Chapter 3 Objectives The fastest known doubling time for a bacterium and under what conditions this occurs The slowest estimated doubling time for a bacterium.

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Presentation transcript:

Chapter 3 Objectives The fastest known doubling time for a bacterium and under what conditions this occurs The slowest estimated doubling time for a bacterium and under what conditions this occurs Calculate a growth rate, u, from the slope of a growth curve Compare and contrast growth in pure culture with growth in the environment The growth curve and the parts of the curve A mathematical equation for each part of the curve as well as the Monod equation At least two electron acceptors that can be used under anaerobic conditions in place of oxygen Whether aerobic or anaerobic metabolism yields more energy and why The mass balance equation for aerobic metabolism

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)?

Growth Curve Log CFU/ml Optical Density Lag

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

Exponential phase Nutrients and conditions are not limiting growth = 2n or X = 2nX0 Where X0 = initial number of cells X = final number of cells n = number of generations 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

This is an increase is 5 orders of magnitude!! 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. This is an increase is 5 orders of magnitude!! How many doublings or generations occurred? Cells grown on salicylate, 0.1% X = 2nX0 3.8 x 109 = 2n(2.2 x 104) 1.73 x 105 = 2n log(1.73 x 105) = nlog2 17.4 = n

How does this compare to growth in the soil? Response of culturable microbial community to addition of a carbon source. Soil Unamended CFU/g soil 1% Glucose 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 Only a 1 to 2 order of increase!!

Now compare how environmental conditions can impact metabolism in soil Degradation of straw under different conditions 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 Nigeria hot all year round with a rainy season and a dry season England is a warm summer and cold winter and plenty of rainfall Saskatoon cold winters and summers in the high 60s. Plenty of rainfall

Calculating growth rate during exponential growth dX/dt = uX where u = specific growth rate (h-1) 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

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

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

What is fastest known doubling time? Slowest? 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?

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?

Growth Curve Stationary

Stationary phase Death phase nutrients become limiting and/or toxic waste products accumulate growth = death (dX/dt = 0) 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. The Monod equation describes the dependence of the growth rate on the 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)

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: dX/dt = uX u = dX/Xdt u = um S Ks + S . Monod equation Exponential growth equation dX/dt = um S X Ks + S . Does not describe death phase!

Ks . There are two special cases for the Monod growth equation At high substrate concentration when S>>Ks, the Monod equation simplifies to: dX/dt = umX growth will occur at the maximal growth rate. Ks 2. 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?

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) Pentachlorophenol (C6Cl5OH) Octadecane (C18H38) 0.4 0.05 1.49

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

Aerobic vs. anaerobic metabolism

Examples of when this knowledge is important?? 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 Some Typical Terminal Electron Acceptors