# User:Jaroslaw Karcz/Sandbox

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And you also solve this with the quadratic equation! And it becomes really messy...

And you also solve this with the quadratic equation! And it becomes really messy...
- $2. \frac{d[GFP]}{dt} = \frac{k_3[AP]^n}{K_m + [A]^n}*\frac{[E]^m}{(K_E)^m + [E]^m}$ + $2. \frac{d[GFP]}{dt} = k_3[AP]*\frac{[E]^m}{(K_E)^m + [E]^m}$ - $3. \frac{d[E]}{dt} = -\alpha_{1}*\frac{k_3[A]^n}{K_m + [A]^n}*\frac{[E]^m}{(K_E)^m + [E]^m} - \alpha_{2}k_1[Po]*\frac{[E]^m}{(K_E)^m + [E]^m}$ + $3. \frac{d[E]}{dt} = -\alpha_{1}k_3[AP]*\frac{[E]^m}{(K_E)^m + [E]^m} - \alpha_{2}k_1[Po]*\frac{[E]^m}{(K_E)^m + [E]^m}$
Another energy loss term can be included here...
Another energy loss term can be included here... - $3. \frac{d[E]}{dt} = -\alpha_{1}*\frac{k_3[A]^n}{K_m + [A]^n}*\frac{[E]^m}{(K_E)^m + [E]^m} - \alpha_{2}k_1[Po]*\frac{[E]^m}{(K_E)^m + [E]^m} - k_r*\frac{[E]^m}{(K_E)^m + [E]^m}$ + $3. \frac{d[E]}{dt} = -\alpha_{1}k_3[AP]*\frac{[E]^m}{(K_E)^m + [E]^m} - \alpha_{2}k_1[Po]*\frac{[E]^m}{(K_E)^m + [E]^m} - k_r*\frac{[E]^m}{(K_E)^m + [E]^m}$
kr will be include three things: promoter strength (like k1 or k6), promoter concentration [Po] and energy consumption αr...
kr will be include three things: promoter strength (like k1 or k6), promoter concentration [Po] and energy consumption αr...

# 20 September 2007

It was suggested that the levelling-off of the fluorescence-time curve for construct 1 (T9002) [1] could more realistically be attributed to the extinction of "energy" in the system. The model for Infector Dectector needed to be amended, to include this phenomenon.

## Microbial growth rate as function of single rate-limiting substrate

Although a study of microbial growth rate is undertaken here, this could be altered to feature the system survival/lifetime, as we are dealing with S30 cell extract. The lifetime of the system would be reflected by the rate of change of system energy (d[nutrient] / dt), which here is decreasing, as there is no source of replenishment of nutrient.

A review of literature suggests that multiple models have been developed to describe this feature of the system. The most widely used models are the Monod, Grau, Teisser, Moser and Contois equations.
These equations describe the functional relationship between the microbial growth rate and essential substrate (nutrient) concentration.

It was proposed, and noted from experimental data, that the behaviour of the system can be described as being limited by the energy of the system, and this could be achieved by using the Hill function.

$\mu_{obs} = \mu_{max}\frac{[S]^n}{K_s^n + [S]^n} \cdots (1)$

where

• S = limiting nutrient/substrate ("energy in system")
• μ = instantaneous (observed) growth rate coefficient
• μmax = maximal growth rate coefficient
• K_s = half-saturation coefficient
• n = positive co-operativity coefficient

For Infector Detector, let $S \rightarrow E$ such that (1) becomes

$\mu = \mu_{max}\frac{[E]^n}{K_s^n + [E]^n} \cdots (2)$

where E represents the energy of the system, and μ, effectively, the efficiency of the system. Also, take μmax equal to 1, such that maximal system efficiency is attained at μ = 1. Ks = KE; the half-saturation coefficient.

Furthermore, the energy term, E, needs to be included in the biochemical network equations of our system as follows:

These network equations hold true for both constructs, 1 and 2; they only differ with respect to k1, a measure of constitutive expression. k1 is non-zero for construct 1; for construct 2, k1 is zero.

$\mu = \frac{[E]^n}{K_E^n + [E]^n}$

$\frac{d[LuxR]}{dt} = k_1\mu + k_3[A] - k2[LuxR][AHL]- \delta_{LuxR}[LuxR]$
$\frac{d[AHL]}{dt} = k_3[A] - k2[LuxR][AHL]- \delta_{AHL}[AHL]$

$\frac{d[A]}{dt} = -k_3[A] + k2[LuxR][AHL]- k_4[A][pLux] + k_5[AP]$

$\frac{d[AP]}{dt} = k_4[A][pLux] - k_5[AP] - k_6[AP]$

$\frac{d[GFP]}{dt} = k_6[AP]\mu - \delta_{GFP}[GFP]$

$\frac{d[E]}{dt} = -k_1\mu - k_6[AP]\mu$

Koch and Schaechter studied the effect of glucose concentration ([E]) on the observed growth rate of E. coli in a pure culture. The value for the co-operativity coefficient, n, was found to be 2.38.

## Simplified System, with unlimited energy

• Assumptions:
• the formation of the AHL/LuxR complex reaches steady-state very quickly, compared to gene expression mechanisms.
• the binding/unbinding of the AHL/LuxR complex, on the pLux promoter, reaches steady-state very quickly, compared to gene expression mechanisms.

$\frac{d[AHL]}{dt} = - \delta_{AHL}[AHL]$

$\frac{d[LuxR]}{dt} = k_1 - \delta_{LuxR}[LuxR]$

[A] = k2 * [AHL][LuxR]

$\frac{d[GFP]}{dt} = \frac{k_3[A]^n}{K_m + [A]^n} - \delta_{GFP}[GFP]$

## Simplified System, with limited energy

• Assumptions:
• the formation of the AHL/LuxR complex reaches steady-state very quickly, compared to gene expression mechanisms.
• the binding/unbinding of the AHL/LuxR complex, on the pLux promoter, reaches steady-state very quickly, compared to gene expression mechanisms.
• Energy is supposed to only be consumed during gene synthesis (transcription and translation)
• The protein synthesis rate is supposed to be a function of the energy available in the system.

$\frac{d[AHL]}{dt} = - \delta_{AHL}[AHL]$

$\frac{d[LuxR]}{dt} = k_1*\frac{[E]^m}{(K_E)^m + [E]^m} - \delta_{LuxR}[LuxR]$

[A] = k2 * [AHL][LuxR]

$\frac{d[GFP]}{dt} = \frac{k_3[A]^n}{K_m + [A]^n}*\frac{[E]^m}{(K_E)^m + [E]^m} - \delta_{GFP}[GFP]$

$\frac{d[E]}{dt} = -\alpha_{1}*\frac{k_3[A]^n}{K_m + [A]^n}*\frac{[E]^m}{(K_E)^m + [E]^m} - \alpha_{2}k_1*\frac{[E]^m}{(K_E)^m + [E]^m}$

## Simplified System, with unlimited energy (edit)

• Assumptions:
• the formation of the AHL/LuxR complex reaches steady-state very quickly, compared to gene expression mechanisms.
• the binding/unbinding of the AHL/LuxR complex, on the pLux promoter, reaches steady-state very quickly, compared to gene expression mechanisms.
• Energy is supposed to only be consumed during gene synthesis (transcription and translation) - this assumption is true anyway, since the limiting factor, amino acids, are only used in translation.
• The protein synthesis rate is supposed to be a function of the energy available in the system.
• The degradation rates are neglectable.
• For k1, we are assuming that all the promoter spots are saturated.

$1. \frac{d[LuxR]}{dt} = k_1*[Po]*\frac{[E]^m}{(K_E)^m + [E]^m}$

$[A] = \frac{k_2}{k_3}*[LuxR][AHL]$

$[A] = \frac{k_2}{k_3}*([LuxR_{tot}]-[A])([AHL_{tot}]-[A])$

$[A]^2 - ([LuxR_{tot}] + [AHL_{tot}] + \frac{k_3}{k_2})[A] + [LuxR_{tot}][AHL_{tot}]= 0$
You solve this with the quadratic equation... only one of the roots will be right! It can be the left or the right.

$[AP]^2 - ([A_{tot}] + [Po_{tot}] + \frac{k_5}{k_4})[AP] + [A_{tot}][Po_{tot}]= 0$
And you also solve this with the quadratic equation! And it becomes really messy...

$2. \frac{d[GFP]}{dt} = k_3[AP]*\frac{[E]^m}{(K_E)^m + [E]^m}$

$3. \frac{d[E]}{dt} = -\alpha_{1}k_3[AP]*\frac{[E]^m}{(K_E)^m + [E]^m} - \alpha_{2}k_1[Po]*\frac{[E]^m}{(K_E)^m + [E]^m}$
Another energy loss term can be included here...

$3. \frac{d[E]}{dt} = -\alpha_{1}k_3[AP]*\frac{[E]^m}{(K_E)^m + [E]^m} - \alpha_{2}k_1[Po]*\frac{[E]^m}{(K_E)^m + [E]^m} - k_r*\frac{[E]^m}{(K_E)^m + [E]^m}$
kr will be include three things: promoter strength (like k1 or k6), promoter concentration [Po] and energy consumption αr...