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1. The fastest known doubling time for a bacterium and under what conditions this occurs
2. The slowest estimated doubling time for a bacterium and under what conditions this occurs
3. Calculate a growth rate, u, from the slope of a growth curve
4. Compare and contrast growth in pure culture with growth in the environment
5. The growth curve and the parts of the curve
6. A mathematical equation for each part of the curve as well as the Monod equation
7. At least two electron acceptors that can be used under anaerobic conditions in place of oxygen
8. Whether aerobic or anaerobic metabolism yields more energy and why
9. The mass balance equation for aerobic metabolism
Chapter 3 Objectives
Lecture 3 – Growth
What are the differences between growth in a flask in pure culture and growth in the environment (e.g. soil, water, skin surfaces, leaf surfaces)?
vs.
Stationary
Turbidity (optical density)9.0
8.0
7.0
6.0
5.0
4.0
Time
Op
tica
l den
sity
Log
CF
U/m
l1
0Growth Curve
Lo
g C
FU
/ml
Op
tica
l D
ensi
ty
Lag
Three causes for lag: physiological lag
low initial numbers
Lag phase
appropriate gene(s) absent
growth approx. = 0 (dX/dt = 0)
Nutrients and conditions are not limiting
Exponential phase
20
21
22
23
24
2n
20
21
22
23
24
2n
20
21
22
23
24
2n
20
21
22
23
24
2n
20
21
22
23
24
2n
20
21
22
23
24
2n
growth = 2n or X = 2nX0
Where X0 = initial number of cells
X = final number of cells
n = number of generations
Time (Hours)
0 20 40 60 80 100
Viab
le C
ount
(CFU
/ml)
1.0e+4
1.0e+5
1.0e+6
1.0e+7
1.0e+8
1.0e+9
1.0e+10
Cells grown on salicylate, 0.1%
Example: An experiment was performed in a lab flask growing cells on 0.1% salicylate and starting with 2.2 x 104 cells. As the experiment below shows, at the end there were 3.8 x 109 cells.
3.8 x 109 = 2n(2.2 x 104)
1.73 x 105 = 2n
log(1.73 x 105) = nlog2
17.4 = n
This is an increase is 5 orders of magnitude!!
How many doublings or generations occurred?
X = 2nX0
Soil
Unamended
CFU/g soil
1% Glucose
CFU/g soil Log Increase
Pima
Brazito
Clover Springs
Mt. Lemmon
5.6 x 105
1.1 x 106
1.4 x 107
1.4 x 106
4.6 x 107
1.1 x 108
1.9 x 108
8.3 x 107
1.9
2.0
1.1
1.7
Response of culturable microbial community to addition of a carbon source.
How does this compare to growth in the soil?
Only a 1 to 2 order of increase!!
Residue
Half-life
Days
u
Days-1 Relative rate
Wheat straw, laboratory
Rye straw, Nigeria
Rye straw, England
Wheat straw, Saskatoon
9
17
75
160
0.008
0.04
0.01
0.003
1
0.5
0.125
0.05
Degradation of straw under different conditions
Now compare how environmental conditions can impact metabolism in soil
dX/dt = uX where u = specific growth rate (h-1)
Rearrange: dX/X = udt
Integrate: lnX = ut + C, where C = lnX0
lnX = ut + ln X0 or X = X0eut
Note that u, the growth rate, is the slope of this straight line
y = mx + b (equation for a straight line)
Time (Hours)
0 20 40 60 80 100
Via
ble
Co
un
t (C
FU
/ml)
1.0e+4
1.0e+5
1.0e+6
1.0e+7
1.0e+8
1.0e+9
1.0e+10
dX/dt = uX where u = specific growth rate (h-1)
Calculating growth rate during exponential growth
Rearrange: dX/X = udt
Integrate: lnX = ut + C, where C = lnX0
lnX = ut + ln X0 or X = X0eut
Note that u, the growth rate, is the slope of this straight line
y = mx + b (equation for a straight line)
Calculating growth rate during exponential growth
dX/dt = uX where u = specific growth rate (h-1)
Time (Hours)
0 20 40 60 80 100
Via
ble
Cou
nt (
CF
U/m
l)
1.0e+4
1.0e+5
1.0e+6
1.0e+7
1.0e+8
1.0e+9
1.0e+10
lnX = ut + ln X0 or u = lnX – lnX0
t – t0
u = ln 5.5 x 108 – ln 1.7 x 105
8.2 - 4.2= 2 hr-1
Find the slope of this growth curve
Now calculate the doubling time
If you know the growth rate, u, you can calculate the doubling time for the culture.
For X to be doubled: X/X0 = 2
or: 2 = eut
From the previous problem, u = 2 hr-1,
2 = e2(t)
t = 0.34 hr = 20.4 min
lnX = ut + ln X0
What is fastest known doubling time? Slowest?
How can you change the growth rate???
When under ideal, nonlimiting conditions, the growth rate can only be changed by changing the temperature (growth increases with increasing temp.). Otherwise to change the growth rate, you must obtain a different microbe or use a different substrate.
In the environment (non-ideal conditions), the growth rate can be changed by figuring out what the limiting condition in that environment is.
Question: Is exponential growth a frequent occurrence in the environment?
Stationary
Turbidity (optical density)9.0
8.0
7.0
6.0
5.0
4.0
Time
Op
tica
l den
sity
Log
CF
U/m
l1
0Growth Curve
Stationary
nutrients become limiting and/or toxic waste products accumulate
growth = death (dX/dt = 0)
Stationary phase
death > growth (dX/dt = -kdX)
Death phase
Monod Equation
The exponential growth equation describes only a part of the growth curve as shown in the graph below.
Time (Hours)
0 20 40 60 80 100V
iabl
e C
ount
(C
FU
/ml)
1.0e+4
1.0e+5
1.0e+6
1.0e+7
1.0e+8
1.0e+9
1.0e+10
u = specific growth rate (h-1)
um = maximal growth rate (h-1)
S = substrate concentration (mg L-1)
Ks = half saturation constant (mg L-1)
u = um S Ks + S
.
The Monod equation describes the dependence of the growth rate on the substrate concentration:
Combining the Monod equation and the exponential growth equation allows expression of an equation that describes the increase in cell mass through the lag, exponential, and stationary phases of growth:
u = um S Ks + S
. dX/dt = uX
u = dX/Xdt
Monod equation Exponential growth equation
dX/dt = um S X Ks + S
. .
Time (Hours)
0 20 40 60 80 100
Via
ble
Cou
nt (
CF
U/m
l)
1.0e+4
1.0e+5
1.0e+6
1.0e+7
1.0e+8
1.0e+9
1.0e+10
Does not describe death phase!
There are two special cases for the Monod growth equation
1. At high substrate concentration when S>>Ks, the Monod equation simplifies to:
dX/dt = umX
2. At low substrate concentration when S<< Ks, the Monod equation simplifies to:
dX/dt = um S X Ks
. .
Which of the above two cases is the norm for environmental samples?
growth will occur at the maximal growth rate.
growth will have a first order dependence on substrate concentration (growth rate is very sensitive to S).
Ks
In this case the growth equation must be expressed in terms of substrate concentration. The equations for cell increase and substrate loss can be related by the cell yield:
Growth in terms of substrate loss
Glucose (C6H12O6) Pentachlorophenol (C6Cl5OH) Octadecane (C18H38)
0.4 0.05 1.49
dS/dt = -1/Y (dX/dt) where Y = cell yield
Y = g cell mass produced g substrate consumed
Combine with: dX/dt = um S X
Ks + S
..
dS/dt = -1/Y (dX/dt) dS/dt = -1/Y (dX/dt)
Combine with: dX/dt = um S X
Ks + S
..
0 1 2 3 4 5 6 7 8
Rem
ain
ing
p
hena
nthr
ene
(%
)
Time (days)
.dS/dt = - um (S X)
Y (Ks + S)
Which parts of this curve does the equation describe?
Growth in terms of substrate loss
Aerobic vs. anaerobic metabolism
Aerobic metabolism
General equation:(C6H12O6) + 6(O2) 6(CO2) + 6(H2O)
Mass balance equation:a(C6H12O6) + b(NH3) + c(O2) d(C5H7 NO2) + e(CO2) + f(H2O) cell mass
The mass balance equation illustrates that some of the carbon in the substrate is used to build new cell mass and some is oxidized completely to CO2 to provide energy for the cell.
Using the mass balance equation and the cell yield, one can calculate the % of the substrate carbon that is used to build new cell mass and the % that is evolved as CO2
Examples of when this knowledge is important??
Anaerobic metabolism
Under anaerobic conditions, the substrate undergoes disproportionation, whereby some of the carbon is oxidized completely to CO2 and some is reduced to CH4 (because CO2) acts as a terminal electron acceptor.
General equation:C6H12O6 + alternate TEA CO2 + CH4 + H2O
F orm o f R esp ira tion T yp ica l R edoxP o tentia l
E lec tronA ccep to r
P roduc ts
A erob ic re sp ira tion + 400 m V O 2 H 2 O
Nitra te re sp i ra tion/ denitr i fica tion - 100 m V N O 3- N O 2
- , N 2
S ulfa te reduc tion - 160 to - 200 m V S O 4-2 H S - , H 2S
M ethanogenes is - 300 m v C O 2 C H 4
Some Typical Terminal Electron Acceptors