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FrostbiteBest 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
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
hm have you taken a look at this page? http://en.wikipedia.org/wiki/Nucleic_acid_thermodynamics
 4 months ago

FrostbiteBest ResponseYou've already chosen the best response.1
Nope... and it sure say it straight forward.
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
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.
 4 months ago

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

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

FrostbiteBest ResponseYou've already chosen the best response.1
A lovely linear line.
 4 months ago

FrostbiteBest ResponseYou've already chosen the best response.1
And I bet you got to use the Van Hoff equation.
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
Thats on the page too, but it's written as the nonreciprocal form
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
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.
 4 months ago

FrostbiteBest ResponseYou've already chosen the best response.1
Wait for that sequence I use [A] = [B]
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
yeah, that seems reasonable, but you have to take into account the definition of \(T_m\), where half is bound, half unbound
 4 months ago

FrostbiteBest 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?
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
hm so you're assuming that Gibbs doesn't change as a function of temp?
 4 months ago

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

FrostbiteBest ResponseYou've already chosen the best response.1
Assume that the change heat capacity is 0
 4 months ago

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

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

FrostbiteBest ResponseYou've already chosen the best response.1
@amistre64 can you evaluate if the integration is correct?
 4 months ago

amistre64Best 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
 4 months ago

FrostbiteBest 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! :)
 4 months ago

amistre64Best ResponseYou've already chosen the best response.0
sithgiggles and a ton of others could prolly do it in thier sleep
 4 months ago

FrostbiteBest ResponseYou've already chosen the best response.1
Now that makes me just sad to hear.
 4 months ago

FrostbiteBest 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) \[\Large [M]=2 \times [D]=C _{T}/2\] \[\Large K(T _{M})=\frac{ [D] }{ [M]^{2} }=\frac{ C _{T}/4 }{ (C_{T}/2)^{2} }=\frac{ 1 }{ C _{T} }\]
 4 months ago

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

aaronqBest ResponseYou've already chosen the best response.2
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?
 4 months ago

FrostbiteBest 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)
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
yeah, out of pure curiosity, i'd like to see it!
 4 months ago

FrostbiteBest 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).
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
hahaha.. i'll give it a read later, thanks dude
 4 months ago

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

FrostbiteBest 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.
 4 months ago

FrostbiteBest ResponseYou've already chosen the best response.1
But I may have done a error
 4 months ago

aaronqBest ResponseYou've already chosen the best response.2
yeah, i think you missed it
 4 months ago

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

FrostbiteBest ResponseYou've already chosen the best response.1
except it is wrong.... I need a minus!
 4 months ago

FrostbiteBest ResponseYou've already chosen the best response.1
@aaronq there is the missing minus?
 4 months ago

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