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Frostbite
 3 years ago
DNAmeltingpoint determination
Frostbite
 3 years ago
DNAmeltingpoint determination

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Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1For a oligonucleotid with a selfcomplementary sequence when base pairing antiparallel, how could we relate the equilibrium constant with the meltingpoint temperature and total amount of nucleotid \(C_{T}\) @aaronq

aaronq
 3 years ago
Best ResponseYou've already chosen the best response.3hm have you taken a look at this page? http://en.wikipedia.org/wiki/Nucleic_acid_thermodynamics

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1Nope... and it sure say it straight forward.

aaronq
 3 years ago
Best ResponseYou've already chosen the best response.3yeah, you should search it on google, a lot of interesting websites came up. I've, myself, never done this, but it sure is interesting.

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1except some of it's logic don't make since to me through their derivation.

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1I wish to derive the expression: \[\LARGE \frac{ 1 }{ T_{m} }=\frac{ R }{ \Delta H^\Theta } \ln(C_{T})+\frac{ \Delta S^\Theta }{ \Delta H^\Theta }\]

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1A lovely linear line.

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1And I bet you got to use the Van Hoff equation.

aaronq
 3 years ago
Best ResponseYou've already chosen the best response.3Thats on the page too, but it's written as the nonreciprocal form

aaronq
 3 years ago
Best ResponseYou've already chosen the best response.3well 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.

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1Wait for that sequence I use [A] = [B]

aaronq
 3 years ago
Best ResponseYou've already chosen the best response.3yeah, that seems reasonable, but you have to take into account the definition of \(T_m\), where half is bound, half unbound

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1I just define it from the gibbs energy and say that K is meassured at the meltingpoint T_m. I guess?

aaronq
 3 years ago
Best ResponseYou've already chosen the best response.3hm so you're assuming that Gibbs doesn't change as a function of temp?

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1Hmmm well I was thinking about to use that: \[\LARGE \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p}=\frac{ \Delta H }{ RT^2 }\]

Frostbite
 3 years ago
Best ResponseYou've already chosen the best response.1Assume that the change heat capacity is 0

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1Okay I give it total new try from the van Holff equation \[\LARGE \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p}=\frac{ \Delta H }{ RT^2 }\] \[\Large \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p} = \frac{ \Delta H }{ RT^2 } \rightarrow \int\limits_{T}^{T _{M}} \partial \ln(K)=\int\limits_{T}^{T _{M}}\frac{ \Delta H }{ RT^2 }\partial T\] \[\Large \left[ \partial \ln(K) \right]_{T}^{T _{M}}=\left[ \frac{ \Delta H }{ R }T^{1} \right]_{T}^{T _{M}}\]

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1\[\Large \ln \left[ K(T _{M}) \right]\ln \left[ K(T) \right]=\frac{ \Delta H }{ R }T_{M}^{1}+\frac{ \Delta H }{ R }T ^{1}\]

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1@amistre64 can you evaluate if the integration is correct?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0sorry, 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

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1Oh, well thanks for looking at it anyway, wish you the best of luck with your finals! :)

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0sithgiggles and a ton of others could prolly do it in thier sleep

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1Now that makes me just sad to hear.

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1dw: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) \[\Large [M]=2 \times [D]=C _{T}/2\] \[\Large K(T _{M})=\frac{ [D] }{ [M]^{2} }=\frac{ C _{T}/4 }{ (C_{T}/2)^{2} }=\frac{ 1 }{ C _{T} }\]

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry it most be \[\Large C_{T}=2 \times [D]+[M]\]

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.3that 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?

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1I 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)

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.3yeah, out of pure curiosity, i'd like to see it!

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1Not mine (sorry I got to disappoint you all, but I did not write articles when I was 4 years old).

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.3hahaha.. i'll give it a read later, thanks dude

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.3i think you made a small mistake in step 9, inside the bracket it should be \(ln(C^{1}_T)\dfrac{\Delta H\color{red}T\Delta S\Delta H }{RT}=ln(C_T)+\dfrac{\Delta S}{R}\)

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1I've remember that minus sign? I started by multiplying into the bracket instead of evaluating it.

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1But I may have done a error

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.3yeah, i think you missed it

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1I still think that this rewriting was awesome: \[\LARGE \ln[K(T)]=\frac{ \Delta G }{ RT }\]

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1except it is wrong.... I need a minus!

Frostbite
 2 years ago
Best ResponseYou've already chosen the best response.1@aaronq there is the missing minus?

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.3\(\dfrac{R}{\Delta H}( ln(C^{1}_T)\dfrac{\Delta HT\Delta S\Delta H }{RT})\) 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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