## Serial dilution — Wikipedia

A **serial dilution** is the stepwise dilution of a substance in solution. Usually the dilution factor at each step is constant, resulting in a geometric progression of the concentration in a logarithmic fashion. A ten-fold serial dilution could be 1 M, 0.1 M, 0.01 M, 0.001 M … Serial dilutions are used to accurately create highly diluted solutions as well as solutions for experiments resulting in concentration curves with a logarithmic scale. A tenfold dilution for each step is called a **logarithmic dilution** or **log-dilution**, a 3.16-fold (10^{0.5}-fold) dilution is called a **half-logarithmic dilution** or **half-log dilution**, and a 1.78-fold (10^{0.25}-fold) dilution is called a **quarter-logarithmic dilution** or **quarter-log dilution**. Serial dilutions are widely used in experimental sciences, including biochemistry, pharmacology, microbiology, and physics.

### In biology and medicine[edit]

In biology and medicine, besides the more conventional uses described above, serial dilution may also be used to reduce the concentration of microscopic organisms or cells in a sample. As, for instance, the number and size of bacterial colonies that grow on an agar plate in a given time is concentration-dependent, and since many other diagnostic techniques involve physically counting the number of micro-organisms or cells on specials printed with grids (for comparing concentrations of two organisms or cell types in the sample) or wells of a given volume (for absolute concentrations), dilution can be useful for getting more manageable results.^{[1]} Serial dilution is also a cheaper and simpler method for preparing cultures from a single cell than optical tweezers and micromanipulators.^{[2]}

### In homeopathy[edit]

Serial dilution is one of the core foundational practices of homeopathy, with «succussion», or shaking, occurring between each dilution. In homeopathy, serial dilutions (called potentisation) are often taken so far that by the time the last dilution is completed, no molecules of the original substance are likely to remain.^{[3]}^{[4]}

### See also[edit]

### References[edit]

**^**K. R. Aneja.*Experiments in Microbiology, Plant Pathology and Biotechnology*. New Age Publishers,**2005**, p. 69.

en.wikipedia.org

## Serial Dilutions

Serial Dilutions

Advantages of «Serial Dilutions»

This section is not a recipe for your experiment. It explains some

**principles for designing dilutions that give optimal results**. Once

you understand these principles, you will be better able to design

the dilutions you need for each specific case.

Often in experimental work, you need to cover a range of

concentrations, so you need to make a bunch of different

dilutions. For example, you need to do such dilutions of the

standard IgG to make the standard curve in ELISA, and then again

for the unknown samples in ELISA.

You might think it would be good to dilute 1/2, 1/3, 1/10, 1/100.

These seem like nice numbers. There are two problems with this series of

dilutions.

- The dilutions are
**unnecessarily complicated to make**. You need to do a different

calculation, and measure different volumes, for each one. It takes a long

time, and it is too easy to make a mistake. - The dilutions cover the range from 1/2 to 1/100
**unevenly**.

In fact, the 1/2 vs. 1/3 dilutions differ by only 1.5-fold in concentration,

while the 1/10 vs. 1/100 dilutions differ by ten-fold. If you are going to

measure results for four dilutions, it is a waste of time and materials

to make two of them almost the same. And what if the half-maximal signal

occurs between 1/10 and 1/100? You won’t be able to tell exactly where it

is because of the big space between those two.

**Serial dilutions** are **much easier to make** and

they **cover the range evenly**.

Serial dilutions are made by making the same dilution step over and over,

using the previous dilution as the input to the next dilution in each step.

Since the **dilution-fold is the same in each step**, the dilutions

are a geometric series (constant ratio between any adjacent dilutions).

For example:

Notice that each dilution is three-fold relative to the previous one.

In four dilutions, we have covered a range of 181/3 = 60-fold.

If that isn’t enough range, consider a series of five-fold dilutions:

Here we’ve covered a (625/5) = 125-fold range.

No matter where the half-max falls in a series of 5-fold dilutions,

it is no more than

2.2-fold («middle» [square root] of a 5-fold step) away

from a data point — so the coverage of the range is thorough and even.

When you need to cover several factors of ten (several «orders of magnitude») with

a series of dilutions, it usually makes the most sense to plot the dilutions

(relative concentrations) on a **logarithmic scale**. This avoids bunching most

of the points up at one end and having just the last point way far

down the scale.

Before making serial dilutions, you need to make **rough estimates of the concentrations in your unknowns**, and your uncertainty in those estimates. For example,

if A

_{280}says you have 7.0 mg total protein/ml, and you think

the protein could be anywhere between 10% and 100% pure, then your

assay needs to be able to see anything between 0.7 and 7 mg/ml.

That means you need to cover a ten-fold range of dilutions, or maybe a bit

more to be sure.

If the half-max of your assay occurs at

about 0.5

mg/ml,

then your minimum dilution fold is

(700

mg/ml)/(0.5

mg/ml) = 1,400.

Your maximum is

(7000

mg/ml)/(0.5

mg/ml) = 14,000.

So to be safe, you might want to cover 1,000 through 20,000.

In general, before designing a dilution series, you need to decide:

- What are the
**lowest and highest concentrations (or dilutions)**

you need to test in order to be certain of finding the half-max? These

determine the range of the dilution series. **How many tests**do you want to make? This determines the size of the

experiment, and how much of your reagents you consume. More tests will cover

the range in more detail, but may take too long to perform (or cost too much).

Fewer tests are easier to do, but may not cover the range in enough detail

to get an accurate result.**What volume of each dilution do you need**to make in order to have

enough for the replicate tests you plan to do?

Now suppose you decide that **six tests** will be adequate (perhaps

each in quadruplicate).

Well, starting at 1/1,000, you need five equal dilution steps (giving you

six total dilutions counting the starting 1/1,000) that end in

a 20-fold higher dilution (giving 1/20,000). You can **decide on a good step size** easily by trial and error. Would 2-fold work? 1/2, 1/4, 1/8, 1/16, 1/32. Yes, in fact

that covers 32-fold, more than the 20-fold range we need. (The exact answer

is the 5th root of 20, which your calculator will tell you is 1.82 fold

per step. It is much easier to go with 2-fold dilutions and gives about the

same result.)

So, you need to make a 1/1,000 dilution to start with. Then you need to

serially dilute that 2-fold per step in five steps. You could make 1/1,000 by

adding 1 microliter of sample to 0.999 ml diluent. **Why is that a poor choice?**

Because you can’t measure 1 microliter (or even 10 microliters) accurately

with ordinary pipeters. So, make three serial 1/10 dilutions

(0.1 ml [100 microliters] into 0.9 ml): 1/10 x 1/10 x 1/10 = 1/1,000.

Now you could add 1.0 ml of the starting 1/1,000 dilution to

1.0 ml of diluent, making a 2-fold dilution (giving 1/2,000).

Then remove 1.0 ml from that dilution (leaving 1.0 ml for your

tests), and add it to 1.0 ml of diluent in the next tube (giving

1/4,000). And so forth for 3 more serial dilution steps (giving

1/8,000, 1/16,000, and 1/32,000). You end up with 1.0 ml of each dilution.

If that is enough to perform all of your tests, this dilution plan

will work. If you need larger volumes, increase the volumes you use

to make your dilutions (e.g. 2.0 ml + 2.0 ml in each step).

www.bio.umass.edu

## Serial Dilution | Science Primer

Many procedures performed in modern biology and chemistry laboratories require sets of solutions that cover a range of concentration^{*}s. These include quantifying the number of bacteria in a sample using plate counts and development of standard curves for quantitative colorimetric, radiometric and enzymatic assays. Sets of solutions over a range of concentrations are prepared via serial dilution^{*}.

Options

Reset

Container | 1 | 2 | 3 | 4 | 5 |

Number of Particles (n) | |||||

Volume (ml) | |||||

Concentration (n/ml) |

To perform a serial dilution, a small amount of a well-mixed solution is transferred into a new container and additional water or other solvent^{*} is added to dilute the original solution. The diluted sample is then used as the base solution to make an additional dilution. Doing this several times results in a range of concentrations.

The initial concentration and target range needed for a given assay determines the size and number of dilution steps required. Often, serial dilutions are performed in steps of 10 or 100. They are described as ratios of the original and final concentrations. For example, a 1:10 dilution is a mixture of one part of a solution and nine parts of additional solvent. To make a 1:100 dilution, one part of the solution is mixed with 99 parts of additional solvent.

Mixing 100 µL of a stock solution with 900 µL of water makes a 1:10 dilution. The final volume of the diluted sample is 1000 µL (1 mL) and the concentration is 1/10 that of the original solution. This is commonly referred to as a 10x dilution.

The illustration above follows the relationship between volume of solvent, number of molecules of solute and concentration of a solution over a set of 4 dilutions. The concentration can be tracked in M which is a common unit for chemistry, or particles per ml, which is common when diluting bacterial cultures to low concentrations. With molar concentrations it is safe to assume that the solute is evenly dispersed through the solution such that the concentrations change predictably with each dilution. With particles, like bacterial cells, the solution becomes patchy at low concentrations and actual concentration can diverge from the expected concentration.

Test your understanding with the serial dilution practice problems

**Video Overview**

**Related Content**

scienceprimer.com

## ChemTeam: Serial Dilution

ChemTeam: Serial Dilution

#### Serial Dilution

Return to Solutions Menu

**Problem #1:** The following successive dilutions are applied to a stock solution that is 5.60 M sucrose:

Solution A = 46.0 mL of the stock solution is diluted to 116 mL

Solution B = 58.0 mL of Solution A is diluted to 248 mL

Solution C = 87.0 mL of Solution B is diluted to 287 mL

What is the concentration of sucrose in solution C?

**Solution:** (the solution will be followed by some discussion)

1) Assign unknowns to the final molarities of each dilution:

molarity of solution A = x

molarity of solution B = y

molarity of solution C = z

2) Set up all three dilutions using M_{1}V_{1} = M_{2}V_{2}

Solution A ⇒ (5.60) (46) = (x) (116)

Solution B ⇒ (x) (58.0) = (y) (248)

Solution C ⇒ (y) (87.0) = (z) (287)

3) Each of these three eqations may now be solved in turn to give the final answer (symbolized by the variable z):

Solution A ⇒ (5.60) (46) = (x) (116)

x = 2.2207 MSolution B ⇒ (2.2207) (58.0) = (y) (248)

y = 0.51936 MSolution C ⇒ (0.51936) (87.0) = (z) (287)

z = 0.157 M (this is the final answer)

Advice: if you use the above technique, make sure to carry several guard digits as you do each calculation. In the above, I carried two guard digits in my values for x and y.

I would like to now reconsider the solution. In place of the specific numbers in step two, I would like to use symbols:

Solution A ⇒ (C

_{i}) (V_{1}) = (x) (V_{2})

Solution B ⇒ (x) (V_{3}) = (y) (V_{4})

Solution C ⇒ (y) (V_{5}) = (C_{f}) (V_{6})x and y = intermediate concentrations

C_{i}= initial concentration

C_{f}= final concentration

I’m now going to form one equation that relates C_{i} and C_{f}. To do this, I will eliminate the intemediate concentrations symbolized by x and y. First x:

(C

_{i}) (V_{1}) = (x) (V_{2})x = (C

_{i}) (V_{1}/ V_{2})

Now, substitue into the equation for solution B:

(C

_{i}) (V_{1}/ V_{2}) (V_{3}) = (y) (V_{4})y = (C

_{i}) (V_{1}/ V_{2}) (V_{3}/ V_{4})

Now, substitue into the equation for solution C:

(C

_{i}) (V_{1}/ V_{2}) (V_{3}/ V_{4}) (V_{5}) = (C_{f}) (V_{6})(C

_{i}) (V_{1}/ V_{2}) (V_{3}/ V_{4}) (V_{5}/ V_{6}) = C_{f}

Let’s see if it works:

C

_{f}= (5.60) (46 / 116) (58 / 248) (87 / 287)C

_{f}= 0.157 MHey! It works!

Notice that each volume ratio represents a dilution. Consequently, each volume ratio is a factor less than 1.

**Problem #2:** What is the molarity of dilution 1?

1) A stock solution of salicylic acid is diluted by transferring 10.0 mL of the stock solution into a beaker and adding 40.0 mL of water. The beaker is labeled dilution 1.

2) 5.00 mL of the solution in the beaker labeled dilution 1 is transferred to another beaker and 15.0 mL of water is added. The beaker is labeled dilution 2.

3) 5.00 mL of the solution in the beaker labeled dilution 2 is transferred to another beaker and 15.0 mL of water is added. The beaker is labeled dilution 3.

4) 1.00 mL of the solution in the beaker labeled dilution 3 is transferred over to another beaker called dilution 4 and enough water is added so that the final volume is 5.00 mL and the concentration of the final dilution is 2.10 x 10

^{-6}M.

**Solution:**

1) from dilution 4 to dilution 3:

(x) (1.00 mL) = (2.10 x 10

^{-6}) (5.00 mL)x = 1.05 x 10

^{-5}M (concentration of dilution 3)

2) from dilution 3 to dilution 2:

(x) (5.00 mL) = (1.05 x 10

^{-5}mol/L) (20.0 mL)x = 4.20 x 10

^{-5}M (concentration of dilution 2)

3) from dilution 2 to dilution 1:

(x) (5.00 mL) = (4.20 x 10

^{-5}mol/L) (20.0 mL)x = 1.68 x 10

^{-4}M (concentration of dilution 1)

4) from dilution 1 to the stock:

(x) (10.00 mL) = (1.68 x 10

^{-4}mol/L) (50.0 mL)x = 8.40 x 10

^{-4}M (concentration of the stock solution)

Note how the problem is worded to avoid giving a final volume. It simply tells you the starting volume (5.00 mL in 2 and 3) and that 15.0 mL was added. Nowhere does it state that the solution was diluted to 20 mL. Also, note the implicit assumtion that the volumes are additive. This is a resonable assumption in this example.

Here is a repeat of the combined equation from above:

(C

_{i}) (V_{1}/ V_{2}) (V_{3}/ V_{4}) (V_{5}/ V_{6}) = C_{f}

Since there are four dilutions in Problem #2, I will add another volume term:

(C

_{i}) (V_{1}/ V_{2}) (V_{3}/ V_{4}) (V_{5}/ V_{6}) (V_{7}/ V_{8}) = C_{f}

Since we are interested in C_{i}, let us move all the volume terms to the other side:

C

_{i}= (C_{f}) (V_{2}/ V_{1}) (V_{4}/ V_{3}) (V_{6}/ V_{5}) (V_{8}/ V_{7})

Now, let us insert values into the revised equation (please note that all even subscripts are the final molarity and the odd subscripts are the starting molarity):

C

_{i}= (2.10 x 10^{-6}) (50 / 10) (20 / 5) (20 / 5) (5 / 1) = 8.40 x 10^{-4}M

**Problem #3:** You have prepared 10 mL of a stock solution of 1.000 M hydrochloric acid, HCl. You serially dilute the stock solution 3 times, such that each step of the serial dilution dilutes the original solution by 1/2. What will the final concentration be in the final test tube?

**Solution #1:**

Start with 10 mL of 1 M solution (I’ll ignore sig figs for the moment).

Take the 10 mL of solution and add 10 mL of water. The solution is now 20 mL of 0.5 M.

Take the 20 mL of solution and add 20 mL of water. It’s now 40 mL of 0.25 M.

Take the 40 mL of solution and add 40 mL of water. It is now 80 mL 0.125 M.

So, the final concentration is 0.125 M.

Note how the concentration is cut in half each time the total volume doubles.

**Solution #2:**

This is a cleaned up version of the answer I copied from Yahoo Answers:

OK, so you’re starting with 10 mL of 1 M solution and you’re diluting it by half.

So take 5 mL of solution and add 5 mL of water, the solution is now 0.5 M.

Take 5 mL of that and add 5 mL of water. It is now 0.25 M

Take 5 mL of that and add 5 mL of water. It is now 0.125 M.

The final concentration is 0.125 M.

Note how only the stock solution is used as well as half of each diluted solution in each step. This is the correct way to do this type of dilution in the lab. If you spill a solution in a given step, you can return to the solution of the previous step and continue (after you clean up the spill!).

I did not use any of the notation used in Problems #1 and #2. I will leave that as an exercise to the reader.

The above type of dilution is called a two-fold dilution. In a two-fold dilution, the volume of the original solution is always doubled, as in going from 1 to 2.

A three-fold dilution would be a tripling of the solution volume, as in going from 1 to 3.

Although not a serial dilution, the below is an example of a two-fold dilution.

**Problem #4:** To make a two-fold dilution of 10 mL of solution, what amount of solvent would you use and how would you do this?

To perform a dilution you can always use the equation:

(Volume of solution)/(vol of solution + vol of solvent)

Since it is a two-fold dilution, you use the same volume of solvent as you have of solution.

In this case, you just add ten milliliters:

(10)/(10 + 10) = 1/2 —> a 1 to 2 dilution (also called two-fold).

If this was a three-fold dilution, we would have this: (10)/(10 + 20) = 1/3 —> a 1 to 3 dilution (also called three-fold).

**Problem #5:** To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to weigh out a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring 0.661 g of K_{2}Cr_{2}O_{7} to a 250.0-mL volumetric flask and adding water to the mark. A sample of this solution of volume 1.000 mL was transferred to a 500.0-mL volumetric flask and diluted to the mark with water. Then 10.0 mL of the diluted solution was transferred to a 250.0-mL flask and diluted to the mark with water.

(a) What is the final concentration of K

_{2}Cr_{2}O_{7}in solution?

(b) What mass of K_{2}Cr_{2}O_{7}is in this final solution?

**Solution:**

1) Calculate the molarity of the first solution thusly:

MV = mass / molar mass

(x) (0.2500 L) = 0.661 g / 294.181 g/mol

x = 0.008987664 M (I won’t round off very much until the end.)

2) Now, use this for the first dilution (1.000 mL to 500.0 mL):

M

_{1}V_{1}= M_{2}V_{2}(0.008987664 mol/L) (1.000 mL) = (x) (500.0 mL)

x = 0.000017975328 M

3) Use M_{1}V_{1} = M_{2}V_{2} again for the second dilution (10.00 mL to 250.0 mL):

(0.000017975328 mol/L) (10.00 mL) = (x) (250.0 mL)

x = 0.00000071901312 M

Rounded to four sig figs, it is 0.0000007190 M (the answer to part (a) of the question)

4) Lastly, use MV = mass / molar mass to get the mass of K_{2}Cr_{2}O_{7} in the final 250 mL of solution.

(0.00000071901312 mol/L) (0.2500 L) = x / 294.181 g/mol

x = 0.00005288 g (or, if you prefer 5.288 x 10

^{-5}g)This answer to part (b) gives the amount that would have had to have been weighed out if the solution had been prepared directly.

Return to Solutions Menu

www.chemteam.info

## Serial dilution

The story on the previous pages has many parallels with life in a microbiology lab. Frequently, you will find it necessary to add water (or some other medium) to a stock (or soup, get it?) with a known concentration to make a more dilute solution.

Why would you want to do this? Let’s say you need to get 1/10,000th of a mL in order to count the bacteria in it. That would be pretty difficult with a pipette. But instead, you could

- Take that 1 mL and put it in 99 mLs of saline solution. You still have the same number of bacteria, but now they’re spread out.
- Now you can take 1 mL of THAT solution, add it to a new container, top off with 99mLs of water, and you’ll have only 1/100th of the original bacteria in the mL.
- Pull out 1 mL of this latest mixture, and you’ll have 1/100th of 1/100th, which is (multiplying the fractions together) 1/10,000th.

When you do serial dilutions, you **multiply together all of the dilution factors**. Make sure you are clear on what constitutes a dilution factor. When I add a small amount of the concentrated stuff to an empty container, «top it off» with saline/water/whatever, then remove some small amount, this constitutes a dilution. So above, even though its broken down into 3 steps, there are really only 2 full dilution steps.

When in doubt, try to think it through logically. **Often it helps to think through the whole process using some concrete number.** For example: «I started with 300,000 cells in a mL, put those into 99 mLs of saline, and took out a mL, so there must have been 3,000 cells in that mL. Then I put those 3,000 cells into 99 mLs of saline, and took out one mL again, so there must have been 30 cells in that mL. Overall my dilution must have been to 30 from 300,000, which is the same as to 1 from 10,000 (1:10,000).» Of course, the idea that you started with 300,000 cells is pure fiction, but it can help you make sure that you’ve done the dilutions correctly.

#### Put one mL of a stock into 99 mLs of water. Take 1 mL of that and put it in 999 mLs of water. What is the total dilution?

(To make this problem interactive, turn on javascript!)

- I need a hint … :Assume that you start with 1,000,000 cells in one mL of solution
- …another hint … :After first dilution, you are left with 10,000 cels in one mL of solution
- …another hint … :After second dilution, you are left with 10 cells in one mL of solution
- …another hint … :Multiply two dilution factors to get the total dilution

##### I think I have the answer: 1:100,000

#### Start with a 1:1000 dilution someone else has made. Take 1 mL and put it in 99 mLs of water. What is the total dilution?

(To make this problem interactive, turn on javascript!)

- I need a hint … :Assume that you start with 10,000 cells in one mL of solution
- …another hint … :After first dilution, you are left with 100 cells in one mL of solution
- …another hint … :First dilution factor is 1:100
- …another hint … :Multiply the dilution factor you obtained with the initial dilution factor to get the total dilution

##### I think I have the answer: 1:100,000

#### Put one mL of a stock into 49 mLs of water. Take 1 mL of that and put it in 49 mLs of water. What is the total dilution?

(To make this problem interactive, turn on javascript!)

- I need a hint … :Assume that you are starting with 20,000 cells in one mL of solution
- …another hint … :After first dilution you are left with 400 cells in one mL of solution (20,000/50)
- …another hint … :After second dilution you are left with 8 cells in one mL of solution (400/50)
- …another hint … :Multiply two dilution factors (1/50 * 1/50) to get the total dilution

##### I think I have the answer: 1:2500

#### Put one mL of a stock into 99 mLs of water. Repeat 3 more times. What is the total dilution?

(To make this problem interactive, turn on javascript!)

- I need a hint … :Assume you are starting with 1,000,000,000 cells in one mL of solution
- …another hint … :After first dilution you are left with 10,000,000 cells in one mL of solution (dilution factor 1:100)
- …another hint … :Dilution factor is the same each time
- …another hint … :Multiply all four dilution factors to get total dilution

##### I think I have the answer: 1:100,000,000

#### Start with a 1:10,000 dilution someone else has made. In order to get a 1:100,000 dilution, you need to put 1 mL into __ mLs of water.

(To make this problem interactive, turn on javascript!)

- I need a hint … :Determine the dilution factor to get 1:100,000 dilution from 1:10,000 dilution
- …another hint … :Dilution factor is 1:10
- …another hint … :You need to have 10 mL of solution to get 1:100,000 of dilution

##### I think I have the answer: 9

#### Starting with a dilution made by a TA, you add 1 mL to 49 mLs of water to get a 1:500,000 dilution. What was the TA’s dilution factor?

(To make this problem interactive, turn on javascript!)

- I need a hint … :Your dilution factor is 1:50
- …another hint … :Divide final dilution with your dilution factor to get initial dilution

##### I think I have the answer: 1:10,000

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## Bio-Resource: Serial Dilution of Bacterial culture / Stock solution

** Serial Dilution** is a series of sequential dilution of a substance in solution.

**How to do serial Dilution?**

To dilute a stock solution, Mix the stock solution with diluent. The ratio of the Final volume to the aliqout volume of the stock is known as Dilution Factor (DF). Dilution is the inverse of Dilution Factor. 10 fold dilution is actually 10^-1 dilution (Inverse of Dilution Factor: 1/10).

**Example:**

Diluting a 1 M (Molar) Stock solution 1000 fold by serial Dilution.

1000 fold serial dilution can be done by three 10 fold dilutions. (10*10*10 = 1000 fold).

Lets see how to serially dilute 1 M stock solution to a final volume of 100 mL.

Dilution Factor = Final Volume / Volume of Stock.

Since we need to do 10 fold dilution, dilution factor is 10.

Final Volume — 100 mL

X — Stock Solution required.

so; 10 = 100 mL /X mL;

10X = 100; X= 100/10 = 10 mL.

Diluent Volume required = [Final Volume — Stock Volume] = 100mL — 10mL = 90mL.

To do a 10 fold dilution of 1M Stock solution to 100mL,

Mix 10mL of 1M stock with 90mL of Diluent. Label the tube as Dilution 1.

From Dilution 1 tube take 10mL and mix 90mL of diluent and Label it as Dilution 2.

From From Dilution 2 tube take 10mL and mix 90mL of diluent and Label it as Dilution 3.

As we have done 10 fold dilutions, the tube 3 (Dilution 3) will be 1000 fold diluted than the stock.

Stock solution is 1 M, after 1000 fold dilution it becomes — 1/1000 = 0.001 M.

Dilution 1 — 0.1 M

Dilution 2 — 0.01 M

Dilution 3 — 0.001 M

**How to do Serial Dilution of Bacterial Culture?**

Dilution of bacterial culture can be performed the same way as explained above. Serially dilute the bacterial culture, Bacterial culture is generally serially diluted for plating (for bacterial isolation), for re-inoculation or for other applications like checking optical density, etc.

Below is the Overnight incubated plate, plated by serial dilution of the sample. Note the plate with lesser number bacterial colonies that is the one which is plated from a more diluted solution.

**To calculate CFU/mL or CFU/g Click here**

__Bacterial Culture Dilution & Calculation__

Example:

You have a bacterial culture, which has 10^6 bacterial cells and you want 100 cells/mL how to dilute the culture to get 100cells/mL

Solution:

Inital Concetration — 10^6 cells

Final Concentration Required — 100 cells/mL or 10^2 cells/mL

You need to do Four 10 fold dilution [10*10*10*10 = 10000 fold dilution] to a volume of 1mL to get 10^2 cells/mL final concentration.

here is how you can do it,

Stock of Bacterial Culture — 10^6 Bacterial Cells

Tube 1 (Dilution 1) — Take 900uL of sterial water in tube 1 and add 100uL of bacterial stock — so tube 1 will have10^5 cells/mL.

Tube 2 (Dilution 2) — Take 900uL of sterial water in tube 2 and add 100uL from Tube 1 — so tube 2 will have10^4 cells/mL.

Tube 3 (Dilution 3) — Take 900uL of sterial water in tube 3 and add 100uL from Tube 2 — so tube 3 will have10^3 cells/mL.

Tube 4 (Dilution 4) — Take 900uL of sterial water in tube 4 and add 100uL from Tube 3 — so tube 4 will have10^2 cells/mL or 100 cells/mL.

**Serial Dilution Calculation: Problems and Solutions**

You do a dilution by combining 100 ml volume of NaCl plus 700 ml unit volumes of Distilled water. What is the dilution factor, i.e, how many more times dilute is it than the original concentration?

Answer:

Final Volume = Volume of Diluent (Distilled Water here) + 100 (Volume of Stock ) = 800 mL

Dilution Factor = Final Volume / Volume of Stock Aliquot

= 800 / 100 = 8

Dilution Factor is 8 or the stock is 1/8 times diluted.

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## Serial dilution

A **serial dilution** is the stepwise dilution of a substance in solution. Usually the dilution factor at each step is constant, resulting in a geometric progression of the concentration in a logarithmic fashion. A ten-fold serial dilution could be 1 M, 0.1 M, 0.01 M, 0.001 M… Serial dilutions are used to accurately create highly diluted solutions as well as solutions for experiments resulting in concentration curves with a logarithmic scale. A tenfold dilution for each step is called a **logarithmic dilution** or **log-dilution** and a 3.16-fold (10^{0.5}-fold) dilution is called a **half-logarithmic dilution** or **half-log dilution**. Serial dilutions are widely used in experimental sciences, including biochemistry, pharmacology, and physics as well as in homeopathy.

** In biology and medicine **

In biology and medicine, besides the more conventional uses described above, serial dilution may also be used to reduce the concentration of microscopic organisms or cells in a sample. As, for instance the number and size of bacterial colonies that grow on an agar plate in a given time is concentration-dependent, and since many other diagnostic techniques involve physically counting the number of micro-organisms or cells on specials printed with grids (for comparing concentrations of two organisms or cell types in the sample) or wells of a given volume (for absolute concentrations), dilution can be useful for getting more manageable results. [*K. R. Aneja. «Experiments in Microbiology, Plant Pathology and Biotechnology». New Age Publishers, 2005, p. 69. ISBN 812241494X*] Serial dilution is also a cheaper and simpler method for preparing cultures from a single cell than optical tweezers and micromanipulators. [

*cite book |last=Booth |first=C. |authorlink= |coauthors=et al |editor= |others= |title=Extremophiles |origdate= |origyear= |origmonth= |url=http://books.google.com/books?id=eMFwInCi_NwC&pg=PA543&dq=serial+dilution+axenic&lr=&sig=PUhi7qQOY3Vi69eZdQcuwatobac#PPA543,M1 |format= |accessdate= |accessyear= |accessmonth= |edition= |series=Methods in microbiology 35 |date= |year=2006 |month= |publisher=Academic Press |location= |language= |isbn=0125215363 |oclc= |doi= |id= |pages=543 |chapter= |chapterurl= |quote=*]

** In homeopathy **

Serial dilution is one of the core foundational practices of homeopathy, with «succussion», or shaking, occurring between each dilution. In homeopathy, serial dilutions (called potentisation) are often taken so far that by the time the last dilution is completed, no molecules of the original substance are likely to remain. [*cite journal|title=Homeopathy: Holmes, Hogwarts, and the Prince of Wales|author=Weissmann, Gerald|journal=The FASEB Journal|date=2006|volume=20|pages=1755–1758|url=http://www.fasebj.org/cgi/content/full/20/11/1755|accessdate=2008-02-01|doi=10.1096/fj.06-0901ufm*] cite journal|author=Ernst, Edzard|title=Is homeopathy a clinically valuable approach?|doi=10.1016/j.tips.2005.09.003|journal=Trends in Pharmacological Sciences|volume=26|issue=11|date=November 2005|pages=547–548]

** References **

* Michael L. Bishop, Edward P. Fody, Larry E. Schoeff. «Clinical Chemistry: Principles, Procedures, Correlations». Lippincott Williams & Wilkins, **2004**, p. 24. ISBN 0781746116.

** External links **

* [*http://abacus.bates.edu/~ganderso/biology/resources/dilutions.html How to Make Simple Solutions and Dilutions*] , Bates College

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