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I don't want to get back into the force just yet, so I added some more to the camera. Now it won't save if the folder it is saving in has more bytes taken up then specified by the user. So there is a maximum amount of bytes that the folder can have. I also copied and pasted the entire .vi into a new .vi to get rid of that mysterious break point from yesterday. Now I am gonna procrastinate and think about adding compression.
Still no compression, but on Andy's request there is now an automated record time. Just put in the on time and the off time so something like Time On: 6:30 PM and Time Off: 7:00 AM. And then switch a button and it'll record between 6:30 and 7:00.
And now there is a compression filter on. I just picked the first one. I don't know which one is the best.
Andy Maloney 00:21, 9 September 2009 (EDT): Oh yeah. You kick ass!
After a quick skimming i found another paper that agrees with what i thought yesterday of just switching the sign and using bell equation for rate constant dependent force. It is Hyeon's "Kinesin's Backsteps under Mechanical Load" page 9. So I am starting to feel a bit better about this idea. This is a long paper but i should come back to it to fully read one day.
So until I talk to Koch, i am feeling more comfortable with this idea. but the question is which transitions get this force dependence. Koch's original idea, which i still believe in, was that just the feet binding and unbinding should matter. I couldn't tell you if the diffusive step should as well. This question'll take some thinking. With the Bell equation there will be a new constant to play around with along with the k values. That is d. If d = 0 then there is no force dependence. So, a low value for d means a low force dependence. A higher value means a greater change with a change in F.
Before I forget there is also another check to make sure everything is working, and that is back stepping and stall force. Apparently there should be a stall force of about (i think) 3-5 pN. This means when the kinesin is pulled back at this force the molecule doesn't move. But when it exceeds this force it can move backwards. I don't know how this'll come about with Bell's equation. Also the stall force can increase with increasing ATP concentration. I am not sure if the rate of back stepping does or not. OK I can kind of see how this can come about with Bell's equation.
So I'll need a system where pulling along the coordinate direction is positive. So this moves the kinesin along faster. While pulling opposite is negative. So the stall force sort of makes sense in that with no force F = 0, the kinesin walks forward. So with a negative force it won't stop immediately. It'll take some extra negative force to slow it down then bring the velocity to 0 and then step backwards. I guess my major question is what does it do during a stall force. I mean the kinesin isn't realy just sitting there. It has to be doing something. It probably is detaching nucleotides. Because if the foot unbound there'd be a good chance to move in a direction. So the unbinding rate constant must've gotten so large that there is a better chance to fuck around with the nucleotide than it woudl be to unbind. or something...i am getting myself confused and have to give it thought
I found another paper "Kinesin: a molecular motor with a spring in its step" by Neil Thomas, and they almost have the same sign change in front of the force, but they have everything that is raised to the exponent squared. So something like . now this is the first i saw something like this (at least since i've started looking). I still haven't spent time looking through these papers i am talking about because i just don't have the energy right now, but at least noting it means i can come back to it in time.