As a specific biological example for Exercise 24.2 and Fig. E24.2(b), the synthesis of tryptophan can be

As a specific biological example for Exercise 24.2 and Fig. E24.2(b),ˆ— the synthesis of tryptophan can be described by the following set of material balances:

where k₁, k₂, k₃, and k₄ represent kinetic rate constants for the synthesis of free operator, mRNA transcription, translation, and tryptophan synthesis, respectively. Parameters O _t, μ, k_d1, and k_d2 refer to total operator site concentration, specific growth rate of E. coli, degradation rate constants of free operator O_R, and mRNA, respectively. E and T represent concentrations of enzyme anthranilate synthase and tryptophan, respectively, in the cell. K_g and g are the half saturation constant and kinetic constant for the uptake of
tryptophan for protein synthesis in the cell. Model parameter values are as follows: k₁ = 50 min^-1; k₂ = 15 min^-1; k₃ = 90 min^-1; k₄ = 59 min^-1; O_t = 3.32 nM; k_d1 = 0.5 min^-1; k _d2 = 15 min^-1; μ = 0.01 min^-1; g = 25 μM ‹… min^-1; K_g = 0.2 μM. Here, controllers C₁(T), C₂(T), and C₃(T) represent repression, attenuation, and inhibition, respectively, by tryptophan and are modeled by a particular form of Michaelis€“Menten kinetics (the Hill equation) as follows:

K_i ,1, K_i,2, and K_i,3 represent the half-saturation constants, with values K_i,1 = 3.53 μM; K_i,2 = 0.04 μM; K_i,3 = 810 μM, whereas sensitivity of genetic regulation to tryptophan concentration, η_H = 1.92.

(a) Draw a block diagram, using one block for each of the four states. Comment on the similarities between this diagram and schematic (b) in Fig. E24.2.

(b) Simulate the response of the system to a step change in the concentration of the medium (change g from 25 to 0 μM).

(c) Calculate the rise time, overshoot, decay ratio, and settling time for the closed-loop response.

(d) Omit the inner two feedback loops (by setting C₂ and C₃ to 0) and change the following rate constants: K_i,1 = 8 × 10^-8 μM; η_H = 0.5. Repeat the simulation described in part (b), and obtain the new closed-loop properties for this network (compared to part (c)).

Transcribed Image Text:

IOR) = k,[0,]C,[T] – kj[OR] – H[Og] dt IMRNA] = k,[OR]C_[T] – k„[MRNA] – µ[MRNA] dt [E] = k½[MRNA] – µ[E] dt [T] [T] + Kg [T] = K¼C3[T][E] 8; - μΤ1 dt K} 72 1,2 |C,(T) = C3(T) = K + T12 i,1 KH + TH C,(T)= i,3 K? + T1.72 ' ri,2 i,1 i,3