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