PROKARYOTIC BIO LAB
PROKARYOTIC BIO LAB MCB 4403L
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This 5 page Class Notes was uploaded by Roma Grimes on Thursday September 17, 2015. The Class Notes belongs to MCB 4403L at Florida State University taught by Katarzyna Chodyla in Fall. Since its upload, it has received 40 views. For similar materials see /class/205411/mcb-4403l-florida-state-university in Microbiology at Florida State University.
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Date Created: 09/17/15
Dilution Theory and Problems Microorganisms are often counted in the laboratory using such methods as the viable plate count where a dilution of a sample is plated onto or into an agar medium After incubation plates with 30300 colonies per standardsized plate are counted This number of colonies 30300 was chosen because the number counted is high enough to have statistical accuracy yet low enough to avoid nutrient competition among the developing colonies Each of the colonies is presumed to have arisen from only one cell although this may not be true if pairs chains or groups of cells are not completely broken apart before plating It is possible but unlikely for an original undiluted sample of microorganisms which is to be counted to have 30300 cells ml so that a pour plate using a 1 ml volume from the sample will give good results More likely a sample will have greater numbers of cells ml sometimes as in the case of unpolluted water samples the sample will have less In either case the sample must be manipulated so that it contains a number of cells in the correct range for plating If the cell number is high the sample is diluted if too low the sample is concentrated Dilutions are performed by careful aseptic pipet ng of a known volume of sample into a known volume of a sterile buffer water or saline This is mixed well and can be used for plating andor further dilutions If the number of cellsml is unknown then a range of dilutions are usually prepared and plated Concentration of samples is usually performed by ltration through a lter with pores small enough to retain the microorganisms This way the volume of the original sample can be reduced so that the number of cellsml increases to the countable range Since this procedure is rarely used in our classes this exercise will focus on dilution theory In order to make the calculation of the number of cells ml in the original sample easier dilutions are designed to be easy to handle mathematically The most common dilutions are 110 1100 and 11000 Looking rst at the 110 or 101 dilution it can be made by mixing 1 ml of sample with 9 ml of sterile dilution buffer This gives the fraction 1ml of sample 1ml 1ml 1 01 101 1ml of sample 9ml ofbuffer 1ml 9ml 10ml 10 Some alternative methods of obtaining a 110 dilution are 10ml of sample 10ml 10ml 1 1 101 10ml of sample 90ml of buffer 10ml 90ml 100ml 10 0r 01ml of sample 01 01 1 01 101 01ml of sample 09ml ofbuffer 01 09 10 10 0r 05ml of sample 05 05 1 01 101 05ml of sample 45ml ofbuffer 05 45 50 10 What if the sample is not liquid for example food samples Since 1 ml ofwater weighs 1 gram under standard conditions you can consider that 1 gram of any sample is equal to 1 ml Therefore 1 gram of pepper 9 ml of dilution buffer a 110 dilution of the pepper It is recognized that 1 gram of pepper might not actually have a volume of exactly 1 ml but if the arbitrary assignment is recognized as standard it can be used conveniently and reproducibly How is a 1100 dilution obtained Usually 1 ml of sample 99 ml buffer or 01 ml of sample 99 ml of buffer are used to give this dilution A 11000 dilution can be obtained by adding 01 ml of sample to 999 ml of buffer Once the dilution is made an aliquot can be plated on an agar plate using the spread plate technique or in an agar medium using the pour plate technique After incubation the colonies are counted How do the colonies on the plate relate to the number of cells colonyforming units or CPU in the original sample Try a problem 0 One ml of a sample was mixed with 99 ml ofbuffer One ml of this was plated using the pour plate method in nutrient agar After incubation 241 colonies were present on the plate How many colonyforming units were present per ml of the original sample State your answer in CPUml The dilution used was 1199 1100 001 102 One ml of the dilution contained 241 colonyforming units How much did one ml of the original sample contain Obviously more than 241 colonyforming units To arrive at the correct number either divide the colonyforming units ml of the dilution by the dilution 241 colony forming unitsml divided by 10392 241 X 102 241 X 104 CPUml of original sample or multiply by the dilution factor The dilution factor is defined as the inverse of the dilution therefore in this example the dilution factor would be 11039Z 102 Multiplying by the dilution factor 241 colonyforming unitsml X 102 241 X 102 241 X 104 CFUml of original sample Either way the answer you get is the same Try another problem 0 One ml of a sample was mixed with 99 ml buffer Onetenth of a ml of this was plated on nutrient agar After incubation 142 colonies were present on the plate How many colonyforming units were present per ml of the original sample In this case the mathematics of the dilutions is the same but the number of colonies counted on the plate the represent number of colonyforming units in only 01 ml of the dilution Remember you should always report your answers as CFU ml Therefore the number of colonyforming units per ml of dilution is 142 colonies divided by 01 ml plated or 142 X 10 1420 colonyforming unitsml of dilution Dividing this number by the 10392 dilution results in a final answer of 1420 X 102 or 142 X 105 colonyforming units ml of original sample Try another One ml of a sample was mixed with 99 ml ofbuffer One ml of this mixture was added to a sterile petri plate and mixed with 25 ml of molted agar cooled to 45 degrees The mixture was allowed to solidify undisturbed After incubation 241 colonies were present on the plate How many colonyforming units were present per ml of the original sample The answer to this problem is identical to the answer to the rst problem The amount of agar used is not relevant because the number of organisms present depends only on the number of ml of the dilution which were added to the agar Whether 25 ml or 35 ml of agar were used the number of organisms added to the agar remains constant so the colony count will be identical What if the sample requires more dilution For example what if the sample to be counted is a culture of Escherichia coli grown overnight in a rich medium The number of cellsml will be around 109 1000000000 Obviously a single dilution would not be enough so successive dilutions must be performed An easy example of a successive dilution is making a 1100 dilution using only two tubes of 90ml buffer You would add 1 ml of sample to the first 9ml tube and mix well This is a 110 dilution Next you would remove 1 ml from this dilution and add it to the second 9ml tube ofbuffer This is also a 110 dilution but not of the original sample Instead this is a 1 10 dilution of a 1 10 dilution To arrive at the nal dilution multiply the second dilution by the rst dilution 110 X 110 1100 If you made another 110 dilution from the second tube you would have a 11000 nal dilution Try a problem 0 An overnight culture of Escherichia coli is used as a sample One ml of this culture is added to a bottle containing 99ml of buffer This dilution is mixed well as all dilutions arel and one ml of this is mixed in 9 ml of buffer This second dilution is diluted by three successive 1 10 dilutions The last fifth dilution is plated ie 01 ml is plated on nutrient agar After incubating the plate 56 colonies are counted How many colony forming units were present per ml of E coli culture To solve this problem try writing out the procedure Then multiply the successive dilutions together here is where scienti c notation comes in handy since it is easy to add the exponents without losing a zero 1 1 1 1 1 1 1 99 1 9 10 10 10 1000000 OR 102 x 101 x 101 x 101 X 101 106 The number of colonyforming units per ml of the dilution can then be divided by the nal dilution 56 colonies divided by 106 560 colonyforming unitsm1 X 106 01 ml plated 56 X 108 colonyforming unitsml Note that the 01 ml that was plated is treated as another 110 dilution Some microbiologists treat it as such in their computations rather than dividing the colony count by the number of milliliters plated as done here Whichever way it is done the answer will be the same The only way to understand dilution theory well is to practice it so you should work the problems given in this supplement The answers to the problems will be given to you by your lab TA in the next lab period Before you start review the basic rules of dilution theory Since dilutions result in physically lowering the number of cells per ml in a solution the calculations must mathematically raise the number of cells per ml Therefore remember that the final answer should be larger than the original colony count not smaller It is not possible to have negative exponents in the final number of colonyforming units ml a negative exponent implies that there is less than one colonyforming unit ml The number of ml of agar into which a sample is added is irrelevant unless of course it is further diluted The number of ml plated is relevant The dilution is determined by dividing the number of ml of sample added to the dilution buffer by the total of the ml of sample plus the ml of buffer Successive dilutions are multiplied to find the total dilution Plates with 30 300 colonies on them should be used for the greatest accuracy in counting The number of colonyforming units ml of original culture is calculated by either dividing the number of colonyforming units ml of the dilution which gave 30300 colonies per plate by that dilution or by multiplying the number of colonyforming units ml or the dilution which gave 30300 colonies per plate by the inverse of that dilution ie the dilution factor If more than one plate of a dilution has been prepared and is counted the colony counts of that dilution only should be averaged QUESTIONS 1 One ml of a water sample is added to 9 ml of sterile water This is mixed well and further diluted by 4 successive 110 dilutions Onetenth of a ml of each dilution is spread on a plate of nutrient agar After incubation the following data were obtained Dilution used for plating Amount plated Colony counts after incubation rst 110 01 ml too many to count second 110 01 ml third 110 01 ml 67 fourth 110 01 ml 5 fth 110 01 ml 0 What was the number of colonyforming units ml of the original water sample that were capable of growing on nutrient agar 2 Three grams of soil were added to 27 ml of sterile water and shaken vigorously After the soil particles settled 01 ml of this was added to 99 ml of sterile water This was further diluted by 4 successive 110 dilutions One ml from the last dilution was used to prepare a pour plate After incubation 289 colonies were present on this plate What was the number of colonyforming units gram of the soil 3 A bacterial culture was diluted and results from duplicate plates were obtained as indicated below What was the number of colonyforming unitsml of the original culture Dilution used Amount Colony counts after incubation for plating plated results from duplicate plates 10392 01 ml too many to count 10393 01 ml too many to count 10394 01 ml 321 39 403 10395 01 ml 34 42 10396 01 ml 6 39 1 10397 01 ml 0 39 0 10398 01 ml 0 39 0 4 Ten grams of hamburger were added to 90 ml of sterile buffer This was mixed well in a blender One tenth of a ml of this slurry was added to 99 ml of sterile buffer After thorough mixing this suspension was further diluted by successive 1100 and 110 dilutions Onetenth of a ml of this nal dilution was plated onto Plate Count Agar After incubation 52 colonies were present How many colonyforming units were present in the total 10 gram sample of hamburger Devise a scheme to prepare a 10396 dilution on a plate using the least number ofsterile water dilution tubes Devise a scheme to prepare 120 140 and 180 dilutions of a disinfectant