Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Similar presentations


Presentation on theme: "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."— Presentation transcript:

1 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

2 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)?

3 Growth Curve Log CFU/ml Optical Density Lag

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

5 Nutrients and conditions are not limiting Exponential phase 202122232420212223242 2n2n 202122232420212223242 2n2n 202122232420212223242 2n2n 202122232420212223242 2n2n 202122232420212223242 2n2n 202122232420212223242 2n2n growth = 2 n or X = 2 n X 0 Where X 0 = initial number of cells X = final number of cells n = number of generations

6 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 10 4 cells. As the experiment below shows, at the end there were 3.8 x 10 9 cells. 3.8 x 10 9 = 2 n (2.2 x 10 4 ) 1.73 x 10 5 = 2 n log(1.73 x 10 5 ) = nlog2 17.4 = n This is an increase is 5 orders of magnitude!! How many doublings or generations occurred? X = 2 n X 0

7 Soil Unamended CFU/g soil 1% Glucose CFU/g soilLog Increase Pima Brazito Clover Springs Mt. Lemmon 5.6 x 10 5 1.1 x 10 6 1.4 x 10 7 1.4 x 10 6 4.6 x 10 7 1.1 x 10 8 1.9 x 10 8 8.3 x 10 7 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!!

8 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

9 d X/dt = uX where u = specific growth rate (h -1 ) Rearrange: dX/X = udt Integrate: lnX = ut + C, where C = lnX 0 lnX = ut + ln X 0 or X = X 0 e ut Note that u, the growth rate, is the slope of this straight line y = mx + b (equation for a straight line) d X/dt = uX where u = specific growth rate (h -1 ) Calculating growth rate during exponential growth

10 Rearrange: dX/X = udt Integrate: lnX = ut + C, where C = lnX 0 lnX = ut + ln X 0 or X = X 0 e ut 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 d X/dt = uX where u = specific growth rate (h -1 )

11 lnX = ut + ln X 0 or u = lnX – lnX 0 t – t 0 u = ln 5.5 x 10 8 – ln 1.7 x 10 5 8.2 - 4.2 = 2 hr -1 Find the slope of this growth curve

12 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/X 0 = 2 or: 2 = e ut From the previous problem, u = 2 hr -1, 2 = e 2(t) t = 0.34 hr = 20.4 min lnX = ut + ln X 0 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 Stationary

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

16 Monod Equation The exponential growth equation describes only a part of the growth curve as shown in the graph below. u = specific growth rate (h -1 ) u m = maximal growth rate (h -1 ) S = substrate concentration (mg L -1 ) K s = half saturation constant (mg L -1 ) u = u m S K s + S. The Monod equation describes the dependence of the growth rate on the substrate concentration:

17 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 = u m S K s + S. dX/dt = uX u = dX/Xdt Monod equation Exponential growth equation dX/dt = u m S X K s + S.. Does not describe death phase!

18 There are two special cases for the Monod growth equation 1.At high substrate concentration when S>>K s, the Monod equation simplifies to: dX/dt = u m X 2. At low substrate concentration when S<< K s, the Monod equation simplifies to: dX/dt = u m S X K s.. 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). KsKs

19 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 (C 6 H 12 O 6 ) Pentachlorophenol (C 6 Cl 5 OH) Octadecane (C 18 H 38 ) 0.40.051.49 dS/dt = -1/Y (dX/dt) where Y = cell yield Y = g cell mass produced g substrate consumed

20 Combine with: dX/dt = u m S X K s + S.. dS/dt = -1/Y (dX/dt) Combine with: dX/dt = u m S X K s + S... dS/dt = - u m (S X) Y (K s + S) Which parts of this curve does the equation describe? Growth in terms of substrate loss

21 Aerobic vs. anaerobic metabolism

22 Aerobic metabolism General equation: (C 6 H 12 O 6 ) + 6(O 2 ) 6(CO 2 ) + 6(H 2 O) Mass balance equation: a(C 6 H 12 O 6 ) + b(NH 3 ) + c(O 2 ) d(C 5 H 7 NO 2 ) + e(CO 2 ) + f(H 2 O) 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 CO 2 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 CO 2 Examples of when this knowledge is important??

23 Anaerobic metabolism Under anaerobic conditions, the substrate undergoes disproportionation, whereby some of the carbon is oxidized completely to CO 2 and some is reduced to CH 4 (because CO 2 ) acts as a terminal electron acceptor. General equation: C 6 H 12 O 6 + alternate TEA CO 2 + CH 4 + H 2 O Some Typical Terminal Electron Acceptors


Download ppt "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."

Similar presentations


Ads by Google