## Frostbite 2 years ago DNA-meltingpoint determination

1. Frostbite

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

3. Frostbite

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

4. aaronq

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

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

6. Frostbite

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

A lovely linear line.

8. Frostbite

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

9. aaronq

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

10. Frostbite

11. aaronq

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

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

13. Frostbite

SO [AB]=2x[A}

14. aaronq

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

15. Frostbite

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

16. aaronq

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

17. Frostbite

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

Assume that the change heat capacity is 0

19. aaronq

seems reasonable :)

20. Frostbite

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

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

22. Frostbite

@amistre64 can you evaluate if the integration is correct?

23. amistre64

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

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

25. amistre64

thnx :)

26. amistre64

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

27. Frostbite

Now that makes me just sad to hear.

28. Frostbite

|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

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

30. aaronq

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

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

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

33. Frostbite

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

34. aaronq

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

35. abb0t

what?

36. Frostbite

Here it is.

37. aaronq

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

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

39. Frostbite

But I may have done a error

40. aaronq

yeah, i think you missed it

41. Frostbite

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

42. Frostbite

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

43. Frostbite

@aaronq there is the missing minus?

44. aaronq

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