kinetics of growth

21
Presented by: 1.Akshay Patil 2.Apoorva Khairnar 3.Shravan Shah

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Kinetics of growth

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Page 1: Kinetics of growth

Presented by:1.Akshay Patil2.Apoorva Khairnar3.Shravan Shah

Page 2: Kinetics of growth

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

Objectives

Page 3: Kinetics of growth

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

Page 4: Kinetics of growth

Three causes for lag: physiological lag

low initial numbers

Lag phase

appropriate gene(s) absent

growth approx. = 0 (dX/dt = 0)

Page 5: Kinetics of growth

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

Page 6: Kinetics of growth

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

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

Page 7: Kinetics of growth

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!!

Page 8: Kinetics of growth

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

Page 9: Kinetics of 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

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

Page 10: Kinetics of 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)

Page 11: Kinetics of growth

Time (Hours)

0 20 40 60 80 100

Via

ble

Cou

nt (C

FU/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

Page 12: Kinetics of growth

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?

Page 13: Kinetics of growth

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?

Page 14: Kinetics of growth

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

Page 15: Kinetics of growth

nutrients become limiting and/or toxic waste products accumulate

growth = death (dX/dt = 0)

Stationary phase

death > growth (dX/dt = -kdX)

Death phase

Page 16: Kinetics of growth

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 80 100

Via

ble

Cou

nt (C

FU/m

l)

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:

Page 17: Kinetics of growth

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!

Page 18: Kinetics of growth

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

Page 19: Kinetics of growth

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

Page 20: Kinetics of growth

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

Page 21: Kinetics of growth