# Deriving fluorescence intensity equations

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.

• 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.
– Ed V
Apr 25, 2020 at 21:45
• 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.
– MaxW
Apr 25, 2020 at 21:56
• 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 $\ce{F_i = ϕ_i*C_i}$ ?
– MaxW
Apr 25, 2020 at 23:50
• dah... I think I figured it out. F is the total intensity and $i$ denotes a particular component. So it should be $$\ce{}F = \sum\limits_{i = 1}^n ϕ_i*C_i$$
– MaxW
Apr 26, 2020 at 0:44
• Sorry about the unclarity. $F$ is in fact the total intensity, $ϕ$ is the fluorescence constant and $i$ denotes a particular component. Apr 26, 2020 at 6:54

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.

• 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? Apr 26, 2020 at 11:28
• $\phi_{A0}C_{A0}$ the measured intensity being proportional to yield and how much is there and similarly for the complex. Apr 26, 2020 at 12:30
• Thanks! Would it also be possible to provide help on how to derive the equation shown in the first question using the given definitions? Apr 26, 2020 at 12:58
• see the last paragraph which explains this in words Apr 26, 2020 at 14:08