This document outlines the objectives and key concepts of bacterial growth. It discusses the fastest and slowest known doubling times for bacteria under different conditions. It also explains the different phases of a bacterial growth curve (lag, exponential, stationary, death), including the mathematical equations that describe each phase. Additionally, it covers factors that influence growth rates like temperature, substrates, and environmental conditions. The Monod equation is presented which models bacterial growth in relation to substrate concentration.

1. Presented by:
1.Akshay Patil
2.Apoorva Khairnar
3.Shravan Shah

2. Objectives
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

3. Growth Curve
S t a t io n a r y
7 .0
De
Expo
nent
ia l
10
Log CFU/ml
L o g C F U /m l
8 .0
6 .0
5 .0
4 .0
Lag
Lag
T im e
a th
O p tic l d e n s it y
Opticala Density
T u r b i d i t y ( o p t i c a l d e n s it y )
9 .0

4. Lag phase
Three causes for lag:
physiological lag
low initial numbers
appropriate gene(s) absent
growth approx. = 0
(dX/dt = 0)

5. Exponential phase
Nutrients and conditions are not limiting
growth = 2n
or
X = 2 nX 0
Where X0 = initial number of cells
X = final number of cells
n = number of generations
20
20
21
21
22
22
23
23
24
24
2n
2n

6. Viable Count (CFU/ml)
1.0e+8
1.0e+7
Example: An experiment was performed in a lab flask growing cells on
0.1% salicylate and starting with 2.21.0e+6 4 cells. As the experiment
x 10
below shows, at the end there were 3.8 x 109 cells.
1.0e+5
This is an increase is 5 orders of magnitude!!
1.0e+4
0
20
How many doublings or generations occurred?
40
60
80
Time (Hours)
X = 2nX0
3.8 x 109 = 2n(2.2 x 104)
1.73 x 10 = 2
5
n
log(1.73 x 105) = nlog2
17.4 = n
Cells grown on salicylate, 0.1%
100

7. How does this compare to growth in the soil?
Response of culturable microbial community to addition of a
carbon source.
Unamended
CFU/g soil
Soil
Pima
Brazito
Clover Springs
Mt. Lemmon
1% Glucose
CFU/g soil
Log Increase
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
Only a 1 to 2 order of increase!!

8. Now compare how environmental conditions
can impact metabolism in soil
Degradation of straw under different conditions
Residue
Wheat straw, laboratory
Rye straw, Nigeria
Rye straw, England
Wheat straw, Saskatoon
Half-life
Days
u
Days-1
Relative rate
9
17
75
160
0.008
0.04
0.01
0.003
1
0.5
0.125
0.05

9. Calculating growth rate during exponential growth
dX/dt = uX
where u = specific growth rate (h-1)
Rearrange: dX/X = udt
Integrate: lnX = ut + C, where C = lnX0
dX/dt = uX
where u = specific growth rate (h-1)
lnX = ut + ln X0 or
X = X0eut
y = mx + b (equation for a straight line)
Note that u, the growth rate, is the slope of this straight line

10. Calculating growth rate during exponential growth
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
y = mx + b (equation for a straight line)
Note that u, the growth rate, is the slope of this straight line

11. 0
20
40
60
80
100
Time (Hours)
Find the slope of this growth curve
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

12. Now calculate the doubling time
If you know the growth rate, u, you can calculate the doubling time
for the culture.
lnX = ut + ln X0
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
What is fastest known doubling time? Slowest?

13. 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?

14. Growth Curve
T u r b i d i t y ( o p t i c a l d e n s it y )
9 .0
7 .0
De
6 .0
5 .0
4 .0
Lag
T im e
a th
O p tic a l d e n s it y
S t a t io n a r y
Stationary
Expo
nent
ia l
Log
10
C F U /m l
8 .0

15. Stationary phase
nutrients become limiting and/or toxic waste products accumulate
growth = death
(dX/dt = 0)
Death phase
death > growth
(dX/dt = -kdX)

16. Viable Coun
1.0e+6
Monod Equation
1.0e+5
The exponential growth equation describes only a part of the growth
curve as shown in the graph below.
1.0e+4
0
20
40
60
80
100
The Monod equation describes the dependence of the growth rate on the
Time (Hours)
substrate concentration:
u =
.
um S
Ks + S
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)

17. 1.0e+10
1.0e+9
Viable Count (CFU/ml)
Combining the Monod equation and the exponential growth equation allows
1.0e+8
expression of an equation that describes the increase in cell mass through
the lag, exponential, and stationary phases of growth:
1.0e+7
u =
um . S
Ks + S
Monod equation
1.0e+6
dX/dt = uX
1.0e+5
u = dX/Xdt
1.0e+4
0
20
40
60
80
Time (Hours)
Exponential growth equation
dX/dt = um. S . X
Ks + S
Does not describe death phase!
100

18. 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
growth will occur at the
maximal growth rate.
2.
Ks
At low substrate concentration
when S<< Ks, the Monod equation
simplifies to:
dX/dt = um. S . X
Ks
growth will have a first order dependence on substrate concentration
(growth rate is very sensitive to S).
Which of the above two cases is the norm for environmental samples?

19. Growth in terms of substrate loss
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:
dS/dt = -1/Y (dX/dt)
where Y = cell yield
Y = g cell mass produced
g substrate consumed
Glucose (C6H12O6)
0.4
Pentachlorophenol (C6Cl5OH)
0.05
Octadecane (C18H38)
1.49

20. Growth in terms of substrate loss
dS/dt = -1/Y (dX/dt)
Combine with: dX/dt = um. S . X
dS/dt = - um (S . X)
Ks + S
Y (Ks + S)
R e m a in in g
p h e n a n th re n e (% )
Which parts of this curve does the equation describe?
0
1
2
3
4
T im e ( d a y s )
5
6
7
8