Team:Cornell/project/drylab/modeling/time response

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Time Response

How do our sensors respond to varying analyte concentration over time?

To make sense of data collected remotely, we must know the time it takes for our sensor to produce a signal in response to a change in analyte concentration. In other words, we need to know whether current output at any time can be taken to represent analyte concentration at that time , or whether our system has memory. From preliminary data presented for our arsenic sensor in our results page, we know the response time of the S.A.F.E. B.E.T. sensor is on the order of 1.5 days. That is, it takes nearly a day and a half for our system to fully respond once the analyte concentration in the reactor changes.

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Problem Setup

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Solution

Concentration of Analyte In Reactor
First, we need to solve for the concentration of analyte in the reactor as a function of time ($c(t)$) given an oscillating input of analyte ($[A](t)$)—with arbitrary frequency and amplitude—in the river. By performing a mass balance around the reactor (and ignoring any contribution from generation or consumption), we may write the following differential equation: $$\frac{dc}{dt} = [A](t)\cdot\frac{F}{V}-c\cdot\frac{F}{V}=D([A](t)-c),$$ $$ \mathrm{where} \ \ [A](t) = A_0\cdot\sin(2 \pi ft)+A_0,$$ and the dilution rate, $D$, equals 3.6 day$^{-1}$ for our system. Using $e^{Dt}$ as an integrating factor, we find $$c(t) = \frac{(A0 (D^2 \sin(2 \pi f t)+D^2-2 \pi D f \cos(2 \pi f t)+4 \pi^2 f^2))}{(D^2+4 \pi^2 f^2)}+k_1 e^{-D t}$$ $$ \mathrm{where} \ \ k_1 = - \frac{(A_0 (D^2-2 \pi D f+4 \pi^2))}{(D^2+4 \pi^2 f^2)} \ \ \mathrm{such\ that} \ c(0) =0 $$

Concentration of Analyte In Reactor
Now that we have an analytical solution for the concentration of analyte in the reactor over time, we wish to model the current response given the input function $c(t)$. To accomplish this, we model the time rate of change of current of our arsenic sensor using a Hill function [1] with a cooperativity coefficient of unity—fitting data presented in Fig. X of our Current Response characterization page—and lumping all transcriptional, translational, and post-translational processes: $$\frac{dI}{dt} = \frac{\beta_1 \cdot c(t)}{c_{1/2}+c(t)}+\beta_0 - \alpha I, $$ where $c(t)$ is as defined above; $\beta_1$ is 1.4 $\mu$A/day, $\beta_0$ is 2.8 $\mu$A/day, $c_{1/2}$, the half-saturating analyte concentration, is 100 $\mu$M, $\alpha$ = 0.69 day$^{-1}$; the saturating current output is $(\beta_1 + \beta_0)/\alpha$.

Using this ordinary differential equation, we may numerically solve for current as a function of time—given the input function derived above. The results of such numerical solutions for various input function—performed using the differential equation solver ode45 in MATLAB—are shown below.

Results: Time-Averaged Output for Rapidly Oscillating Analyte Concentrations

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Nam ac pulvinar felis. Mauris vitae erat at orci semper aliquet vitae quis urna. Donec sit amet tortor porttitor diam bibendum viverra. Nam dui nulla, viverra sed lacinia lobortis, ullamcorper et neque. Etiam rhoncus nibh a lacus varius vehicula convallis mi rutrum. Vestibulum vel nunc sit amet ipsum feugiat consectetur. Nulla nec ante vitae dui tristique accumsan. Morbi felis est, ornare a vestibulum vel, vulputate a eros. Nulla facilisi.

References

[1] Hill, A. (1910) The possible effects of the aggregation of the molecules of haemoglobin on its dissociation curves. J Physiol 40: 4–7.

[2] Reference 2

[3] Reference 3