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 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 = 92
MW of Biomass (Klebsiella aerogenes), CH1.73O0.43N0.24 = 23.97/0.92 = 26.1
Degree of reduction for substrate i.e. glycerol (ᵞ ) = 4.7
s
Degree of reduction for biomass (ᵞ ) = 4x1+1x1.73-2x0.43-3x0.24 = 4.15
B
No. of carbon atoms in glycerol, C3H8O3 (w) = 3
Yield of biomass (Yxs) = 0.40 gg-1
MW substrate
Now, c = Yxs = 0.4x92/26.1 = 1.41 g mol biomass/g mol substrate
MWcells
And oxygen requirement (a) = ᵞ)=
B = 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?

2. Solution:
MW of acetic acid = 60
MW of Biomass (methane bacteria), CH1.4O0.40N0.20 = 22.6
Degree of reduction for substrate i.e. acetic acid (ᵞ ) = 4.0
s
Degree of reduction for biomass (ᵞ ) = 4x1+1x1.4-2x0.40-3x0.20 = 4.0
B
No. of carbon atoms in acetic acid, CH3COOH (w) = 2
Degree of freedom of product (ᵞ ) = 8
p
No. of carbon atoms in the product i.e. methane (j) = 1
Since 0.67 g CO2 is produced from each g acetic acid,
0.67gCO2 1m olCO2 60gaceticacid
d= = 0.91 g moles of CO2 per g mol of substrate
1gAceticaci
d 44gCO2 1m olacetica
cid
Now, elemental balance when a product is also formed:
N balance: z + bi = cᵟ + fm
Or, 0 + b*1 = c*0.20 + f*0
Or, c = b/0.2
Or, c = 5b .................... (i)
Carbon balance: w = c + 0.91 + fj
Or, 2 = 5b + 0.91 + f*1 (b = 5c from eq (i))
Or, 5b + f = 2-0.91
Or, 5b + f = 1.09 .................... (ii)
O balance: y + 2a + bh = cβ + 2d +e + fl
Or, 2 + 0 + b*0 = c*0.40 + 2*0.91 + e + f*0
Or, 2 = 0.40c + 1.82 + e
Or, 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 + fk
Or, 4 + b*3 = c*1.4 + 2e + f*4
Or, 4 + 3b = 1.4c + 2e + 4f
From, eq (i) 4 + 3b = 1.4*5b + 2e +4f

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 = 0.036
From (i): c = 5* 0.036 = 0.18
From (iii): e = 0.18 – 2* 0.036
Or, e = 0.108
From (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*(ᵞ )/j*(ᵞ )
s p
= 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 + H2 O
(Biomass)

4. Molecular weight of glucose =180
Degree of reduction of substrate (ᵞ )
s = 4.0
No. of carbon atoms in the substrate (w) = 6 (for glucose)
Molecular weight of biomass with ash = 24.44/0.95 = 25.73
Degree of reduction of biomass (ᵞ )B =4
Part (a)
By elemental balance we have to find c,
C balance: w = c +d
Or, 6 = c + d
Or, c = 6 – d ....................... (i)
N balance: z + bi = cᵞ
Or, 0 + b*1 = c*0.16
Or, 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 +e
Or, 6 + 0 + b*0 = c*0.54 + 2*d + e
Or, 6 = 0.54c + 2d + e .................... (iv)
H balance: x + bg = cα + 2e
Or, 12 + b*3 = c*4 + 2*1.56
Or, 12 + 3b = 4c + 3.12 ....................... (v)
From (ii) and (v) we have: 12 + 3*0.16c = 4c + 3.12
Or, c = 2.52 g mol of biomass/ g mol of glucose
MW cells
Now, Biomass Yield (Yxs ) = c
MWsubstrat e = 2.52 *25.73/180 = 0.36 g g-1
Now,
cmax= w*(ᵞ )/ (ᵞ ) = 6*4/4 = 6 g mol of biomass / g mol of glucose
s B
Converting to mass basis,
Yxs, max = [(cmax)* (Molecular weight of cells)/ (Molecular weight of substrate)]
= 6*25.73/180
= 0.86 gg-1

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) = 32
Degree of reduction (ᵞ )
s =6
No. of carbon atoms in the substrate (w) = 1 (for methanol)
Molecular weight of biomass including ash = 22.52 / 0.94 = 23.95
Degree of reduction of biomass (ᵞ )B = 4.3
By 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 ᵞ – c ᵞ )
s B
= 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?

6. SOLUTION:
no O
C6H12O6 + NH3 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) = 24.6
Molecular weight of (Z. Mobilis) = 24.6
Degree of reduction (ᵞ ) for glucose = 4.0
s
Degree of reduction (ᵞ for ethanol = 6.0
p)
Degree of reduction (ᵞ ) for ethanol = 6.0
B
w for glucose = 6.0
w 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*0
Or, b = 0.16
Carbon balance: w = c + d + fj
Or, 6 = 0.8 + d + f*2
Or, d + 2f = 5.2 .................... (i)
O balance: y + 2a + bh = cβ + 2d +e + fl
Or, 6 = 0.8*0.50 + 2*d + e + f*1
Or, 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 + fk
Or, 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 = 8.32

7. Or, 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*0
Or, b = 0.074
Carbon balance: w = c + d + fj
Or, 6 = 0.8 + d + f*2
Or, d + 2f = 5.2 .................... (i)
O balance: y + 2a + bh = cβ + 2d +e + fl
Or, 6 = 0.8*0.50 + 2*d + e + f*1
Or, 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 + fk
Or, 12 + 0.074*3 = 0.8*4 + 2*e + f*6
Replacing e by 3f - 4.8 we have 2*(3f - 4.8) + 6f = 8.32
Or, 6f - 9.6 + 6f = 9.022
Or, 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

8. 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 = 180
Molecular 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) = 4
Degree of reduction of biomass(γB) =(4)(1)+(1)(1.79)+(-2)(0.56)+(-3)(0.17) = 4.16
w 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 biomass
32
c( MWofcells)
Y XS
MWofsubstrate
Y XS ( MWofsubstrate) 0.37 180
c
( MWofcells) 25.13

9. = 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) = 26.13
Final Concentration of biomass = 25g l-1

11. 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

12. 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 )s
Y XS
MWofsubstrate
YXS ( MWofsubstrate) 0.48 180
c 3.51
( MWofbiom as )
s 24.6
f =20% of c = 0.2(3.51) = 0.702
The general formula for product stoichiometry is
CwHxOyNz + aO2 +b HgOhNi cCHαOβNδ +dCO2+eH2O + fCjHkOlNm…… (b)
Comparing equation (a) and (b), balancing for nitrogen, we get
bi = cδ + fm
or, 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.

13. 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

14. 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,
c( MWbiom ass)
YXS
MWsubstrate
c(24.6)
0.14
46
0.14 4.6
c 0.26
46
f ( MWproduct)
YPS
MWsubstrate
f (60)
0.92
46
0.92 46
f 0.71
60
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.