# Chapter 15 Microbial enumeration.

OBJECTIVEIn this chapter you will learn

* basic calculations for estimating microbial populations.

Overview

You can count large organisms in soil by several physical extraction procedures; not so with microorganisms and the smaller soil animals. Even if it is possible to physically extract the organisms, the populations are so high that counting is excessively tedious and time-consuming. For microorganisms, it is also generally impossible to extract all members of the population, and physically impossible to count the enormous numbers that are present in soil. Consequently, methods of counting the smaller organisms in soil (protozoa-sized or smaller) generally rely on some sort of dilution procedure followed by direct counting, plating on selective media, or cultivation to elicit a characteristic population response. These procedures are the topic of this chapter.

Serial Dilution

Serial dilution is one of the most basic procedures in soil biology studies. The principle is very simple (Figure 15-1). A soil sample is suspended in buffer and subsamples are resuspended in additional buffer until an appropriate amount of dilution is achieved. Serial dilution has the same effect as diluting soil in a very large volume of buffer all at once, but is much more convenient and also gives a range of dilutions from which samples can be drawn.

[FIGURE 15-1 OMITTED]

The most critical step in serial dilution is calculating how much soil was originally diluted, and by how much it was diluted. It is important to remember that everything is usually based on grams of oven-dry soil equivalent, even though you typically dilute moist soil. So, you must take into account the volume of water added with soil and the volume contributed by the soil itself. Subsequent dilutions are typically done in 10-fold steps, which simplifies calculations.

The formula for calculating populations after serial dilution is straightforward but requires careful attention to units. First, calculate the dilution factor (DF).

g dry soil x dilution ratio, x dilution [ratio.sub.2] x ... dilution [ratio.sub.n]/ diluent vol. + soil vol. + soil [H.sub.2]O

= g dry soil/vol. diluent = DF [15-1]

The diluent is the solution into which the soil is being diluted and serially transferred. The dilution ratio is

volume transferred/ total volume after transfer [15-2]

Example 15-1

What is the dilution ratio if 5 mL is added to 22 mL?

Solution

The dilution ratio is 5 mL/22 mL + 5 mL = 0.185.

The most complicated step is the initial dilution, because you have to take into account the change in the diluent volume that occurs when you add the soil and the water that is in a moist soil, which also contributes to the total volume.

For example, a prior chapter on soil water (Chapter 7) indicated that 10 g of moist soil at 10 percent gravimetric water content actually contains 0.91 g (or mL) of [H.sub.2]O and 9.09 g of dry soil ([0.91/9.09] x 100 = 10%). Furthermore, the 9.09 g of soil has a volume of 3.43 [cm.sup.3] (or mL) if we assume a particle density of 2.65 g [cm.sup.-3] (9.09 g/2.635 g [cm.sup.-3] = 3.43 [cm.sup.3]).

Example 15-2

How much volume does 15 g of moist soil at 35 percent gravimetric water content contribute to 100 mL of diluent?

Solution

15 g moist soil / [1 + gravimetric water content (35%)] = 15g / 1.35=11.1g dry soil

15 g moist soil - 11.1 g dry soil = 3.9 g [H.sub.2]O

11.1 g dry soil / 2.65 g [cm.sup.-3] = 4.2 [cm.sup.3] soil

3.9 g [H.sub.2]O + 4.2 [cm.sup.3] soil = 3.9 mL [H.sub.2]O + 4.2 mL soil = 8.1 mL

The total diluent volume after the addition of soil is 100 mL + 8.1 mL = 108.1 mL. When all is said and done the number of organisms per gram soil is

(number of organisms/vol. analyzed)/ DF [15-3]

Example 15-3

If 10 g of soil at 25 percent moisture content was suspended in 95 mL of water and serially diluted three more times in 10-fold steps, what is the total population if 23 organisms were present when 0.5 mL of the last sample was analyzed?

Solution

The total dilution ratio is 0.001 since each of the three dilution steps was 10-fold, or 1/10. The initial dilution was 10g/95 mL. However, the 10g of moist soil contained 2g of water and only 8g of soil (2g [H.sub.2]O/8g soil = 25% gravimetric water). The 8g of soil contributes 3.0 mL to volume (8g/2.65g [cm.sup.-3] = 3.0 [cm.sup.3] or mL). So, the total volume of diluent was 95 mL + 2 mL + 3 mL = 100 mL. (This is why so many procedures for serially diluting soil start with an initial dilution of 10g/95 mL; it's usually close to a 10-fold dilution.) The concentration of soil in each sample analyzed or the dilution factor is therefore

(8 g x 0.001)/100 = 8 x [10.sup.5] g [mL.sup.-1]

(23 organisms/0.5 mL) / 8 x [10.sup.-5] g [mL.sup.-1] = 5.75 x [10.sup.5] organisms per g soil

Selective Plating

Counting bacteria, fungi, or even viruses typically involves serial dilution of a soil or water sample and then applying the diluted soil or water to a plate of solid media, which is then incubated. For bacteria and fungi, you usually want to have between 20 and 200 colonies growing on the media at the end of incubation. Each colony is assumed to originally represent one bacterial cell, spore, or fragment of fungi. Thus, the organisms you end up counting are termed colony forming units, or CFUs.

Example 15-4

How much will you have to dilute a soil sample that you suspect contains 1 x [10.sup.6] CFU per gram if you want to plate 1.0 mL and count less than 200 colonies?

Solution

We can estimate the dilution we might need by the following:

200 CFU [mL.sup.-1]/ ? g [mL.sup.-1] = 1 x [10.sup.6] CFU [g.sup.-1]

Solving for "?" gives 2 x [10.sup.-4] g [mL.sup.-1] as the final soil concentration. Thus, suspending 10 g of soil in 95 mL of buffer and conducting three more 10-fold dilutions will give an overall dilution of the soil sample of approximately 10,000, which is required for this anticipated soil concentration.

Most Probable Number

Not all soil organisms grow on solid media. In addition, soil biologists sometimes want to enumerate separate physiological groups from among a host of similarly looking organisms. One approach to this problem is a technique called most probable number (MPN) enumeration (Alexander, 1982). In MPN enumeration, a soil or water sample is serially diluted and then subsamples from each dilution level are dispensed into replicated tubes of growth media. If at least one viable organism is present, it will grow, or at least cause some characteristic biochemical change in the media that can be detected. The pattern of tubes demonstrating these changes for each dilution is then used to calculate, based on statistical methods, the most probable number of cells that were in the original sample (Figure 15-2).

Once the pattern of positive and negative responses is obtained, a computer program or table is used to calculate the most probable number per gram of soil or milliliter of sample. Most probable number tables are specific to the dilution series and number of replicate tubes that were used for analysis. A table for 10-fold dilutions with five replicate tubes per dilution level is located in Appendix 5. A sample Microsoft Excel program for calculating MPN by spreadsheet is in Appendix 6.

Find the most diluted sample in which all of the inoculated containers give a positive response. Call this dilution [p.sub.1]. Call the next two highest dilutions [p.sub.2] and [p.sub.3]. From a prepared table of MPN values for the same dilution series and number of replicated samples, read down the column for [p.sub.1] until you reach the value for the number of positive samples you observed. Read across for the value of [p.sub.2] corresponding to the number of positive samples for [p.sub.2]. Then read across into the body of the table until you reach the column for [p.sub.3] that corresponds to the number of positive samples in that dilution. The number listed is the MPN of organisms in the inoculum you added.

To calculate the MPN in the original sample, divide the tabular MPN by the dilution factor corresponding to [p.sub.2]. Remember that the dilution factor is

g soil or mL sample/vol. of diluent

and in the MPN table corresponds to the initial dilution times the dilution step.

Example 15-5

In attempting to enumerate nitrifiers from a soil sample, 10 g of field moist soil at 25 percent gravimetric water content was serially diluted in 10-fold steps for each dilution and these dilutions were used to inoculate five replicate tubes of growth media with 1.0 mL from each dilution. The original dilution was 8 g soil/95 mL diluent. The appropriate dilutions for [p.sub.1], [p.sub.2], and [p.sub.3] and the results of the incubation are given in Figure 15-3.

Solution

The number of positive samples for [p.sub.1] = 5, [p.sub.2] = 4, and [p.sub.3] = 2. The corresponding MPN for this sequence in Appendix 5 is 2.2. This number is divided by (initial dilution X the dilution step) = (8 g/95 mL) X [10.sup.-3]. The final value is 2612 nitrifiers per gram soil.

A 95 percent confidence limit can be put on the MPN estimate by multiplying and dividing the calculated MPN by a unique factor for each combination of dilution ratio and number of replicate samples. A table of these factors is located in Appendix 5. As a general rule, unless MPN values differ by a factor of 10, there is not a significant difference between them.

[FIGURE 15-2 OMITTED]

[FIGURE 15-3 OMITTED]

Example 15-6

What is the 95 percent confidence interval for Example 15-5 above?

Solution

It is given that a 10-fold serial dilution with five replications was used. The corresponding factor for this treatment combination is 3.3. Therefore, the 95 percent confidence interval is 2612 nitrifiers per g x 3.3 and 2612 nitrifiers per g / 3.3. The true MPN should therefore lie between 792 and 8620 nitrifiers per g.

Direct Count Calculations

It is well appreciated by most soil microbiologists that an optimistic estimate of the true number of microbes in soil determined by plate counting is only 1 to 10 percent. Furthermore, although the MPN technique overcomes some of the limitations of plate counting, it suffers from an inherent lack of precision. The MPN technique likewise suffers from the limitation that an organism may be present in a sample, yet not be counted because the growth media is wrong.

As long as you can distinguish microbes from a background of organic debris and mineral material, you can use direct microscopy for enumeration. Vital stains, such as acridine orange or fluorescein diacetate, which mark specific cell components such as the cell wall or proteins, can also be used to flag microbes. Direct microscopy will yield a total count of all cells, living and dead, unless the stain distinguishes between the two.

To calculate direct counts, an involved, but relatively simple, equation is used.

cell number/[W.sub.s] soil = ([N.sub.f])(A)/([a.sub.m)([V.sub.f])DF [15-4]

where

[N.sub.f] = average number of cells per microscope field

A = filter area ([mm.sup.2] or [cm.sup.2])

[a.sub.m] = area of microscope field ([mm.sub.2] or [cm.sub.2])

[V.sub.f] = volume filtered (mL)

DF = dilution factor

[W.sub.s] = dry weight of soil (g)

The area of the microscope field is the actual surface area of the filter that is visible when you look through the lens. Because of magnification, it only represents a fraction of the total surface area of the filter. Consequently, 10 or more microscope fields must be examined to get a reasonable approximation of the average cell density. A special slide called a stage micrometer is used to calibrate the field area. A stage micrometer is simply a microscopic ruler incised into a glass slide. Many microscopes are also conveniently equipped with an ocular lens that contains a superimposed grid of squares. Once the length of each square is determined with a stage micrometer, it becomes relatively easy to determine the length of any organism in the microscope field.

Example 15-7

After an initial dilution of 10 g oven-dry soil per 95 mL diluent, the sample is diluted 1000-fold more and 5 mL is dispensed onto a filter. The diameter of the filter is 60 mm and the diameter of the microscope field is 10 [micro]m. If an average of five cells were counted per field, what is the original population?

Solution

The dilution factor is (8g/95 mL) x 0.001 = 8.42 x [10.sub.-5] g [mL.sub.-1.] The area of the filter which, as you remember from basic geometry is the surface area of a circle, is [pi] [(60 mm/2).sup.2] = 2827 [mm.sup.2]. Likewise, the surface area of the microscope field is [pi] [(10 [micro]m/2).sup.2], or [pi] [(0.01 mm/2).sup.2] which is equivalent to 7.854 x [10.sup.-5] [mm.sup.2]. Entering these values into the appropriate equation gives

cell number/g soil = ([N.sub.f])(A)/ ([a.sub.m)([V.sub.f]) DF = (5 cells)(2827 [mm.sup.2])/(7.854 x [10.sup.-5] [mm.sup.2])(5 mL)(8.42 x [10.sup.-5] g [mL.sup.-1])

= 4.27 X [10.sup.11] cells [g.sup.-1]

Determining the length of microorganisms in soil is particularly applicable to estimating fungal numbers. Because fungi form long hyphae, a more precise measure of the fungi in soil is usually the length rather than the number. By assuming the fungi are approximately the same shape as a cylinder, you can estimate the volume they occupy as well as the length. There is no fundamental difference in the calculation of fungal length as opposed to bacterial number by direct microscopy. The only change is that the average length of fungi is recorded rather than the number of hyphal strands.

Sample Problems

Serial Dilution

1. Ten grams of soil was suspended in 95 mL of buffer and serially diluted in 10-fold steps a total of four times. When 1.0 mL was analyzed, there were only two organisms counted. If the starting moisture content was 15 percent, how many organisms were there per gram of soil?

2. Five grams of soil at 20 percent moisture were resuspended in 40 mL of buffer. The suspension was diluted in twofold steps five more times. What was the final soil concentration?

3. Ten grams of soil at 30 percent moisture was suspended in 50 mL of buffer. The suspension was serially diluted in fivefold steps an additional six times. Two milliliters from the final suspension were analyzed and found to have six organisms. What was the original concentration of organisms in the soil?

4. Five grams of oven-dry soil was suspended in 95 mL of water and serially diluted four times in 10-fold steps. What was the final concentration of soil in the last dilution?

5. A microbiologist wanted to dilute exactly 10 g of oven-dry soil in 95 mL of buffer and serially dilute the subsequent suspension 10,000-fold more. If he was starting with moist soil at 25 percent gravimetric water content, what is his final soil concentration?

Plate Counts

1. How many bacteria were present if 0.1 mL of a [10.sup.5] dilution of 10 g of soil yielded 32 bacteria?

2. How many bacteria were initially present if two plates, each inoculated with 1.0 mL of a [10.sup.6] dilution of soil yielded 150 and 75 CFU, respectively?

3. How much will a soil sample have to be diluted if it is suspected to contain [10.sup.4] bacteria per gram soil and you want to inoculate each plate with 0.15 mL of sample?

Most Probable Number

1. Ten milliliters of water was initially diluted in 90 mL of buffer and serially diluted 10-fold in six more steps to determine algae numbers. The [p.sub.1], [p.sub.2], and [p.sub.3] values were 4, 2, and 1, respectively, beginning at the [10.sup.-3] dilution step. What is the MPN of algae assuming five replicate tubes were used at each dilution?

2. Five grams of moist soil at 25 percent gravimetric water content was diluted in 95 mL of buffer. Three more 10-fold serial dilutions were performed and 2 mL of inoculum was dispensed onto five replicate tubes of growth media. After two weeks the results were recorded. The [p.sub.1], [p.sub.2], and [p.sub.3] values at the [10.sup.-2], [10.sup.-3] , and [10.sup.-4] dilution steps were 5, 1, and 1, respectively. What is the MPN for this sample?

3. Ten grams of soil at 33 percent moisture were diluted in 95 mL of buffer and examined for the population of mycorrhizal fungi by serial dilution in 10-fold steps and inoculation into replicate containers of sterile soil. The results are shown below.

Dilution step [10.sup.-1] [10.sup.-2] [10.sup.-3] Sample 1 + + - Sample 2 + + + Sample 3 + - + Sample 4 + + + Sample 5 + + - [10.sup.-4] [10.sup.-5] [10.sup.-6] Sample 1 - - - Sample 2 - - - Sample 3 + - - Sample 4 - - - Sample 5 - - -

What is the MPN for mycorrhizae in the original soil sample?

4. What is the 95 percent confidence interval for an MPN analysis if the calculated MPN was 2.5 x [10.sup.4] cells per milliliter and the MPN design employed fivefold serial dilutions with seven replications?

Direct Count Calculations

1. One gram of soil was diluted into the equivalent of 1 L, then 2 mL were spread onto the surface of a 60 mm filter. The filter was examined by microscopy and 15 cells were counted per microscope field. Each microscope field had a diameter of 100 [micro]m. What was the total number of cells in the original sample?

2. Ten grams of soil at 33 percent gravimetric water content was diluted in 95 mL water, and four additional fivefold serial dilutions were carried out. Ten milliliters of the resulting suspension were filtered on a 75 mm diameter filter, and 15 cells were counted per microscope field. Each microscope field was 50 [micro]m in diameter. What was the total cell count in the original sample?

3. If the concentration of cells in a water is presumed to be [10.sup.6] [mL.sup.-1] and the sample was diluted 1000-fold before filtering, how much material should be filtered to obtain 25 cells per microscope field, assuming that the filterable area is 5 [cm.sup.2] and each microscope field is 100 [micro][m.sup.2]?

Reference

Alexander, M. (1982). Most probable number methods for microbial populations. In A. L. Page et al. (Eds.), Methods of soil analysis, Part 2. Microbiological and biochemical properties (2nd ed., pp. 815-820). Madison, WI: Soil Science Society of America.

Printer friendly Cite/link Email Feedback | |

Title Annotation: | Section IV Problem Solving in Soil Biology |
---|---|

Publication: | Math for Soil Scientists |

Geographic Code: | 1USA |

Date: | Jan 1, 2006 |

Words: | 3286 |

Previous Article: | Chapter 14 Microbial growth, yield, and mortality. |

Next Article: | Chapter 16 Microbial biomass. |

Topics: |