## transition probabilities in 3B

Bayesian analysis of blinking and bleaching

### transition probabilities in 3B

Hi, I have some questions about transition probabilities in 3B.

1. Here, transition probabilities are given by P1=0.16,P2=0.84,P3=P4=0.495. How can you get those values?
2. "a fluorophore was more likely to remain in the on state than to transition to the off state." I think this can lead that the probability of P1 is larger than P2, but why the value of P2 is larger.
3. "The values of P3 and P4 were taken with a variety of typical off-state lifetimes (10-3–-10–2 s) assuming a monoexponential decay. This led to values of P4 between 0.1 and 0.8." I don't understand why off-state lifetimes led this values of P4.

xufan

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Joined: Wed May 29, 2013 9:08 am

### Re: transition probabilities in 3B

Hi Fan Xu,

The probability calculations were done using figures for lifetimes from the paper I cited. However I would note at the start that I don't think anymore that the dark state that we see is really what would be classically thought of as the triplet state. The photochemistry of these dyes and proteins is pretty complex and requires a lot of approximations. Having said that, the results we get are only weakly dependent on the transition probabilities (as long as they all add up correctly).

The next thing to note is that the states which I use, 'emitting', 'not emitting', and 'bleached', don't correspond directly to electronic states, and I think this has led to some confusion, as I've used the word state to discuss both the electronic states and the named states, such as 'emitting', which don't correspond directly to electronic states. If we are in the emitting state, then the fluorophore is repeatedly transitioning from the ground state to the singlet state and emitting a photon, over and over again. So in answer to:
(2) "a fluorophore was more likely to remain in the on state than to transition to the off state." I think this can lead that the probability of P1 is larger than P2, but why the value of P2 is larger.
So if a fluorophore starts from the ground state, the most likely thing to happen is for it to be excited into the singlet state, and then decay and emit a photon. So for each point at which a different outcome was possible (i.e. each time the fluorophore absorbed a photon), the most likely thing is for it to remain in the emitting state. However, the singlet state has a much much shorter lifetime than the triplet state (or whatever the dark state is, as discussed above). So even though, for each photon absorption, the most likely thing is that the fluorophore remains in the emitting state, because the lifetime of that state is so much shorter than the lifetime of the triplet state, the probability of P1 comes out smaller.

So we have transition probabilities P1, P2, P3, P4 and P5.
From the state diagram in the paper we can get the following relationships:
P1+P2 = 1 (a fluorophore in the emitting state must either stay in the emitting state or go into the non-emitting state)
P3+P4+P5 = 1 (a fluorophore in the non-emitting state must either stay in the non-emitting state, go into the emitting state, or enter the bleached state)
P6 = 1 (once a fluorophore has entered the bleached state, it cannot leave)

We then have to know something about the lifetimes of each of these states, and how likely they are to happen relative to each other.
Let's call the lifetimes tau1 for the singlet state and tau2 for the so-called triplet state.
We would also need to know the probablility to go from ground to singlet rather than ground to triplet - generally estimated to be 1000 - 10000.
We also have to estimate the probability of triplet to ground versus triplet to permanently bleached - generally estimated to be around 100 (though for this and the above quite a range of values have been reported.

We model the fluorescence emission as a monoexponential decay process, so N = N0 exp (-t/tau)
Consider a fluorophore which is in the singlet state at t=0, then at a later time t
P(still in singlet state) = exp (-t/tau1)
P(decayed to ground state) = 1 - exp(-t/tau1)

For a fluorophore in the triplet state at t=0
P(still in triplet state) = exp(-t/tau2)
P(decayed to ground state = (1-exp(-t/tau2))* 0.99 (if you take the value of 100 for the relative likelihood of going from triplet to ground vs. triplet to permanently bleached)
P(decayed to bleached state) = (1 -exp(-t/tau2)*0.01

Now let's consider what we're doing in a typical experiment. We take a camera image, which is say 20ms long, and what we want to know is what state the fluorophore is at the end of that. We know the relative probabilities of each transition, and we know their lifetimes, so now we just need to do the calculation of how likely the fluorophore is to be in each state at the end of a single frame.

I hope that's made it a bit clearer, do ask if you'd like any further details.
susancox

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Joined: Wed Jul 18, 2012 3:34 pm 