28
Tema 2: Growth and cell division Paper asignado: MreB.pdf
| Date post: | 16-Dec-2015 |
| Category: |
Documents |
| Upload: | felix-cole |
| View: | 215 times |
| Download: | 0 times |
Tema 2: Growth and cell division
Paper asignado: MreB.pdf
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 absorbancia a 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 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)
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 aclimatación a condiciones de crecimiento, síntesisde RNA, duplicación DNA
Fase Exponencial: número de células se 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 de división celular y muerte estan en balance
Fase de Muerte: las condicionesprevalecientes no pueden sostener más 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?
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)
[Sub
stra
te]
OD
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
Adenylatecyclase
IIIGlc
Pyr
P O operon lactoseBlocked
PEP
PTS proteins
Adenylatecyclase
IIIGlc
Pyr
P O operon
ATPcAMP
lactose
Activated
CAP
No Glucose
P
PP
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 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 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: exponential growth
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
Log x
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 por unidad 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
