lecture_17_TurbidostatChemostat - Lecture 17 GROWTH AT LOW...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 17 GROWTH AT LOW CONCENTRATIONS OF LIMITING NUTRIENTS. Correction • • • • • • • ln(N2/N1) = k(t2‐t1), substitute ln(2N1/N1) = kg Ln2 = kg g = ln2/k or 0.693/k k = ln2g k = ln2/g So in the example above, k=1.92hr‐1 , g=0.693/1.92hr‐1= 0.36 hr or the culture doubles every 21.7 min. Continuous culture • TURBIDOSTAT: is an apparatus that monitors the culture density and when the densisty reaches a preset point, the device lets in new media and allows an equal amount of culture to leave. Thereby keeping a constant volume and letting the growth rate be set by the bacteria. In a turbidistat the growth rate is a function of the organism, it is internally controlled. At any given temperature, if you want to change the growth rate you must change the medium. Used to grow bacteria at maximal rates. Turbidostats • Work when cells are growing unrestricted. • k = kmax, maximum growth rate. Life at limited growth rates Remember this? What if you keep decreasing the [x]? [x] ln Cell # [x/2] [x/4] [x/8] [ X] Plot growth rate, k versus [x] (which we will now call [c]) 1 Growth Rate, k 1 0 0 x/8 0 0 1 5 Limiting nutrient concentration [c] What happens as we decrease [x]? • At some concentration of x, k begins to decrease. • Above this concentration k is independent of [x] • k does not change, called kmax, and it is the maximum growth rate at those conditions. 1 Growth behaves like an enzymatic reaction 0.8 kmax Growth Rate, k 0.6 0.4 0.2 Ks, Substrate concentration where k= kmax/2 0 0 1 5 Limiting nutrient concentration [c] • Growth rate and substrate concentration follow first‐order kinetics; where some component of the system becomes saturated (limiting) at low substrate concentrations. • k=kmax c/(Ks+c) • k is the specific growth rate at a concentration, c of limiting nutrient; Ks is a constant that this numerically equal to the substrate concentration at which k=kmax/2 • At unrestricted growth k=kmax Look familiar? k=kmax c/(Ks+c) v=vmax S/(Km+S) So, we can now begin to look at growth under nutrient limiting conditions Chemostat • A device used to grow bacteria at submaximal growth rates and at high concentrations. • It can maintain a culture indefinitely in the range where k is limited by the substrate concentration, [c]. Therefore k can be varied by the amount of nutrients in the chemostat. Here is how it works: ‐control addition of fresh medium ‐control volume by over flow ‐mix culture to ensure rapid equilibration of the fresh medium. Mathematical basis of the Chemostat. • V is the volume of culture and F the flow rate, measured in units of culture volume per hour, then the • The average time a cell spends in the culture vessel is called the MEAN RESIDENT TIME. • MRT = V/F, the reciprocal of MRT is the dilution rate, D = F/V A Chemostat Sets the Flow Rate V, volume The first fundamental principle of the chemostat: • If the chemostat is running at steady state, the number (N) or mass (x) of cells removed through the overflow is matched exactly by the number of or mass of cells produced in the culture, that is: rate of production of cells through growth = rate of loss of cells through overflow. • this can be described by: dx/dt = kx‐xF/V and dx/dt = kx‐Dx In the chemostat the number of cells does not change so dx/dt = 0 and therefore kx=Dx or k=D Growth rate is equal to the dilution rate. Therefore you can simply change the growth rate by changing the dilution rate of the specific nutrient and the volume of the vessel. This works because at low nutrient concentration the growth rate is related to the nutrient concentration Remember • k=kmax c/(Ks+c), if you solve for c and substitute D for k, you find • • c=KsD/(kmax‐D) giving the relationship between Dilution rate, D, and c, the concentration of the nutrient. CELL MASS IN A CHEMOSTAT In steady state the substrate concentration in the culture vessel remains constant, therefore: Substrate added from the reservoir = substrate used for growth + substrate lost through overflow • or Dcr = dc/dt + Dc, • where cr is the nutrient concentration in the reservoir and dc/dt the rate of utilization of the nutrient. Substitute (dx/dt)(dc/dx), so, dx/dt= kx and dc/dx =1/Y (yield coefficient). • We get Dcr = kx/Y + Dc, and because in steady state D=k • x=Y(cr‐c) Which relates biomass (x) to the dilution rate (D) and limiting nutrient concentration ([c]) Key Conditions at steady state • Two key conditions of a chemostate operating in steady state. • 1. N = Nmax, cell mass does not change. Cell division is equal to loss of cells due to wash out • 2. k = D. the growth rate is set by the dilution rate. How do you change D • D = flow rate (f)/ culture volume (v) or D=f/v • So, to increase the dilution rate you either decrease V or increase F, and, to decrease D you either decrease F or increase V • As D increases, c increases remember, that biomass is related to the concentration and Y, the growth yield: • x = Y(cr‐c) So if Y and cr are constant, how does x vary with c? • as c increases x decreases So, to increase the biomass, you either increase cr, or decrease D Relation between biomass, x, and substrate concentration to dilution rate c = cr [x] [c] D kmax Setting up a Chemostat • If we inoculate a chemostat with a culture and we are using a medium that is limiting for a single nutrient then the following should happen: Remember D is constant! • 1. Initially S>>>>Ks so k=kmax and all cells will be dividing faster then lost in the overflow, Not steady state! • 2. As the cell density increases, the nutrient becomes limiting, S decreases and therefore k decreases. • 3. k decreases until the rate of growth equals the rate of loss due to overflow • 4. At this point the chemostat is in equilibrium, x=xmax and S and k are stable. Chemostat is self correcting • Suppose you are running a at steady state‐ now increase the flow rate, what happens: • 1. Cells are transiently lost by overflow, due to an increase in their rate of formation, • 2. The number of cells, x, in the vessel decreases, • 3. The less dense culture will therefore utilize the limiting nutrient at a slower rate, • 4. Before the nutrient concentration will rise, • 5. k will increase until it is sufficient to match the rate of loss of cells through the overflow. WHAT PROBLEMS DO YOU NEED TO AVOID IN A CHEMOSTAT • You are growing cells at a constant slow rate by limiting a vital nutrient. What type of potential problems can you see? • 1. What happens if cells stick to the side of vessel? No washout, so the chemostat will collapse. • 2. What about the stability of the population? What cool things can you do with a chemostat? • Look for specific classes of mutants 1 Utilization mutants 0.8 Growth Rate, k 0.6 0.4 Wild type strain 0.2 Mutant able to uptake c, better than wild type 0 0 1 5 Limiting nutrient concentration [c] ...
View Full Document

Ask a homework question - tutors are online