Tema 2: Growth and cell division

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

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

3. Turbidity Beer-Lambert law I / Io=10xl Si tomamos el logaritmo de: Entonces Log (Io / 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/Io=10xl

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

5. Cámara de conteo Petroff-Hausser cubre objetos • pozo central (0.02mm profundidad) • contiene una rejilla que se divide en cuadrantes microscópicos Total cell counts Limitations 1) Are the cells dead or alive? 2) Make right dilution (<106 cells/ml)

6. 12 + 16 + 20 + 18 +14 = 16 25 cuadrados subdivididos en 16 unidades 5 • varios cuadros se cuentan • se promedian y se multiplica por 25 • (# de cuadros en la rejilla) 16 x 25 = 400 400 células/ 0.02 mm3 • la cifra resultante es el # de células en • un volumen que corresponde a las • dimensiones de la rejilla: • (1mm2 x 0.02mm) = 0.02mm3) 400 x = 0.02 mm3 10mm3 pozo central (0.02mm profundidad) • resultados pueden ser expresados • en # de células por ml (cm3) 2 x105células /cm3 = 2 x105células /ml Total cell counts

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

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

10. Results Culture Cell growth by protein quantification

11. Crecimiento Poblacional Bacteriano El ciclo de la curva de crecimiento Fase Estacionaria: se agotan nutrientes, se acumulan desperdicios dañinos, procesos de división celular y muerte estan en balance 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 de Muerte: las condiciones prevalecientes no pueden sostener más crecimiento y las células mueren

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

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. Figure 2.4 10 RNA mass Relative amount 5 protein DNA 1 0.6 1.0 1.5 2.0 2.5 μ (doublings/ hour)

15. Faster growing cells • More RNA, more ribosomes (65% RNA) • More DNA • More mass Why is that?

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

17. Glucose PTS (phosphotransferase) proteins IIIGlc Adenylate cyclase Pyr PEP Glucose-6-P P O operon 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 Blocked lactose

18. P P P PTS proteins No Glucose IIIGlc Adenylate cyclase ATP Pyr PEP cAMP CAP lactose P O operon Activated

19. 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. Molar growth yield constant weight dry cell Y = weight of nutrient weight dry cell (g) Ym In aerobic bacteria = moles of substrate Cell mass Y glucose = 0.5 Means that 50% of the glucose CO2 ???? Growth yields Y is the growth yield constant Y= amount of dry weight of cells produced per weight of nutrient used.

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

21. log g = generation time g = t /y y = t /g Where t is the time elapsed log x = log x0+0.301y log x = log x0+ 0.301t g Growth kinetics: exponential growth x = anything that doubles each generation (cells, protein ,DNA) x0 = starting value y = number of generations x = x02y

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

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

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

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

26. 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. Steady State Growth and Continuous Growth 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. Dx Dx = rate of loss of cells x = # of cells in the growth chamber If you have 3.67 X 107, What is the value of Dx? Dx = (0.01h-1) 3.67 X 107 Dx =3.67 X 105 cells h-1 The chemostat: Equations D Dilution rate D = F / V F= flow rate V= volume of medium in the growth chamber 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

28. 108= x028 x0= 3.9 X 105cells/mL C1V1= C2V2 108 (X) = 3.9 X 105 (1000) X = 3.9 mL Problem: How to choose the proper size inoculum. You have a culture with a density of 108 cells/mL and you would like To subculture it so that 16 hr later the density is is 108 cells/mL again. If g = 2 hr, what should x0 be? x = x02y y = t /g y = t /g = 16/2= 8