IGEM:UNAM/2009/Notebook/Modeling logbook Claudia/2010/06/07

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Objective: Research the photon flux emitted by Vibrio fisheri
In the free-living state, Vibrio Fischeri emits < 0.8/photons/sec/cell and is present in the seawater at a concentration of 10 cells/ liter.

In the light organ of the Hawaiian sepiolid squid the bacteria grows to densities of approximately 10^10 cells/liter and the bacteria emit more than 800  photons/sec/cell.

So,

(10^10) cells * 800 photons = 8.0 × 10^12 photons/sec.

According to other data the maximum light emission of a bacterial cell is about 10^4 quanta/sec, meaning that to be seen, the cell density must be high, namely about 10^9 to 10^10 cells /ml.

(10^10) cells * (10^4)quanta/sec = 1.0 × 10^14 quanta/sec.

Thus, we expect a maximum light emission between 10^12 photons/sec and 10^14 quanta/sec.

Calculating the light intensity of the last photon flux, we have the next formula:

The photon flux density (Φ) at a given light intensity is: Φ= P/ E

where P is the light intensity in Watts and E the energy of a photon in Jules.

As we want to obtain P, the formula gives:

P= Φ * E

In this case Φ = 1 × 10^14 quanta/sec and to get the value of E we use the formula:

E = hc / λ

With λ = 495 x 10^-9 m (495 nm), which is the wavelength emitted by the genes luxAB. The values of h and c were described in the previous entry (2010/04/06). click here

Substituting:

E= ((6.63 * (10^(-34))) * (3 * (10^8))) / (495 * (10^(-9))) E= 4.01818182 × 10^-19 J

Finally, we calculate the value of P:

P= (1 × 10^14 ) *( 4.01818182 × 10^-19) P= 4.01818182 × 10^-5 W ~ 0.04mW per milliliter.

With 500ml at a cell density of 10^10 cells/ml, we could obtain 20mW.

0.04mW/ml*500ml= 20mW.

And, LovTAP is saturated with 20 mW/cm^2 irradiance at 470 nm with 4.72599297 × 10^16 photons/s cm2. click here