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Frostbite Group TitleBest ResponseYou've already chosen the best response.1
For 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
 11 months ago

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
Nope... and it sure say it straight forward.
 11 months ago

aaronq Group TitleBest ResponseYou've already chosen the best response.3
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.
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
except some of it's logic don't make since to me through their derivation.
 11 months ago

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
A lovely linear line.
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
And I bet you got to use the Van Hoff equation.
 11 months ago

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
What about the 2?
 11 months ago

aaronq Group TitleBest ResponseYou've already chosen the best response.3
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.
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
Wait for that sequence I use [A] = [B]
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
SO [AB]=2x[A}
 11 months ago

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

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

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

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
Assume that the change heat capacity is 0
 11 months ago

aaronq Group TitleBest ResponseYou've already chosen the best response.3
seems reasonable :)
 11 months ago

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

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
@amistre64 can you evaluate if the integration is correct?
 11 months ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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
 11 months ago

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
thnx :)
 11 months ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
sithgiggles and a ton of others could prolly do it in thier sleep
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
Now that makes me just sad to hear.
 11 months ago

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

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

aaronq Group TitleBest ResponseYou've already chosen the best response.3
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?
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
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)
 11 months ago

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

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

aaronq Group TitleBest ResponseYou've already chosen the best response.3
hahaha.. i'll give it a read later, thanks dude
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
Here it is.
 11 months ago

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
I've remember that minus sign? I started by multiplying into the bracket instead of evaluating it.
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
But I may have done a error
 11 months ago

aaronq Group TitleBest ResponseYou've already chosen the best response.3
yeah, i think you missed it
 11 months ago

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
except it is wrong.... I need a minus!
 11 months ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.1
@aaronq there is the missing minus?
 11 months ago

aaronq Group TitleBest 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)
 11 months ago
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