Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Frostbite

  • one year ago

DNA-meltingpoint determination

  • This Question is Closed
  1. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    hm have you taken a look at this page? http://en.wikipedia.org/wiki/Nucleic_acid_thermodynamics

  3. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Nope... and it sure say it straight forward.

  4. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    except some of it's logic don't make since to me through their derivation.

  6. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A lovely linear line.

  8. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    And I bet you got to use the Van Hoff equation.

  9. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Thats on the page too, but it's written as the non-reciprocal form

  10. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What about the 2?

  11. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Wait for that sequence I use [A] = [B]

  13. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    SO [AB]=2x[A}

  14. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    yeah, that seems reasonable, but you have to take into account the definition of \(T_m\), where half is bound, half unbound

  15. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I just define it from the gibbs energy and say that K is meassured at the meltingpoint T_m. I guess?

  16. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    hm so you're assuming that Gibbs doesn't change as a function of temp?

  17. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Assume that the change heat capacity is 0

  19. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    seems reasonable :)

  20. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\Large \ln \left[ K(T _{M}) \right]-\ln \left[ K(T) \right]=-\frac{ \Delta H }{ R }T_{M}^{-1}+\frac{ \Delta H }{ R }T ^{-1}\]

  22. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @amistre64 can you evaluate if the integration is correct?

  23. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, well thanks for looking at it anyway, wish you the best of luck with your finals! :)

  25. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx :)

  26. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sithgiggles and a ton of others could prolly do it in thier sleep

  27. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Now that makes me just sad to hear.

  28. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Sorry it most be \[\Large C_{T}=2 \times [D]+[M]\]

  30. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    yeah, out of pure curiosity, i'd like to see it!

  33. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  34. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    hahaha.. i'll give it a read later, thanks dude

  35. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what?

  36. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Here it is.

  37. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I've remember that minus sign? I started by multiplying into the bracket instead of evaluating it.

  39. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But I may have done a error

  40. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    yeah, i think you missed it

  41. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I still think that this rewriting was awesome: \[\LARGE \ln[K(T)]=\frac{ \Delta G }{ RT }\]

  42. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    except it is wrong.... I need a minus!

  43. Frostbite
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @aaronq there is the missing minus?

  44. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \(-\dfrac{R}{\Delta H}( ln(C^{-1}_T)-\dfrac{\Delta H-T\Delta S-\Delta H }{RT})\) 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)

  45. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.