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Author Topic: How do I calculate Fluoresence resonant energy transfer (FRET)?  (Read 4653 times)

Offline Variola

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Hello all  :)


Does anyone here know anything about FRET? ( Fluoresence resonant energy transfer??
I am doing some coursework,  and despite my best efforts I am struggling with the equation. We had half a hurried lecture on FRET and my lecture slides are not clear!

This is the question I have  been given:

The fluorescence resonance energy transfer efficiency measured between a rhodamine dye molecule and a cyanine dye on either end of a 15 base strand of DNA is E = 75%

a. Given that the 50% transfer distance between the two dyes is R0 = 60 (Angstroms) what is the separation of the two dye labels on the DNA.


And this is the equation we were given to work it out.

E=  R06            
  R6 + R06
While I understand what Ro is, I cannot grasp what Ro6 is! So I have no idea how to rearrange the equation to get the separation number Google has so far been quite unhelpful and now I am really quite stuck. Any help would be really appreciated!!  :)

EDIT: Just realised I hadn't set this up as a question. Oops!
« Last Edit: 26/05/2009 22:55:04 by Variola »


 

Offline DrN

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Ah, a few years ago I spent what at the time felt like a lifetime in a darkened room doing FRET and the like on my little fibroblasts. And although the confocal was programmed to do much of the grunt work, I really should be able to answer that. Will go away and try and remember what we did! 
 

Offline DrN

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Fluoresence resonant energy transfer (FRET)
« Reply #2 on: 26/05/2009 23:16:01 »
OK, so I mostly did FRAP and FLIP, but did do some FRET. Can't really remember a lot, but my incomplete garbling may help fill some gaps somewhere!

R0 is the distance between the donor and acceptor probes where teh energy transfer is 50%. The equation is generally written with the '0's being subscript and the '6' being superscript. I guess this is something to do with the fact that 10 Angstrom = 1 nm, so the 60 Angstrom distance between the probes is 6 nm (but don't quote me - I'm working on it!).

E = R06/(R06+R6)

This graph is quite good:




So, I think essentially, in this case:
R06= 60
E = 0.75

You can re-arrange the equation as follows: (I'm not doing sub and superscript anymore, its getting annoying!)

E(R06+R6) = R06             (multiply both sides by (R06+R6))

(E x R06) + (E x R6) = R06  (multiply out)

R06 + R6 = R06/E            (divide both sides by E)

R6 = (R06/E) - R06          (minus R06)

Substitute the numbers:

R6 = (60/0.75)-60 = 20 Angstrom.

Does this make sense? The two probes must be closer than 60 A to get a transfer efficiency above 50%.

 

Offline Variola

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You are an absolute star thankyou so much!!!

Yes it makes much more sense now I understand what they mean!!  [:o)]  :) :) I can get on with it rather than staring blankly at the page!!! I dont usuall like posting work related things up here but google was no help...unless I wanted to know the FRET spacing on a guitar!!! ;D

 

Offline DrN

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I'm a bit (OK, a lot) rusty, but it makes sense to me. Don't blame me if I've missed something though please!!  ;D

Actually brings back how interesting I found all this. FRAP especially was lovely to do - fluorescence recovery after photobleaching. Basically we used the confocal to bleach a tiny patch of fluorescence in a cell, then calculated how long it took the fluorescence to recover as a measure of how mobile the fluorescently-labelled proteins in the cell were. Got the paper published in JBC. (O'Toole et al, JBC 2003; 278(46):45770-6 if you're interested)

 

Offline DrN

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