A common issue for the microbiology lab is the determination of starting inoculum concentration. If the inoculum concentration is determined by plating, the inoculum is several days old before use. This essay describes the use of turbidity to estimate microbial concentration in a suspension, using the Antimicrobial Efficacy Test as the example.
Determination of Inoculum for the AET
The compendial antimicrobial efficacy test (AET) requires inoculation of the product with microorganisms to a final concentration of approximately 106 CFU/mL. Although this seems to be a minor point, it does serve to illustrate some of the inherent difficulties in microbiological testing and the need for experienced and academically trained microbiologists to head the laboratory.
Let’s look at the compendial guidance. The Pharm Eur (1) instruction on preparing the inoculum for the AET states:
“To harvest the … cultures, use a sterile suspending fluid … Add sufficient suspending fluid to reduce the microbial count to about 108 micro-organisms per milliliter…Remove immediately a suitable sample from each suspension and determine the number of colony-forming units per milliliter in each suspension by plate count or membrane filtration (2.6.12). This value serves to determine the inoculum and the baseline to use in the test. The suspensions shall be used immediately.”
There are, of course, two problems with these instructions. The first is that the technician is instructed to use an inoculum of about 108 microorganisms per milliliter and then instructed to determine this by plate count. Colony forming units (CFU) and cells are two different measures and this will inevitably lead to difficulties as the unfortunate lab worker cannot guarantee the number of cells in the suspension, only the number of CFU found. However, we can accept the scientific inaccuracy as the numbers will generally work out. The more serious problem is the instruction to use the plate count CFU for determination of the inoculum for the test, and that the suspension shall be used immediately. This quite frankly cannot be done. If you use the suspension immediately, the plate counts are unavailable, if you use the plate counts to set the inoculum, then the suspension is at least a day old.
Contrast these instructions with those in the USP (2) for the same exercise:
“To harvest the … cultures, use sterile saline … Add sufficient … to obtain a microbial count of about 1 x 108 cfu per mL…[Note: The estimate of inoculum concentration may be performed by turbidimetric measurements for the challenge organisms. Refrigerate the suspension if it is not used within 2 hours].
Determine the number of cfu per mL in each suspension …to confirm the initial cfu per mL estimate. This value serves to calibrate the size of the inoculum used in the test.”
These USP instructions have the advantage of being physically possible to perform, an obvious advantage to the lab worker. However, the turbidometric measure of the cells is also only an approximation of CFU. Thus the instruction to confirm the numbers (after the test is underway) with the plate count is an important control on the test.
This article will explore the turbidometric approximation for cell numbers, and important controls on the process as well as potential pitfalls to the method.
Theory
Light scattering techniques to monitor the concentration of pure cultures have the enormous advantages of being rapid and nondestructive. However, they do not measure cell numbers nor do they measure CFU. Light scattering is most closely related to the dry weight of the cells (3).
Light is passed through the suspension of microorganisms, and all light that is not absorbed is re-radiated. There is a significant amount of physics involved in this, and those interested are referred to optical treatises, particularly those discussing Huygens’ Principle (a good choice is Light Scattering by Small Particles by H C Van De Hulst). For our purposes it is enough to say that light passing through a suspension of microorganisms is scattered, and the amount of scatter is an indication of the biomass present in the suspension. In visible light, this appears “milky” or “cloudy” to the eye (3). It follows from this that if the concentration of scattering particles becomes high, then multiple scattering events become possible.
Methods
McFarland Turbidity Standards
McFarland standards can be used to visually approximate the concentration of cells in a suspension. The McFarland Scale represents specific concentrations of CFU/mL and is designed to be used for estimating concentrations of gram negative bacteria such as E. coli. Note that this estimate becomes uncertain with organisms outside the normal usage as different species of bacteria differ in size and mass, as do yeast and mold. Use of this method would require calibration and validation.
McFarland Standards are generally labeled 0.5 through 10 and filled with suspensions of Barium salts. (Note - latex bead suspensions are now also available which extend the shelf life of the material). The standards may be made in the lab by preparing a 1% solution of anhydrous BaCl2 and a 1% solution of H2SO4 – mix them in the proportions listed in the table. They should be stored in the dark, in a tightly sealed container at 20-25oC and should be stable for approximately 6 months (4).
The advantage of the use of these standards is that no incubation time or equipment is needed to estimate bacterial numbers. The disadvantage is that there is some subjectivity involved in interpreting the turbidity, and that the numbers are valid only for those microorganisms similar to E. coli. In addition, the values are not in the appropriate range for the AET inoculum and so further dilutions may be required.
Approximate E. coli concentrations on McFarland Scale
McFarland Scale
CFU (x106/mL)
1% BaCl2/ 1% H2SO4 (mL)
0.5
<300
0.05/9.95
1
300
0.1/9.9
2
600
0.2/9.8
3
900
0.3/9.7
4
1200
0.4/9.6
5
1500
0.5/9.5
6
1800
0.6/9.4
7
2100
0.7/9.3
8
2400
0.8/9.2
9
2700
0.9/9.1
10
3000
1.0/9.0
Spectrophotometer
The spectrophotometer method measures turbidity directly. The best case (i.e. most sensitive) would be to have a narrow slit and a small detector so that only the light scattered in the forward direction would be seen by the detector. This instrument would give larger apparent absorption readings than other instruments.
As should be obvious, each spectrophotometer used must be independently calibrated for use in estimating microbial concentrations. Not only is the apparent absorption affected by the width of the instrument’s slit, the condition of the filter, and the size and condition of the detector, but also each time the lamp is changed the calibration needs to be repeated as different bulbs may vary in total output.
The correlation of absorption to dry weight is very good for dilute suspensions of bacteria (5), and this relationship seems to hold regardless of cell size (although the relationship of absorption to CFU does not). However, in more concentrated suspensions this correlation (absorption to dry weight) no longer holds. The linear range of absorption to estimated CFU is of limited scope and for this reason the calibration study must demonstrate the linear range of the absorbance vs CFU values and the relevant values.
Procedure
As there are a variety of different instruments, there cannot be one single procedure. In general, the spectrophotometer can be set at a wavelength of 420 – 660 nm. This wavelength must be standardized and may need to be adjusted specifically to the material being tested. Different vegetative cells, bacterial spores and spores of Aspergillus niger may not have the same maximal absorbance wavelength.
It is important to have the cells in known physiological state of growth. That is to say, as the cell size varies with phase of growth (lag, log, stationery) the approximate relationship between absorbance and CFU will also vary. A recommended practice might be to pass a single well-isolated colony twice on overnight cultures surface streaks from the refrigerated stock, harvesting the rapidly growing culture from the second passage for preparation of vegetative cells. This also will serve to minimize a source of variability for the AET (6).
A second source of concern might be the cuvette used for the measurement – care must be taken to maintain the correct orientation of the cuvette, and to protect it from damage that could affect the passage of light. Finally, it is necessary to blank the spectrophotometer (adjust the absorbance reading to zero) using a standard, either water or the suspending fluid, and maintain this practice.
Calibration
It must be stressed that this calibration should be done for all organisms. The size of the organism, any associated pigments, the preparation of the suspension, and other factors all influence the readings. This calibration study should also be rechecked after changing the bulb on the light source, and should be reevaluated throughout the life of the light bulb.
The calibration itself is simple to perform. Prepare a concentrated solution of the organism, grown under the conditions that will be used for the test. Make a series of dilutions to cover the range of absorption measurements of interest; 5 to 8 dilutions are recommended. Immediately take the spectrophotometer readings in sequence, and then take a confirmatory reading of the first in series to confirm that no growth has occurred. The dilutions are then immediately plated for viable count (serial dilution of the suspensions will be necessary). Graph the relationship between the absorbance and the CFU/mL after the plate counts are available and use values in the linear range of this graph.
As there are several factors that can affect this curve (quality of lamp output, size of slit, condition of filter, condition of detector, microorganism characteristic, etc) this calibration should be confirmed when the conditions of the assay change.
Conclusions
The use of optical density to estimate CFU in a suspension is possible, if basic precautions are taken. It is important to control:
The physiological state of the organism
The species of the organisms (i.e. don’t calibrate the instrument using E. coli and expect the numbers to work for Candida albicans)
The nature and condition of the equipment
Despite the inherent inaccuracy of the method, if the procedure is adequately controlled and calibrated the estimation of microbial numbers by optical density (either by McFarland Standards or spectrophotometrically) is sufficiently accurate for use in preparing inocula for QC testing and offers the overwhelming advantages of being rapid, low cost and non-destructive
Saturday, May 2, 2009
The Gram Stain
Gram staining is an empirical method of differentiating bacterial species into two large groups (Gram-positive and Gram-negative) based on the chemical and physical properties of their cell walls. The method is named after its inventor, the Danish scientist Hans Christian Gram (1853-1938), who developed the technique in 1884 (Gram 1884). The importance of this determination to correct identification of bacteria cannot be overstated as all phenotypic methods begin with this assay.
The Basic Method
1. First, a loopful of a pure culture is smeared on a slide and allowed to air dry. The culture can come from a thick suspension of a liquid culture or a pure colony from a plate suspended in water on the microscope slide. Important considerations:
· Take a small inoculum – don’t make a thick smear that cannot be completely decolorized. This could make gram-negative organisms appear to be gram-positive or gram-variable.
· Take a fresh culture – old cultures stain erratically.
2. Fix the cells to the slide by heat or by exposure to methanol. Heat fix the slide by passing it (cell side up) through a flame to warm the glass. Do not let the glass become hot to the touch.
3. Crystal violet (a basic dye) is then added by covering the heat-fixed cells with a prepared solution. Allow to stain for approximately 1 minute.
4. Briefly rinse the slide with water. The heat-fixed cells should look purple at this stage.
5. Add iodine (Gram's iodine) solution (1% iodine, 2% potassium iodide in water) for 1 minute. This acts as a mordant and fixes the dye, making it more difficult to decolorize and reducing some of the variability of the test.
6. Briefly rinse with water.
7. Decolorize the sample by applying 95% ethanol or a mixture of acetone and alcohol. This can be done in a steady stream, or a series of washes. The important aspect is to ensure that all the color has come out that will do so easily. This step washes away unbound crystal violet, leaving Gram-positive organisms stained purple with Gram-negative organisms colorless. The decolorization of the cells is the most “operator-dependent” step of the process and the one that is most likely to be performed incorrectly.
8. Rinse with water to stop decolorization.
9. Rinse the slide with a counterstain (safranin or carbol fuchsin) which stains all cells red. The counterstain stains both gram-negative and gram-positive cells. However, the purple gram-positive color is not altered by the presence of the counter-stain, it’s effect is only seen in the previously colorless gram-negative cells which now appear pink/red.
10. Blot gently and allow the slide to dry. Do not smear.
What’s Going On?
Bacteria have a cell wall made up of peptidoglycan. This cell wall provides rigidity to the cell, and protection from osmotic lysis in dilute solutions. Gram-positive bacteria have a thick mesh-like cell wall, gram-negative bacteria have a thin cell wall and an outer phospholipid bilayer membrane. The crystal violet stain is small enough to penetrate through the matrix of the cell wall of both types of cells, but the iodine-dye complex exits only with difficulty (Davies et al. 1983)
The decolorizing mixture dehydrates cell wall, and serves as a solvent to rinse out the dye-iodine complex. In Gram-negative bacteria it also dissolves the outer membrane of the gram-negative cell wall aiding in the release of the dye. It is the thickness of the cell wall that characterizes the response of the cells to the staining procedure. In addition to the clearly gram-positive and gram-negative, there are many species that are “gram-variable” with intermediate cell wall structure (Beveridge and Graham 1991). As noted above, the decolorization step is critical to the success of the procedure.
Gram’s method involves staining the sample cells dark blue, decolorizing those cells with a thin cell wall by rinsing the sample, then counterstaining with a red dye. The cells with a thick cell wall appear blue (gram positive) as crystal violet is retained within the cells, and so the red dye cannot be seen. Those cells with a thin cell wall, and therefore decolorized, appear red (gram negative).
It is a prudent practice to always include a positive and negative control on the staining procedure to confirm the accuracy of the results (Murray et al 1994) and to perform proficiency testing on the ability of the technicians to correctly interpret the stains (Andserson, et al. 2005).
Excessive Decolorization
It is clear that the decolorization step is the one most likely to cause problems in the gram stain. The particular concerns in this step are listed below (reviewed in McClelland 2001)
Excessive heat during fixationHeat fixing the cells, when done to excess, alters the cell morphology and makes the cells more easily decolorized.
Low concentration of crystal violetConcentrations of crystal violet up to 2% can be used successfully, however low concentrations result in stained cells that are easily decolorized. The standard 0.3% solution is good, if decolorization does not generally exceed 10 seconds.
Excessive washing between stepsThe crystal violet stain is susceptible to wash-out with water (but not the crystal violet-iodine complex). Do not use more than a 5 second water rinse at any stage of the procedure.
Insufficient iodine exposureThe amount of the mordant available is important to the formation of the crystal violet - iodine complex. The lower the concentration, the easier to decolorize (0.33% - 1% commonly used). Also, QC of the reagent is important as exposure to air and elevated temperatures hasten the loss of Gram’s iodine from solution. A closed bottle (0.33% starting concentration) at room temperature will lose >50% of available iodine in 30 days, an open bottle >90%. Loss of 60% iodine results in erratic results.
Prolonged decolorization95% ethanol decolorizes more slowly, and may be recommended for inexperienced technicians while experienced workers can use the acetone-alcohol mix. Skill is needed to gauge when decolorization is complete.
Excessive counterstainingAs the counterstain is also a basic dye, it is possible to replace the crystal violet—iodine complex in gram- positive cells with an over-exposure to the counterstain. The counterstain should not be left on the slide for more than 30 seconds.
Alternatives to the Gram Stain
Gram’s staining method is plainly not without its problems. It is messy, complicated, and prone to operator error. The method also requires a large number of cells (although a membrane-filtration technique has been reported; Romero, et al 1988). However, it is also central to phenotypic microbial identification techniques.
This method, and it’s liabilities, are of immediate interest to those involved in environmental monitoring programs as one of the most common isolates in an EM program, Bacillus spp., will frequently stain gram variable or gram negative despite being a gram-positive rod (this is especially true with older cultures). The problems with Gram’s method have lead to a search for other tests that correlate with the cell wall structure of the gram-positive and the gram-negative cells. Several improvements/alternatives to the classical gram stain have appeared in the literature.
KOH String Test
The KOH String Test is done using a drop of 3% potassium hydroxide on a glass slide. A visible loopful of cells from a single, well-isolated colony is mixed into the drop. If the mixture becomes viscous within 60 seconds of mixing (KOH-positive) then the colony is considered gram-negative. The reaction depends on the lysis of the gram-negative cell in the dilute alkali solution releasing cellular DNA to turn the suspension viscous. This method has been shown effective for food microorganisms (Powers 1995), and for Bacillus spp (Carlone et al 1983, Gregersen 1978), although it may be problematic for some anaerobes (Carlone et al 1983, but also see Halebian et al 1981).
This test has the advantage of simplicity, and it can be performed on older cultures. False negative results can occur in the test by using too little inoculum or too much KOH (DNA-induced viscosity not noticeable). False positive results can occur from too heavy an inoculum (the solution will appear to gel, but not string), or inoculation with mucoid colonies. This can serve as a valuable adjunct to the tradition gram stain method (von Graevenitz and Bucher 1983).
Aminopeptidase Test
L-alanine aminopeptidase is an enzyme localized in the bacterial cell wall which cleaves the amino acid L-alanine from various peptides. Significant activity is found almost only in Gram-negative microorganisms, all Gram-positive or Gram-variable microorganisms so far studied display no or very weak activity (Cerny 1976, Carlone et al. 1983). To perform the test, the reagent is used to make a suspension (with the bacteria). Aminopeptidase activity of the bacteria causes the release of 4-nitroaniline from the reagent, turning the suspension yellow. The test is especially useful for non-fermenters and gram-variable organisms, and is a one step test with several suppliers of kits. Results of the test are available in 5 minutes.
Fluorescent Stains
A popular combination of fluorescent stains for use in gram staining (particularly for flow-cytometry) involves the use of the fluorescent nucleic acid binding dyes hexidium iodide (HI) and SYTO 13. HI penetrates gram-positive but not gram-negative organisms, but SYTO 13 penetrates both. When the dyes were used together in a single step, gram-negative organisms are green fluorescent by SYTO 13 while gram-positive organisms are red-orange fluorescent by HI which overpowers the green of SYTO 13 (Mason et al 1998). There are commercial kits available for this procedure, which requires a fluorescent microscope or a flow cytometer.
Sizemore et al (1990) developed a different approach to fluorescent labeling of cells. Fluorescence-labeled wheat germ agglutinin binds specifically to N-acetylglucosamine in the outer peptidoglycan layer of gram-positive bacteria. The peptidoglycan layer of gram-negative bacteria is covered by a membrane and is not labeled by the lectin. A variant of this method has also been used to “gram stain” microorganisms in milk for direct measurement by flow cytometry.
LAL-based Assay
Charles River Laboratories has just released a product to be used with their PTS instrument – the PTS Gram ID (Farmer 2005). This methodology makes use of the same reaction used for the chromogenic LAL test. Gram-negative organisms, with bacterial endotoxin, initiate the LAL coagulase cascade which results in activation of the proclotting enzyme, a protease. In the LAL test, this enzyme cleaves a peptide from the horseshoe crab coagulen, resulting in a clot. It can also cleave a peptide from a synthetic substrate, yielding a chromophore (p-nitroaniline) which is yellow and can be measured photometrically at 385 nm (Iwanaga 1987). Gram-positive organisms, lacking endotoxin, do not trigger the color change in this method, while gram-negative organisms do trigger it. Results are available within 10 minutes.
Summary
The differentiation of bacteria into either the gram-positive or the gram-negative group is fundamental to most bacterial identification systems. This task is usually accomplished through the use of Gram’s Staining Method. Unfortunately, the gram stain methodology is complex and prone to error. This operator-dependence can be addressed by attention to detail, and by the use of controls on the test. Additional steps might include confirmatory tests, of which several examples were given. As with all microbiology assays, full technician training and competent review of the data are critical quality control steps for good laboratory results.
The Basic Method
1. First, a loopful of a pure culture is smeared on a slide and allowed to air dry. The culture can come from a thick suspension of a liquid culture or a pure colony from a plate suspended in water on the microscope slide. Important considerations:
· Take a small inoculum – don’t make a thick smear that cannot be completely decolorized. This could make gram-negative organisms appear to be gram-positive or gram-variable.
· Take a fresh culture – old cultures stain erratically.
2. Fix the cells to the slide by heat or by exposure to methanol. Heat fix the slide by passing it (cell side up) through a flame to warm the glass. Do not let the glass become hot to the touch.
3. Crystal violet (a basic dye) is then added by covering the heat-fixed cells with a prepared solution. Allow to stain for approximately 1 minute.
4. Briefly rinse the slide with water. The heat-fixed cells should look purple at this stage.
5. Add iodine (Gram's iodine) solution (1% iodine, 2% potassium iodide in water) for 1 minute. This acts as a mordant and fixes the dye, making it more difficult to decolorize and reducing some of the variability of the test.
6. Briefly rinse with water.
7. Decolorize the sample by applying 95% ethanol or a mixture of acetone and alcohol. This can be done in a steady stream, or a series of washes. The important aspect is to ensure that all the color has come out that will do so easily. This step washes away unbound crystal violet, leaving Gram-positive organisms stained purple with Gram-negative organisms colorless. The decolorization of the cells is the most “operator-dependent” step of the process and the one that is most likely to be performed incorrectly.
8. Rinse with water to stop decolorization.
9. Rinse the slide with a counterstain (safranin or carbol fuchsin) which stains all cells red. The counterstain stains both gram-negative and gram-positive cells. However, the purple gram-positive color is not altered by the presence of the counter-stain, it’s effect is only seen in the previously colorless gram-negative cells which now appear pink/red.
10. Blot gently and allow the slide to dry. Do not smear.
What’s Going On?
Bacteria have a cell wall made up of peptidoglycan. This cell wall provides rigidity to the cell, and protection from osmotic lysis in dilute solutions. Gram-positive bacteria have a thick mesh-like cell wall, gram-negative bacteria have a thin cell wall and an outer phospholipid bilayer membrane. The crystal violet stain is small enough to penetrate through the matrix of the cell wall of both types of cells, but the iodine-dye complex exits only with difficulty (Davies et al. 1983)
The decolorizing mixture dehydrates cell wall, and serves as a solvent to rinse out the dye-iodine complex. In Gram-negative bacteria it also dissolves the outer membrane of the gram-negative cell wall aiding in the release of the dye. It is the thickness of the cell wall that characterizes the response of the cells to the staining procedure. In addition to the clearly gram-positive and gram-negative, there are many species that are “gram-variable” with intermediate cell wall structure (Beveridge and Graham 1991). As noted above, the decolorization step is critical to the success of the procedure.
Gram’s method involves staining the sample cells dark blue, decolorizing those cells with a thin cell wall by rinsing the sample, then counterstaining with a red dye. The cells with a thick cell wall appear blue (gram positive) as crystal violet is retained within the cells, and so the red dye cannot be seen. Those cells with a thin cell wall, and therefore decolorized, appear red (gram negative).
It is a prudent practice to always include a positive and negative control on the staining procedure to confirm the accuracy of the results (Murray et al 1994) and to perform proficiency testing on the ability of the technicians to correctly interpret the stains (Andserson, et al. 2005).
Excessive Decolorization
It is clear that the decolorization step is the one most likely to cause problems in the gram stain. The particular concerns in this step are listed below (reviewed in McClelland 2001)
Excessive heat during fixationHeat fixing the cells, when done to excess, alters the cell morphology and makes the cells more easily decolorized.
Low concentration of crystal violetConcentrations of crystal violet up to 2% can be used successfully, however low concentrations result in stained cells that are easily decolorized. The standard 0.3% solution is good, if decolorization does not generally exceed 10 seconds.
Excessive washing between stepsThe crystal violet stain is susceptible to wash-out with water (but not the crystal violet-iodine complex). Do not use more than a 5 second water rinse at any stage of the procedure.
Insufficient iodine exposureThe amount of the mordant available is important to the formation of the crystal violet - iodine complex. The lower the concentration, the easier to decolorize (0.33% - 1% commonly used). Also, QC of the reagent is important as exposure to air and elevated temperatures hasten the loss of Gram’s iodine from solution. A closed bottle (0.33% starting concentration) at room temperature will lose >50% of available iodine in 30 days, an open bottle >90%. Loss of 60% iodine results in erratic results.
Prolonged decolorization95% ethanol decolorizes more slowly, and may be recommended for inexperienced technicians while experienced workers can use the acetone-alcohol mix. Skill is needed to gauge when decolorization is complete.
Excessive counterstainingAs the counterstain is also a basic dye, it is possible to replace the crystal violet—iodine complex in gram- positive cells with an over-exposure to the counterstain. The counterstain should not be left on the slide for more than 30 seconds.
Alternatives to the Gram Stain
Gram’s staining method is plainly not without its problems. It is messy, complicated, and prone to operator error. The method also requires a large number of cells (although a membrane-filtration technique has been reported; Romero, et al 1988). However, it is also central to phenotypic microbial identification techniques.
This method, and it’s liabilities, are of immediate interest to those involved in environmental monitoring programs as one of the most common isolates in an EM program, Bacillus spp., will frequently stain gram variable or gram negative despite being a gram-positive rod (this is especially true with older cultures). The problems with Gram’s method have lead to a search for other tests that correlate with the cell wall structure of the gram-positive and the gram-negative cells. Several improvements/alternatives to the classical gram stain have appeared in the literature.
KOH String Test
The KOH String Test is done using a drop of 3% potassium hydroxide on a glass slide. A visible loopful of cells from a single, well-isolated colony is mixed into the drop. If the mixture becomes viscous within 60 seconds of mixing (KOH-positive) then the colony is considered gram-negative. The reaction depends on the lysis of the gram-negative cell in the dilute alkali solution releasing cellular DNA to turn the suspension viscous. This method has been shown effective for food microorganisms (Powers 1995), and for Bacillus spp (Carlone et al 1983, Gregersen 1978), although it may be problematic for some anaerobes (Carlone et al 1983, but also see Halebian et al 1981).
This test has the advantage of simplicity, and it can be performed on older cultures. False negative results can occur in the test by using too little inoculum or too much KOH (DNA-induced viscosity not noticeable). False positive results can occur from too heavy an inoculum (the solution will appear to gel, but not string), or inoculation with mucoid colonies. This can serve as a valuable adjunct to the tradition gram stain method (von Graevenitz and Bucher 1983).
Aminopeptidase Test
L-alanine aminopeptidase is an enzyme localized in the bacterial cell wall which cleaves the amino acid L-alanine from various peptides. Significant activity is found almost only in Gram-negative microorganisms, all Gram-positive or Gram-variable microorganisms so far studied display no or very weak activity (Cerny 1976, Carlone et al. 1983). To perform the test, the reagent is used to make a suspension (with the bacteria). Aminopeptidase activity of the bacteria causes the release of 4-nitroaniline from the reagent, turning the suspension yellow. The test is especially useful for non-fermenters and gram-variable organisms, and is a one step test with several suppliers of kits. Results of the test are available in 5 minutes.
Fluorescent Stains
A popular combination of fluorescent stains for use in gram staining (particularly for flow-cytometry) involves the use of the fluorescent nucleic acid binding dyes hexidium iodide (HI) and SYTO 13. HI penetrates gram-positive but not gram-negative organisms, but SYTO 13 penetrates both. When the dyes were used together in a single step, gram-negative organisms are green fluorescent by SYTO 13 while gram-positive organisms are red-orange fluorescent by HI which overpowers the green of SYTO 13 (Mason et al 1998). There are commercial kits available for this procedure, which requires a fluorescent microscope or a flow cytometer.
Sizemore et al (1990) developed a different approach to fluorescent labeling of cells. Fluorescence-labeled wheat germ agglutinin binds specifically to N-acetylglucosamine in the outer peptidoglycan layer of gram-positive bacteria. The peptidoglycan layer of gram-negative bacteria is covered by a membrane and is not labeled by the lectin. A variant of this method has also been used to “gram stain” microorganisms in milk for direct measurement by flow cytometry.
LAL-based Assay
Charles River Laboratories has just released a product to be used with their PTS instrument – the PTS Gram ID (Farmer 2005). This methodology makes use of the same reaction used for the chromogenic LAL test. Gram-negative organisms, with bacterial endotoxin, initiate the LAL coagulase cascade which results in activation of the proclotting enzyme, a protease. In the LAL test, this enzyme cleaves a peptide from the horseshoe crab coagulen, resulting in a clot. It can also cleave a peptide from a synthetic substrate, yielding a chromophore (p-nitroaniline) which is yellow and can be measured photometrically at 385 nm (Iwanaga 1987). Gram-positive organisms, lacking endotoxin, do not trigger the color change in this method, while gram-negative organisms do trigger it. Results are available within 10 minutes.
Summary
The differentiation of bacteria into either the gram-positive or the gram-negative group is fundamental to most bacterial identification systems. This task is usually accomplished through the use of Gram’s Staining Method. Unfortunately, the gram stain methodology is complex and prone to error. This operator-dependence can be addressed by attention to detail, and by the use of controls on the test. Additional steps might include confirmatory tests, of which several examples were given. As with all microbiology assays, full technician training and competent review of the data are critical quality control steps for good laboratory results.
Microbial Recovery
Introduction
The PMFList is a source of great ideas for review and for further thought. One that keeps coming up on the list is the question of 70% recovery (as described in USP chapter <1227> Validation of Microbial Recovery from Pharmacopeial Articles) and 50% recovery as described in the harmonized chapter <61> Microbiological Examination Of Nonsterile Products: Microbial Enumeration Tests.
The questions and discussion seem to fall into two distinct groups – the first a discussion about when to apply 70% and when to apply 50% as your recovery criteria (with frequent complaints about the inferred lack of consistency in USP) and the second a discussion of what types of tests we are talking about. We will look at these issues separately.
What are we talking about?
The first thing to do is to establish the scope of the discussion. For starters, let’s begin by stating that the compendial chapters are, by definition, validated. This refers to those chapters in the USP that number under 1000. We therefore cannot really “validate” the test method, instead we are trying to demonstrate the suitability of the recovery method. This has been referred to as “verification” (Porter 2007) and in the harmonized Microbial Limits chapters as “method suitability.”
The point of a method suitability study in microbiology is not to validate the assay, but rather to demonstrate that our specific test method is suitable; that the recovery scheme allows recovery of viable microorganisms. In other words, microorganisms are not prevented from growing in the experimental system by residual antimicrobial activity of the product
This demonstration is critical in accurate determination of disinfecting efficacy, bioburden, sterility or any test that requires determination of surviving microorganisms in a product containing antimicrobial properties. Failure to confirm adequate neutralization and recovery could result in under-reporting of surviving microorganisms. This expectation of 70% recovery can also be applied to media growth promotion studies, where a new batch of media is compared to a previously qualified batch for its ability to support at least 70% of a standard inoculum.
A convenient method for this neutralization is through the use of recovery diluents designed to neutralize commonly used antimicrobials. A number of reagents are used in this regard (reviewed by Russell 1981; Furr & Rogers 1987). However, some of these compounds may be toxic to the test organisms (Reybrouck 1978) and so it is also important to determine the potential toxicity of the neutralizing medium (recovery diluent). These two activities, neutralizer efficacy and growth promotion (or neutralizer toxicity), are equally important in this consideration. A schematic of a design for this type of study is presented below, where a consistent inoculum is added to the product in the recovery diluent, peptone in the recovery diluent (use the same volume of peptone as that of the product), and into peptone. These are then plated 5-6 times to provide a good estimate of the number of organisms present (Wilson and Kullman 1931). The Neutralizer Efficacy is determined by comparing the recovery in the peptone suspension to that in the Product + Recovery Diluent suspension, Neutralizer Toxicity by comparing the Peptone suspension to the Peptone + Recovery Diluent (USP 2007a).
What is not part of this discussion?
It should be obvious from the previous discussion that the “method suitability” study is highly controlled. A standard inoculum is added to three tubes, and then replicate aliquots are removed and immediately plated. In a perfect world the numbers would be in agreement 100% of the time, but we work in microbiology. Even in such a simple design the opportunity for variability is enormous, and there are workers in the field who are vehement that no better than 50% should be expected between replicates of this type. One wonders if this is a limitation of the test system or of their laboratory training program. In any event, the discussion of 50% to 70% between the populations applies only to this design (and those closely related to it).
The recommendation in USP of 70% recovery was never meant to apply to studies of microbial recovery from solid surfaces. These studies are extremely complicated, and are confounded by issues of recovery efficacy of swabs, contact plates, and other methods (Buggy, et al 1983, Rose et al 2004, Whyte 1989). In addition, if vegetative cells are used for the study, there is the additional problem of die-off due to dessication (Potts 1994).
Recovery studies looking at bioburden of solid surfaces (facility, equipment, medical device or personnel) are not part of the 50% to 70% debate. They have their own set of issues and will be discussed in a later newsletter.
Is there any support for these numbers?
There are two studies which directly support the 70% recovery acceptance criterion.
Proud and Sutton (1992) describe the development of a “universal” diluting fluid for membrane filtration sterility testing using a modification of the design described above. The product was placed in a filtration apparatus containing 100 mL of the diluting fluid, and then passed through the membrane, followed by two additional 100 mL rinses. The membrane was then removed and placed on the surface of a nutrient agar plate for incubation and enumeration. Each treatment was performed at least three times. CFU were converted to their log10 values, and ANOVA analysis performed on the replicates. When all was said and done, a recovery of 75% of the inoculum count (raw CFU – untransformed) passed the ANOVA analysis.
Sutton, et al. (2002) conducted a large study on methods to recover microorganisms in the presence of surface disinfectants. “Neutralizer efficacy (NE) ratios were determined [in this study] by comparing the recovery of identical inocula from the neutralizing solution in the presence, or the absence, of a 1:10 dilution of the biocide. Neutralizer toxicity (NT) ratios were determined between recovery of viable microorganisms incubated for a short period in peptone, and in the neutralizing medium without the biocide. An effective and non-toxic neutralizer was initially identified by NE and NT ratios of ≥ 0.75. Statistical evaluation of the data was performed by ANOVA, with Dunnett’s test for multiple comparisons used to confirm failures. By this analysis, 239/244 identified failures were confirmed by ANOVA of 588 NT and NE comparisons (5 presumptive failures were not confirmed by statistical analysis). We therefore conclude that recovery of 75% is a suitable criterion (2% false negative rate) for neutralizer evaluations.”
A side issue to this discussion is the occasional use of 70-130% recovery as the acceptance criteria. I have trouble with this one – would you really disqualify a method because it improves your recovery over expectations? In my opinion the acceptance criteria should be that the test treatment should recover at least 70%, with no consideration of recovery by the test in excess of the comparator treatment.
Which should you use?
I am of the opinion that 70% is easily attainable if the technicians are proficient and the recovery method works. This may require 5-6 replicates, rather than the usual duplicate plates per sample. However, this is a “verification” study or a “method suitability” study (or whatever we wish to call it) and so may be worth a bit more work.
So, how did they get different criteria in the USP? Chapter <1227> was developed to address a specific concern – that of providing information on microbial recovery studies (not limited to neutralizer efficacy) for use in the pharmaceutical industry. This work was well in progress by 1996 (USP 1996). The harmonization program discussed this point much later, and after negotiation agreed to the 50% so that agreement could occur. No data was presented to support the assertion that 50% was appropriate (by my records), it was, however, the number that could be accepted.
The harmonized USP chapter <61> (USP 2006b) cites a 50% recovery frequency and so this is the official acceptance criteria for this test. If you wish to use 50% for the acceptance criteria for all method suitability studies (non-compendial bioburden tests, method suitability studies for disinfectancy tests, Antimicrobial Efficacy tests, media growth promotion, etc) I would strongly urge a solid rationale for failing to observe the recommendation of chapter <1227>. In addition, I would be prepared to answer questions of technician proficiency as the suspicion may be that your lab is not confident of reproducibility to 70% even between identical samples.
The PMFList is a source of great ideas for review and for further thought. One that keeps coming up on the list is the question of 70% recovery (as described in USP chapter <1227> Validation of Microbial Recovery from Pharmacopeial Articles) and 50% recovery as described in the harmonized chapter <61> Microbiological Examination Of Nonsterile Products: Microbial Enumeration Tests.
The questions and discussion seem to fall into two distinct groups – the first a discussion about when to apply 70% and when to apply 50% as your recovery criteria (with frequent complaints about the inferred lack of consistency in USP) and the second a discussion of what types of tests we are talking about. We will look at these issues separately.
What are we talking about?
The first thing to do is to establish the scope of the discussion. For starters, let’s begin by stating that the compendial chapters are, by definition, validated. This refers to those chapters in the USP that number under 1000. We therefore cannot really “validate” the test method, instead we are trying to demonstrate the suitability of the recovery method. This has been referred to as “verification” (Porter 2007) and in the harmonized Microbial Limits chapters as “method suitability.”
The point of a method suitability study in microbiology is not to validate the assay, but rather to demonstrate that our specific test method is suitable; that the recovery scheme allows recovery of viable microorganisms. In other words, microorganisms are not prevented from growing in the experimental system by residual antimicrobial activity of the product
This demonstration is critical in accurate determination of disinfecting efficacy, bioburden, sterility or any test that requires determination of surviving microorganisms in a product containing antimicrobial properties. Failure to confirm adequate neutralization and recovery could result in under-reporting of surviving microorganisms. This expectation of 70% recovery can also be applied to media growth promotion studies, where a new batch of media is compared to a previously qualified batch for its ability to support at least 70% of a standard inoculum.
A convenient method for this neutralization is through the use of recovery diluents designed to neutralize commonly used antimicrobials. A number of reagents are used in this regard (reviewed by Russell 1981; Furr & Rogers 1987). However, some of these compounds may be toxic to the test organisms (Reybrouck 1978) and so it is also important to determine the potential toxicity of the neutralizing medium (recovery diluent). These two activities, neutralizer efficacy and growth promotion (or neutralizer toxicity), are equally important in this consideration. A schematic of a design for this type of study is presented below, where a consistent inoculum is added to the product in the recovery diluent, peptone in the recovery diluent (use the same volume of peptone as that of the product), and into peptone. These are then plated 5-6 times to provide a good estimate of the number of organisms present (Wilson and Kullman 1931). The Neutralizer Efficacy is determined by comparing the recovery in the peptone suspension to that in the Product + Recovery Diluent suspension, Neutralizer Toxicity by comparing the Peptone suspension to the Peptone + Recovery Diluent (USP 2007a).
What is not part of this discussion?
It should be obvious from the previous discussion that the “method suitability” study is highly controlled. A standard inoculum is added to three tubes, and then replicate aliquots are removed and immediately plated. In a perfect world the numbers would be in agreement 100% of the time, but we work in microbiology. Even in such a simple design the opportunity for variability is enormous, and there are workers in the field who are vehement that no better than 50% should be expected between replicates of this type. One wonders if this is a limitation of the test system or of their laboratory training program. In any event, the discussion of 50% to 70% between the populations applies only to this design (and those closely related to it).
The recommendation in USP of 70% recovery was never meant to apply to studies of microbial recovery from solid surfaces. These studies are extremely complicated, and are confounded by issues of recovery efficacy of swabs, contact plates, and other methods (Buggy, et al 1983, Rose et al 2004, Whyte 1989). In addition, if vegetative cells are used for the study, there is the additional problem of die-off due to dessication (Potts 1994).
Recovery studies looking at bioburden of solid surfaces (facility, equipment, medical device or personnel) are not part of the 50% to 70% debate. They have their own set of issues and will be discussed in a later newsletter.
Is there any support for these numbers?
There are two studies which directly support the 70% recovery acceptance criterion.
Proud and Sutton (1992) describe the development of a “universal” diluting fluid for membrane filtration sterility testing using a modification of the design described above. The product was placed in a filtration apparatus containing 100 mL of the diluting fluid, and then passed through the membrane, followed by two additional 100 mL rinses. The membrane was then removed and placed on the surface of a nutrient agar plate for incubation and enumeration. Each treatment was performed at least three times. CFU were converted to their log10 values, and ANOVA analysis performed on the replicates. When all was said and done, a recovery of 75% of the inoculum count (raw CFU – untransformed) passed the ANOVA analysis.
Sutton, et al. (2002) conducted a large study on methods to recover microorganisms in the presence of surface disinfectants. “Neutralizer efficacy (NE) ratios were determined [in this study] by comparing the recovery of identical inocula from the neutralizing solution in the presence, or the absence, of a 1:10 dilution of the biocide. Neutralizer toxicity (NT) ratios were determined between recovery of viable microorganisms incubated for a short period in peptone, and in the neutralizing medium without the biocide. An effective and non-toxic neutralizer was initially identified by NE and NT ratios of ≥ 0.75. Statistical evaluation of the data was performed by ANOVA, with Dunnett’s test for multiple comparisons used to confirm failures. By this analysis, 239/244 identified failures were confirmed by ANOVA of 588 NT and NE comparisons (5 presumptive failures were not confirmed by statistical analysis). We therefore conclude that recovery of 75% is a suitable criterion (2% false negative rate) for neutralizer evaluations.”
A side issue to this discussion is the occasional use of 70-130% recovery as the acceptance criteria. I have trouble with this one – would you really disqualify a method because it improves your recovery over expectations? In my opinion the acceptance criteria should be that the test treatment should recover at least 70%, with no consideration of recovery by the test in excess of the comparator treatment.
Which should you use?
I am of the opinion that 70% is easily attainable if the technicians are proficient and the recovery method works. This may require 5-6 replicates, rather than the usual duplicate plates per sample. However, this is a “verification” study or a “method suitability” study (or whatever we wish to call it) and so may be worth a bit more work.
So, how did they get different criteria in the USP? Chapter <1227> was developed to address a specific concern – that of providing information on microbial recovery studies (not limited to neutralizer efficacy) for use in the pharmaceutical industry. This work was well in progress by 1996 (USP 1996). The harmonization program discussed this point much later, and after negotiation agreed to the 50% so that agreement could occur. No data was presented to support the assertion that 50% was appropriate (by my records), it was, however, the number that could be accepted.
The harmonized USP chapter <61> (USP 2006b) cites a 50% recovery frequency and so this is the official acceptance criteria for this test. If you wish to use 50% for the acceptance criteria for all method suitability studies (non-compendial bioburden tests, method suitability studies for disinfectancy tests, Antimicrobial Efficacy tests, media growth promotion, etc) I would strongly urge a solid rationale for failing to observe the recommendation of chapter <1227>. In addition, I would be prepared to answer questions of technician proficiency as the suspicion may be that your lab is not confident of reproducibility to 70% even between identical samples.
Friday, May 1, 2009
Manipulating Carbohydrates
The application of manipulating carbohydrates while preparing for a contest is done with intent on the muscle glycogen and water content. The thought is for you to have as much glycogen in your muscles as you can on the show day of the competition. From every gram of glycogen in your body, you have another three grams of water added into your muscle cells also. That means that a weight gain of two to four pounds could be predicted if you do it correctly.
Using Creatine before a contest
Loading yourself with creatine is done in an attempt to saturate and control a larger amount of creatine in your muscle cells. The creatine you take in helps you out with the creatine phosphate section of the kreb cycle to create energy for your working cells.
Side note: Kreb cycle – The kreb cycle is what is found in all plants and animals, a variety of enzymatic reactions in the mitochondria of cells, used to create a lot of energy phosphate compounds that are a main source of cellular energy.
Protocols for Creatine:
- First off, 20-25 grams per day for the period of 5-7 days
- Afterwards, 2-5 grams per day for 21-37 days
It has also been shown that creatine raises your total body water and glycogen while not using fluid distribution. A two week study was done to show the results that combining both the carbohydrate loading after you use creatine supplementation has lead to prove that it gives a super concentration of glycogen and water in your muscle cells.
Competitive Bodybuilders: Usage Suggestion
10 Days out: Only have an intake of a high carbohydrate based diet.
7 Days out: Put five grams of creatine into your body daily.
5 Days out: Start the period of carbohydrate depletion for the process of three days which is followed with a cessation of creatine.
2 Days of carbohydrate loading and gradual water intake decreases.
Another suggestion is for you to takea potassium supplement which could help out with any cramping you may be having.
While more research is needed on this topic to come to a conclusion, we all know that for certain people, certain choices will be made. While professional bodybuilders use a certain type of training and preperation to make sure their muscles are as good as possible for performing at the show they are featured in, other less competitive bodybuilders may not worry as much about this issue.
Either way, it is best for you to stick around and find out any conclusion that is discovered.
Using Creatine before a contest
Loading yourself with creatine is done in an attempt to saturate and control a larger amount of creatine in your muscle cells. The creatine you take in helps you out with the creatine phosphate section of the kreb cycle to create energy for your working cells.
Side note: Kreb cycle – The kreb cycle is what is found in all plants and animals, a variety of enzymatic reactions in the mitochondria of cells, used to create a lot of energy phosphate compounds that are a main source of cellular energy.
Protocols for Creatine:
- First off, 20-25 grams per day for the period of 5-7 days
- Afterwards, 2-5 grams per day for 21-37 days
It has also been shown that creatine raises your total body water and glycogen while not using fluid distribution. A two week study was done to show the results that combining both the carbohydrate loading after you use creatine supplementation has lead to prove that it gives a super concentration of glycogen and water in your muscle cells.
Competitive Bodybuilders: Usage Suggestion
10 Days out: Only have an intake of a high carbohydrate based diet.
7 Days out: Put five grams of creatine into your body daily.
5 Days out: Start the period of carbohydrate depletion for the process of three days which is followed with a cessation of creatine.
2 Days of carbohydrate loading and gradual water intake decreases.
Another suggestion is for you to takea potassium supplement which could help out with any cramping you may be having.
While more research is needed on this topic to come to a conclusion, we all know that for certain people, certain choices will be made. While professional bodybuilders use a certain type of training and preperation to make sure their muscles are as good as possible for performing at the show they are featured in, other less competitive bodybuilders may not worry as much about this issue.
Either way, it is best for you to stick around and find out any conclusion that is discovered.
Thursday, April 30, 2009
Pharmacokinetics
Pharmacokinetics is the study of what the body does to a drug.
Pharmacodynamics is the study of what a drug does to the body.
Routes of Drug Administration:
Intravenous
Oral
Buccal
Sublingual
Rectal
Intramuscular
Transdermal
Subcutaneous
Inhalational
Topical
Of all of these routes you are most likely to be asked about the transdermal, as it is fashionable.
Otherwise, most other basic pharmacology questions tend to concern the pharmacology of intravenous agents; that is what is discussed below.
First Order Kinetics:
A constant fraction of the drug in the body is eliminated per unit time. The rate of elimination is proportional to the amount of drug in the body. The majority of drugs are eliminated in this way.
What follows concerns drugs which follow first order kinetics.
The Volume of Distribution (Vd) is the amount of drug in the body divided by the concentration in the blood. Drugs that are highly lipid soluble, such as digoxin, have a very high volume of distribution (500 litres). Drugs which are lipid insoluble, such as neuromuscular blockers, remain in the blood, and have a low Vd.
The Clearance (Cl) of a drug is the volume of plasma from which the drug is completely removed per unit time. The amount eliminated is proportional to the concentration of the drug in the blood.
The fraction of the drug in the body eliminated per unit time is determined by the elimination constant (kel). This is represented by the slope of the line of the log plasma concentration versus time.
Cl = kel x Vd
Rate of elimination = clearance x concentration in the blood.
Elimination half life (t1/2): the time taken for plasma concentration to reduce by 50%. After 4 half lives, elimination is 94% complete.
It can be shown that the kel = the log of 2 divided by the t1/2 = 0.693/t1/2.
Likewise, Cl = kel x Vd, so, Cl = 0.693Vd/t1/2.
And t1/2 = 0.693 x Vd / cl
The rate of elimination is the clearance times the concentration in the plasma
Roe = Cl x Cp
Fraction of the total drug removed per unit time = Cl/Vd.
If the volume of distribution is increased, then the kel will decrease, the t1/2 will increase, but the clearance won't change.
Confused?
Example: You have a 10ml container of orange squash. You put this into a litre (ok 990ml!) of water. The Vd of the orange squash is 1000ml. If, each minute, you empty 10ml of the orange liquid into the 10ml container, discard this, and replace it with 10ml of water. The clearance is 10 ml per minute. The elimination half life is: 70 minutes . The kel is Cl/Vd = 10/1000 = 0.01. Shown the other way, 0.693/50 = 0.01.
If the volume of the container is increased to 2000ml, then the clearance remains the same, but the Vd, and consequently the t1/2, increases (to 140 minutes).
Simple, isn't it?
What is described above is a single compartment model, what would occur if the bloodstream was the only compartment in the body (or if the Vd = the blood volume). But the human body is more complex than this: there are many compartments: muscle, fat, brain tissue etc. In order to describe this, we use multicompartment models.
Multicompartment Models:
Why does a patient wake up after 5 minutes after an injection of thiopentone when we know that it takes several hours to eliminate this drug from the body? What happens is that, initially the drug is all in the blood and this blood goes to "vessel rich" organs; principally the brain. After a few minutes the drug starts to venture off into other tissues (fat, muscle etc) it redistributes, the concentration in the brain decreases and the patient wakes up! The drug thus redistributes into other compartments.
If you were to represent this phenomenon graphically, you would follow a picture of rapid fall in blood concentration, a plateau, and then a slower gradual fall. The first part is the rapid redistribution phase, the alpha phase, the plateau is the equilibrium phase (where blood concentration = tissue concentration), and the slower phase, the beta phase, is the elimination phase where blood and tissue concentrations fall in tandem. This is a simple two compartment model and is as much as you need to know.
An couple of interesting pieces of information can be derived from the log concentration versus time graph. If you extrapolate back the elimination line to the y axis, then you get to a point called the CP0 - a theoretical point representing the concentration that would have existed at the start if the dose had been instantly distributed (dose/Vd). From this new straight line you can figure out how long it takes for the concentration to drop by 50%: the elimination half life. Likewise, a similar procedure can be performed on the α phase: the redistribution half life.
While it is very important that you understand these concepts, the reality is that most drugs are infinitely more complicated that this, and computer calculations are required to derive this data.
Bioavailability
This is the fraction of the administered dose that reaches the systemic circulation. Bioavailability is 100% for intravenous injection. It varies for other routes depending on incomplete absorption, first pass hepatic metabolism etc. Thus one plots plasma concentration against time, and the bioavailability is the area under the curve.
Zero Order Elimination
Why if I have 10 pints of beer before midnight will I fail a breathalyser test at 8 am the following morning? Either this is due to alcohol having a very long half life (which it does not) or that alcohol is cleared in a different way.
What happens is that the metabolic pathways responsible for alcohol metabolism are rapidly saturated and that clearance is determined by how fast these pathways can work. The metabolic pathways work to their limit. This is known as zero order kinetics: a constant amount of drug is eliminated per unit time. This form of kinetics occours with several important drugs at high dosage concentrations: phenytoin, salicylates, theophylline, and thiopentone (at very large doses). Because high dose thio is very slow to clear, we no longer use it in infusion for status epilepticus (as it takes ages for the patient to wake up!).
Dosage regimens
The strategy for treating patients with drugs is to give sufficient amounts that the required theraputic effect arises, but not a toxic dose.
The maintenance dose is equal to the rate of elimination at steady state (i.e.at steady state, rate of elimination = rate of administration):
Dosing rate = clearance x desired plasma concentration.
Drugs will accumulate within the body if the drug has not been fully eliminated before the next dose. Steady state concentration is thus arrived at after four half lives. This is all very well if you are willing to wait 4 half lives for the drug to be fully effective, but what if you are not? What you may need to do is to "load" the volume of distribution with the drug to achieve target plasma concentrations rapidly: the loading dose.
The loading dose = the volume of distribution x the desired concentration (i.e. the concentration at steady state).
You can figure this out by: Loading dose = usual maintenance dose / usual dosage interval x kel (t1/2/0.693).
Hepatic Drug Clearance
Many drugs are extensively metabolised by the liver. The rate of elimination depends on 1) The liver's inherent ability to metabolise the drug, 2) the amount of drug presented to the liver for metabolism. This is important because drugs administered orally are delivered from the gut to the portal vein to the liver: the liver gobbles up a varying chunk of the administered drug (pre-systemic elimination) and less is available to the body for theraputic effect. This is why you have to give a higher dose of morphine, for examole, orally, than intravenously.
Hepatic drug clearance (i.e. the amount of each drug gobbled up by the liver) depends on:
1) The Intrinsic clearance (Cl int).
2) Hepatic blood flow.
These two factors are independent of one another, and their combined effect is the proportion of drug gobbled up: the extraction ratio.
For drugs that have a low intrinsic clearance, this effect can be increased by giving a second agent that boosts the effect of the liver's enzyme system; these are enzyme inducers. Examples of such drugs are cigarrettes, antiepileptics (carbamazepine & phenytoin), rifampicin, griseofulvin, alcohol and spironolactone (CAR GAS) [also barbiturates]. Consequnetly if a drug addict is given rifampicin or tuberculosis, a higher dose of heroin is required for the same effect. Enzyme inhibitors have the opposite effect: examples are flagyl, allopurinol, cimetidine, erythromycin, dextropropoxyphene, imipramine, (the) pill (FACE DIP).
Likewise, if the blood flow increases, the liver has less chance to gobble up the drug, and the extraction ratio falls. This is particularly the case, as you would expect, of the intrinsic clearance is low.
Illustration: Think of factory workers picking bad apples out of a pile on a conveyor belt, if only one person (low intrinsic clearance) is doing the picking and the speed of the conveyor belt is increased, more bad apples get through. If there are several pickers (high intrinsic clearance) then they are much more able to cope with an increase in the speed of the conveyor belt, but there will come a rate at which they will become overwhelmed, and bad apples will get through.
From this example you can take home this message: for drugs with a low extraction ratio, the kinetics (the body's ability to deal with the drug) depends on enzymatic activity (giving an enzyme inducer effectively gives the single picker 4 arms!). For high extraction ratio drugs, kinetics depends on liver blood flow - the slower the flow the higher the extraction, the higher the flow the lower the extraction ratio.
Drug distribution
When a drug is introduced into the body, where it ends up depends on a number of factors:
1) blood flow, tissues with the highest blood flow receive the drug first, 2) protein binding, drugs stuck to plasma proteins are crippled, they can only go where the proteins go (and that's not very far!), 3) lipid solubility and the degree of ionisation, this describes the ability of drugs to enter tissues (highly lipid soluble / unionised drugs can basically go anywhere).
Protein Binding
Most drugs bind to proteins, either albumin or alpha-1 acid glycoprotein (AAG), to a greater or lesser extent. Drugs prefer to be free, it is in this state that they can travel throughout the body, in and out of tissues and have their biological effect. The downside of this is that they are easy prey for metabolising enzymes.
As you would expect, more highly bound drugs have a longer duration of action and a lower volume of distribution. Generally high extraction ratio drugs' clearance is high because of low protein binding and, conversely, low extraction ratio drugs' clearance is strongly dependent on the amount of protein binding.
Why is this important? If a drug is highly protein bound, you need to give loads of it to get a theraputic effect; as so much is stuck to protein. But what happens if another agent comes along and starts to compete with the drug for the binding site on the protein? Yes, you guessed it, the amount of free drug is increased. This is really important for drugs that are highly protein bound: if a drug is 97% bound to albumin and there is a 3% reduction in binding (displaced by another drug), then the free drug concentration doubles; if a drug is 70% bound and there is a 3% reduction in binding, this will make little difference.
The drugs that you really need to keep an eye on are: warfarin, diazepam, propranolol and phenytoin. For example, a patient on warfarin is admitted with seizures, you treat the patient with phenytoin, next thing you know - his INR is 10.
The amount of albumin does not appear to be hugely relavent. In disease states such as sepsis, the serum albumin drops drastically, but the free drug concentration does not appear to increase
Degree of ionisation
This is really important with regard to local anaesthetics. The essential fact to know is that highly ionized drugs cannnot cross lipid membranes (basically they can't go anywhere) and unionised drugs can cross freely. Morphine is highly ionised, fentanyl is the opposite. Consequently the latter has a faster onset of action. The degree of ionisation depends on the pKa of the drug and the pH of the local environment. The pKa is the the pH at which the drug is 50% ionised. Most drugs are either weak acids or weak bases. Acids are most highly ionised at a high pH (i.e. in an alkaline environment). Bases are most highly ionised in an acidic environment (low pH). For a weak acid, the more acidic the environment, the less ionised the drug, and the more easily it crosses lipid membranes. If you take this acid, at pKa it is 50% ionised, if you add 2 pH points to this (more alkaline), it becomes 90% ionised, if you reduce the pH (more acidic) by two units, it becomes 10% ionised. Weak bases have the opposite effect.
Local anaesthetics are weak bases: the closer the pKa of the local anaesthetic to the local tissue pH, the more unionised the drug is. That is why lignocaine(pKa 7.7) has a faster onset of action than bupivicaine (pKa 8.3). If the local tissues are alkalinised (e.g. by adding bicarbonate to the local anaesthetic), then the tisssue pH is brought closer to the pKa, and the onset of action is hastened.
Pharmacodynamics is the study of what a drug does to the body.
Routes of Drug Administration:
Intravenous
Oral
Buccal
Sublingual
Rectal
Intramuscular
Transdermal
Subcutaneous
Inhalational
Topical
Of all of these routes you are most likely to be asked about the transdermal, as it is fashionable.
Otherwise, most other basic pharmacology questions tend to concern the pharmacology of intravenous agents; that is what is discussed below.
First Order Kinetics:
A constant fraction of the drug in the body is eliminated per unit time. The rate of elimination is proportional to the amount of drug in the body. The majority of drugs are eliminated in this way.
What follows concerns drugs which follow first order kinetics.
The Volume of Distribution (Vd) is the amount of drug in the body divided by the concentration in the blood. Drugs that are highly lipid soluble, such as digoxin, have a very high volume of distribution (500 litres). Drugs which are lipid insoluble, such as neuromuscular blockers, remain in the blood, and have a low Vd.
The Clearance (Cl) of a drug is the volume of plasma from which the drug is completely removed per unit time. The amount eliminated is proportional to the concentration of the drug in the blood.
The fraction of the drug in the body eliminated per unit time is determined by the elimination constant (kel). This is represented by the slope of the line of the log plasma concentration versus time.
Cl = kel x Vd
Rate of elimination = clearance x concentration in the blood.
Elimination half life (t1/2): the time taken for plasma concentration to reduce by 50%. After 4 half lives, elimination is 94% complete.
It can be shown that the kel = the log of 2 divided by the t1/2 = 0.693/t1/2.
Likewise, Cl = kel x Vd, so, Cl = 0.693Vd/t1/2.
And t1/2 = 0.693 x Vd / cl
The rate of elimination is the clearance times the concentration in the plasma
Roe = Cl x Cp
Fraction of the total drug removed per unit time = Cl/Vd.
If the volume of distribution is increased, then the kel will decrease, the t1/2 will increase, but the clearance won't change.
Confused?
Example: You have a 10ml container of orange squash. You put this into a litre (ok 990ml!) of water. The Vd of the orange squash is 1000ml. If, each minute, you empty 10ml of the orange liquid into the 10ml container, discard this, and replace it with 10ml of water. The clearance is 10 ml per minute. The elimination half life is: 70 minutes . The kel is Cl/Vd = 10/1000 = 0.01. Shown the other way, 0.693/50 = 0.01.
If the volume of the container is increased to 2000ml, then the clearance remains the same, but the Vd, and consequently the t1/2, increases (to 140 minutes).
Simple, isn't it?
What is described above is a single compartment model, what would occur if the bloodstream was the only compartment in the body (or if the Vd = the blood volume). But the human body is more complex than this: there are many compartments: muscle, fat, brain tissue etc. In order to describe this, we use multicompartment models.
Multicompartment Models:
Why does a patient wake up after 5 minutes after an injection of thiopentone when we know that it takes several hours to eliminate this drug from the body? What happens is that, initially the drug is all in the blood and this blood goes to "vessel rich" organs; principally the brain. After a few minutes the drug starts to venture off into other tissues (fat, muscle etc) it redistributes, the concentration in the brain decreases and the patient wakes up! The drug thus redistributes into other compartments.
If you were to represent this phenomenon graphically, you would follow a picture of rapid fall in blood concentration, a plateau, and then a slower gradual fall. The first part is the rapid redistribution phase, the alpha phase, the plateau is the equilibrium phase (where blood concentration = tissue concentration), and the slower phase, the beta phase, is the elimination phase where blood and tissue concentrations fall in tandem. This is a simple two compartment model and is as much as you need to know.
An couple of interesting pieces of information can be derived from the log concentration versus time graph. If you extrapolate back the elimination line to the y axis, then you get to a point called the CP0 - a theoretical point representing the concentration that would have existed at the start if the dose had been instantly distributed (dose/Vd). From this new straight line you can figure out how long it takes for the concentration to drop by 50%: the elimination half life. Likewise, a similar procedure can be performed on the α phase: the redistribution half life.
While it is very important that you understand these concepts, the reality is that most drugs are infinitely more complicated that this, and computer calculations are required to derive this data.
Bioavailability
This is the fraction of the administered dose that reaches the systemic circulation. Bioavailability is 100% for intravenous injection. It varies for other routes depending on incomplete absorption, first pass hepatic metabolism etc. Thus one plots plasma concentration against time, and the bioavailability is the area under the curve.
Zero Order Elimination
Why if I have 10 pints of beer before midnight will I fail a breathalyser test at 8 am the following morning? Either this is due to alcohol having a very long half life (which it does not) or that alcohol is cleared in a different way.
What happens is that the metabolic pathways responsible for alcohol metabolism are rapidly saturated and that clearance is determined by how fast these pathways can work. The metabolic pathways work to their limit. This is known as zero order kinetics: a constant amount of drug is eliminated per unit time. This form of kinetics occours with several important drugs at high dosage concentrations: phenytoin, salicylates, theophylline, and thiopentone (at very large doses). Because high dose thio is very slow to clear, we no longer use it in infusion for status epilepticus (as it takes ages for the patient to wake up!).
Dosage regimens
The strategy for treating patients with drugs is to give sufficient amounts that the required theraputic effect arises, but not a toxic dose.
The maintenance dose is equal to the rate of elimination at steady state (i.e.at steady state, rate of elimination = rate of administration):
Dosing rate = clearance x desired plasma concentration.
Drugs will accumulate within the body if the drug has not been fully eliminated before the next dose. Steady state concentration is thus arrived at after four half lives. This is all very well if you are willing to wait 4 half lives for the drug to be fully effective, but what if you are not? What you may need to do is to "load" the volume of distribution with the drug to achieve target plasma concentrations rapidly: the loading dose.
The loading dose = the volume of distribution x the desired concentration (i.e. the concentration at steady state).
You can figure this out by: Loading dose = usual maintenance dose / usual dosage interval x kel (t1/2/0.693).
Hepatic Drug Clearance
Many drugs are extensively metabolised by the liver. The rate of elimination depends on 1) The liver's inherent ability to metabolise the drug, 2) the amount of drug presented to the liver for metabolism. This is important because drugs administered orally are delivered from the gut to the portal vein to the liver: the liver gobbles up a varying chunk of the administered drug (pre-systemic elimination) and less is available to the body for theraputic effect. This is why you have to give a higher dose of morphine, for examole, orally, than intravenously.
Hepatic drug clearance (i.e. the amount of each drug gobbled up by the liver) depends on:
1) The Intrinsic clearance (Cl int).
2) Hepatic blood flow.
These two factors are independent of one another, and their combined effect is the proportion of drug gobbled up: the extraction ratio.
For drugs that have a low intrinsic clearance, this effect can be increased by giving a second agent that boosts the effect of the liver's enzyme system; these are enzyme inducers. Examples of such drugs are cigarrettes, antiepileptics (carbamazepine & phenytoin), rifampicin, griseofulvin, alcohol and spironolactone (CAR GAS) [also barbiturates]. Consequnetly if a drug addict is given rifampicin or tuberculosis, a higher dose of heroin is required for the same effect. Enzyme inhibitors have the opposite effect: examples are flagyl, allopurinol, cimetidine, erythromycin, dextropropoxyphene, imipramine, (the) pill (FACE DIP).
Likewise, if the blood flow increases, the liver has less chance to gobble up the drug, and the extraction ratio falls. This is particularly the case, as you would expect, of the intrinsic clearance is low.
Illustration: Think of factory workers picking bad apples out of a pile on a conveyor belt, if only one person (low intrinsic clearance) is doing the picking and the speed of the conveyor belt is increased, more bad apples get through. If there are several pickers (high intrinsic clearance) then they are much more able to cope with an increase in the speed of the conveyor belt, but there will come a rate at which they will become overwhelmed, and bad apples will get through.
From this example you can take home this message: for drugs with a low extraction ratio, the kinetics (the body's ability to deal with the drug) depends on enzymatic activity (giving an enzyme inducer effectively gives the single picker 4 arms!). For high extraction ratio drugs, kinetics depends on liver blood flow - the slower the flow the higher the extraction, the higher the flow the lower the extraction ratio.
Drug distribution
When a drug is introduced into the body, where it ends up depends on a number of factors:
1) blood flow, tissues with the highest blood flow receive the drug first, 2) protein binding, drugs stuck to plasma proteins are crippled, they can only go where the proteins go (and that's not very far!), 3) lipid solubility and the degree of ionisation, this describes the ability of drugs to enter tissues (highly lipid soluble / unionised drugs can basically go anywhere).
Protein Binding
Most drugs bind to proteins, either albumin or alpha-1 acid glycoprotein (AAG), to a greater or lesser extent. Drugs prefer to be free, it is in this state that they can travel throughout the body, in and out of tissues and have their biological effect. The downside of this is that they are easy prey for metabolising enzymes.
As you would expect, more highly bound drugs have a longer duration of action and a lower volume of distribution. Generally high extraction ratio drugs' clearance is high because of low protein binding and, conversely, low extraction ratio drugs' clearance is strongly dependent on the amount of protein binding.
Why is this important? If a drug is highly protein bound, you need to give loads of it to get a theraputic effect; as so much is stuck to protein. But what happens if another agent comes along and starts to compete with the drug for the binding site on the protein? Yes, you guessed it, the amount of free drug is increased. This is really important for drugs that are highly protein bound: if a drug is 97% bound to albumin and there is a 3% reduction in binding (displaced by another drug), then the free drug concentration doubles; if a drug is 70% bound and there is a 3% reduction in binding, this will make little difference.
The drugs that you really need to keep an eye on are: warfarin, diazepam, propranolol and phenytoin. For example, a patient on warfarin is admitted with seizures, you treat the patient with phenytoin, next thing you know - his INR is 10.
The amount of albumin does not appear to be hugely relavent. In disease states such as sepsis, the serum albumin drops drastically, but the free drug concentration does not appear to increase
Degree of ionisation
This is really important with regard to local anaesthetics. The essential fact to know is that highly ionized drugs cannnot cross lipid membranes (basically they can't go anywhere) and unionised drugs can cross freely. Morphine is highly ionised, fentanyl is the opposite. Consequently the latter has a faster onset of action. The degree of ionisation depends on the pKa of the drug and the pH of the local environment. The pKa is the the pH at which the drug is 50% ionised. Most drugs are either weak acids or weak bases. Acids are most highly ionised at a high pH (i.e. in an alkaline environment). Bases are most highly ionised in an acidic environment (low pH). For a weak acid, the more acidic the environment, the less ionised the drug, and the more easily it crosses lipid membranes. If you take this acid, at pKa it is 50% ionised, if you add 2 pH points to this (more alkaline), it becomes 90% ionised, if you reduce the pH (more acidic) by two units, it becomes 10% ionised. Weak bases have the opposite effect.
Local anaesthetics are weak bases: the closer the pKa of the local anaesthetic to the local tissue pH, the more unionised the drug is. That is why lignocaine(pKa 7.7) has a faster onset of action than bupivicaine (pKa 8.3). If the local tissues are alkalinised (e.g. by adding bicarbonate to the local anaesthetic), then the tisssue pH is brought closer to the pKa, and the onset of action is hastened.
Xenobiotic metabolism
Xenobiotic metabolism is the set of metabolic pathways that modify the chemical structure of xenobiotics, which are compounds foreign to an organism's normal biochemistry, such as drugs and poisons. These pathways are a form of biotransformation present in all major groups of organisms, and are considered to be of ancient origin. These reactions often act to detoxify poisonous compounds; however, in some cases, the intermediates in xenobiotic metabolism can themselves be the cause of toxic effects.
Xenobiotic metabolism is divided into three phases. In phase I, enzymes such as cytochrome P450 oxidases introduce reactive or polar groups into xenobiotics. These modified compounds are then conjugated to polar compounds in phase II reactions. These reactions are catalysed by transferase enzymes such as glutathione S-transferases. Finally, in phase III, the conjugated xenobiotics may be further processed, before being recognised by efflux transporters and pumped out of cells.
The reactions in these pathways are of particular interest in medicine as part of drug metabolism and as a factor contributing to multidrug resistance in infectious diseases and cancer chemotherapy. The actions of some drugs as substrates or inhibitors of enzymes involved in xenobiotic metabolism are a common reason for hazardous drug interactions. These pathways are also important in environmental science, with the xenobiotic metabolism of microorganisms determining whether a pollutant will be broken down during bioremediation, or persist in the environment.
Xenobiotic metabolism is divided into three phases. In phase I, enzymes such as cytochrome P450 oxidases introduce reactive or polar groups into xenobiotics. These modified compounds are then conjugated to polar compounds in phase II reactions. These reactions are catalysed by transferase enzymes such as glutathione S-transferases. Finally, in phase III, the conjugated xenobiotics may be further processed, before being recognised by efflux transporters and pumped out of cells.
The reactions in these pathways are of particular interest in medicine as part of drug metabolism and as a factor contributing to multidrug resistance in infectious diseases and cancer chemotherapy. The actions of some drugs as substrates or inhibitors of enzymes involved in xenobiotic metabolism are a common reason for hazardous drug interactions. These pathways are also important in environmental science, with the xenobiotic metabolism of microorganisms determining whether a pollutant will be broken down during bioremediation, or persist in the environment.
MTT assay
The MTT assay and the MTS assay are laboratory tests and standard colorimetric assays (an assay which measures changes in color) for measuring the activity of enzymes that reduce MTT or MTS + PMS to formazan, giving a purple color. It can also be used to determine cytotoxicity of potential medicinal agents and other toxic materials, since those agents would result in cell toxicity and therefore metabolic dysfunction and therefore decreased performance in the assay.
Yellow MTT (3-(4,5-Dimethylthiazol-2-yl)-2,5-diphenyltetrazolium bromide, a tetrazole) is reduced to purple formazan in living cells.[1] A solubilization solution (usually either dimethyl sulfoxide, an acidified ethanol solution, or a solution of the detergent sodium dodecyl sulfate in diluted hydrochloric acid) is added to dissolve the insoluble purple formazan product into a colored solution. The absorbance of this colored solution can be quantified by measuring at a certain wavelength (usually between 500 and 600 nm) by a spectrophotometer. The absorption maximum is dependent on the solvent employed.
MTS is a more recent alternative to MTT. MTS (3-(4,5-dimethylthiazol-2-yl)-5-(3-carboxymethoxyphenyl)-2-(4-sulfophenyl)-2H-tetrazolium), in the presence of phenazine methosulfate (PMS), produces a water-soluble formazan product that had an absorbance maximum at 490-500 nm in phosphate-buffered saline.[2] It is advantageous over MTT in that (1) the reagents MTS + PMS are reduced more efficiently than MTT, and (2) the product is water soluble, decreasing toxicity to cells seen with an insoluble product.
These reductions take place only when reductase enzymes are active, and therefore conversion is often used as a measure of viable (living) cells. However, it is important to keep in mind that other viability tests (such as the CASY cell counting technology) sometimes give completely different results, as many different conditions can increase or decrease metabolic activity. Changes in metabolic activity can give large changes in MTT or MTS results while the number of viable cells is constant. When the amount of purple formazan produced by cells treated with an agent is compared with the amount of formazan produced by untreated control cells, the effectiveness of the agent in causing death, or changing metabolism of cells, can be deduced through the production of a dose-response curve
Yellow MTT (3-(4,5-Dimethylthiazol-2-yl)-2,5-diphenyltetrazolium bromide, a tetrazole) is reduced to purple formazan in living cells.[1] A solubilization solution (usually either dimethyl sulfoxide, an acidified ethanol solution, or a solution of the detergent sodium dodecyl sulfate in diluted hydrochloric acid) is added to dissolve the insoluble purple formazan product into a colored solution. The absorbance of this colored solution can be quantified by measuring at a certain wavelength (usually between 500 and 600 nm) by a spectrophotometer. The absorption maximum is dependent on the solvent employed.
MTS is a more recent alternative to MTT. MTS (3-(4,5-dimethylthiazol-2-yl)-5-(3-carboxymethoxyphenyl)-2-(4-sulfophenyl)-2H-tetrazolium), in the presence of phenazine methosulfate (PMS), produces a water-soluble formazan product that had an absorbance maximum at 490-500 nm in phosphate-buffered saline.[2] It is advantageous over MTT in that (1) the reagents MTS + PMS are reduced more efficiently than MTT, and (2) the product is water soluble, decreasing toxicity to cells seen with an insoluble product.
These reductions take place only when reductase enzymes are active, and therefore conversion is often used as a measure of viable (living) cells. However, it is important to keep in mind that other viability tests (such as the CASY cell counting technology) sometimes give completely different results, as many different conditions can increase or decrease metabolic activity. Changes in metabolic activity can give large changes in MTT or MTS results while the number of viable cells is constant. When the amount of purple formazan produced by cells treated with an agent is compared with the amount of formazan produced by untreated control cells, the effectiveness of the agent in causing death, or changing metabolism of cells, can be deduced through the production of a dose-response curve
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