Assume $ v_1 = k_1 S $ and $ v_2 = k_2 P $ Then $\frac{dP}{dt} = k_1 S - k_2 P $. At steady-state this equals 0 so that $$ P = \frac{k_1 S}{k_2} $$ In other words P is a linear function of S. In terms of logarithmic senstivities or the reponse coefficient: $$ R^P_S = \frac{dP}{dS} \frac{S}{P} = \frac{k_1}{k_2} \frac{k_2}{k_1} = 1 $$ That is a 1% change in S will lead to a 1% change in the steady-state level of P. This is important because in their pathway (d, shown below) they use this property to ensure that when S also regulates a secondary pathway that in turn up regulates the consumption step v2, the level of P remains unchanged. This property is not very robust however and any change to the rate law on $v_1$ will result in the pathway failing to maintain P constant. For example if $v_1 = k_o + k_1 S$, where we've added a basal rate of $k_o$, then the response becomes: $$ R^P_S = = \frac{k_1 S}{k_o + k_1 S} $$ that is the response is no longer proportional. Change S in this case will result in P changing. The code below will model this system. One important point to make is that this isn't an example of integral control.

```
import tellurium as te
import matplotlib.pyplot as plt
r1 = te.loada("""
-> P; k0 + k1*S
P ->; k2*X*P
-> X; k3*S
X ->; k4*X
k1 = 1; k2 = 1
k3 = 1; k4 = 1
# Set basal rate to zero
k0 = 0; S = 1
# Change signal, P won't change
at time > 10: S = S*2
# Change basal rate and set S back to what it was
at time > 25: k0 = 0.3, S = 1;
# Change signal, this time P will change
at time > 40: S = S*2
""")
m = r1.simulate(0, 60, 200, ['time', 'P', 'S'])
plt.plot (m['time'], m['P'], label='P')
plt.plot (m['time'], m['S'], label='S')
plt.text(2, 0.75, "Basal = 0")
plt.text(9, 0.9, "Change S")
plt.text(14, 1.2, "P restored")
plt.text(20, 0.75, "Set basal > 0")
plt.text(35, 0.9, "Change S, P not restored")
plt.legend()
```

## No comments:

Post a Comment