Avalekander Week 11: Difference between revisions

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Results: added in calculations
Results: added values for the 2L concentration
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==Results==
==Results==


*Parameters were plugged into y = qK/(r-q) and x = (u-y)/E to solve for cell biomass and nutrient mass (g) at equilibrium
#Parameters were plugged into y = qK/(r-q) and x = (u-y)/E to solve for cell biomass and nutrient mass (g) at equilibrium
**Nutrient mass- y=(0.1(8))/(0.8-0.1)=1.14 g
#*Nutrient mass- y=(0.1(8))/(0.8-0.1)=1.14 g
**Cell biomass- x=(5-1.14)/1.5 = 2.6 g
#*Cell biomass- x=(5-1.14)/1.5 = 2.6 g
##Concenration in a 2L chemostat:
#*Nutrient mass = 1.14g/2L = 0.55 g/L
#*Cell biomass = 2.6g/L = 1.3 g/L


==Conclusion==
==Conclusion==

Revision as of 20:56, 9 April 2019

Purpose

  • The purpose of this assignment is again familiarizing ourselves with MatLab and using it to perform calculations on the ongoing study being done with data collected from the Dahlquist Lab.

Methods

  • The MATLAB files named chemostat_script.m and chemostat_dynamics.m were used to simulate a chemostat and compare the computations to a steady state outcome. *The zip file contained the files used here
  1. The parameter values used were:
    • q = 0.10 (1/hr)
    • u = 5 (g/L)
    • E=1.5
    • r=0.8 (1/hr)
    • K = 8 (g)
  • The data was used to formulate the steady states of cell biomass and nutrient mass.
  1. Assuming a 2 liter chemostat, the steady state concentrations of cells and nutrient was then calculated.
  2. The system dynamics were simulated using the MATLAB files and the parameters of (1). The following questions were then answered:
    • Do the graphs show the system going to steady state?
    • Do the steady states match your (1) calculations?
    • The graphs wre uploaded to our individual journals.
    • BONUS: can you get two y-axes, with the second one to the right of the picture like in the journal articles you’ve read?

Results

  1. Parameters were plugged into y = qK/(r-q) and x = (u-y)/E to solve for cell biomass and nutrient mass (g) at equilibrium
    • Nutrient mass- y=(0.1(8))/(0.8-0.1)=1.14 g
    • Cell biomass- x=(5-1.14)/1.5 = 2.6 g
    1. Concenration in a 2L chemostat:
    • Nutrient mass = 1.14g/2L = 0.55 g/L
    • Cell biomass = 2.6g/L = 1.3 g/L

Conclusion

Acknowledgements

  • I would like to acknowledge my homework partner, Edward with whom I worked with to complete the assignment.
  • I would also like to acknowledge Dr. Dahlquist, as well as Dr. Fitzpatrick.

Except for what is noted above, this individual journal entry was completed by me and not copied from another source.Avalekander (talk) 19:54, 5 April 2019 (PDT)

References

  • Dahlquist, K. & Fitpatrick, B. (2019). "BIOL388/S19: Week 11" Biomathematical Modeling, Loyola Marymount University. Accessed from:Week 11 Assignment Page

Biology 388 Assignments