A community for students.
Here's the question you clicked on:
 0 viewing
Frostbite
 3 years ago
DNAmeltingpoint determination
Frostbite
 3 years ago
DNAmeltingpoint determination

This Question is Closed

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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1@amistre64 can you evaluate if the integration is correct?

amistre64
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0sithgiggles and a ton of others could prolly do it in thier sleep

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

Frostbite
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry it most be \[\Large C_{T}=2 \times [D]+[M]\]

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

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

aaronq
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1But I may have done a error

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

Frostbite
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1except it is wrong.... I need a minus!

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

aaronq
 3 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)
Ask your own question
Sign UpFind 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
 Engagement 19 Mad Hatter
 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.