Team:Imperial College London/Modelling/Output/Parameters and Constants

Let

• k₁ = rate constant for E+S → ES
• k₂ = rate constant for E+S ← ES</math>
• k_cat = rate constant for ES → E+P

We know that K_m = (k_cat + k₂)/k₁

Assuming that k_cat << k₂ << k₁, we can rewrite K_m ≈ k₂/k₁
From this paper <a href="http://peds.oxfordjournals.org/cgi/reprint/14/12/993">[1]</a> the constants for TEV can be found:
For example, for wildtype TEV: K_m = 0.061±0.010mM and k_cat = 0.16±0.01s^-1
These values correspond with our assumption that k_cat = 0.1 s^-1 and K_m = 0.01 mM.
Hence, we can estimate the following orders of magnitude for the rate constants:
k₁ = 10⁸M^-1s^-1
k₂ = 10³s^-1
Using these values should be a good approximation for our model.

Growth rate, gr (divisions/h): 0.53 ≤ gr ≤ 2.18 <a href="http://www.ncbi.nlm.nih.gov/pmc/articles/PMC235685/pdf/jbacter00326-0019.pdf">[2]</a>
Hence on average, gr = 1.5 divisions per hour, which gives one division every 40mins
To deduce degradation rate we use the following formula:
τ_1/2 = ln2/k, where τ_1/2 = 0.667 hours and k = degradation rate
k = ln2/τ_1/2 = 0.000289s^-1

We have found production rates for two arbitrary proteins in E.Coli. We want to get estimates of production rates by comparing the lengths of the proteins (number of amino-acids).
As this approach is very vague, it is important to realise its limitations and inconsistencies:

• Values are taken from E.Coli not B.sub.
• The two production rates are of the same value for quite different amino-acid number which indicates that protein folding is limiting the production rates.

LacY production = 100 molecules/min<a href="http://bionumbers.hms.harvard.edu/bionumber.aspx?s=y&id=100738&ver=0&hlid=29205">[3]</a> (417 Amino Acids<a href="http://www.uniprot.org/uniprot/P02920">[4]</a>)
LacZ production = 100 molecules/min<a href="http://bionumbers.hms.harvard.edu/bionumber.aspx?s=y&id=100737&ver=0&hlid=29206">[5]</a> (1024 AA<a href="http://www.uniprot.org/uniprot/P00722">[6]</a>)
Average production ≈ 100molecules/min 720 AA
This gives us: TEV production ≈ 24 molecules/min = 0.40 molecules/s (3054 AA<a href="http://www.uniprot.org/uniprot/P04517">[7]</a>)
As production rate needs to be expressed in concentration units per unit volume, the above number is converted to mols/s and divided by the cell volume: 2.3808×10^-10 mol/dm³/s
C23D production ≈ 252 molecules/min = 4.2 molecules/s (285 AA<a href="http://www.uniprot.org/uniprot/P54721#section_x-ref">[8]</a>) → 2.4998×10^-9 mol/dm³/s
We will treat these numbers as guiding us in terms of range of orders of magnitudes. We will try to run our models for variety of values and determine system’s limitations.

Wild type (used for our simulations): K_m = 10 μM; k_cat = 52s^-1
Mutated type: K_m = 40 μM; k_cat = 192s⁻¹
Consequently, the ratio of K_m/k_cat of the mutant (K_m/k_cat = 4.8) is slightly lower than the ratio of the wild type (K_m/k_cat = 5.2), indicating that the mutation has little effect on the catalytic efficiency <a href="http://www.springerlink.com/content/e3718758m5052214/fulltext.pdf">[9]</a>.

diameter: d = 0.87μm; length: l = 4.7μm
This gives: Volume= πd²l/4 = 2.793999μm³ ≈ 2.79×10^-15 dm³

The length of the coil is 90 AA.
The whole construct is then: 1617 AA
Therefore, split TEV production rate ≈ 1.2606×10^-10 mol/dm³/s

For a concentration of 0.1 M with built up levels of dioxygenase the colour change happens within seconds.
We will run our models for 0.1M ± several orders of magnitude to determine the smallest catechol concentration that will give a significant difference between the simple production response and the amplified response.