How can this problem be started? Serial diution problem using DNA stock solution

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Generate standard curves. One example is to make a standard curve of absorbances from known amounts of BSA (0.25 mg; 0.5 mg; 1.0 mg and 2.0 mg) to determine protein content of an unknown protein using Beer's Law (A=ecl).

2. Working with bacteria to get a reasonable number of bacteria to count on a plate.

This can be done in two ways

1. Independently dilute each sample. Pipette out the appropriate amount by removing a decreasing amount of from the stock solution. Note that you need to open the stock solution 4x in the example below:

Example- Make a 1/2 dilution series of BSA ranging from 0.25 - 2 mg of protein

Attachment

This works wells for some situations but not always. It would be difficult to dilute bacteria to 1/1000,000 to get a reasonable dilution to count on a plate.

2. Serial dilutions of consecutive dilutions. These dilutions are dependent on one another. This strategy is useful if you only want to use a small amount of your original stock concentration. It is also handy if measuring out dilutions independently is not possible (such as taking out 0.000001 ml of your stock or original solution).

Example- make 10 ml series to dilute a bacterial stock 1/ 100,000. First you prepare 10 ml of dH20 into 3 test tubes. Remove 0.1 ml of water from each 3 tubes - now you have 9.9 ml of water (diluent)

Original Broth -> Take out 0.1 ml add to Test Tube #1- 1/100 dilution (MIX WELL!)

Tube #1 -> Take out 0.1 ml add to Test Tube #2- 1/100 dilution of a 1/100 dilution = 1/10,000 total dilution relative to the original broth . (MIX WELL)

Tube #2 -> Take out 0.1 ml (100 µl) and add to Tube #3- 1/100 dilution of a 1/10,000 dil = 1/100,000 dilution in Tube #3 relative to the orig. solution d(MIX WELL)

Note that you end up w/ 10 ml of solution in Tube 3, and 9.9 ml in Tube 1 & 2

Practice - Set up a series of 3 test tubes containing 10 ml of 1/10 diluted bacteria. What is your final dilution?

Answer:

Tube #1 Add 1 ml of original broth to 9 ml of water, Mix well (1/10 dilution)

Tube #2 Take 1 ml from Tube #1 to 9 ml of water Mix well ( 1/100 dilution)

Tube #3 Take 1 ml from Tube #2 to 9 ml of water. Mix well ( 1/1000 dilution- Final)

Dr. Sheperd, a wonderful prof who is now retired from SDSU, showed me this neat trick (that I think that he came up with himself!) to get all the solutions to equal the same exact volume in each test tube. Simply use this equation & simplify using algebra (Ah Hah- Something you learned from high school math is applicable!):

X/ (X + Y) = Z

X= volume to be transferred to successive tubes

Y= volume needed in each tube at the end of the dilution scheme

Z = the dilution factor.

Example: So lets say you need 3 ml of a 1/2 dilution.

X=?

Y= 3 ml

Z = 1/2

X/(X + 3ml) = 1/2Do some algebra times (X + 3) on both sides

X = (X + 3 ml)/2Do some algebra tricks to simplify (just multiple both sides x 2 to get rid of that pesky 1/2 fraction)

2X= X + 3ml(Do some more algebra tricks (subtract 1X on each side to get the Xs to be on the same side of the equation)

X = 3 ml

Add 3 ml of diluent (water) to each tube and transfer 3.0 ml successively until you reach the last tube- then just discard the last 3.0 ml transfer. Now each tube will have the same amount of the dilution series. Just to check that over

Original tube- Take out 1.5 ml + 3 ml of water (Mix Well) Tube #1

Hum is that right? 3.0 ml/ (3 ml + 3.0 ml) total = 3.0 ml / 6.0 ml = 1/2 dilution

Ok- then take out 3 ml from Tube #1 + 3 ml of water Tube 2 Mix well

Then take out 3 ml from Tube #2 + 3 ml of water Tube #3 Mix well.

Note that there is a 1/2 dilution that increases at each step.

Practice Problem. What volume do you need to transfer to make 3 ml of a set of 3 serial dilutions of 1/3?

Thank you so much!! really helped!! lol

so I have starting stock of the target dna concentration as 3.2 x10^21 DNA copies/uL.

how did you go from 3.15x10^21 to 3.15x10^20?

Well, I'm not trying to be a spoiler but it's hardly a trick. This equation is how it works, purely. The diluted factor represent the concentration of the broth in its diluted solution, simple as that.

I haven't come across that kind of exercise before but I assume it requires us to do 2 things. (1) dilute the given solution to a required stock solution (2x10^7 copies/ul) and (2) find a diluted factor to make a serial dilution. The first part should be easy for you, i suppose. Second part is somehow troublesome. So here the trick (real trick) that would help if you are not a fan of algebra. This trick is extremely helpful if you work in a lab and want to diluted some solutions to a concentration needed. This trick works with any kind of concentration (%, mol/ml. ug/ml, copies/ml, cells/ml...)

Lets say you have a solution that have A concentration, and you want to dilute it to B concentration (A>B of course). The solvent denoted as C which is  generally water (but you can dilute with less concentrated solution too).

Draw a diagram as my attachment:
You could see that the result come up instantly (feel free to double check it using algebra lol). It's too simple to explain, yet too hard to prove.

Back to your problem, I was using your problem as example. As you continue to do it on and on with next dilutions, dilution factor 1/10 (ratio 1/9) is repeated in pattern.

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Somebody may feel hard to believe if this trick works but it does, check it if you want to. In routine work, it would save us lots of time, especially in labs where we have to dilute solution almost every days. 
It also come in handy if you want to concentrate a very diluted solution to a more concentrated solution. For example, if you accidentally diluted a DNA solution to 2x10^3 copies/ul instead of 2x10^4 copies/ul, and you have some stock (2x10^7) left. In that case, A would be 2x10^7, B=2x10^4, C=2x10^3.

Thanks for sharing this.