Modeling the Competitive Growth of Saccharomyces Cerevisiae and Schizosaccharomyces Kefir by Joshua Francois Joon Park BENG 221 10/7/112 !Introduction and Background Yeasts are eukaryotic organisms that serve many practical purposes in today’s world. One well known type of yeast, Saccharomyces cerevisiae, is especially useful. One such purpose is for biological research. Saccharomyces cerevisiae can easily be grown in large quantities, which makes the yeast cells readily available for laboratory experiments.1 Also, its genome is known, making it a great cell type for genetic engineering. Saccharomyces cerevisiae is also useful for its ability to produce ethanol. At the end of glycolysis, pyruvate is formed1. If oxygen is present in the environment (aerobic conditions), this molecule will be turned into acetyl-CoA and go to the citric acid cycle to eventually make CO2 and H2O. If oxygen is not present in the environment (anaerobic conditions), pyruvate is reduced with the help of lactate dehydrogenase to lactate. In yeast cells, pyruvate is not converted to acetyl-CoA. Instead, it is first converted to acetaldehyde and then ethanol with the help of pyruvate decarboxylase and alcohol dehydrogenase, respectively. This process, known as fermentation, is used industrially to make alcoholic beverages, but also alternative fuel. In both research and industrial uses, the function of yeast cells can be enhanced by having more yeast cells. A factor that can set back the functionality of one type of yeast can possibly be contamination by another type of yeast. When placed in the same environment, two or more yeast species may be in competition for resources.2 In the case of yeast cells, different species may compete for nutrients. As one species grows, it consumes and therefore can deplete nutrients in the environment. This depletion can then hinder the growth of the other species. Another environmental change that can affect the other species may be the waste products created. The scenario presented above is a short description of conditions laid out in Georgii Frantsevich Gause’s 1932 article “Experimental Studies on the Struggle for Existence.” In this article, Gause used the logistic equation to model growth of a mixed population and applied it to two different types of yeast, Saccharomyces cerevisiae and Schizosaccharomyces kefir. A similar scenario was examined in this project. The numerical solutions were compared with the results found by Gause.! Problem Statement Write and solve, analytically and numerically, a system of equations that appropriately modeled the growth of the two yeast species, growing in a common environment, given the initial mass of both species. Analytical Solution Two functions, y1 and y2, were defined such that: y1(t) = mass of yeast 1 (Saccharomyces cerevisiae) at time t (in hours), and!!y2(t) = mass of yeast 2 (Schizosaccharomyces kefir) at time t (in hours). To model the growth of each species of yeast individually, the logistic equation was used: dy1dt= b1y1K1! y1K1"#$%&'3 !where b1 = intrinsic growth rate of the species and K1 = carrying capacity of the environment. However, this logistic equation does not take into account the presence of the other species in the environment. In order to model the competitive nature of their growth, a particular term in the equation was further examined: K1! y1K1"#$%&' It was observed that this term in the differential equation represents the limiting nature of the environment as the population reaches its capacity. It can be shown that as y1 K1, dy1/dt  0. Taking the presence of y2 into account, it is assumed that the presence of y2 will accelerate this process. To reflect this assumption, the following term is used instead:! K1! (y1+!y2)K1"#$%&' where α = proportionality constant. Applying this term to the original logistic equation, the following set of differential equations was obtained: dy1dt= b1y1K1! (y1+!y2)K1"#$%&' (1) !!!!!! ! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!dy2dt= b2y2K2! (y2+ ßy1)K2"#$%&'! ! (2) It can be shown that this set of differential equations is difficult to solve using the methods covered in class. In order to simplify the equations, the following assumptions were made: y1>> y2, !! 1,!ß ! 1. The first assumption would be valid if y1 represents the yeast species of interest and y2 represents contamination. The assumptions regarding α and ß were based on experimental values determined by Gause. Then, equations (1) and (2) simplify to: dy1dt= b1y1K1! y1K1"#$%&' (3) dy2dt= b2y2K2! ßy1K2"#$%&' (4) Now, (3) can be solved analytically, in the following way:!dy1dt= b1y1(K1! y1K1)!K1dy1y1(K1! y1)"= b1dt"4 !Using partial fractions,!Ay1+BK1! y1=K1y1(K1! y1) AK1! Ay1+ By1= K1AK1= K1! ! > A = 1.(B ! A )y1= 0 ! ! > B = 1. Integrating, 1y1+1K1! y1"#$%&'(dy1= b1dt(ln(y1) ! ln(K1! y1) = b1t + C 'lnK1! y1y1"#$%&'= !b1t + C K1! y1y1= e!b1t+C= Ae!b1ty1=K11+ Ae!b1t Using initial conditions to find A: y1(0) = y10=K11+ AA =K1! y10y10y1=K11+K1! y10y10"#$%&'e!b1t Final solution for y1(t) is: y1=K1y10y10+ (K1! y10)e!b1t (5) Now, the analytical solution for y1 can be used to find the solution for y2: dy2dt= b2y2K2!!y1K2"#$%&'= b2y2K2!!K1y10y10+ (K1! y10)e!b1tK2"#$$$$%&''''= b2!b2!K1K2y101y10+ (K1! y10)e!b1t"#$%&'"#$$%&''y2 Substituting for variables to simplify the equation,5 !M = b2!K1K2y10;!B = y10;! A = K1! y10;!"= !b1 !dy2y2= ! b2dt " MdtB + Ae!t! A table of integral was used to integrate the right hand side:3 ln(y2) = b2t ! M!t ! ln(Ae!t+ BB!"#$%&'+ C!!!!!!!!!!= b2t ! ßb2b1K1K2(b1t + ln((K1! y10)e!b1t+ y10) + C N = ßb2b1K1K2y2= eb2te! Nb1te! N ln (K1!y10)e! b1t+y10( )eCA = eCy2=Aeb2teb1t(K1! y10)e!b1t+ y10( )( )N=Aeb2tK1+ y10(eb1t!1)( )N Using initial conditions to find A: y2(0) = y20=AK1NA = y20K1N Finally, y2= y20eb2tK1K1+ y10(eb1t!1)"#$%&'N,!! N = ßb2b1K1K2 (6) Numerical Plots (Note: Matlab code included in Appendix) The analytical and numerical solutions to equations (1) and (2) were plotted in Matlab for the purpose of comparing the analytical and numerical solutions. The blue curve represents the solution curve for the growth of species 1 (Saccharomyces cerevisiae or Sce) and the green curve represents the solution for the growth of species 2 (Schizosaccharomyces cerevisiae or