Lkelly9 Week 4

From OpenWetWare
Jump to navigationJump to search

Navigation Links

Electronic Lab Notebook

Purpose

The purpose of this assignment is to create and manipulate three different nutrient/cell population models. The population is growing inside of a chemostat. Different constants and parameters were manipulated in order to model the effect on the population and the amount of food.

Methods/Workflow

  • Week 4 Assignment PDF was referenced for the directions of this assignment.
  • Created the function 'ChemostatLK.m'
function dx = ChemostatLK(t,x)
global q u V K r
dx = zeros(size(x));
dx(1) = (q*u) - (q*x(1)) - x(2)*((V*x(1))/(x(1)+K));
dx(2) = -(q*x(2)) + ((r*x(2))*(V*x(1))/(x(1)+K));
end
  • Created the script 'ChemostatscriptLK.m'
global q u V K r
q= 0;
u= 1;
V= 56;
K= 40;
r= 6;
tt = 0:0.1:10;
x0 = [1;2];
[t,x] = ode45('ChemostatLK',tt,x0);
plot(t,x)
xlabel('time')
ylabel('abundance')
title('Population and food source over time')
legend('Population','Food')
  • Put in various values for the variables to see how the population reacted.
  • Added a linear mortality term, (-dx(2)), to the function
function dx = ChemostatLK(t,x)
global q u V K r d
dx = zeros(size(x));
dx(1) = (q*u) - (q*x(1)) - x(2)*((V*x(1))/(x(1)+K));
dx(2) = -(q*x(2)) + ((r*x(2))*(V*x(1))/(x(1)+K)) - (d*x(2));
end
  • Put in various values for the variables to see how the population reacted.
  • Added a second nutrient that is required for the population to grow
function dx = ChemostatLK(t,x)
global q u V K r
dx = zeros(size(x));
dx(1) = (q*u) - (q*x(1)) - x(2)*((V*x(1))/(x(1)+K));
dx(2) = -(q*x(2)) + ((r*x(2))*(V*x(1))/(x(1)+K));
dx(3) = (q*u) - (q*x(3)) - x(2)*((V*x(1)+x(3))/(x(1)+x(3)+K));
end

Results

Consider the nutrient/cell population model

  • First, make sure you understand which variables are the state variables (dependent variables that determine the concentrations) and which variables are parameters (e.g., rate constants).
    • The state variables are:
      • consumption rate = -yVmax(c/K+c)
        • Variables are y and c
    • The parameters are
      • inflow rate = q*u
      • outflow rate = q*c(t)
      • outflow rate = -qy
      • r = net growth rate
  • Find values of c and y that hold the system in equilibrium: that is, find values of c and y for which constant functions at those values are actually solutions of the differential equation system.
    • y = 0, u = 1
  • Simulate this system with different values for the constants and the initial concentrations of nutrients and cells. The initial nutrient level can be =0, but the constants and the initial cell population size need to be positive. Can you make any observations about how the system behaves?
    • In all of the plots below, the population increases as the amount of food decreases, and then levels out after a short period of time.
  1. Q2 u3 V10 K50 r5 lkellyplot.tif
    • Figure 1. The variables are Q=2, u=3, V=10, K=50, r=5
  2. Q5 u1 V5 K30 r6lkellyplot.tif
    • Figure 2. The variables are Q=5, u=1, V=5, K=30, r=6
  3. Q10 u12 V10 K60 r4lkellyplot.tif
    • Figure 3. The variables are Q=10, u=12, V=10, K=60, r=4
  • What changes if we return a mortality term to the population equation? Repeat b and c for a linear mortality term.
    • The system is at equilibrium at the same values.
    • In the graphs below, the population showed a significant decrease due to the added linear mortality term.
  1. Q10 u34 V10 K3 r5 d30lkellyplot.tif
  • Figure 4. The variables are Q=10, u=34, V=10, K=3, r=5 and d=30.
  1. Q2 u16 V15 K65 r5 d4lkellyplot.tif
  • Figure 5. The variables are Q=2, u=16, V=15, K=65, r=5, and d=4.
  • Create a model with two nutrients, both of which are required for the population to grow.
  1. Q2 u0.5 V15 K60 r5lkellyplot.tif
  • Figure 6. The variables are Q=2, u=0.5, V=15, K=60, and r=5

Conclusion

I fulfilled the purpose of this assignment, which was to create and manipulate three different models of population growth and nutrients. One of the main findings of today's project was that in each of the scenarios, the population size and the amount of food available leveled out to a consistent rate. This is because the chemostat adds nutrients continuously at a fixed flow rate and concentration. The flow out of the chemostat is also set at a fixed rate. The amount of food was never at zero because more food is always being brought in. With an unlimited amount of food, the population is free to grow as much as it wants. The population decreased for a period of time when a linear mortality term was added, but both the population and the amount of food leveled out as time went on. When a second nutrient was added, both nutrients decreased as the population increased and then leveled out.

Acknowledgements

Lauren M. Kelly 21:54, 8 February 2017 (EST)

References

Dahlquist, Kam D. (2017) BIOL398-05/S17:Week 4. Retrieved from http://www.openwetware.org/wiki/BIOL398-05/S17:Week_4 on 8 February 2017.