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 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 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. 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 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 14.87 w = reciprocal of slope, ie - = 0.067 g mmol-' = 67 g mol-' F = 335 g mol-' r$ = 13.9 g m1-I 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 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. 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 349 4.6 Your drawing should look similar to ours. Make sure you have all of the inputs in the right place. 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 m- h-' . Responses to SAQs 351 b) The yield = 05, and productivity = 9. From Figure 4.11 at yield = 05 and estimatin~ productivity = 9, the minimum oxy n-transfer rate is about 05 moles oxygen 1" h- . This represents 32 x 0.5 = 16 g 02 1- h or 16 kg 02 m-3 h". $4 3) Heat Evolution Rate (again we have at least two ways to calculate this). a) From Figure 4.12, with yield 05, the heat produced is about 27 k Jou$s-pr gram biomass (or 27,000 k Joules per kg biomass). With productivity 9 kg m h , the heat evolution rate is 27,000 x 9 = 243,000 k Joules m-3 h-'. k Joules I-' h-I, or 251,OOO k Joules m-3 h-I. b) From Figure 4.13, with productivity 9 and yield 0.5, the heat produced is about 251 4) Methanol Concentration To support a biomass concentration of 30 kg m-3 at yield 0.5, the methanol concmtration in the incoming medium would need to be 30/05 = 60 kg m-3. Thus at 90% utilisation, the concentration in the incoming medium would need to be 60 x 100/90 = 66.7 kg m-3. 4.1 0 Your flow diagram might not look exactly like the one shown: for instance you will probably have represented the fermentation step by 'fermentation', rather than drawing the bioreador. Nevertheless you should have all the unit processes and inputs shown above in your own diagram. 352 Responses 4.1 1 4.1 2 4.1 3 1) productivity = biomass concentration x D. 2) Percentage utilisation = 100 - [(concentration in outgoing medium/ concentration in 3) Minimum OTR = productivity x 1.6 (see SAQ 4.9). incoming medium) x 1001. D fi-') PrOdUCtiVity Minimumm Methanol (kg biomass m-3 h-') (kg a requked m3 h-') Utilisation (%) 0.1 3.43 5.49 98.0 0.2 6.62 10.59 94.8 0.3 9.33 14.93 88.9 0.4 10.52 16.83 75.3 0.5 1.25 2.0 7.2 Dilution rate of 0.5 h-' produces the fastest growth rate (since p = D), but gives pr productivity and substrate utilisation. This is a feature of chemostat cultures, when p approaches thy.- Dilution rate 0.4 h-' 'ves highest productivity, but 2596 of the substrate remains unused. Dilution rate 0.3 h- gives better substrate utilisation but with reduced productivity. Both these dilution rates require minimum OTRs greater than the bioreactor can supply. This means that such biomass concentrations could not be produced in practice. Dilution rate 0.2 h" produces efficient substrate conversion and has an OTR that the system can just about provide. Dilution rate 0.1 h-' gives better substrate conversion but lower productivity. The chosen dilution rate would be just below 0.2 h-'. In practice dilution rates of between 0.1 h-' and 02 h-' are used to operate the production system. F 1) output With 1500 m3 at D = 0.2 h-', volume of culture produced is 300 m3 h-' (ie 1,500 x 02 m3 h" ) At 36 kg biomass m3, the output is 10,800 kg h" or 259,200 lcg day (259 tonnes day-') or 2) Biomass production of 94,608 tonnes would require 94,608/05 = 1B,216 tomes 94#608 tonnes yr-'. methanol. The most likely order is as outlined below. 1) Establish the size of the market. 2) Isolate organisms that can use the substrate 3) Measure temperature optimum. 4) & 5) Measure affinity for substrate/Measure protein mntent. 6) Perform feeding trials in animals. Responses to SAQs 353 7) Apply for sales licence. 8) Marketing 9) Full-scale production. The reasons for this sequence are: 1) The size of the market must be established first of all. There is no point in developing a product you cannot sell. 2) Once a substrate is chosen you must select organisms that can grow on it. 3) The temperature optima of isolated organisms would be measured early on in 4) & 5) The affinity for the substrate would be measured in small-scale continuous culture. The protein content of organisms would probably be measured at about the same time, or perhaps earlier with 3). 6) Feeding trials would only be performed on selected organisms, perhaps on just one organism that appears suitable for production. There is no point performing such expensive trials until the charaderistics of the organism in culture and chemical profiles (i.e. the operations in 3),4) and 5)) have been carried out. 7) A sales licence can only be applied for when extensive feeding and toxicity trials have been carried out. development, usually in shakeflask culture. 8) There is no point marketing a product until you have a sales licence. However you need to market the product at pilot-scale, to establish the market before you bep full-scale production. 9) Full-scale production is the last step. 4.1 4 1) The most significant production cost in SCP production is the cost of raw materials, ranging from 55-7796 of the total production cost. See Table 4.9. 2) The substrate contributes least to production cost when it is a waste. See Table 4.9 - Sulphite waste liquor is the waste and contributes only 17% to total production cost. 3) For liquid substrates the most significant equipment cost is fermentation, ranging from 43 to 51% of the total equipment costs. See Table 4.10. 4) The most significant running cost is fermentation, ranging from 53 to 77% of the total running costs. See Table 4.11. 5) The most significant running cost of the fermentation is aeration, ranging from 7l to 92% of fermentation running costs. See Table 4.12. 6) The most sigruficant cost of fermentation equipment is usually the bioreactor. However, for fungal SCP processes from sulphite waste liquor, the cooling system is the most signifmnt cost. See Table 4.13. 7) The most sigruficant cost of harvesting equipment is drytng. See Table 4.14. 8) The most significant running cost of harvesting and drying is drying. See Table 4.15. 354 Responses 4.1 5 The cost of soya protein is $0.29 kg-’ (from Table 4.7). The cost of yeasts from n-alkanes is $0.42 kg (from Table 4.7). By comparison to yeasts from n-alkanes, bacteria from methanol cost 0.98 in proqortion (Table 4.81, and so the cost of bacterial SCP must cost 0.42 x 0.98 = $0.41 kg -. At 60% protein, the cost of bacterial protein is 0.41 x 100/60 = $0.68 kg’ Methanol as a substrate contributes 47% of the total production cost of bacterial SCP (and therefore of bacterial protein) (Table 4.9). Therefore of the $0.68 kg that bacterid protein costs 0.68 x 0.47 = $0.32 kg’ is accounted for by the cost of methanol, and 0.68 x 0.53 = $0.36 kg is accounted for by the remaining production costs. If methanol costs are discounted altogether, the remaining production cost of bacterial protein is still $0.36 kg-’, ie more than the cost of soya protein. At these levels the particular SCP process in question is clearly not competitive. 4.1 6 Yeast from n-alkanes has a production cost of $0.42 kg’ (Table 4.7). 1) Aeration comprises 88.9% of the running costs of fermentation (Table 4.12). Running costs of fermentation comprise 67.9% of the Total running costs (Table 4.1 1). Running costs comprise 23.8% of the total production cost (Table 4.9). Aeration thus contributes 0.42 x 88.9/100 x 67.9/100 x 23.8/100 = $0.060 per kg biomass 2) Cooling comprises 7.9% of running cost of fermentation (Table 4.12). Cooling thus contributes 0.42 x 7.9/100 x 67.9/100 x 23.8/100=$0.005 per kg biomass. 3) Ammonia comprises 11.1% of total production costs (Table 4.9). Ammonia thus contributes 0.42 x 11.1/100 = $0.046 per kg biomass. 4) Labour comprises 8.4% of total production costs (Table 4.9). Labour thus contributes 0.42 x 8.4/100 = $0.035 per kg biomass. 5) Drying comprises 77.7% of the running costs of harvesting and drying (Table 4.15). Harvesting and drying costs comprise 5.9% and 21% respectively (=26.9%) of total running costs (Table 4.11). Running costs reprrsent 23.8% of total production costs (Table 4.9). Drylng thus contributes 0.42 x 77.7/100 x 26.9/100 x 23.8/100 = $0.U20 per kg biomass. Aeration contributes most to the production cost of the yeast, at $0.060 per kg. Cooling comprises least, at $0.005 per kg. 4.1 7 To produce 1 kg Candida sp. would require 1/0.48 = 2.08 kg substrate. At $0.2 kg-’ this represents a cost of $0.417 per kg biomass. With the additional fixed cost of $0.205, the total fixed cost of producing Candida sp. would be 0.417 + 0.205 = $0.622 per kg biomass. Responses to SAQs 355 To produce lkg Fusarium sp. would require 1 /0.45 = 2.22 kg substrate. At $0.2 kg -' this represents a cost of $0.444 per kg biomass. With the additional fixed cost of $0.205, the total fixed cost of producing Fusarium sp. would be 0.444 + 0.205 = $0.649 per kg biomass. 1) To produce Candida sp. as feed, the cheapest option would be non-aseptic fermentation, centrifugation and drying. These add only $0.005 and 0.02 to the cost, for centrifugation and drymg. The cost of feed from Candida sp. is thus 0.622 + 0.005 + 0.02 = $0.647 per kg biomass or = 0.647 x 100/60 = $1.078 per kg protein. 2) To produce Candida sp. as food additive the above process could be operated, except that fermentation would have to be aseptic (with sterilisation costs). The cost would be 0.647 + 0.04 = $0.687 per kg biomass or $1.145 per kg protein. 3) To produce Fusarium sp. as feed, the cheapest option would be non-aseptic fermentation, filtration and drying. The cost is thus 0.649 + 0.001 + 0.02 = $0.670 per kg biomass. or = 0.670 x 100/45 = $1.489 per kg protein. 4) To produce Fusarium sp. as high-protein food additive, the above system could be used, with aseptic fermentation and grinding the product after drying. The cost is thus 0.670 + 0.04 + 0.01 = $0.720 per kg biomass. or 0.72 x 100/45 = $1.600 per kg protein. 5) To produce Fusarium sp. for meat substitutes the aseptic system must be operated with recovery by filtration followed by deep-freezing. The cost is thus 0.649 + 0.04 + 0.001 + 0.04 = $0.730 per kg biomass. With competing feeds at $0.80 per kg protein, and Cdz& sp. and Fusarium sp. at $1.078 and $1.489 per kg protein respectively, neither SCP process is profitable for producing feed. With high-protein food additives at $1.55 per kg protein the process producing Candida sp. at $1.45 per kg protein would make a small profit of $0.10 per kg protein (or 0.1 x 60/100 = $0.06 per kg biomass). The process producing Fusmium sp. at $1.600 perkg protein would not be profitable. With meat substitutes at $1.05 per kg, and Fusm'um sp. at $0.730 per kg biomass, this product would make a profit of 1.050 - 0.730 = $0.32 per kg. This would be the most profitable product. 356 Responses Responses to Chapter 5 SAQs 5.1 raw material 4 4 medium preparation medium I sterilisation roduction 4 ioreactor separation I i cell biomass effluent and waste disposal stored stock culture P"mW cuture seed vessel m C rn 0 Q .- $ e !! 5 c rn Q. 3 produd ' extraction I cell free supernatant I .- P 8 5 g rn e !! Q c rn C U final ure prosuct - purification Responses to SAQs 357 5.2 pyruvate v acetvl CaA - oxaloacetate citrate malate isocitrate aconitase lowered activity due to lack of ferrous ions isocitrate dehydrogenase very low activity; inhibited by citric acid fumarate a-oxog lutarate a-oxoqlutarate dehydrogenase totally inhibited by glucose and ammonium; inactive in these cells T I succinate succinyl CoA ~ 5.3 The missing words in the paragraph are: a) ammonium. b) manganese. c) ATP. d) constitutive. e) citrate. r) isocitrate. 358 Responses 5.4 balanced growth atric acid accumulation Inhibits Stimulate Inhibits StimnlakS ATP ADP Pyruvate kinase glucose& phospate ATP Citrate Pyruvate carboxylase GTP AMP Pyruvate decarboxylase Citrate synthase ATP Citrate Isocitrate dehydrogenase ADP AMP Citrate 5.5 The medium must contain: 0 0 a suitable carbon source in high concentration; a nitrogen source, preferably ammonia; Aeration is mandatory and must be constant. It is essential to produce and maintain highly acidic conditions with a pH of 2.0 or below to minimise potential problems caused by contamination. low levels or the absence of trace metals, particularly manganese and ferrous ions. 5.6 Statement 1, the submerged culture process. Statement 2, the Koji process. Statement 3, submerged culture process. Statement 4, the submerged culture process. Statement 5, the Koji process. Statement 6, the surface culture process. Statement 7, the submerged culture process. Responses to SAQs 359 5.7 The overall reaction for this cycle is: 2 acetyl CoA + succinate + 2 GASH Using this as our starting point and then taking another set of reactions around the pathway starting from oxaloacetate and acetyl CoA we have: 2 acetyl CoA + succinate + 2 GASH SUCCiMte + OXaloacetate oxaloacetate + acetyl GA + citrate + GASH citrate + isocitrate isocitrate + succinate + glyoxylate glyoxylate + acetyl CoA + malate + Co ASH su~ci~te + malate 4 acetyl CoA + 2 malate + 4 GASH There are in fact several reasons, the main ones being: 1) The presence of a condensing enzyme which combines pyruvate and acetyl CoA to 2) Conversion of itaconic acid via citraconic acid and itatartaric acid has been 3) Copper ions inmase the yield of itaconic acid. 5.8 yield citramalic acid has been confirmed. demonstrated. 5.9 1) True. 2) True. 3) False. The reverse is true, because sodium gluconate is far more soluble than calcium gl~mnate high glucose amcentrations can be used to produce higher production yields of sodium gluconate. 360 Responses 4) False. D-gluconolactone is produced directly from glucose via glucose oxidase. 6-phosphogluconolactone is an intermediate in the hexose monophosphate pathway. 5) True. 6) False. Glucose oxidase is specific for 0-D-glucose. 5.10 fermentation mixture filtration mycelium and suspended solids disposal to 7.5 with sodium hydroxide active carbon treatment drum dried active carbon treatment filtrate Pq concentrated ------_ to 45% solids pH evaporation (20% soh) cooling /\ calcium crystals Iconate of T centrifugation washing in exchange dried at 80°C filtration calcium aqueous 7 solution sulphate of gluconic acid gluconic Kl dried crystals of calcium gluconate add H2S0, seed with required crystal A crystallise at <30°C 30-75OC >70°C 61 h 61 Responses to SA& 361 Responses to Chapter 6 SAQs 6.1 Both penicillin G and amoxycillin confurm to the general structure. In the case of penicillin G, R = whilst with amoxycillin, R = When written in this foxm, it should be apparent that penicillins can be diversified by substituting different groups in place of R. The penicillin class of antibiotics does, in fact, contain members that have different R groups. We will examine these later in the chapter and describe the strategies used to generate penicillins with different Substitutions. 1) Penicillin G production can be encouraged by adding &phenylacetic acid to the 2) The obvious addition to produce penicillin X is pOH&phenyladc acid. 6.2 medium. This was described in the text and will not be elaborated upon here. - This could be derived from tyrosine via the amine derivative, in a process similar to the production of &phenylacetic acid in CSL. 3) We would expect you to suggest the addition of: HOOC - C& - CH = CH - C& - U+ to produce penicillin F. 362 Responses 4) Similarly, we would anticipate that the addition of HOOC - CH2 - (05 - cH3 would 5) Penicillin V was mentioned in the text. The addition of phenoxyacetic acid lead to the production of penicillin K. encourages the production of this penicillin. 6) The compound we would anticipate you would suggest to produce penicillin N is /Pmnfig uration HOOC - CH2 - (CH2)3 - CH - NH2 I COOH This is Da-aminoadipic acid. You will learn a little later in the chapter that the normal biosynthetic pathway of pencillin production involves the incorporation of L-a aminoadipic acid. The product is called isopenicillin N. Its side chain has the structure: JL-mnfiguration - C - (CH2)4 - CH - NH2 It I 0 COOH Conversion to penicillin N is achieved by an epimerase that converts the L configuration amino group into the D configuration. Thus, in the case of penicillin N, we may not have to add a special precursor providing an active epimerase was present. With glucose as substrate, substrate consumption is fast causing rapid growth and a depression of the pH. Penicillin production only begms towards the end of the growth phase. At this stage the pH is below the optimum (pH 657.0) for penicillin production. Thus penicillin production only proceeds for a short period and at a slow rate. Yields are, therefore, low. Note also that the biomass is unstable and, in the absence of external carbohydrate, begins to lyse. With X as a substrate, substrate consumption is quite slow. Growth is slower on this substrate and even when maximum biomass levels are reached, there is still substantial amounts of substrate left. Thus the cells are able to utilise this to maintain themselves and also to support penicillin production. Because of this there is no violent swing in pH towards the acid end during the growth phase, nor to the alkaline end because of cell lysis. In this case the pH is maintained within the optimum range for penicillin production. Penicillin production is therefore extended, resulting in greater yields. Carbohydrates which give this type of result include lactose. 6.3 Responses to SAQS 363 6.4 No, the enzyme IPN-acyltransferase is not highly specific. We learnt in section 6.3 that a range of penicillins auld be produced by incubating penicillin-producing cells in the presence of analogues of acetic acid. Inclusion of gphenoxyacetic acid in the media leads to the production of penicillin V, while inclusion of gphenylacetic acid leads to the production of penicillin G. The implication is that IPN-acyltransferase will add a range of acyl groups onto the amino penicillinic acid moiety. If you reexamine Table 6.2 you will see that the transferase appeaxx to be capable of using a wide range of acyl groups- This was a fairly open ended question. The advantages we hoped you would identify 6.5 Were: 0 the biocatalytic process is carried out in water, whereas the chemical deacylation techniques require halogenated solvents. This has environmental ansequenm. Obviously using water is cheaper, safer and less polluting than using halogenated organic solvents; the biocatalytic process yields acyl residues (for example phenylacetic acid) which may be reused as precursors for p-lactam production; the biocatalytic process is cheaper if the enzyme can be used many times; the biocatalytic process is specific and yields few by-products; the desired product is easy to separate from the biocatalytic reaction mixture. With the chemical catalytic process several other components (PQ5; (CH3)3 Sia; ROW may also be present. 0 0 0 0 6.6 1) a) Yes, the 6-0-methyldeoxystreptamine would be incorporated in place of 2 deoxystreptamine. Its structure would be: 364 Responses b) Yes, the 3-N-methyl4eoxyskptamine would be incorporated to form c) No, the methyl group at position 5 would block addition of the 2) 5-0-methylstreptomycin into the neomycin moiety. The assumptions we have made in coming to our conclusions in 1) are that: 0 0 the compounds are taken up by the cells; the enzymes responsible for adding the streptamine derivatives to the rest of the neomycin will work with a range of substrates and can work providing the hydroxyl groups in position 4 and 5 are available; the cells do not further metabolise the added analogues prior to their incorporation into the neomycin product. 0 Responses to SAQs 365 Responses to Chapter 7 SAQs 7.1 1) False. D-xylose is a pentose sugar which are very rarely found in microbial 2) False. Pyruvate ketals contribute to the anionic nature of exopolysaccharides. 3) True. 4) False. However, it is true that only relatively few yeasts and filamentous fun@ exopolysaccharides. produce exopolysaccharides. 5) True. Uronic acids contribute to the anionic nature of exopolysaccharides. 7.2 7.3 1) Glucose: source of carbon and energy for growth and exopolysaccharide production. Succinic acid: source of carbon and energy; improves metabolic balance between carbon flux from glucose and oxidation throgh TCA cycle. Ammonium chloride: source of nitrogen (required for synthesis of proteins, nucleic acids and co-enzymes). Yeast extrack source of growth factors (amino acids, vitamins, nucleotides). Potassium: the principal inorganic cation; cofactor for some enzymes. Phosphate: constituent of nucleic acids, phospholipids and coenzymes. Magnesium sulphate: magnesium is an important cellular cation; inorganic cofactor for many enzymatic reactions, including those involving ATP; functions in binding enzymes to substrate. Iron: constituent of cytochromes and other heme or nonheme protreins; cofactor for a number of enzymes. 366 Responses Calcium: important cellular cation; cofactor for some enzymes; important in enzyme Trace elements: inorganic constituents of special enzymes. 2) Biomass concentration = 10 x 2.5 = 25 g 1". 3) Biomass concentration = 25 g 1-' (ammonia is the limiting nutrient). Stability. 7.4 1) Xanthan is synthesised during exponential growth of the culture, whereas succinoglycan is synthesised after growth has ceased (biomass constant). 2) Since succinoglycan synthesis commences after the growth phase there would be no carbon available for succinoglycan formation. 3) From Figure 3.6 since growth and xanthan synthesis ceases simultaneously, we can deduce that the limiting nutrient is glucose. The rapid decrease in glucose concentration during the growth phase also suggests this. 4) Glucose limits succinoglycan production (ammonia limits growth). It is necessary to limit succinoglycan production to avoid an excessively viscous fermentation liquor which would affect mixing and subsequent downstream pn>cessing to mover exopoly saccharide. 2.5 - osg 1-' 140 m mole I-' Y--, = biomass produced/ammonia consumed = 5) = 0.0121 g m mole-' = 12.1 g mole-' = 205.7 g g" 7.5 An appropriate order is: Addition of cations Heat treatment Addition of solvent Distillation Centrifugation Vacuum drying Propylene oxide treatment Sterile nitrogen gas treatment Milling cool dry storage Aseptic packaging 7.6 1) The missing words are: a) pseudoplastic; Responses to SAQs 367 b) viscoelasticity; c) ordered, melting d) above. 2) Alginates with a high content of L-guluronic acid have a relatively high capacity for cross-linking, involving the binding of divalent cations, and thus form brittle gels. D-mannuronic acid does not bind divalent cations, so the gels are more flexible (less cross linlung). 7.7 7.0 The missing words, in order, are: intracellularly, bisphosphates, energy, lipid, subunits, enzyme, homopolysaccharides. Deficiency in gumD will mean that synthesis is not started and there will be no label incorporated either from the UDP[14C] glucose or the GDP["CI mannw. A deficiency in gumM will @ve labelled lipid carrier if glucose is used but not if mannose is the labelled sugar nucleotide. A mutant with a non-functional gumK gene will incorporate label into the lipid and soluble fractions from UDP[14Cl glucose and GDP["CI mannose, producing the polytrimeric polysaccharide lacking the terminal mannose. The genes responsible for encoding the proteins specifically involved in exopolysaccharide synthesis are clustered in one large operon. The genes encoding the proteins for sugar nucleotide phosphates, which are not necessarily specifically used for exopolysaccharide synthesis, also tend to be clustered. 7.9 7.1 0 7.1 1 A, sensor B, effectorprotein C, degradingenzyme D, activator protein E, activated complex F, catabolite activator protein G, pathogenicity regulating genes H, exopolysaccharide gene cassette I, pathogenicity genes 7.1 2 Fxopolysaccharide Application Important Physical Properties welan enhanced oil high viscosity over broad tem erature range; stable to 149 c. CP emulsan dextrans curdlan cleaning agent stabilises hydrocarbon/water emulsions at low concentrations; low toxicity. plasma substitute equivalent viscosity and osmotic properties to blood plasma; non-irritant and non-toxic. food additive retains shape in cooked food and gels at relatively low temperatures (55800C) polyethylene glycol/dextran purification of biological aqueous two-phase with low water mixtures materials interfacial tension alginate gellan immobilising enzymes gel formation under mild conditions use in glazes, jellies and icings heat stable and gives a very clear thermostable gel which sets at lower concentrations and more rapidly than most other polysaccharides; superior flavour release characteristics. Responses to SAQs 369 Responses to Chapter 8 SAQs 8.1 1) Cysteine, cystine, methionine. 2) The eight essential amino acids are listed immediately after Table 8.2. 3) L-Proline and L-tryptophan (see Table 8.2). 4) Glycine (see Table 8.2). 5) Cysteine. 6) L-Methionine (see Table 8.1). 7) L-Cysteine (see Table 8.1). 8.2 Possible advantages of using enantiomerically pure compounds as drugs include: 1) avoids detrimental side effects 2) enables more specific drug action 3) reduces amount of drug administered 4) avoids work involved in metabolism of inactive enantiomer. 8.3 1) Enzymes in the pathway to L-phenylalanine are subject to feedback inhibition by products (amino acids) arising from pathway intermediates. 2) Auxotrophic mutant lack one or more enzymes involved in the synthesis of amino acids (such as tyrosine). This prevents accumulation of the amino acid and thus avoids feedback inhibition of enzymatic steps in the L-phenylalanine pathway. 3) Regulatory mutants are not subject to feedback inhibition, even by L-phenylalanine itself. 4) A Try- mutant would not be subject to feedback inhibition by overproduction of tryptophan. Also, the mutation may allow more chonsmate to proceed to prephenate via E3 (see Figure 8.4) and thus through to L-phenylalanine. 8.4 1) In fed-batch mode residual substrate concentration may be maintained at very low levels. This would reduce substrate costs, may remove catabolic repressive effects and may avoid possible toxic effects of the substrate. The fed-batch mode of operation may also avoid oxygen depletion of the culture during rapid growth. 2) Monoseptic fermentations are easier to characterise and to repeat, compared to fermentation involving more than one micro-organism 370 Responses 3) Addition of antibiotics to the fermentation broth may be used to: avoid problems associated with growth revertants (eg auxotrophic back mutation); ensure that genetic material (eg plasmid DNA) is maintained within the process micmrganism. 4) Use of fresh starting material for each fermentation run may avoid problems of back mutation and loss of genetic material. 5) Addition of chelating agents to the fermentation medium may help to inhibit phage multiplication by prevention of phage adsorption to the cell wall. 8.5 1) Solvent extraction. 2) Crystallisation, ion exchange, electrodialysis. 3) Crystallisation. 4) Decoloration. 5) Evaporation. 8.6 1) From the data 6 g 1-’ glucose is used = 0.0333 mol 1-I. 2, During this period 1.4 g 1” of phenylalanine is produced = - = 0.0085 mol 1-’. 1.4 165 Number of moles of glucose needed to provide the carbon for this phenylalanine is 0.0085 x 2 mol 1-’ = 0.0170 mol le’. During this period 0.3 g 1-’ acetate are produced = 0.005 mol 1”. - 0.005 Number of moles of glucose needed to provide the carbon for this acetate is 7 - 0.0025 mol 1”. Thus, to produce the acetate and phenylalanine, 0.0170 + 0.0025 mol 1-’ glucose are carried = 0.0195 mol 1”. But 0.0333 mol 1” of glucose is consumed. - 0.333 x 100% = (41%) of the consu& glucose is not used This means that to produce phenylalanine and acetic acid but is used for maintenance (note the cells are in stationary phase and thus there is no cell growth over this period). 8.7 Total sales = 500,000 kg x $50 = $25 million a year Net sales = $25 million - 20% = $20 million Gross profit = $20 million - (500,000 kg x 24.5$) = $7.75 million Responses to SAQs 371 Taking into account administration and R and D costs. Profit = $7.75 - (12.5% of �20 million) = $5.25 million Profit for the first year = $5.25 million - (10% of $40 million) = $1 -25 million Net earnings in first year (after taxes) = 50% of $1.25 million = $0.625 million (A in Equation 8.4) = 50% of $5.25 million = $2.625 million (B in Equation 8.4) (C in Equation 8.4) Net earnings per annum in the period to come Working capital = 25% of $20 million = $5 million Total fixed capital = $40 million (D in Equation 8.4). Using Equation 8.4, for the period of investment (15 years) we have: 0.0 1) phenylpyruvic acid: L-amino acid aminotransferase 2) N-acetyl-DL-phenylalanine: acylase 3) acetamidocinnamic acid: acylase L-amino acid aminotransferase 4) phenylpuruvic acid: L-amino acid dehydrogenase 5) trm-cinnamic acid: phenylalanine-ammonia-lyase 6) phenyllactic acid: L-hydroxy acid dehydrogenase L-amino acid dehydrogenase 8.9 1) D- or L-phenylalanine induces the enzyme phenylalanine aminoaminotransferase, which is required for L-phenylalanine synthesis. 2) An alkaline pH (- pH 11) is desirable in order to: achieve high conversion rates; inmase solubility of L-phenylalanine; inhibit enzymes catalysing degradation of L-phenylalanine and formation of byproducts; reduce inhibition of the reaction by the keto form of phenylpyruvic acid. 3) Concentrations of 4% phenylpyruvic acid leads to end product inhibition by the L-phenylalanine produced, resulting in lower final yields. 4) Cell immobilisation is relatively costly because of the cost of the immobilisation procedure and of supplying cofactor. 372 Responses 8.1 0 Cost price of L-phenylalanine by direct fermentation is 28m kg-’. From Table 8.8: 1) Cost price of L-phenylalanine by precursor feeding is 35.05 $ kg-’ and substrate costs are 18.05$ kg-’. It follows that for precursor feeding to be competitive, substrate costs would have to be reduced by: 35.05 - 28.50 = 655$ kg-’. So, the percentage reduction in substrate costs would be: - 6S5 x100=36%. 18.05 2) A similar approach for bioconversion, based on the date given in Table 8.8, shows that the percentage reduction in substrate costs required is 11%. (30.22 - 28.5$ kg-1 = 1.m kg-1 reduction is required. Substrates cost 15.- kg-1. Thus % reductase is x 100 = 11%). Responses to SA@ 373 Responses to Chapter 9 SAQs 9.1 a) 1) The organism would probably grow quite well. The structure shown is simply analogous to the acetyl ester of cholesterol except that the side chain has been removed. The organism would, presumeably, degrade the sterol nucleus by its normal route. 2) The organism would probably only show slight growth. It would be able to degrade the side chain in the normal way, but the hydroxyl group in position 19 blocks the metabolism of the sterol nucleus. Thus only a small portion of the molecule can be catabolised. 3) The modification to positions 19,5 and 6 prevents metabolism of the ring structure, thus the organism would not be able to utilise this substrate (no growth). There is no side chain to catabolise. 4) The organism would not be able to grow on this substrate as the hydroxyl in position 19 blocks metabolism of the ring structure (see 2). There is no side chain to catabolise. b) The likely metabolic products from compound 1 are COZ and HD, as this organism can completely metabolise this substrate. Incubation with compound 2 is likely to lead to removal of the side chain, so a mapr product might be: However, there may be some further, minor modifications to this basic structure. Compounds 3 and 4 are likely to remain unaltered. The isolate appears to produce the enzymes for the complete catabolism of lithacholic acid. However, in the presence of Pb” ions, some of these catabolic enzymes are inhibited, leading to the accumulation of partial breakdown products. It appears that enzymes involved in catabolism of the ring structure are more susceptable to inhibition by Pb2’ ions than are the enzymes involved in side chain catabolism. 9.2 9.3 1) hydrolysis (ester hydrolysis) 2) oxidation (alcohol oxidation) 3) isomerisation - the double bond has been moved from position 56 to position 4-5. We would normally denote such an isomerisation by As ~ > A4. 4) hydroxylation - a hydroxyl group is added to position 11. 374 Responses 9.4 1) You should have identified this reaction as an lla-hydroxylation. The most commonly used organism to carry out lla-hydrolylations is lUnwpus nigmns. The substrate in this case is p”geSter0ne. 2) This is an ll&hydroxylation, usually carried out using Cumlmia lunata The substrate in this case is known as substrate S. 3) This is a 16a-hydroxylation, usually carried out using Streptatyces rosaochramogenes. 1) This is a hydroxylation. It may also be regarded as an oxidation. 2) This is also a hydroxylation. 3) This is an epoxidation. 4) This is a reduction in which the aldehyde group is reduced to an alcohol. The way to achieve resolution is to use the lipase to selectively esterify the L-menthol. Thus: 9.5 9.6 DL-menthol Lmenthyl-5-phenyhmlera@ Dmenthol We can, therefore, isolate D-menthol from the mixture. The L-menthol could be recovered by hydrolysing the L-menthyl-5-phenylvalerate. You might anticipate using an esterase to carry out this hydrolysis such as the enzyme from Rhodotomla minuta described in the text. This is a fairly open-ended question and there are many options available. Here we will give some illustrative examples. 9.7 Responses to SAQs 375 First, the chemical difference between the ring structures of PGE and PGF prostaglandins is that in PGE there is a keto group at position 9, whilst in PGF this position is taken up by a hydroxyl (secondary alcohol) group. The conversion of a ketone to a secondary alcohol is, biochemically, fairly straightforward involving a hydrogenation. Hydrogenases tend to be stereospecific. Thus, incubating portions of PGEl with appropriate hydrogenating enzymes would yield samples of either PGFa or PGFB. An altemative approach would be to chemically reduce the keto group at position 9, forming a racemic mixture of PGFa and PGFB. These could then be resolved enzymatically by, for example, esterfymg the 9-hydroxy and using speciiic esterases to selectively deesterify PGFa or PGFB.