# Dwell time equations for ATP-sythase?

I have read that every 120 degree rotation of the F1 complex of ATP-synthase can be split into a 30 degree rotation and a 90 degree rotation. In between these two are dwell times, the one before the 90 degree rotation been called the ATP-binding dwell time and the one after been called the interim dwell time and that both of these fall of exponentially. I cannot, however, find a derivation of the probability density function of these dwell times, or there exact form. If anyone knows such a derivation or where to find one this would be much appreciated.

• Are you asking for the mathematical model or the experimental measurements ? – WYSIWYG Sep 7 '15 at 6:00
• @WYSIWYG The mathematical model – Quantum spaghettification Sep 7 '15 at 6:01
• Usually, the dwell time between independent reactions is exponentially distributed. You can have a look at this paper for a biophysics based mathematical model. – WYSIWYG Sep 7 '15 at 8:09
• @WYSIWYG I am pretty sure that one depends on the concentration of atp present though – Quantum spaghettification Sep 8 '15 at 13:00
• The rate of binding depends on the concentration of ADP (the substrate) but not the rate of conversion to ATP. – WYSIWYG Sep 8 '15 at 13:03

ATP Binding dwell time

Here is a very dodgy derivation of the binding dwell time (note I am the OP). For this we have $$E+S\rightarrow ES$$ Where $E$ is the enzyme and $S$ is the substrate (keeping this general). We can write our rate equation as: $$rate=k[E][S]$$ But we can also write the rate as: $$rate=-\frac{d[E]}{dt}$$ so that: $$\frac{d[E]}{dt}=-k[E][S]$$ Now here is the bit that I think is a bit dodgy, we are going to assume that $[S]$ is approximately constant. So that: $$[E]=Ae^{-k[S]t}$$ We can take $n=[E]V$ to represent the number of enzymes left, where $V$ is the volume, then letting $n_0=VA=constant$: $$n=n_0e^{-k[S]t}$$ The probability of been within the time $t$ is then given by: $$F(t)=1-e^{-k[S]t}$$ Which is our commutative distribution function. Differentiating this gives our probability density function: $$f(t)=k[S]e^{-k[s]t}$$

Interim dwell time

The interim dwell time (also called the catalytic dwell time) involves two steps: 1. The cleavage of the enzyme bound ATP. 2. The release of the hydrolysed products.

Each of which is going to follow the distribution (analogous to the above) of:

$$p_i(t)=k_i e^{-k_i t_i}$$ For $i=1,2$ respectively. The joint probability distribution is found by the convolution of these two : $$p_T(t)=\int^{\tau}_0 p_1(t) p_2(\tau-t)dt$$ Which gives: $$p_T(t)=\frac{k_1 k_2}{k_1-k_2} (e^{-k_2 \tau}-e^{-k_1\tau})$$

Edit

Although starting from deterministic equations (which actually doesn't really matter), this method does hold and is simple. The type of reaction is called a pseudo-1st-order reaction (or 2nd order class 2) which explains my approximation for $[S]$ been constant. See  and 

• @WYSIWYG Please can you explain, the equation $n=n_0 e^{-k[s]t}$ is equivalent to that for nuclear decay, which gives an equivalent probability density to that I have given. – Quantum spaghettification Sep 8 '15 at 18:18