Tema 2: Growth and cell division

Growth is an increase in mass

Total and viable cell counts

Dry weight

Protein

Can be measured by:

Turbidity

I / Io=10xl

Beer-Lambert lawSi 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

Turbidity

Turbidity

600nm

incidentlight detected

light

absorbancia a 600nmaumenta según aumenta el# de células

se miden lecturas de absorbanciaa 600nm

Io I

X=cell density

Total cell counts

pozo central (0.02mm profundidad)• contiene una rejilla que se divide en cuadrantes microscópicos

Cámara de conteo Petroff-Hausser

cubre objetos

Limitations1) Are the cells dead or alive?2) Make right dilution (<106 cells/ml)

• varios cuadros se cuentanse promedian y se multiplica por 25(# de cuadros en la rejilla)

• la cifra resultante es el # de células en un volumen que corresponde a lasdimensiones de la rejilla:(1mm2 x 0.02mm) = 0.02mm3)

• resultados pueden ser expresadosen # de células por ml (cm3)

12 + 16 + 20 + 18 +145

= 16

16 x 25 = 400

400 células/ 0.02 mm3

400 0.02 mm3

= x10mm3

2 x105 células /cm3 = 2 x105 células /ml

Total cell counts

25 cuadradossubdivididos en 16 unidades

pozo central (0.02mm profundidad)

Total cell counts

Electronic cell count

Chamber 1 Chamber 2

Measure electrical conductivity

Count of cells in the cultureand the size distribution of

the cells

Viable cell counts

Limitations1) must avoid clumps.2) Some bacteria platewith poor efficiency.3) It will always dependon the medium used.

Dry weight: Is the most direct way to measure growth

Drying oven3 -7 days

1-10ml Samplethru time

Culture

Results

Cell growth by protein quantification

Results

Culture

Crecimiento Poblacional BacterianoEl ciclo de la curva de crecimiento

Fase Lag : periodo de aclimatacióna condiciones de crecimiento, síntesisde RNA, duplicación DNA

Fase Exponencial : número de célulasse duplica a intérvalos regulares de tiempo, ocurre bajo condiciones idealesde crecimiento (ej .abundancia de nutrientes)

Fase Estacionaria : se agotan nutrientes,se acumulan desperdicios dañinos, procesos dedivisión celular y muerte estan en balance

Fase de Muerte : las condicionesprevalecientes no pueden sostenermás crecimiento y las células mueren

Adaptive responses to nutrient limitations

1) Starvation conditions2) Growth no growth3) Generation time days or months4) 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

Stationary phase

1) Changes is size 5-10 µm to 1-2 nm.2) Changes in morphology form rod-shape to coccoid-shape3) Changes in cell surface from hydrophilic to hydrophobic.4) Formation of fibrils and aggregates5) Changes in phospholipids from unsaturated to cyclopropane6) 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.

1

5

10

Rel

ativ

e am

ount

µ (doublings/ hour)

0.6 1.0 1.5 2.0 2.5

RNA

mass

protein

DNA

Figure 2.4

Faster growing cells

1) More RNA, more ribosomes (65% RNA)2) More DNA3) More mass

Why is that?

Como la célula regula la síntesis de RNA?

1) Growth rate dependent2) Amino acid starvation

Guanosine tetraphosphate (ppGpp) slows down the synthesisof rRNA and tRNA.

inactivatedRelA Less rRNA, tRNA

3) Carbon and energy limitationSpoT bifunctional protein synthesizes and degrades ppGpp

Positive regulationFis (DNA-binding protein) its synthesis increase in rich mediathus increasing the synthesis of rRNA.

lactose

glucosegrowth

Time

Diauxic growth

Catabolite repression by glucose

1) Repression in synthesis of degrading enzymes for the 2nd compound 2) Inhibits the uptake of other sugars (Inducer exclusion)

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 proteins

Adenylatecyclase

IIIGlc

Pyr

P O operon lactoseBlocked

PEP

PTS proteins

Adenylatecyclase

IIIGlc

Pyr

P O operon

ATPcAMP

lactose

Activated

CAP

No Glucose

P

PP

40 minutes

20 minutes

Cell Division:is the splitting of a mother cell into 2 daughter cells separated by a septum

A.

B.

A: Contraste de faseB: Tinción de región nucleoide (azul) y anillo FtsZ (rojo)

Anillo FtsZ y División Celular

Proteína FtsZ :

• universalmente distribuida en procariotas incluyendo Archaea, mitocondrias y cloroplastos

•estructuralmente similar a tubulina (proteína presente en microtúbulos eucariotas)

The cell division in E. coli requires the following proteins:

FtsZ- forms the ring perhaps with the help of ZipA.

FtsA- is recruited by FtsZ and joins the ring followed by FtsK.

Then FtsQ joins in, followed by FtsL and B (L and B codependent); follwed by FtsW, I, N.

In Bacillus subtilis: all the proteins are recruited in a concerted manner.

DIVISOMA: conjunto de proteínas (Fts) que lleva a cabo la división celular en procariotas

Proteínas Fts : “filamentous temperature sensitive”

FtsZ : proteína que se polimeriza (10,000 unidades)en forma de un anillo alrededordel plano de división

•también atrae otros componentesdel divisoma (FtsA, I, K y Zip A)

•hidrolisa GTP para polimerizar y depolimerizar el anillo

ZipA : proteína de anclaje que fijael anillo FtsZ a la membrana citoplásmica

FtsA : hidrolisa ATP para ensamblarlos componentes del divisoma

FtsQ, LB,W, I, N : síntesis de peptidoglícano, es inactivada por el antibiótico penicilina

FtsK : separación de los cromosomas

E. coli

How is the site of septum formation determined?

Nucleoid occlusion- FtsZ ring does not form in parts of the cell containinga dense nucleoid mass.

- B. subtilis has a DNA-binding protein that prevents ring formation.

Min system (E. coli)- the system is a group of cell-membrane-associatedproteins that prevent the formation of the FtsZ ring inlocations other than the cell center.

Operon minB: is minC, minD, and minE- where MinD a membrane-associated protein recruits and MinC forming a complex that is less abundant at thecenter of the cell.- MinE accumulates as a ring near mid cell and stimulatesthe release of MinDC complex.

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 bythe quantity of a single nutrient that is the sole source of energy and carbon.

Y =weight dry cellweight of nutrient Ym =

weight dry cell (g)

moles of substrate

Molar growth yield constant

In aerobic bacteria

Y glucose = 0.5 Means that 50% of the glucose Cell mass

CO2 ????

In fermenting bacteria (glucose):

Streptococcus faecalisYm = 22 but this organism generates 2 ATP/ mole of glucose

How many cells per mole of ATP??

YATP = 22 / 2 = 11g cells per mole of ATP.

Zymomonas mobilis8.6g cells per mole of ATP.

Calculate its Ym ?This organism generates 1 ATP / mole of glucose.

Ym = 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

Growth kinetics

x = x02y

x = anything that doubles each generation (cells, protein ,DNA)

x0 = starting value

y = number of generationslog

log x = log x0+ 0.301yg = generation time

g = t /y y = t /gWhere t is the time elapsed

log x = log x0+ 0.301tg

Slope = 0.301/g

log x0

time

log x = log x0+ 0.301tg

Y = b + mx

g (tiempo de generación):= t(Af)-t(Ai)

90-60=30 min

Matemática de crecimiento microbiano

k ==== (logNf – log No) / t

k= instantaneous growth rate,número de generaciones porunidad de tiempo

No=número de células inicial

Nf = No=número de células final

30 60 90 120 150 180 210 240

min

log

bact

eria

l num

bers

/ml

log

Abs

orba

nce

550-

600n

m

0.1

0.2

0.3

0.4

0.5

0.6

0.7

How g relates to k?

Slope = 0.301/g

log x0

time

log x = log x0+ 0.301tg

Y = b + mx

log x = kt/2.303 + log x0

slope

k/2.303 = 0.301/g k = 0.301/g (2.303) k = 0.693/g

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

k max

2

Ks [S]

k

kmax S

Ks + Sk =

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.

The chemostat: Equations

D

If a chemostat is operated at F = 10 ml/h and V= 1LWhat is the value of D.

D = 10 ml/h / (1000mL)D = 0.01h-1

Dx

Dx = rate of loss of cellsx = # 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

Dilution rate D = F / V

F= flow rateV= volume of medium in the growth chamber

Problem: How to choose the proper size inoculum.

You have a culture with a density of 108 cells/mL and you would likeTo 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

108= x028 x0= 3.9 X 105 cells/mL

C1V1= C2V2

108 (X) = 3.9 X 105 (1000)X = 3.9 mL