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# Using a Serial Dilution to Count Bacteria in a Sample

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Consider the following dilution scheme:

A. 1oomL lake water
B. 5mL lake water into 45 mL diluent 0.1 ml - colonies TNTC
C. 1mL from B into 9ml diluent 0.1 ml - colonies 425
D. 0.5mL from C into 4.5 mL diluent 0.1 ml - colonies 58
E. 2mL from D into 19mL diluent 0.1 ml - colonies 12
F. 0.1mL from E into 0.9 mL diluent 0.1 ml - colonies 1

a. What dilution is achieved by adding 2mL of incoulum to 19mL of a diluent?
b. Would you expect any change in the answer of the above problem if the firat dilution was made by adding 1 mL of sample to 9 mL of diluent? Why or why not?
c. What are the total number of CFU's in the entire 100mL amount of the original lake water sample?

HELP! I'm stumped on how to do this!!!!

https://brainmass.com/biology/human-biology/serial-dilution-scheme-count-bacteria-sample-347495

#### Solution Preview

Dilution schemes are used to decrease the number of bacteria in a sample to a countable level. This is an example of a dilution series that will lower the bacterial numbers enough so you can count colonies on a petri plate. Each colony is assumed to come from one bacterium so from the number on the plate you can calculate how many was in your original sample. If you plate your lake water directly on a petri plate then there will be billions of bacteria and they will completely cover the plate. This does not help you determine how many cells were in the water. I will walk you through the dilution so you understand what is occurring.

In order to successfully understand dilutions you can use the following formulas.
? To find a dilution of a single tube the dilution = volume of sample/(volume of sample + volume of diluent).
? To find a total dilution after doing a dilution series using multiple tubes (this is called a serial dilution) you multiply the dilutions of the individual tubes. For instance when a sample diluted 1/10 is added to the next tube to make a 1/100 dilution the total dilution for the dilution series is 1/10 x 1/100 = 1/1000
? To find the number of bacteria/ml in the original sample after growing bacteria on a plate you use the following formula; CFU/ml = (# colonies on a plate)(1/volume used to plate bacteria)(1/dilution of that tube)

I will walk through the dilution series provided in this problem and you can see how these formulas are used to determine the dilution in each tube, the total dilution of the lake water, and the final count of bacteria.

The first dilution was to take 5 ml of lake water and put it into 50 ml of diluent. This gives you a 5/50 dilution which can be reduced down to 1/10. When you look at the 5/50 this was determine because the top number is the number of ml that you transferred and the bottom number is the number of ml total now in the tube (5ml lake water + 45 ml diluent = 50 ml total). 5/50 can then be reduced to 1/10 because 5 divides into 50 10 times. A 1/10 ...

#### Solution Summary

When determining the number of bacteria in a sample you must perform a serial dilution followed by plating bacteria onto a petri plate, which is called determining the colony forming units. This explanation will walk you through the steps of a serial dilution. It explains how to do the calculations that must be performed to dilute a bacterial sample. It also explains how to calculate the number of bacteria in an original sample after the dilutions have been done and the number of bacterial colonies on each petri plate have been counted.

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## Estimating microbial growth

Describe the various methods used to estimate the amount of microbial growth that has taken place in a culture:
a. viable/standard plate count
b. serial dilutions
c. turbidity measurement
d. direct microscopic count
e. coulter counter
f. membrane filtration

a. viable/standard plate count

The viable count is done by plating the samples onto appropriate culture medium and counting the number of cell or colonies formed from the plates. (1) First you have to dilute your sample to 10 fold and if you count that there is 20 colonies or bacteria cells from the colonies. You can calculate how many viable cell by multiplying the number of colonies 20 X 10 = 200 colonies. So the total number of viable cells is 200 colonies forming units. The units for the viable cell count is cfu, which stands for colony forming units.

b. serial dilutions

A serial dilutions is performed in order to determine the how many bacteria colonies that are formed from the original cultures. It is usually performed before the viable plate cell count method. For example, the original culture containing bacteria are diluted by a serial dilutions series by taking 10ul of the original culture and add to 90 ul water. This is a 10-fold dilution (Dilution1) because it is dilutied 10X times. From the 10fold (dilution 1), you can take another 10 ul from dilution1 and and 90 ul of water, so you have 10X10 fold dilution, which is 100X Fold dilution (dilution2) because you diluted it two times serially. From the dilution 2, you can make another dilution by taking 10 ul from dilution 2 and add 90 ul water; this will be 10X10X10 dilution (dilution 3). You can make further dilution from dilution 3 like, 10 ul (dilution 3) and 90 ul water to yield dilution 4 ( 10 X 10 X10 X10). If you stop here, then you will have a 1 X 10e4 dilution. If you plate the aliquot of this dilution 4, and you get 20 colonies or cells. You can determine the number of original bacterial cells in your culture by multiplying 20 X 10X10X10X10 = 200000 colonies in your bacteria cultures.

c. turbidity measurement

Turbidity method is a quick and easy method to measure the turbidity or "cloudiness" of a culture and " translate the cloudiness into cell numbers". (1) This method is prefered when a large number of cultures are counted. The turbidity of the culture must be correlated with the initial cell number. You can do this by determining the cloudiness of different concentrations of a given bacteria in the medium and use a standare plate count to determine how many viable colonies contribute to the amount of turbidity. If you have 20 colonies, then you measure how turbid that is by the absorbance(nm). For example, if 20 colonies measured a turbidity of 500 nm, the 10 colonies should give you smaller number such as 250 nm. 5 colonies should give you a turbidity of 150 nm. You will construct a standard curve correlating the amount of colonies on the x axis and the amount of turbidity by the y axis. From this standard curve, you can extrapolate how many cells you have by measuring the turbidity. The turbidity can be measured using an instrument called the calorimeter or sphectrophotometer to measure the light source that travel through the bacteria cells. (1)

d. direct microscopic count

A "Petroff-Hauser" counting chambers is used to count the number of bacteria cells in a culture or liquid medium. You inoculate an aliquot of your culture onto the counting chambers and using a microscope, you count the number of cells in each chamber. There are four chambers in the counting chamber. So you just count how many cells you have.
If you have 5 per chambers, than you would have 5 X4 = 20 cells or colonies. It is not possible to distinguish dead cells from live cells so the viable cell count is not reliable in this method. (1) The advantage is that it is a fast method. (1)

e. coulter counter

The coulter counter is a instrument that uses the " Electrical Sensing Zone" method for counting cells. It works by using two electrodes; one inside the aperature and one outside the aperture. The two electrodes create an electrical field and a disturbance between the two electric fields is measured when the cells or particles pass through the aperture. (2) When the bacteria particles pass through the aperture, they cause a change in the electric field as a "voltage pulse". (2) The concentration of particles can be counted this way by the measurement of the voltage pulse.

f. Membrane filtration

Membrane filtration is really a method to "separate particles from a liquid form for the purpose of purifying bacteria from the liquid solution. " (3) The membrane filter has a pores of a particular size that can separate bacteria of a particular size from the cultures. The membrane pores has sizes that is small enough to filter out larger particles than bacteria. So the pores of the membrane only allow the bacteria to filter through, but left particles larger than bacteria to stay behind and not filter through. The outside of the solutions contain larger particles than bacteria and the filtrate only contain small particles like bacteria .(3)

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