Responses to SAQ Responses to sAQs Responses to Chapter 2 SAQs 1 The reasons are: 1)True: Growing and dividing cells need to use substrate to provide energy and materials for growth, maintenance and product formation. In immobilised (non-growing) systems the energy and materials are only required for cell maintenance and product formation 2)True: Separation of the biocatalyst(enzyme or cells)from the product stream is not required and relatively few other substances are present. 3)True: The concentration of immobilised biocatalyst in a reactor can be much greate than for systems using soluble of free cells 4)True: Higher substrate concentrations can be used, which often enables higher product concentrations to be achieved, thus reducing volume of effluent for a give product yield 2.2 The biotransformation should be carried out using whole cells because: 1)The multi-componentenzyme systemis likely to havemuch lower activity in enzyme preparations satisfy when using whol乌 2)The reaction has a cofactor requirement(see Table 2. 1)which is relatively easy to 2.3 1)Batch Fermentation run times are relatively short and the same bioreactor can be used for making several different products 2)Continuous. Long fermentation run times, with many generations, increase the chance of mutation or loss of plasmid DNA 3)Continuous. Difficult to maintain sterile conditions over very long periods Contaminants may grow faster (out compete)process organisms and take over the 4)Batch Fermentation run times are relatively short. 5)Batch. All waste products(metabolites) accumulate in batch culture; in continuous culture much is lost through the outflow 6)Batch. Batch cultures have astationary phase during which little or no growth
Responses to SAQs 341 Responses to SAQs Responses to Chapter 2 SAQs 2.1 The reasons are: 2.2 2.3 1) True: Growing and dividing cells neec. to use substrate to provide energy and materials for growth, maintenance and product formation. In immobilised (non-growing) systems the energy and materials are only required for cell maintenance and product formation. 2) True: Separation of the biocatalyst (enzyme or cells) from the product stream is not required and relatively few other substances are present. 3) True: The concentration of immobilised biocatalyst in a reactor can be much greater than for systems using soluble of fxw cells. 4) True: Higher substrate concentrations can be used, which often enables higher product concentrations to be achieved, thus reducing volume of effluent for a given product yield. The biotransformation should be carried out using whole cells because: 1 ) The multicomponent enzyme system is likely to have much lower activity in enzyme 2) The reaction has a cofactor requirement (see Table 2.1) which is relatively easy to preparations. satisfy when using whole cells. 1) Batch. Fermentation run times are relatively short and the same bioreador can be 2) Continuous. Long fermentation run times, with many generations, increase the used for making several different products. chance of mutation or loss of plasmid DNA. 3) Continuous. Difficult to maintain sterile conditions over very long periods. Contaminants may grow faster (out compete) process organisms and take over the vessel. 4) Batch. Fermentation run times are relatively short. 5) Batch. All waste products (metabolites) accumulate in batch culture; in continuous 6) Batch. Batch cultures have a 'stationary phase' during which little or no growth culture much is lost through the outflow. occurs
342 Res 7 Continuous. Unlike batch it is not possible to identify all of the materials involved in the fermentation run 8)Continuous. When in steady state the culture does not change with time and is, therefore, relatively easy to operate and control 2.4 Costs of downstream processing for bioprocesses are increased by 1)low concentrations of products, 2)numerous impurities at low concentration and 3)intracellular materials Gif cell disruption is necessary. However, the high specificity of biocatalysts is a benefit to downstream processing since products closely related to the desired product are less likely to be present. Waste products of bioprocesses are likely to be less environmentally damaging, which also reduces downstream processing costs Costs of downstream processing for purely chemical synthesis would be increased by 1)low specificity of reactions(giving rise to chemical contaminants closely related to the desired product) and 2)the toxic/corrosive nature of the chemicals 2.5 Process A: fed-batch mode free cells vacuum fermentation Process B batch or fed-batch modes nmobilised cells Process C: continuous mode ultrafiltration with enzyzme recycling
342 Responses 2.4 7) Continuous. Unlike batch it is not possible to iden* all of the materials involved in the fermentation run. 8) Continuous. When in steady state the culturr does not change with time and is, therefore, relatively easy to operate and control. Costs of downstream processing for bioprocesses are inmased by 1) low concentrations of products, 2) numerous impurities at low concentration and 3) intracellular materials (if cell dis~ption is necessary). However, the high specificity of biocatalysts is a benefit to downstream processing since products closely related to the desired product are less likely to be present. Waste products of bioprocesses are likely to be less environmentally damaging, which also reduces downstream processing costs. Costs of downstream processing for purely chemical synthesis would be ind by 1) low specificity of reactions (giving rise to chemical contaminants closely related to the desired product) and 2) the toxic/corrosive nature of the chemicals. 2.5 Process A: Process B Process c fed-batch mode free cells vacuum fermentation batch or fed-batch modes immobilisedcells solvent extraction continuous mode free enzyme ultrafiltration with enzyme recycling
Responses to SAQs Responses to Chapter 3 SAQs 3.1 Reaction equation d CHr O+bo+C Hnn >CHa OB Na+ d hio+e cO2 Balances for the four elements: C: a=1+e H:a'x+cl=α+2d O:ay+2b=β+d+2e Six moles of glucose are used for each mole of biomass produced, thus KC HO+b0+CHN > CH166 O z7 No20+d Ho+e coz ie unknown coefficient are b'c d ande C:6=1+e N:c1=0.20 c’=0.2 H:6(2)+0.2(4)=1666+2d O6(1)+2b=027+5567+25 b=4918 The reaction equation then becomes 6CHO+4918O2+0.2HN >CH1666 O.7 No20+5. 567 HiO+ 5CO2 3.2 Decrease. Decrease in degree of reductance of substrate increases the demand for NADH. See.7 2)Increase. Increased efficiency of oxidative phosphorylation increases the P/0 quotient. See E-3.3. 3)Increase. A lowered energy demand for biomass synthesis increases YATP 3.3 2) Types 3 and 4 3) Type 3
Responses to SAQs 343 Responses to Chapter 3 SAQs 3.1 Reaction equation: Balances for the four elements: Ca’=l+e’ H a’x + CI = 01 + 2d’ O: a’y + 2b’ = + d’ + 2e‘ N. c‘n = 6 Six moles of glucose are used for each mole of biomass produced, thus 6C Ha + b’02 + c’H4” > C HI.& 00~ N0.m + d‘ HzO + e’ C02 ie unknown coefficient are b’, c’, d‘ and e‘ C:6=1 +e‘ e’ = 5 N dl = 0.20 H. 6(2) + 0.2(4) = 1 .& + 2 d‘ 0.6(1) + 2 b’ = 0.27 + 5.567 + 2.5 The reaction equation then becomes: c’ = 0.2 d’ = 5567 b’ = 4.918 6 CHD + 4.918 02 + 0.2 I-€4N > C HI.= Oon No20 +5567 Ha + 5CO2 3.2 1) Decrease. Decrease in degree of reductance of substrate increases the demand for NADH.SeeE -3.7. 2) Increase. Increased efficiency of oxidative phosphorylation increases the P/O 3) Increase. A lowered energy demand for biomass synthesis increases YXE. quotient. See E - 3.3. See E - 3.4. 3.3 1) Typel. 2) Types 3 and 4. 3) Type3. 4) Typel
R 3.4 Productivities at low and high dilution rates are 2.40 and 1.92 kg product m"h".The process should therefore be operated at the low dilution rat Since biomass productivity is Dx, product productivity(P)can be calculated as follows: kg product m h m=0.013 C-mol substrate/ C-mol biomass h- Yw=0.48 C-mol biomass/c-mol substrate These values were obtained as follows DOh 2) 0.426 We will plot y, against n since at steady state l_I 2.5 2,4 22 2.1 0 Slope=0.013C-mol substrate/C-mol biomass h=m
344 Responses 3.4 Productivities at low and high dilution rates are 2.40 and 1.92 kg product m3 h-'. The pmcess should therefore be operated at the low dilution rate. Since biomass productivity is IX, product productivity (P) can be calculated as follows: p=- . x . D = kg product m" h" m = 0.013 C-mol substrate/C-mol biomass h-' Y r = 0.48 C-mol biomass/C-mol substrate These values were obtained as follows. YPF - YX/S 3.5 1 1 0.9 1.11 0.475 2.10 0.4 2.50 0.470 2.13 0.2 5.00 0.465 2.15 0.1 J 10.00 0.455 220 0.05 20.00 0.426 2.35 - yx/s D &-'I jj (h) Yx/. 1 1 11 We will plot - against - since at steady state - = - Y?US D DP 2.5 - 2.4 - 2.3 - 1 2.2 - y XIS 2.1 - 2.0 - - I-- I I I 0 6 10 15 20 1 D - Slope = 0.013 C-mol substrate/C-mol biomass h-' = m
Responses to SAQs Intercept= 2.09 C-mol substrate/C-mol biomass So, y m=0.48 C-mol biomass/C-mol substrate YYo, is determined from Yx/s using (E-3.15) Yx/=0.475 y=4+2-2=+4 Y=4+1666-(3x02)-(2x0.27=452 0.4x4526 4 4526 4526 1-0475x 0.80 C-mol biomass/ mol O2 3.6 P/O This is obtained using: Yo= YATP.P/O Firstly, Yo, is determined graphically: (mmol g1h1) 12 (h1) aA=reciprocal of slope, ie 14.87=0.067g mmol Yo=335g mor ATP= 13.9 g mo
Responses to SAQs 345 1 Intercept = 2.09 C-mol substrate/C-mol biomass = - So, Y r = 0.48 C-mol biomass/C-mol substrate. Y%, is determined from Yx/, using (E - 3.15): y,s Yx/. = 0.475 yo = -2 x 2 = 4 ys = 4 + 2 - 2= 4 yx = 4 + 1.666 - (3 x 0.2) - (2 x 0.27) = 4526 4526 0.475 x - 4 4526 4 X 4 Y& = 4526 1 - 0.475 x - = 0.80 C-mol biomass/molG 3.6 P/O quotient = 2.4 This is obtained using: YF = m . P/O Firstly, w is determined graphically: (mrnol g-1 h-I) 12 ' 10 ' 8, 902 6 4- 2- c 0 oI2 0:4 0:s D (h-') 1 w 14.87 = reciprocal of slope, ie - = 0.067 g mmol-' = 67 g mol-' F = 335 g mol-' r$ = 13.9 g m1-I