Beer’s law is possible because of the concept of transmittance. As depicted in Figure 7–1, a cuvette containing the solution to be analyzed is placed within a basic spectrophotometer so that light can be shown through it. The solution will absorb some of the light (absorbance) and the remainder will be transmitted through the cuvette to the photodetector. Notice that the arrow to the right of the cuvette, which represents the transmitted light, is smaller than the arrow to the left of the cuvette, which represents the incident light. The ratio of the amount of transmitted light divided by the amount of incident light is known as transmittance. This is demonstrated mathematically as follows:

FIGURE 7–1 Light path through a spectrophotometer.

Transmittance=(LTLI)

where:

LT=TransmittedlightLI=Incidentlight

Transmittance ratios range from 0.000 to 1.000. By multiplying by 100, percent transmittance values can be obtained, ranging from 0.00 to 100.00 on a linear scale. If all of the light is transmitted through a solution—that is, no light is absorbed by the solution—the transmittance ratio is 1.00 and the percent transmittance is 100%. Conversely, if no light is transmitted through the solution (i.e., there is 100% absorbance of the light by the solution), the transmittance ratio is 0.000, or 0% transmittance. Mathematically, transmittance and absorbance are related by the following formula:

A=−logT

This can be manipulated to the following formula:

A=log(1T)

The formula can be converted from T to % T:

A=log(1T)100%100%=A=log100%%T

Solving this equation:

A=log100%−log%TORA=2.00−log%T

In contrast to transmittance ratios, absorbance values range from 0.000 to infinity on a logarithmic scale. Figure 7–2 is a picture of an absorbance and percent transmittance meter. Notice how much easier it is to interpret the linear percent transmittance scale versus the logarithmic absorbance scale.

By convention, absorbance values are reported to the third decimal place—0.000—whereas percent T values are reported to the nearest tenth: 0.0% T.

Using the formula from Example 7–5, and substituting into it the values from this problem, the following formula is derived:

The concentration of QC level 1 is 72 mg/dL.

Given the absorbance of a 250 mg/dL standard is 0.750, and the absorbance of QC level 2 is 0.680, what is the concentration of QC level 2?

Using the same formula, and substituting into it the values of this new problem, the concentration of QC level 2 is 227 mg/dL.

If a single-point standard is used in a manually performed assay and a number of unknowns need concentrations calculated, the mathematics involved can become quite tedious. Notice that within the formula, once the assay is performed, there are two constants: the concentration of the standard and the absorbance of the standard. By mathematically manipulating the formula, a “factor” can be derived. The concentration of an unknown can be determined by multiplying the absorbance of the unknown by the factor. Using the factor simplifies the mathematics involved when calculating the concentrations of multiple unknowns.

Concofunknown=(concentrationofstandard)(absorbanceofunknown)absorbanceofstandard

By dividing the concentration of the standard by the absorbance of the standard, a factor is derived. The remaining unknown concentrations in the run can be calculated quickly by multiplying this factor by each individual absorbance.

Concentrationofunknown=(Factor)(absorbanceofunknown)

Example 7–6

Glucose was analyzed in three patient samples and two control samples using a manual method. The single standard used had a concentration of 200 mg/dL and an assayed absorbance of 0.448. Calculate the concentration of the five unknowns given the following absorbance readings for each:

Sample1:Absorbance=0.295Sample2:Absorbance=0.210Sample3:Absorbance=0.300Sample4:Absorbance=0.350Sample5:Absorbance=0.270

Each of these unknowns could be calculated as in Example 7–5. However, calculating six unknowns by this method is time consuming. Using a factor speeds up the calculations involved in multiple samples. To calculate the factor for this assay, divide the concentration of the standard by the obtained absorbance of the standard:

Factor=2000.448=446

Next, use the factor 444 and multiply each sample absorbance to obtain the sample concentration:

Additional Examples

Using the factor method and the information given in the table, complete the table by calculating the concentrations of three unknowns that were analyzed using the single-point standard method.

Sample No.	Standard Concentration	Absorbance of Standard	Factor	Absorbance of Unknown	Concentration of Unknowns
Standard	12.0 mg/dL	1.455	8.2
Unknown No. 1			8.2	1.215
Unknown No. 2			8.2	0.637
Unknown No. 3			8.2	0.283

Sample No.	Standard Concentration	Absorbance of Standard	Factor	Absorbance of Unknown	Concentration of Unknowns
Standard	12.0 mg/dL	1.455	8.2
Unknown No. 1			8.2	1.215	10.0 mg/dL
Unknown No. 2			8.2	0.637	5.2 mg/dL
Unknown No. 3			8.2	0.283	2.3 mg/dL

Using the factor method and the information given in the table, complete the table by calculating the concentrations of three unknowns that were analyzed using the single-point standard method.

Sample No.	Standard Concentration	Absorbance of Standard	Factor	Absorbance of Unknown	Concentration of Unknowns
Standard	80 mg/dL	0.850	94
Unknown No. 1			94	0.515
Unknown No. 2			94	0.780
Unknown No. 3			94	0.625

Sample No.	Standard Concentration	Absorbance of Standard	Factor	Absorbance of Unknown	Concentration of Unknowns
Standard	80 mg/dL	0.850	94
Unknown No. 1			94	0.515	48 mg/dL
Unknown No. 2			94	0.780	73 mg/dL
Unknown No. 3			94	0.625	59 mg/dL

Using the factor method and the information given in the table, complete the table by calculating the concentrations of three unknowns that were analyzed using the single-point standard method.

Sample No.	Standard Concentration	Absorbance of Standard	Factor	Absorbance of Unknown	Concentration of Unknowns
Standard	500 mg/dL	1.280	391
Unknown No. 1			391	1.145
Unknown No. 2			391	1.080
Unknown No. 3			391	0.849

Sample No.	Standard Concentration	Absorbance of Standard	Factor	Absorbance of Unknown	Concentration of Unknowns
Standard	500 mg/dL	1.280	391
Unknown No. 1			391	1.145	448 mg/dL
Unknown No. 2			391	1.080	422 mg/dL
Unknown No. 3			391	0.849	332 mg/dL

Standard Curve Method for Determining the Concentration of Unknowns

Most assays do not use a single standard to quantify unknowns. Instead, four to six standards of different concentrations are used that are spread evenly across the linear range of the method. In this manner, more of the linear range of the assay is covered by the standards compared with the single-standard method. On many automated instruments, a series of standards, or calibrators, are used to “set” the standard curve that is stored in the instrument’s computer. The electronic signal generated by the transmittance value of a sample is compared with the computer’s standard curve for the analyte being measured in the sample. The computer then generates a quantitative result for that sample’s analyte that can then be sent to the instrument’s computer and/or printer.

Students of clinical laboratory science often learn the concept of a standard curve by graphing on linear graph paper the concentrations of standards on the x axis and the absorbance of the standards on the y axis. A “standard curve” is shown in Figure 7–3. The following brief discussion of manually produced standard curves is designed to assist the students of clinical laboratory science to better understand what is happening inside the “black box” of an automated instrument.

In a manual assay that follows Beer’s law, and using linear graph paper, the standard curve of the absorbances obtained from the standards should be a straight line. The concentrations of the standards should range between the highest and lowest linear range and two to three concentrations evenly spaced between the two outermost limits. The standard curve line should be a “best fit” line, not “point to point” or “connect the dots.” The line should also never extend beyond the highest standard. Patient samples that have absorbances higher than the highest concentration of standard must be diluted so the absorbance obtained falls within the linear range and reanalyzed. Standard curves are not limited to the chemistry laboratory; they are also used in the hematology laboratory.

A standard curve of the percent transmittance values of the standards can also be plotted. However, if plotted on linear graph paper, the % T standard curve is curvilinear, as shown in Figure 7–4. To form a straight line, % T values must be plotted on semilogarithm graph paper as shown in Figure 7–5.

FIGURE 7–4 % T standard curve on linear paper.

Example 7–7

A manual glucose assay was performed using 50 mg/dL, 100 mg/dL, 200 mg/dL, 300 mg/dL, and 400 mg/dL standards. The following is a list of the absorbance value of each standard.

Standard Concentration	Absorbance
50 mg/dL	0.150
100 mg/dL	0.300
200 mg/dL	0.600
300 mg/dL	0.900

Using the absorbance values and the concentration of each standard, the following standard curve (Figure 7–6, A) can be produced. Notice that the line does not extend beyond the 300 mg/dL standard concentration and intersects 0,0.

FIGURE 7–6 A to C, Standard curve of glucose assay.

If a patient sample was also analyzed at the same time that the standards were analyzed and had an absorbance value of 0.400, what would the glucose concentration of the sample be?

To solve this problem, use the graph of the standard curve (see Figure 7–6, A). Find on the y axis scale 0.400 absorbance (Figure 7–6, B).

Next, using a ruler or straightedge, a line is drawn across the graph until it intersects the standard curve line. Now a line is drawn straight down to the x axis (Figure 7–6, C).

The point where the vertical line intersects the x axis is the concentration of the patient sample with an absorbance of 0.400. This glucose value from Figure 7–6, C is 135 mg/dL.

Additional Examples

Using the same glucose graph as in Example 7–7, what would the concentration be if the patient’s absorbance value had been 0.690?

By following the absorbance readings up to 0.690, then going across the graph until the line intersects with the standard curve line, then by reading down to the glucose concentration, the glucose concentration would have been 230 mg/dL.

Again by using the same graph, what if the absorbance had been 0.200?

The result would have been 70 mg/dL.

Example 7–8

A total protein assay was performed using four protein standards, two quality control specimens, and five patient specimens. The absorbance values for each of the standards, quality control specimens, and patient specimens are listed below.

Sample	Absorbance
3.00 g/dL standard	0.195
4.50 g/dL standard	0.290
7.50 g/dL standard	0.480
15.00 g/dL standard	0.960
QC level 1	0.270
QC level 2	0.440
Patient 1	0.580
Patient 2	0.170
Patient 3	0.260
Patient 4	0.490
Patient 5	0.900

Figure 7–7 is a graph of the standards. From the standard curve line obtained, what are the concentrations of the quality control materials and each of the patient specimens?

FIGURE 7–7 Standard curve of total protein assay.

The absorbance of QC level 1 is 0.270. Find the absorbance on the y axis, and draw a horizontal line until it intersects the standard curve line. Next, draw a vertical line from the intersection of the horizontal line with the standard curve to the x axis. The concentration of QC level 1 is found at the point on the x axis at which the vertical line intersects the x axis. In this example, the total protein concentration of QC level 1 is 4.20 g/dL. QC level 2 is found in the same manner. QC level 2 has an absorbance of 0.440 that corresponds from the graph to a concentration of 6.80 g/dL. The total protein concentration of each patient is listed below:

Patient No.	Concentration
1	9.00 g/dL
2	2.70 g/dL
3	4.00 g/dL
4	7.70 g/dL
5	14.00 g/dL

Additional Examples

Using the total protein graph in Example 7–8, what would the protein value be for an unknown with an absorbance of 0.160?

The protein value would be 2.40 g/dL.

Using the protein graph again, what would the protein value be if the unknown had a value of 0.840?

The result is 13.00 mg/dL.

Preparing Working Standards for Standard Curves Used for Manual Methods

The standards used for standard curves are obtained usually either by buying them directly from a vendor or obtaining one concentration of standard and, by performing dilutions, preparing a series of working standards. When performing dilutions of the standard, it is important to remember that there is a limited supply of the standard. Therefore, the limiting factor for the dilutional series is the total quantity of standard available to dilute. When the standard that must be used or that is available contains only 3 mL, it serves no useful purpose to devise a dilutional scheme that requires 5 mL of standard. To prepare a series of working standards, the calculation of C₁V₁ = C₂V₂ is used. (Refer to Chapter 6 for additional information.)

Example 7–9

A technologist needs to prepare a standard curve for a total protein assay. The stock standard has a concentration of 10 g/dL. Five working standards are necessary with concentrations of 8.0 g/dL, 6.0 g/dL, 4.0 g/dL, and 2.0 g/dL and a total volume of 2 mL each. Deionized water is used as the diluent. What is the technologist’s next step?

The formula C₁V₁ = C₂V₂ is used to solve this problem.

Where: C₁ = original concentration (stock concentration of 10 g/dL)

V1=unknownvolumeneededC2=workingstandardconcentrationV2=volumeofworkingstandard

To determine how to prepare the 8.0-g/dL working standard, substitute the data that are known into the equation:

Therefore, to prepare the 8.0-g/dL working standard, add 1.6 mL of the 10-g/dL stock standard to 0.4 mL of deionized water to make a total volume of 2.0 mL.

How is the 6.0-g/dL working standard prepared?

Use the same formula and substitute into it the 6.0-g/dL instead of the 8.0-g/dL:

Therefore, add 1.2 mL of the 10-g/dL stock standard and 0.8 mL of deionized water together to prepare the 6.0-g/dL working standard.

How is the 4.0-g/dL standard prepared?

Use the same formula as above:

Therefore, 0.8 mL of the 10-g/dL stock standard is added to 1.2 mL of deionized water to prepare the 4.0-g/dL working standard.

How is the 2.0-g/dL working standard prepared?

Again, use the same formula:

To prepare the 2.0-g/dL standard, add 0.4 mL of the 10-g/dL stock standard to 1.6 mL of deionized water.

Notice that the total volume of the 10-g/dL standard that is used to prepare the working standards is only 4 mL.

Additional Examples

An MLT student had to make four working standards with a total volume of 3.0 mL from a 300 mg/dL standard. The concentrations of the working standards were 200, 150, 75, and 50 mg/dL. How would this student prepare these standards?

By using the C₁V₁ = C₂V₂ formula, the standard volumes and diluent volumes for each working standard are listed in the following table.

Standard Concentration	Amount of Standard Needed	Amount of Diluent Needed
300 mg/dL
200 mg/dL	2 mL of 300 mg/dL standard	1 mL diluent
150 mg/dL	1.5 mL of 300 mg/dL standard	1.5 mL diluent
75 mg/dL	0.75 mL of 300 mg/dL standard	2.25 mL diluent
50 mg/dL	0.5 mL of 300 mg/dL standard	2.5 mL diluent

A series of four working standards with a 2.0 mL volume had to be prepared from a 40.0 mg/dL standard. The concentrations of the working standards are 30, 20, 10, and 5 mg/dL. How are these working standards prepared?

By using the C₁V₁ = C₂V₂ formula, the standard volumes and diluent volumes for each working standard are listed in the following table.

Standard Concentration	Amount of Standard Needed	Amount of Diluent Needed
40 mg/dL (2 mL total volume)
30 mg/dL	1.5 mL of 40 mg/dL standard	0.5 mL diluent
20 mg/dL	1.0 mL of 40 mg/dL standard	1.0 mL diluent
10 mg/dL	0.5 mL of 40 mg/dL standard	1.50 mL diluent
5 mg/dL	0.25 mL of 40 mg/dL standard	1.75 mL diluent

A series of four working standards with a 3.0 mL volume had to be prepared from a 15.0 mg/dL standard. The concentrations of the working standards are 12, 9, 7, and 5 mg/dL. How are these working standards prepared?

By using the C₁V₁ = C₂V₂ formula, the standard volumes and diluent volumes for each working standard are listed in the following table.

Standard Concentration	Amount of Standard Needed	Amount of Diluent Needed
15 mg/dL
12 mg/dL	2.4 mL of 15 mg/dL standard	0.6 mL diluent
9 mg/dL	1.8 mL of 15 mg/dL standard	1.2 mL diluent
7 mg/dL	1.4 mL of 15 mg/dL standard	1.6 mL diluent
5 mg/dL	1.0 mL of 15 mg/dL standard	2.0 mL diluent