ¿¿

¿0 .14=c (24 . 6 )46

¿c=0 . 14×4 . 646

=0 . 26 ¿ ¿¿Y PS=

f ( MWproduct )MWsubstrate

0 .92=f (60 )46

f =0 . 92×4660

=0 .71

Oxygen demand is given by the equation

a = 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)

=1.307

Hence oxygen demand is 1.307 g per g of substrate consumed.