# Clinical Chemistry Laboratory

*At the end of this chapter, the reader should be able to do the following:*

1. Convert between absorbance and transmittance values.

2. Calculate the concentration of unknown samples using Beer’s law.

3. Calculate the concentrations of unknown samples by single-standard, multiple standards, and millimolar absorptivity methods.

4. Calculate the concentration of enzyme unknown samples using the principles of kinetic analysis.

5. Calculate the pH of solutions using the Henderson-Hasselbalch equation.

6. Calculate bicarbonate, pH, and pCO_{2} of unknown samples using the derivation of the Henderson-Hasselbalch equation.

7. Interpret a patient’s acid-base status.

8. Calculate the anion gap of unknown samples.

9. Calculate serum and urine osmolality concentrations.

10. Calculate the osmol gap in patient samples.

11. Calculate the concentration of LDL and VLDL cholesterol.

## SPECTROPHOTOMETRY

It is beyond the scope of this book to go into great detail about spectrophotometry and spectrophotometric techniques. In general, spectrophotometry is used to quantify the concentration of various analytes based on the amount of light that the analyte absorbs. It is based on the theory of Beer’s law, which states that the amount of absorbance of a solution is directly proportional to the solution’s concentration. In practical terms, the darker the solution, the higher the absorbance and the more concentrated the solution.

Beer’s law: | A = abc | where: | A = absorbance |

a = absorptivity coefficient | |||

b = pathlength | |||

c = concentration |

The absorptivity coefficient (a) is the amount of light absorbed by an analyte at a specific wavelength and is constant for a particular analyte at a particular wavelength if certain conditions such as temperature, solvent, and pH remain constant. The pathlength (b) is the distance that light travels through the solution and (if the analysis is performed correctly) is also a constant. Thus by removing the two constants in the equation, the following equation is derived:

which states that absorbance is directly proportional to concentration. By using Beer’s law, many instruments can rapidly calculate the concentration of many different analytes.

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:

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:

This can be manipulated to the following formula:

The formula can be converted from T to % 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.

###### Example 7–1

A manual creatinine assay was performed using the Jaffé method in a MLT student lab. A 45% T reading was obtained for the cuvette containing the level 1 creatinine control and reagent. What is the absorbance value?

To solve this problem, use the following formula:

Substituting in the % T value, the following equation is derived:

Using a calculator or the logarithm table in Appendix 2–A, determine the log of 45.

Therefore, the % T value of 45% is equal to an absorbance value of **0.347**.

###### Example 7–2

Given a transmittance ratio of 0.400, what is the absorbance value?

To solve this problem, first convert the transmittance ratio to a percent transmittance value.

Next, substitute 40.0% T into the conversion formula:

Using a calculator or the logarithm table in Appendix 2–A, determine the log of 40:

Therefore, the transmittance ratio of 0.400 is equal to an absorbance value of **0.398**.

To solve, first convert the ratio of 0.125 into percentage by multiplying by 100. The % T result is 12.5%. Next, use the absorbance formula to convert the % T value to absorbance:

Therefore, a transmittance ratio of 0.125 is equal to an absorbance value of **0.903**.

Using the formula A = 2.00 – log % T, the absorbance value is **0.111**.

###### Example 7–3

Given an absorbance value of 0.875, what is the % T value?

Using the formula and substituting into it the given values, the following equation is derived:

Solving this equation leads to the following:

The negative signs in this equation can be removed because they are found on both sides of the equation.

Next, find the antilog of 1.125 from the logarithm table or by using a calculator. The antilog of a number is determined on many calculators by pressing INV and then LOG. Some calculators require another key be pressed before pressing the LOG key. Follow the manufacturer’s directions for determining antilogarithms for your calculator.

To verify this result, use 13.34 as % T and determine the absorbance.

This problem can be solved just like Example 7–3.

Can you work this problem backwards to prove that 7.50 is the correct answer?

Just measuring the absorbance or % T values of solutions will not result in any meaningful results. When analytes are measured, “standards” are used within the analysis. Standards are known quantities of the chemical that is to be analyzed. By including standards within an analysis, the patient specimens can be compared with the standards for quantification. We know before we analyze the standards what the concentrations of the standards are and within the analysis we measure the absorbance of the standards. Once the standards and patient specimens’ absorbance values are measured, there are three techniques to determine the concentration of the analyte in the patient’s specimen. These techniques can be used when performing a manual method and are also used by automated instruments to determine the concentration of analytes and controls.

### Single-Standard Method for Determining the Concentration of Unknowns

Beer’s law states that

###### Example 7–4

A manual glucose assay is performed using a single 150 mg/dL standard. Upon analysis, the absorbance of the standard is 0.465. The absorbance of a patient’s sample is 0.338. What is the concentration of the patient’s sample?

To solve this problem, use ratio and proportion:

Using algebra and crossmultiplying:

Therefore, the patient specimen has a glucose concentration of **109** mg/dL. By convention, glucose values tend to be reported to the nearest whole number.

A manual serum creatinine is performed in a MLT student laboratory experiment. A 5.0 mg/dL standard is used and the following results were obtained: Absorbance of the standard = 1.142, absorbance of the unknown = 0.778. What is the concentration of the unknown?

Using ratio and proportion, the following formula is derived:

Therefore, the concentration of the unknown is **3.4** mg/dL. By convention, creatinine values tend to be reported to the nearest one tenth.

Given a standard concentration of 300 mg/dL with an absorbance equal to 1.262 and an unknown concentration’s absorbance of 0.715, what is the concentration of the unknown?

Using ratio and proportion, the concentration of the unknown is **170** mg/dL.

By manipulating the ratio and proportion calculation, the following formula can be derived to determine the concentration of unknowns using a single-point standard:

Dilute the unknown and reanalyze or use a standard with a higher concentration. It is assumed that with a single-point standard assay, the reaction is linear up to the standard concentration. However, absorbances beyond the absorbance of the standard cannot be assumed to be linear.

###### Example 7–5

The absorbance of a 40.0 mg/dL blood urea nitrogen (BUN) standard is 0.758. The absorbance of the patient’s serum specimen is 0.220. What is the patient’s serum BUN concentration?

Use the formula to determine the concentration of BUN in the patient’s serum:

Given the absorbance of a 150 mg/dL standard is 0.425, and the absorbance of QC level 1 is 0.205, calculate the concentration of QC level 1.

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.

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.

###### 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:

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:

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

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.

###### 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.

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*).

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.

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.

###### 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?

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 |

#### 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 C_{1}V_{1} = C_{2}V_{2} 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 C_{1}V_{1} = C_{2}V_{2} is used to solve this problem.

Where: C_{1} = original concentration (stock concentration of 10 g/dL)

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?

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.

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 C_{1}V_{1} = C_{2}V_{2} 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 C_{1}V_{1} = C_{2}V_{2} 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 C_{1}V_{1} = C_{2}V_{2} 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 |

### Molar Absorptivity Method for Determining the Concentration of Unknowns

Another method for determining the concentration of analytes is by using the molar absorptivity of the analyte. Some automated methods may calculate the concentration of analytes using this method, but the concept is presented here for the benefit of students of clinical laboratory science. The molar absorptivity for a compound is constant at a given wavelength and when conditions, such as temperature, remain constant. Remember that Beer’s law states that

###### Example 7–10

The molar absorptivity of NADH is

Using Beer’s law that A = abc, substitute the data that is known into it:

Therefore, the concentration of the compound is **2.65 × 10 ^{−4} M**.

## END-POINT VERSUS KINETIC REACTIONS

Chemical reactions in the clinical chemistry laboratory may be end-point or kinetic. End-point assays measure the absorbance of the reaction at the completion of the reaction. Many manual wet chemistry assays performed by students of clinical laboratory science are end-point assays with an incubation period from 15 to 20 minutes to allow for completion of the reaction.

End-point assays can use a single standard, a standard curve, or the molar absorptivity method to calculate the concentration of patient samples. End-point assays may use enzymes within the reaction as part of the reagent, but analyses of enzymes are not performed as end-point reactions. In many laboratories, analytes are measured by large multichannel random access analyzers. In the quest for increased speed, kinetic reactions were developed. A kinetic reaction differs from an end-point assay in that the reaction does not go to completion; rather, absorbances are taken at certain intervals for short periods. Some analyzers continuously monitor the absorbance readings instead of monitoring at fixed intervals to improve accuracy. In any reaction, kinetic or end-point, there may be three phases: the lag phase in which the reactant and reagents are first reacting together, the reacting phase during which the product is formed, and the reagent depletion phase as shown in Figure 7–8.

In the lag phase, the absorbance of the product is not constant; in the reacting phase the absorbance steadily changes. If there is more reactant than the reagent can react with, the third phase—reagent depletion—occurs. In this phase, the absorbance values remain the same. In an end-point assay, the reagents are in such excess that the reagent depletion phase should not occur. Rather, the reaction is allowed to proceed past the reacting phase until all of the reactant is exhausted. At this point, the absorbance is stable. In a kinetic reaction, the absorbance readings are taken during the second phase. In a kinetic reaction, the change in absorbance, or delta absorbance, is measured.

## ENZYME KINETICS

This section will feature a broad overview of enzyme kinetics and the associated calculation. This section is designed to help the student of clinical laboratory science rather than the practitioner as manual enzyme assays are no longer (thankfully!) performed in the clinical laboratory but may be performed by students during their didactic training.

Kinetic reactions may be “first order” or “zero order” as shown in Figure 7–9.

### First Order Reactions

First order reactions are reactions in which the enzyme is in excess and the substrate concentration (the concentration of the analyte to be measured) is the limiting factor. In a first order reaction, the substrate concentration is low relative to the enzyme concentration. At low substrate concentrations, the rate of the reaction is dependent on the substrate concentration. First order reactions tend to be used when nonenzyme analytes are measured and enzymes are used in the reaction sequence as a reagent.

### Zero Order Kinetics

When the activity of an enzyme needs to be measured, conditions of the assay are maintained to allow zero order kinetics to occur. In zero order kinetics, the rate of the reaction is directly proportional to the enzyme concentration and independent of the substrate concentration. The following reaction occurs when measuring enzyme activity:

In a zero order kinetic assay, the substrates are kept in excess; the rate limiting factor is the concentration of the enzyme being analyzed. All other secondary enzymes used in the reaction are also kept in excess. Temperature, pH, and other variables are kept constant during the reaction. The rate of the reaction can be calculated by the Michaelis-Menten equation: