## Frostbite Group Title DNA-meltingpoint determination 8 months ago 8 months ago

1. Frostbite Group Title

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 Group Title

3. Frostbite Group Title

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

4. aaronq Group Title

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 Group Title

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

6. Frostbite Group Title

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 Group Title

A lovely linear line.

8. Frostbite Group Title

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

9. aaronq Group Title

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

10. Frostbite Group Title

11. aaronq Group Title

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 Group Title

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

13. Frostbite Group Title

SO [AB]=2x[A}

14. aaronq Group Title

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

15. Frostbite Group Title

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

16. aaronq Group Title

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

17. Frostbite Group Title

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 Group Title

Assume that the change heat capacity is 0

19. aaronq Group Title

seems reasonable :)

20. Frostbite Group Title

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 Group Title

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

22. Frostbite Group Title

@amistre64 can you evaluate if the integration is correct?

23. amistre64 Group Title

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 Group Title

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

25. amistre64 Group Title

thnx :)

26. amistre64 Group Title

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

27. Frostbite Group Title

Now that makes me just sad to hear.

28. Frostbite Group Title

|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 Group Title

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

30. aaronq Group Title

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 Group Title

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 Group Title

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

33. Frostbite Group Title

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

34. aaronq Group Title

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

35. abb0t Group Title

what?

36. Frostbite Group Title

Here it is.

37. aaronq Group Title

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 Group Title

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

39. Frostbite Group Title

But I may have done a error

40. aaronq Group Title

yeah, i think you missed it

41. Frostbite Group Title

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

42. Frostbite Group Title

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

43. Frostbite Group Title

@aaronq there is the missing minus?

44. aaronq Group Title

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