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5. Consider this reaction: where the dashed lines mean that s and q do not get consumed in the corresponding reactions (they both behave as enzymes). There are also constants ki for each of the rates (not shown). Make sure that you understand then why these are the reasonable mass-action equations to describe the system: dtdp=k1sk2pqdtdq=k3sk4q (or one could have used, instead, a more complicated Michaelis-Menten model). (a) Find the steady state, written as a function of (nozero) s. Note that p() (though not q() ) is independent of s. This is an example of adaptation, meaning that the system transiently responds to a "signal" s (assumed a constant), but, after a while, it returns to some "default" value which is independent of the stimulus s (and hence the system is ready to react to other signals). (b) Graph (using a computer) the plot of p(t) versus t, assuming that k1=k2=2 and k3=k4=1, and p(0)=q(0)=0, for each of the following three values of s:s=0.5,3,20. You should see that p(t)1 as t (which should be consistent to your answer to part (b)) but that the system initially reacts quite differently (in terms of "overshoot") for different values of s