Question

In a murder trial in the 1990 s, the defendant's blood was found at the crime scene. The prosecutor argued that blood was left by the defendant during the crime. The defense argued that police "planted" the defendant's blood from a sample collected later. Blood is normally collected in a vial containing the metal-binding compound EDTA (as an anticoagulant) at a concentration of $\mathbf{~}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{mM}$after the vial is filled with blood. At the time of the trial, procedures to measure EDTA in blood were not well established. Even though the amount of EDTA found in the crime-scene blood was orders of magnitude below $\mathbf{~}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{mM}$

, the jury acquitted the defendant. This trial motivated the development of a new method to measure EDTA in blood.

(a) Precision and accuracy. To measure accuracy and precision of the method, blood was fortified with EDTA to known levels.

$$\begin{gathered} \mathbf{Accuracy}\mathbf{=}\mathbf{100}\mathbf{\times }\frac{\mathbf{mean}\mathbf{}\mathbf{value}\mathbf{}\mathbf{found}\mathbf{-}\mathbf{known}\mathbf{}\mathbf{value}}{\mathbf{known}\mathbf{}\mathbf{value}} \\ \mathbf{Precision}\mathbf{=}\mathbf{100}\mathbf{\times }\frac{\mathbf{standard}\mathbf{}\mathbf{deviation}}{\mathbf{mean}}\mathbf{=}\mathbf{coeffcient}\mathbf{}\mathbf{of}\mathbf{}\mathbf{variation} \end{gathered}$$

For each of the three spike levels in the table, find the precision and accuracy of the quality control samples.

(b) Detection and quantitation limits. Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175,104,164,193,131,189,155,133,151, and 176. Ten blanks had a mean reading of 45 - 1 . The slope of the calibration curve is$\mathbf{1}\mathbf{.}\mathbf{75}\mathbf{\times }{\mathbf{19}}^{\mathbf{9}}{\mathbf{M}}^{\mathbf{-}\mathbf{1}}$. Estimate the signal and concentration detection limits and the lower limit of quantitation for EDTA.

Short answer

(a) The required solution is $22.2\mathrm{ng}/\mathrm{mL}:$localid="1654937944575" $\mathrm{precision}\mathrm{is}23.8\%$; accuracy(b)(

(b) $9\mathrm{is}6.6\%$and $88.2\mathrm{ng}/\mathrm{mL}:\mathrm{precision}\mathrm{is}7.8\%;\mathrm{accuracy}\mathrm{and}314\mathrm{ng}/\mathrm{mL}:\mathrm{precision}\mathrm{is}7.8\%;\mathrm{accuracy}\mathrm{is}-3.6\%$

(c) Signal detection Limit $129.6$

Concentration detection limit $\mathrm{is}4.8\times {10}^{-8}\mathrm{M}$

Lower detection of quantitation$\mathrm{is}1.6\times {10}^{-7}\mathrm{M}$

Step-by-step solution

Definition of precision, accuracy and detection and quantitation limits

• Precision refers to how closely a set of measurements are related to one another. Even if the results are not close to the right value, precise measurements are extremely reproducible.
• The degree of similarity between a measurement and its true value is known as accuracy.
• The lowest concentration of analyte that can be consistently detected and quantified are defined as the limits of detection (LOD) and quantitation (LOQ), respectively.

Determine the accuracy and precision for 22.2ng/mL 88.2ng/mL 314ng/mL

(a)

Find the accuracy and precision of the analytical procedure utilised in the first portion of the issue (part a). To do so, we must first calculate the sample mean and standard deviation.

For $$\begin{gathered} 22.2\mathrm{ng}/\mathrm{mL} \\ \overline{\mathrm{x}}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{\left(33.3+19.5+20.8+20.8\right)\mathrm{ng}/\mathrm{mL}}{5} \\ =\frac{118.3\mathrm{ngmL}}{5} \\ \overline{\mathrm{x}}=23.66\mathrm{ng}/\mathrm{mL} \\ \mathrm{s}=\sqrt{\frac{{\mathrm{x}}_1-\overline{\mathrm{x}})}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(33.3-23.66\right)}^2+{\left(19.5-23.66\right)}^2+{\left(20.8-23.66\right)}^2+{\left(20.8-23.66\right)}^2}{4}}\mathrm{ng}/\mathrm{mL} \\ =\sqrt{31.663\mathrm{ng}/\mathrm{mL}} \\ \mathrm{s}=5.626988537\mathrm{ng}/\mathrm{mL} \end{gathered}$$

Now solve for precision and accuracy once it accounted for the sample mean and standard deviation:

$$\begin{gathered} \mathrm{precision}=\frac{\mathrm{standard}\mathrm{deviation}}{\mathrm{mean}}\times 100 \\ =\frac{5.626988537\mathrm{ng}/\mathrm{mL}}{23.66\mathrm{ng}/\mathrm{mL}}\times 100 \\ =23.78270726\% \\ \mathrm{precision}=23.8\% \\ \mathrm{accuracy}=\frac{\mathrm{mean}\mathrm{value}-\mathrm{known}\mathrm{value}}{\mathrm{known}\mathrm{value}}\times 100 \\ =\frac{\left(23.66-22.2\right)\mathrm{ng}/\mathrm{mL}}{22.2\mathrm{ng}/\mathrm{mL}}\times 100 \\ =6.576576577\% \\ \mathrm{accuracy}=6.6\% \end{gathered}$$

Let's go on to the second spike, 88.2ng/ML. We'll just do the same thing we did with the prior spike: we'll solve for sample mean standard deviation first, then precision and accuracy.

For localid="1654939640907" $88.2\mathrm{ng}/\mathrm{mL};$

$$\begin{gathered} \overline{\mathrm{x}}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{\left(83.6+69.0+83.4+100.0+76.4\right)\mathrm{ng}/\mathrm{mL}}{5} \\ =\frac{412.4\mathrm{ng}/\mathrm{mL}}{5} \\ \overline{\mathrm{x}}=82.48\mathrm{ng}/\mathrm{mL} \\ \mathrm{s}=\sqrt{\frac{{\left({\mathrm{x}}_1-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(83.6-82.48\right)}^2+{\left(69.0-82.48\right)}^2+{\left(83.4-82.48\right)}^2+{\left(100.0-82.48\right)}^2+{\left(76.4-82.48\right)}^2}{4}}\mathrm{ng}/\mathrm{mL} \\ =\sqrt{131.932\mathrm{ng}/\mathrm{mL}} \\ \mathrm{s}=11.48616559\mathrm{ng}/\mathrm{mL} \end{gathered}$$

Now solve for precision and accuracy by moving on to the following step:

$$\begin{gathered} \mathrm{precision}=\frac{\mathrm{standard}\mathrm{deviation}}{\mathrm{mean}}\times 100 \\ =\frac{11.48616559\mathrm{ng}/\mathrm{mL}}{82.48\mathrm{ng}/\mathrm{mL}}\times 100 \\ =13.926200096\% \\ \mathrm{precision}=13.9\% \\ \mathrm{accuracy}=\frac{\mathrm{mean}\mathrm{value}-\mathrm{known}\mathrm{value}}{\mathrm{known}\mathrm{value}}\times 100 \\ =\frac{\left(11.48616559-88.2\right)\mathrm{ng}/\mathrm{mL}}{88.2\mathrm{ng}/\mathrm{mL}}\times 100 \\ =6.48526077\% \\ \mathrm{accuracy}=6.5\% \end{gathered}$$

For the final sub-part of letter (a), we simply solve for the sample mean and standard deviation first, followed by precision and accuracy.

For $314\mathrm{ng}/\mathrm{mL}$

$$\begin{gathered} \overline{\mathrm{x}}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{\left(322+305+282+329+276\right)\mathrm{ng}/\mathrm{mL}}{5} \\ =\frac{1514\mathrm{ng}/\mathrm{mL}}{5} \\ \overline{\mathrm{x}}=302.8\mathrm{ng}/\mathrm{mL} \\ \mathrm{s}=\sqrt{\frac{{\mathrm{i}}^{{\left({\mathrm{x}}_1-\overline{\mathrm{x}}\right)}^{\mathrm{z}}}}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(322-308.2\right)}^2+{\left(305-308.2\right)}^2+{\left(282-308.2\right)}^2+{\left(329-308.2\right)}^2+{\left(276-308.2\right)}^2}{4}}\mathrm{ng}/\mathrm{mL} \\ =\sqrt{552.7\mathrm{ng}/\mathrm{mL}} \\ \mathrm{s}=23.50957252\mathrm{ng}/\mathrm{mL} \end{gathered}$$

Precision and accuracy in problem solving:

$$\begin{gathered} \mathrm{precision}=\frac{\mathrm{standard}\mathrm{deviation}}{\mathrm{mean}}\times 100 \\ =\frac{2350957272\mathrm{ng}/\mathrm{mL}}{302.8\mathrm{ng}/\mathrm{mL}}\times 100 \\ =7.76405957272\% \\ \mathrm{precision}=7.8\% \\ \mathrm{accuracy}=\frac{\mathrm{mean}\mathrm{value}-\mathrm{known}\mathrm{value}}{\mathrm{known}\mathrm{value}}\times 100 \\ =\frac{\left(302.8-314\right)\mathrm{ng}/\mathrm{mL}}{314\mathrm{ng}/\mathrm{mL}}\times 100 \\ =3.566878981\% \\ \mathrm{accuracy}=3.6\% \end{gathered}$$

Determine the single dectection limit, concertation dectection limit and lower limit of quantitation

(b)

Recall that we are given the following for part b:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{hlank}}=45\times 0\mathrm{m} \\ =1.75\times {10}^9{\mathrm{M}}^{-1} \end{gathered}$$

10 instrumental readings: 175, 104, 164, 193, 131, 189, 155, 133, 151, and

Before we can solve the required— single detection limit, concentration detection limit, and lower limit of quantitation— we must compute mean and standard deviation, as we did in the previous part.

How to calculate the sample mean and standard deviation:

$$\begin{gathered} \overline{\mathrm{x}}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{\left(175+104+164+193+131+189+155+133+151+176\right)}{5} \\ =\frac{1571}{10} \\ \overline{\mathrm{x}}=157.1 \\ \mathrm{s}=\sqrt{\frac{{{\mathrm{i}}_{\mathrm{i}}}^{{\left({\mathrm{x}}_1-\overline{\mathrm{x}}\right)}^{\mathrm{z}}}}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(175-157.1\right)}^2+{\left(104-157.1\right)}^2+L{\left(151-157.1\right)}^2+{\left(176-157.1\right)}^2}{9}} \\ =\sqrt{794.9888889} \\ \mathrm{s}=28.2 \end{gathered}$$

Find the required values now that we have obtained the values for the sample mean and standard deviation. After we've written down the formulas, it may apply them to the given and computed values.

$$\begin{gathered} \mathrm{signal}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s}=45.0+\left(3\times 28.2\right) \\ =45.0+84.6 \\ \mathrm{signal}\mathrm{detection}\mathrm{limit}=129.6 \end{gathered}$$

$$\begin{gathered} \mathrm{concentration}\mathrm{detection}\mathrm{limit}=\frac{3\mathrm{s}}{\mathrm{m}} \\ =\frac{\left(3\right)\left(28.2\right)}{1.75\times {10}^9{\mathrm{M}}^{-1}} \\ =\frac{84.6}{1.75\times {10}^9{\mathrm{M}}^{-1}} \\ =4.834285714\times {10}^{-8}\mathrm{M} \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=4.8\times {10}^{-8}\mathrm{M} \end{gathered}$$

$$\begin{gathered} \mathrm{lower}\mathrm{limit}\mathrm{of}\mathrm{quantitation}=\frac{10\mathrm{s}}{\mathrm{m}} \\ =\frac{\left(10\right)\left(28.2\right)}{1.75\times {10}^9\mathrm{M}} \\ =\frac{282.0}{1.75\times {10}^9{\mathrm{M}}^{-1}} \\ =1.611428571\times {10}^{-7}\mathrm{M} \end{gathered}$$