Tema 2: Growth and cell division
Paper asignado: MreB.pdf
Growth is an increase in mass
Can be measured by:
Dry weight
Turbidity
Protein
Total and viable cell counts
Turbidity
The amount of light scattered by a bacterial cell is
proportional to its mass. Beer-Lambert law
I/Io=10xl
Si tomamos el logaritmo de:
Beer-Lambert law
I / Io=10xl
Entonces
Log (Io / I)= -xl
Turbidity, absorbance, optical density
OD = A = xl
Turbidity
se miden lecturas de absorbancia
a 600nm
Io
incident
light
600nm
I
detected
light
X=cell density
absorbancia a 600nm
aumenta según aumenta el
# de células
Total cell counts
Cámara de conteo Petroff-Hausser
cubre objetos
pozo central (0.02mm profundidad)
• contiene una rejilla que se divide en cuadrantes microscópicos
Limitations
1) Are the cells dead or alive?
2) Make right dilution (<106 cells/ml)
Total cell counts
25 cuadrados
subdivididos en
16 unidades
12 + 16 + 20 + 18 +14 = 16
5
16 x 25 = 400
400 células/ 0.02 mm3
400
x
=
0.02 mm3 10mm3
2 x105 células /cm3
= 2 x105 células /ml
pozo central
(0.02mm profundidad)
• 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:
(1mm2 x 0.02mm) = 0.02mm3)
• resultados pueden ser expresados
en # de células por ml (cm3)
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
Viable cell counts
Limitations
1) must avoid clumps.
2) Some bacteria plate
with poor efficiency.
3) It will always depend
on the medium used.
Dry weight: Is the most direct way to measure growth
1-10ml Sample
thru time
Culture
Results
Drying oven
3 -7 days
Cell growth by protein quantification
Results
Culture
Crecimiento Poblacional Bacteriano
El ciclo de la curva de crecimiento
Fase Lag: periodo de aclimatación
Fase Estacionaria: se agotan nutrientes,
a condiciones de crecimiento, síntesis
de RNA, duplicación DNA
se acumulan desperdicios dañinos, procesos de
división celular y muerte estan en balance
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
Adaptive responses to nutrient limitations
1)
2)
3)
4)
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
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.
Figure 2.4
Relative amount
10
RNA
mass
5
protein
DNA
1
0.6
1.0
1.5
2.0
μ (doublings/ hour)
2.5
Faster growing cells
1) More RNA, more ribosomes (65% RNA)
2) More DNA
3) More mass
Why is that?
Diauxic growth
lactose
growth
OD
[Substrate]
glucose
Time
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
Glucose
PTS (phosphotransferase) proteins
IIIGlc
Adenylate
cyclase
Pyr
PEP
Glucose-6-P
Blocked
P
O
operon
lactose
PTS proteins
No Glucose
P
IIIGlc
P
P Adenylate
cyclase
cAMP
ATP
Pyr
PEP
CAP
P
Activated
O
operon
lactose
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 nutrient
Molar growth yield constant
Ym
=
weight dry cell (g)
moles of substrate
In aerobic bacteria
Y glucose = 0.5
Means that 50% of the glucose
Cell mass
CO2
????
In fermenting bacteria (glucose):
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
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: exponential growth
x = anything that doubles each generation
(cells, protein ,DNA)
x = x02y
log
x0 = starting value
y = number of generations
g = generation time
log x = log x0+ 0.301y
log x = log x0+ 0.301t
g
g = t /y
y = t /g
Where t is the time elapsed
Log x
Slope = 0.301/g
log x0
log x = log x0+ 0.301t
g
Y = b + mx
time
Matemática de crecimiento microbiano
g (tiempo de generación):
= t(Af)-t(Ai)
k  (logNf – log No) / t
No=número de células inicial
Nf = No=número de células final
log bacterial numbers/ml
k= instantaneous growth rate,
número de generaciones por
unidad de tiempo
log Absorbance 550-600nm
90-60=30 min
0.7
0.6
0.5
0.4
0.3
0.2
0.1
30
60 90 120 150 180 210 240
min
How g relates to k?
log x = kt/2.303 + log x0
slope
Slope = 0.301/g
log x0
log x = log x0+ 0.301t
g
Y = b + mx
time
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
k max
2
k=
Ks
[S]
kmax S
Ks + S
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
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.
Dx
D = 10 ml/h / (1000mL)
D = 0.01h-1
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
Dx = rate of loss of cells
x = # of cells 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 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
108= x028 x0= 3.9 X 105 cells/mL
C1V1= C2V2
108 (X) = 3.9 X 105 (1000)
X = 3.9 mL