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Contents |
Introduction
Why?
Hockberger et al (1999) suggest that blue light up to 470 nm can have some phototoxic effect on mammalian cells, from 2 to 6 J/cm². In our experiments, we have observed a much higher cell death rate in the cultures exposed for 24h to blue light than in the control. 24h at 20 mW/m² is actually more than 1700 J/cm². To stay on the safe side, and give the cells 6 J/cm² during 4 hours would mean an average of 0.4 mW/cm².
- How much activation can we get with this rate?
- Will we have to push it up?
- What lighting pattern will give best results?
What?
To optimize the number of experiments to perform, we have built a simple model to predict the proportion of photoproduct to be expected at every time point when a sample with LovTAP-VP16 when it's illuminated with a time varying light input.
How?
Photoactivation
When light with the appropriate wavelength goes through a solution with photoactive molecules, a number of photons will be absorbed and transfer their energy to these molecules. In the case of LovTAP-VP16, this energy will favor the conformational change into the “active” state. To simplify the model, we will suppose that the protein will be active once it has received enough energy.
The molar absorptivity, also know as extinction coefficient, ε, is directly related to the absorption cross section, σ, through the formula:
If ε is in L mol⁻¹cm⁻¹ we get σ in m². From it we can calculate the number of photons absorbed by a material per unit of distance traveled by the light, using the expression:
where N is the number of photons, m is the number of absorbing molecules per unit volume and x is distance. We can write m in terms of the amount of molecules in the differential volume dV:
and, being dV=Adx, with A the area seen by the light:
Now, the quantum yield of the LOV2 domain, Q, is defined as the number of photons of a particular wavelength required to trigger the photoactivation of one protein. We can use it to relate dN with dMphot, the variation of the amount of photoproduct:
substituting in the previous expression and defining F as the photon flux:
Activation parameters
From Kasahara et al (2002) (Table I) we took the values of ε = 14000 L mol⁻¹cm⁻¹ and Q = 0.34 for the LOV2 domain in phot1 in Arabidopsis and rize. They are very similar to the LOV2 domain in phot1 from Avena sativa, according to Salomon et al (2000). Kasahara used a 446 nm light source, at 80 µmol m⁻²s⁻¹, what is 22 mW/m². Using those parameters provides the same dynamics they observed in their experiments for the first 10 s at 4ºC.
To adapt this parameters to the wavelength of the LEDs we are using, the molar absorption has to be modified. Absorbance at 470 nm, the wavelength of the LEDs we use, is some 75% (Kasahara et al (2002), Fig. 1) of the peak absorbance at 450 nm. Since the extinction coefficient, or molar absorption, is proportional to the absorbance, we can expect ε to be between 8000 and 10000 L mol⁻¹cm⁻¹.
Deactivation
Since the ground state, or dark state, is more stable, active LovTAP-VP16 proteins will spontaneously release a photon and go back to the dark state. This rate is parametrized with the half life of the photoproduct.
Deactivation parameters
For 4ºC, Kasahara et al (2002) don't provide the value for the half life of the photoproduct, but just the proportion of photoproduct in the equilibrium: 0.91. The half life that allows for this is τ1/2 = 200 s. To verify whether this setup makes sense, we compared it to the experimental results presented in Salomon et al (2000). The only experiment they did with monochromatic light was at 445 nm and 55 µmol m⁻²s⁻¹, or 15 mW/m². In ice, they measured a kppf/kreg of 17, being kreg = ln(2)/τ1/2 and kppf = F Q σ. Using the previously selected parameters, our model yields exactly 17 for that value.
In both references it's said that the half life at room temperature is around 30 s. That would mean the half life being divided by 3.33 every 10ºC. It would then be reasonable to think that at physiological temperature, more than 10ºC over room temperature, the half life would drop to around 9 s.
Results
