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Chapter 4 - Quantitative Microbiology

4 - 1 Why enumerate microbes

Many interesting chemical processes take place in the environment, in our bodies an in the food we eat. In almost all cases microbes play an important roll in these chemical conversions. Here are a few examples

  • When mines are abandoned they leave underground tunnels exposed to air and water. This encourages the growth of sulfur-oxidizing bacteria that catabolize iron sulfide present in the rock into sulfuric acid. The high concentration of acid produced will drop the pH of the water to below 2.0, creating a very corrosive liquid that then exists the mine.

  • One result of urban pollution from cars and smokestacks is the releases ammonia and other reduced nitrogen compounds into the air. This can be used as a substrate by ammonia-oxidizing bacteria, forming nitric acid, which corrodes the surface where theses microbes live. Ammonia-oxidizing bacteria are often found on the surface, and deep in the crevasses, of stone, and are thus found with statues and natural stone buildings. The build up of nitric acid causes the decay of these statues and structures. An example of this type of stone decay is found in the Cölonge cathedral in Germany.

  • Microbes contribute to the global cycling of elements, converting carbon, oxygen, nitrogen, sulfur and phosphorous into their various inorganic and organic forms. In some cases only microbes are capable of certain conversions. For example, only microbes can convert nitrogen gas into ammonia using a process called nitrogen fixation.

  • The decay of dead animals and plants depends upon the action of microorganisms, releasing the elements lock in the deceased so that they can be reused by other life forms.

It is worthwhile to understand all of these processes and the microbes involved in them. However, it is also just as important to know the population of microbes carrying out these conversions. Just because a microbe has been found that carries out a process does not mean that it is the major actor on any chemical conversion. It may be at too low a concentration to contribute significantly. Therefore, it is often vital to determine the number of microbes as well as their identity.

There are also many other reasons one might need to known the population of microorganisms in a given sample. For example, determining the rate at which a microbe is killed by UV light or heat requires analysis of the number of viable cells before, and at various times during treatment. In other experiments, it is important to know that you have the right density for a procedure, such as required for transforming a plasmid constructed in vitro into an E. coli strain. Assessment of microbial populations in applied industries is also important. Many food-processing plants will measure the level and type of microorganisms present in their food by doing counts on selective medium. In addition, viable cell counts will be performed when optimizing heat treatments for processing food. Sewage treatment plants will routinely sample and count the microbes present in their treatment systems to insure the correct type and numbers of bacteria are present. The microbial count can be determined using a wide array of techniques. Note that these assays all require somewhat different information and in different time frames, as explained below. In this chapter we summarize some of the more common methods and then do some virtual experiments performing them.


4 - 2 Viable plate counts

Viable plate counts

One of the most common methods of determining cell number is the viable plate count. A sample to be counted is diluted in a solution that will not harm the microbe, yet does not support its growth (so they do not grow during the analysis). In most cases a volume of liquid (or a portion of solid) from the sample is first diluted 10-fold into buffer and mixed thoroughly. In most cases, a 0.1-1.0 ml portion of this first dilution is then diluted a further 10-fold, giving a total dilution of 100-fold. This process is repeated until a concentration that is estimated to be about 1000 cells per ml is reached. In the spread-plate technique some of the highest dilutions (lowest bacterial density) are then taken and spread with a sterile glass rod onto a solid medium that will support the growth of the microbe. It is important that the liquid spread onto the plate soaks into the agar. This prevents left over liquid on the surface from causing colonies to run together and the need for dry plates restricts the volume to 0.1 ml or less. A second method for counting viable bacteria is the pour plate technique, which consists of mixing a portion of the dilution with molten agar and pouring the mixture into a petri plate. In either case, sample dilution is high enough that individual cells are deposited on the agar and these give rise to colonies. By counting each colony, the total number of colony forming units (CFUs) on the plate is determined. By multiplying this count by the total dilution of the solution, it is possible to find the total number of CFUs in the original sample.

Dilution plating and viable plate counts

Figure 4.1. Dilution plating and viable plate counts. (A) A demonstration of a decimal series of dilutions. The 100 sample is a concentrated solution of methylene blue. A 0.2 ml portion of this was added to 1.8 ml (1:9 ratio) of 0.85% saline to create the 1:10 dilution. After mixing, 0.2 ml of the 10-1 dilution was added to a second tube containing 1.8 ml to create the 10-2 dilution. This was continued to generate the dilution series.
(B) A series of pour plates demonstrating the appearance of a viable plate count. The 3 plates show a 10-7, 10-8, and 10-9 dilution of a natural sample. Note how the number of colony forming units decreases 10 fold between the plates.

One major disadvantage of the viable plate count is the assumption that each colony arises from one cell. In species where cells grow together in clusters, a gross underestimation of the true population results. One example of this are species of Staphylococcus, which is known to form clumps of microorganisms in solution. Each clump is therefore counted as one colony. This problem is why the term CFUs per ml is used instead ofâ bacteria per mlâ for the results of such an analysis. It is a constant reminder that one colony does not equal one cell. Great care must also be taking during dilution and plating to avoid errors. Even one error in dilution can have large effects on the final numbers. The rate at which bacteria give rise to an observable colony can also vary. If too short an incubation time is used, some colonies may be missed. The temperature of incubation and medium conditions must also be optimized to achieve the largest colonies possible so that they are easily counted. Finally, this technique takes time. Depending on the organism, one day to several weeks might be necessary to determine the number of CFUs that were present when the experiment started. Such information may no longer be useful for many experiments.

Despite its shortcomings, the viable plate count is a popular method for determining cell number. The technique is sensitive and has the advantage of only counting living bacteria, which is often the important issue. Any concentration of microorganism can be easily counted, if the appropriate dilution is plated. It is even possible to concentrate a solution before counting, as is often done in water analysis, where bacterial populations are usually at low density. The equipment necessary for performing viable plate counts is readily available in any microbiology lab and is cheap in comparison to other methods. Finally, by using a selective medium it is possible to determine the number of bacteria of a certain class, even in mixed populations. These advantages have made viable plate counts a favorite of food, medical, aquatic and research laboratories for the routine determination of cell number.


4 - 3 The mechanics of dilution plating

Typical bacterial populations in a 1 ml or 1 g sample can be as high as 1011 CFU's. If most samples were spread directly onto a plate, there would be so many organisms that individual colonies would not form and a viable count would be impossible. Therefore, samples almost always need to be diluted. Using simple 1/10 or 1/100 dilutions is most convenient and makes later calculations easier. For a 1/10 dilution 1 part sample is added to 9 parts diluent. Note this is not 1 part to 10 parts (that would be a 1/11 dilution)! For a 1/100 dilution, 1 part is added to 99 parts.

When a known amount of a sample is added by pipette to a known amount of diluent (called a dilution blank) the resulting suspension must be thoroughly mixed. Rolling a tube between the hands is not as effective as giving the tube a series of firm flicks with the index finger. (End-to-end mixing is not done with the cotton plugged or capped tubes because they will leak.) A series of dilutions may be made as desired, and from one or more of these dilutions, known amounts are inoculated onto the surface of agar media in plates. Care must be taken to open the plates just enough to admit the pipette, in order to decrease the risk of contamination.

Following the placement of the inoculum on the surface of the plate, a sterile, L-shaped glass rod (often called a "hockey stick") is used to spread the inoculum over the entire surface of the medium. The part of the rod which enters the plate is sterilized by immersion in 95% ethanol followed by passing the rod through a flame such that the ethanol can burn off. (It should be noted that one such flaming may not always kill all of the endospores which may be present on the rod!) After a short period of cooling, the rod is ready for use.

Once diluents are prepared and then added to a plate, the medium is incubated at an appropriate temperture to allow growth of the microorganisms. Colonies arising on the plate can then be enumerated, and this count used to calculate the total number microbe present in the original sample.


4 - 4 Calculating CFU from dilution plating results

How des a count on a plates get converted to CFUs per gram or ml of sample? Let's illustrate the procedure with an example. Imagine that we perform the following experiment:

Five ml of milk are added to 45 ml of sterile diluent. From this suspension, two serial, 1/100 dilutions are made, and 0.1 ml is plated onto Plate Count Agar from the last dilution. After incubation, 137 colonies are counted on the plate.

This problem may be illustrated as follows:

A drawing of the dilution problem

Figure 4.2. A drawing of the dilution problem . It is normally a good idea to draw out dilution problems until you are comfortable doing them.

Note that it is often a good idea to draw out dilution problems until you are comfortable doing them. It will help you to develop a clear picture of what is being done.

The first step in solving this problem is to work out the total dilution of the sample. First 5 ml is added to 45 ml; This is a 1/10 dilution.

Initial dilution

Figure 4.3. Initial dilution. The initial dilution is a 1 to 10 dilution.

Remember, there are many ways to make 1/10 and 1/100 dilutions. A 0.1 ml to 0.9 ml dilution is the same as a 1 ml to 9 ml dilution and a 13 ml to 117 ml dilution. Next, 1 ml of the first dilution is added to 99 ml to make the second dilution, that is a 1/100 dilution. This is repeated with third dilution giving another 1/100 dilution. Then 0.1 ml of the third dilution is plated out on a plate of PCA. The total dilution of the sample is cumulative and can be represented mathematically as....

Calulating total dilution

Figure 4.4. Calulating total dilution. The total dilution for the problem

Notice that the amount put on the plate is also a dilution. Normally CFUs are reported per ml or per gram. In some cases less than 1 ml is put on the plate and this must be taken into account. One way to solve this, is to factor it into the total dilution. In this problem 0.1 ml was added to the plate, or 1/10th of a ml. So multiply the total dilution by 1/10 for the amount added to the plate. This leaves the total dilution as one-one millionth. The next step is to work out the dilution factor. The dilution factor is the reciprocal of the total dilution. In this case it would be......

Dilution factor

Figure 4.5. Dilution factor. A mathematical representation of the diluction factor.

Finally, multiply the total dilution by the average number of colonies in the plate(s) and report your answer in CFUs/ml or CFUs/gram depending upon where the sample came from; in this case ml because we used milk as a sample.

Total colony forming units

Figure 4.6. Total colony forming units. A calculation of the total number of CFUs in the original milk sample.

With enough practice, dilution problems can be worked out quite easily and rapidly. The method described above is just a suggested approach, if you find another way to do these problems which is more intuitive for you, use it. When doing dilution problems, remember the following:

  • Note that using this method, the answer in CFUs per one milliliter or per one gram is derived. Answers may need to be adjusted if the number of CFUs per sample (other than a milliliter or gram) is requested. Assume 1 gram = 1 ml. (1 ml of water does indeed weigh 1 gram. That is actually how the ml is defined.)

  • Use only those plates with colony counts between 30 and 300. With duplicate or triplicate plating from the same dilution, take the average of the plate counts and then proceed.

  • Note that all individual dilutions and the amount plated are multiplied together.

  • The initial dilution is often different from the subsequent dilutions. This is generally due to the nature of the sample available for analysis.

  • Decimal (1:10) dilutions can be made by adding 1 ml to 9 ml. Proportional amounts can be utilized such as by adding 0.1 ml to 0.9 ml or 11 ml to 99 ml.

  • Centimal (1:100) dilutions can be made by adding 1 ml to 99 ml. This can also be done by adding 0.1 ml to 9.9 ml.

  • Note that plating 0.1 ml of a 10-4 dilution results in the same dilution factor (105) as plating 1 ml of a 10-5 dilution.

Here is another sample problem. Using any method you choose, solve the problem.

One ml of a bacterial culture is pipetted into a 9 ml dilution blank. One-tenth ml of this dilution is pipetted into a 9.9 ml dilution blank. From this dilution one-tenth ml is plated using 25 ml of Plate Count Agar. 219 colonies arise after incubation. How many colony-forming units were present per ml of the original culture?

The correct answer: 2.19 X 106 CFUs / ml. (Note, the 25 ml of Plate Count Agar plated is irrelevant. Why?)


4 - 5 A dilution plating protocol

Before beginning the dilution plating exercise, it is necessary for you to learn how to use micropipettors. We will be using them in this experiment and throughout the semester. Each micropipette costs $200 so please treat them with extreme care. Always adhere to the following rules when using micropipettors:

  • Never set the pipettor to above the upper limit or below the lower limit. Figure 4-19 lists the limits for the micropipettes that we use in this course
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    Figure 4.19. Limits of micropipettes. The limits of the micropipettes. Never go above or below these ranges as it may damage the micropipette.

  • Never point a pipettor up. This can cause liquid to run down into the pipettor and destroy it's parts.
  • When withdrawing liquids with the pipettor, always release the plunger slowly. This prevents liquid from rushing into the end of the pipette and clogging it up and is especially important with large volume pipettors (200-1000 µl).
  • Be sure you use the proper size tip for each pipettor.
  • Always use a new tip for each different liquid.
  • Use the correct pipettor for the volume that is to be dispensed. Never use the 200-1000 µl pipette to dispense volumes below 200 µl.

Protocol for Experiment 2

Period 1

Materials

Test tube containing solution to pipette (practice solution)

4 1.5-ml microcentrifuge tubes

20-200 µl micropipette

200-1000 µl micropipette

Yellow pipette tips

Blue pipette tips

Microcentrifuge

  1. Remove four 1.5-ml microcentrifuge tubes from the beaker. Label each tube 1 through 4 on the frosted labeling spot on the side. Place each in a microcentrifuge rack.
  2. Add the volumes of water listed in Figure 4-20 to tubes 1 through 4.
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    Figure 4.20. Volumes to add. Add these volumed to tubes 1 through 4 as listed.

  4. To add the first volume (Measurement 1), take the 20-200 µl micropipette and set the dial to 20 µl. Follow the protocol below to dispense the volume into the tube:
  5. While holding the pipettor, open the box of yellow tips and firmly press the end of the pipettor into a tip. Remove the pipettor from the box, the tip should come along for the ride. Close the pipette tip box.
    • Press the plunger down to the first stop and place the end of the pipettor tip into the practice solution. The end of the pipette tip should be below the surface of the liquid, but not touching the bottom of the tube. Slowly release the plunger until it stops and then remove the tip from the liquid dragging it along the side of the test tube as you do. This will remove any excess liquid clinging to the side of the pipette tip.
    • Take the tube you labeled 1 and hold it at eye level. Place the pipette tip against the side of the microcentrifuge tube and expel the liquid by pushing the plunger down until the first stop. As you drag the tip along the side, push the plunger to the second stop to blow out any remaining liquid. Remove the micropipette from the tube.
    • To remove the pipette tip, hold the pipettor over the tip discard tray and press the white tip eject button located near your thumb. This will eject the tip.
    • Repeat step 3 for each measurement that you add to the tubes. Continue until both tube 1 and tube 2 have been completed. Take your tubes and place them opposite one another in the microcentrifuge. It is very important that tubes are balanced in the microcentrifuge, so make sure you place directly across from one another. If you have questions, please ask your instructor. Centrifuge the tubes for 2-3 seconds to force all liquid to the bottom of the tubes. Remove the tube and return to your lab bench.
    • Since you added a total of 170 µl to tubes 1 and 2, set the 20-200 µl micropipettor for 170 µl and withdraw the tubes contents. If the tube volume exactly fills the micropipette tip, it is time to celebrate! You did it right!.
    • Perform the additions for tubes 3 and 4 in a similar manner. Use the 200-1000 µl pipettor to dispense the larger volumes when necessary. Next, mix each tube briefly on a vortex mixer, and pulse 2-3 seconds in a microcentrifuge. Since a total of 1000 µl (or 1 ml) was added to tubes 3 and 4, set the 200-1000 µl pipettor to 1000 µl and remove the contents of each tube to check the accuracy of your pipetting.
    • Feel free to practice more until you are comfortable with the micropipettes.

A drawing of the dilution

Figure 4.7. A drawing of the dilution. A schematic for the dilution plating to be performed. Measure pipetting volumes carefully and mix tubes thoroughly. Eacg agar plate is inoculated with 0.1 ml.

Before beginning Period 2 you may find it useful to watch the following video on dilution plating

Period 2

Materials

A 1/10 dilution of hamburger

4 saline or 0.85% saline dilution blanks (9 ml)

4 plates of Plate Count Agar (PCA)

4 plates of MacConkey Agar (MAC)

Pipettors and sterile tips

  1. Label one plate of PCA and one plate of MAC for each of the following plated dilutions (as we defined them on page 121, q.v.): 10-3, 10-4, 10-5 and 10-6.
  2. 3. Label the four 9 ml dilution blanks with the dilutions to make of the hamburger as follows: 10-2, 10-3, 10-4 and 10-5.
  3. With the P1000 pipettor and a blue pipettor tip, aseptically transfer 1 ml of the 1/10 hamburger dilution to the dilution blank labeled 10-2; discard the tip into the disinfectant. (Alternatively, if the hamburger dilution is provided in a 1 ml amount, you can dump the contents of the dilu-tion blank into the hamburger tube.)
  4. Mix this dilution well by holding the tube on the Vortex mixer* for about 5-10 seconds as demonstrated by the instructor; an actual vortex must be achieved for proper mixing. Note that this tube is a 1/100 (i.e., 10-2) dilution of the original, undiluted hamburger.
  5. Referring to the diagram on the next page, continue making serial dilutions of the hamburger, using a new pipettor tip for each new dilution. (Be sure you are discarding the tips into disinfec-tant!) Thus, by making 1 ml inoculations into the 9 ml dilution blanks, we will achieve the ham-burger dilutions which you have labeled on the tubes.
  6. With the P200 pipettor and a yellow tip, aseptically transfer 0.1 ml of the 10-2 dilution onto each of the plates marked 10-3. (Why are these plates marked 10-3 and not 10-2?) With new tips, continue in like manner to inoculate the remaining plates.
  7. If such a device is not available, hold the tube between thumb and forefinger in one hand and flick the bottom of the tube with the fore-finger of the other hand in order to achieve a vortex. (Do not simply roll the tube between the hands or lightly tap the tube. Don't be too gentle, yet don't shake the tubes end-to-end!)

  8. Perform the following with care! (See page 118 including the safety precaution.)
    • Sterilize a hockey stick according to the directions near the bottom of page 118, making sure that the glass hockey stick has cooled sufficiently before immersing it into the ethanol.
    • Spread the inoculum over the entire surface of the 10-6 plates. Proceed to do the same with the remaining plates, moving from the more dilute to the more concentrated inocula. During the course of spreading the plates, the hockey stick need not be re-flamed, as long as you proceed from the more dilute to the more concentrated inocula. When all plates have been spread, resterilize the hockey stick and let it cool before returning it to the drawer.
  9. Incubate the plates (inverted!) and incubate at 30°C for 2 days. If the next period is 3 or more days away, bring the plates to the tray on the stage for 2-day incubation (followed by refrigera-tion which will arrest growth and development of the colonies, preventing overgrowth).

Period 3

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Figure 4.8. Dilution plating results. Example results from a dilution plating. One gram of fresh hamburger hours was diluted as described in the procedure for this expeirment and then plated on MAC agar and PCA Row 1, MacConkey Agar; Row 2 PCA. If you are doing this virtually, you can count the colonies on the plate and determine the CFU per gram of hamburger. Record your values in your notebook.

  1. . Total Aerobic Plate Count: Choose one PCA plate which appears to have between 30 and 300 colonies on it and count the colonies on the plate. As in the example above, determine the number of colony-forming units per gram of the hamburger. Note that this method is technically neither total nor aerobic with regard to microorganisms. Not all microorganisms are able to grow on this medium even though it is termed an all-purpose medium. Aerobic refers to the incubation conditions, not the oxygen relationship of the organisms. Figure 4-8 shows a picture of typical results. The PCA plate has been diluted to 10-4. This is the total dilution. This would be an appropriate plate to count.
  2. . Total Gram-Negative Plate Count: (Note the selective and differential aspects of this medium as discussed on page 126.) Choose one MAC plate which appears to have between 30 and 300 colonies on it and count the colonies on the plate. Determine the number of gram-negative colony-forming units per gram of the hamburger. If distinctly red or pink colonies can be differentiated clearly, what appears to be the relative proportion of lactose fermenters? (E. coli is among the gram-negative organisms which can ferment lactose and thus produce red colonies on MacConkey Agar.) The MAC plate has been diluted to 10-3. This is the total dilution.

4 - 6 Direct particle counts

Counting chambers

The most direct method of counting microorganism is by the use of a microscope and a slide with special chambers of known volume. These slides allow the counting of a small number of cells in a small volume and extrapolating the result to determine the population. An example of such a device is shown in Figure 4-10. A culture is placed on the slide marked with precise grids. The number of cells present in each grid is counted and an average determined. Conversion using a formula gives the number of cells per milliliter in the culture. This method is rapid, a result can be known in just a few minutes, and is easy to perform. However, it is impossible to distinguish living cells from dead ones. If this distinction is important, direct microscopic counts are not the solution. Finally, cultures containing less than 1 million cells per ml are actually too dilute for direct counts since there will be too few cells in the very small volume that is actually examined under the microscope for an accurate count.

The Petroff-Hauser counting chamber

Figure 4.10. The Petroff-Hauser counting chamber. The center of the slide contains a precisely machined grid, with each square (1/20 mm x 1/20 mm) having a known area. The coverslip also rests above the slide a known distance (typcially 1/50 of a mm). Since each of these dimenstions is known, it is possible to calculate the number of cells in each square and pluggin it into a formula. On the left of the figure is a photograph of a Petroff-Hauser slide. On the right is a grid at 100 x magnification showing the size of the squares.

a picture of the petroff-hauser cell counter and then what it looks like on a slide.

Electronic particle counters

Electronic particle counters are useful if the number of bacteria in a sample needs to be counted on a routine basis. The method is based on the property that nonconductive particles, such as bacteria, will cause a disruption in an electric field as they pass through it. A Coulter counter is a type of electronic particle counter in which there is a small opening between electrodes through which suspended particles pass, see Figure 4-11. In this sensing zone, each particle displaces its own volume of electrolyte, causing a current pulse. The pulse is noted and recorded as one particle count. By precisely controlling the rate at which solution passes through the opening, it is possible to get exact, reproducible counts at a rate of up to several thousand bacteria per second. Coulter counters are highly dependent upon particle size and those dependent upon changes in current are near their detectable limits with microorganisms. Particle counters that use light diffraction as a means of sizing and counting particles are also manufactured and can detect particle less than 1 µm in diameter.

The coulter counter

Figure 4.11. The coulter counter. A picture of a Coulter Counter. The diagram at the right demonstrates the opening that the microbes must pass through during counting. When microbes pass through the aperture the electrical potential across the electrodes is distrubes, which the dataprocessing system records as a count. In the lower right is an actual picture of a Multisizerâ„¢ 3 by Beckmann instruments that uses the Coulter counter priciple.

The advantage of this method is the simplicity of its operation and it reproducibility. As in microscopic counts, the machine cannot distinguish between living or dead cells or even between dust and bacteria. Any reasonably sized particle in the solution will be counted. There is also the expense of buying the counter, which can cost many thousands of dollars.


4 - 7 Using a counting chamber

The direct cell counts will be demonstrated using Petroff-Hauser cytometers. These are glass slides with precisely machined chambers and coverslips, so that cells in very small volumes can be counted using the microscope.

The chamber is formed by laying the coverslip on the elevated borders of the central well area. See the diagram below:

Arrangement of the Petroff-Hauser counting chamber

Figure 4.12. Arrangement of the Petroff-Hauser counting chamber. The drawing shows the correct placement of the cover slip. Once the coverslip is placed on the slide, a small amount of culture (about 10 µl) is placed in the well on the left or right of the viewing area. Capillary action then brings the liquid onto the viewing area with the grid.


  1. Using a capillary pipette, place a drop of the broth culture at the edge of the coverslip. Capillary action will draw the liquid under the coverslip. Wait 1-2 min. for the movement to stop and the cells to settle. Then with the low-power objective focus on the grid in the center of the slide; you should see a crosshatched area containing 25 squares each containing 16 squares:

  2. You will not be able to see many bacteria at this magnification so turn the high dry power lens (40X) into line. do not use the oil immersion lens with these chamber, as it just goomers the chamber and coverslip.

  3. Count the number of bacteria in 10-15 of the 1/400 mm2 squares and calculate an average cell number.

  4. To be statistically correct you should use a dilution such that there are no more than three cells in each small (1/400 mm2) square and the total number of cells counted is at least 100.

Figure 4-13 shows two images of a Petroff-Hauser counting chamber. Use this to see what it looks like and for performing a count of the microbes.

To determine the concentration of bacteria in the original culture use the following formula

Formula for the counting chamber

Figure 4.14. Formula for the counting chamber. Use this formula for calculating the number of cells per ml from the count obtained using a counting chamber. Nc is the average number of cells counted per square and D is the dilution of the samples placed on the slide.

For example:

The 103 is there as a conversion factor from mm3 as measured by the chamber to cm3 (a.k.a. ml) as typically expressed for culture density. Here is a more detialed explanation of that conversion factor:

1 ml = 1 cm3 = 1 cm x 1 cm x 1 cm

1 cm = 10 mm

so 1 ml = 10 mm x 10 mm x 10 mm

or 1 ml = 103mm3

If you use one drop (without dilution) from a broth culture: and find an average of 2.31 squares per cell, your results would be:

Calculation of cell number from a counting chamber

Figure 4.15. Calculation of cell number from a counting chamber. If an average of 2.31 cells if found in a 10-1 dilution, the formula would appear as shown here with a result of 4.62 x 108 cells per ml of culture.


4 - 8 Indirect methods

Change in the amount of a cell component

In situations where determining the number of microorganisms is difficult or undesirable for other reasons, the use of indirect methods can be an excellent alternative. These methods measure some quantifiable cell property that increases as a direct result of microbial growth.

The simplest technique of this sort is to measure the weight of cells in a sample. Portions of a culture can be taken at particular intervals and centrifuged at high speed to sediment bacterial cells to the bottom of a vessel. The sedimented cells (called a cell pellet) are then washed to remove contaminating salt, and dried in an oven at 100-105 °C to remove all water, leaving only the mass of components that make up the population of cells. An increase in the dry weight of the cells correlates closely with cell growth. However, this method will count dead as well as living cells. There might also be conditions where the dry weight per cell changes over time or under different conditions. For example, some bacteria that excrete polysaccharides will have a much higher dry weight per cell when growing on high sugar levels (when polysaccharides are produced) than on low. If the species under study forms large clumps of cells such as those that grow filamentously, dry weight is a better measurement of the cell population than is a viable plate count.

It is also possible to follow the change in the amount of a cellular component instead of the entire mass of the cell. This method may be chosen because determining dry weights is difficult or when the total weight of the cell is not giving an accurate picture of the number of individuals in a population. In this case, only one component of the cell is followed such as total protein or total DNA. This has some of the same advantages and disadvantages listed above for dry weight. Additionally, the measurement of a cellular component is more labor-intensive than previously mentioned methods since the component of interest has to be partially purified and then subjected to an analysis designed to measure the desired molecule. The assumption in choosing a single component such as DNA is that that component will be relatively constant per cell. This assumption has a problem when growth rates are different because cells growing at high rates actually have more DNA per cell because of multiple initiations of replication.

Turbidity

A final widely used method for the determination of cell number is a turbidometric measurement or light scattering. This technique depends on the fact that as the number of cells in a solution increases, the solution becomes increasingly turbid (cloudy). The solution looks turbid because light passing through it is scattered by the microorganisms present and the turbidity is proportional to the number of microorganisms in the solution. The turbidity of a culture can be measured using a photometer or a spectrophotometer. The difference between these instruments is the type of light they pass through the sample. Photometers, such as the Klett-Summerson device, use a red, green or blue filter providing a broad spectrum of light. Spectrophotometers use prisms or diffraction gratings supplying a narrow band of wavelengths to the sample. Both instruments measure the amount of transmitted light, the light that makes it from the light source through the sample to the detector.

Measuring the turbidity of a culture

Figure 4.16. Measuring the turbidity of a culture. A spectrophotometer or photometer quantifies the amount of turbidity of a culture. The amount of light scattered from a solution is proportional to cell number. The instruments measures light that is not scattered by the sample.

When measuring light scattering it is important to consider the wavelength of light used a bacterial culture. Microorganisms may contain numerous macromolecules that will absorb light, including DNA (254 nm), proteins (280 nm), cytochromes (400-500 nm), and possible cell pigments. When measuring bacteria by light scattering it is best to pick a wavelength where absorption is at a minimum and for most bacterial cultures wavelengths around 600 nm are a good choice. However, the exact wavelength chosen is species specific.

The amount of light transmitted through a sample is inversely proportional to cell number and can be expressed in the following equation.

The transmittance equation

Figure 4.17. The transmittance equation. The ratio of light hitting the sample (I0) to light passing through a sample (I) is the transmittance.

Where T is the light transmitted, I0 is the light entering the sample and I is the light passing through to the detector.

Due to the nature of light scattering, transmittance decreases geometrically as the cell numbers increase. It is more intuitive to think of the units increasing as growth increases and for most bacterial analysis, transmittance is converted into absorbance using the following equation.

Absorbance

Figure 4.18. Absorbance. The absorbance of a sample it the negative log base 10 of the transmittance.

Absorbance increases in a linear fashion as the cell number increases. When measuring growth of a culture the term optical density (OD) is normally used to more correctly represent the light scattering that is occurring; under optimal conditions, little light is actually absorbed by the culture so the term absorbance is misleading. For most unicellular organisms changes in OD are proportional to changes in cell number (within certain limits) and therefore can be used as a method to follow cell growth. If a precise cell number for a given OD is desired, a standard curve can be generated, where viable plate count or cell mass is plotted as a function of OD. It also wise to develop a standard curve to verify that the OD is actually an accurate portrayal of cell growth. After the standard curve is made, it is then possible to simply measure the OD of the culture and read the cell number from the curve.

The turbidity of a culture is dependent upon the shape and internal light-absorbing components of the microorganism and therefore turbidity readings are species-specific and cannot be compared between different microbes or even between different strains of the same species. As above, there are microbes that change cell size or shape at different stages of growth, which introduces some inaccuracy to this method of cell counting. Also both living and dead cells scatter light and are therefore counted. However, the method is very rapid and simple to perform and provides reliable results when used with care, so it is an extremely common method of real time analysis of prokaryotic populations. In fact it is one of the methods we will use for measuring cell number in the experiment on bacterial growth. Turbidometric measurements also do not destroy the sample.