Supplemental Lecture (97/12/07 update) by Stephen T. Abedon (abedon.1@osu.edu)
 Chapter title: Diluting and Concentrating
 A list of vocabulary words is found toward the end of this document
 Quantification of microorganisms is often necessary. This might be because a certain number of organisms, no more, no less, are required to perform a given experiment, or because one is testing a sample to determine the degree of microbial contamination present, etc. Many methods of quantification have the drawback that only a certain range of concentrations of microorganisms may be enumerated. How does one adjust organismal concentration? The answer is by employing methods of concentrating or diluting. In this lecture we will briefly discuss some of these methods but, of at least equal importance, we will spend a significant amount of time discussing how one keeps track of the degree to which one concentrates or dilutes.
 Overview [of diluting and concentrating]
 D = Dilution Factor
V = Volume
# = Absolute amount of solute or substance suspended or dissolved
C = Concentration
old = Prior to dilution step
new = Following dilution step
plating = Done or obtained in the course of plating
D = V_{new} / V_{old}
D = (#_{old} / V_{old}) / (#_{new} / V_{new})
D = C_{old} / C_{new}
D = C_{old} / #_{plating}
D_{plating} = 1 / V_{plating}
C_{old} = D * C_{new}
C_{old} = D * #_{plating}
C_{plating} = C_{old} / D
 Concentration does not change if V and # change proportionally [(# * k) / (V * k) = # / V].
 Deviations from unit volumes during plating effectively dilute with a dilution factor equal to the inverse of the volumed used (D_{plating} = 1 / V_{plating}).
 Overall dilution factor equals the product of the dilution factors associated with all dilutions done in serial which separate the original, undiluted culture from the final, diluted culture (e.g., D_{overall} = D_{first dilution} * D_{second dilution} * D_{third dilution} * . . . * D_{plating}).
 Methods of concentration [concentrating]
 Removal of solvent:
 The concentration of microorganisms may be accomplished by a number of methods.
 All of these methods involve the removal of solvent (i.e., water) as well as the various dissolved substances found in an organism's extracellular (or extraorganismal) environment.
 Various methods:
 These methods include:
 filtration
 centrifugation
 precipitation
 Which method you employ depends upon what you are trying to accomplish as well as the time and equipment you possess to accomplish the task.
 Diluting back up to desired concentration:
 Regardless of which method you employ, what you end up with upon employing any of these methods, ideally, is a concentrated mass of organisms.
 Typically, organisms must subsequently be properly diluted in order to arrive at a desired suspended concentration.
 Diluting
 Adding solution:
 Diluting is simply the addition of a solution (or plain solvent) to a substance in order to decrease the concentration of the latter substance.
 The substance being diluted may already be suspended or dissolved, or might not yet be suspended or dissolved.
 If it is suspended or dissolved, it may or may not be suspended or dissolved in the same kind of solution or solvent you intend to dilute it with.
 Reducing organism concentration
 Reducing solute concentration:
 Diluting may be employed to reduce the concentration of a solute surrounding an organism.
 For example, an organism may be kept in the presence of an agent which inhibits its metabolism, and then metabolism may be reinitiated in that organism through the dilution of the inhibiting substance.
 Here diluting allows one to avoid a concentration step, i.e., the removal of the inhibitor by first concentrating, and then resuspending the organisms in inhibitorless solution. However, note that the "diluting out" of an inhibitor by this method is at the cost of reducing the concentration of the organism.
 Note that you may have to keep track of the degree to which both the organism and the inhibitor have been diluted.
 Increasing solute concentration:
 Diluting may also be employed as a means of introducing new solutes to a solution.
 For example, an organism might be kept in a solution which is free of carbon and energy sources, and then diluted into a broth in order to initiate metabolism.
 This, too, reduces organism concentration, though note that concentration reduction can be minimized by employing diluents which contain a multifold excess of the ingredients you are introducing. In this way concentrations may be reduced to a lesser extent than if lower concentrations were added.
 For example, an organism suspended in a salt solution can be introduced to broth by adding an equal volume of a 2x broth (i.e., twofold greater concentration broth ingredients than that normally employed) but 1x salt solution to a 1x salt solution. The resulting solution consists of 1x broth and 1x salt solution. However, the organism's concentration will have been reduced twofold. Less organismal dilution may be possible if one can get away with using a greater than 2x broth.
 Algebra of concentration
 # / V = concentration:
 Understanding diluting and concentrating first requires an understanding of the algebra of concentration.
 The term concentration refers simply to a measure of the absolute amount of a substance divided by volume, where the volume represents a sum of (i) the volume of the dissolved or suspended substance and (ii) the volume of the dissolving or suspending medium.
 That is, concentration = # / V where V = V_{substance} + V_{solvent}.
 Concentration may be changed through changes in the value of # or through changes in the value of V.
 Changing #:
 Note that changing concentration through changes in the value of # simply involves the addition of the substance of concern (e.g., the microorganism).
 Various means may be employed to keep track of the amount of this substance added (e.g., grams, moles, absolute numbers, etc.).
 However, beyond noting that one has to be careful when adding substance to change concentration, because this addition typically will increase the value of both # and V, here we will nevertheless not dwell upon this procedure.
 Changing V:
 Changing concentration through changes in V are accomplished through diluting or concentrating.
 A dilution involves increasing the value of V.
 Concentrating involves decreasing the value of V (though, keeping in mind, as noted above, that concentrating often is a two step procedure in which a substance is concentrated too far, and then diluted back to the desired concentration).
 Increasing V = proportional dilution:
 Note that the degree of concentrating or diluting is proportional to the value of V (so long as # remains constant).
 Thus, if V is increased 10fold, that represents a 10fold change in concentration, specifically a 10fold decrease in concentration.
 Similarly, a 10fold decrease in V (i.e., through concentrating) represents a 10fold increase in concentration.
 Changing # and V simultaneously:
 Note also that # and V may be changed simultaneously.
 If this occurs, it is the resulting proportion of # to V which determines concentration.
 Particularly, be aware that simply changing the volume of solution via the removal of an aliquot changes # and V proportionally, and consequently does not change concentration at all!
 Dilution factor
 Volume to volume ratio:
 Assuming that dilution occurs in only a single step, a dilution factor can be defined as the ratio of the final volume to the original volume.
 Thus, if the original volume was 1 ml and the final volume was 10 ml, then so long as # is held constant this gives a dilution factor of 10 to 1 or 10.
 Calculating dilution factors:
 New concentration is equal to the old concentration divided by the dilution factor.
 For example, if the dilution factor equals 10, then the new concentration is 10fold lower.
 Additional examples:
old volume 
new volume 
Dilution factor 
old concentration 
New concentration 
5 
40 
8 
80 
10 
20 
600 
30 
1 
1/30 = 0.0333 
21 
734 
34.95 
100 
2.86 
1 
25 
25 
100 
4 
1 
10^{8} 
10^{8} 
10^{9} 
10^{1} 
1 
10^{8} 
10^{8} 
C 
C*10^{8} 
0.1 
1.7 
17 
3.4 x 10^{9} 
2 x 10^{8} 
 Concentrating:
 Note that dilution factors can also be employed when concentrating. Here, however, the resulting volume is less than the original volume.
 This simply means that the dilution factor has a value of less than one (though greater than 0).
 For example, if the original volume was 10 and the final volume was one, then the dilution factor would be equal to 1 / 10 or 0.1 (or 10^{1})
 Dilution factor calculation from concentration information
 Concentration to concentration:
 Very often one will employ dilutions in order to reduce a concentration from one known concentration to a second, specific concentration.
 To achieve such a dilution, the first step is to calculate the necessary dilution factor.
 Note that we will be able to calculate the resulting volume only once we know the dilution factor. Therefore, we will not able to calculate the dilution factor solely from volume information.
 Old concentration / new concentration:
 Concentration values contain volume information. Specifically, concentration is inversely proportional to volume (i.e., C = # / V, where C = concentration).
 If dilution factor (D) = V_{new} / V_{old} then D must also equal C_{old} / C_{new} since: C_{old} / C_{new} = (# / V_{old}) / (# / V_{new}) = (1 / V_{old}) / (1 / V_{new}) = V_{new} / V_{old}
 In other words, divide your old concentration by your new concentration to obtain your desired dilution factor.
 Below are calculations of dilution factors simply from concentration information:
old concentration 
New concentration 
dilution factor 
24 
6 
4 
10 
1 
10 
10^{8} 
10^{6} 
10^{2} 
3.4 x 10^{9} 
2 x 10^{8} 
17 
3.4 x 10^{11} 
2 x 10^{8} 
1700 
 Turning dilution factors into dilutions
 Dilution factor less one:
 Once you know what dilution factor you need to employ (in order to reduce your concentration some specific amount either from a known to a known, or an unknown to a proportionally unknown), you now need to actually perform your dilution. To determine how this is done we will continue to assume that all dilutions are being made in a single step.
 Note that in making your dilution you will essentially be taking a volume equal to one part of your original solution (or sample and which may or may not consist of your entire sample) and diluting that one part to a final volume equal to the value of your dilution factor multiplied by the volume contained in your one part (i.e., final volume = dilution factor times volume in one part).
 To do this, you will be adding a volume equal to whatever the final volume will be, less one part.
 To make your dilution, given a certain dilution factor, you will add one part sample to a diluent volume consisting of your dilution factor parts minus one. This will reult in a final volume which is your dilution factor times greater than the volume of your diluted sample:
sample volume employed 
dilution factor 
diluent volume employed 
resulting volume 
1 
10 
9 
10 
0.1 
100 
9.9 
10 
0.1 
10 
0.9 
1 
1 
17 
16 
17 
0.1 
24 
2.3 
2.4 
0.2 
52 
10.2 
10.4 
 Diluting while plating [calculating absolutes from volumes]
 Calculating absolute amounts:
 The absolute amount of a suspended or dissolved material (#) is equal to the product of the volume and the concentration (# = V * C = V * #/V).
 Consequently, changes in volume in the absence of concentrating or diluting do not result in changes in concentration.
 Changes in volume do result in absolute changes in #, however (recall that changing volume, in the absence of changes in concentration or dilution, changes # proportionally).
 Deviations from unit volumes:
 Note that typically concentrations are expressed in terms of a unit volume. This means that rather than stating that a solution has a concentration of 10 per 2 ml, the concentration value is instead stated as 5 per ml (i.e., 5 / ml).
 As a consequence, concentration terms and absolute terms are equal so long as it is a unit volume which is being dealt with (i.e., the absolute amount found in 1 ml of a 5 / ml solution is 5).
 Therefore, when a volume other than the unit volume is employed, the absolute amount is increased or reduced in proportion to the volume employed (i.e., a 5 ml volume of a 5 / ml solution has 25 units of dissolved or suspended material, while an 0.5 ml volume of the same solution has 2.5 units of dissolved or suspended material).
 Plate counts:
 When taking a plate count, the volume of sample (or diluted sample) added to the plate has bearing on the absolute number of organisms added to a plate.
 For example, if a twofold greater volume is added, then twofold greater numbers of organisms are added.
 If twofold less volume is added, then twofold fewer organisms are added.
 Plating dilution factor:
 Note that algebraically this change in number of organisms is treated exactly as a dilution (or concentration) is treated whereby the dilution factor here is equal to the unit volume divided by the volume added.
 For example:
sample volume added 
solid media volume added 
dilution factor 
sample concentration 
number of colonies 
1 
20 
1 
400 
400 
0.1 
20 
10 
400 
40 
0.5 
20 
2 
400 
200 
0.1 
20 
10 
4000 
400 
1 
20 
1 
400,000 
400,000 
0.1 
20 
10 
400,000 
40,000 
0.01 
20 
100 
400,000 
4,000 
0.001 
20 
1000 
400,000 
400 
 Upper and lower volume limits:
 Note that typically there are both upper and lower limits on the volume that may be added to plates.
 The upper limit is defined by:
 the volume the plate can hold
 the degree to which the solid media employed can be diluted and still solidify
 the volume that may be employed in a spread plate
 The lower limit is defined by your ability to accurately and precisely handle very small volumes.
 There also, of course, are upper (and lower) limits on the number of organisms one may successfully enumerate per plate.
 Diluting then plating:
 Consequently, one must typically either concentrate or dilute (typically the latter) organisms prior to plating, and then employ some typical volume of diluted sample which is then added to the plate.
 Given dilution prior to plating, note that the overall dilution factor is equal to the dilution factor of the dilution multiplied by the dilution factor of the plating.
 For example:
dilution factor 
sample volume added 
New dilution factor 
original concentration 
colonies on plate 
5 
1 
5 
1000 
200 
5 
0.1 
50 
1000 
20 
50 
0.1 
500 
1000 
2 
10^{8} 
0.1 
10^{9} 
4.0 x 10^{11} 
400 
 Calculating concentration from plate counts
 Once you know how to calculate your overall dilution factor by taking into account both the sample dilution and the volume of diluted sample added to the petri dish, it is actually a trivial exercise to determine the concentration of the original sample.
 It is simply equal to the product of the plate count and the overall dilution factor (with consideration for the error inevitably found in the plate count, of course).
 For example:
plate count 
dilution factor 
volume added to plate 
overall dilution factor 
sample concentration 
237 
10^{6} 
0.1 
10^{7} 
2.37 x 10^{9} 
342 
10^{8} 
1 
10^{8} 
3.42 x 10^{10} 
42 
10^{8} 
0.1 
10^{9} 
4.2 x 10^{10} 
42 
10^{8} 
0.5 
2 x 10^{8} 
8.4 x 10^{9} 
76 
10^{8} 
0.5 
2 x 10^{8} 
1.52 x 10^{10} 
76 
10^{8} 
0.8 
1.25 x 10^{8} 
9.5 x 10^{9} 
 Serial dilution
 Convenient volumes:
 All of the calculations made above assumed that dilutions would be done in one step.
 Note, however, that in a world in which volumes can only be so large or so small, doing one step dilutions can be rather limiting.
 Twostep dilutions:
 We have already considered twostep dilutions, however, in which we diluted and then essentially diluted further by plating less than a unit volume.
 Recall that we multiplied the associated dilution factors (or equivalent) to arrive at an overall dilution factor.
 Recall also that we multiplied the overall dilution factor by the plate count to obtain our estimate for the viable count in the original sample.
 Nstep dilutions:
 There is no reason that one must limit these chains of dilutions (i.e., serial dilutions) to just two steps.
 Note that while there is an inherent limit on the volume which you would add to a plate (e.g., 1 ml), there is no limit to the number of dilutions one can make prior to adding that volume to a plate.
 Note additionally that each time you dilute a diluted sample, you are introducing a new dilution factor, but that each of these dilution factors multiply up to an overall dilution factor.
 Thus, large dilution factors may be achieved without resorting to the use of either very large diluent volumes, or very small transferred volumes (though at the cost of increased sample handling and therefore greater potential for error).
 Example:
 A typical serial dilution might go as follows:
 A sample is suspected to have a concentration in the range of 10^{8}. To determine this concentration you are limited to plate counts of between 30 and 300 colonies per plate. In order to obtain colony counts in this range you need an overall dilution factor of between 10^{8}/30 (= 3.33 x 10^{6}) and 10^{8}/300 (= 3.33 x 10^{5}).
 You are not sure of the sample concentration so, to be on the safe side, you choose to use dilution factors of 10^{5}, 10^{6}, and 10^{7}. You do this by removing 0.1 ml of sample and diluting it into 9.9 ml of buffer to obtain a 100fold dilution. You make sure you mix this diluted sample well with the buffer and then remove 0.1 ml and mix that with a second 9.9 ml buffer. This gives you a second 100fold dilution or 100 x 100 = 10^{4}fold overall dilution. You repeat this a third time to achieve a 100 x 100 x 100 = 10^{6}fold overall dilution (at the expense of only 30 ml buffer!!!).
 You plate 0.1 ml from the 10^{4}fold dilution (second dilution) to achieve a 10^{5}fold overall dilution. You then plate 1.0 ml from the 10^{6}fold dilution, and finally 0.1 from the same dilution to give you your 10^{6} and 10^{7} overall dilution factor, respectively.
 You incubate your plates and obtain colony counts of "too numerous to count", 234, and 21 for your 10^{5}, 10^{6}, and 10^{7}fold overall dilutions, respectively.
 The 234 count is within your 30 to 300 colony range so you use only this plate. Ideally, of course, you would have done the entire dilution series and platings in duplicate or triplicate to be more sure that you haven't introduced any serious errors, but we'll ignore that for this example.
 The concentration of organism in the original sample, therefore, is equal to 234 x 10^{6} = 2.34 10^{8}. This is your viable count, i.e., the number of living organisms you estimate to be in the original sample.
 Illustration, serial dilution
 Minimizing serial dilution errors
 Errors in technique:
 Obvious errors in technique while serial diluting can include:
 improperly measuring diluent and to be diluted volumes
 potential for some volume carry over on the outside of pipets
 dilutions may not be fully mixed prior to your removing volumes from them
 making more than one plating per dilution
 Minimizing the total number of dilutions you employ can help you to minimize many of the above errors.
 Independent duplication:
 With time, practice, and conscientious selfcriticism of your technique, you likely will gain impressive diluting skill. Nevertheless, random variation, independent of your technique, can render individual enumerations less than fully precise.
 To guard against such imprecision it is always a good idea to do your measurements in duplicate.
 Note that in performing duplicate measurements it is important to keep your duplications as independent as possible.
 Ideally this would mean that measurements would be made by different people, in different laboratories, at different times. Obviously this extreme approach toward independence is impractical.
 One dilution series per duplicate:
 A less impractical, minimal approach to the independence which you should attempt to maintain is to repeat enumerations using fully independent dilution series.
 In fact, it is probably the most reasonable approach to never duplicate a plating from one dilution series (i.e., one plate per dilution factor, though potentially more than one dilution factor's plating per dilution series) but instead to initiate a new dilution series to repeat measurements (repeats due to known operator error at the plating step excepted).
 Such an approach guards against the pseudoindependence which results when plating a single dilution factor twice from a given dilution series (i.e., any error the series prior to the final dilution step is ignored when employing such an approach thus potentially resulting in low error but simultaneously low certainty).
 Links
 Yeast plate count lab making a serial dilution
 Vocabulary
 Calculating absolutes
 Calculating concentrations
 Concentrating
 Concentration algebra
 Diluting
 Dilution factor
 Diluting while plating
 Dilution factor from concentration
 Dilutions from dilution factors
 Minimizing errors
 Serial dilution
 Serial dilution, illustration
 Practice questions
 You have a suspension of Escherichia coli with a known concentration of 2.4 x 10^{7} viable cells per ml. You want to dilute this cell suspension down to 1 x 10^{3} cells per ml. Calculate the dilution factor. Design a practical dilution scheme which limits volumes to greater than or equal to 0.1 ml and less than or equal to 10 ml. [PEEK]
 A sample of Staphilococcus aureus has a total cell count of 2.3 x 10^{7} cells per ml. You wish to determine viable count. Use only power of 10 dilutions (dilution factors from 10 to 100, inclusive) and 0.1 ml aliquots added to agar for pour plate. Assume that you want between 30 and 300 colonies per plate. What would be the greatest dilution you employ? [PEEK]
 In a serial dilution you employ the dilutions 0.1:9.9, 1.0:9.0, 0.1:5.5, and then plate out 1.0 ml. What is your overall dilution factor? [PEEK]
 You have three plates with colony counts of 105, 106, and 107. The original sample was diluted 10^{4}fold and 0.1 ml was plated. What is the best, single estimate of the cell concentration in the original sample? [PEEK]
 Name two reasons why serial dilutions are often preferable to large single dilutions. [PEEK]
 Your friend is a lazy kind of guy. Rather than duplicating dilution series in his cell enumeration protocol, he instead simply duplicates dilution factor platings from a single dilution series. Tell me why this technique is problematic. [PEEK]
 By total count you determine that the concentration of an organism in a culture is 5 x 10^{7} cells/ml. Assuming that total count equals viable count, propose an overall dilution factor which would result in the formation of 250 colonies per plate. [PEEK]
 You are determining the viable count of a culture. Your lab assistant has a wicked sense of humor. She dilutes an 0.7 ml volume of culture to a 25 ml volume of diluent. She then removes 0.25 ml from this 25.7 ml volume and mixes it with 9.75 ml diluent. Finally, she plates 0.3 ml from this second dilution. The plate count comes to 169. What is the concentration of the original culture? [PEEK]
 The original volume of your culture is 10 ml. Following dilution and plating with an overall dilution factor of 10^{7} you count an average of 43 colonies per plate. How many cells, in total, were in the original 10 ml of culture? [PEEK]
 By total cell count a culture has 1 x 10^{8} cells/ml. You take 100 ml of this culture and concentrate it to 10 ml. What is your new cell concentration? [PEEK]
 Which is inversely proportional to dilution factor (i.e., is found in the denominator). [PEEK]
 The new volume of a diluted culture.
 The old concentration of a diluted culture.
 The volume of culture (diluted or otherwise) added to a plate.
 The volume of agar added to a plate.
 all of the above.
 none of the above.
 A culture is found to contain 4.3 x 10^{8} cells per ml. To make cheese your recipe (written, no doubt, by that mad microbiologist, Dr. MicroDude) instructs you to add 1.0 ml of culture containing a total of 2.2 x 10^{6} cells. Calculate the necessary dilution factor to achieve this concentration. [PEEK]
 Given a culture which has a concentration of 3.2 x 10^{6} cells / ml and which, upon dilution, yields an 0.1 ml plate count of 42. What dilution factor was employed to produce the diluted culture from which the plate count was made (i.e., the plated 0.1 ml was removed)? [PEEK]
 You have a culture of unknown concentration. A 10^{8}fold overall dilution yields a plate colony count with mean 257. What is the dilution factor employed? [PEEK]
 You have a culture containing a total of 3.4 x 10^{7} cells and which has a volume of 10 ml. Following one dilution scheme you obtain 170 colonies on a plate. What was the dilution factor employed? [PEEK]
 Given a mean of 232 colonies per plate and the following dilution scheme to obtain that colony count, what was the concentration of the original culture? The dilution scheme is as follows: Starting with the original culture, dilute 0.1 ml into 9.9 ml diluent, then 0.1 ml of this diluted culture into 9.9 ml of diluent, then 0.2 ml of this second diluted culture are plated via the pour plate method. [PEEK]
 Given 5 x 10^{11} cells / ml, give me a convenient overall dilution factor (i.e., stick to powers of 10) which would net you between 30 and 300 colonies per plate. [PEEK]
 A cell culture was concentrated by centrifugation and then resuspended in 10 ml (final volume) of sterile normal saline to give a final concentration of 10^{8} cells / ml. What volume of 2x broth would you add to this culture to result in a final broth concentration of 1x? What would the cell concentration subsequently be? [PEEK]
 To get a plate count of a 100,000fold dilution you could dilute a culture 100,000fold then plate 1.0 ml of the diluted culture. If you diluted the culture 0.1 to 9.9, then diluted this now diluted culture 0.1 to 9.9, what volume of diluted culture would you then plate to give a 100,000fold dilution? [PEEK]
 You have a culture containing 1,000,000 organisms which display 20 minute generation times and which are undergoing exponential growth. You let the culture continue to grow for one hour. You then dilute the culture 10,000fold and plate 1 ml. How many colonies do you expect to find on the plate? [PEEK]
 8
 80
 800
 8000
 none of the above
 Dilution #1: You dilute a culture 100fold (10^{2} dilution factor). Dilution #2: You dilute that dilution another 100fold (a 10^{2}fold dilution of a 10^{2}fold diluted culture). Dilution #3: Finally, you dilute that second dilution only 10fold (10^{1}fold dilution of the now twice diluted culture; i.e., you've just done a three step serial dilution). You plate 1.0 ml and 0.1 ml from both the second and third dilutions. You come up with 25, 180, 210, and 1423 colonies per plate. Unfortunately, you recorded your results on the plate tops but the plate data on the plate bottoms. Now, after dropping the plates, you no longer know which dilution went with which colony count (all of the agar from each plate also managed to fly out during this accident and, incredibly, adhere to all four walls of your laboratory, with not a drop landing on the floor). Assuming only minimal dilution errors, what's your best estimate for the concentration of the original culture? [PEEK]
 You think you have in a culture containing between 10^{7} and 10^{9} organisms per ml. You employ the plate count method to refine this estimate. Design an efficient dilution and plating strategy to accomplish this, keeping in mind the concern that usable plates may contain only between 30 and 300 colonies per. Note that the first step in calculating what dilutions to employ is to work out what plate counts you would expect from various dilutions and use these numbers to refine what dilutions and plate volumes to use. This may be done by trial and error (e.g., if you have 10^{9} organisms/ml then a 10^{5} dilution would yield 10^{4} colonies per plate, and so on until you come up with a workable dilution series to employ). However, a more efficient way to accomplish this is to first determine the largest dilution you must use. This dilution is represented by the dilution of 10^{9} organisms per ml down to 30 organisms per plate. That is, 30/10^{9} = 3 x 10^{8}. However, you would increase this dilution to the nearest convenient dilution which is a dilution of 10^{9}. Next determine the smallest dilution you must use. This dilution is represented by the dilution of 10^{7} organisms per ml down to 300 organisms per plate. That is, 300/10^{7} = 3 x 10^{5}. Because this is your minimum dilution, instead of rounding up to a larger dilution, you round down to a smaller dilution to calculate the most convenient minimum dilution to use. Rounding down from 3 x 10^{5} works out to a 10^{5}fold dilution. Now just make sure you plate out all the dilutions ranging from the minimum to the maximum, inclusive. You can use the following sketch if you think that would be helpful (note, the numbers next to or near the boxes are simply names of boxes which I use to present the answer on the exam key and may be ignored). [PEEK]
 Practice question answers
 dilution factor = 2.4 x 10^{7} / 1 x 10^{3} = 2.4 x 10^{4}. A practical dilution scheme would have the following serial dilutions (units are milliliters): 0.1:9.9, 0.1:9.9, 1:1.4.
 First, note that 2.3 x 10^{7} = 230 x 10^{5}. Next, divide this number by 10 in order to incorporate the dilution due to the 0.1 ml plated aliquot. This gives you a value of 230 x 10^{5}. Since 230 is between 30 and 300, this implies a needed dilution factor of 10^{4}. The easiest way to achieve this is to employ a serial dilution series consisting of two 100fold dilutions such as 0.1:9.9 followed by 0.1:9.9.
 The overall dilution factor = 100 x 10 x 56 x 1 = 5.6 x 10^{4}.
 (105 + 106 + 107)/3 = 106. 106 x 10^{4} / 0.1 = 1.06 x 10^{7} cells/ml.
 Large, individual dilutions can waste diluent and/or are impractical due to low end liquid handling limits (i.e., it is tough and often imprecise to transfer very small volumes). In addition you have to keep in mind the size of the vessel necessary to employ very large volumes of diluent.
 Your friend is failing to properly control for errors throughout his dilution series and instead is checking only for errors in the final fluid handling step. Only under unusual circumstances should such a strategy be employed since it can result in a high degree of confidence in a significantly erroneous measurement.
 2 x 10^{5} = 5 x 10^{7} / 250.
 (25.7 / 0.7) * (10 / 0.25) * (1 / 0.3) * 169 = 36.7 * 40 * 3.33 * 169 = 8.3 x 10^{5} cells / ml.
 The answer is concentration times 10. The concentration is equal to 43 x 10^{7}, so the total number of cells is 10 x 43 x 10^{7} = 4.3 x 10^{9}.
 10 x 10^{8} = 1 x 10^{9}.
 iii, the volume of culture (diluted or otherwise) added to a plate.
 195.4.
 3.2 x 10^{6} / (42 * 1 / 0.1) = 7619
 10^{8}
 (3.4 x 10^{7} / 10) / 170 = 2 x 10^{4}. Don't forget that 3.4 x 10^{7} was not given as a concentration value!
 The dilution factor is 100 * 100 * 1 / 0.2 (= 5) = 5 x 10^{4}. The original culture therefore contained 232 x 5 x 10^{4} bacteria(?) / ml = 1.16 x 10^{7} / ml.
 5 x 10^{11} / 30 = 1.7 x 10^{10}. 5 x 10^{11} / 300 = 1.7 x 10^{9}. The only power of 10 between them is 1 x 10^{10}, which is the answer. The "easy" way to do this is to realize that 5 colonies is too few, 500 too many, and 50 just right. 5 x 10^{11} / 50 = 10^{10}.
 10 ml, 5 x 107
 0.1; That is, 10^{2} x 10^{2} x ? = 10^{5} x 1; ? = 10^{1} = 0.1 ml
 iii, 800
 1.95 x 10^{7} cells/ml. Why? The 25 count plate must have been from a 10^{2} * 10^{2} * 10^{1} * 10^{1} dilution (10^{6}). The 25 colonies, however, are too few to consider significant so we ignore this plate though, had we not, this plate would have indicated a concentration of 2.5 x 10^{7}/ml. The 1423 count plate, too, we ignore because the count is too high. Nevertheless, we hypothesize that this count is the product of a 10^{2} * 10^{2} * 10^{0} (10^{4}) dilution. Consequently, had we allowed this plate, we would have concluded that it gives an estimation of the concentration of the original culture of 1.423 x 10^{7}/ml. Note that this appears to be an under count. Finally, the 180 and 210 counts are products the 10^{2} * 10^{2} * 10^{1} and 10^{2} * 10^{2} * 10^{1} * 10^{0} dilutions, that is, 10^{5} for both reached in slightly different steps: 195 = (180 + 210) / 2; 195 x 10^{5} = 1.95 x 10^{7} cells/ml. The answer is 2 x 10^{7} cells / ml (with a little rounding).
 You want to obtain dilutions from between 10^{5} and 10^{9}, inclusive. Given the figure presented above, using the following values will give you this result:

x.1 
x.2 
x.3 
x.4 
x.5 
x.6 
x.7 
1.y 
0.1 
9.9 
 
 
10^{2} 
 
 
2.y 
0.1 
9.9 
0.1 
 
10^{2} 
10^{5} 
 
3.y 
0.1 
9.9 
1.0 
0.1 
10^{2} 
10^{6} 
10^{7} 
3.y 
0.1 
9.9 
1.0 
0.1 
10^{2} 
10^{8} 
10^{9} 
 References
 No entry.