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
346 Respo Therefore: 33.5=13.9. P/O E-3.16 P/O=335 2. 3.7 Statements a), b)and c)are applicable to class 2 metabolites. Since Ya, is a measure of growth efficiency, statement d)is the converse of statement b)and is therefore not 3,8 more reduced +5.00 glucose Glucose +4.00 4.00 more oxidised Gluconate +3.67 Refer to section 3. 1. 2 if you were unable to calculate degrees of reductance 2)The specific rate of exopolysaccharide production( a )is inversely related to growth Yo) 3)From Table 3. 1: Yx/s=80 g dry wt mol" O,Yx/5-180 048 We can see from the units of qs that Y,H=qs So,q=044.02=0.039881h 3.9 Class 1 or 2, depending on the substrate used. We can see from Table 2, for example, that succinoglycan biosynthesis leads to a net production of ATP(Class 2)with ethanol as substrate, but the biosynthesis is energy requiring Class 1)with glucose as substrate 3.10 1)The organism, as we have already seen, has a relatively high P/o quotient ahigl growth efficiency). It would therefore seem to have only a limited capacity for energy dissipation. During citric acid production, energy dissipation is desirable to ensure continued operation of glycolysis leading to citric acid formation 2)A high growth efficiency (high YO )is desirable because sophorolipid production has a high demand for atP
346 Responses Therefore: 33.5 = 13.9 . P/O (E - 3.16) 33.5 13.9 P/O=-=2.4 3.7 Statements a), b) and c) are applicable to class 2 metabolites. Since y4" is a measure of growth efficiency, statement d) is the converse of statement b) and is therefore not applicable. 3.8 1) Degrees of reductance more reduced Ethanol + 5.00 Glycerol + 4.67 Sorbitol + 4.33 Glucose + 4.00 Xylose + 4.00 Gluconate + 3.67 Succinate + 3.50 than [ glucose t more oxidised than glucose Refer to section 3.1.2 if you were unable to calculate degrees of redudance. 2) The specific rate of exopolysaccharide production (q,,) is inversely related to growth 3) From Table 3.1: Yx/s = 80 g dry wt mol-' efficiency (yswX). p = 0.2 h-' (p = D) 80 So,Yxfs=~=o.44gg-' We can see from the units of q, that Y, . p = q,. So, qs = 0.44 .0.2 = 0.039 g g-' h-'. 3.9 Class 1 or 2, depending on the substrate used. We can see from Table 2, for example, that succinoglycan biosynthesis leads to a net production of ATP (Class 2) with ethanol as substrate, but the biosynthesis is energy requiring (Class 1) with glucose as substrate. 1) The organism, as we have already seen, has a relatively high P/O quotient (high growth efficiency). It would therefore seem to have only a limited capacity for energy dissipation. During citric acid production, energy dissipation is desirable to ensure continued operation of glycolysis leading to citric acid formation. 2) A high growth efficiency (high ys") is desirable because sophorolipid production 3.1 0 has a high demand for ATP
Responses to SAQs Responses to Chapter 4 sAQS 4.1 Items (2)and(3)are the main factors encouraging development of sCP as alternatives to plant proteins. This is a method of converting waste organic materials to produce a valuable product. The speed with which micro-organisms can do this surpasses any Items(4)and(5)can be true of both plants and micro-organisms and are thus not a relative advantage to either. Item (1)is not always true. Some micro-organisms are easier to digest than plants, whereas others(such as algae)are more difficult to digest than many plant foods 4.2 1)Incorporating a proportion of SCP into manufactured foods can disguise unpleasant flavours or textures. Food technologists have a wide array of flavourings at their disposal, which can be used to produce particular flavours. If this cannot be done it might be possible to use the sCP as feed 2)Processing the sCP to break up the cells(by milling or some other means), then ncorporating it into prepared foods may overcome this problem. Otherwise it might be possible to use the sCP for feed 3)Blending the sCP with foods containing high levels of the deficient amino acids, or adding the deficient amino acid to the sCP can overcome this problem. Genetic engineering could be used to induce the organism to manufacture deficient amino acids. Otherwise it might be possible to use the sCP for feed 4)Toxicity of SCP is usually due to high RNA content Processing the sCP can reduce the RNA content. Otherwise it should be possible to use the sCP for feed 4.3 3)is the correct response. Algal cultures grown on molasses in open lagoons would become overgrown with bacteria, making the product useless as SCP. Responses 1 and 2 do not apply, as on molasses the organism grows as a heterotroph and requires neither sunlight or CO. Response 4 is incorrect as the yield would not be affected solely by the types of system used In lagoons, yields of algae would in fact probably be less than in bioreactors but the reason for this is 3)-contaminants would use much of the substrate, leaving less available for algal growth
Responses to SAQs 347 Responses to Chapter 4 SAQS 4.1 Items (2) and (3) are the main factors encouraging development of SCP as alternatives to plant proteins. This is a method of converting waste organic materials to produce a valuable product. The speed with which micro-organisms can do this surpasses any form of agriculture. Items (4) and (5) can be true of both plants and micro-organisms and are thus not a relative advantage to either. Item (1) is not always true. Some micro-organisms are easier to digest than plants, whereas others (such as algae) are more difficult to digest than many plant foods. 4.2 1) Incorporating a proportion of SB into manufactured foods can disguise unpleasant flavours or textures. Food technologists have a wide array of flavourings at their disposal, which can be used to produce particular flavours. If this cannot be done it might be possible to use the SCP as feed. 2) Processing the SB to break up the cells (by milling or some other means), then incorporating it into prepared foods may overcome this problem. Otherwise it might be possible to use the SCP for feed. 3) Blending the SCP with foods containing high levels of the deficient amino acids, or adding the deficient amino acid to the SCP can overcome this problem. Genetic engineering could be used to induce the organism to manufacture deficient amino acids. Otherwise it might be possible to use the SB for feed. the RNA content. Otherwise it should be possible to use the SCP for feed. 4) Toxicity of SCP is usually due to high RNA content. Processing the SCP can reduce 4.3 3) is the correct response. Algal cultures grown on molasses in open lagoons would become overgrown with bacteria, making the product useless as SCP. Responses 1 and 2 do not apply, as on molasses the organism grows as a heterotroph and reQuires neither sunlight or COZ. Response 4 is incorrect as the yield would not be affeded solely by the types of system used. In lagoons, yields of algae would in fact probably be less than in bioreactors, but the reason for this is 3) - contaminants would use much of the substrate, leaving less available for algal growth
48 Response 4.4 carbohydrate substrate minerals edium blending water continuous heat sterilisation inoculum filtration continuous fermentatio cell separation water biomass slury dried SCP 4.5 Your diagram may not look exactly like ours-but you can use ours to see if you have the stages in the process in the correct order and if you have omitted an essential step
348 Responses 4.4 4.5 Your diagram may not look exactly like ours - but you can use ours to see if you have the stages in the process in the correct order and if you have omitted an essential step
Responses to SAQs fermentation 0-r ammonia cyclone blast chilling free 4.6 Your drawing should look similar to ours. Make sure you have all of the inputs in the solid starch steamin M steam water moisture content mentation inoculum protein-enriched feed
Responses to SAQs 349 4.6 Your drawing should look similar to ours. Make sure you have all of the inputs in the right place
espouses 4.7 gave 42.oTR. The oxygen requirement for the metabolism of carbohydrates we a) minimum .7 kg of substrate during our discussion of carbohydrates. Did you remember? The oxygen requirement for biomass production on n-alkanes compared to carbohydrate is in the ratio 2.2/0.7=3. 14. In other words, the oxygen requirement for biomass production on n-alkanes exceeds that for production on carbohydrate by a factor of 3. 14. For the system described, the minimum OTR for biomass from carbohydrate is 1.89 kg O2"". The minimum OTR for a similar system based on n-alkanes would be 1.89 x 3 14=5.94 kg On mh b) Heat evolution rate. The heat evolution for biomass production on n-alkanes compared to carbohydrate is in the ratio of 27, 100/12, 300=2.2. In other words, the heat evolution (or cooling requirement if the operating temperatures are the same for biomass production on n-alkanes exceeds that for production on carbohydrate by a factor of 2. 2. For the system described, the heat evolution for production from carbohydrate is 33 210 k Joules mh. The heat evolution for a similar system based on n-alkanes would be 33 210 x 2.2=73, 062 k Joules m]h 4.8 i)Waste paper Trichoderma viride.Solid substrate fermentation. Only this fungus amongst those listed is capable of using the cellulose of which paper is composed. Solid substrate fermentation would be the easiest and cheapest production system. ii)Exhaust gas emissions CHlorella regularis.open lagoons Only Chlorella spp can use the CO in such exhaust gas emissions. Open(sunlit)lagoons would be necessar iii)Molasses. All the organism listed. bioreactors All the organisms listed would grow on the sucrose in molasses. Candida utilis or Kluyveromyces fragilis would be the best organisms to use, as they are food-grade yeasts with high growth rates iv)Effluent. Aspergillus niger. bioreactors or open lagoons Aspergillus niger is strongly amylolytic and capable of using the starch in the effluent. Open lagoon systems operating at low pH, if effective, would be a cheaper method of production than aseptic bioreactors 4.9 1)Output The productivity is 30 x0.3=9 kg h". Thus the output is 9 x 36=324 kg biomass h 2)Minimum otr (we can determine this in at least two different ways a) The yield is 0.5 kg biomass per kg methanol. From Figure 4.10 at yield =0.5, oxygen requirement is about 0.05 moles/g cells. As 1 mole O2=32 g, 0.05 moles = 1.6 g herefore the oxygen requirement is 1.6 g O per g biomass, or 1.6 biomass With productivity 9 kg m h", the minimum OTR is 1.6x9=14.4 kg Om h
350 Responses 4.7 a) Minimum OTR. The oxygen requirement for the metabolism of carbohydrates we gave as 0.7 kg-' of substrate during our discussion of carbohydrates. Did you remember? The oxygen requirement for biomass production on n-alkanes compared to carbohydrate is in the ratio 2.2/0.7 = 3.14. In other words, the oxygen requirement for biomass production on n-alkanes exceeds that for production on carbohydrate by a factor of 3.14. For the system described, the minimum OTR for biomass from carbohydrate is 1.89 kg Q m-3 h". The minimum OTR for a similar system based on n-alkanes would be 1.89 x 3.14 = 5.94 kg Q m-3 h". b) Heat evolution rate. The heat evolution for biomass production on n-alkanes compared to carbohydrate is in the ratio of 27,100/12,300 = 2.2. In other words, the heat evolution (or cooling requirement if the operating temperatures are the same) for biomass production on n-alkanes exceeds that for production on carbohydrate by a factor of 2.2. For the system described, the heat evolution for production from carbohydrate is 33,210 k Joules m-3 h-I. The heat evolution for a similar system based on n-alkanes would be 33,210 x 2.2 = 73,062 k Joules m-3 h-I. 4.8 i) Waste paper ... Tni-ma oiride ... Solid substrate fermentation. Only this fungus amongst those listed is capable of using the cellulose of which paper is composed. Solid substrate fermentation would be the easiest and cheapest production system. ii) Exhaust gas emissions ... CWorella regulmis ... open lagoons. Only Chlorella spp. can use the CQ in such exhaust gas emissions. Open (sunlit) lagoons would be necessary. iii) Molasses ... All the organism listed ... bioreactors All the organisms listed would grow on the suuose in molasses. Candida utilis or Klupvces fragilis would be the best organisms to use, as they are food-grade yeasts with high growth rates. iv) Effluent ... Aspergilhs nigu ... bioreactors or open lagoons. Only Aspergillus niger is strongly amylolytic and capable of using the starch in the effluent. Open lagoon systems operating at low pH, if effective, would be a cheaper method of production than aseptic bioreactors. 4.9 1) output The productivity is 30 x 0.3 = 9 kg m" h-'. Thus the output is 9 x 36 = 324 kg biomass h-'. 2) Minimum OTR (we can determine this in at least two different ways). a) The yield is 0.5 kg biomass per kg methanol. From Figure 4.10 at yield = 05, oxygen requirement is about 0.05 moles/g cells. As 1 mole 02 = 32 g, 0.05 moles = 1.6 g. Therefore the oxygen requirement is 1.6 g Q per g biomass, or 1.6 kg Q per % biomass. With productivity 9 kg m-3 h-', the minimum OTR is 1.6 x 9 = 14.4 kg 02 mh-'