Part I. Boolean modelling exercises. Glucose repression of Icl in yeast In yeast Saccharomyces cerevisiae, expression of enzyme Icl (isocitrate lyase-, involved in the gluconeogenesis pathway) is important for the production of industrially important chemical Succinic acid. Icl is known to be repressed by glucose and understanding of this regulatory phenomenon is essential to increase the Succinic acid production. In a (very very) simple model it can be assumed that: in the absence of glucose, the expression of Icl is induced only in the presence of ethanol and/or acetate. Draw a logic gate representation of the above-described model assuming that all variables are binary. Provide the truth table as well, i.e. a table summarizing all possible input conditions and corresponding output. Truth table: Glucose Ethanol Acetate ICL 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2. Boolean model for LAC operon in E. coli When exposed to a mixture of different nutrient sources, most microorganisms show a preferential utilization pattern, i.e., certain food sources are preferred over the others and consumed before others. This allows them to utilize the cellular resources optimally towards achieving high growth rate and thereby increase the fitness in a competitive environment. One such example is preferential utilization of glucose over lactose in bacterium E. coli (see the following figure). This sequential consumption is regulated at the molecular level by repressing the lactose consumption genes (LAC genes) in the presence of glucose. In fact, expression of lactose consumption genes is regulated extensively in E. coli. In a simple model, we can state that the expression of LAC genes is turned on only in the absence of glucose and in the presence of lactose. Draw a binary logic gate circuit summarizing this model. Also provide the truth table. Biomass!-galactosidase
Growth of E. coli on a mixture of glucose and lactose. The squares show biomass growth while circles show the expression of an enzyme essential for lactose utilization. A shift in the growth regime can clearly be seen after first exponential growth phase. Truth table: Glucose Lactose LAC 0 0 0 0 0 0 0 Glucose NOT Lactose AND LAC 3. Select circuit Using only basic logic gates (i.e. AND, OR & NOT) it is possible to build very complex circuits. This is because any statement/rule that describes a binary output in terms binary inputs can be coded using these three logic operators. Thus, it is also possible to build complex cellular networks in Boolean space by using these basic logic gates. One example of a complex circuit is select circuit, as shown in the following figure. The output is equal to A if C is ON, and equal to B if D is ON. It is assumed that C & D are never ON together. One biological example of such scheme could be a transporter-mediated substrate uptake where C & D code for the transporters for substrates A & B respectively. Your task now is to design a select circuit using the basic logic gates. A B C D Output A C B D 4. Discussion What in your opinion are advantages & disadvantages of Boolean modelling of signal transduction/regulatory pathways? Under what conditions such models should be used? List a signal-response curve that may be suitable for making a binary behaviour assumption. Open question. Sample answer: Advantages: Simple to represent and simulate. No parameters. Easy to interpret the results. Disadvantages: Over-simplification of the reality, hence limited usefulness.
Such models should be used only in the conditions where binary behaviour can be approximated and/or in the conditions where no other detailed information is available about the system.
Part II. Break (x min) Think of something else than systems biology. Part III. Overview of signalling paradigm Example signalling pathways in the yeast Saccharomyces cerevisisae. The signalling scheme shown in above figures is quite conserved across eukaryotes. Understanding the operation of such signalling pathways has many implications in disease research. For example, more than one third of the cancer drugs present in the market target the receptors leading to kinase cascades. New generation of drugs need to target the signal transmission inside the cells in order to achieve better results. In the following set of exercises, you will have a little flavour of the some of the components in these pathways. Although, the
realistic models are much more complex, usually the only difficult part is to identify the components and links between them. Simulation of the resulting set of equations (e.g. set of nonlinear ODEs) is no longer a bottleneck, thanks to high-speed computers. 5. Receptor ligand interaction K D L. R = = LR K equilibrium Calculate equilibrium and dissociation constants for receptor-ligand interaction shown in the above figure. Assume (all concentrations in arbitrary units): [R] = 0-8 [L] = 0-3 [LR] = 0-8 K eq = [LR] / [R][L] = 000 K d = /K eq = 0-3
6. Mitogen Activated Protein Kinase (MAPK) cascade example, longer pathways can give faster response! i) Write differential equation model for a signal transduction cascade shown in the above figure. For simplicity assume that there is only one kinase element downstream of the receptor. You can also neglect the involvement of ATP in the reaction, which contributes the Phosphate group. In a special case of weakly activated pathways, it can be assumed that the fraction of the phosphorylated kinase (i.e. ratio of phosphorylated kinase to the sum of both phosphorylated & unphosphorylated kinase) is much less than. Simplify your model for this new assumption. What happens to the signal output (i.e. concentration of phosphorylated kinase) at the steady state? ii) The activity of the cascade is usually proportional to the concentration of the phosphorylated form of the final kinase in the cascade. Thus the overall effect of cascade can be seen as signal amplification. What effect (positive or negative) will be of the parameters β on this amplification? iii) One interesting fact about the MAPK signalling cascades is that, with a right set of kinetic parameters, the longer cascades can give a faster response (see the figure below, where n is number of kinase steps in the cascade and the total signal amplification is kept constant)! This of course comes at an expense, as each phosphorylation reaction consumes more ATP. Discuss possible biological implication (/s) of such behaviour of the system.
i) Balance over receptor R: dr = "! R Note that there is only degradation term and hence the concentration of R can only decrease over time. Balance over the phosphorylated kinase (X ): dx = " X R #! X * X * is the concentration of the unphosphorylated X. Lets define C as the total concentration of kinase, i.e. C = X + X *. Since the total concentration of kinase will be constant in this model (there is no degradation or synthesis) we can treat C *α as a constant (lets call it α as well for simplicity). dx ) X & = " R ' # $ #! ( % X C $ If C >> X, i.e. weakly activated pathway, we get dx = " R #! X Now, if we assume a steady state (which is not so correct assumption in light of the fact that R is decreasing with time), time derivative becomes zero and we get: X " =! R
ii) We can easily see from the above balance equations that beta will be inversely related with the activity. iii) Open question. 7. Adaptation motif in signalling Combination of a simple linear response element with a second signalling pathway can be described with the following equations. dr = k S! k XR 2 dx = k S! k X 3 4 R and X are responses of two sensing circuits for the input signal S. Show that this combination of two sensors results in an adapted response, i.e. the steady state value of R is independent of signal strength S. The behaviour of this system is similar to the sense of smell and hence also called as sniffer. At steady state the differential terms in the equations are equal to zero. Hence we get: 0 = k S! k XR 0 = k S! k X 3 from () k S R = k X 2 from (2) k S 3 X = k 2 4 4 3 2 4 () (2) (3) Using this value of X in (3) k k R = k k Thus, the steady-state response R is independent of the signal S.