Take a spoonful of Cheerios and add it to a spoonful of milk, stir it up well, take a spoonful of this new mix and add it to a spoonful of fresh milk, stir that up & add a spoonful of it to another spoonful of new milk and… You’ve got yourself a SERIAL cereal DILUTION!

In a serial dilution you perform a series of stepwise dilutions where you’re diluting by the same factor each time (e.g. 1:2 in this example) so the concentration (relative amount of the thing to the non-thing) decreases exponentially. So each mix has a dilution-factor times less than the one before it. So in this case, where our dilution factor is 2, each new Cheerio mix has 1/2 as many Cheerios per spoonful than the one before it.

If instead of mixing 1 spoonful to 1 spoonful each time we’d mixed 1 spoonful to 9 spoonfuls of milk (a 1:10 dilution) our Cheerios would dilute out a lot faster. We’d have a dilution factor of 10, so each new spoonful would have 10-times-less Cheerios than the one before.

Serial dilutions are kinda like pyramid schemes with the opposite effect – if you start out with a certain amount of money it gets distributed to more and more people – but you can’t give out more money than you started with so people end up with less and less money.

When we have something changing by a constant multiplier (like 1/2 or 1/10) we call it geometric, or exponential, growth/decay. And it gives us shortcuts to work with as we’ll see…

So, how do serial dilutions work & why do we use them? You keep diluting the same thing over and over but in a constant-stepwise fashion, stepwise being key & they’re useful for lots of things like helping with accuracy & creating “standard curves” which are a series of things on known concentration that allow you to compare yours to (I was doing this today, hence the post…) More on these why’s after we discuss the how (and I’ll probably do a follow-up post on standard curves because people might find that helpful too).

But let’s get back to the how for now, skipping the part where you milk the cow – say you want to do 1:2 dilutions of Cheerios, we’ll call them C for shortness. So take the starting solution, which is all C. And you mix it with an equal amount of milk. e.g. if you took 1 spoonful of C + 1 spoonful of milk you’d get 2 spoonfuls of of a 50% C solution

Then you take that 1st dilution and do the same thing with it – but now since you’re starting with a 50% solution, when you half it, you get a 25% solution. And if you do the same thing with the 25% solution, you’ll get a 25/2= 12.5% solution, and you can keep doing this, each time getting a solution half as dilute as the one before.

100% -> 50% -> 25% -> 12.5% -> 6.25%

So in the span of 5 samples you can cover a spread from 100% to 6.25%

If that’s not big enough for you, you can keep on diluting (e.g 6.25% -> 3.125%…) But you’ll still only be decreasing by 1/2 each time. If you want to “move faster” you can choose a higher dilution factor.

Dilution factor refers to the amount by which you’re diluting by each step. So in this example, the dilution factor was 2/1 = 2. What if instead of 2 you chose a dilution factor of 10? Now over the span of 5 samples you go from 100% -> 10% -> 1% -> 0.1% -> 0.01%

So, for example, if you started with 100 spoonfuls of X, the 5th tube would have 1 spoonful-worth of Cheerios and 99 spoonfuls-worth of milk. But you didn’t get there by mixing 1 spoonful of Cheerios of X with 99 spoonfuls of milk – you got there stepwise.

  • 100 C
  • 10 C + 90 milk
  • 10(10 C + 90 milk) + 90 milk
  • 10(10(10 C + 90 milk) + 90 milk)
  • 10(10(10(10 A + 90 milk) + 90 milk)) + 90 milk

    If you were to take samples of equal volumes (like equal spoonfuls), each sample will have 1/10 of the molecules of B as the one before it -> so if solution 1 has 100 Cheerios, solution 2 has 10, 3 has 1, 4 has 0.1? I guess you can cut a Cheerio, but how do you have 0.1 molecules? Well, we’re usually dealing with waayyy more than 0.1 molecules because 1 usually has way more than 100 molecules.

    In fact, molecules are so tiny and abundant that we often speak of their concentrations in “moles” and moles are 6.02×10^23 of something (anything – like how a dozen can be used for donuts or bagels). Molarity (M) is a unit of concentration corresponding to 1 mol per L. So if our solution of C is 1M that would mean that 1L of it has 6.02×10^23 Cheerios in it and 100mL has 6.02×10^22 Cheerios, so 4 will still have plenty! http://bit.ly/2r4RnrX

    Even when we speak in terms of milimolar (mM) you have 1/1000 of a mol/L, micromolar (μM) you still have 1/1000,000 and nanomolar (nM) still has 1/1000,000,000 mol/L which is 6.02×10^13 Cheerios per L which is still a lot! (definitely more than physically possible but that’s never stopped the bumbling biochemist!)

    Say you want to know the Cheerios concentration without having to count them out. Serial dilutions follow the same dilution “rule” as “normal” dilutions -> M1V1 = M2V2

    The initial molarity (or whatever concentration unit you’re using – could be Cheerios per spoonful it just has to be the same on both sides of the equation) times the initial volume equals the final molarity times the final volume. It’s sometimes written C1V1 = C1V2 to reflect the fact that it doesn’t have to be molarity but I learned it this way. Lots more on it here: http://bit.ly/2yogj1i

    With a serial dilution, M1V1 still equals M2V2 but for each dilution your M1 is the M2 from the previous dilution. Instead of having to calculate out the M2 stepwise, you can take advantage of the “geometric progression” of the dilution series -> each time you dilute, M2 gets divided by the dilution factor, which is the same as multiplying by (1/DF) So, starting with some initial dilution (M2)
  • M2
  • M2*(1/DF)
  • (M2(1/DF))(1/DF)
  • ((M2(1/DF))(1/DF))*(1/DF)
  • (((M2(1/DF))(1/DF))(1/DF))(1/DF)

    That’s supposed to be helpful?! Bear with me – that’s all just multiplication, so we can order/group it however we want, so we can rearrange things a bit and get…
  • M2
  • M2((1/DF))
  • (M2)((1/DF)*(1/DF))
  • (M2)((1/DF)(1/DF)(1/DF))
  • (M2)((1/DF)(1/DF)(1/DF)*(1/DF))

    A little better, but here’s where it really clicks – Exponent form
  • M2
  • M2((1/DF))
  • (M2)((1/DF)2)
  • (M2)((1/DF)^3)
  • (M2)((1/DF)^4)

    Now we don’t even have to worry about calculating any of the concentrations of the previous dilutions in order to find the concentration of a later one. We just need to know the starting concentration, the dilution factor, and the number or dilutions

    Say you want to graph the average # of Cheerios per spoonful of each of the mixtures. If you were to try to graph that on a “normal”, linear scale (where the tick marks are evenly spaced by addition (e.g. 1, 2, 3, 4, 5,….) you’d have a bunch of dots squished together near the origin (0,0) and then that 100% way out on the right. If you want to get the values in a better seeable form, you can instead plot it on a logarithmic scale. Logarithmic scales use tick marks that are spaced apart based on exponents that you’re raising something to. So if you have a DF of 2 you’d use a log2 scale to get a line & w/a DF of 10 you’d use log10. more on exponents & logs: http://bit.ly/2TCM2UE

    Normally, instead of diluting Cheerios, I’m diluting things like RNA or protein and instead of spoonfulling them I pipet them. Today, in addition to the serial dilutions I did for my stuff which involve diluting things that are “invisible” to us, I did a serial dilution of a dye to show you.

    Depending on the volumes you want it’s helpful to do them in strip tubes (like PCR strip tubes) or deep well blocks, etc. – something where the liquid-holders are connected so you don’t mix them up & you don’t have to worry about numbering tons of tubes, etc.

    I start by pipetting the diluent (water in this case) into all the tubes except the first one which I save for the starting one that doesn’t need to get diluted. And then I stick the starting one into the first tube and the 2nd tube. Usually what I do is I transfer then pipet up-down-up-down-up-down… (usually I do 5 times) ending on “up” then transfer to the next tube for the down

    When you’re making dilutions, you have to remember that you’re taking out part of your “V2” each time – so if you do 10uL + 10uL = 20uL and then take 10uL of that for the next one you only have 10uL of that one left – not 20. So if you need 20, you should be doing 20 + 20 (actually if you need 20 you should be doing 25+25 or something because it’s always good to have a little extra to account for evaporation, tube/pipet sticking-to-ness, etc.)

    And speaking of that stickiness, another reason to do serial dilutions is that they aid with accuracy. We don’t want one of those Cheerios boxes that’s “BONUS – 10% more FREE!”

    As I talked about in a previous post, each instrument & measurement tool you use has a bit of error (e.g. if you pipet out 1mL it could really be 1.01mL or 0.99mL instead of 1.00mL). And the smaller the thing you’re measuring the bigger a “small” error can make. (think of a drop in a pool vs a drop on a ladybug). http://bit.ly/33hkCd8

    And if you’re pipetting a concentrated solution, 1 drop packs a lot of punch. So that little drop of liquid stuck to the outside of the pipette holds a lot of molecules that you aren’t accounting for. So it’s better to be pipetting larger volumes of more dilute things.

    But sometimes you need to make a really dilute solution. Say you want to make a million-fold dilution of something (e.g. you have something that’s 1M and you want to get it to 1uM). Even if you pipetted just 1uL, you’d have to dilute that in a whole L of water to get 1uM. Instead of doing it that way you can do it stepwise.

    Another time serial dilutions are great is for making “Standard curves.” These are where you take solutions of known concentration and you make a plot of them and then you compare a sample of unknown concentration to the plot of known concentrations to see where it falls along the line.

    more on topics mentioned (& others) #365DaysOfScience All (with topics listed) 👉 http://bit.ly/2OllAB0

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