Calibration curve. (You can do this exercise with your calculator, but it is more easily done by the spreadsheet in Figure.) In the Bradford protein determination, the color of a dye changes from brown to blue when it binds to protein. Absorbance of light is measured.

(a) Find the equation of the least-squares straight line through these points in the form $\mathbf{y}\mathbf{=}\mathbf{\left[}\mathbf{m}\mathbf{\right(}\mathbf{\pm }{\mathbf{u}}_{\mathbf{m}}\mathbf{\left)}\mathbf{\right]}\mathbf{x}\mathbf{+}\mathbf{\left[}\mathbf{b}\mathbf{\right(}\mathbf{\pm }{\mathbf{u}}_{\mathbf{b}}\mathbf{\left)}\mathbf{\right]}$with a reasonable number of significant figures.

(b) Make a graph showing the experimental data and the calculated straight line.

(c) An unknown protein sample gave an absorbance of0.973. Calculate the number of micrograms of protein in the unknown and estimate its uncertainty.

(a)

The figure shows the measured light absorption.

The least-squares straight line equation is as follows:

The equation of the straight line will be as $y(\pm s_y)=\left[m\right(\pm u_m\left)\right]x+\left[b\right(\pm u_b\left)\right]$$y(\pm s_y)=\left[m\right(\pm u_m\left)\right]x+\left[b\right(\pm u_b\left)\right]$

The given data, the value of $x_i^2$, $d_i$and $d_i^2$are tabulated as follows,

$$\begin{gathered} D=\left|\begin{array}{cc}\sum _{}{x}_i^2& \sum _{}{x}_i\\ \sum _{}{x}_i& n\end{array}\right| \\ =\left(2628.29\right)\left(5\right)-\left(93.60\right)\left(93.60\right) \\ =4380.5 \end{gathered}$$

$$\begin{gathered} m=\left|\begin{array}{cc}\sum _{}\left(x_i{y}_i\right)& \sum _{}{x}_i\\ \sum _{}{y}_i& n\end{array}\right|÷D \\ =\frac{\left(101.275\right)\left(5\right)-\left(93.60\right)\left(4.391\right)}{D} \\ =95.377:4380.5 \\ =0.021773 \\ b=\left|\sum _{}\left(x_i^2\right)\sum _{}\left(x_i{y}_i\right)\right|÷D \\ =2061.48÷4380.5 \\ =0.47060 \\ {s}_y^2=\frac{\sum _{}\left(d_i^2\right)}{n-2} \\ =\frac{9.64\times {10}^{-5}}{3} \\ =3.21\times {10}^5 \\ {s}_y=0.00567 \end{gathered}$$

$$\begin{gathered} s_y^2=\frac{{\displaystyle \sum _{}^2}\left(d_i^2\right)}{n-2} \\ =\frac{9.64\times {10}^{-5}}{3} \\ =3.21\times {10}^{-5} \\ {s}_y=0.00567 \\ {u}_m=\sqrt{\frac{s_y^{2}n}{D}} \\ =\sqrt{\frac{\left(3.21\times {10}^{-5}\right)5}{4380.5}} \\ =0.000191 \\ {u}_b=\sqrt{\frac{s_y^2\sum _{}{x}_i^2}{D}} \\ =\sqrt{\frac{\left(3.21\times {10}^{-5}\right)\left(2628.29\right)}{4380.5}} \\ =0.00439 \end{gathered}$$

The below figure shows the spread sheet along with the graph that was used to calculate the straight line for the provided locations

As a result, the straight line equation is expressed as follows:

$$\begin{gathered} y=\left[m\left(\pm u_m\right)\right]x+\left[b\left(\pm u_b\right)\right] \\ =\left[0.0218\left(\pm 0.0002\right)\right]x+\left[0.471\left(0.004\right)\right] \end{gathered}$$

The equation of the least-squares straight line passing through the given points is written as: $y=[0.0218(\pm 0.0002\left)\right]x+[0.471(\pm 0.004\left)\right]$

(b)

The graph below depicts the experimental data as well as the estimated straight line.

(c)

The figure shows the measured light absorption.

Micrograms of unidentified protein sample:

Rewrite the straight-line equation as follows:

$$\begin{gathered} \mathrm{x}=\frac{\mathrm{y}-\mathrm{b}}{\mathrm{m}} \\ =\frac{0.973-0.471}{0.0218} \\ =23.0\mathrm{\mu g} \\ =\mathrm{or}23.07\mathrm{\mu g} \end{gathered}$$

The uncertainty in x is calculated as follows,

Uncertainty in$x\left(s_x\right)$

$$\begin{gathered} {\mathrm{u}}_{\mathrm{x}}=\frac{{\mathrm{s}}_{\mathrm{y}}}{\left|\mathrm{m}\right|}\sqrt{\frac{1}{\mathrm{k}}+\frac{1}{\mathrm{n}}+\frac{{\left(\mathrm{y}-\overline{\mathrm{y}}\right)}^2}{{\mathrm{m}}^2\sum _{}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2}} \\ =\frac{0.00567}{\left|0.02177\right|}\sqrt{\frac{1}{1}+\frac{1}{5}+\frac{\left(0.973-0.8782\right)}{{\left(0.02177\right)}^2\left(876.1\right)}} \\ =0.29\mathrm{\mu g} \end{gathered}$$

The spread sheet used to calculate the amount of micrograms of unknown protein sample and its uncertainty is shown above.

As a result, the spreadsheet calculates the number of micrograms of unknown protein sample and its uncertainty as $23.1\pm 0.3\mu g$.