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Mr_Saturn512's avatar

Did I do this serial dilution wrong?

Asked by Mr_Saturn512 (558points) April 13th, 2014

Math was never really my strong point.

Each solution has to be 10mL. I have to make a serial dilution from 4mg/mL to 2mg/mL, 1, .8, .6, .4, .2, and lastly .1

The first is 4mg/mL, so I took 40 mg of BSA (the protein I’m using) and mixed it with 10mL of PBS (the buffer I’m using)

For 2mg/mL, I took 5mL of the above and mixed it with 5mL of PBS.

For 1mg/mL, I took 2.5mL of the above and mixed it with 7.5mL of PBS.

For .8mg/mL, I took 2mL of the above, mix it with 8mL of PBS.

You can see where this is going, as each time I make it more dilute, I’m taking .5 less from the previous solution and mixing it with .5 more of PBS. (Except for the .1mg/mL, there it’s .25mL of .2mg/mL with 9.75mL of PBS)

I have to record the absorbances of these solutions. The 4mg/mL was fine, as was the 2mg/mL.

However, EVERYTHING below that is too dilute. By the time I observe .6mg/mL, I get 0 absorbance reading, like nothing is there.

Did I do this wrong? Did I have a brain fart and think something about this was logical and it’s really not? The thing is, this experiment has been done before, so I know something is wrong because the previous experiment had all these numbers for all these solutions. I got virtually nothing beneath 2mg/mL.

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3 Answers

BhacSsylan's avatar

Yeah, this is wrong. So, first of all, if you’re doing dilutions always remember this equation:

M1 x V1= M2 x V2

Where M1 is the initial concentration, V1 is the volume of that solution you used, M2 is the final volume, and V2 is the volume of that final solution. Rearranging that algebraically will give you any of the four values if you have the other three.

Next, your writing is a little unclear. When you say “of the above”, do you mean of the top solution, or the one directly before? Because if you mean the top one, your first three are right but that’s parallel, not serial. If you mean the one directly above, yeah, that’s really wrong. So, lets test a few with the equation:

For the first, M1 is 4 mg/ml, V1 is 5 mg/ml, and V2 is 10 mg/ml (5+5). So it’s 4×5 = M2×10 so M2 = (4×5)/10 = 2 mg/ml. So, that one’s fine. But then things get tricky.

So the next, assuming parallel dilutions, you’re diluting 2.5 ml of the 4 mg/ml into 10, so it’s

4 mg/mL x 2.5 mL = M2×10 mL; M2 = (4×2.5)/10 = 1 mg/ml. Correct, but again that’s parallel.

If you did it serially, it would be 2.5 mL of 2 mg/mL to a final 10 mL, so

2 mg/mL x 2.5 mL = M2×10 mL; M2 = (2×2.5)/10 = .5 mg/ml. Half what you expected. And this error would compound if it’s what you did.

And why i expect that’s probably the case, the last example you give is .25mL of .2mg/mL with 9.75mL of PBS, which would be

.2 mg/mL x .25 mL = M2×10 mL; M2 = (.2 x .25)/10 = .005. Rather a bit lower than you expected.

Seriously, M1 x V1 = M2 x V2. One of the best equations ever for chemistry.

BhacSsylan (9534points)“Great Answer” (0points) Flag as…
Mr_Saturn512's avatar

Yea. It was a mouthful to try to explain. But when I said “above” I meant the previous solution directly before.

So in other words, I was applying numbers from a parallel dilution to a serial dilution? I should redo this now with this equation to get the correct numbers for a SERIAL dilution.

So. . . it should look something like this now, correct?:

4 mg/ml x 5 ml = 2 mg/ml x 10 ml

2 mg/ml x 5 ml = 1 mg/ml x 10 ml

1 mg/ml x 8 ml = 0.8 mg/ml x 10 ml

0.8 mg/ml x 7.5 ml = 0.6 mg/ml x 10 ml

0.6 mg/ml x 6.66666666666667 ml = 0.4 mg/ml x 10 ml

0.4 mg/ml x 5 ml = 0.2 mg/ml x 10 ml

0.2 mg/ml x 5 ml = 0.1 mg/ml x 10 ml

Mr_Saturn512 (558points)“Great Answer” (0points) Flag as…
BhacSsylan's avatar

Yep, that all looks better now.

BhacSsylan (9534points)“Great Answer” (0points) Flag as…

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