BTF2042 Industrial Microbiology Lecture 5 PDF

BTF2042 Industrial Microbiology Dr. L.J.S Undugoda Senior Lecturer Department of Biosystems Technology University of Sri Jayewardenepura Generation time Time required for the population to double in size Varies depending on species o...

BTF2042 Industrial Microbiology Dr. L.J.S Undugoda Senior Lecturer Department of Biosystems Technology University of Sri Jayewardenepura Generation time Time required for the population to double in size Varies depending on species of microorganism and environmental conditions Range is from 10 minutes for some bacteria to several days for some eukaryotic microorganisms  b = B x 2n Where, B0 = number of organisms at zero time b = number of organisms after n generations. n= number of generations G (generation time) = (time, in minutes or hours)/n(number of generations) G = t/n t = time interval in hours or minutes B = number of bacteria at the beginning of a time interval b = number of bacteria at the end of the time interval n = number of generations (number of times the cell population doubles during the time interval) b = B x 2n (This equation is an expression of growth by binary fission) Solve for n: logb = logB + nlog2 n = logb - logB log2 n = logb - logB.301 n = 3.3 logb/B G = t/n Solve for G G= t 3.3 log b/B Q1 What is the generation time of a bacterial population that increases from 10,000 cells to 10,000,000 cells in four hours of growth  Calculating growth rate during exponential growth 1.0e+10dX/dt = uX where u = specific growth rate (h-1) Rearrange: 1.0e+9 dX/X = udt Viable Count (CFU/ml) Integrate: 1.0e+8 lnX = ut + C, where C = lnX0 dX/dt = uX where u = specific growth rate (h-1) 1.0e+7 lnX = ut + ln X0 or X = X0eut 1.0e+6y = mx + b (equation for a straight line) 1.0e+5 Note that u, the growth rate, is the slope of this straight line 1.0e+4 0 20 40 60 80 100 Time (Hours)  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 Question 4 : Yeast culture fermentation is occurred in a fermenter which is running as a batch culture. Initial cell biomass is 5x108 cells and it shows 10 cell/h growth rate. Calculate the cell biomass after another 30 min of incubation period? Effect of Substrate Concentration on Growth So far we have discussed each of the growth phases and have shown that each phase can be described mathematically One can also write equations to allow description of the entire growth curve. Such equations become increasingly complex. For example, one of the first and simplest descriptions is the Monod equation, which was developed by Jacques Monod in the 1940s.  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 1.0e+10 Ks + S 1.0e+9 Viable Count (CFU/ml) 1.0e+8 u = specific growth rate (h-1) 1.0e+7 um = maximal growth rate (h-1) 1.0e+6 1.0e+5 S = substrate concentration (mg L-1) 1.0e+4 0 20 40 60 80 100 Ks = half saturation constant (mg L-1) Time (Hours) 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 = um. S Ks + S u = dX/Xdt Monod equation Exponential growth equation 1.0e+10 1.0e+9 dX/dt = um. S. X Viable Count (CFU/ml) 1.0e+8 Ks + S 1.0e+7 1.0e+6 1.0e+5 Does not describe death phase! 1.0e+4 0 20 40 60 80 100 Time (Hours)  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. Ks 2. At low substrate concentration when S