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Frostbite
 one year ago
DNAmeltingpoint determination
Frostbite
 one year ago
DNAmeltingpoint determination

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

aaronq
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1A lovely linear line.

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

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

aaronq
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Wait for that sequence I use [A] = [B]

aaronq
 one year 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
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Assume that the change heat capacity is 0

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

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

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

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

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

Frostbite
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.3hahaha.. i'll give it a read later, thanks dude

aaronq
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1But I may have done a error

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

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

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

aaronq
 one year 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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