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Tema 2: Growth and cell division Paper asignado: MreB.pdf.

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Presentation on theme: "Tema 2: Growth and cell division Paper asignado: MreB.pdf."— Presentation transcript:

1 Tema 2: Growth and cell division Paper asignado: MreB.pdf

2 Growth is an increase in mass Total and viable cell counts Dry weight Protein Can be measured by: Turbidity

3 I / I o =10 xl Beer-Lambert law Si tomamos el logaritmo de: Entonces Log (I o / I)= -xl Turbidity, absorbance, optical density OD = A = xl The amount of light scattered by a bacterial cell is proportional to its mass. Beer-Lambert law I/I o =10 xl Turbidity

4 600nm incident light detected light absorbancia a 600nm aumenta según aumenta el # de células se miden lecturas de absorbancia a 600nm IoIo I X=cell density

5 Total cell counts pozo central (0.02mm profundidad) contiene una rejilla que se divide en cuadrantes microscópicos Cámara de conteo Petroff-Hausser cubre objetos Limitations 1) Are the cells dead or alive? 2) Make right dilution (<10 6 cells/ml)

6 varios cuadros se cuentan se promedian y se multiplica por 25 (# de cuadros en la rejilla) la cifra resultante es el # de células en un volumen que corresponde a las dimensiones de la rejilla: (1mm 2 x 0.02mm) = 0.02mm 3) resultados pueden ser expresados en # de células por ml (cm 3 ) 12 + 16 + 20 + 18 +14 5 = 16 16 x 25 = 400 400 células/ 0.02 mm 3 400 0.02 mm 3 = x 10mm 3 2 x10 5 células /cm 3 = 2 x10 5 células /ml Total cell counts 25 cuadrados subdivididos en 16 unidades pozo central (0.02mm profundidad)

7 Total cell counts Electronic cell count Chamber 1Chamber 2 Measure electrical conductivity Count of cells in the culture and the size distribution of the cells

8 Viable cell counts Limitations 1) must avoid clumps. 2) Some bacteria plate with poor efficiency. 3) It will always depend on the medium used.

9 Dry weight: Is the most direct way to measure growth Drying oven 3 -7 days 1-10ml Sample thru time Culture Results

10 Cell growth by protein quantification Results Culture

11 Crecimiento Poblacional Bacteriano El ciclo de la curva de crecimiento Fase Lag: periodo de aclimatación a condiciones de crecimiento, síntesis de RNA, duplicación DNA Fase Exponencial: número de células se duplica a intérvalos regulares de tiempo, ocurre bajo condiciones ideales de crecimiento (ej.abundancia de nutrientes) Fase Estacionaria: se agotan nutrientes, se acumulan desperdicios dañinos, procesos de división celular y muerte estan en balance Fase de Muerte: las condiciones prevalecientes no pueden sostener más crecimiento y las células mueren

12 Adaptive responses to nutrient limitations 1)Starvation conditions 2)Growth no growth 3)Generation time days or months 4)Depletion of an essential nutrient Responses: 1) PO 4 may induce the high affinity i- PO4 uptake systems and/or enzymes that degrade o- PO 4 (phosphatases). Stationary phaseSporulate

13 Stationary phase 1) Changes is size 5-10 μm to 1-2 nm. 2) Changes in morphology form rod-shape to coccoid-shape 3) Changes in cell surface from hydrophilic to hydrophobic. 4) Formation of fibrils and aggregates 5) Changes in phospholipids from unsaturated to cyclopropane 6) The metabolic rate slow down but increase turnover of protein and RNA. 7) May synthesize 50 to 70 or more new proteins. 8) Cells may become more resistant to environmental stress Temperature, osmotic stress, high salt, ethanol, solvents, pH.

14 1 5 10 Relative amount μ (doublings/ hour) 0.6 1.0 RNA mass protein DNA Figure 2.4

15 Faster growing cells 1)More RNA, more ribosomes (65% RNA) 2)More DNA 3)More mass Why is that?

16 lactose glucose growth Time Diauxic growth Catabolite repression by glucose 1)Repression in synthesis of degrading enzymes for the 2 nd compound 2)Inhibits the uptake of other sugars (Inducer exclusion) [Substrate] OD

17 Catabolite repression In Rhizobium: It grows first in C-4 acids of the TCA cycle then in Glucose In Pseudomonas aeruginosa: It grows first on organic acids then on carbohydrates PEP Glucose Glucose-6-P PTS (phosphotransferase) proteins Adenylate cyclase III Glc Pyr POoperon lactoseBlocked

18 PEP PTS proteins Adenylate cyclase III Glc Pyr POoperon ATP cAMP lactose Activated CAP No Glucose P P P

19 Growth yields Y is the growth yield constant Y= amount of dry weight of cells produced per weight of nutrient used. This is determined when the limitation of the production of cells is controlled by The quantity of a single nutrient that is the sole source of energy and carbon. Y = weight dry cell weight of nutrientYmYm = weight dry cell (g) moles of substrate Molar growth yield constant In aerobic bacteria Y glucose = 0.5Means that 50% of the glucose Cell mass CO 2 ????

20 In fermenting bacteria (glucose): Streptococcus faecalis (actualmente Enterococcus faecalis) Y m = 22 but this organism it generates 2 ATP/ mole of glucose How many cells per mole of ATP?? Y ATP = 22 / 2 = 11g cells per mole of ATP. Zymomonas mobilis 8.6g cells per mole of ATP. Calculate its Y m ? This organism generates 1 ATP / mole of glucose. Y m = 8.6g X 1 = 8.6 As an average you could expect that an organism fermenting glucose would form 10.5 g of cells per mole of ATP produced

21 Growth kinetics: exponential growth x = x 0 2 y x = anything that doubles each generation (cells, protein,DNA) x 0 = starting value y = number of generations log log x = log x 0 + 0.301y g = generation time g = t /y y = t /g Where t is the time elapsed log x = log x0+ 0.301t g

22 Slope = 0.301/g log x 0 time log x = log x0+ 0.301t g Y = b + mx Log x

23 g (tiempo de generación): = t(A f )-t(A i ) 90-60=30 min Matemática de crecimiento microbiano k  (logN f – log N o ) / t k= instantaneous growth rate, número de generaciones por unidad de tiempo N o =número de células inicial N f = No=número de células final 30 60 90 120 150 180 210 240 min log bacterial numbers/ml log Absorbance 550-600nm 0.1 0.2 0.3 0.4 0.5 0.6 0.7

24 How g relates to k? Slope = 0.301/g log x 0 time log x = log x0+ 0.301t g Y = b + mx log x = kt/2.303 + log x 0 slope k/2.303 = 0.301/g k = 0.301/g (2.303) k = 0.693/g

25 Relationship between growth rate (k) and the nutrient concentration (S) Natural environments nutrients[ ] How is k affected by this fact? K is limited by the rates of nutrient uptake k max 2 KsKs [S] k k max S K s + S k =

26 Steady State Growth and Continuous Growth Quimiostato : artefacto para regular el crecimiento de un cultivo durante un periodo prolongado. Se usa en aplicaciones industriales para la derivación de productos de origen microbiano. The chemostat relieves the insufficiency of nutrients, the accumulation of toxic substances, and the accumulation of excess cells in the culture, which are the parameters that initiate the stationary phase of the growth cycle. The bacterial culture can be grown and maintained at relatively constant conditions depending on the flow rate of the nutrients.

27 The chemostat: Equations D If a chemostat is operated at F = 10 ml/h and V= 1L What is the value of D. D = 10 ml/h / (1000mL) D = 0.01h -1 Dx Dx = rate of loss of cells x = # of cells in the growth chamber If you have 3.67 X 10 7, What is the value of Dx? Dx = (0.01h -1 ) 3.67 X 10 7 Dx = 3.67 X 10 5 cells h -1 Dilution rate D = F / V F= flow rate V= volume of medium in the growth chamber

28 Problem: How to choose the proper size inoculum. You have a culture with a density of 10 8 cells/mL and you would like To subculture it so that 16 hr later the density is is 10 8 cells/mL again. If g = 2 hr, what should x 0 be? x = x 0 2 y y = t /g y = t /g = 16/2= 8 10 8 = x 0 2 8 x 0 = 3.9 X 10 5 cells/mL C 1 V 1 = C 2 V 2 10 8 (X) = 3.9 X 10 5 (1000) X = 3.9 mL

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