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I've been having trouble with deriving the equations in the following problem.

The interaction between DNA and AO to form the AO–DNA complex can be expressed by the following reaction:

$$\ce{AO + DNA <=> AO-DNA},$$

whose equilibrium constant is $$K = \frac{\ce{[AO-DNA]}}{\ce{[AO][DNA]}} \tag{1}$$

Consider that initially there is only AO in the measuring cell displaying an emission at $\lambda_{em} = \pu{520 nm}$, and finally at equilibrium both AO and AO–DNA complex have emission at the same wavelength.

Additionally, the binding equilibrium constant for AO intercalation to DNA (ignore AO self-aggregation and dimerization) can be determined from the equation:

$$\frac{C_\ce{AO}}{ΔF} = \frac{1}{Δϕ}+\frac{1}{ΔϕK}\frac{1}{\ce{[DNA]}} \tag{2}$$

$F - ϕ_\ce{AO} \cdot C_\ce{AO} = ΔF$, $F = ϕ_i \cdot C_i$, and $ϕ_\ce{AO} - ϕ_\ce{AO-DNA} = Δϕ$ is given.

$F$ is the overall intensity, $ϕ$ is the fluorescence constant and $i$ denotes a particular component.

1) Show that $ΔF = [\ce{AO-DNA}]Δϕ$

2) Derive equation (2) starting from equation (1)

In terms of the first part, I tried working out formulas for both initial and final fluorescence intensities $F$ using $F = ϕ_i \cdot C_i$ but didn't really get anywhere after that. In terms of the second part, I tried substituting $ΔF = [\ce{AO-DNA}]Δϕ$ into the equilibrium equation but didn't really get further from that either. I would really like to get some tips on where to get started.

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    $\begingroup$ It would help if you would edit to define all the terms. I can guess what the concentrations and quantum efficiencies are, but you can eliminate the guessing. $\endgroup$
    – Ed V
    Apr 25, 2020 at 21:45
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    $\begingroup$ I'm guessing that $\phi$ = quantum efficiency, F = final fluorescence intensity. Something that disturbs me is that the molar attenuation coefficient isn't considered which makes this more complicated. $\endgroup$
    – MaxW
    Apr 25, 2020 at 21:56
  • $\begingroup$ I must admit being totally baffled as to what the subscript $i$ means in $\ce{F = ϕ_i*C_i}$. Maybe initial? If so then why isn't it $\ce{F_i = ϕ_i*C_i}$ ? $\endgroup$
    – MaxW
    Apr 25, 2020 at 23:50
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    $\begingroup$ dah... I think I figured it out. F is the total intensity and $i$ denotes a particular component. So it should be $$\ce{}F = \sum\limits_{i = 1}^n ϕ_i*C_i$$ $\endgroup$
    – MaxW
    Apr 26, 2020 at 0:44
  • $\begingroup$ Sorry about the unclarity. $F$ is in fact the total intensity, $ϕ$ is the fluorescence constant and $i$ denotes a particular component. $\endgroup$ Apr 26, 2020 at 6:54

1 Answer 1

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Concentrating on fluorescence intensity for the moment the equation becomes $I_O = I_D $ where $I_O$ is that for the free dye and $I_D$ for that bound to the DNA. If $\alpha$ dissociates the equilibrium is then

$$\begin{align} &I_O \quad \rightleftharpoons &I_D \\ &1-\alpha &\alpha \end{align}$$

and the equilibrium constant $K=I_D/I_O=\alpha/(1-\alpha)$.

The total emission is $\displaystyle F=I_O(1-\alpha)+I_D\alpha=I_O\frac{1}{1+K}+I_D\frac{K}{1+K}$

Rearranging this gives $\displaystyle \frac{1}{I_O+KI_D}+\frac{1}{I_D}=\frac{1}{F}$

Replacing intensity with product of fluorescence yield and concentration and using the rather odd definitions given should give the equation you seek. Note that $F-\phi_{AO}C_{AO}$ is the intensity of the bound dye.

The first question states that the intensity due to the bound dye $F-\phi_{AO}C_{AO}$ is equal to the amount bound times the difference in yields, this because the dye fluoresces at the same wavelength whether bound or not.

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  • $\begingroup$ I don't quite understand how $I_O$ and $I_D$ are defined using the definitions given. In other words, I'm not sure with what I should replace the intensities $I_O$ and $I_D$. If I understood correctly $I_D = {F - ϕ_{AO}C_{AO}}$. If, so how should the $I_O$ be replaced? $\endgroup$ Apr 26, 2020 at 11:28
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    $\begingroup$ $\phi_{A0}C_{A0}$ the measured intensity being proportional to yield and how much is there and similarly for the complex. $\endgroup$
    – porphyrin
    Apr 26, 2020 at 12:30
  • $\begingroup$ Thanks! Would it also be possible to provide help on how to derive the equation shown in the first question using the given definitions? $\endgroup$ Apr 26, 2020 at 12:58
  • $\begingroup$ see the last paragraph which explains this in words $\endgroup$
    – porphyrin
    Apr 26, 2020 at 14:08

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