DNA-meltingpoint determination

- Frostbite

DNA-meltingpoint determination

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Frostbite

For a oligonucleotid with a self-complementary 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

hm have you taken a look at this page? http://en.wikipedia.org/wiki/Nucleic_acid_thermodynamics

- Frostbite

Nope... and it sure say it straight forward.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- aaronq

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.

- Frostbite

except some of it's logic don't make since to me through their derivation.

- Frostbite

I wish to derive the expression:
\[\LARGE \frac{ 1 }{ T_{m} }=\frac{ R }{ \Delta H^\Theta } \ln(C_{T})+\frac{ \Delta S^\Theta }{ \Delta H^\Theta }\]

- Frostbite

A lovely linear line.

- Frostbite

And I bet you got to use the Van Hoff equation.

- aaronq

Thats on the page too, but it's written as the non-reciprocal form

##### 1 Attachment

- Frostbite

What about the 2?

- aaronq

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.

- Frostbite

Wait for that sequence I use [A] = [B]

- Frostbite

SO [AB]=2x[A}

- aaronq

yeah, that seems reasonable, but you have to take into account the definition of \(T_m\), where half is bound, half unbound

- Frostbite

I just define it from the gibbs energy and say that K is meassured at the meltingpoint T_m. I guess?

- aaronq

hm so you're assuming that Gibbs doesn't change as a function of temp?

- Frostbite

Hmmm well I was thinking about to use that:
\[\LARGE \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p}=\frac{ \Delta H }{ RT^2 }\]

- Frostbite

Assume that the change heat capacity is 0

- aaronq

seems reasonable :)

- Frostbite

Okay I give it total new try from the van Holff equation
\[\LARGE \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p}=\frac{ \Delta H }{ RT^2 }\]
\[\Large \left( \frac{ \partial \ln(K) }{ \partial T } \right)_{p} = \frac{ \Delta H }{ RT^2 } \rightarrow \int\limits_{T}^{T _{M}} \partial \ln(K)=\int\limits_{T}^{T _{M}}\frac{ \Delta H }{ RT^2 }\partial T\]
\[\Large \left[ \partial \ln(K) \right]_{T}^{T _{M}}=-\left[ \frac{ \Delta H }{ R }T^{-1} \right]_{T}^{T _{M}}\]

- Frostbite

\[\Large \ln \left[ K(T _{M}) \right]-\ln \left[ K(T) \right]=-\frac{ \Delta H }{ R }T_{M}^{-1}+\frac{ \Delta H }{ R }T ^{-1}\]

- Frostbite

@amistre64 can you evaluate if the integration is correct?

- amistre64

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

- Frostbite

Oh, well thanks for looking at it anyway, wish you the best of luck with your finals! :)

- amistre64

thnx :)

- amistre64

sithgiggles and a ton of others could prolly do it in thier sleep

- Frostbite

Now that makes me just sad to hear.

- Frostbite

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

- Frostbite

Sorry it most be
\[\Large C_{T}=2 \times [D]+[M]\]

- aaronq

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?

- Frostbite

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)

- aaronq

yeah, out of pure curiosity, i'd like to see it!

- Frostbite

Not mine (sorry I got to disappoint you all, but I did not write articles when I was 4 years old).

##### 1 Attachment

- aaronq

hahaha.. i'll give it a read later, thanks dude

- abb0t

what?

- Frostbite

Here it is.

##### 1 Attachment

- aaronq

i think you made a small mistake in step 9, inside the bracket it should be
\(ln(C^{-1}_T)-\dfrac{\Delta H\color{red}-T\Delta S-\Delta H }{RT}=-ln(C_T)+\dfrac{\Delta S}{R}\)

- Frostbite

I've remember that minus sign? I started by multiplying into the bracket instead of evaluating it.

- Frostbite

But I may have done a error

- aaronq

yeah, i think you missed it

- Frostbite

I still think that this rewriting was awesome:
\[\LARGE \ln[K(T)]=\frac{ \Delta G }{ RT }\]

- Frostbite

except it is wrong.... I need a minus!

- Frostbite

@aaronq there is the missing minus?

- aaronq

\(-\dfrac{R}{\Delta H}( ln(C^{-1}_T)-\dfrac{\Delta H-T\Delta S-\Delta H }{RT})\)
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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.