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Frostbite

  • 2 years ago

DNA-meltingpoint determination

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  1. Frostbite
    • 2 years ago
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    For a oligonucleotid with a self-complementary sequence when base pairing antiparallel, how could we relate the equilibrium constant with the meltingpoint temperature and total amount of nucleotid \(C_{T}\) @aaronq

  2. aaronq
    • 2 years ago
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    hm have you taken a look at this page? http://en.wikipedia.org/wiki/Nucleic_acid_thermodynamics

  3. Frostbite
    • 2 years ago
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    Nope... and it sure say it straight forward.

  4. aaronq
    • 2 years ago
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    yeah, you should search it on google, a lot of interesting websites came up. I've, myself, never done this, but it sure is interesting.

  5. Frostbite
    • 2 years ago
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    except some of it's logic don't make since to me through their derivation.

  6. Frostbite
    • 2 years ago
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    I wish to derive the expression: \[\LARGE \frac{ 1 }{ T_{m} }=\frac{ R }{ \Delta H^\Theta } \ln(C_{T})+\frac{ \Delta S^\Theta }{ \Delta H^\Theta }\]

  7. Frostbite
    • 2 years ago
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    A lovely linear line.

  8. Frostbite
    • 2 years ago
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    And I bet you got to use the Van Hoff equation.

  9. aaronq
    • 2 years ago
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    Thats on the page too, but it's written as the non-reciprocal form

  10. Frostbite
    • 2 years ago
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    What about the 2?

  11. aaronq
    • 2 years ago
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    well they said they're assuming no partial binding states, \(T_m\) being defined as 1/2 of \(C_T\) being unhybridized, it makes sense that the total amount is divided by 2.

  12. Frostbite
    • 2 years ago
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    Wait for that sequence I use [A] = [B]

  13. Frostbite
    • 2 years ago
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    SO [AB]=2x[A}

  14. aaronq
    • 2 years ago
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    yeah, that seems reasonable, but you have to take into account the definition of \(T_m\), where half is bound, half unbound

  15. Frostbite
    • 2 years ago
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    I just define it from the gibbs energy and say that K is meassured at the meltingpoint T_m. I guess?

  16. aaronq
    • 2 years ago
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    hm so you're assuming that Gibbs doesn't change as a function of temp?

  17. Frostbite
    • 2 years ago
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    Hmmm well I was thinking about to use that: \[\LARGE \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p}=\frac{ \Delta H }{ RT^2 }\]

  18. Frostbite
    • 2 years ago
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    Assume that the change heat capacity is 0

  19. aaronq
    • 2 years ago
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    seems reasonable :)

  20. Frostbite
    • 2 years ago
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    Okay I give it total new try from the van Holff equation \[\LARGE \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p}=\frac{ \Delta H }{ RT^2 }\] \[\Large \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p} = \frac{ \Delta H }{ RT^2 } \rightarrow \int\limits_{T}^{T _{M}} \partial \ln(K)=\int\limits_{T}^{T _{M}}\frac{ \Delta H }{ RT^2 }\partial T\] \[\Large \left[ \partial \ln(K) \right]_{T}^{T _{M}}=-\left[ \frac{ \Delta H }{ R }T^{-1} \right]_{T}^{T _{M}}\]

  21. Frostbite
    • 2 years ago
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    \[\Large \ln \left[ K(T _{M}) \right]-\ln \left[ K(T) \right]=-\frac{ \Delta H }{ R }T_{M}^{-1}+\frac{ \Delta H }{ R }T ^{-1}\]

  22. Frostbite
    • 2 years ago
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    @amistre64 can you evaluate if the integration is correct?

  23. amistre64
    • 2 years ago
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    sorry, but i cant focus on that. occupied with finals this week and its messing up my concentration :/ That and id have to read up on it to make sure

  24. Frostbite
    • 2 years ago
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    Oh, well thanks for looking at it anyway, wish you the best of luck with your finals! :)

  25. amistre64
    • 2 years ago
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    thnx :)

  26. amistre64
    • 2 years ago
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    sithgiggles and a ton of others could prolly do it in thier sleep

  27. Frostbite
    • 2 years ago
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    Now that makes me just sad to hear.

  28. Frostbite
    • 2 years ago
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    |dw:1386772266680:dw| Lets look at the reaction: \[\LARGE \sf M+M \leftrightharpoons D\] with the belong equlibriumconstant: \[\Large K=\frac{ \left[ D \right] }{ \left[ M \right]^{2} }\] The total concentration most be: \[\Large C _{T}=[D]+2 \times [M]\] From the drawing we define the meltingpoint ( @aaronq as you suggested) \[\Large [M]=2 \times [D]=C _{T}/2\] \[\Large K(T _{M})=\frac{ [D] }{ [M]^{2} }=\frac{ C _{T}/4 }{ (C_{T}/2)^{2} }=\frac{ 1 }{ C _{T} }\]

  29. Frostbite
    • 2 years ago
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    Sorry it most be \[\Large C_{T}=2 \times [D]+[M]\]

  30. aaronq
    • 2 years ago
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    that seems right, dude. Do you think writing the equation as \(D\rightleftharpoons M +M \) then using the \(K_D=\dfrac {[M]^2}{[D]}\) would make a difference?

  31. Frostbite
    • 2 years ago
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    I suppose, I only know that if you substitute the expression for the equilibrium constant into the evaluated integral you get to derive the equation I wanted. (I can upload the results if any interest)

  32. aaronq
    • 2 years ago
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    yeah, out of pure curiosity, i'd like to see it!

  33. Frostbite
    • 2 years ago
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    Not mine (sorry I got to disappoint you all, but I did not write articles when I was 4 years old).

  34. aaronq
    • 2 years ago
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    hahaha.. i'll give it a read later, thanks dude

  35. abb0t
    • 2 years ago
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    what?

  36. Frostbite
    • 2 years ago
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    Here it is.

  37. aaronq
    • 2 years ago
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    i think you made a small mistake in step 9, inside the bracket it should be \(ln(C^{-1}_T)-\dfrac{\Delta H\color{red}-T\Delta S-\Delta H }{RT}=-ln(C_T)+\dfrac{\Delta S}{R}\)

  38. Frostbite
    • 2 years ago
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    I've remember that minus sign? I started by multiplying into the bracket instead of evaluating it.

  39. Frostbite
    • 2 years ago
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    But I may have done a error

  40. aaronq
    • 2 years ago
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    yeah, i think you missed it

  41. Frostbite
    • 2 years ago
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    I still think that this rewriting was awesome: \[\LARGE \ln[K(T)]=\frac{ \Delta G }{ RT }\]

  42. Frostbite
    • 2 years ago
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    except it is wrong.... I need a minus!

  43. Frostbite
    • 2 years ago
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    @aaronq there is the missing minus?

  44. aaronq
    • 2 years ago
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    \(-\dfrac{R}{\Delta H}( ln(C^{-1}_T)-\dfrac{\Delta H-T\Delta S-\Delta H }{RT})\) distribute the minus \(-\dfrac{R}{\Delta H}( ln(C^{-1}_T)+\dfrac{-\Delta H+T\Delta S + \Delta H }{RT})\) \(-\dfrac{R}{\Delta H}( ln(C^{-1}_T)+\dfrac{T\Delta S}{RT})\) \(-\dfrac{R}{\Delta H}(- ln(C_T)+\dfrac{\Delta S}{R})\) \(\dfrac{R}{\Delta H}ln(C_T)-\dfrac{\Delta S}{\Delta H}\) (10)

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