Question

The protein bovine serum albumin can bind several molecules of the dye methyl orange. To measure the binding constant K for one dye molecule, solutions were prepared with a fixed concentration $\left(x_0\right)$of dye and a larger, variable concentration of protein (P). The equilibrium is Reaction 19-18, with X 5 methyl

orange.

Experimental data are shown in cells A16-D20 in the spreadsheet on the next page. The authors report the increase in absorbance$\mathbf{∆}\mathbf{A}$at 490 nm as P is added to X. X and PX absorb visible light, but P does not. Equilibrium expression 19-20 applies and [PX] is given by Equation 19-21. Before P is added, the absorbance is. The increase in absorbance when P is added is

The spreadsheet uses Solver to vary K and $\mathbf{∆}\mathbf{E}$in cells B10:B11 to minimize the sum of squares of differences between observed and calculatedin solutions with different amounts of P. Cell E16 computes [PX] from Equation 19-21, which is Equation A on line 6 of the spreadsheet. Cells F16 and G16 find [X] and [P] from mass balances. Cell H16 computes ${\mathbf{\Delta A}}_{\mathbf{\text{calc}}}\mathbf{=}\mathbf{\Delta E}\mathbf{\left[}\mathbf{PX}\mathbf{\right]}$which is Equation B on line 7.

To estimate a value of K in cell B10, suppose that 50% of X has reacted in row 20 of the spreadsheet. The total concentration of X is. If half is reacted, then$\mathbf{\left[}\mathbf{X}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\mathbf{PX}\mathbf{\right]}\mathbf{=}\mathbf{2}\mathbf{.85}\mathbf{\mu M}\mathbf{\text{and}}\mathbf{\left[}\mathbf{P}\mathbf{\right]}\mathbf{=}{\mathbf{P}}_{\mathbf{0}}\mathbf{-}\mathbf{\left[}\mathbf{PX}\mathbf{\right]}\mathbf{=}\mathbf{40}\mathbf{.4}\mathbf{-}\mathbf{2}\mathbf{.85}\mathbf{=}\mathbf{37}\mathbf{.55}\mathbf{\mu M}$The binding

constant is $\mathbf{K}\mathbf{=}\mathbf{\left[}\mathbf{PX}\mathbf{\right]}\mathbf{\wedge }\mathbf{P}\mathbf{\left]}\mathbf{\right[}\mathbf{X}\mathbf{\left]}\mathbf{\right)}\mathbf{=}\mathbf{[}\mathbf{2}\mathbf{.85}\mathbf{\mu M}\mathbf{]}\mathbf{\wedge }\mathbf{37}\mathbf{.55}\mathbf{\mu M}\mathbf{\left]}\mathbf{\right[}\mathbf{2}\mathbf{.85}\mathbf{\mu M}\mathbf{\left]}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.7}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}$which we enter as our guess for K in cell B10. We estimatein cell B11 by supposing that 50% of X has reacted in row 20. In Equation B on line 7, $\mathbf{\Delta A}\mathbf{=}\mathbf{\Delta }\mathit{\epsilon }\mathbf{\left[}\mathbf{PX}\mathbf{\right]}$.The measured

value ofin row 20 is 0.0291 and we just estimated that. Therefore, our guess for localid="1668328314124" $$\begin{gathered} \mathbf{\Delta \epsilon }\mathbf{\text{in cell}}\mathbf{B}\mathbf{11}\mathbf{\text{is}}\mathbf{\Delta \epsilon }\mathbf{=}\mathbf{\Delta A}\mathbf{/} \\ \mathbf{\left[}\mathbf{PX}\mathbf{\right]}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.0291}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{.85}\mathbf{\mu M}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}} \end{gathered}$$in cell B11 is

Your assignment is to write formulas in columns E through J of the spreadsheet to reproduce what is shown and to find values in cells E17:J20. Then use Solver to find K andin cells B10:B11 to minimize$\mathbf{\Sigma }{\left({\mathbf{A}}_{\text{oths}}-{\mathbf{A}}_{\text{calc}}\right)}^2$in cell I21.

Short answer

The value of$∆\epsilon$ and K which is in cell B10:B11 will minimize $\mathrm{\Sigma }{\left({\mathrm{A}}_{\text{oths}}-{\mathrm{A}}_{\mathrm{calc}}\right)}^2$in cell I21. We get the spreadsheet as,

Step-by-step solution

Find K:

The value of$∆\epsilon$ and K which is in cell B10:B11 will minimize $\mathrm{\Sigma }{\left({\mathrm{A}}_{\text{oths}}-{\mathrm{A}}_{\mathrm{calc}}\right)}^2$in cell I21. We get the spreadsheet as,