6. 발효공학

1. 6. 발효공학

2. 대표적인 식품발효 미생물 • 식초: 세균 (Acetobacter aceti) • 간장: 곰팡이 (Aspergillus oryzae, A. sojae) • 빵, 주류: 효모 (Saccharomyces cerevisiae) • 치즈: 곰팡이 (Penicillium rogueforti) • 유산균 음료: 세균 (Lactobacillus bulgaricus) • L. acidophiluc) • 김치: 세균 (Lactobacillus plantarum, Leuconostoc mesenteroides) • 청국장: 세균 (Bacillus natto)

3. 미생물의 배양 • 기본 발효공정 • 배지 조제 • 설비 살균 • 종균 준비 • 균의 증식 • 생산물의 추출과 정제 • 발효폐기물의 처리

4. (2) 발효공정 종류 균체 생산 (yeast, SCP) 효소 생산 대사산물 (metabolites) 생산 - primary metabolites during trophophase - secondary metabolites during stationary phase (idiophase) transformation or modification of precursors

5. (3) 미생물 배양법  배지 형태 - 액체배지 표면배양 (surface culture, standing culture) 심부배양 (submerged culture) - 고체배지: agar 농산물 or 그 폐기물 (쌀, 보리, 밀, 콩) Baffle

6. C C C A A0 A Q Q 0 0 V  기질공급 방법 - 회분식 배양 (batch culture) 초기 배지 공급 closed system culture condition: not fixed - 연속배양 (continuous culture) 계속적인 배지 공급 & 배양액 제거 open system chemostat, turbidostat 농도 CA 체적 V ( 일정 )

7. F (S0) Vt V0 X, S, P  기질공급 방법 - 유가식 배양 (fed-batch culture) 간헐적 배지 공급 culture condition: not fixed

8. 다음 중 산소공급이 가장 잘 될 것 같은 조건은? Wider flask Use of baffled flasks Higher stirrer speed Larger liquid volume    

9. Exit air port Agitator inlet for nutrients Drive motor Mechanical seal baffle probe Shaft Inlet air port sparger impeller Impeller baffles fill/drainline (4) 발효조  통기교반형 발효조 (stirred tank reactor)

10. For liquid volumes greater than 3 liters, air sparging is required for effective oxygen transfer. Agitation is used to break up bubbles and thus further increase air/liquid surface area. 300 rpm 450 rpm 750 rpm

11. Airlift fermenter with external draft tube Airlift fermenter with internal draft tube Bubble column better • better mass and heat transfer • more uniform shear conditions  기포탑형 발효조 (airlift fermenter) culture of shear sensitive organisms such as moulds and plant cells Disengagement zone • add volume to the reactor, • reduce foaming and • minimize recirculation of • bubbles • removal of excess CO2

12. Major disadvantages of airlift fermenters over stirred tank bioreactors • comparatively higher energy requirements • higher levels of foam formation • cell damage, particularly with animal cell cultures • (due to the shear forces which arise bubble at the surface burst)

13. 다음 기포형 반응기에 관한 설명 중 가장 옳지 않은 것은? They are designed to be tall to encourage low shear conditions The absence of an impeller reduces shear conditions The presence of a draft tube in airlift reactors increases mass and heat transfer rates The large height to diameter ratio increases oxygen transfer rates    

14. draft tube의 역할은 increase mixing efficiency reduce bubble coalescence increase the saturation concentration of oxygen equalize shear throughout the reactor    

15. Disengagement zone effluent Small particles containing cells or enzymes Gravity causes rising particles to fall balanced (Air) Feed Energy from feed and air causes particles to rise  유동층 발효조 (fluidized fermenter) • high biomass concentrations • good mass transfer rates

16. - small and light particles: easy to flow with the liquid - 2 phase system: no aeration which are not aerated 3 phase system: aerated by sparging - used widely in wastewater treatment - used for animal cell culture : animal cells are trapped in gels or on the surface of special particles known as "microcarriers" (example of perfusion culture technology)

17. Large non-moving particles  고정형 발효조 (fixed bed reactor) - Cells: immobilized by absorption on or entrapment in solid, non-moving solid surfaces. - Liquid feed: pumped through or allowed to trickle over the surface of the solids - Industrial applications waste water treatment production of enzymes and amino acids steroid transformations

18. 2. 미생물의 생육속도론 • product 생산을 위해 미생물 성장에 대한 이해 필요 • 미생물의 생장 • 환경에 대한 반응  복제, 세포 크기의 변화 • substrate + cells  extracellular product + cells •  S + X   P + nX • 미생물생장과 효소반응과의 유사성 • X; 자가 촉매 반응 (효소 역할)

19. (2) 세포 농도의 정량 • 미생물 생장과 관련된 반응속도와 양론적 관계 결정에 필수적 • 세포 수, 밀도 결정 • - 직접계수; Petroff-Hausser, Hemocytometer • - viable count • - particle counter • 세포 질량, 농도 결정 • - 직접법; 건조중량, turbidity • - 간접법; DNA, protein, LPS, cell volume, ATP, chitin

20. (3) 회분식 생장      

21.  Lag phase - adaptation - stimulation of synthesis of biomolecules - depends on X0, trace element, growth condition status of inoculum  Log or exponential phase - balanced growth doubling time (td) : 일정

22. d[P] d[X] d[X] dt dt dt 1 X  S + X   P + nX - X; 자가촉매 - 단위: g-dry wt./l - 효소와 차이점: [X]; dependent on time In enz reaction, v = = k2[ES] In growth, growth rate = = •X ; exponential growth model • X = X0 at t = 0 • = specific growth rate (h-1) =

23.  =  • X = ; 양변 적분 lnX – lnX0 =  • t X = X0 • e  t Doubling time, td lnX – lnX0 =  • t lnX – lnX0 = lnX/X0 = ln2 =  • td td = ln2/  d[X] d[X] dt dt 1 X

24. 1. exponential phase에서 미생물 수가 매 2시간마다 2배로 된다. 이기간 중 specific growth rate은? 2. 한 세균의specific growth rate이 0.5 h-1일 때, generation time은? 0.347 h-1 1.386 h

25. lnX Time (h) 한 세균이 30분마다 분열한다. 여러분이 100 cells/ml로 실험을 시작하였다면 다음과 같이 시간이 경과했을 때 반응액에 존재하는 세균 수는 얼마나 될까? X = X0 • e  t Time (h) Number of cells td = ln2/  = 0.5 h 1599 25569 408877 6538185 104549393  = 1.38 h-1 2 4 6 8 10

26.  deceleration phase 필수 영양소, space 고갈, 유해산물 축적 대사조건, 속도 조절, 구조 변화  stationary phase net growth rate = 0 growth rate = death rate 2차 대사물 생산 저장물질  단량체  에너지원으로 사용 (endogenous metabolism)

27. d[N] dt  death phase - = kd´• N N: population number at time t kd ´: death rate constant Ns: population number at stationary phase - dN = kd´• dt ; 양변 적분 1 N lnNs – lnN = kd´ • t N = Ns •e- kd´• t

28. 1 N d[X] d[N] dt dt X S 1 X • (4) Growth parameters  , specific growth rate (h-1) =  td, doubling time = ln2/   kd ´, death rate constant = - •  Yield coefficient based on product formation • Yx/s = - ; 겉 보기 생장수율 S = Sassimilation into biomass + Sassimilation into extracellular products Sgrowth E + S maintenanceE

29. X P P O2 S S • Oxygen yield coefficient (Yx/O2) • Yx/O2 = - •  Yield coefficient based on product formation • YP/S = - • YP/X = -

30. glucose 20 fructose 15 10 Bacterial dry weight (g/ml) sucrose 5 0 0 4 8 12 Substrate concentration (mM) 미생물을 이용한 배양 결과 다음과 같은 결과를 얻었다. 1g Maltose => 0.6 g Lactate + 0.2 g Acetate + 0.2 g of cellsYP/S (yield of acetate from maltose), Yx/s (yield of cell from maltose) 0.2 g.g-1 0.6 g.g-1 0.1 g.g-1 0.05 g.g-1    

31. 1 M  X dt m dS  유지계수 (maintenance coefficient, m) 세포기능 유지 (손상부위 수리, 운반, 운동, 삼투압 조절…)를 위한 세포단위 질량 당 기질 섭취 속도를 나타냄 specific rate of substrate uptake for cellular maintenance

32. Time Concentration Time dP d[P] dP dX 1 X dt dx dx dt 1  YP/X X  미생물 산물 관련 계수 a. growth-associated product • the rate of product formation is proportional to the rate of biomass increase • When growth ceases, product formation ceases. qp (specific rate of product formation)   qp = = = 

33. Concentration Time dP 1 X dt  미생물 산물 관련 계수 b. nongrowth-associated product • disproportionate to the cell growth rate • product formation will continue to occur after growth ceases. qp = =  Why ?

34. Time Concentration Time  미생물 산물 관련 계수 c. mixed growth-associated product qp =   +  If  = 0, qp = . If  = 0,  = YP/X

35. Thermal death activation growth d[X] d[X] dt dt Temperature (5) 온도가 성장에 미치는 영향 The Arrhenius equation  energy of activation  • X = -Ea/RT  = A •e  energy of inactivation = - kd ´ • X -Ed/RT Ad ´•e kd ´ =

36.  is not dependent on DO. DO limits growth  DO Critical oxygen concentration (6) DO가 성장에 미치는 영향 oxygen: serve as a substrate in aerobic metabolism has limited water solubility (8.84 ppm at 20oC, 1 atm) To get maximum growth, DO > critical conc. bac, yeast; 포화농도의 5-10% fungi; 포화농도의 10 – 50%

37. Organism Temperature(oC) CO,cr (mM O2)Azotobacter vinelandii 30 0.018 Aspergillus oryzae 30 0.020 Escherichia coli 37 0.008 Penicillium chrysogenum 30 0.009 Pseudomonas denitrificans 30 0.009 Pseudomonas ovalis 30 0.034 Serratia marcescens 31 0.015 Saccharomyces cerevisiae 30 0.004 Torula utilis 30 0.063

38. 다음 중 옳은 것은? Oxygen is a growth limiting substrate when the dissolved oxygen concentration is less than Co,crit The critical oxygen concentration is equal to the saturation concentration of oxygen. Oxygen is a growth limiting substate when the dissolved oxygen concentration is less than Co* Oxygen becomes a growth limiting nutrient when the dissolved oxygen concentration falls below 8ppm.    

39. Phase I Phase II O2 rich bubble

40. 1. Phase I air bubble  liquid oxygen transfer rate (OTR, mg-O2/L • h) = kL• a (C* - CL) kL : oxygen transfer coefficient (m/h) a : air-liquid interfacial area (cm2/cm3) kL• a : volumetric oxygen transfer coefficient (h-1) C*: saturation DO concentration (mg/L) CL: DO in the medium (mg/L) O2

41. dO2 dO2 dX 1 1 X X dX dt dt X 1 Yx/O2 Yx/O2 2. Phase II liquid  cells oxygen uptake rate (OUR) = qO2 • X = • X = • X =  • O2  qO2 = specific rate of O2 uptake

42. O2 growth + maintenance OUR = qO2 • X + mO2 • X = + mO2 • X mO2= m dO2 1  • X  • X X dt Yx/O2 Yx/O2 When OTR limits the growth, it is possible to ignore mO2 . OTR = OUR kL• a (C* - CL) = ;  • X = Yx/O2 • kL• a (C* - CL)

43. Eschericia coli를 다음과 같은 조건에서 배양할 때 성장은 산소에 의해 제한됨을 알 수 있었다. 이 때 배양액 내의 세포 농도를 구하라. Yx/O21.1 mg cells/mg O2 CL 0.256 mg O2/L kLa 121 h-1 C* 8 mg/L  0.651 h-1 1583 mg cells/L

44. Calculate value for the yield (Yx/O2) and maintenance (mO2) parameters from the following data: Hint: mO2 is negligible during the exponential phase and Yx/O2 is negligible during the stationary phase.

45. O2 growth + maintenance OUR = qO2 • X + mO2 • X = + mO2 • X LnX y = 0.3262x - 1.2193 y = 0.1598x + 0.0277 OUR  • X Yx/O2 Time (h) X

46. (7) 증식과정의 정량화 속도론적 해석을 통한 성장의 kinetic study a. 회분식 배양기 가정  세포 구조적 측면 세포조성 불변 가정 균형생장 가정 (exponential phase)  기질 제한 성장 가정 : 단 하나의 화학물질 S에 의한 성장 제한 (S의 대사에 관여하는 enzyme system에 의해 전체 생장 결정)

47.  m m• S  = Ks + S [S] Limiting factor; S Enz: M-M kinetics Chemical rxn: Langmir-Hinshelwood kinetics Growth: Monod kinetics µm and Ks are dependent on - the organism, - the growth limiting nutrient, - fermentation medium, - environmental factors µm; 0.01 - 3 h-1 Ks; typically less than 0.1 g.l-1 m : maximum growth rate Ks : saturation constant half velocity constant Monod constant

48. 어느 세균이 glucose를 탄소원으로 하여 성장할 때 다음과 같은 growth parameter 를 얻었다. µm = 0.65 h-1 Ks = 0.05 g.l-1 glucose의 농도가 다음과 같을 때 이 세균의 specific growth rate은? [glucose] (g.l-1) µ (h-1) 0.01 0.1 0.5 1.0 20 90 0.108 0.433 0.591 0.619 0.645 0.65

49. Headspace volume Working Volume (VR) b. 연속배양에서의 세포성장 Culture medium Fresh medium at time t; Steady state 도달 - constant X, S, P volume taken up by the medium, microbes, and gas bubbles 70-80% of the total volume

50. 연속배양의 장점 • Cells at a constant physiological state and growth rate • Shorter turnaround time • No segregation into growth phases • Easier to optimize productivity • Smaller reactors and lower capital cost • 연속배양의 단점 • Large risk • use of continuous cultures to produce many therapeutic products is not accepted by the US FDA • Many important products are released from different phases of batch culture growth • contamination