Question

DNA-meltingpoint determination

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

A lovely linear line.

Answer 8

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

Answer 9

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

Answer 10

What about the 2?

Answer 11

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.

Answer 12

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

Answer 13

SO [AB]=2x[A}

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

Assume that the change heat capacity is 0

Answer 19

seems reasonable :)

Answer 20

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

Answer 21

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

Answer 22

@amistre64 can you evaluate if the integration is correct?

Answer 23

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

Answer 24

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

Answer 25

thnx :)

Answer 26

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

Answer 27

Now that makes me just sad to hear.

Answer 28

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