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.
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$\begingroup$ Are you asking for the mathematical model or the experimental measurements ? $\endgroup$– WYSIWYGCommented Sep 7, 2015 at 6:00
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$\begingroup$ @WYSIWYG The mathematical model $\endgroup$– Quantum spaghettificationCommented Sep 7, 2015 at 6:01
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$\begingroup$ Usually, the dwell time between independent reactions is exponentially distributed. You can have a look at this paper for a biophysics based mathematical model. $\endgroup$– WYSIWYGCommented Sep 7, 2015 at 8:09
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$\begingroup$ @WYSIWYG I am pretty sure that one depends on the concentration of atp present though $\endgroup$– Quantum spaghettificationCommented Sep 8, 2015 at 13:00
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1$\begingroup$ The rate of binding depends on the concentration of ADP (the substrate) but not the rate of conversion to ATP. $\endgroup$– WYSIWYGCommented Sep 8, 2015 at 13:03
1 Answer
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[7]: 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 [8]: $$p_T(t)=\int^{\tau}_0 p_1(t) p_2(\tau-t)dt$$ 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 [9] and [10]
Sources for additional info
- https://youtu.be/X_YXTWU2maY?list=PLbKSbFnKYVY3j6ubaW1zgTXj5C4443v8s
- http://www.nature.com/articles/srep08773
- http://www.ks.uiuc.edu/Research/atp_hydrolysis/
- http://www.pnas.org/content/85/17/6314.full.pdf
- https://books.google.co.uk/books?id=uIxwICNmLKEC&pg=PA199&lpg=PA199&dq=catalytic+dwell+time+distribution+%5Batp%5D&source=bl&ots=u8oBcVNCDc&sig=PKvdVAwT0WxdqTuVwMxhU3RfSwM&hl=en&sa=X&ved=0CCAQ6AEwADgKahUKEwiGhKDalufHAhXkadsKHc54ADs#v=onepage&q=catalytic%20dwell%20time%20distribution%20%5Batp%5D&f=false
- http://crystal.harvard.edu/PDFs/floyd_biophysj.2010.pdf
- http://www.ncbi.nlm.nih.gov/pubmed/16258036
- http://crystal.harvard.edu/PDFs/floyd_biophysj.2010.pdf
- http://glutxi.umassmed.edu/grad/GradKinetics.pdf
- http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions/Pseudo-1st-order_reactions
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$\begingroup$ You cannot obtain probability distributions from these deterministic equations. $\endgroup$– WYSIWYGCommented Sep 8, 2015 at 18:09
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$\begingroup$ @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. $\endgroup$ Commented Sep 8, 2015 at 18:18
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1$\begingroup$ It will not give you probability distribution because this rate is based on deterministic mass action kinetics.. You have to represent your equations in terms of probability instead of reaction rates to obtain those equations... Have a look at chemical master equations $\endgroup$– WYSIWYGCommented Sep 8, 2015 at 18:26
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$\begingroup$ @WYSIWYG I asked (and then when I came up with an answer, answered) the following question on Chemistry.SE I think my answers provides a replacement for the first part of the derivation using the master equation: chemistry.stackexchange.com/questions/37074/… $\endgroup$ Commented Sep 9, 2015 at 9:27
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